An Introduction to Sobolev Spaces [1 ed.] 9781681089133, 9781681089157, 9781681089140

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An Introduction to Sobolev Spaces Authored by

Erhan Pis¸kin Department of Mathematics Dicle University Diyarbakir Turkey

&

Baver Okutmus¸tur Department of Mathematics Middle East Technical University (METU) Ankara, Turkey

An Introduction to Sobolev Spaces Authors: Erhan Piskin and Baver Okutmustur ISBN (Online): 978-1-68108-913-3 ISBN (Print): 978-1-68108-914-0 ISBN (Paperback): 978-1-68108-915-7 © 2021, Bentham Books imprint. Published by Bentham Science Publishers Sharjah, UAE. All Rights Reserved.

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Contents

Preface 1

2

i

Preliminaries 1.1 Metric Spaces . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . 1.2.1 Banach Space . . . . . . . . . . . . . . . . . . . . 1.2.2 Inner Product Space . . . . . . . . . . . . . . . . 1.2.3 Hilbert Space . . . . . . . . . . . . . . . . . . . . 1.3 Some Important Theorems . . . . . . . . . . . . . . . . . 1.3.1 Lebesgue and Monotone Convergence Theorems . 1.3.2 Fatou Lemma and Lebesgue Dominated Convergence Theorem . . . . . . . . . . . . . . . . . . . 1.3.3 Fixed Point Theorems . . . . . . . . . . . . . . . 1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

33 34 38

Lp (Ω) Spaces 2.1 Basic Definitions . . . . . . . . . . . 2.2 Some Important Inequalities . . . . . 2.3 Green’s Identities . . . . . . . . . . . 2.4 L∞ Space . . . . . . . . . . . . . . 2.5 Embedding of Lp (Ω) Spaces . . . . 2.6 Lploc (Ω) Space . . . . . . . . . . . . 2.7 Space of Continuous Functions . . . 2.8 C ∞ 0 (Ω) Space and Compact Support 2.9 H¨older Space . . . . . . . . . . . . . 2.10 Lp (a, b; X) Space . . . . . . . . . .

41 41 45 68 71 75 78 80 81 82 83

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1 2 16 24 29 31 33 33

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2.11 C m ([0, T ]; X) Space . . . . . . . . . . . . . . . . . . . 2.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . .

83 84

Weak Derivative 3.1 Introduction . . . . . . . . . . . . 3.2 Examples . . . . . . . . . . . . . 3.3 Properties of the Weak Derivative 3.4 Exercises . . . . . . . . . . . . .

86 86 87 94 95

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Sobolev Spaces 97 4.1 Basic Definitions and Examples . . . . . . . . . . . . . . 97 m,p 4.2 Weighted Lebesgue Lpw (Ω) and Sobolev W w (Ω) Spaces 113 s n 4.3 H (R ) Sobolev Space . . . . . . . . . . . . . . . . . . . 114 4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 118

5

Sobolev Embedding Theorems 5.1 Embedding Property . . . . . . . . . 5.2 Domains and Properties . . . . . . . . 5.3 Embedding Theorems . . . . . . . . . 5.4 Compact Embedding . . . . . . . . . 5.5 Embedding in H s (Rn ) Sobolev Space 5.6 Exercises . . . . . . . . . . . . . . .

6

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Variable Exponent Lebesgue and Sobolev Spaces 6.1 Basic Concepts . . . . . . . . . . . . . . . . 6.2 Lp(x) (Ω) Space . . . . . . . . . . . . . . . . 6.3 W m,p(x) (Ω) Space . . . . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . . .

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119 119 125 126 131 132 135

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138 138 140 151 153

Applications on Differential Equations 154 7.1 Weak Solutions in Ordinary Differential Equations . . . . 154

7.2 7.3

Local Existence of Solutions of Nonlinear Timoshenko Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 Existence of Solutions of Sixth Order Boussinesq Equation 170

Bibliography

184

List of Symbols

188

Subject Index

190

i

Preface

Sobolev spaces were firstly defined by the Russian mathematician S. L. Sobolev (1908-1989) in the 1930s. Several properties of these spaces have been studied by mathematicians till today. Especially existence and uniqueness, asymptotic behavior, blow up, stability and instability of the solution of many differential equations that occur in applied and in engineering sciences are carried out with the help of Sobolev spaces and embedding theorems in these spaces. This book provides a brief introduction to Sobolev spaces at a simple level with illustrated examples. Some of their applications might be relevant both to undergraduate and graduate students, mathematicians, and engineers who have an interest in getting a quick, but carefully presented, mathematically sound basic knowledge in this domain. In this regard the book fills an important gap in that field. There are seven chapters in the book. The first chapter is devoted to basic concepts consisting of metric spaces, normed spaces, inner product spaces, Hilbert spaces and some important results on fixed point theorems. The second chapter is based on Lp spaces. Several important inequalities, embedding property of Lp and some other spaces are presented in this part. In chapter three, we introduce the notion of the weak derivative. There are many numerical example in order the reader to distinguish the weak derivative from the classical one. Sobolev spaces are presented and widely analyzed in chapter four. Chapter five is devoted to Sobolev embedding theorems. The variable exponent Lebesgue and Sobolev spaces are investigated in chapter six. Finally, in chapter seven, we present the importance of Sobolev spaces in light of their application to some differential equations.

ii

ACKNOWLEDGEMENTS: None. CONSENT FOR PUBLICATION: Not Applicable. CONFLICT OF INTEREST: The authors confirm that this book contents have no conflict of interest

An Introduction to Sobolev Spaces, 2021, 1-40

1

Chapter 1 Preliminaries

Abstract:The main objective of this chapter is to remind the reader of some basic notion and fundamental facts about real analysis and functional analysis required for the comprehension of the following chapters. It is assumed that readers are familiar with the concept of metric and normed spaces. The definitions of a metric space, convergence of a sequence in a metric space, completeness, compactness and the Heine Borel theorem are introduced in the first part. Here there are some well–known properties of topological concepts needed to recall. Next the notion of norm and normed spaces, equivalent norms, compactness and relatively compactness, Banach space, dual space, weak and weak* convergence are presented. Before the Hilbert spaces, inner products and inner product spaces are briefly expressed. Here the relation between the Banach and Hilbert spaces is given by some examples. Most of the theorems of that part are stated without proof since they can easily be found from any real or functional analysis book given in the reference part. Following the Fatou lemma and the Lebesgue dominated convergence theorem, the chapter ends with some important theorems based on fixed point properties on Banach spaces. For instance the Tychonoff fixed point theorem is an extension of the Schauder’s fixed point theorem and the Schauder’s theorem is an extension of the Brouwer fixed point theorem. Since the proofs of these theorems require additional knowledge, we refer the reader to the book by Papageorgiou and Winkert [1] and reference therein. In addition [2–7] may also be functional.

Keywords: Metric, normed space, completeness, convergence in a metric space, weak convergence, Banach space, dual, Hilbert space, fixed point theorems. Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

2

An Introduction to Sobolev Spaces

1.1

Erhan Pis¸kin and Baver Okutmus¸tur

Metric Spaces

Definition 1.1.1. Let X be a non-empty set. A metric on X is a function d : X × X → R+ ∪ {0} which satisfies the following conditions: (m1 ) d(x, y) > 0 for all x, y ∈ X with x 6= y; (m2 ) d(x, y) = 0 ⇔ x = y for all x, y ∈ X; (m3 ) d(x, y) = d(y, x) for all x, y ∈ X (Symmetry); (m4 ) d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ X (Triangle inequality). Here d(x, y) is the distance between x and y; moreover the ordered pair (X, d) is called a metric space. Remark 1.1.2. (m1 ) can be obtained from the other three properties. Considering the triangle inequality d(x, x) ≤ d(x, y) + d(y, x),

∀x, y ∈ X,

and using the properties (m2 ) and (m3 ), it follows that 0 = d(x, x) ≤ 2 d(x, y) ⇒ 0 ≤ d (x, y) . Thus for x 6= y we have 0 < d (x, y) which is the desired result. Remark 1.1.3. Applying n-times the inequality (m4 ) for all x, y, z1 , z2 , · · · , zn ∈ X, we obtain d(x, y) ≤ d(x, z1 ) + d(z1 , y) ≤ d(x, z1 ) + d(z1 , z2 ) + d (z2 , y) .. . ≤ d(x, z1 ) + d(z1 , z2 ) + · · · + d (zn , y) which is said to be the generalized triangle inequality.

An Introduction to Sobolev Spaces 3

Preliminaries

Example 1.1.4. Show that the function d : R × R → R+ ∪ {0} defined as d(x, y) = |x − y| satisfies metric conditions. This metric is known as absolute metric or Euclid metric. Solution. The conditions (m1 )–(m4 ) are satisfied as follows: (m1 )–(m2 ): By using the property of absolute value, it follows that d(x, y) = |x − y| ≥ 0,

∀ x, y ∈ R.

Notice that if x 6= y, the above relation yields d(x, y) = |x − y| > 0. Thus the (m1 ) is satisfied. On the other hand, d(x, y) = |x − y| = 0 ⇔ x − y = 0 ⇔ x = y which verifies (m2 ). (m3 ): For all x, y ∈ R, it follows that d(x, y) = = = = =

|x − y| |(−1) (y − x)| |(−1)| · |(y − x)| |y − x| d(y, x)

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Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

(m4 ) Finally, for all x, y, z ∈ R, we have d(x, y) = = ≤ =

|x − y| |x − z + z − y| |x − z| + |z − y| d(x, z) + d(z, y).

As a result, d is a metric and (R, d) is a metric space.

In a similar way, we can show that the functions defined in the following exercise satisfy metric properties. The proof is left as an exercise to the reader. Exercise 1.1.5. Let X = Rn (or Cn ) with x = (x1 , x2 , · · · , xn ) and y = (y1 , y2 , · · · , yn ) be given. Show that (i) the function d1 : X × X → R+ ∪ {0} defined as d1 (x, y) =

n X

|xi − yi |

i=1

is a metric. (ii) the function dp : X × X → R+ ∪ {0} defined as dp (x, y) =

n X

! p1 |xi − yi |p

( with 1 ≤ p < ∞)

i=1

is a metric. 1)

1)

Here if p = 1, then d1 is obtained. For p > 1, Minkowski inequality can be used in order to show the triangle inequality.

An Introduction to Sobolev Spaces 5

Preliminaries

(iii) the function d∞ : X × X → R+ ∪ {0} defined as d∞ (x, y) = max |xi − yi | 1≤i≤n

is a metric. Example 1.1.6. Determine whether the function d : N × N → N defined by d(x, y) = |x − y|2 is a metric or not? Solution. Consider the triangle inequality d(x, y) ≤ d(x, z) + d(z, y) which is equivalent to |x − y|2 ≤ |x − z|2 + |z − y|2 . Observe that this inequality is not valid for all x, y, z. As an example, for x = 5, y = 1 and z = 3, it fails. Hence, the function is not a metric. We leave the proof of the following as an exercise to the reader. Exercise 1.1.7. Let C [a, b] be the set of continuous functions defined by C [a, b] = {f | f : [a, b] → R and f is continuous on [a, b]} . Show that the following functions Z b i. d1 (f, g) = |f (x) − g(x)| dx a

Z ii. d2 (f, g) = a

b

 12 |f (x) − g(x)|2 dx

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Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

iii. d∞ (f, g) = sup {|f (x) − g(x)| : x ∈ [a, b]} defined on C [a, b] are metrics.

Example 1.1.8. Let the functions f (x) = x2 − 4x and g(x) = −x2 be given. Evaluate i. d1 (f, g) for (C [0, 3] , d1 ) . ii. d2 (f, g) for (C [0, 3] , d2 ) . iii. d∞ (f, g) for (C [0, 3] , d∞ ) . Solution. The graphs of these functions are illustrated in Figure 1.1. Observe that f and g intersect at x = 0 and x = 2. i. Z d1 (f, g) =

3

x2 − 4x + x2 dx

Z0 2 4x − 2x

=

2




3

Z


2x2 − 4x dx

dx + 2

0

16 = . 3 ii. Z d2 (f, g) = 0

3

2x2 − 4x 2 dx

 12

√ 6 10 = . 5

iii.  d∞ (f, g) = sup 2x2 − 4x : x ∈ [0, 3] = sup {0, 2, 6} = 6.

Preliminaries

An Introduction to Sobolev Spaces

Figure 1.1: Graphs of functions f and g.

In the following, we deal with some fundamental concepts. Definition 1.1.9. Let (X, d) be a metric space, x0 ∈ X and r > 0. Then Br (x0 ) = {x ∈ X : d (x, x0 ) < r} is said to be an open ball having radius r with center x0 , Br (x0 ) = {x ∈ X : d (x, x0 ) ≤ r} is said to be a closed ball having radius r with center x0 , Sr (x0 ) = {x ∈ X : d (x, x0 ) = r} is said to be the boundary of the ball having radius r with center x0 .

7

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Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Open ball, closed ball and the boundary of a ball having radius r with center x0 are denoted by Br (x0 ) = B (x0 , r) , B r (x0 ) = B (x0 , r) and Sr (x0 ) = S (x0 , r) , respectively. Example 1.1.10. Describe the open balls Br (0) for the metrics d1 (x, y) =

2 X

|xi − yi | ,

i=1

d2 (x, y) =

2 X

! 21 |xi − yi |2

,

i=1

and d∞ (x, y) = max |xi − yi | 1≤i≤2

in R2 . Solution. Consider the points x = (x1 , x2 ) , y = (y1 , y2 ) and 0 = (0, 0) in R2 . It follows that, for the metric d1 we have

Figure 1.2: Graph of region described by Br (0) for the metric d1 .

 Br (0) = x ∈ R2 : d1 (x, 0) < r  = x ∈ R2 : |x1 − 0| + |x2 − 0| < r  = x ∈ R2 : |x1 | + |x2 | < r

An Introduction to Sobolev Spaces 9

Preliminaries

Figure 1.3: Graph of region described by Br (0) for the metric d2 .

which is illustrated in Figure 1.2. Similarly, for the metric d2 the open ball is described by  Br (0) = x ∈ R2 : d2 (x, 0) < r   q |x1 − 0|2 + |x2 − 0|2 < r = x ∈ R2 : n o 2 2 2 2 = x ∈ R : |x1 | + |x2 | < r and it is given in Figure 1.3. Finally, for the metric d∞ , we have

Figure 1.4: Graph of region described by Br (0) for the metric d∞ .

 Br (0) = x ∈ R2  = x ∈ R2  = x ∈ R2  = x ∈ R2

: : : :

d∞ (x, 0) < r max {|x1 − 0| , |x2 − 0|} < r max {|x1 | , |x2 |} < r |x1 | < r and |x2 | < r .

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Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

It is illustrated in Figure 1.4. Definition 1.1.11. i. Let (X, d) be a metric space with x0 ∈ X and ε > 0 be a real number. Then the set B (x0 , ε) = {x ∈ X : d (x, x0 ) < ε} is said to be the ε neighborhood of x0 . ii. Let (X, d) be a metric space and A ⊂ X. For any ε > 0 and x0 ∈ A, if B (x0 , ε) ⊂ A is satisfied, then the point x0 is said to be an interior point of A. The set of all interior points of A is called the interior of A and it is denoted by Ao . iii. If all points of A are interior points, i.e. A = Ao , then A is said to be an open set. iv. A point x of a set A ⊂ (X, d) is a boundary point of X if every neighborhood of x contains both points of A and points not belonging to A. Note that x may or may not belong to A. v. A set A is said to be closed if and only if it contains all of its boundary points. Equivalently, if the complement set of A (denoted by X\A) is open, then A is a closed set. vi. Let (X, d) be a metric space and A ⊂ X. If all ε neighborhood of a point x0 ∈ X with A includes at least one different point other than x0 , i.e. ∀ ε > 0 (B (x0 , ε) − {x0 }) ∩ A 6= ∅, then x0 is said to be an accumulation point of A. An accumulation point of a set A may or may not be an element of A.

Preliminaries

An Introduction to Sobolev Spaces

11

If there is no element of A in the neighborhood of the point x0 ∈ A, then x0 is an isolated point. In other words, an isolated point of A is a point in A that is not a limit point. vii. All limit points are accumulation points; however, the reverse may always not be true. viii. The union of all accumulation points of a set A (denoted by A0 ) and A itself is called the closure of A and denoted by A; that is, A = A ∪ A0 . Note that A is the smallest closed set containing A. Example 1.1.12. Consider the set A = (1, 5] . All points (except the end points) of A are interior points. For instance, the point x0 = 4 ∈ A is an interior point of A since for all ε > 0, it follows that B (4, ε) ⊂ A. On the other hand, the end point x0 = 5 ∈ A is not an interior point since for all ε > 0, it follows that B (5, ε) A. As a result, A is not an open set because there exists a point of A that is not interior. Example 1.1.13. Consider the set A = (3, 7) . Since all points of A are interior points, A is an open set. Similarly, any open interval (a, b) is an open set. Example 1.1.14. Consider the set A = (1, 4) ∪ {5} . Here all points in the interval [1, 4] are accumulation points. On the other hand the point 5 is an isolated point. Example 1.1.15. The set of all accumulation points of A = (5, 9] is A0 = [5, 9] . Hence the closure of A is A = A ∪ A0 = [5, 9] .

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Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Definition 1.1.16 (Dense Set, Separable Space). Let (X, d) be a metric space and Y ⊂ X be given. If the relation Y =X is satisfied, then the set Y is dense in the set X. If X has a dense and countable subspace, then X is said to be separable. Example 1.1.17. The set of rational numbers Q is countable and it is dense in the set of real numbers R. Thus, it is separable. Definition 1.1.18. A sequence (xn ) in a metric space (X, d) is said to converge to a point x0 ∈ X if for every ε > 0 there exists an integer N such that n ≥ N implies that d (xn , x0 ) < ε. We say that the sequence (xn ) is convergent and x0 is the limit of (xn ) . In other words we write xn → x0

or

lim xn = x0

n→∞

or equivalently lim d (xn , x0 ) = 0.

n→∞

On the other hand, if (xn ) is not convergent, it is divergent . Definition 1.1.19. A sequence (xn ) in a metric space (X, d) is said to be a Cauchy sequence if for every ε > 0 there exists an integer N such that d (xn , xm ) < ε whenever n ≥ N and m ≥ N. In other words, (xn ) is a Cauchy sequence if d (xn , xm ) → 0 as n, m → ∞, or equivalently lim d (xn , xm ) = 0.

n,m→∞

Preliminaries

An Introduction to Sobolev Spaces

13

Theorem 1.1.20. Every convergent sequence is a Cauchy sequence. Proof. Let (xn ) be a convergent sequence in X with limn→∞ xn = y ∈ X. This implies that for any ε > 0, there is a number N ∈ N such that d(y, xn ) < ε for all n ≥ N. Then, for any m, n ≥ N , it follows that d(xm , xn ) ≤ d(xm , y) + d(y, xn ) < 2ε by the triangular inequality. Hence (xn ) is a Cauchy sequence. Definition 1.1.21. A metric space (X, d) is said to be complete if every Cauchy sequence in X converges to an element of X. Example 1.1.22. Show that the set of rational numbers Q is not complete with respect to the absolute (value) metric d (x, y) = |x − y| on R. Solution. If Q is complete, then all Cauchy sequences (xn ) in Q must converge to a number in Q. Consider a sequence in Q given by  n  1 (xn ) = 1+ n which is a Cauchy sequence Q. The limit of that sequence is  n 1 lim xn = lim 1 + =e∈ / Q. n→∞ n→∞ n As a result Q is not complete. Example 1.1.23. Let f and g be functions defined on the set of continuous functions C [a, b] . Consider the metrics Z b i. d1 (f, g) = |f (x) − g(x)| dx, a

14

An Introduction to Sobolev Spaces

Erhan Pis¸kin and Baver Okutmus¸tur

ii. d∞ (f, g) = sup {|f (x) − g(x)| : x ∈ [a, b]} . Although the metric space (C [a, b] , d∞ ) is complete, (C [a, b] , d1 ) is not a complete metric space. Here the integral is (described as) Riemann integral. The completeness of a space depends on the metric defined on it. Definition 1.1.24. Let (X, d) and (Y, d0 ) be metric spaces, f : X → Y be a function and x0 be a point in X. For every ε > 0, if there exists a δ > 0 such that whenever x ∈ B (x0 , δ) , we have f (x) ∈ B (f (x0 ) , ε), i.e., x ∈ B (x0 , δ) =⇒ f (x) ∈ B (f (x0 ) , ε) , then the function f is said to be continuous at the point x0 . Figure 1.5 describes the map of this continuous function f from X to Y .

Figure 1.5: Map of a continuous function f from X to Y .

Definition 1.1.25. A set A is bounded if it can be enclosed by a ball, i.e., A ⊆ B(x; r) for some x ∈ X and r > 0. Definition 1.1.26. A set A in a metric space X is said to be compact if every S open covering of A has a finite subcovering; that is, if A ⊆ i Oi , where Oi are open sets, then A ⊆ Oi1 ∪ Oi2 ∪ · · · ∪ Oin for some i1 , i2 , · · · , in .

Preliminaries

An Introduction to Sobolev Spaces

15

Definition 1.1.27. Let (X, d) be a metric space and Y ⊂ X be given. If each sequence (xn ) ∈ Y has a convergent subsequence with a limit in Y , then Y is said to be sequentially compact. If X is compact, the metric space (X, d) is called sequentially compact metric space. Moreover, Y is relatively sequentially compact if it satisfies the same condition except that the limit of convergent subsequence need not be in Y . Theorem 1.1.1. Let (X, d) be a metric space and Y ⊂ X be given. Then the following properties are equivalent: 1. Y is a sequentially compact set. 2. Y is a (topologically) compact set. 3. Y is closed and bounded. Theorem 1.1.2. If S is a compact subset of the metric space Ω, then S is closed and bounded. Theorem 1.1.3 (Heine-Borel). A subset S of R is compact if and only if it is closed and bounded. Theorem 1.1.28. A subset S of Rn is compact if and only if it closed and bounded. The last theorem is also known as the Bolzano-Weierstsrass theorem. We address the reader to [8, 9] and the references therein for the proofs of the above theorems. Further details and application of these results on functional analysis can also be found in [10–12]. Example 1.1.29. The closed interval [0, 1] is compact. The set of real numbers R is not compact. Example 1.1.30. Rn is not compact (for n ≥ 1). However, by the HeineBorel theorem, every closed and bounded subset of Rn is compact.

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Definition 1.1.31. Let (X, d) be a metric space and Y ⊂ X be given. If the closure of Y (denoted by Y ) is compact, then Y is said to be relatively compact. Moreover it is said to be locally compact if each point y of Y has a neighborhood whose closure is compact. Example 1.1.32. Any bounded set in Rn is relatively compact by the HeineBorel theorem. 1.2

Normed Spaces

Definition 1.2.1. Let X be a vector space over the real (or complex) numbers. A function, k·k, from X into the non-negative real numbers is called − − a norm on X if for all → x ,→ y ∈ X and λ ∈ R, the conditions → − − − (n ) k→ x k > 0 if → x 6= 0 1

− − (n2 ) k→ xk=0⇔→ x =0 − − (n ) kλ→ x k = |λ| · k→ xk 3

− − − − (n4 ) k→ x +→ y k ≤ k→ x k + k→ yk − − are satisfied. The norm of → x is denoted by k→ x k . Moreover, the vector space X with norm is called normed vector space or normed space. k·k is said to be a semi norm if only the conditions (n3 ) and (n4 ) are − satisfied. Notice that for a semi norm the condition k→ x k = 0 may be satisfied for non-zero x. − − Theorem 1.2.2. Let (X, k·k) be a normed space. Then for all → x ,→ y ∈ X, the function d defined by − − − − d(→ x ,→ y ) = k→ x −→ yk is a metric on X. This metric is called normed metric.

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Preliminaries

Proof. We show that metric conditions (m1 )–(m4 ) are satisfied for the function d. (m1 ): Using the norm property (n1 ), it follows that − − − − − − d(→ x ,→ y ) = k→ x −→ y k > 0 if → x = 6 → y. (m2 ): The norm property (n2 ) implies − − − − − − d(→ x ,→ y ) = k→ x −→ yk=0⇔→ x =→ y. (m3 ): This property can be verified by the following: − − − − d(→ x ,→ y ) = k→ x −→ yk − − = k(−1) (→ y −→ x )k − − = |(−1)| · k→ y −→ xk − − = k→ y −→ xk − − = d(→ y ,→ x) (m4 ): By the help of property (n4 ), it follows that − − − − d(→ x ,→ y ) = k→ x −→ yk − − − − = k→ x −→ z +→ z −→ yk → − → − → − − ≤ kx − z k+kz −→ yk → − → − → − → − = d( x , z ) + d( z , y ) Hence metric conditions are satisfied. Example 1.2.3. Let C [a, b] be the set of all continuous real valued functions on the interval [a, b] . Show that the function k·k∞ : C [a, b] → R defined as kuk∞ = sup {|u(x)| : x ∈ [a, b]} , describes a norm on the set C [a, b] .

u ∈ C [a, b]

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Solution. We verify the norm conditions (n1 )–(n4 ): (n1 )–(n2 ): We need to show that kuk > 0 if u 6= 0, and kuk = 0 ⇔ u = 0. For all x ∈ [a, b], it follows that kuk∞ = sup |u(x)| > 0

if u 6= 0.

Besides, using the property of absolute value, we get kuk∞ = sup {|u(x)| : x ∈ [a, b]} = 0 ⇔ |u(x)| = 0 ⇔ u(x) = 0 which is the desired result. (n3 ): For all λ ∈ R the property kλuk = |λ| · kuk should be satisfied. Using the property of the given function, we get kλuk∞ = = = =

sup {|λu(x)| : x ∈ [a, b]} sup {|λ| · |u(x)| : x ∈ [a, b]} |λ| sup {|u(x)| : x ∈ [a, b]} |λ| · kuk∞ .

(n4 ): The triangle inequality property ku + vk ≤ kuk + kvk must be satisfied. Using the property of absolute value, it follows that ku + vk∞ = ≤ ≤ =

sup {|u(x) + v(x)| : x ∈ [a, b]} sup {|u(x)| + |v(x)| : x ∈ [a, b]} sup {|u(x)| : x ∈ [a, b]} + sup {|v(x)| : x ∈ [a, b]} kuk∞ + kvk∞ .

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As a result, kuk∞ is a norm and (C [a, b] , kuk∞ ) is a normed space. Example 1.2.4. Given that the function u : [1, 5] → R, is defined by u (x) = x2 − 4x + 7. Evaluate kuk∞ =? Solution. In order to calculate the norm kuk∞ = sup {|u(x)| : x ∈ [1, 5]} , we first check the critical points of the function u. Differentiating u, we obtain u0 (x) = 2x − 4 = 0 =⇒ x = 2. It follows that kuk∞ = sup {|u (1)| , |u (2)| , |u (5)|} = sup {4, 3, 12} = 12.

Definition 1.2.5. Let k·k1 and k·k2 norms be given on the vector space X. For every x ∈ X if there exists c1 , c2 > 0 such that c1 kxk1 ≤ kxk2 ≤ c2 kxk1 , then k·k1 and k·k2 are called equivalent norms. As a result of this definition, we can deduce that different norms can determine the same topology. The following result is based on the fact that any two norms on a finite dimensional vector space determine the same topology.

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Theorem 1.2.6. All the norms defined on finite dimensional normed (vector) spaces are equivalent. As a consequence of Theorem 1.2.6, all the norms defined on finite dimensional normed spaces describe the same topology on these spaces. For instance, if a sequence (xn ) in a norm space X is convergent, bounded or Cauchy sequence with respect to the norm k·k1 (k·k2 ), then it is also convergent, bounded or Cauchy sequence with respect to the norm k·k2 (k·k1 ). Example 1.2.7. Prove that the inequalities (i) kxk∞ ≤ kxk1 ≤ n kxk∞ √ (ii) kxk∞ ≤ kxk2 ≤ n kxk∞ are satisfied for the norms kxk1 =

n X

|xi | ,

i=1

kxk2 =

n X

! 21 x2i

,

i=1

kxk∞ = max |xi | 1≤i≤n

defined in Rn . Notice that, since Rn is finite dimensional, then by Theorem 1.2.6 these three norms are equivalent. Solution. Without loss of generality, let |xk | be the maximum of |x1 | , |x2 | , |x3 | , · · · , |xk | , · · · , |xn | . The proofs are as follows.

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21

(i) By definition of the given norms, we calculate

kxk∞ = max |xi | 1≤i≤n

= max {|x1 | , |x2 | , |x3 | , · · · , |xk | , · · · , |xn |} = |xk | ≤ |x1 | + |x2 | + |x3 | + · · · + |xk | + |xk+1 | + · · · + |xn | n X = |xi | i=1

= kxk1

(1.1)

and

kxk1 =

n X

|xi |

i=1

= ≤ = =

|x1 | + |x2 | + · · · + |xn | |xk | + |xk | + · · · + |xk | n |xk | n kxk∞ .

Combining the results of (1.1) and (1.2), it follows that

kxk∞ ≤ kxk1 ≤ n kxk∞ which is the desired result.

(1.2)

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(ii) We get two inequalities from kxk∞ = max |xi | 1≤i≤n

= |xk | q = |xk |2 q ≤ |x1 |2 + |x2 |2 + · · · + |xn |2 ! 12 n X = |xi |2 i=1

= kxk2

(1.3)

and kxk2 = = ≤ = = =

n X

! 12 |xi |2

qi=1 |x1 |2 + |x2 |2 + · · · + |xn |2 q |xk |2 + |xk |2 + · · · + |xk |2 + · · · + |xk |2 q n |xk |2 √ n |xk | √ n kxk∞ .

Using (1.3) and (1.4), it can be deduced that √ kxk∞ ≤ kxk2 ≤ n kxk∞. This completes the proof.

(1.4)

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23

Example 1.2.8. The norms defined by 1

Z kf k1 =

0

Z 1  21  21 |f (x)|2 dx and kf k2 = (x + 1) |f (x)|2 dx 0

are equivalent in the C [0, 1] space. This can be verified by Z

1

|f (x)|2 dx

 12

 1  21 Z ≤  (x + 1) |f (x)|2 dx

0

0

Z ≤

1

 12 (2) |f (x)|2 dx

0

=

√

Z 2

1

|f (x)|2 dx

 12

0

which implies that kf k1 ≤ kf k2 ≤

√

2 kf k1 .

Definition 1.2.9. A subset of a normed space X is said to be compact if every sequence of points in A has a subsequence which converges in X to an element of A. Definition 1.2.10. Let X and Y be normed spaces with corresponding norms k · kX and k · kY , respectively. A continuous linear operator L ∈ L(X; Y ) is said to be compact if it maps every bounded subset of X onto a relatively compact subset of Y . The following result can be viewed as a consequence of the Heine-Borel theorem for finite dimensional normed spaces. Theorem 1.2.1. Let Y ⊂ X be given such that X is a finite dimensional normed space. The set Y is compact if and only if Y is closed and bounded.

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1.2.1

Banach Space

Every convergent sequence is Cauchy by the theory of metric spaces. The reverse of this statement may not always be true; that is, there are metric spaces and normed spaces in which a Cauchy sequence does not converge. In the following we provide completeness property of a normed space which result a Banach space. Here only fundamental properties of Banach spaces are introduced. Further details can be found in real and functional analysis books given in the reference part. Definition 1.2.11. A normed space (X, k·k) is said to be complete if every Cauchy sequence in X converges to a limit in X. Example 1.2.12. The space R is a complete metric space with respect to its natural metric. Example 1.2.13. Every finite dimensional normed space is complete. Definition 1.2.14. A normed space (X, d) is said to be a Banach space if it is complete as a metric space. Example 1.2.15. The space `∞ (N) = {(xi )∞ i=1 : supi |xi | < ∞} is a complete metric space under the norm k·k∞ . Similarly, `p (N) is complete under the `p -norm. These normed spaces are Banach spaces. Definition 1.2.16. Let X be a normed space. The set of all bounded linear functions on X forms dual space of X denoted by X 0 or X ∗ . The dual space of X is a Banach space with respect to the norm kf kX 0 =

|f (x)| < ∞. x∈X,x6=0 kxkX sup

The dual of a dual space is called second dual of X and it is a linear vector space of the form (X 0 )0 = X 00 . If X = X 00 , then the space X is said to be a reflexive space.

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25

Theorem 1.2.17. The dual X 0 of a normed space X is a Banach space. This theorem is quite important from the following point of view. No matter if the norm space X is a Banach space or not; by the consequence of this theorem, the dual space X 0 is a Banach space. We address the reader to the functional analysis books in the reference for the proof and further details. Example 1.2.18. (The dual space of Lp (Ω) space) The space of p−integrable functions (in the Lebesgue sense) with the norm Z
1/p kukLp (Ω) = |u(x)|p dx , (1 ≤ p < ∞), Ω

is called Lp (Ω) space. The dual of Lp (Ω) space is Lq (Ω) space where 1 1 + = 1. p q Here the numbers p and q are called conjugate numbers. Example 1.2.19. The dual space of L1 (Ω) is the space L∞ (Ω) which is the space of essentially bounded functions on Ω with respect to the norm kukL∞ (Ω) = ess sup |u(x)|. x∈Ω

Example 1.2.20. (The dual space of `p , 1 ≤ p < ∞) The space of all p−summable sequences (xj ) of scalars with the norm kxj kp =

∞ X

p

|xj |

1/p

,

(1 ≤ p < ∞),

j=1

is called `p space. The dual space of `p is `q if 1 ≤ p < ∞ and

1 1 + = 1. p q

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Example 1.2.21. (The dual space of c0 ) The space of all sequences of scalars converging to zero, denoted by c0 , is a subspace of `∞ . The dual space of c0 is `1 . The reader can reach further details on dual spaces and the proofs of the above examples in [9]. In the following, we deal with some different types of convergence results for normed spaces. Theorem 1.2.22. If a normed space X is reflexive, it is complete and hence it is a Banach space. Theorem 1.2.23. Every finite dimensional normed space is reflexive. The proofs of the above theorems are also addressed to [9]. Rn , `p and Lp [a, b] with 1 < p < ∞ are some examples of reflexive spaces. On the other hand `1 , `∞ , L1 [a, b], C[a, b], and c0 are nonreflexive spaces. Definition 1.2.24. A sequence (xn ) in a normed space X is said to be convergent or strongly convergent if there exists x0 ∈ X such that lim kxn − x0 kX = 0.

n→∞

Strongly convergence is denoted by xn → x0 . Definition 1.2.25. A sequence (xn ) in a normed space X is said to be weakly convergent if there exists x0 ∈ X such that lim u (xn ) = u (x0 )

n→∞

for every u ∈ X 0 . Weakly convergence is denoted by w

xn * x0 or xn → x0 .

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27

Definition 1.2.26. A sequence (un ) in X 0 is said to be weak* convergent if for every x ∈ X there exists u ∈ X 0 such that un (x) → u (x) holds. Weak* convergence is denoted by w∗

un −→ u. The results of the above statements on convergence can be listed briefly in the following: xn → x0 ⇐⇒ xn → x0 , w

xn → x0 ⇐⇒ ∀u ∈ X 0 , u (xn ) → u (x0 ) , w∗

un −→ u ⇐⇒ ∀x ∈ X, un (x) → u (x) . Remark 1.2.27. Strong convergence, weak convergence and weak* convergence are equivalent in finite dimensional spaces. Definition 1.2.28. A sequence (xn ) in a normed space X is said to be a Cauchy sequence if and only if for every ε > 0 there exists an integer N such that kxn − xm k < ε is satisfied whenever m, n > N . Theorem 1.2.29. Every finite dimensional vector space is a Banach space. This theorem is a consequence of Theorem 1.2.22 and Theorem 1.2.23 since every finite dimensional normed space is reflexive and thus every reflexive normed space is a Banach space. Example 1.2.30. Let C [a, b] be the space of all real valued continuous functions on [a, b] . C [a, b] is a Banach space with respect to the norm kuk∞ = sup |u (x)| ; x∈[a,b]

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however, it is not a Banach space with respect to the norm 

Zb

kukp = 

 p1 |u (x)|p dx .

a

As a result of that example, a space may or may not be a Banach space depending on the norm defined on it. Theorem1.2.29 may not be valid for infinite dimensional Banach spaces. Some important Banach spaces are given in the following examples. Example 1.2.31. (`p space, 1 ≤ p < ∞). The space `p is a normed space for 1 ≤ p < ∞ since it satisfies all properties of normed spaces. Proof of completeness is left as an exercise. Notice that p = ∞ is a particular case of `p which is a normed space with respect to the norm kxj k∞ = sup{|xj | : j ∈ N}.

(1.5)

The resulting space is called `∞ space which is also a Banach space. Example 1.2.32. (c0 space) The collection of all sequences of scalars converging to 0, denoted by c0 , is a Banach space. Since it is a closed subspace of `∞ , the result can easily be verified if the norm is chosen the same as the one of `∞ space. Notice that in the previous example we used the fact that every closed subset of a complete metric space is also a complete metric space; thus, as Banach spaces are complete, a closed subspace of a Banach space is also a Banach space. An other immediate consequence is that, a Banach subspace of a metric space is closed.

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1.2.2

Inner Product Space

Inner product spaces are special normed spaces. Inner product and inner product spaces are defined in the following: Definition 1.2.33. Let X be a real vector space. If the function h·, ·i : X × X → R satisfies the following properties (i) hu, ui ≥ 0 and hu, ui = 0 ⇐⇒ u = 0, (ii) hu, vi = hv, ui , (iii) hu + v, wi = hu, wi + hv, wi , (iv) hcu, vi = hu, cvi = c hu, vi , then this function is said to be an inner product. Moreover, an inner product space is a vector space X with an inner product defined on X. An inner product space is also known as a pre-Hilbert space. Example 1.2.34. The function h·, ·i : Rn × Rn → R defined by hu, vi =

n X

ui vi

i=1

satisfies inner product properties. This inner product is said to be the standard inner product on Rn . Definition 1.2.35. Two vectors u and v in an inner product space are orthogonal if hu, vi = 0 and denoted by u ⊥ v. Example 1.2.36. For any pair of orthogonal vectors u and v, the identity ku + vk2 = kuk2 + kvk2

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holds. Indeed if u and v are orthogonal, then hu, vi = hv, ui = 0. The result then follows. This property is known as the Pythagorean formula. More generally, if u1 , u2 , · · · , un are orthogonal, one can verify that k

n X k=1

2

uk k =

n X

kuk k2 .

k=1

Theorem 1.2.37. The inner product space (X, h·, ·i) satisfies kx + yk2 + kx − yk2 = 2 kxk2 + 2 kyk2

(1.6)

and i 1h 2 2 2 2 (1.7) kx + yk − kx − yk + i kx + iyk − i kx − iyk hx, yi = 4 p with respect to the norm kxk = hx, xi for all x, y ∈ X. The relations (1.6) and (1.7) are called parallelogram and polar equivalence properties, respectively. Proof. Using the definition of inner product, it follows that kx + yk2 = kxk2 + 2 hx, yi + kyk2 and kx − yk2 = kxk2 − 2 hx, yi + kyk2 . Adding these identities, the relation (1.6) is obtained. On the other hand, to prove (1.7), we calculate kx + yk2 , − kx − yk2 , i kx + iyk2

and

− i kx − iyk2 .

Finally adding these terms together, we obtain the desired result. Theorem 1.2.38. A normed space X is an inner product space if and only if the equation (1.6) is satisfied for all x, y ∈ X.

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n

31



Example 1.2.39. Determine whether the normed space R , k·kp is an inner product or not with respect to the norm ! p1 n X kxkp = xpi i=1

where 1 ≤ p < ∞ and p 6= 2. Solution. We first check the parallelogram rule. Consider the vectors x = (1, 1, 0, 0, · · · , 0) and y = (1, −1, 0, 0, · · · , 0) in Rn . It follows that x + y = (2, 0, 0, 0, · · · , 0) , x − y = (0, 2, 0, 0, · · · , 0) . Then since

1

kxkp = kykp = 2 p , kx + ykp = kx − ykp = 2,   the parallelogram rule fails. As a result the normed space Rn , k·kp is not an inner product space for 1 ≤ p < ∞ and p 6= 2. 1.2.3

Hilbert Space

Every inner product space can be completed to a Hilbert space which has both analytically and geometrically based properties. Because of the existence of a scalar product determining the norm, Hilbert spaces have several applications. Recall that Banach spaces have also fundamental importance in applications, however, there are Banach spaces which are not Hilbert spaces due to the norm that cannot be defined by a scalar product. On the other hand, every Hilbert space is also a Banach space.

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Definition 1.2.40. A complete inner product space (as a metric space) is called a Hilbert space. In other words, if each Cauchy sequence in an inner product space (X, h·, ·i) has a limit in X, then the inner product space (X, h·, ·i) is a Hilbert space. Theorem 1.2.41. Every finite dimensional inner product space is a Hilbert space. Example 1.2.42. The spaces Rn , Cn , `2 , L2 [a, b] and L2 (R) are Hilbert spaces. Notice that `2 and L2 spaces are Hilbert spaces whereas `p and Lp with p 6= 2 are not Hilbert spaces; however, they are Banach spaces. Definition and further properties of Lp spaces are provided in Chapter 2. Example 1.2.43. Given x = (x1 , x2 , · · · , xn ) and y = (y1 , y2 , · · · , yn ) on Rn such that the inner product hx, yi = x1 y1 + x2 y2 + · · · + xn yn be defined. Then Rn space is a Hilbert space with respect to this inner product. Remark 1.2.44. A normed space may not be an inner product space. However, an inner product space is a normed space. For instance, if we take p x = y in the inner product hx, yi , we get hx, xi = kxk . Consequently every Hilbert space is a Banach space; but there are many examples of Banach spaces which are not Hilbert spaces, e.g. l∞ is a Banach space but not a Hilbert space. Example 1.2.45. The space of continuous functions on the unit interval, C[0, 1], with the inner product Z 1 hf, gi = f (t)g(t) dt 0

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is not a Hilbert space. However, C[0, 1] is an inner product space with respect to the given inner product. Example 1.2.46. Lp is not a Hilbert space for p 6= 2. Theorem 1.2.47. Every Hilbert space is reflexive. The proof of this theorem and further details on Hilbert spaces can be found in [13, 14] given in the reference part. 1.3 1.3.1

Some Important Theorems Lebesgue and Monotone Convergence Theorems

Theorem 1.3.1 (Lebesgue Theorem). Let fn : Ω → R+ (n = 1, 2, · · · ) be a sequence of non-negative measurable functions. Then Z X ∞ ∞ Z X fn dµ = fn dµ. Ω n=1

n=1

Ω

Theorem 1.3.2 (Monotone Convergence Theorem-Levi Theorem). Let (Ω, A, µ) be a measure space and suppose that (fn )n≥1 is a monotone increasing sequence of non-negative extended real-valued measurable functions. Then Z Z lim fn dµ = lim fn dµ. n→∞

Ω

Ω n→∞

Notice that each of the integrals in the monotone convergence theorem is defined and not necessarily finite. 1.3.2

Fatou Lemma and Lebesgue Dominated Convergence Theorem

Lemma 1.3.1 (Fatou Lemma). Let (Ω, A, µ) be a measure space and suppose that (fn )n≥1 is a sequence of non-negative extended real-valued measurable functions. Then Z Z lim inf fn dµ ≤ lim inf fn dµ. Ω n→∞

n→∞

Ω

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Fatou’s lemma is a consequence of monotone convergence theorem. On the other hand an important result of Fatou’s lemma is the following dominated convergence theorem. Theorem 1.3.3 (Lebesgue Dominated Convergence Theorem). Let (Ω, A, µ) be a measure space and (fn )n≥1 be a sequence of real-valued measurable functions. Suppose that fn (x) → f (x) and there exists a non-negative integrable function g such that |fn (x)| ≤ g(x) for all x ∈ Ω. Then Z Z Z lim fn dµ = lim fn dµ = f dµ. n→∞

Ω

Ω n→∞

Ω

As a result of this theorem, we have sufficient conditions so that almost everywhere convergence of a sequence of functions implies convergence in L1 norm. The proofs of this subsection and further details can be found in [2, 3, 6, 7] given in the reference part. 1.3.3

Fixed Point Theorems

Definition 1.3.4. Let X 6= ∅ be a set and T : X → X be a function. Then a point x is called a fixed point of T if T (x) = x. Example 1.3.5. Fixed points of the function T : R → R defined by T (x) = x2 are obtained by T (x) = x, x2 = x, which implies x = 0 or x = 1 Example 1.3.6. Fixed points of the function T : R → R defined by T (x) = 4x are obtained by T (x) = 4x, 4x = x,

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which implies x = 0. Example 1.3.7. The function T : R → R defined by T (x) = x + 3 does not have any fixed point because T (x) = x + 3, x+3 = x has no solution. As a result of these examples, a function may have one or more than one fix point or it may not have any fixed point. There are some important fixed point theorems related to existence and uniqueness of fixed points. In the following, after defining contraction mappings, we state the fixed point theorems given by Banach in 1922, by Brouwer in 1912, by Schauder in 1930 and by Tychonoff in 1935. Definition 1.3.8. Let (X, d) be a metric space and T : X → X be a function. If there exists α ≥ 0 such that for all x, y ∈ X, d (T (x) , T (y)) ≤ α d (x, y) then T is called a Lipschitz continuous function. Moreover, • if 0 ≤ α < 1 then T is said to be contraction mapping. • if 0 ≤ α ≤ 1 then T is said to be non-expansive mapping. Theorem 1.3.9 (Banach Fixed Point Theorem). If (X, d) is a complete metric space and T is contraction mapping, then T has a unique fixed point. Proof. We first prove the existence of fixed point. Since T is a contraction mapping, there exists α with 0 ≤ α < 1 such that for all x, y ∈ X, the relation d (T (x) , T (y)) ≤ α d (x, y)

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is satisfied. Let x0 ∈ X be an arbitrary point and xn = T xn−1 be given. Then x1 = T x0 , x2 = T x1 = T 2 x0 , .. . xn−1 = T xn−2 = · · · = T n−1 x0 , xn = T xn−1 = · · · = T n x0 and d (xn+1 , xn ) = d (T xn , T xn−1 ) ≤ αd (xn , xn−1 ) ≤ α2 d (xn−1 , xn−2 ) .. . ≤ αn d (x1 , x0 ) are satisfied. Hence using the generalized triangle inequality for m ≥ n, we get d (xn , xm ) ≤ d (xn , xn+1 ) + d (xn+1 , xn+2 ) + · · · + d (xm−1 , xm ) ≤ αn d (x0 , x1 ) + αn+1 d (x0 , x1 ) + · · · + αm−1 d (x0 , x1 )
= αn + αn+1 + · · · + αm−1 d (x0 , x1 )
= αn 1 + α + α2 + · · · + αm−n−1 d (x0 , x1 ) αn − αm d (x0 , x1 ) . = 1−α The relation 0 ≤ α < 1 implies αn − αm d (x0 , x1 ) → 0. 1−α

An Introduction to Sobolev Spaces 37

Preliminaries

That is, the sequence (xn ) is a Cauchy sequence in X space and hence the space (X, d) is complete. As X is complete, then there is x ∈ X such that lim xn = x.

n→∞

Besides, the relation   T x = T lim xn = lim T (xn ) = lim xn+1 = x n→∞

n→∞

n→∞

shows that x is a fixed point of T . In order to show uniqueness of fixed point, let x and x0 be two unique points. It follows that d (x, x0 ) = d (T x, T x0 ) ≤ α d (x, x0 ) . Using 0 ≤ α < 1, we have d (x, x0 ) = 0. Therefore x = x0 . Remark 1.3.10. Banach fixed point theorem is sufficient condition but not necessary; in other words, there are mappings having fixed points which do not have contraction property. The first fixed point theorem given of the topological form deals with a space X that is a topologically simple subset of Rn and a continuous mapping of X to X. This is Brouwer’s fixed point theorem which guarantees the existence of a fixed point whenever X is the unit ball in Rn and T is continuous. Theorem 1.3.11 (Brouwer Fixed Point Theorem). Let B be a closed ball in Rn given by B = {x ∈ Rn : kx − x0 k ≤ r} , that is B is a compact and convex subset of Rn . Then the continuous mapping T : B → B has at least one fixed point.

8

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Remark 1.3.12. Brouwer fixed point theorem is not valid for the infinite dimensional Banach spaces. Brouwer fixed point theorem was extended by Schauder to the case where X is a compact and convex subset of a normed linear space. Theorem 1.3.13 (Schauder Fixed Point Theorem). Let X 6= Ø be a closed, bounded, convex subset of a Banach space B. Then the continuous mapping T : X → X has at least one fixed point. For a proof of this theorem, refer to [15] where Dugundji’s extension theorem is used. We finalize the chapter by the Tychonoff fixed point theorem which may be considered as an extension of the Schauder fixed point theorem to locally convex topological vector spaces. Theorem 1.3.14 (Tychonoff Fixed Point Theorem). Let A be a locally convex topological (vector) space and X be non-empty, compact and convex subset of this space. Then the continuous mapping T : X → X has at least one fixed point. For the proofs of the above fixed point theorems we address the reader to [2–7] and the reference therein. In addition, many problems of fixed point theorems in multifunctions are widely analyzed in [15].

1.4

Exercises

1. Show that

( 0, d(x, y) = 1,

if x = y if x 6= y

is a metric on any set. This metric is known as the discrete metric.

An Introduction to Sobolev Spaces 39

Preliminaries

2. Show that the vector space E(= R or C) with the discrete metric d (given in the 1st exercise) is a metric space but not a normed space. 3. Prove that the real line R is a metric space. 4. Let (X, d) be a metric space. Show that ρ(x, y) =

d(x, y) 1 + d(x, y)

is a metric on X for all x, y ∈ X. 5. Let p ≥ 1 be an arbitrary real number. Show that Rn is a Banach space with respect to the norm kxkp = (|x1 |p + |x2 |p + · · · + |xn |p )1/p for (x1 , x2 , · · · , xn ) ∈ Rn . 6. Let 0 < p < 1 in the previous example. Show that Rn is not a normed space for that case. 7. Show that Rn is a Banach space with respect to the norm kxk∞ = sup{|x1 |, |x2 |, · · · , |xn |} n

for (x1 , x2 , · · · , xn ) ∈ Rn . 8. Prove that `p space is a Banach space for 1 ≤ p ≤ ∞ with corresponding norms. 9. Prove that R with the usual absolute value is a Banach space. 10. Prove that `p space is not an inner product space and therefore it is not a Hilbert space if p 6= 2. 11. Prove that the space C[a, b] is not a Hilbert space

40

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An Introduction to Sobolev Spaces

12. Let E be the vector space of continuous functions x on [a, b]. If its norm is defined by Z b kxk1 = |x(t)| dt, a

then show that E is a normed space but it is not a Banach space. 13. Let E be the vector space of continuous functions x on [a, b] and p > 1 be a real number. If its norm is defined by Z b 1/p p kxkp = |x(t)| dt , a

then show that E is a normed space but it is not a Banach space. 14. Let X = (0, 1) and d(x, y) = |x − y| be a metric on X. Let ρ(x, y) =

|x − y| 1 + |x − y|

be another metric on X for all x, y ∈ (0, 1). Show that d and ρ are equivalent on (0, 1). 0

15. Show that the dual of `2 space is itself; that is (`2 ) = `2 .

An Introduction to Sobolev Spaces, 2021, 41-85

41

Chapter 2 Lp(Ω) Spaces

Abstract:In this chapter, we deal with the spaces of Lp (Ω). Here p may be a positive finite number or equal to ∞ and Ω is a measurable set. We start by introducing Lp (Ω) space with its norm and provide some simple examples for different choices of Ω. Next part is devoted to several important inequalities which are stated in order as the Young, H¨older and Minkowski inequalities with their corresponding reverse inequalities, the interpolation, Gronwall, Komornik and Nakao inequalities. Moreover, two types of the Green’s identities are presented. After investigating particular cases of finite p, we then consider the infinite case p = ∞, namely L∞ space. The Riesz-Fischer theorem is then analyzed with its proof in detail. In the remaining part, embedding property of Lp (Ω) space, L1loc (Ω) space, the space of continuous functions, C0∞ (Ω) space, H¨older space and C m ([0, T ]; X) space with further theorems and properties are also given out in this chapter. Most of the theorems and propositions in that chapter are expressed with proofs. The statements without proof are addressed to the references given at the end of the book. Particularly [16–20] may be worthwhile.

Keywords: Lp (Ω) spaces, Young inequality, H¨older inequality, Minkowski inequality, Gronwall inequality, Komornik inequality, Nakao inequality, Green’s identities, Riesz-Fischer theorem, L1loc (Ω) space, C0∞ (Ω) space, H¨older space. 2.1

Basic Definitions

In several books (Ω, F, µ) (or (X, F, µ)) notation is used to denote a σfinite measure space. Here Ω (or X) denotes the underlying space, F is the Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

42

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An Introduction to Sobolev Spaces

σ-algebra of measurable sets, and µ is the measure. Whenever 1 < p < ∞, Lp (Ω, F, µ) space consists of all measurable functions on Ω satisfying Z |u (x)|p dµ(x) < ∞. (2.1) Ω

If the underlying space is identified, one can use the notations Lp (Ω, F, µ), Lp (Ω), or only Lp . If u ∈ Lp (Ω, F, µ), then Lp norm of u is defined by Z  p1 p kukLp (Ω,F,µ)) = kukp = |u (x)| dµ(x) . Ω

Remark 2.1.1. If the norm kukLp = 0, this does not imply that u = 0; however, it means that u = 0 almost everywhere (a.e.) for the measure µ. For that reason, the necessity of providing a more accurate definition of Lp yields to introduce an equivalence relation so that if u = v a.e., then u and v are equivalent. Accordingly, Lp space consists of all equivalence classes of functions satisfying (2.1). Notice that as a consequence of the above remark, an error term may appear in practice if the elements in Lp are considered as functions rather than equivalence classes of functions. Throughout this chapter we use either Lp (Ω) or simply Lp notation rather than Lp (Ω, F, µ). Moreover Ω is a measurable set on Rn and µ is equal to Lebesgue measure. Below we provide a formal definition of Lp and its corresponding norm according to the remark and notation given above. Definition 2.1.2. Let Ω be a measurable set on Rn . If u is measurable and 1 < p < ∞ such that |u (x)|p is Lebesgue integrable, that is, Z |u (x)|p dx < ∞, (2.2) Ω

then the functions u (x) are of the class of p−th degree integrable functions and this class is denoted by Lp (Ω) or Lp . The norm in this space is defined

An Introduction to Sobolev Spaces 43

Lp(Ω) Spaces

by Z kukLp (Ω) = kukp =

|u (x)|p dx

 p1 .

(2.3)

Ω

If p = 1, then the space L1 (Ω) consists of all integrable functions on Ω. As a result L1 is a complete normed vector space with the normk · kL1 . On the other hand, if p = 2, the space L2 (Ω) gives a Hilbert space. Remark 2.1.3. Lp space is an example of normed spaces. The triangle inequality fails in general when 0 < p < 1, which implies that k · kLp is not a norm on Lp for this range of p, hence it is not a Banach space. On the other hand the family of Lp spaces with 1 < p < ∞ are important examples of Banach spaces. Example 2.1.4. If Ω = (0, 1) and u (x) = x2 , then evaluate the norm kukL2 (Ω) =? Solution. It follows by the definition of norm that kukL2 (Ω) =

√  12  Z1  12  x5 1  12 5 x2 2 dx = |u (x)|2 dx = = 5 0 5 Ω

Z

0

which is the desired result. Example 2.1.5. If Ω = (0, 4) and u (x) = e3x , then evaluate the norm kukL4 (Ω) =? Solution. The result follows by the definition of norm; that is kukL4 (Ω) =

Z Ω

|u (x)|4 dx

 14

 Z4  14  14  Z4 3x 4 12x = e dx = e dx 0

r  e12x 4  14 48 − 1 4 e = = . 12 0 12

0

44

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Remark 2.1.6. Consider the space Lp (Ω). i. If p = 1, then L1 (Ω) = L (Ω) is the space of integrable functions, ii. If p = 2, then L2 (Ω) is the space of square-integrable functions. Theorem 2.1.7. Lp (Ω) is a linear space. Proof. Let u, v ∈ Lp (Ω) and α ∈ R be given. It follows that i.

Z

p

p

Z

|u (x)|p dx < ∞,

|αu (x)| dx = |α| Ω

Ω

and ii. Z

Z

p

|u (x) + v (x)| dx ≤ Ω

(|u| + |v|)p dx

ZΩ

[2. max {|u| , |v|}]p dx ΩZ ≤ 2p (|u|p + |v|p ) dx ΩZ  Z p p p = 2 |u| dx + |v| dx ≤

< ∞

Ω

are satisfied. Thus it is a linear space. 1 Example 2.1.8. Let Ω = (0, 9) and u (x) = √ . x i. Show that u (x) ∈ L (Ω). ii. Show that u (x) ∈ / L2 (Ω) .

Ω

Lp(Ω) Spaces

An Introduction to Sobolev Spaces

45

Solution. i. Using the definition of L (Ω) it follows that Z Z 9 1 √ √ dx = 2 x 9 = 6 < ∞ |u (x)| dx = x 0 Ω 0 which means the function u ∈ L (Ω) = L (0, 9) . ii. Similarly, we have Z Z 2 |u (x)| dx = Ω

0

2 1 √ dx = ln x |9 = ∞ 0 x

9

which implies that u does not belong to the space L2 (Ω) = L2 (0, 9) . Theorem 2.1.9. Let u, v ∈ Lp (Ω). Then the distance function given by d (u, v) = ku − vkp Z  p1 p = |u − v| dx Ω

is a metric. Remark 2.1.10. The space L2 (Ω) with the inner product Z hu, vi = uv dx Ω

is a Hilbert space. 2.2

Some Important Inequalities

Inequalities are very important in all branches of mathematics. For instance, the inequalities are quite useful while determining whether a given space is metric or normed space and showing the existence and uniqueness of differential equations.

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An Introduction to Sobolev Spaces

Theorem 2.2.1. Let a, b ≥ 0 and 1 ≤ p < ∞ be given. Then the inequality (a + b)p ≤ 2p−1 (ap + bp ) holds. Proof. The proof is obvious for the case p = 1. For the case p > 1, the function f (t) = tp is convex on the interval [0, ∞) . Since the function is convex, then the

Figure 2.1: Graph of the function f (t) = tp .

image of midpoint of two points (a, ap ) and (b, bp ) on this function    p a+b a+b f = 2 2 is smaller than or equal to the midpoint of their images ap + bp ; 2

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

that is, the inequality 

a+b 2

p

ap + bp ≤ 2

is satisfied. This can also be observed by the Figure 2.1. It follows that (a + b)p ≤ 2p−1 (ap + bp ) which is the result.

Theorem 2.2.2 (Young Inequality). Let a, b ≥ 0 and p > 1 be given with 1 1 the identity + = 1. Then the inequality p q ap bq + ab ≤ p q is satisfied. Proof. The graph of function f (x) = xp−1 is given in the Figure 2.2. Here

Figure 2.2: Graph of the function f (x) = xp−1 .

there are three conditions for f (a) and b: i) b > f (a) , ii) b < f (a) and iii) b = f (a). (In the Figure 2.2, it is clear



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

that b > f (a)). In all three cases, the product ab is smaller than the area of shaded region. That is, ab ≤ A + B holds. Thus, the shaded regions are Za A=

Za ydx =

p−1

x

0

0

Zb

Zb

a ap xp dx = = p 0 p

and B=

xdy = 0

0

As a result the inequality

q b y bq y q−1 dx = = . q 0 q

ap b q ab ≤ + p q

is satisfied. Y Remark 2.2.3. Letting a = δX and b = in the Young inequality where δ δ > 0 is a real number, it follows that δ p X p δ −q Y q XY ≤ + . p q 1

Remark 2.2.4 (Young Inequality with ). Substituting a = (εp) p X and Y b= 1 in the Young inequality gives p (εp) XY ≤ εX p + C (ε) Y q q

where C (ε) = (εp)− p q −1 .

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

Theorem 2.2.5 (H¨older Inequality). Let 1 ≤ p ≤ ∞ be given with the 1 1 relation + = 1. If u ∈ Lp (Ω) and v ∈ Lq (Ω) , then uv ∈ L (Ω) and p q kuvk1 ≤ kukp kvkq holds. Proof. The inequality is obvious if u or v is zero. Consider the case u 6= 0, v 6= 0. Then letting |v| |u| ve b = a= kukp kvkq in the Young inequality

ap bq ab ≤ + , p q

it follows that  |u| |v| . ≤ kukp kvkq

|u| kukp

p



|v| kvkq

q

+ p q |u|p |v|q = + p kukpp q kvkqq

holds. Integrating both sides of inequalities over the region Ω yields ! Z Z p q |u| |v| |u| |v| . dx ≤ dx. p + p kukp q kvkqq Ω kukp kvkq Ω The last inequality implies Z Z Z 1 |u|p |v|q |uv| dx ≤ p dx + q dx kukp kvkq Ω Ω p kukp Ω q kvkq Z Z 1 1 p = |u| dx + |v|q dx. p q p kukp Ω q kvkq Ω



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Since

Z

p

|u| dx = Ω

Z

kukpp ,

and Ω

|v|q dx = kvkqq ,

it follows that 1 kukp kvkq Therefore

Z |uv| dx ≤ Ω

1 1 + = 1. p q

Z Ω

|uv| dx ≤ kukp kvkq ,

or equivalently kuvk1 ≤ kukp kvkq holds, which is the desired result. 1 1 1 + = . If u ∈ Lp (Ω) and p q r q r v ∈ L (Ω) , then uv ∈ L (Ω) and the inequality Conclusion 2.2.6. Given p, q, r > 0 with

kuvkr ≤ kukp kvkq holds. Proof. Since the equality equality to the integration Z

r r + = 1 is satisfied, applying the H¨older inp q Z

r

|uv| dx = Ω

|u|r |v|r dx,

Ω

it follows that Z  pr Z  rq Z p q |u|r |v|r dx ≤ (|u|r ) r dx (|v|r ) r dx Ω Ω Ω Z  pr Z  rq p q = |u| dx |v| dx Ω

Ω

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

1 is satisfied. Here taking the −th power of the inequality gives r  p1 Z Z  1r Z  1q |u|p dx |uv|r dx ≤ |v|q dx , Ω

Ω

Ω

and hence kuvkr ≤ kukp kvkq holds. The H¨older inequality can also be written for more than two functions. n Y pi ui Conclusion 2.2.7. Let ui ∈ L (Ω) (1 ≤ i ≤ n) be given for u = i=1

and pi > 0. If

n X 1 1 = , p q i=1 i

then u ∈ Lq (Ω) and the inequality kukq ≤

n Y

kui kpi

i=1

holds. Theorem 2.2.8 (Minkowski Inequality). Let 1 ≤ p < ∞ be given. If u, v ∈ Lp (Ω) , then u + v ∈ Lp (Ω) and the inequality ku + vkp ≤ kukp + kvkp is satisfied. Proof. If p = 1, then we have Z ku + vk1 =

ZΩ

≤

|u + v| dx Z |u| dx + |v| dx

Ω

= kuk1 + kvk1

Ω

2

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

which is the required result. If p > 1, it follows that Z Z p |u + v| dx = |u + v|p−1 · |u + v| dx Ω ZΩ Z p−1 ≤ |u + v| · |u| dx + |u + v|p−1 · |v| dx Ω

Ω

holds. Next we estimate the integrations for the right hand side of the inequality. Using the H¨older inequality gives Z  1q  p1 Z Z p−1 p (p−1)q |u| · |u + v| dx ≤ |u| dx · |u + v| dx . Ω

Since

Ω

Ω

1 1 + = 1 holds, then (p − 1) q = p. This yields p q Z  1q  p1 Z Z |u| · |u + v|p−1 dx ≤ |u + v|p dx |u|p dx · Ω

Ω

Ω p q

= kukp ku + vkp . The inequality Z Ω

p

|v| · |u + v|p−1 dx ≤ kvkp ku + vkpq

can also be obtained in a similar way. Substituting these last two relations on the above inequality, we get Z Z Z p p−1 |u + v| dx ≤ |u| · |u + v| dx + |v| · |u + v|p−1 dx Ω

Ω

Ω p q

p

≤ kukp ku + vkp + kvkp ku + vkpq   p = kukp + kvkp ku + vkpq

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 3

and ku +

vkpp



p q



≤ kukp + kvkp ku + vkp . p q

Finally dividing the last inequality by ku + vkp yields   p− pq ku + vkp ≤ kukp + kvkp , and thus, by the relation p −

p q

= 1, we reach

ku + vkp ≤ kukp + kvkp which is the desired result. Theorem 2.2.9 (Reverse H¨older Inequality). Let 0 < p < 1 be given with Z 1 p

+

1 q

= 1. If u ∈ Lp (Ω) and 0 < |v|q dx < ∞, then the inequality Ω Z Z 1/p  Z 1/q p q |u · v| dx ≥ |u| dx · |v| dx Ω

Ω

Ω

holds. Proof. The product uv ∈ L1 (Ω) is clearly valid; otherwise the left hand side of the inequality can not be finite. Let ϕ = |v|−p and ψ = |uv|p be given. Then ϕψ = |u|p p0 1 0 0 , it implies that q = −pq 0 so that and if we take p = > 1 and q = 0 p p −1 p0 q0 ψ ∈ L (Ω) and ϕ ∈ L (Ω) . Next, by using the H¨older inequality, we get Z Z p |u (x)| dx = ψ (x) ϕ (x) dx Ω

Ω

≤ kψkp0 kϕkq0  10 Z  10 Z p q q0 p0 . |ϕ (x)| dx |ψ (x)| dx = Ω Ω 1−p Z p Z q = |u (x) v (x)| dx . |v (x)| dx . Ω

Ω



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Z Dividing each side of this inequality by

|v (x)|q dx

1−p gives

Ω

 Z −1+p Z p q |u (x)| dx . |v (x)| dx ≤ |u (x) · v (x)| dx ,

Z

p

Ω

Ω

Ω

1 and taking −th power of each side, we get p

Z Ω

Z  −1+p   p1 Z p q p ≤ |v (x)| dx |u (x) · v (x)| dx , |u (x)| dx . Ω

where we used the relation

Ω

−1 + p 1 = . This completes the proof. p q

Theorem 2.2.10 (Reverse Minkowski Inequality). Let 0 < p < 1 be given. If u, v ∈ Lp (Ω) , then the inequality k|u| + |v|kp ≥ kukp + kvkp holds.

Proof. The inequality is valid for u = v = 0. Let the left side of the inequality be greater than zero. Then by the reverse H¨older inequality, it

Lp(Ω) Spaces

An Introduction to Sobolev Spaces

55

follows that k|u| + |v|kpp =

Z

(|u| + |v|)p−1 · (|u| + |v|) dx ZΩ Z p−1 = |u| . (|u| + |v|) dx + |v| · (|u| + |v|)p−1 dx Ω

Ω

 p1 Z  q  1q p p−1 |u| dx · (|u| + |v|) dx

Z ≥ Ω

Ω

Z +

 p1 Z  q  1q |v|p dx · (|u| + |v|)p−1 dx

Ω

Ω

 1q   p kukp + kvkp ≥ (|u| + |v|) dx Ω  p  q = k|u| + |v|kp kukp + kvkp . Z

p q

Next dividing each side of the above inequality by k|u| + |v|kp gives k|u| + Since

p− pq |v|kp

≥ kukp + kvkp .

1 1 p + = 1, it follows that p − = 1 and hence the inequality p q q k|u| + |v|kp ≥ kukp + kvkp

is satisfied. Theorem 2.2.11 (Interpolation Inequality1) ). Let 1 ≤ p < q < r and 0 < θ < 1 be given with 1 θ 1−θ + = . p r q 1)

For 0 < θ < 1 and ∀u ∈ X, the inequalities of the form θ

1−θ

kukY ≤ c kukX kukZ

are called the interpolation inequalities. These inequalities have important roles in analysis. Furthermore they can be used in order to show the existence of solutions of ordinary differential equations.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

If u ∈ Lp (Ω) ∩ Lr (Ω) then u ∈ Lq (Ω) and the inequality kukq ≤ kukθp kuk1−θ r holds. Proof. Let s =

p 1 1 and + 0 = 1 be satisfied. Then s ≥ 1 and θq s s s s = = s−1

p θq

0

p θq

−1

=

p p − θq

are satisfied. Moreover 1 p r θ 1−θ + = =⇒ = p r q p − θq (1 − θ) q yields s r = s − 1 (1 − θ) q

s0 =

so that r < ∞. Next applying the H¨older inequality, it follows that Z kukqq = |u|q dx ZΩ = |u|θq |u|(1−θ)q dx Ω

Z ≤ =

θqs

|u| Ω kukθq p

 1s Z

(1−θ)qs0

|u|

dx

 10 s dx

Ω

kuk(1−θ)q . r

1 Finally taking -th power of each side of the above relation gives q kukq ≤ kukθp kuk1−θ . r The proof can similarly be done for the case r = ∞.

An Introduction to Sobolev Spaces 

Lp(Ω) Spaces

Definition 2.2.12 (Convolution). The convolution of the functions u and v is defined by Z u (x − y) v (y) dy.

(u ∗ v) (x) = Rn

The commutativity property is satisfied for convolution function, that is, for all x ∈ Rn , (u ∗ v) (x) = (v ∗ u) (x) . The convolution function (u ∗ v) is (locally) more regular than u and v.

Theorem 2.2.13 (Young). Let 1 ≤ p < ∞, u ∈ L1 (Rn ) and v ∈ Lp (Rn ) be given. Then u ∗ v ∈ Lp (Rn ) and ku ∗ vkp ≤ kuk1 kvkp holds. Conclusion 2.2.14. Let u ∈ Lp (Rn ) and v ∈ Lq (Rn ) be given with the 1 1 1 relation + = + 1. Then u ∗ v ∈ Lr (Rn ) and p q r ku ∗ vkr ≤ kukp kvkq holds. Conclusion 2.2.15. If p, q, r ≥ 1 with

1 1 1 + + = 2, then p q r

Z (u ∗ v) (x) w (x) dx ≤ kukp kvkq kwkr Rn

holds.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Theorem 2.2.16 (Gronwall Inequality-Differential Form). Let u (t) be a non-negative absolutely continuous function on the interval [0, T ], ϕ (t) and ψ (t) be non-negative summable functions so that the inequality u0 (t) ≤ ϕ (t) u (t) + ψ (t) holds. Then the relation u (t) ≤ e

Rt 0

ϕ(τ )dτ

  Z t u (0) + ψ (τ ) dτ 0

is valid for 0 ≤ t ≤ T . Proof. Differentiating the product of u(t) and exponential function with integral power, it follows that  Rt Rt d  − 0 ϕ(τ )dτ u (t) e = e− 0 ϕ(τ )dτ (u0 (t) − ϕ (t) u (t)) dt Rt ≤ e− 0 ϕ(τ )dτ ψ (t) . Then integrating the above relation on the interval (0, t) yields Z t R Rt t u (t) e− 0 ϕ(τ )dτ − u (0) ≤ e− 0 ϕ(τ )dτ ψ (τ ) dτ, 0

which is equivalent to −

u (t) e

Rt 0

ϕ(τ )dτ

Z ≤ u (0) +

t

e

−

Rt 0

ϕ(τ )dτ

ψ (τ ) dτ,

0

so that −

u (t) e

Rt 0

ϕ(τ )dτ

t

Z ≤ u (0) +

ψ (τ ) dτ 0

is satisfied. Therefore the inequality   Z t Rt ϕ(τ )dτ u (t) ≤ e 0 u (0) + ψ (τ ) dτ 0

is obtained for 0 ≤ t ≤ T .

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

Theorem 2.2.17 (Gronwall Inequality-Integral Form). i. Let v (t) be a non-negative summable function on [0, T ] satisfying the integral inequality Z t v (t) ≤ C1 v (τ ) dτ + C2 0

for almost all 0 ≤ t ≤ T and constants C1 , C2 ≥ 0. Then the inequality
v (t) ≤ C2 1 + C1 teC1 t holds. ii. Specially, if the inequality t

Z v (t) ≤ C1

v (τ ) dτ 0

holds, then it implies that v (t) = 0. Proof. i. Let Z u (t) =

t

v (τ ) dτ 0

be given. Then the inequality Z v (t) ≤ C1

t

v (τ ) dτ + C2 0

yields u0 (t) ≤ C1 u (t) + C2 . By the help of the differential form of the Gronwall inequality, it follows that u (t) ≤ eC1 t (u (0) + C2 t) = C2 teC1 t .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

As a result Z

t

v (t) ≤ C1

v (τ ) dτ + C2 0

= C1 u (t) + C2
≤ C2 1 + C1 teC1 t holds. ii. The result follows by substituting C2 = 0 in part i. Theorem 2.2.18 (Komornik Inequality I). Let h : [0, ∞) → [0, ∞) be a non-increasing function and assume that there exists a constant c > 0 such that Z ∞ h (τ ) dτ ≤ ch (t) (2.4) t

holds for all t ≥ 0. Then, we have t

h (t) ≤ h (0) e1− c ,

∀t ≥ c.

(2.5)

(Komornik 1994, [21]) Proof. For all x ≥ 0, set a function f (x) = e

x c

Z

∞

h (τ ) dτ.

(2.6)

x

Here f is local absolutely continuous and by the inequality (2.4), it is nonincreasing. Next differentiating (2.6) by taking into account (2.4), it follows that Z x 1 x ∞ 0 f (x) = e c h (τ ) dτ + e c (−h (x)) c xZ ∞  1 x h (τ ) dτ − c h (x) = ec c x ≤ 0.

Lp(Ω) Spaces

An Introduction to Sobolev Spaces

61

Since f is non-increasing, f (x) ≤ f (0) is satisfied for all x ≥ 0. Using this fact with the inequality (2.4), we have Z ∞ x ec h (τ ) dτ = f (x) x

≤ f (0) Z ∞ = h (τ ) dτ 0

≤ c h (0) . It follows that

Z

∞

x

h (τ ) dτ ≤ c h (0) e− c .

(2.7)

x

On the other hand, as h is non-negative and non-increasing, we have Z ∞ Z x+c h (τ ) dτ ≥ h (τ ) dτ x x Z x+c ≥ min h (τ ) dτ τ ∈[x,x+c]

x

= c h (x + c) .

(2.8)

Next, by (2.7) and (2.8) the inequality x

h (x + c) ≤ h (0) e− c

holds for all x ≥ 0. If we set x + c = t, it follows that t

h (t) ≤ h (0) e1− c which is the desired result. Theorem 2.2.19 (Komornik Inequality II). Let h : [0, ∞) → [0, ∞) be a non-increasing function and assume that there are two constants α > 0 and c > 0 such that the inequality Z ∞ hα+1 (τ ) dτ ≤ chα (0) h (t) (2.9) t



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

holds for all t ≥ 0. Then, we have  − 1 c + αt α h (t) ≤ h (0) , c + αc

∀t ≥ c.

(2.10)

(Komornik 1994, [21]) Proof. If h (0) = 0, then h (t) = 0. Thus the inequality (2.10) is obviously satisfied. h (t) is defined. Next instead of h (t), Suppose that h (0) 6= 0. Then h (0) h (t) take by assigning h (0) = 1. Thus, for all t ≥ c, the inequality h (0)  h (t) ≤

c + αt c + αc

− α1

should be proven. Let us define a function F : [0, ∞) → [0, ∞) by Z ∞ F (t) = hα+1 (τ ) dτ.

(2.11)

(2.12)

t

So F is local absolutely continuous and non-increasing. By deriving the inequality (2.12), we obtain F 0 (t) = −hα+1 (t) .

(2.13)

On the other hand, taking into account (2.9) and by the assumption h (0) = 1, it follows that Z ∞ F (t) = hα+1 (τ ) dτ t

≤ chα (0) h (t) = ch (t) .

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

This yields h (t) ≥

F (t) . c

(2.14)

Then by the relations (2.13) and (2.14), for almost all interval (0, ∞), the inequality −F 0 ≥ c−α−1 F α+1 holds. Here for B = sup {t : h (t) > 0} , for almost all interval (0, B), the inequality
0 F −α ≥ αc−α−1 (2.15) can be written. (Here notice that F −α (t) is defined for t < B). Next for all s ∈ [0, B), integrating the inequality (2.15) on the interval [0, s] , it follows that F −α (s) − F −α (0) ≥ αc−α−1 s. Then for all s ∈ [0, B) , we obtain F (s) ≤ F −α (0) + αc−α−1 s


− α1

.

(2.16)

Since F (s) = 0 for s ≥ B, the inequality (2.16) is satisfied for all s ≥ 0. Next by (2.9), it follows that F (0) ≤ chα+1 (0) = c. Then the right hand side of the inequality (2.16) becomes
− 1 F −α (0) + αc−α−1 s α ≤ = c

c−α + αc−α−1 s α+1 α

1


− α1

(c + αs)− α .

(2.17)

In addition, since h is non-negative and non-increasing function, the left



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

hand side of the inequality (2.16) becomes Z ∞ F (s) = hα+1 (τ ) dτ s Z c+(α+1)s Z α+1 = h (τ ) dτ + s

Z

∞

hα+1 (τ ) dτ

c+(α+1)s c+(α+1)s

≥

hα+1 (τ ) dτ

s

≥ (c + αs) hα+1 (c + (α + 1) s) .

(2.18)

Then by (2.16)-(2.18), we obtain (c + αs) hα+1 (c + (α + 1) s) ≤ c

α+1 α

1

(c + αs)− α .

It implies that α+1

1

hα+1 (c + (α + 1) s) ≤ c α (c + αs)− α −1  αs − α+1 α . = 1+ c Thus if we set c + (α + 1) s = t, it follows that  − 1 c + αt α h (t) ≤ c + αc which completes the proof. Note: The inequalities (2.5) and (2.10) are satisfied for 0 ≤ t < c. Thus h (t) ≤ h (0). Theorem 2.2.20 (Generalized Komornik Inequality). Let h : [0, ∞) → [0, ∞) be a non-increasing function, ϕ : [0, ∞) → [0, ∞) be an increasing function of class C 1 and ϕ (t) → ∞ for

t → ∞ and

ϕ (0) = 0

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An Introduction to Sobolev Spaces 

be satisfied. Moreover for α ≥ 0, ω > 0 and for all S ≥ 0, the inequality Z ∞ 1 hα+1 (t) ϕ0 (t) dt ≤ hα (0) h (S) (2.19) ω S holds. Then for all t ≥ 0, if α = 0 then h (t) ≤ h (0) e1−ωϕ(t) ,   α1 1+α if α > 0 then h (t) ≤ h (0) . 1 + ωαϕ (t)

(2.20) (2.21)

(Martinez 1999, [22]) Proof. Define a function f : [0, ∞) → [0, ∞) such that
f (τ ) = h ϕ−1 (τ ) . Observe that f is non-increasing and for ∀ 0 ≤ S ≤ T < ∞, we have Z ϕ(T ) Z ϕ(T )
f 1+α (τ ) dτ = h1+α ϕ−1 (τ ) dτ. ϕ(S)

ϕ(S)

Applying the transformation ϕ−1 (τ ) = t gives ϕ (t) = τ and ϕ0 (t) dt = dτ. It follows that Z T Z ϕ(T ) 1+α f (τ ) dτ = h1+α (t) ϕ0 (t) dt ϕ(S)

S

1 α h (0) h (S) ω 1 α f (0) f (ϕ (S)) . = ω

≤

Next setting s = ϕ (S) and using lim ϕ (T ) = ∞, it follows for all s ≥ 0 T →∞ that Z ∞ 1 f 1+α (τ ) dτ ≤ f α (0) f (s) . (2.22) ω s



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

The last inequality (2.22) is a Gronwall type inequality. As a result, by using the Komornik inequalities, for all s ≥ 0 we obtain if

α = 0

then

if

α > 0

then

f (s) ≤ f (0) e1−ωs ,  1 1+α α f (t) ≤ f (0) . 1 + ωαs

The relations (2.20) and (2.21) follow by h (t) = f (ϕ (t)) which completes the proof.

Note: Taking ϕ (t) = t for all t ≥ 0 in the generalized Komornik inequality gives the classical Komornik inequalities. Theorem 2.2.21 (Nakao Inequality). Let ϕ (t) be a non-increasing, nonnegative function defined on [0, ∞) so that for w0 > 0 and α ≥ 0 the inequality ϕ1+α (t) ≤ w0 (ϕ (t) − ϕ (t + 1)) , t ≥ 0 (2.23) holds. Then for each t ≥ 0 ( + ϕ (t) ≤ ϕ (0) e−w1 [t−1] ,

α = 0, 1
− ϕ (t) ≤ ϕ (0)−α + w0−1 α [t − 1]+ α , α > 0,

+

where [t − 1] = max {t − 1, 0} and w1 = ln (Nakao 1977, [23]; Nakao 1978, [24])



Proof. We divide the proof into two parts. i. For α > 0, set y (t) = ϕ−α (t) .

w0 w0 −1



.

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An Introduction to Sobolev Spaces 

It follows that y(t + 1) − y(t) Z 1 d = [θϕ (t + 1) + (1 − θ) ϕ (t)]−α dθ dθ Z0 1 n o −α−1 −α [θϕ (t + 1) + (1 − θ) ϕ (t)] . [ϕ (t + 1) − ϕ (t)] dθ = 0 Z 1 ≥α [θϕ (t + 1) + (1 − θ) ϕ (t)]−α−1 w0−1 ϕ1+α (t) dθ 0 Z 1 −1 1+α [θϕ (t + 1) + (1 − θ) ϕ (t)]−α−1 dθ = αw0 ϕ (t) ≥ =

0 −1 1+α αw0 ϕ (t) ϕ−1−α (t) αw0−1 .

Then y (t + 1) ≥ y (t) + αw0−1 and y (t + 1) ≥ y (0) + αw0−1 t can be deduced. Next by definition of y (t) we obtain ϕ−α (t + 1) ≥ ϕ−α (0) + αw0−1 t and ϕ−α (t) ≥ ϕ−α (0) + αw0−1 [t − 1]+ . As a result we obtain the inequality ϕ (t) ≤ ϕ−α (0) + αw0−1 [t − 1]+ ii. Let α = 0. By the inequality (2.23), we have   1 ϕ (t + 1) ≤ 1 − ϕ (t) w0


− α1

.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Moreover if t ≥ 1 then we have n ≤ t ≤ n+1 for integer n. It follows that   1 ϕ (t) ≤ 1 − ϕ (t − 1) . w0 Then we get  2 1 ϕ (t) ≤ 1 − ϕ (t − 2) w0 .. . n 1 ϕ (t − n) . ≤ 1− w0 Using the last inequality, we obtain  n 1 ϕ (t) ≤ 1 − ϕ (t − n) w0  [t−1]+ 1 ≤ ϕ (0) 1 − w0 = ϕ (0) e

 −1 + −[t−1] ln 1− w1 0

+

= ϕ (0) e−w1 [t−1] .   w0 Notice that here w1 = ln w0 −1 .

2.3

Green’s Identities

For simplicity, we consider elements in R3 . Let u = u (x, y, z) be a function and F = (F1 , F2 , F3 ) be a vector in R3 . In the following we provide some useful notations: • ∇u = grad u = (ux , uy , uz ) ,

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

• ∇ · F = div F = F1x + F2y + F3z , • 4u = div grad u = ∇ · ∇ u = uxx + uyy + uzz , • |∇u|2 = u2x + u2y + u2z , • ∇ · ∇ = ∇2 = ∆.
Theorem 2.3.1 (Green’s First Identity). For u ∈ C 2 (Ω) ∩ C Ω , Z Z Z ∂u v∆u dx = v ds − ∇v∇u dx Ω ∂Ω ∂n Ω where n is the unit vector in the outward normal direction and Proof. In the one-dimensional case, we have (vux )x = vx ux + vuxx Similarly, for higher dimensions, we have ∇ (v∇u) = ∇v∇u + v∆u, v∆u = ∇ (v∇u) − ∇v∇u. Integrating the latter one gives Z Z Z v∆u dx = ∇ (v∇u) dx − ∇v∇u dx. Ω

Ω

Ω

Note that here we used ZZZ

Z · · · dx dy dz =

· · · dx. Ω

Then by divergence theorem, since we have Z Z ∇F dx = F · n ds, Ω

∂Ω

∂u ∂n

= n · ∇u.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

it follows that Z

Z

Z

v∇u · nds − ∇v∇u dx, Z Ω ZΩ Z∂Ω ∂u ds − ∇v∇u dx, v∆u dx = v Ω Ω ∂Ω ∂n v∆u dx =

which is the desired result.
Theorem 2.3.2 (Green’s Second Identity). For u ∈ C 2 (Ω) ∩ C Ω ,  Z Z  ∂v ∂u (u∆v − v∆u) dx = u −v ds. ∂n ∂n Ω ∂Ω Proof. Let F = u∇v − v∇u. It follows that ∇F = = = =

divF ∇ (u∇v − v∇u) ∇u∇v + u∇2 v − ∇v∇u − v∇2 u u∆v − v∆u.

Using this result in the divergence theorem Z Z ∇F dx = F · nds Ω

∂Ω

gives Z

Z (u∆v − v∆u) dx =

Ω

(u∇v − v∇u) · n ds Z∂Ω

(u∇v · n − v∇u · n) ds  Z∂Ω  ∂u ∂v −v ds = u ∂n ∂n ∂Ω =

which is the result.

Lp(Ω) Spaces

2.4

An Introduction to Sobolev Spaces 

L∞ Space

Definition 2.4.1 (L∞ Space). For a measurable function u in a domain Ω, if there exists a constant K such that |u (x)| ≤ K is satisfied almost everywhere (a.e.)2) , then the function u is said to be bounded almost everywhere. The maximum bound for such K constant is called essential supremum of |u| in the region Ω and denoted by ess sup |u (x)| . x∈Ω

The space of almost everywhere bounded functions in Ω is said to be L (Ω) space. In other words,   u : Ω → R : u is measurable and ∞ L (Ω) = . ∃ a constant K such that |u| ≤ K ∞

The norm in this space is kukL∞ (Ω) = ess sup |u (x)| x∈Ω

= inf {K : |u| ≤ K } . Example 2.4.2. a. Is the function u (x) = ex an element of L∞ (R)? b. Is the function v (x) = sin x an element of L∞ (R)? Solution. a. Since the function u (x) = ex is not bounded, it follows that u∈ / L∞ (R). 2)

Let B ⊂ A and measure of B be zero (|B| = 0). Then a property which is satisfied at all points of the set A\B is said to be satisfied almost everywhere of the set A.

2

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

b. Since |v (x)| = |sin x| ≤ 1, it implies that v ∈ L∞ (R). In addition, we have ksin xkL∞ (R) = 1. Theorem 2.4.3 (Riesz-Fischer). Lp (Ω) is a Banach space if 1 ≤ p ≤ ∞. Proof. Case 1. 1 ≤ p < ∞. Let (un ) be a Cauchy sequence in
Lp (Ω) space. Then for j = 1, 2, · · · , there is a subsequence unj of (un ) such that

un − un ≤ 1 . j+1 j 2j If we set

m X un (x) − un (x) , vm (x) = j+1 j j=1

then follows that m X 1 kvm k ≤ 2j j=1

1 1 1 + 2 + ··· + m 2  2  2 m 1 = 1− 2 < 1. =

Taking v (x) = lim vm (x) and using the monotone convergence them→∞ orem, we obtain Z Z |v (x)|p dx = lim |vm (x)|p dx ≤ 1. Ω

m→∞

Ω

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An Introduction to Sobolev Spaces 

So v (x) < ∞ almost everywhere over the region Ω and by Lebesgue’s dominated convergence theorem the series un1 (x) +

∞ X 

 unj+1 (x) − unj (x)

(2.24)

j=1

converges to u (x) almost everywhere in Ω. If we set u (x) = 0, the limit of (2.24) cannot be defined. Whenever the series (2.24) is convergent, the limit lim unm (x) = u (x) m→∞

almost everywhere over the region Ω. Since (un ) is a Cauchy sequence, we have for all ε > 0 there exist m, n ≥ n0 such that kum − un k < ε. Then by Fatou’s lemma, we get Z Z p p |u (x) − un (x)| dx = lim unj (x) − un (x) dx Ω Ω j→∞ Z p ≤ lim inf unj (x) − un (x) dx j→∞ p

Ω

≤ ε.

It follows that for n → ∞, we have u = (u − un ) + un ∈ Lp (Ω)

and

ku − un kp → 0.

Namely, Lp (Ω) is a complete space and hence it is a Banach space. Case 2. p = ∞. If (un ) is a Cauchy sequence in L∞ (Ω) space, then there exists a set A ⊂ Ω having measure zero such that x ∈ / A. In this case, for all m, n = 1, 2, · · · , the relations |un (x)| ≤ sup |un (x)| = kun k∞



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

and |un (x) − um (x)| ≤ kun − um k∞ are satisfied. Thus the sequence (un ) uniformly converges to u over the region Ω\A. Next, if we set u (x) = 0 for x ∈ A, then u ∈ L∞ (Ω) and kun − uk∞ → 0 for n → ∞. As a result, L∞ (Ω) is complete.

Theorem 2.4.4. If u, v ∈ L∞ (Ω) then u + v ∈ L∞ (Ω) and the inequality ku + vk∞ ≤ kuk∞ + kvk∞ is satisfied. Proof. It is clear that u + v ∈ L∞ (Ω) . To prove the inequality, we use the definition of L∞ norm. It follows that ku + vk∞ = ess sup |u (x) + v (x)| x∈Ω

≤ ess sup |u (x)| + ess sup |v (x)| x∈Ω

x∈Ω

= kuk∞ + kvk∞ .

Theorem 2.4.5. If u ∈ L1 (Ω) and v ∈ L∞ (Ω) , then kuvk1 ≤ kuk1 kvk∞ .

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

Proof. The result follows by Z kuvk1 =

|uv| dx ZΩ |u| |v| dx

= ZΩ ≤

|u| ess sup |v (x)| dx ZΩ

= Ω

x∈Ω

|u| dx · kvk∞

= kuk1 kvk∞ .

Some additional properties of Lp (Ω) space are briefly provided in the following theorem. Theorem 2.4.6. Let

1 1 + = 1 and 1 < p < ∞ be satisfied. Then p q

i. Lp (Ω) space is separable, reflexive and its dual space is Lq (Ω) space. ii. L1 (Ω) space is not separable, but it is reflexive and its dual space is L∞ (Ω) space. iii. L∞ (Ω) space is neither separable nor reflexive, but it covers L1 (Ω) space. 2.5

Embedding of Lp (Ω) Spaces

R Theorem 2.5.1. Let vol (Ω) = Ω dx < ∞ and 1 ≤ p ≤ q ≤ ∞ be given. If u ∈ Lq (Ω) , then u ∈ Lp (Ω) and 1

1

kukp ≤ (vol (Ω)) p − q kukq .

(2.25)



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

In other words, the embedding Lq (Ω) ,→ Lp (Ω) holds. If u ∈ L∞ (Ω) , then lim kukp = kuk∞ .

p→∞

Consequently, for 1 ≤ p ≤ ∞ if there exists a constant K such that u ∈ Lp (Ω) and kukp ≤ K are satisfied, then u ∈ L∞ (Ω) and kuk∞ ≤ K. Proof. For the case p = q or q = ∞, the relation (2.25) is obviously satisfied. If 1 ≤ p < q < ∞ and u ∈ Lq (Ω) , then by the H¨older inequality Z

|u|p dx ≤

1− pq  pq Z . |1| dx |u|q dx

Z

Ω

Ω

Ω

Next taking p1 −th power of both sides, it follows that Z

|u|p dx

 p1

Z

|u|q dx

≤

Ω

Ω

Z kukp ≤

Ω

 p1 − 1q , |1| dx

 1q Z Ω

 p1 − 1q 1dx kukq , 1

1

kukp ≤ (vol (Ω)) p − q kukq . The inequality (2.25) yields lim kukp ≤ kuk∞

p→∞

(2.26)

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

On the other hand, let u ∈ L∞ (Ω) and kuk∞ = M > 0 be satisfied. Then for all ε > 0, by definition of supremum it follows that

Z

|u (x)| ≥ M − ε,  p1 Z  p1 |u|p dx ≥ |M − ε|p dx ,

Ω

Ω

 p1

Z kukp ≥ |M − ε|

1dx

.

Ω

So we get lim

p→∞,ε→0

kukp ≥ M = kuk∞ .

(2.27)

As a consequence, from the relations (2.26) and (2.27), we obtain lim kukp = kuk∞

p→∞

which completes the proof. Conclusion 2.5.2. If the region Ω is of finite measure and 1 ≤ p ≤ q ≤ ∞ is satisfied, then L∞ (Ω) ,→ Lq (Ω) ,→ Lp (Ω) ,→ L1 (Ω) . Notice that if the region is not finite, embedding property may always not be true. 1 Example 2.5.3. Let Ω = (1, ∞) and the function u (x) = be given. Then x u ∈ L2 (1, ∞), but u ∈ / L1 (1, ∞) . Example 2.5.4. Let Ω = (1, 3). If u ∈ L2 (Ω) then show that u ∈ L1 (Ω) and √ kuk1 ≤ 2 kuk2 ;



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

that is, show that the embedding L2 (1, 3) ,→ L1 (1, 3) is satisfied. Solution. By the H¨older inequality we have Z  pq Z 1− pq Z . |u|p dx ≤ |u|q dx |1| dx Ω

Ω

Ω

As p = 1, q = 2 and Ω = (1, 3) , it follows that Z 3 Z 3 |u| dx = (|u| .1) dx 1

1

Z ≤

3

|u|2 dx

 12 Z

1− 12

3

|1| dx

1

.

1

Then we get Z3

 1

Z3

|u| dx ≤ 2 2  1

 12 |u|2 dx ,

1

kuk1 ≤

√

2 kuk2 .

As a result L2 (1, 3) ,→ L1 (1, 3) holds. 2.6

Lploc (Ω) Space

In order to define Lploc (Ω) space, we first need the definition of an exact interior region given in the following.

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

Definition 2.6.1. Let Ω and Ω1 be two regions in Rn with Ω1 ⊂ Ω. If Ω1 ⊂ Ω then Ω1 is an exact interior region of Ω and denoted by Ω1 ⊂⊂ Ω. Namely, both Ω1 and the closure of Ω1 are subsets of Ω. Example 2.6.2. Let Ω1 = (1, 7) and Ω2 = (2, 4) . The closure of Ω2 is Ω2 = [2, 4] . Since Ω2 ⊂ Ω1 and Ω2 ⊂, Ω1 we have Ω2 ⊂⊂ Ω1 . Definition 2.6.3. Lploc (Ω) space is defined by Lploc (Ω) = {u : u ∈ Lp (Ω0 ) , ∀Ω0 ⊂⊂ Ω} . Conclusion 2.6.4. By Definition 2.6.3, we have Lp (Ω) ⊂ Lploc (Ω) . 1 Example 2.6.5. Given Ω = (0, 1) and the function u (x) = . Show that x u∈ / L1 (Ω) but u ∈ L1loc (Ω). Solution. u ∈ / L1 (Ω) because we have Z

Z1 1 |u (x)| dx = dx = (ln |x|) |10 = ∞. x 0

Ω

Next consider an interval (a, b) on (0, 1) such that 0 < a < b < 1. It follows that Zb Zb 1 b |u (x)| dx = dx = ln |x| |ba = ln < ∞, x a a

a

which means u ∈ L1loc (Ω) .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

2.7

Space of Continuous Functions

If α = (α1 , α2 , · · · , αn ) is the n-union of non-negative αi ’s, then α is called multiple indices. Let |α| = α1 + α2 + · · · + αn n X = αi i=1

be satisfied. If x = (x1 , x2 , · · · , xn ) ∈ Rn , then xα = xα1 1 xα2 2 · · · xαnn and ∂ |α| u α1 α2 αn D u = α1 α2 αn = D uD u · · · D u. ∂x1 ∂x2 · · · ∂xn α

Example 2.7.1. If α = (2, 1, 3) then |α| = 2 + 1 + 3 = 6 and hence D(2,1,3) u =

∂ 6u = D2 u1 Du2 D3 u3 = ux1 x1 x2 x3 x3 x3 . 2 3 ∂x1 ∂x2 ∂x3

Definition 2.7.2. C (Ω) is said to be the space of continuous functions defined on the region Ω. Moreover, the norm on this space is kukC(Ω) = sup |u (x)| . x∈Ω

Definition 2.7.3. C m (Ω) is the space of m-times continuously differentiable functions (i.e. Dα u are all continuous) on Ω. In particular, C 0 (Ω) = C (Ω) is the space of continuous functions on Ω. The norm on this space is X kukC m (Ω) = sup |Dα u (x)| . |α|≤m

x∈Ω

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

Furthermore, the space C ∞ (Ω) is the space of infinitely all orders differentiable and continuous functions on Ω; that is, ∞

C (Ω) =

∞ \

C m (Ω) .

m=0

Example 2.7.4. a. If u (x) =

√

x, then u ∈ C ([0, 1]) , but u ∈ / C 1 ([0, 1]) .

b. If u (x) = sin x, then u ∈ C ∞ (R) . c. If u (x) = ex , then u ∈ C ∞ (R) . Definition 2.7.5. CB (Ω) is the space of bounded functions on C (Ω) . Moreover, CBm (Ω) is the space of m-times continuously differentiable and bounded functions (i.e. Dm u are continuous and bounded). Namely n o m m CB (Ω) = u ∈ C (Ω) : kukC m (Ω) < ∞ . The following two examples are left as exercises to the reader. Example 2.7.6. If u (x) = sin x, then u ∈ CB∞ (R) . Example 2.7.7. If u (x) = ex , then u ∈ / CB∞ (R) . 2.8

C∞ 0 (Ω) Space and Compact Support

Definition 2.8.1. The set of all functions u defined in Rn so that u and derivatives of u of all order are zero except on a bounded region is said to be functions with compact support. That is, supp u = {x : u (x) 6= 0} If supp u ⊂⊂ Ω, then the function u has compact support on Ω and denoted by C0∞ (Ω) . The elements of C0∞ (Ω) space are called test functions.

2

An Introduction to Sobolev Spaces

Erhan Pis¸kin and Baver Okutmus¸tur

Example 2.8.2. The function u(x) defined by ( 1 e− 4−x2 , x ∈ (−2, 2) u (x) = 0, x∈ / (−2, 2) has compact support; that is, u ∈ C0∞ (Ω) and supp u = [−2, 2] . The graph of u(x) is illustrated in Figure 2.3.

Figure 2.3: Graph of the function u(x).

2.9

H¨older Space

Definition 2.9.1. Let Ω ⊂ Rn , 0 ≤ α ≤ 1, m, c be non-negative real numbers and u (x) is a real or complex valued function defined on Rn . If the inequality |u (x) − u (y)| ≤ c |x − y|α holds, then the function u (x) is called the H¨older continuous and the constant α is called the H¨older exponent. If α = 1, then the function u (x) is said to be Lipschitz continuous. Moreover, if α = 0, u (x) is said to be bounded.

An Introduction to Sobolev Spaces 

Lp(Ω) Spaces


The space of functions u (x) defined on C m Ω space satisfying the H¨older condition |Dα u (x) − Dα u (y)| ≤ c |x − y|α (where 0 ≤ |α| ≤ m
and c is constant) is called the H¨older space and is m,α denoted by C Ω .
Theorem 2.9.2. The H¨older space C m,α Ω with the norm kukC m,α (Ω) = kukC m (Ω) + max sup 0≤|α|≤m x6=y

|Dα u (x) − Dα u (y)| |x − y|α

is a Banach space. 2.10 Lp (a, b; X) Space

Definition 2.10.1. Let −∞ ≤ a < b ≤ ∞ be given. The space of measurable functions defined from (a, b) to X satisfying the condition kukX ∈ Lp (a, b) is called Lp ((a, b) ; X) space. This space is a Banach space with the norm   p1 b  Z      kukp dt , 1 ≤ p < ∞ X kukLp (a,b; X) = a     ess sup kukX , p = ∞.  t∈(a,b)

2.11 C m ([0, T ]; X) Space

Definition 2.11.1. The space of m times continuously differentiable functions defined for all t ∈ [0, T ] from [0, T ] to X is called C m ([0, T ] ; X) space.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

This space is a Banach space with the norm kukC m ([0,T ]; X) = max sup kDα ukX . |α|≤mt∈[0,T ]

Theorem 2.11.2. If 1 ≤ p ≤ ∞, u ∈ Lp (0, T ; X) and ut ∈ Lp (0, T ; X) , then u ∈ C 1 (0, T ; X) .

2.12

Exercises

1. If Ω = (1, 2) and u (x) = x3 , then calculate kukL4 (Ω) =? 2. If Ω = (0, 1) × (0, 2) and u (x, y) = x.ey , then calculate kukL2 (Ω) =? 1 3. Let Ω = (0, 8) and u (x) = √ be given. 3 x (a) Determine if the function u is in L2 (Ω) space? (b) Determine if the function u is in L3 (Ω) space? 4. Given that a, b ≥ 0 and δ > 0. Show that ab ≤ δa2 +

b2 . 4δ

5. Show that u ∈ CB∞ (R) if u (x) = sin x. 6. Show that u ∈ / CB∞ (R) if u (x) = ex .

Lp(Ω) Spaces

An Introduction to Sobolev Spaces 

7. Let u and v be two functions. If p ≥ 1, δ > 0, then prove that p+1

δ p+1 p.δ − p

p−1 p+1 . kukp+1 + kvkp+1

uv |v| ≤ p+1 . 1 p+1 p+1

8. Prove that

Z

u∆udx = − k∇uk2

Ω

1 9. Let the function u (x) = √ and Ω = (0, 4) be given. Show that x 2 2 u∈ / L (Ω) but u ∈ Lloc (Ω) 10. If α = (1, 3, 2) and u (x, y, z) = x3 e2y sin z, then evaluate Dα u =? 11. Let the function f : [0, 1] → R be defined by  0, if x = 0,  f (x) = √ 1 n, if x ∈ n+1 , n1 where n ∈ N. Show that (a) f ∈ Lp (0, 1) for 1 ≤ p < 2; (b) f ∈ / Lp (0, 1) for 2 ≤ p ≤ ∞. 12. Prove that Lp is not a Banach space if 0 < p < 1. 13. If f ∈ L1 (0, 1), prove that Z lim

n→∞

1

xn f (x) dx = 0.

0

14. Show that the function f (x) = sin x+cos x is not Lebesgue integrable for 0 < x < ∞.

An Introduction to Sobolev Spaces, 2021, 86-

Chapter 3 Weak Derivative

Abstract: This chapter is dedicated to applications of the weak derivatives. We extend the classical knowledge of derivative and provide basic properties of weak derivatives in this part. The concept of the weak derivative can be considered as a generalization of the classical derivative. More precisely there are functions which are not differentiable in the classical sense but weakly differentiable. Further properties and analysis are given in the examples of this section. For the applications of weak derivatives and further details, we refer the reader to [25–28].

Keywords: Weak derivative, classical derivative, α-th weak derivative, L1loc (Ω) space, multivariable derivative, generalized weak derivative, Heaviside function, Dirac-delta function. 3.1

Introduction

The notion of weak derivative can be considered as a generalization of the classical derivative. There are weakly differentiable functions which are not differentiable in the classical sense. For instance, the function u (x) = |x| has no classical derivative at x = 0. However, it has a weak derivative. Indeed, if u ∈ C 1 (a, b) and ϕ ∈ C01 (a, b) , by integration by parts we obtain Zb a

u (x) ϕ0 (x) dx = u (x) ϕ (x) |ba −

Zb

u0 (x) ϕ (x) dx,

a Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

An Introduction to Sobolev Spaces 

Weak Derivative

which is equal to Zb

u (x) ϕ0 (x) dx = u (b) ϕ (b) − u (a) ϕ (a) −

a

Zb

u0 (x) ϕ (x) dx,

a

As ϕ ∈ C01 (a, b) , we have ϕ (b) = ϕ (a) = 0. Hence it follows that Zb a

u (x) ϕ0 (x) dx = −

Zb

u0 (x) ϕ (x) dx.

a

Similarly, for higher order and multivariable derivatives, the following definition is useful. Definition 3.1.1. Let the function u ∈ L1loc (Ω) and multi indices α be given. If for every ϕ ∈ C0∞ (Ω) the relation Z Z |α| α u(x)D ϕ(x) dx = (−1) ϕ(x) v(x) dx (3.1) Ω

Ω

is satisfied, then the function v ∈ L1loc (Ω) is called the α-th weak (generalized) derivative of u and denoted by v = Dα u. In several sources, v = vα = Dα u is also used to denote the weak derivative. Remark 3.1.2. Weak derivatives are unique. 3.2

Examples

We provide several examples in this section in order to strengthen the notion of the weak derivative. Example 3.2.1. Find the weak derivative of u (x) = |x| at the point x = 0 where u : R → R.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Solution. Let ϕ ∈ C0∞ (R) and Du (x) = u0 (x) = v (x) . Then we have Z∞

Z∞

uϕ0 dx =

−∞

−∞ Z0

=

|x| ϕ0 dx

(−x) ϕ0 dx +

−∞

Z∞

xϕ0 dx.

0

If we apply integration by parts to the above integration, (i.e. if x = U, ϕ0 dx = dV then dx = dU, ϕ = V ), it follows that Z∞

|x| ϕ0 dx = −xϕ |0−∞ +

−∞

ϕdx + xϕ |∞ 0 −

−∞ ∞ Z

Z0 ϕdx −

=

Z0

−∞

Z∞ ϕdx 0

ϕdx 0

Z∞ sgnx · ϕ dx.

= − −∞

As a result the weak derivative of u (x) = |x| at the point x = 0 is v (x) = sgnx.

Example 3.2.2. Let n = 1 (one-variable) and Ω = (0, 2) be given. Show that the weak derivative of the function  x; 0 < x ≤ 1, u (x) = 1; 1 < x < 2,

An Introduction to Sobolev Spaces 

Weak Derivative

is

 v (x) =

1; 0 < x ≤ 1, 0; 1 < x < 2.

Solution. Let ϕ ∈ C0∞ (0, 2) and Du (x) = v (x) . Then we have Z2

Z1 uD ϕ dx =

Z2 uD ϕ dx +

0

uD ϕ dx

0

1

Z1

Z2

=

xD ϕ dx + 0

D ϕ dx. 1

Applying integration by parts to the first integral on the right hand side of the above equation, we get (If x = U, Dϕdx = dV then dx = dU, ϕ = V ) Z

2

Z

1

ϕ dx + ϕ|21 0 Z 1 = 1 · ϕ (1) − 0.ϕ (0) − ϕdx + ϕ (2) − ϕ (1) 0 Z 1 = − ϕdx + ϕ (2) .

u Dϕ dx = 0

x ϕ|10

−

0

Since ϕ ∈ C0∞ (0, 2), it implies that ϕ (2) = ϕ (0) = 0. Then we obtain Z 2 Z 1 uDϕ dx = − ϕ dx 0 0 Z 1 Z 2 = − 1.ϕ dx − 0.ϕ dx 0 1 Z 2 = − vϕ dx. 0



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

As a result  Du = v =

1; 0 < x ≤ 1, 0; 1 < x < 2

which is the weak derivative of u. Example 3.2.3. If Ω = (−1, 1) , find the weak derivative of the function  1 + x; −1 < x ≤ 0, u (x) = 1 − x; 0 < x < 1. Solution. Let ϕ ∈ C0∞ (−1, 1) and Du (x) = v (x) . It follows that Z0

Z1 uDϕ dx = −1

Z1 uDϕdx +

uDϕ dx

−1 Z0

0

Z1 (1 − x) Dϕ dx

(1 + x) Dϕ dx +

= −1

0

= [(1 + x) ϕ] |0−1 −

Z0

ϕdx + [(1 − x) ϕ] |10 −

−1

Z0 = −

Z1 ϕ dx −

−1 Z1

= −

(−1) ϕdx 0

vϕ dx. −1

Thus we have  v (x) =

1; x ∈ (−1, 0] , −1; x ∈ (0, 1) .

Z1 (−1) ϕ dx 0

An Introduction to Sobolev Spaces 

Weak Derivative

Example 3.2.4. If Ω = (0, 2) , show that the function  x; x ∈ (0, 1] , u (x) = 2; x ∈ (1, 2) , has no weak derivative. Solution. Let ϕ ∈ C0∞ (0, 2) and Du (x) = v (x). It follows that Z2 −

Z2 vϕdx =

0

uDϕ dx 0

Z1 =

Z2 uDϕdx +

uDϕ dx

0

1

Z1

Z2

=

xDϕdx + 0

2Dϕ dx. 1

Applying integration by parts to the first integral of the right hand side of the above equation gives Z2 Z1 uDϕdx = xϕ |10 − ϕdx + 2 ϕ |21 0

0 Z1

ϕdx + 2ϕ (2) − 2ϕ (1)

= ϕ (1) − 0

Z1 = −

ϕdx − ϕ (1) . 0

Then we have

Z2 −

Z1 vϕdx = −

0

ϕdx − ϕ (1) 0

2

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

which is equivalent to Z2

Z1 vϕ dx −

ϕ (1) = 0

ϕ dx. 0

Suppose that the sequence of continuously differentiable functions (ϕm ) satisfies 0 ≤ ϕm ≤ 1, ϕm (1) = 1 and ϕm (x) → 0 for all x 6= 1. If we substitute the sequence of functions (ϕm ) in place of the function ϕ into the above relation, we get  2  Z Z1 1 = lim ϕm (1) = lim  v (x) ϕm (x) dx − ϕm (x) dx = 0 m→∞

m→∞

0

0

which is a contradiction. As a result, the function has no weak derivative. Example 3.2.5. Find the weak derivative of the function  1; x < 0, u (x) = x + 1; x ≥ 0. Solution. Let ϕ ∈ C0∞ (Ω) and Du (x) = v (x). Using the definition of the weak derivative, it follows that Z∞

Z∞

Z0 uDϕdx =

−∞

uDϕ dx + −∞ Z0

=

0 ∞ Z

Dϕ dx + −∞

uDϕ dx

(x + 1) Dϕ dx. 0

Applying integration by parts to the second integral on the right hand side of the above equation,

An Introduction to Sobolev Spaces

Weak Derivative

93

that is, if x + 1 = U, Dϕ dx = dV then dx = dU, ϕ = V ; it follows that Z∞

Z0

Dϕ dx + (x + 1) ϕ |∞ 0 −

uDϕ dx = −∞

−∞

Z∞ ϕ dx 0

Z∞ = ϕ (0) − ϕ (0) −

ϕ dx 0

Z∞ = −

ϕ dx 0

Z∞ =

H (x) ϕ dx. −∞

Here the function  H (x) =

0; x < 0, 1; x ≥ 0,

is the Heaviside function. Now we should find the weak derivative of H (x) . We have Z∞

Z∞

Z0 H (x) Dϕ dx =

−∞

H (x) Dϕ dx + −∞ Z∞

=

H (x) Dϕ dx 0

Dϕ dx 0

= −ϕ (0) Z∞ = − δ (x) ϕ dx. −∞



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Observe that

 δ (x) =

∞; x = 0, 0; x = 6 0,

is the Dirac-Delta function. 3.3

Properties of the Weak Derivative

We terminate the chapter by introducing the following theorem where one can find some basic properties of the weak derivative and compare it with the classical one, particularly by the first two statements. Theorem 3.3.1. i. If a function has (classical) derivative, then it has also weak derivative. However, the reverse may not always be true. That is, a function which has weak derivative may not have (classical) derivative. ii. Classical derivative is pointwise; however, the weak derivative is defined globally. iii. Weak derivative is linear. That is, if u1 , u2 ∈ L1loc (Ω) and c1 , c2 ∈ R, we have Dα (c1 u1 + c2 u2 ) = c1 Dα (u1 ) + c2 Dα (u2 ) . iv. If u has a weak derivative v = Dα u and if v has a weak derivative w = Dβ v = Dβ (Dα u), then Dβ+α u = Dβ (Dα u) = Dβ v = w. An other crucial property of the weak derivative which distinguishes it from the classical derivative is that, for mixed derivative, the order of differentiation is not important. In other words, the weak derivative does not depend on the order of differentiation.

An Introduction to Sobolev Spaces 

Weak Derivative

As a final remark in the following, continuity, differentiability and integrability properties for the weak and the classical derivatives are taken into account. Remark 3.3.2. In the classical sense, a function must be continuous in order to be differentiable. However, the weak derivative does not need neither continuity nor differentiability of the function; integrability is sufficient.

3.4

Exercises

1. Find the weak derivative of the function f : R → R given by f (x) = |x − 2| at the point x = 2. 2. Find the weak derivative of the function u : R → R given by  3x, if 0 < x ≤ 2 u (x) = 2x + 2, if 2 < x < 3 3. Find the weak derivative of the function u : R → R given by  2, if 0 < x ≤ 1 u (x) = x, if 1 < x < 2 4. Let u : R → R be defined as    cos x, x 1− , u (x) = π   0, Calculate the weak derivative of u.

if

−π ≤x≤0 if 0 < x ≤ π otherwise



An Introduction to Sobolev Spaces

Erhan Pis¸kin and Baver Okutmus¸tur

5. Give an example and counterexample for the statements (i) of the Theorem 3.3.1. 6. Prove the statement (iv) of the Theorem 3.3.1. 7. If u ∈ L1loc (a, b) and its weak derivative v(x) = D u(x) = 0, then show that u(x) = c ∈ R for almost all x ∈ (a, b). 8. Let n = 1 (one variable) and Ω = (−1, 1). If u(x) = |x| for all x ∈ (−1, 1), then calculate its weak derivative. 9. Let the function f : R → R be given by f (x) = 21 x|x|, x ∈ R. Find the n-th weak derivative of f for n = 1, 2. 10. Prove that mixed derivatives of the weak derivatives do not depend on the order of differentiation. 11. Let the function f : R → R be given by f (x) = H(x) cos x, x ∈ R where the function H is the Heaviside function. Find the 1st and the 2nd weak derivatives of f .

An Introduction to Sobolev Spaces, 2021, 97-8

Chapter 4 Sobolev Spaces

Abstract: Sobolev spaces were defined by the Russian mathematician Sergei Lvovich Sobolev (1908-1989) in the 1930s. Denoted by W m,p (Ω) , Sobolev spaces are the space of functions whose all m-th order generalized derivatives are in Lp (Ω) space and partial derivatives of these spaces satisfy certain integrability conditions. Notice that generalized derivative refers to the weak derivative which is defined in the previous chapter. In this part we present fundamental properties of Sobolev spaces with several examples. For further reading on Sobolev spaces we cite [29–32].

Keywords:Sobolev space, Banach space, dual space, Sobolev space of real order, Sobolev space of negative order, weighted Lebesgue space, Schwartz space, Plancherel theorem, embedding, weight function. 4.1

Basic Definitions and Examples

Definition 4.1.1. Let Ω be an open set in Rn and m be an arbitrary nonnegative integer such that 1 ≤ p ≤ ∞ be satisfied. Then the space defined by W m,p (Ω) = {u ∈ Lp (Ω) : Dα u ∈ Lp (Ω) , 0 ≤ |α| ≤ m}

(4.1)

is called the Sobolev space. In other words, W m,p (Ω) is the collection of all functions in Lp (Ω) such that all distribution derivatives upto order m are also in Lp (Ω) . The norm in the Sobolev space is described as follows: If 1 ≤ p < ∞, Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

then  p1

 kukW m,p (Ω) = kukm,p = 

X

k Dα ukpLp (Ω) 

0≤|α|≤m

 p1

 = 

X Z 0≤|α|≤m

Ω

|Dα u|pLp (Ω) 

Notice that the Sobolev space W m,p (Ω) is the closure of C ∞ (Ω) with respect to the norm above. On the other hand for p = ∞, we have kukW m,∞ (Ω) = kukm,∞ = max k Dα ukL∞ (Ω) . 0≤|α|≤m

Theorem 4.1.2. W m,p (Ω) space is a Banach space. Proof. Let (un ) be a Cauchy sequence in W m,p (Ω) space. Then for 0 ≤ |α| ≤ m, (Dα u) is a Cauchy sequence in Lp (Ω) space. As Lp (Ω) is exact, for 0 ≤ |α| ≤ m, there exist functions u and uα and in Lp (Ω) space for n→∞ un → u and Dα un → uα . By definition of the weak derivative, it follows that Z Z α uD ϕ dx = lim un Dα ϕ dx n→∞ Ω Ω Z |α| Dα un ϕ dx = lim (−1) n→∞ Ω Z = (−1)|α| Dα uϕ dx Ω

As a result, Dα u = uα , and hence by definition of Sobolev space, we have u ∈ W m,p (Ω) .

An Introduction to Sobolev Spaces 

Sobolev Spaces

Example 4.1.3. Let u = u (x1 , x2 , · · · , xn ) . In order to have u ∈ W 1,2 (Ω) , ∂u ∈ L2 (Ω) u ∈ L2 (Ω) and ∂xi should be satisfied. Here the norm is Z  12 Z kukW 1,2 (Ω) = |u|2 dx + |∇u|2 dx . Ω

Ω

Definition 4.1.4. W0m,p (Ω) space is the set of functions in W m,p (Ω) space for which all derivatives up to the (m−1)-th order are zero at the boundary ∂Ω of Ω. That is, n o m,p m,p 0 (m−1) W0 (Ω) = u ∈ W (Ω) : u|∂Ω = u |∂Ω = · · · = u =0 . ∂Ω

Remark 4.1.5. i. The closure of C0∞ (Ω) space on W m,p (Ω) space is W0m,p (Ω). That is, u ∈ W0m,p (Ω) if and only if there exists a sequence of functions (uk ) ⊂ C0∞ (Ω) such that lim uk (x) = u is satisfied. k→∞

ii. W0m,p (Ω) space is a subspace of W m,p (Ω) space. That is, W0m,p (Ω) ⊂ W m,p (Ω) . Theorem 4.1.6. If 1 ≤ p < ∞ and m is an arbitrary nonnegative integer, then W0m,p (Rn ) = W m,p (Rn ) . Remark 4.1.7. In W m,p (Ω) space we have the following: if m = 0, then W 0,p (Ω) = Lp (Ω) if p = 2, then W m,2 (Ω) = H m (Ω) and furthermore we have kukH01 (Ω) w k∇ukL2 (Ω) .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Definition 4.1.8. W m,2 (Ω) = H m (Ω) space is an inner product (Hilbert) space with the inner product X Z hu, viH m (Ω) = hu, vim,2 = Dα u (x) · Dα v (x) dx |α|≤m

Ω

which is equivalent to Z X Z hu, viH 1 (Ω) = uv dx + Dα u (x) · Dα v (x) dx Ω

1≤|α|≤m

Ω

Example 4.1.9. If Ω = [0, 2π] × [0, 2π] and given that the function u (x) = u (x1 , x2 ) = cos x1 + sin x2 , determine whether u ∈ W 1,2 (Ω)? Solution. If u ∈ W 1,2 (Ω) , then u and its first order derivatives must be in L2 (Ω) ; that is, u ∈ L2 (Ω) ,

∂u ∈ L2 (Ω) ∂x1

and

∂u ∈ L2 (Ω) ∂x2

must be satisfied. In other words, the integrations Z

|u|2 dx =

Ω

Z2π Z2π 0

|cos x1 + sin x2 |2 dx1 dx2 ,

0

Z2π Z2π Z ∂u 2 |− sin x1 |2 dx1 dx2 ∂x1 dx = Ω 0

and

0

Z Z2π Z2π ∂u 2 |cos x2 |2 dx1 dx2 ∂x2 dx = Ω 0

0

An Introduction to Sobolev Spaces 

Sobolev Spaces

must be bounded. It follows that since sine and cosine functions are bounded, i.e., | cos x1 | ≤ 1 and | sin x2 | ≤ 1, the above integrations will also be bounded. Indeed, Z

|u|2 dx =

Ω

Z2π Z2π 0 0 Z2π Z2π

≤

|cos x1 + sin x2 |2 dx1 dx2

|1 + 1|2 dx1 dx2

0 0 Z2π Z2π

4dx1 dx2

= 0

0 2

= 16π < ∞, Z2π Z2π Z ∂u 2 |− sin x1 |2 dx1 dx2 ∂x1 dx = Ω 0 0 Z2π Z2π

≤ 0

|1|2 dx1 dx2

0 2

= 4π < ∞ and Z Z2π Z2π ∂u 2 |cos x2 |2 dx1 dx2 ∂x2 dx = Ω 0 0 Z2π Z2π

≤ 0

0 2

|1|2 dx1 dx2

= 4π < ∞.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Hence, we have u ∈ L2 (Ω) , u ∈ W 1,2 (Ω) .

∂u ∂u ∈ L2 (Ω) and ∈ L2 (Ω) . That is, ∂x1 ∂x2

Example 4.1.10. If Ω = (0, 1) × (0, 2) , show that the function √ u (x) = u (x1 , x2 ) = x1 is not in W 1,2 (Ω) space. Solution. If u ∈ W 1,2 (Ω) , then u and its first order derivatives must be in L2 (Ω) ; that is, u ∈ L2 (Ω) ,

∂u ∂u ∈ L2 (Ω) and ∈ L2 (Ω) ∂x1 ∂x2

should be satisfied. We have Z Z2 Z1 ∂u 2 1 2 √ dx1 dx2 = ∞; 2 x1 ∂x1 dx = Ω 0

which implies that

0

∂u ∈ / L2 (Ω) . Therefore u ∈ / W 1,2 (Ω) . ∂x1

Example 4.1.11. Decide if the function u (x) = x3 is in W 2,2 (0, 1) space or not? Solution. If u ∈ W 2,2 (Ω) , then u and its first and second order derivatives must be in L2 (0, 1) ; that is, ∂u ∂ 2u 2 ∈ L (0, 1) and ∈ L2 (0, 1) u ∈ L (0, 1) , ∂x ∂x2 2

An Introduction to Sobolev Spaces 3

Sobolev Spaces

should be satisfied. Observe that Z

|u|2 dx =

Ω

Z1 0

7 1 3 2 x x dx = = 1 < ∞, 7 0 7

Z 2 Z1 5 1 ∂u 2 2 9x dx = 3x dx = = 9 −2. Then the function Z 2π |u| dx = α+2 Ω is in L2 (Ω) . As a result α > −2 =⇒ u ∈ W −1,2 (Ω) .

4.2

Weighted Lebesgue Lpw (Ω) and Sobolev W m,p w (Ω) Spaces

Definition 4.2.1. The locally integrable function w (x) is called the weight function if w (x) > 0 for almost all x ∈ Rn .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Definition 4.2.2. Let Ω ⊂ Rn be an open region, 0 < p < ∞ and w be a weight function. Then the class of measurable functions u satisfying Z w (x) |u (x)|p dx < ∞ Ω

is said to be the weighted Lebesgue space and it is denoted by Lpw (Ω) . Furthermore the norm on this space is described by Z  p1 p kukp,w = w (x) |u (x)| dx . Ω

Definition 4.2.3. Let Ω be a region in Rn , m be a non-negative integer and 1 ≤ p ≤ ∞ is satisfied. Then the space defined by Wwm,p (Ω) = {u ∈ Lpw (Ω) : Dα u ∈ Lpw (Ω) , 0 ≤ |α| ≤ m} is said to be the weighted Sobolev space. In this Sobolev space, the norm (for 1 ≤ p < ∞) is described by   p1 X kukWwm,p (Ω) = kukm,p,w =  k Dα ukpLpw (Ω)  .

(4.2)

0≤|α|≤m

4.3

H s (Rn ) Sobolev Space

In this subsection, we introduce Sobolev spaces under the Fourier transformations. Definition 4.3.1. The strictly decreasing function space   S = u ∈ C ∞ (Rn ) : lim xβ Dα u (x) = 0, ∀ α, β ∈ Nn |x|→∞

is called the Schwartz space.

An Introduction to Sobolev Spaces 

Sobolev Spaces

Example 4.3.2. The function 2

u (x) = e−|x|

is in S (Rn ) space. More generally, for any polynomial P (x), the function defined by 2 g (x) = P (x) e−|x| is in S (Rn ) space. Example 4.3.3. If k ∈ N, then the function u (x) = 

1 1 + |x|

2

k

is not in S (Rn ) space because h i 2k lim |x| · u (x) 6= 0.

|x|→∞

Definition 4.3.4. The Fourier transform of f : Rn → C denoted by u b is defined by Z 1 u (x) e−iξx dx u b (ξ) = n (2π) 2 n R

n

where u ∈ S (R ). Moreover, the inverse Fourier transform denoted by u e is defined by Z 1 u e (x) = u (ξ) eiξx dξ. n (2π) 2 n R

Theorem 4.3.5 (Plancherel Theorem). If u ∈ L1 (Rn ) ∩ L2 (Rn ), then u b, u e ∈ L2 (Rn ) and kukL2 (Rn ) = kb ukL2 (Rn ) = ke ukL2 (Rn ) are satisfied.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

We are now ready for introducing H m (Rn ) Sobolev space. In the Plancherel theorem, since we have kukL2 (Rn ) = kb ukL2 (Rn ) , then it follows that

2 X X

d 2 α α kD ukL2 (Rn ) =

D u 2 |α|≤m

L (Rn )

|α|≤m

=

X

2

k(iξ)α u bkL2 (Rn )

|α|≤m

=

X Z |α|≤m

=

|iξ|2α |b u (ξ)|2 dξ

Rn

X Z |α|≤m

|ξ|2α |b u (ξ)|2 dξ

Rn



 Z = Rn

X 

u (ξ)|2 dξ. |ξ|2α  |b

|α|≤m

Let X

|ξ|2α = 1 + |ξ|2 + |ξ|4 + · · · + |ξ|2m = Pm (ξ)

|α|≤m

be satisfied. Since

Pm (ξ) m = 1 |ξ|→±∞ (1 + |ξ 2 |) lim

and

Pm (ξ) > 0, (1 + |ξ 2 |)m

then there exist constants c1 , c2 > 0 such that the inequality
m
m c1 1 + ξ 2 ≤ Pm (ξ) ≤ c2 1 + ξ 2

An Introduction to Sobolev Spaces 

Sobolev Spaces

holds. Thus we obtain Z X kDα uk2L2 (Rn ) =


m 1 + ξ 2 |b u (ξ)|2 dξ.

Rn

|α|≤m

From the above result, we reach the following definition. Definition 4.3.6. H m (Rn ) Sobolev space is defined by n
m m n H (R ) = u ∈ L2 (Rn ) : 1 + ξ 2 2 u b (ξ) ∈ L2 (Rn ) ,

+

m ∈ Z ∪ {0} .

Moreover the norm on this space is described by Z  21 2
m 2 kukH m (Rn ) = 1 + ξ |b u (ξ)| dξ n R

2
m2

= 1 + ξ u b 2 n . L (R )

If the integers m ≥ 0 are replaced by real numbers s ≥ 0, we obtain the space n o 2
2s s n 2 n 2 n H (R ) = u ∈ L (R ) : 1 + ξ u b (ξ) ∈ L (R ) and the associated norm is described by Z  21 2
s 2 kukH s (Rn ) = 1 + ξ |b u (ξ)| dξ . Rn

2
1 If we denote 1 + ξ 2 = hξi , then we have  H s (Rn ) = u ∈ L2 (Rn ) : hξis u b (ξ) ∈ L2 (Rn ) , and the corresponding norm is Z kukH s (Rn ) =

s

2s

2

hξi |b u (ξ)| dξ Rn

 12 .

o



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

As 1 + ξ 2 ≥ 1, s1 ≤ s2 implies that
s
s 1 + ξ 2 1 ≤ 1 + ξ 2 2 . Multiplying both sides of this inequality by |b u (ξ)|2 and then taking integration, we obtain the inequality kukH s1 (Rn ) ≤ kukH s2 (Rn ) . In other words, for s1 ≤ s2 , the continuous embedding H s2 (Rn ) ,→ H s1 (Rn ) holds. Furthermore, if s1 < s < s2 , then we have H s2 (Rn ) ,→ H s (Rn ) ,→ H s1 (Rn ) . Theorem 4.3.7. Let s > 0 be a real number. Then n o 2
− 2s −s n 0 n 2 n H (R ) = u ∈ S (R ) : 1 + ξ u b (ξ) ∈ L (R ) .

4.4

Exercises

1. Suppose that Ω = [0, π] and u (x) = sin x. Determine whether the function u is in the W 2,3 (Ω) space? 2. Let Ω = (1, 3) × (0, 4) and u (x1 , x2 ) = x21 x2 be given. Determine if the function u is in the W 1,3 (Ω) space? 3. If Ω = (0, 1) and u (x) = x3 , then evaluate the norm kukW 2,2 (Ω) . 4. Determine if the function u (x) = cos space?

 xπ  2

is in the W01,2 (0, 3)

An Introduction to Sobolev Spaces, 2021, 119-137

119

Chapter 5 Sobolev Embedding Theorems

Abstract: We start this chapter by introducing the embedded property of normed spaces. Next the Sobolev-Gagliardo-Nirenberg inequality and its proof are given. We then define the cone, Lipschitz and C m class conditions and relation between them is analyzed by an example. The remaining part is dedicated to Sobolev embedding theorems with further inequalities such as Sobolev Poincar´e theorem, Poincar´e inequality and logarithmic Sobolev inequalities. Compact embedding and embedding in H s (Rn ) Sobolev spaces are investigated in the final part of the chapter.

Keywords: Embedding property of normed spaces, Sobolev embedding theorems, Sobolev-Gagliardo-Nirenberg inequality, Poincar´e inequality, C m condition, Sobolev Poincar´e theorem, compact embedding, cone condition, logarithmic Sobolev inequality, interpolation inequality. 5.1

Embedding Property

Definition 5.1.1. Let X and Y be two normed spaces and c be a constant not depending on u. Then i. if all elements of X are in Y , (X ⊂ Y ) and ii. if for all u ∈ X, the inequality kukY ≤ c kukX holds, then X is said to be embedded in Y and denoted by X ,→ Y . The notation ,→ replaces by ,→,→ if the embedding is compact. Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

 0

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Further details on compact embedding are given in Section 5.4. Lemma 5.1.1. For all u ∈ C0∞ (Rn ) , the inequality Z∞ u (x) ≤

|u0 (t)| dt

−∞

is satisfied. Proof. u ∈ C0∞ (Rn ) implies that u (−∞) = u (∞) = 0. Then Zx u (x) =

u0 (t)dt

−∞



Zx

≤ 

|u0 (t)| dt +

−∞ ∞ Z

Z∞

 |u0 (t)| dt

x

|u0 (t)| dt

= −∞

which is the desired result. Theorem 5.1.2 (Sobolev-Gagliardo-Nirenberg Inequality). Let n ≥ 2, 1 ≤ p < n and q =

np 1 1 1 ,( = − ) n−p q p n

be given. Then W 1,p (Rn ) ,→ Lq (Rn ) and for all u ∈ C0∞ (Rn ) , there exists a constant c = c (n, p) > 0 such that the inequality kukq ≤ c k∇ukp is satisfied.

An Introduction to Sobolev Spaces  

Sobolev Embedding Theorems

Proof. If p = 1: For u ∈ C0∞ (Rn ), x = (x1 , x2 , · · · , xn ) and integers 1 ≤ i ≤ n, we can write Z∞ |u (x)| ≤ |∂i u (x1 , · · · , xi−1 , t, xi+1 , · · · , xn )| dt. −∞

Here ∂i u denotes the partial derivative with respect to the variable i. Let Z∞ |∂i u (x1 , · · · , xi−1 , t, xi+1 , · · · , xn )| dt

Fi (x) = −∞

and Fi,m (x) =

 ∞ R R∞   ··· |∂i u (x)| dx1 · · · dxm ; if i ≤ m    

−∞ R∞

−∞ R∞

···

−∞

|Fi (x)| dx1 · · · dxm ; if i > m

−∞

be given. Then if the above inequality is written for each variable, we get |u (x)| ≤ F1 (x) , |u (x)| ≤ F2 (x) , .. . |u (x)| ≤ Fn (x) . Multiplying all of these inequalities side by side |u|n ≤ (F1 · F2 · · · Fn ) , and taking

1 n−1 −th

power of both sides yields n

1

|u| n−1 ≤ (F1 · F2 · · · Fn ) n−1 .

 2

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Next integrating the above inequality, we get Z Z n 1 |u| n−1 dx ≤ (F1 · F2 · · · Fn ) n−1 dx. Rn

Rn

Then for k = n − 1 and p1 = p2 = · · · = pk = n − 1, (

1 1 1 + + ··· + = 1) p1 p2 pk

if we apply the H¨older inequality, it follows that Z  1 Z 1 1  n n−1 n−1 n−1 n−1 F1 · F2 · · · F n |u| dx ≤ dx Rn

Rn

1 Z  n−1

Z ≤

F1 dx

1  n−1

Z ···

F2 dx

Rn

Rn

Rn

= (F1,m (x) · F2,m (x) · · · Fn,m (x))

1 n−1

.

Therefore, we get Z

 n−1 n n ≤ |u| n−1 dx

Rn

n Y

! n1 k∂i uk1

i=1

Applying the relation of arithmetic-geometric mean ! n1 n n Y 1X ai ai ≤ n i=1 i=1 for ai > 0, n ∈ N to the above inequality, we reach Z  n−1 n n n 1X n−1 |u| dx ≤ k∂i uk1 , n i=1 Rn and

n

kuk

n n−1

1X ≤ k∂i uk1 . n i=1

.

1  n−1 Fn dx

An Introduction to Sobolev Spaces  

Sobolev Embedding Theorems

Then applying the Cauchy-Schwarz inequality (a1 b1 + a2 b2 + · · · + an bn )2 ≤ a21 + a22 + · · · + a2n




b21 + b22 + · · · + b2n




to the right hand side of the inequality yields r n   X 2 2 2 2 2 2 k∂i uk1 ≤ (1 + 1 + · · · + 1 ) (∂1 u) + (∂2 u) + · · · (∂n u) i=1

√

=

n |∇u| .

Consequently we obtain kuk

n n−1

1 ≤ √ k∇uk1 . n

If p > 1: Substituting uα in place of u in the last inequality gives ∇uf α = α |u|α−1 |∇u| , and hence we get α

ku k It follows that Z |u|

αn n−1

n n−1

α  n−1 n .α

dx

Rn

"Z |u| Rn

αn n−1

1

α−1 ≤ √ α |u| |∇u| . n

 n−1 #α αn

dx

1 ≤ √ α n

Z

1 ≤ √ α n

Z

|u|α−1 |∇u| dx,

Rn

|u|α−1 |∇u| dx.

Rn

Next applying the H¨older inequality for p1 + p10 = 1; it follows that  p1 Z  10 Z p 1 (α−1)p0 p α |u| dx |∇u| dx , kuk αn ≤ √ α n−1 n Rn Rn 1 kukααn ≤ √ α kukα−1 (α−1)p0 k∇ukp . n−1 n

 

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Here if we take

αn (n − 1) p = (α − 1) p0 , then α = and we get n−1 n−p (n−1)p n(p−1) 1 (n − 1) p n−p kuk np ≤ √ kuk npn−p k∇ukp . n−p n−p n n−p

Then dividing each side of the inequality by kuk kuk

(n−1)p n(p−1) n−p − n−p np n−p

Next considering the relation

≤

n(p−1) n−p np n−p

implies

(n − 1) p √ k∇ukp . (n − p) n

n (p − 1) (n − 1) p − = 1, we obtain the n−p n−p

inequality kuk where c =

np n−p

≤ c k∇ukp ,

(n − 1) p √ . (n − p) n

Remark 5.1.3. The constant (n − 1) p √ (n − p) n which depends on n and p is not the best constant. The best constant for p = 1 is 1 h  n i n1 c (n, 1) = √ Γ 1 + . 2 n π  1− np p p−1 , the best If 1 < p < n and a, b > 0 such that u (x) = a + b |x| constant is   n1
 1− p1 n Γ 1 + Γ (n) 1 p−1    2  . c (n, p) = 1 √ n n n − p p n π Γ p Γ 1+n− p c (n, p) =

Here Γ (n) is the Gamma function (Talenti 1976, [32]).

Sobolev Embedding Theorems

5.2

An Introduction to Sobolev Spaces 

Domains and Properties

In the n-dimensional Euclidean space Rn , an open and connected set Ω ∈ Rn is called a domain. In order to use embedding theorems in the domain Ω and for further properties, regularity of the boundary ∂Ω of Ω should be analyzed. For this reason, there are some classifications depending on properties of ∂Ω. In the following some of them are introduced. Definition 5.2.1 (Cone Condition). Let Br2 (y) be an open ball which does not contain the open ball Br1 (x) and the point x in Rn . The set defined by Kx = Br1 (x) ∩ {x + λ(z − x) : z ∈ Br2 (y) , λ > 0} is said to be a finite cone with the top point x. For an open domain Ω in Rn , if all x points of Ω are top points of the cone Kx ⊂ Ω and all Kx cones can be obtained from a cone K by isomorphic and isometric transforms, then Ω said to satisfy the cone condition. Definition 5.2.2 (Lipschitz Condition). Let Ω be a domain and Ux be an open ball of x. If for all point x in Ω, there exists a Lipschitz continuous open ball Ux on the boundary of Ux ∩ Ω, then Ω is said to satisfy the Lipschitz condition. Definition 5.2.3 (C m Class Condition). Let Ω be a domain and Bx be a unit open ball whose center is origin. If ∂Ω is the closure of open sets Ωi and the transformation ψi : Ωi → Bx satisfies the conditions i. ψi (Ωi ∩ ∂Ω) = Bx ∩ ∂Rn+ ii. ψi (Ωi ∩ Ω) = Bx ∩ Rn+

iii. ψi ∈ C m Ωi ve ψi−1 ∈ C m Bx then Ω is a class of C m .

 

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Conclusion 5.2.4. If a domain satisfies the C m condition, then it also satisfies the Lipschitz condition. A domain satisfying the Lipschitz condition will also satisfy the cone condition. In other words C m =⇒ Lipschitz condition =⇒ Cone condition. Example 5.2.5. Consider the three domains given in the Figure 5.1, Figure 5.2 and Figure 5.3.

Figure 5.1: Regular domain

Figure 5.2: C m domain

Figure 5.3: Domain not satisfying cone condition

The domain in Figure 5.1 is regular. In other words the C m , Lipschitz and cone conditions are satisfied. In Figure 5.2, the domain satisfies the Lipschitz and cone conditions; however, C m condition is not satisfied. Finally, the domain in Figure 5.3 does not satisfy the cone condition; and thus, it does not satisfy the Lipschitz and C m conditions. 5.3

Embedding Theorems

We address the reader to [2] and [7] for the proof of the following Sobolev embedding theorem. Theorem 5.3.1 (Sobolev Embedding Theorem). Let Ω satisfies the cone condition in Rn , m ≥ 1 and j ≥ 0 be integers and 1 ≤ p < ∞ such that

An Introduction to Sobolev Spaces  

Sobolev Embedding Theorems

i. If mp > n, the embedding W j+m,p (Ω) ,→ CBj (Ω) is obtained. ii. If mp = n, the embedding W j+m,p (Ω) ,→ W j,q (Ω) ,

p≤q mp, n ≤ mp,

is satisfied if W0m,p (Ω) ,→ Lq (Ω) . If we take m = 1 and p = 2, then  2≤q≤

2n n−2

n>2 +∞ n ≤ 2

implies W01,2 (Ω) ,→ Lq (Ω) , that is, kukq ≤ C kukW01,2 (Ω) is satisfied. Moreover, since we have kukW01,2 (Ω) ≡ k∇ukL2 (Ω) = k∇uk , the proof is done. Theorem 5.3.3. If the embedding W m,p (Ω) ,→ Lq (Ω) is compact for some p ≤ q, then |Ω| < ∞. Theorem 5.3.4. If the embedding W m,p (Ω) ,→ Lq (Ω) exists for some p and q values satisfying 1 ≤ q < p, then |Ω| < ∞.

An Introduction to Sobolev Spaces  

Sobolev Embedding Theorems

Notice that we neglect the proof of Sobolev embedding theorem since the it is quite long and requires further details which are out of scope of this work. Curios readers are addressed to the related sources in the reference part. The rest of this chapter is dedicated for some important inequalities for particular cases. Theorem 5.3.5 (Poincar´e Inequality). If u ∈ W01,2 (a, b) , then kuk2 ≤ c ku0 k2 where c =

b−a √ . 2

Proof. If u ∈ W01,2 (a, b), then u (a) = u (b) = 0. Then since Zx u (x) = u0 (τ ) dτ, a

by the H¨older inequality it follows that  21  x  12  x Z Z 2 u (x) ≤  12 dτ  .  (u0 (τ )) dτ  a

a

which is equivalent to u (x) ≤

√

x − a ku0 k2 .

Next taking square of both sides and integrating on (a, b) , we get Zb a

u2 (x) dx ≤

Zb

2

(x − a) ku0 k2 dx

a 2

= ku0 k2

Zb (x − a) dx a

(b − a)2 0 2 ku k2 . = 2



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Thus we have

(b − a)2 0 2 ≤ ku k2 . 2 If we take square-root of each side, it follows that kuk22

b−a kuk2 ≤ √ ku0 k2 2 which accomplished the proof. Theorem 5.3.6. If u ∈ C01 [a, b] , then √ kukC[a,b] ≤ b − a ku0 kL2 [a,b] . Proof. If u ∈ C01 [a, b] , then we have u (a) = u (b) = 0. Since Zx u (x) =

u0 (τ ) dτ,

a

by the H¨older inequality we obtain  x  12  x  21 Z Z 2 u (x) ≤  12 dτ  ·  (u0 (τ )) dτ  , a

which implies u (x) ≤

a

√

x − a ku0 k2 .

Next taking maximum of both sides on [a, b] for the last inequality, it follows that √ max |u (x)| ≤ ku0 k2 · max x − a, x∈[a,b]

x∈[a,b]

which is equivalent to kukC[a,b] ≤ This completes the proof.

√

b − a ku0 kL2 [a,b] .

An Introduction to Sobolev Spaces 

Sobolev Embedding Theorems

Lemma 5.3.1 (Logarithmic Sobolev Inequality). If u ∈ H01 (Ω), then the inequality Z a2 n 2 2 |u| ln |u| dx ≤ kuk ln kuk + k∇uk2 − (1 + ln a) kuk2 2π 2 Ω

is satisfied (Gross 1975 [5]). Here a > 0 is an arbitrary number and n refers dimension. 5.4

Compact Embedding

Definition 5.4.1. Given an embedding X ,→ Y . If all bounded sequences in X have a convergent subsequence in Y , then the space X is said to be compactly embedded to the space Y and generally this relation is denoted by X ,→,→ Y. Theorem 5.4.2 (Rellich-Kondrachov). Let Ω be an open domain satisfying the cone condition in Rn , Ω0 ⊂ Ω be a bounded domain, m ≥ 1 and j ≥ 0 be integers and 1 ≤ p < ∞. Then the following compact embeddings are satisfied: i. If mp < n, then for all q ∈ [1, p∗ ) , with

1 1 m = − , then p∗ p n

W j+m,p (Ω) ,→,→ W j,q (Ω0 ) is satisfied. ii. If mp = n then for all q ∈ [1, ∞) W j+m,p (Ω) ,→,→ W j,q (Ω0 ) is satisfied.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

iii. If mp > n, then for all q ∈ [1, ∞) W j+m,p (Ω) ,→,→ CBj (Ω0 ) W j+m,p (Ω) ,→,→ W j,q (Ω0 ) are satisfied. If Ω is bounded, we can take Ω0 = Ω. If Ω is an arbitrary domain in Rn , then by taking W0j+m,p (Ω) instead of W j+m,p (Ω) , the embeddings above are again satisfied. If moreover, Ω is an open, bounded and C 1 class domain in Rn , the above embeddings will still be valid. 5.5

Embedding in H s (Rn ) Sobolev Space

Theorem 5.5.1. If s > n2 , then H s (Rn ) ,→ C (Rn ) holds. That is, there exists a constant c > 0 such that for all u ∈ H s (Rn ) the inequality kukC(Rn ) ≤ c kukH s (Rn ) is satisfied. Proof. As we have u (x) =

Z

1 (2π)

n 2

u b (ξ) eiξx dξ,

Rn

it follows that Z 1 iξx |u (x)| = u b (ξ) e dξ n (2π) 2 n Z R ≤ c1 |b u (ξ)| dξ Rn

= c1

Z h Rn

i
− s 2
2s 1+ ξ |b u (ξ)| 1 + ξ 2 2 dξ,

An Introduction to Sobolev Spaces 

Sobolev Embedding Theorems

where we used eiξx = |cos ξx + i sin ξx| = 1. Applying the H¨older inequality to the above relation yields  21    12 Z Z
s
−s 1 + ξ 2 |b u (ξ)|2 dξ  ·  |u (x)| ≤ c1  1 + ξ 2 dξ  Rn

Rn

 Z = c1 kukH s (Rn ) · 

 21 1 dξ  . (1 + |ξ 2 |)s

Rn

Observe that, if we use polar coordinates for the integration   12 Z 1  dξ  , (1 + |ξ 2 |)s Rn

it can be rewritten as

Z∞

rn−1 dr . (1 + r2 )s

0

n This integration is convergent whenever 2s > n =⇒ s > . Hence the 2 inequality |u (x)| ≤ c1 c2 kukH s (Rn ) must be satisfied. Finally taking the maximum of both sides gives kukC(Rn ) ≤ c kukH s (Rn ) . This completes the proof. Theorem 5.5.2. If s >

n 2

+ m, then H s (Rn ) ,→ C m (Rn ) .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Theorem 5.5.3 (Interpolation Inequality in H s (Rn ) Space). Let 1 ≤ s1 < s2 , θ ∈ (0, 1) and s = θs1 + (1 − θ) s2 be given. Then if u ∈ H s2 (Rn ) , the inequality kukH s (Rn ) ≤ kukθH s1 (Rn ) kuk1−θ H s2 (Rn ) holds. Proof. Since u ∈ H s2 (Rn ) , we have u ∈ H s (Rn ) and u ∈ H s1 (Rn ) . It follows that Z
s 1 + ξ 2 |b u (ξ)|2 dξ Rn

Z =


θs +(1−θ)s2 1 + ξ 2 1 |b u (ξ)|2θ+2(1−θ) dξ

Rn

Z =


(1−θ)s2
θs |b u (ξ)|2(1−θ) dξ. 1 + ξ 2 1 |b u (ξ)|2θ 1 + ξ 2

Rn

If we use
θs 1 1 + ξ 2 1 |b u (ξ)|2θ ∈ L θ (Rn ) ,
(1−θ)s2 1 1 + ξ 2 |b u (ξ)|2(1−θ) ∈ L 1−θ (Rn ) , and

1 1 + =1 θ 1−θ in the H¨older inequality, we obtain Z
s 1 + ξ 2 |b u (ξ)|2 dξ Rn  θ  1−θ Z Z
s
s ≤ 1 + ξ 2 1 |b u (ξ)|2  ·  1 + ξ 2 2 |b u (ξ)|2 dξ  , Rn

Rn

An Introduction to Sobolev Spaces 

Sobolev Embedding Theorems

and next taking square-root of each side yields Z  21 2
s 2 1 + ξ |b u (ξ)| dξ Rn

 1−θ 2

 θ2 

 Z ≤


s 1 + ξ 2 1 |b u (ξ)|2  · 

Rn

Z


s 1 + ξ 2 2 |b u (ξ)|2 dξ 

.

Rn

As a result we have kukH s (Rn ) ≤ kukθH s1 (Rn ) kuk1−θ H s2 (Rn ) which is the desired result.

5.6

Exercises

1. Let Ω be a non-empty open and bounded subset of Rn . If f, g ∈ W 1,2 (Ω) = H 1 (Ω), show that f g ∈ W 1,1 (Ω) and ∂ ∂f ∂g (f g) = ·g+f · , ∂xi ∂xi ∂xi

i = 1, 2, · · · , n

in D0 (Ω) and almost everywhere in Ω. 2. Let the Sobolev space W 2,p (0, 1) is equipped with the usual norm ! p1 2 Z 1 X kuk2,p = |Dα u|pLp (Ω) α=0

0

and the space C 1 [0, 1] is equipped with the norm kf kC 1 = max |f (t)| + max |f 0 (t)|, ∀f ∈ C 1 [0, 1]. t∈[0,1]

t∈[0,1]



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Show that W 2,p (0, 1) is compactly embedded into C 1 [0, 1] for p ∈ (1, ∞). 3. If f (x) = and

1 , then show that f ∈ W 1,1 (0, 1) with f (0) = 0 1 + | log x|

f (x) ∈ / L1 (0, 1). x

4. Show that g(x) =

f (x) ∈ W 1,p (0, 1) with g(0) = 0. x

5. Let f ∈ W 2,1 (0, 1) with f (0) = f 0 (0) = 0 and   f (x) , if x ∈ (0, 1], g(x) = x  0, if x = 0. Show that g ∈ W 1,1 (0, 1) 6. Let 1 < p < ∞, I = (0, 1) and f ∈ W01,p (I) be given. Show that the function f (x) g(x) = ∈ Lp (I) x(1 − x) and kgkp ≤ Cp kf 0 kp .

7. Let the function g be defined on (0, 1) such that  f (x), if x ∈ (0, 1), g(x) = 0, if x ∈ R − (0, 1). If f ∈ W 1,p (0, 1) with 1 ≤ p < ∞, then show that g ∈ W 1,p (0, 1)

Sobolev Embedding Theorems

An Introduction to Sobolev Spaces

8. In the previous question, if f ∈ Lp (0, 1) with 1 ≤ p < ∞ and g ∈ W 1,p (R), then show that f ∈ W01,p (0, 1) 9. Let Ω = {(x, y) : 0 < x < 1, 0 < y < x2 } ⊂ R2 . Prove that H 1 (Ω) 6⊂ Lp (Ω) if p > 6. 10. For all f ∈ W 1,1 (R) ⊂ C0 (R) prove that

1

df kf k∞ ≤ . 2 dx 1 1 be defined for x ∈ (0, 1). Prove that x f is H¨older continuous of exponent α ∈ (0, 1); but it is not Lipschitz.

11. Let the function f (x) = x log



An Introduction to Sobolev Spaces, 2021, 18 153

18

Chapter 6 Variable Exponent Lebesgue and Sobolev Spaces

Abstract: In this chapter, we present the variable exponent Lebesgue spaces defined by Orlicz1) in 1931. Although the variable exponent Lebesgue spaces are introduced in 1931, it began to be actively studied in 1990s. We especially used the resources by Diening, Harjulehto, H¨ast¨o, R˚uzˇ iˇcka 2011; Fan, Zhao 2001; Kov´acˇ ik, R´akosnik 1991; Cruz-Uribe, Fiorenza 2013; Radulescu and Repovs 2015 for this section. The notion variable exponent Lebesgue and Sobolev spaces is directly related to the classical Lebesgue and Sobolev spaces where the constant p is replaced with the function p(·) which may depend on a variable. Further properties of these spaces are introduced and analyzed in that chapter.

Keywords: Variable exponent, exponent function, locally log-H¨older continuous, variable exponent Lebesgue space, Luxemburg norm, H¨older inequality, variable exponent Sobolev space, reflexive space, unit ball property, Lipschitz-continuous function.

6.1

Basic Concepts

Definition 6.1.1. Let Ω be a domain in Rn with positive measure (|Ω| > 0). Then the measurable function p : Ω → [1, ∞] 1)

Wladyslaw R. Orlicz (24.05.1903-09.08.1990), Polish mathematician

Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

An Introduction to Sobolev Spaces 39

Variable Exponent Spaces

is called the variable exponent. The family of variable exponent functions is denoted by P (Ω) . Moreover, for p− = essinf p (x) and p+ = esssup p (x) x∈Ω

x∈Ω

the relation 1 ≤ p− ≤ p (x) ≤ p+ ≤ ∞ holds. The domain Ω is composed of three sets p(·) Ω∞ = {x ∈ Ω : p (x) = ∞} , p(·)

Ω1 = {x ∈ Ω : p (x) = 1} , Ωp(·) = {x ∈ Ω : 1 < p (x) < ∞} . ∗ These sets can also be denoted without writing p (·) by Ω∞ , Ω1 , Ω∗ . Definition 6.1.2. The conjugate of the exponent function p (·) is denoted by p0 (·) where 1 1 + 0 =1 p (x) p (x) is satisfied for x ∈ Ω. Furthermore, we can write
0 + (p0 (·)) = p− ,
0 − (p0 (·)) = p+ . Definition 6.1.3. Let Ω ⊂ R with r : Ω → R be given. Then for all x, y ∈ Ω and |x − y| < 12 , if there exists a constant c0 such that the inequality c0 |r (x) − r (y)| ≤ − ln |x − y| is satisfied, then the function r (·) is said to be locally log-H¨older continuous and denoted by r (·) ∈ LH0 (Ω) . If for all x ∈ Ω, there exist constants c0 and r∞ such that c∞ |r (x) − r∞ | ≤ ln (e + |x|)



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

is satisfied, then the function r (·) is said to be log-H¨older continuous at infinity and denoted by r (·) ∈ LH∞ (Ω) . If the function r (·) is both locally log-H¨older continuous and log-H¨older continuous at infinity, then it is denoted by r (·) ∈ LH (Ω) . 6.2

Lp(x) (Ω) Space

Variable exponent Lebesgue space is obtained from the classical Lebesgue space by replacing the constant p with the function p(·) which may depend on a variable. Definition 6.2.1. Let Ω be a domain in Rn such that |Ω| > 0. Then if Z ρp(·) (u) = |u (x)|p(x) dx + ess sup |u (x)| x∈Ω∞ Ω\Ω∞

Z =

|u (x)|p(x) dx + ku (x)kL∞ (Ω∞ ) < ∞

Ω\Ω∞

or p+ < ∞ holds (as |Ω∞ | = 0 will be satisfied), then the class of measurable functions u (x) satisfying Z ρp(·) (u) = |u (x)|p(x) dx < ∞ Ω

is called the variable exponent Lebesgue space and denoted either by Lp(x) (Ω) or by Lp(·) (Ω). Here ρp(·) (u) (or ρ (u)) is called modular functional. Example 6.2.2. If Ω = [1, 4] , p (x) = u ∈ Lp(·) (Ω) .

1 2 and u (x) = 2x , then show that x

An Introduction to Sobolev Spaces 

Variable Exponent Spaces

Solution. Since the function p (x) = have |Ω∞ | = 0. Then Z

p(x)

|u (x)|

ρp(.) (u) = Ω

1 is bounded on the domain Ω, we x

4 Z4 1 14 2x x2 x = dx = 2 dx = 0 : ρp(.) ≤1 λ   p(x) Z   u (x) dx ≤ 1 . = inf λ > 0 :   λ Ω

This norm is called the Luxemburg norm. If we take p (·) = p as a constant, then it turns to be a classical norm in the Lebesgue space. Theorem 6.2.4. Lp(x) (Ω) space is a Banach space. Moreover, if 1 < p− ≤ p+ < ∞ is satisfied, then Lp(x) (Ω) space is reflexive. Theorem 6.2.5. Let u, v ∈ Lp(x) (Ω) and if |u (x)| ≤ |v (x)| is satisfied almost everywhere, then kukp(x) ≤ kvkp(x) . Proof. The inequality |u (x)| ≤ |v (x)| implies |u (x)| |v (x)| ≤ λ λ



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

for all λ > 0. Then using the definition of the function ρp(·) , it follows that     |u (x)| |u (x)| ρp(·) ≤ ρp(·) . λ λ Then we get         |u (x)| |v (x)| λ > 0 : ρp(·) ≤ 1 ⊂ λ > 0 : ρp(·) ≤1 , λ λ which implies         |u (x)| |v (x)| inf λ > 0 : ρp(·) ≤ 1 ≤ inf λ > 0 : ρp(·) ≤1 . λ λ As a result we have kukp(x) ≤ kvkp(x) .

Notice that the modular functional may always not satisfy the triangle inequality ρp(·) (u + v) ≤ ρp(·) (u) + ρp(·) (u) . A similar inequality other than the triangle inequality is given in the following theorem. Theorem 6.2.6. If 1 ≤ p (·) < ∞ and u, v ∈ Lp(x) (Ω) , then the inequality  +  ρp(·) (u + v) ≤ 2p ρp(·) (u) + ρp(·) (u) holds. Proof. We can write |u + v|p(·) ≤ 2p(·) [sup {u, v}]p(·) h i p(·) p(·) p(·) ≤ 2 |u| + |v| h i p(·) p(·) p+ ≤ 2 |u| + |v| .

An Introduction to Sobolev Spaces 

Variable Exponent Spaces

Next integration each side of the above inequality over the domain Ω gives   Z Z Z + |u + v|p(·) dx ≤ 2p  |u|p(·) dx + |v|p(·) dx , Ω

Ω

Ω

which is equivalent to ρp(·) (u + v) ≤ 2p

+



 ρp(·) (u) + ρp(·) (u) .

Theorem 6.2.7. If 1 ≤ p (·) < ∞ and

1 p−

≤ s < ∞, then

k|u|s kp(·) = kukssp(·) Proof. In the definition of norm  Z  s k|u| kp(·) = inf λ > 0 :  Ω

 p(x) s  |u (x)| dx ≤ 1 , λ 

if we substitute λ = µs , it follows that   p(x) Z s   |u (x)| s s k|u| kp(·) = inf µ > 0 : dx ≤ 1   µs Ω   sp(x) Z   u (x) s = inf µ > 0 : dx ≤ 1   µ Ω s   sp(x) Z   u (x) = inf µ > 0 : dx ≤ 1    µ Ω     s u = inf µ > 0 : ρsp(·) ≤1 µ = kukssp(·) which is the desired result.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Theorem 6.2.8. If u, v ∈ Lp(·) (Ω) , then ρp(·) satisfies the following properties. i. ρp(·) (u) ≥ 0 and ρp(·) (|u|) = ρp(·) (u) , ii. ρp(·) (u) = 0 ⇐⇒ u = 0 a.e., iii. If ρp(·) (u) < ∞ then for x ∈ Ω a.e., u (x) < ∞. iv. ρp(·) is convex. In other words, for α, β ≥ 0 and α + β = 1, the inequality ρp(·) (αu + βv) ≤ αρp(·) (u) + βρp(·) (v) holds. Proof. All the statements i.-iv. can be verified easily by using the definition of ρp(·) . Theorem 6.2.9. If u ∈ Lp(·) (Ω) and 0 < kukp(·) < ∞, then ! u ρp(·) ≤ 1. kukp(·) Theorem 6.2.10. If u ∈ Lp(·) (Ω) and kukp(·) ≤ 1, then ρp(·) (u) ≤ kukp(·) . Proof. Since the function ρp(·) (u) = ρp(·) kukp(·) ·

u kukp(·)

!

u kukp(·)

!

is convex, it follows that the inequality ρp(·) (u) ≤ kukp(·) · ρp(·)

An Introduction to Sobolev Spaces 

Variable Exponent Spaces

holds. We know by Theorem 6.2.9 that ρp(·)

u kukp(·)

! ≤ 1;

so by using this fact in the above inequality gives ρp(·) (u) ≤ kukp(·) . This accomplishes the proof. Theorem 6.2.11. Let u ∈ Lp(·) (Ω) . Then kukp(·) ≤ 1

if and only if

ρp(·) (u) ≤ 1.

Proof. (=⇒) If kukp(·) ≤ 1, by Theorem 6.2.10. we have ρp(·) (u) ≤ kukp(·) . Then it is clear that ρp(·) (u) ≤ 1. (⇐=) If ρp(·) (u) ≤ 1, then by using the definition of Luxemburg norm o n u ≤1 , kukp(x) = inf λ > 0 : ρp(·) λ we can conclude that λ = 1. As kukp(·) is the infimum of all λ values, we have kukp(·) ≤ 1.

Theorem 6.2.12. Let u ∈ Lp(·) (Ω). If either kukp(·) > 1 or ρp(·) (u) > 1 is satisfied, then kukp(·) < ρp(·) (u)



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Proof. By Theorem 6.2.11 we have ρp(·) (u) > 1 ⇐⇒ kukp(·) > 1. For that reason it is sufficient to check one of the cases kukp(·) > 1 or ρp(·) (u) > 1. Let ρp(·) (u) > 1 be satisfied. Then we have 1 ≤ 1. ρp(·) (u) Using the convex property of ρp(·) , we can write   u 1 < · ρp(·) (u) = 1. ρp(·) ρp(·) (u) ρp(·) (u) Then the definition of Luxemburg norm implies kukp(·) < ρp(·) (u) which is the desired result. Theorem 6.2.13. If u ∈ Lp(·) (Ω) and for x ∈ Ω a.e. u (x) 6= 0, then ! u = 1. ρp(·) kukp(·) Theorem 6.2.14. Let 1 ≤ p (·) < ∞ and u ∈ Lp(·) (Ω) be given. Then i. If λ > 1, −

+

−

+

ρp(·) (u) ≤ λρp(·) (u) ≤ λp ρp(·) (u) ≤ ρp(·) (λu) ≤ λp ρp(·) (u) holds. ii. If 0 < λ < 1, ρp(·) (u) ≥ λρp(·) (u) ≥ λp ρp(·) (u) ≥ ρp(·) (λu) ≥ λp ρp(·) (u) holds.

An Introduction to Sobolev Spaces 

Variable Exponent Spaces

Proof. i. If λ > 1, then ρp(·) (u) ≤ λρp(·) (u) . Then since −

+

λp ≤ λp(·) ≤ λp , it follows that p−

p−

Z

λ ρp(·) (u) = λ

|u (x)|p(x) dx

Ω

Z = Ω Z

≤

−

λp |u (x)|p(x) dx λp(x) |u (x)|p(x) dx

Ω

= ρp(·) (λu) Z p+ ≤ λ |u (x)|p(x) dx Ω p+

= λ ρp(·) (u) . Finally we obtain −

+

ρp(·) (u) ≤ λρp(·) (u) ≤ λp ρp(·) (u) ≤ ρp(·) (λu) ≤ λp ρp(·) (u) . ii. If 0 < λ < 1, then ρp(·) (u) ≥ λρp(·) (u) . As we have −

+

λp ≥ λp(·) ≥ λp ,



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

it follows that p−

p−

Z

λ ρp(·) (u) = λ

|u (x)|p(x) dx

Ω

Z ≥

λp(x) |u (x)|p(x) dx

Ω

= ρp(·) (λu) Z p+ |u (x)|p(x) dx ≥ λ Ω +

= λp ρp(·) (u) . As a result we get −

+

ρp(·) (u) ≥ λρp(·) (u) ≥ λp ρp(·) (u) ≥ ρp(·) (λu) ≥ λp ρp(·) (u) .

Theorem 6.2.15 (Unit Ball Property). Let 1 ≤ p (·) < ∞ and u ∈ Lp(·) (Ω) be given. Then i. kukp(·) < 1 (= 1; > 1) ⇐⇒ ρp(·) (u) < 1 (= 1; > 1) −

+

ii. If kukp(·) > 1, then kukpp(·) ≤ ρp(·) (u) ≤ kukpp(·) −

+

iii. If kukp(·) < 1, then kukpp(·) ≥ ρp(·) (u) ≥ kukpp(·) .

Proof. i. It follows by Theorems 6.2.10, 6.2.11 and 6.2.12. ii. If kukp(·) > 1 then 1 1 then

< 1. By Theorem 6.2.14 (i.), we have

−

+

λp ρp(·) (u) ≤ ρp(·) (λu) ≤ λp ρp(·) (u) . Letting λ =

1 kukp(·)

in the above inequalities, we get ! 1 u 1 ρ (u) ≤ ρ ≤ p(·) p(·) − + ρp(·) (u) . kukp(·) kukpp(·) kukpp(·)   u Moreover since ρp(·) kuk = 1 by Theorem 6.2.13, it follows that p(·)

1 p−

kukp(·)

ρp(·) (u) ≤ 1 ≤

1 +

kukpp(·)

ρp(·) (u) ,

which is equivalent to −

+

kukpp(·) ≥ ρp(·) (u) ≥ kukpp(·) .

 

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Theorem 6.2.16. Let p (·) ∈ P (Ω) and |Ω∞ | = 0 be given. i. If kukp(·) > 1 then 1

1

1

1

ρp(·) (u) p+ ≤ kukp(·) ≤ ρp(·) (u) p− . ii. If 0 < kukp(·) < 1 then ρp(·) (u) p− ≤ kukp(·) ≤ ρp(·) (u) p+ . Proof. Proof follows by Theorem 6.2.14. In the next theorem, the H¨older inequality for the variable exponent Lebesgue space is given. Comparing with the classical H¨older inequality, notice that there is an additional term in this inequality. Theorem 6.2.17 (H¨older Inequality). Let u ∈ Lp(·) (Ω) , v ∈ Lq(·) (Ω) be given and the inequalities 1 ≤ p− ≤ p (·) ≤ p+ < ∞, 1 ≤ q − ≤ p (·) ≤ q + < ∞,

1 1 + =1 p (·) q (·)

holds. Then uv ∈ L1 (Ω) and   Z 1 1 |uv| dx ≤ + kukp(·) kvkq(·) . p− q − Ω

If u ∈ Lp(·) (Ω), v ∈ Lq(·) (Ω) and the equality 1 1 1 + = p (·) q (·) r (·) holds, then the generalized H¨older inequality kuvkr(·) ≤ c kukp(·) kvkq(·) is satisfied.

An Introduction to Sobolev Spaces  

Variable Exponent Spaces

Theorem 6.2.18 (Embedding in Lp(·) Space). Let Ω be a bounded domain 2) in Rn , 0 < |Ω| < ∞ and p (·) , q (·) ∈ L∞ + (Ω) be given. Then the embedding Lq(·) (Ω) ,→ Lp(·) (Ω) is satisfied if and only if for x ∈ Ω a.e., the inequality p (·) ≤ q (·) holds. In this case, there exists a constant c > 0 such that kukp(·) ≤ c (1 + |Ω|) kukq(·) . 6.3

W m,p(x) (Ω) Space

The variable exponent Sobolev space W m,p(.) (Ω) is obtained from W m,p (Ω) space by replacing the constant term p with the function p(·) that may depend on a variable. Definition 6.3.1. Let Ω ⊂ Rn , m be a positive integer, α = (α1 , α2 , · · · , αn ) be multi-index and p : Ω → [1, ∞) be a measurable function. Then the space defined by n o m,p(.) p(.) α p(.) W (Ω) = u ∈ L (Ω) : D u ∈ L (Ω) , 0 ≤ |α| ≤ m is called the variable exponent Sobolev space. This space is a Banach space with respect to the norm X kukW m,p(·) (Ω) = kukm,p(·) = kDα ukp(·) . 0≤|α|≤m

If 1 < p− ≤ p+ < ∞ is satisfied, then the space W m,p(·) (Ω) is reflexive. m,p(·) Moreover the closure of C0∞ (Ω) space on W m,p(·) (Ω) space is W0 (Ω) . 2)

If 1 ≤ p− ≤ p+ < ∞ then L∞ +

  ∞ (Ω) = u ∈ L (Ω) : essinfu ≥ 1 Ω



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

It is clear that m,p(·)

W0

(Ω) ⊂ W m,p(·) (Ω) .

The last part of this section is devoted to embedding theorems of the variable exponent Sobolev spaces. We skip the proofs of these theorems and address the interested reader to [2, 4, 26, 29, 33]. Theorem 6.3.2. Let Ω be a bounded domain in Rn , 0 < |Ω| < ∞ and p (·) , q (·) ∈ L∞ + (Ω) be given. If p (·) ≤ q (·) , then we have continuous embedding W m,q(·) (Ω) ,→ W m,p(·) (Ω) . Theorem 6.3.3. Let Ω be an open domain satisfying cone condition in Rn , 0 < m < np , the function p (·) be a Lipschitz-continuous function satisfying n 1 < p− ≤ p+ < m over the domain Ω and q (·) : Ω → R for x ∈ Ω a.e. the relation np (·) p (·) ≤ q (·) ≤ p∗ (·) = n − mp (·) be satisfied. Then we have continuous embedding W m,p(.) (Ω) ,→ Lq(·) (Ω) . Theorem 6.3.4. Let Ω be an open domain satisfying cone condition in Rn , 0 < m < np , the function p (·) be a Lipschitz-continuous function satisfying n 1 < p− ≤ p+ < m over the domain Ω and q (·) : Ω → R for x ∈ Ω a.e. the inequality p (·) ≤ q (·) and essinf [p∗ (x) − q (x)] > 0 x∈Ω

be satisfied. Then we have continuous embedding W m,p(·) (Ω) ,→ Lq(·) (Ω)

An Introduction to Sobolev Spaces  

Variable Exponent Spaces

Theorem 6.3.5. Let Ω be a bounded domain in Rn and p (·) : Ω → [1, n) 1 be a continuous function. For 0 < ε < n−1 and for all x ∈ Ω, if 1 ≤ q (·) ≤ p∗ (·) − ε then kukq(·) ≤ c k∇ukp(·) . Theorem 6.3.6 (Poincar´e Inequality). Let Ω be a smooth domain in Rn and m,p(·) p (·) ∈ LH (Ω) be given. Then for all u ∈ W0 (Ω) , the inequality kukp(·) ≤ c k∇ukp(·) holds where c = c (n, p (·) , Ω) .

6.4

Exercises

1. Let 2 ≤ m ∈ N and 1 ≤ p < ∞ with mp < n. Then for all j ∈ N np and for all p ≤ q ≤ , prove that W m+j,p (Rn ) is continuously n − kp embedded in W j,q (Rn ). 2. Consider the previous exercise (Ex 1); prove that W m,p (Rn ) is continuously embedded in Lq (Rn ). 3. Prove that for all n ≤ q < ∞, the space W 1,n (Rn ) is continuously embedded in the space Lq (Rn ). 4. Let 2 ≤ m ∈ N and 1 ≤ p < ∞ with mp = n. Then for all j ∈ N and for all p ≤ q < ∞, prove that W m+j,p (Rn ) is continuously embedded in W j,q (Rn ). 5. Consider the previous exercise (Ex 4); prove that W m,p (Rn ) is continuously embedded in Lq (Rn ).

An Introduction to Sobolev Spaces, 2021, 154-183

Chapter 7 Applications on Differential Equations

Abstract:In this chapter we analyze the solution spaces of some differential equations. Mainly the local existence of solutions of nonlinear Timoshenko equation and the existence of solutions of sixth order Boussinesq equation are analyzed. The main results of this part are based on the papers [34, 35]. For further details we also cite [36, 37] and the reference therein.

Keywords: Cauchy problem, weak solution, sixth-order Boussinesq equation, Timoshenko Equation, Gronwalls inequality, blow up, contraction mapping, Young inequality, H¨older inequality, Sobolev-Poincar´e inequality. 7.1

Weak Solutions in Ordinary Differential Equations

We consider the boundary value problem 0

(−p (x) y 0 ) + q (x) y = f (x) , a < x < b, y (a) = y (b) = 0.

(7.1) (7.2)

of the time-independent heat conduction along a metal rod. Here p (x) is the thermal conductivity coefficient, q (x) is heat transfer coefficient between the rod and the outdoor, f (x) is the heat source density and y (x) is the temperature function. Definition 7.1.1. Let the function y = y (x) be continuous on [a, b] with continuous second derivative on (a, b), satisfying the equation (7.1) at the Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

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Applications on Differential Equations

points x = a and x = b on the interval (a, b) , that is, y (x) ∈ C 2 (a, b) ∩ C [a, b] , then y (x) is the classical solution of the problem (7.1)-(7.2). In the following theorem, the existence and uniqueness of this boundary value problem is given. Theorem 7.1.2. If the inequalities 0 < c0 ≤ p (x) ≤ c1 , 0 ≤ q (x) ≤ c2 , and p (x) ∈ C 1 [a, b] and q (x) , f (x) ∈ C [a, b] are satisfied, then the problem (7.1)–(7.2) has classical solution. Example 7.1.3. If p (x) = 1, q (x) = 0, f (x) = e−x and a = 0, b = 2, then write and solve the linear differential equation. Solution. Substituting the given values in (7.1)–(7.2), we get linear differential equation −y 00 = e−x , 0 < x < 2, y (0) = y (2) = 0. The solution of this equation is y = −e−x +



1 − e2 2e2

 x + 1.

Furthermore since y ∈ C 2 (0, 2) ∩ C [0, 2] , it is a classical solution.

 6

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Notice that the smoothness condition of the above theorem may not always be satisfied. In the following example the heat density is not smooth and hence the classical solution does not exist. Example 7.1.4. If p (x) = 1, q (x) = 0, a = 0, b = 1 and the function is given by  1, 0 ≤ x < x0 < 1, f (x) = 4, x0 < x ≤ 1, then we get 00



−y =

1, 0 ≤ x < x0 < 1. 4, x0 < x ≤ 1.

Since f (x) ∈ / C [a, b] , there is no classical solution. If the rod is made of two different material, then the heat conductivity constant may not be a smooth function. Example 7.1.5. If f (x) = 1, q (x) = 0, a = 0, b = 1 and the conductivity constant is given by  3, 0 ≤ x < x0 < 1, p (x) = 4, x0 < x ≤ 1, then p (x) ∈ / C 1 [a, b] . Therefore there is no classical solution of this problem. As we observed in these examples, many sort of problems the function may not be smooth (continuous, differentiable) enough. For that reason, we extend the notion of classical solutions to weak (generalized) solutions and thus, Sobolev spaces get involved at that point.

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If we take into account the problem (7.1)–(7.2) with the condition y ∈ C 2 (a, b) not necessarily be satisfied. Multiplying both sides of the equation (7.1) with an infinitely differentiable smooth function ϕ ∈ C0∞ (a, b) and integrating the result equation on the interval [a, b] , it follows that Zb h i 0 0 (−p (x) y ) + q (x) y − f (x) ϕ (x) dx = 0, a

or equivalently Zb h

i (−p (x) y ) ϕ (x) + q (x) yϕ (x) − f (x) ϕ (x) dx = 0. 0 0

a

Applying integration by parts to the first term of the integration gives b

−p (x) y 0 ϕ (x) |a +

Zb

[p (x) y 0 ϕ0 (x) + q (x) yϕ (x) − f (x) ϕ (x)] dx = 0.

a

Since ϕ ∈ C0∞ (a, b) , we have ϕ (a) = ϕ (b) = 0. Therefore we get Zb

[p (x) y 0 ϕ0 (x) + q (x) yϕ (x)] dx =

a

Zb f (x) ϕ (x) dx.

(7.3)

a

Now suppose that the constants p (x) and q (x) in (7.3) satisfy 0 < c0 ≤ p (x) ≤ c1 , 0 ≤ q (x) ≤ c2 . It follows that Zb

Zb f (x) ϕ (x) dx ≤ c1

a

a

y 0 ϕ0 (x) dx + c2

Zb yϕ (x) dx. a

 

Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Then by the H¨older inequality, we get 

Zb

Zb

f (x) ϕ (x) dx ≤ c1  a

 12  2

|y 0 | dx . 

 12 2 |ϕ0 (x)| dx

a

a



Zb

Zb

+c2 

 12  |y|2 dx . 

Zb

 12 |ϕ (x)|2 dx . (7.4)

a

a

One can easily observe by the right hand side of the inequality (7.4) that the function y and its generalized derivative are in L2 (a, b) space. That is, y ∈ W 1,2 (a, b) and ϕ (x) ∈ W 1,2 (a, b) . Notice that we do even not need the smoothness of the coefficients p (x) and q (x) . As a result, by the inequality (7.4), we can extend the result as in the following theorem for the coefficients and solution. Definition 7.1.6. For all ϕ ∈ W01,2 (a, b) , the function y ∈ W01,2 (a, b) satisfying the equation Zb a

[p (x) y 0 ϕ0 (x) + q (x) yϕ (x)] dx =

Zb f (x) ϕ (x) dx

(7.5)

a

is called the weak (generalized) solution of the boundary value problem (7.1)–(7.2). Here the equation (7.5) is called integral identity. Next we introduce the existence and uniqueness theorem for weak solutions.

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Theorem 7.1.7. If the inequalities 0 < c0 ≤ p (x) ≤ c1 and 0 ≤ q (x) ≤ c2 , are satisfied and f (x) ∈ L2 (a, b) , then the boundary value problem (7.1)–(7.2) has a weak solution and it is unique. 7.2

Local Existence of Solutions of Nonlinear Timoshenko Equation

In this section, we analyze the locally existence of the problem given by    2 p−1 q−1 2   utt + 4 u − M k∇uk 4 u + |ut | ut = |u| u, (x, t) ∈ Ω × (0, T ) , u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) ,   ∂ u (x, t) = ∂ν u (x, t) = 0,

x ∈ Ω, x ∈ ∂Ω, (7.6)

[Pis¸kin 2015, [34]]. (H1) Let M (s) = m1 + m2 sγ be locally positive Lipschitz function and L be Lipschitz constant. Moreover, for s ≥ 0, the inequality M (s) ≥ m0 > 0 be satisfied. (H2) Let p and q satisfy the following conditions  1 < p < ∞, n≤2 n+2 n > 2, 1 < p ≤ n−2 ,  1 < q < ∞, n≤2 n 1 < q ≤ n−2 , n > 2.



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Now we can demonstrate the locally existence of the problem (7.6). Theorem 7.2.1. Let (H1) and (H2) be satisfied. Suppose that (u0 , u1 ) ∈ H02 (Ω) × L2 (Ω) . Then the equation (7.6) has a unique solution in the space described by
u ∈ C [0, T ) ; H02 (Ω) ,
ut ∈ C [0, T ) ; L2 (Ω) ∩ Lp+1 (Ω × (0, T )) . Moreover, at least one of the following statements is valid. i. T = ∞ (Global Existence), ii. If t → T − , then kut k2 + k∆uk2 → ∞ (Blow up). Proof. Let T > 0 and R0 > 0 be two parameters such that the space depending on these parameters is defined by ( v ∈ C([0, T ); H02 (Ω)), vt ∈ C([0, T ); L2 (Ω) ∩ Lp+1 (Ω × (0, T ))) XT,R0 = e(v(t)) ≤ R02 , t ∈ [0, T ], v(0) = u0 , vt (0) = u1 where e (v (t)) = kvt k2 + k∆vk2

(7.7)

Thus XT,R0 space is a complete metric space with respect to the metric d (u, v) = sup e (u (t) − v (t)) . 0≤t≤T

Next define a non-linear transformation v ∈ XT,R0 , u = Sv where the function u = Sv

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be the unique solution of    2 p−1 q−1 2   utt + 4 u − M k∇vk 4 u + |ut | ut = |v| v, u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) ,   ∂ u (x, t) = ∂ν u (x, t) = 0.

(7.8)

Now considering T >0 and R0 > 0, we want to show the following statements: i. S : XT,R0 → XT,R0 , (S is an onto mapping form the space XT,R0 to XT,R0 space). ii. S is a contraction mapping in XT,R0 space associated to the metric d (·, ·) . Firstly we prove the statement i.. Multiply the equation   2 2 utt + 4 u − M k∇vk 4 u + |ut |p−1 ut = |v|q−1 v by 2ut , we get 

2

2ut utt + 2ut 4 u − M k∇vk

2



2ut 4 u + 2ut |ut |p−1 ut

= 2ut |v|q−1 v. Then integrating the last equation on Ω gives Z Z 2ut utt dx + 2ut 42 udx Ω

Ω

Z −



M k∇vk

2



Ω

Z = Ω

2ut |v|q−1 v dx,

Z 2ut 4 udx + Ω

2ut |ut |p−1 ut dx



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

which implies    d  2 2 2 2 kut k + k∆uk + M k∇vk k∇uk + 2 kut kp+1 p+1 dt   Z  d  2 = M k∇vk k∇uk2 + 2ut |v|q−1 v dx dt Ω

and d e1 (u (t)) + 2 kut kp+1 p+1 dt   Z  d  2 2 k∇uk + 2ut |v|q−1 vdx M k∇vk = dt Ω

= I1 + I2 ,

(7.9)

where 2



2

2

e1 (u (t)) = kut k + k∆uk + M k∇vk



k∇uk2 .

(7.10)

We next try to get estimates for I1 and I2 . If L is Lipschitz constant, then it follows that Z  2 0 ∇v∇vt dx k∇uk2 I1 = 2M k∇vk Ω

≤ 2L k∇vk kvt k k∇uk2 ≤ LR02 e1 (u (t))

(7.11)

On the other hand using the H¨older and Sobolev-Poincar´e inequalities, I2 can be estimated by I2 ≤ 2 kvkq2q kut k ≤ 2C∗q k∆vkq kut k 1

≤ 2C∗q R0q [e1 (u (t))] 2 .

(7.12)

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Applications on Differential Equations

If we write (7.11) and (7.12) in (7.9), we obtain d e1 (u (t)) + 2 kut kp+1 p+1 = I1 + I2 dt 1 . ≤ LR02 e1 (u (t)) + 2C∗q R0q [e1 (u (t))] 2(7.13) Integrating this inequality on the interval [0, T ] yields ZT 

 d e1 (u (t)) + 2 kut kp+1 p+1 dt dt

0

≤

ZT 

LR02 e1 (u (t))

+

1 2C∗q R0q [e1 (u (t))] 2



dt

0

Then applying the Gronwall’s inequality to the last inequality, it follows that ZT   1 2 p+1 q q 2 e1 (u (t)) + 2 kut kp+1 dt ≤ e1 (u0 ) + 2C∗ R0 T eLR0 T 0 2

= C (u0 , u1 , R0 , T ) eLR0 T

(7.14)

Since by (7.7) and (7.10) we have e (u (t)) ≤ e1 (u (t)) , it follows by (7.14) that ZT e (u (t)) + 2

2

LR0 T . kut kp+1 p+1 dt ≤ C (u0 , u1 , R0 , T ) e

(7.15)

0

Next writing the value e (u (t)) in this inequality results kut k2 + k∆uk2 + 2

ZT 0

2

LR0 T kut kp+1 p+1 dt ≤ C (u0 , u1 , R0 , T ) e



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Hence we have
u ∈ L∞ [0, T ) ; H02 (Ω) , and
ut ∈ C [0, T ) ; L∞ (Ω) ∩ Lp+1 (Ω × (0, T )) . Finally, the last statement implies that
ut ∈ C [0, T ) ; H02 (Ω) . On the other hand, in order the transformation S to satisfy the property (i), it is sufficient to chose parameters T and R0 so that the inequality 2

C (u0 , u1 , R0 , T ) eLR0 T ≤ R02

(7.16)

holds. At this stage, we aim to show that
ut ∈ C [0, T ) ; L2 (Ω) . For all t0 ∈ [0, T ], we define w (t) = u (t) − u (t0 ) = S (v (t)) − S (v (t0 )) . Then w satisfies    2 p−1 p−1 2  w + 4 w − M k∇v (t)k 4 w + |u (t)| u (t) − |u (t )| ut (t0 )  tt t t t 0  h    i     = M k∇v (t)k2 − M k∇v (t0 )k2 4 t0    q−1 q−1 + |v (t)| v (t) − |v (t0 )| v (t0 ) ,      u (x, 0) = 0, ut (x, 0) = 0,    ∂ u (x, t) = 0. u (x, t) = ∂ν (7.17)

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If we multiple both sides of this equation by 2wt and integrate it on Ω we get Z   d p−1 p−1 e2 (w (t)) + 2 |ut (t)| ut (t) − |ut (t0 )| ut (t0 ) (ut (t) − ut (t0 )) dx dt Ω h    i Z 2 2 = 2 M k∇v (t)k − M k∇v (t0 )k 4u (t0 ) wt dx Ω





d M k∇v (t)k2 k∇wk2 dt Z   q−1 q−1 +2 |v (t)| v (t) − |v (t0 )| v (t0 ) wt dx 

+

Ω

= I3 + I4 + I5

(7.18)

where 

2

2



2

e2 (w (t)) = kwt k + k∆wk + M k∇v (t)k



k∇wk

2



and I3

h    i Z 2 2 = 2 M k∇v (t)k − M k∇v (t0 )k 4u (t0 ) wt dx, Ω



I4 I5

 d  2 = M k∇v (t)k k∇wk2 , dt Z   q−1 q−1 = 2 |v (t)| v (t) − |v (t0 )| v (t0 ) wt dx.

Next integrating (7.18) on the interval [t0 , t] and considering e2 (w (t)) = 0,

(7.19)



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

it follows that e2 (w (t)) + 2

Zt Z 

p−1

|ut (t)|

p−1

ut (t) − |ut (t0 )|



ut (t0 ) (ut (t) − ut (t0 )) dx dt

t0 Ω

Zt =

(I3 + I4 + I5 ) dt. t0

Then by taking into account (7.16), it follows that I3 ≤ 4L k∆vk k∆vt k k∆wk2 ≤ 4LR04 ,

(7.20)

I4 ≤ 4LR04 ,

(7.21)

and I5

 

q−1 q−1 ≤ 2 |v (t)| v (t) − |v (t0 )| v (t0 ) kwt k   q q ≤ 2 kvk2q + kv (t0 )k2q kwt k ≤ 2 (k∇vkq + k∇v (t0 )kq ) kwt k ≤ 4R0q+1 .

(7.22)

By the last inequalities, we can write I3 + I4 + I5 ≤ 8LR04 + 4R0q+1 = C (R) . Since |u (t)|p−1 u (t) is monotone for t > t0 , we have e2 (w (t)) ≤ C (R) (t − t0 ) → 0 ift → t+ 0. On the other hand, since t < t0 and ut ∈ Lp+1 ((0, T ) × Ω), then for t → t− 0,

(7.23)

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Applications on Differential Equations

it follows that e2 (w (t)) ≤ C (R) (t − t0 ) Zt Z  +2 |ut (t)|p−1 ut (t) − |ut (t0 )|p−1 ut (t0 ) (ut (t) − ut (t0 )) dxdt) t0 Ω

→ 0. Then for t → t0 we have e (w (t)) ≤ Ce2 (w (t)) → 0. As a result we obtain
ut ∈ C [0, T ) ; L2 (Ω) . Thus, it can easily be concluded that u (t) ∈ XT,R0 . It remains to proof the statement (ii), that is, to show S is a contraction mapping in XT,R0 space associated to the metric d (·, ·) . Let v1 , v2 ∈ XT,R0 and u1 = Sv1 and u2 = Sv2 be given. Our purpose is to show the inequality d (Sv1 , Sv2 ) = d (u1 , u2 ) ≤ Cd (v1 , v2 ) .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

Observe that, if (7.16) is valid, then we have u1 , u2 ∈ XT,R0 . Next if we denote w as w = u1 − u2 , it implies that    2 2  wtt + 4 w − M k∇v1 k 4 w + |u1t |p−1 u1t − |u2t |p−1 u2t   h    i     2 2 q−1 q−1 = M k∇v1 k − M k∇v2 k 4 u2 + |v1 | v1 − |v2 | v2 ,   u (x, 0) = 0, ut (x, 0) = 0,    ∂ u (x, t) = ∂ν u (x, t) = 0. (7.24) If we multiply (7.24) by 2wt and integrate it on Ω × (0, t) , we obtain i   d h 2 2 2 2 kw (t)k + k∆wk + M k∇v1 k k∆wk dt Z   p−1 p−1 +2 |u1t | u1t − |u2t | u2t (u1t − u2t ) dx Ω

i Z   h  2 2 4u2 w (t) dx = 2 M k∇v1 k − M k∇v2 k  +

d M k∇v1 k2 dt 



2

k∇wk + 2

ZΩ 

q−1

|v1 |

q−1

v1 − |v2 |



v2 wt dx

Ω

= I6 + I7 + I8

(7.25)

To proceed the estimates of I6 , I7 and I8 ; we observe that h    i Z 2 2 I6 = 2 M k∇v1 k − M k∇v2 k 4u2 w (t) dx Ω





≤ 2L k∇v1 k2 + k∇v2 k2 k∇v1 − ∇v2 k kwt k 1

1

≤ 4LC0 R02 [e (v1 − v2 )] 2 [e (wt )] 2 ,

(7.26)

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Applications on Differential Equations



 d  2 = M k∇v1 k k∇wk2 dt ≤ 2L k∇v1 k k∇v1t k k∇wk2 ≤ LR02 e (w (t))

I7

(7.27)

and I8 = 2

Z 

q−1

|v1 |

q−1

v1 − |v2 |



v2 wt dx

Ω

Z ≤ 2q

(|v1 |q + |v2 |q ) |v1 − v2 | wt dx

Ω

≤ 2q k|v1 | + |v2 |kq−1 2n kwt k n(q−1) kv1 − v2 k n−2   q−1 q−1 ≤ 2qC∗ k|v1 |kn(q−1) + k|v2 |kn(q−1) k∇ (v1 − v2 )kq−1 n(q−1) 1

1

≤ 2qC∗ R0q−1 [e (v1 − v2 )] 2 [e (wt )] 2 .

(7.28)

Now we substitute these inequalities in (7.25) to get i   d h 2 2 2 2 kw (t)k + k∆wk + M k∇v1 k k∆wk dt Z   p−1 p−1 +2 |u1t | u1t − |u2t | u2t (u1t − u2t ) dx = I6 + I7 + I8 1 1 = 4LC0 R02 [e (v1 − v2 )] 2 [e (wt )] 2 + LR02 e (w (t)) 1

1

+2qC∗ R0q−1 [e (v1 − v2 )] 2 [e (wt )] 2 and integrate both sides on (0, t) by also considering the initial data, it follows that h i2 2 q−1 2 T 2 eLR0 T sup e (v1 − v2 ) . e (w (t)) ≤ 4LC0 R0 + 2qC∗ R0 0≤t≤T

Then by definition of d (u, v) , we get d (u1 , u2 ) ≤ C (T, R0 ) d (v1 , v2 )



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

where C (T, R0 ) = C 2

h

4LC0 R02 + 2qC∗ R0q−1

i2

LR02 T

T 2e

 .

Finally if C (T, R0 ) < 1, then S is a contraction mapping. Consequently, if T and R0 are chosen sufficiently small, then (i) and (ii) are satisfied. Thus, by Banach contraction mapping theorem, we have local solution. 7.3

Existence of Solutions of Sixth Order Boussinesq Equation

This section is devoted to the proof of the existence of solution of a sixth order non-linear damping term Cauchy problem given by utt − uxx − uxxtt − uxxxxxx − αuxxt = f (u)xx , x ∈ R, t > 0 u(x, 0) = ϕ(x), ut (x, 0) = ψ(x)

(7.29) (7.30)

by using contraction mapping. Here α is a constant, ϕ(x) and ψ(x) are the given initial value functions and f (u) is a non-linear function (Polat and Pis¸kin 2015, [35]). Lemma 7.3.1. Suppose that for some T > 0 and s ∈ R, ϕ ∈ H s , ψ ∈ H s−2 and h ∈ L1 ([0, T ]; H s−2 ) are satisfied. Then for the linear equation utt − uxx − uxxtt − uxxxxxx = (h(x, t))xx , x ∈ R, t > 0 u(x, 0) = ϕ(x), ut (x, 0) = ψ(x)

(7.31) (7.32)

the Cauchy problem has a unique solution u ∈ C([0, T ], H s )∩C 1 ([0, T ], H s−2 ) with the initial condition given as in (7.30) and   Z t kh(τ )kH s−2 dτ ku(t)kH s +kut (t)kH s−2 ≤ 4(1+T ) kϕkH s + kψkH s−2 + 0

is satisfied.

(7.33)

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Proof. Applying Fourier transform to the equation (7.31) with respect to x gives ˆ t), uˆtt − (iξ)2 uˆ − (iξ)2 uˆtt − (iξ)6 uˆ = (iξ)2 h(ξ, ˆ t), uˆtt + ξ 2 uˆ + ξ 2 uˆtt + ξ 6 uˆ = −iξ 2 h(ξ, ξ2 + ξ6 ξ2 ˆ uˆ = − uˆtt + h(ξ, t), 1 + ξ2 1 + ξ2 ξ2 ˆ 2 uˆtt + ρ (ξ) uˆ = − h(ξ, t) 1 + ξ2 ˆ uˆ(ξ, 0) = ϕ(ξ), ˆ uˆt (ξ, 0) = ψ(ξ)

(7.34) (7.35)

where

ξ2 + ξ6 . 1 + ξ2 Observe that the partial differential equation is turned to be a non-homogeneous ordinary differential equation with parameter ξ. The solution of this ordinary differential equation is of the form ρ2 (ξ) =

uˆ(ξ, t) = uˆh (ξ, t) + uˆp (ξ, t). Here the solution of the homogeneous part uˆtt + ρ2 (ξ) uˆ = 0 is uˆh (ξ, t) = c1 (ξ) cos(tρ (ξ)) + c2 (ξ) sin(tρ (ξ)). The particular solution of (7.34) can be found by the method of variation of parameters. It follows that uˆp (ξ, t) = uˆ1 (ξ) cos(tρ (ξ)) + uˆ2 (ξ) sin(tρ (ξ)) which yields (

cos(tρ (ξ))(ˆ u1 )t + sin(tρ (ξ))(ˆ u2 )t = 0 ξ2 ˆ −ρ (ξ) sin(tρ (ξ))(ˆ u1 )t + ρ (ξ) cos(tρ (ξ))(ˆ u2 )t = − 1+ξ 2 h (ξ, t)



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

If we solve the system of these equations, we find ξ2 ˆ (ξ, t) h (ˆ u1 )t = sin(tξ) ρ (ξ) (1 + ξ 2 ) and

ξ2 ˆ (ξ, t) . h (ˆ u2 )t = − cos(tξ) 2 ρ (ξ) (1 + ξ ) Next, integrating these equations on the interval [0, t] it follows that Z t ξ2 ˆ (ξ, τ ) dτ uˆ1 = sin(τ ξ) h 2 ρ (ξ) (1 + ξ ) 0 and

t

Z uˆ2 = − 0

ξ2 ˆ (ξ, τ ) dτ. h cos(τ ξ) ρ (ξ) (1 + ξ 2 )

Then we have uˆp (ξ, t) = uˆ1 (ξ) cos(tρ (ξ)) + uˆ2 (ξ) sin(tρ (ξ)) Z t ξ2 ˆ (ξ, τ ) dτ. cos(tρ (ξ)) = sin(τ ξ) h 2) ρ (ξ) (1 + ξ 0 Z t ξ2 ˆ (ξ, τ ) dτ. sin(tρ (ξ)) − cos(τ ξ) h 2) ρ (ξ) (1 + ξ 0 and

t

ξ 2 sin((t − τ )ρ (ξ)) ˆ h (ξ, τ ) dτ. 2) ρ (ξ) (1 + ξ 0 Therefore the general solution is Z

uˆp (ξ, t) = −

uˆ(ξ, t) = c1 (ξ) cos(tρ (ξ)) + c2 (ξ) sin(tρ (ξ)) Z t 2 ξ sin((t − τ )ρ (ξ)) ˆ h (ξ, τ ) dτ. − 2) ρ (ξ) (1 + ξ 0 Next, applying initial data to (7.35) yields c1 (ξ) = ϕ(ξ), ˆ c2 (ξ) =

ˆ ψ(ξ) . ρ (ξ)

An Introduction to Sobolev Spaces 

Applications on Differential Equations

Thus, the solution is ˆ ψ(ξ) uˆ(ξ, t) = ϕ(ξ) ˆ cos(tρ (ξ)) + sin(tρ (ξ)) ρ (ξ) Z t 2 ξ sin((t − τ )ρ (ξ)) ˆ − h (ξ, τ ) dτ. ρ (ξ) (1 + ξ 2 ) 0

(7.36)

We can find a valid estimation of u(x, t) in H s norm by using the norm related to the Fourier transformation

2

s kwk2H s = (1 + ξ 2 ) 2 w(ξ) ˆ Z 2 = (1 + ξ 2 )s |w(ξ)| ˆ dξ. R

If we take vˆ1 = ϕ(ξ) ˆ cos(tρ (ξ)), ˆ ψ(ξ) vˆ2 = sin(tρ (ξ)) ρ (ξ) Z vˆ3 = − 0

t

ξ 2 sin((t − τ )ρ (ξ)) ˆ h (ξ, τ ) dτ ρ (ξ) (1 + ξ 2 )

with u = v1 + v2 + v3 , then we obtain kukH s ≤ kv1 kH s + kv2 kH s + kv3 kH s .

(7.37)

We next derive the terms on the right hand side of (7.37) one by one. The first term follows by Z 2 kv1 k2H s = (1 + ξ 2 )s |ϕ(ξ)| ˆ cos2 (tρ (ξ))dξ ZR 2 (1 + ξ 2 )s |ϕ(ξ)| ˆ dξ ≤ R

= kϕk2H s .



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

If we use the inequality sin(tξ) |ξt| ξ ≤ |ξ| = t for the second one, as ξ disappears, it effects the estimation. Considering the inequality sin(tξ) 1 ξ ≤ ξ, we have a singularity at ξ = 0. For that reason we separate the integral into two parts. Z 2 2 2 2 s sin (tξ) ˆ kv2 kH s = (1 + ξ ) 2 ψ(ξ) dξ ρ (ξ) R
2 Z 2 1 + ξ sin (tξ) ˆ 2 2 s = (1 + ξ ) ψ(ξ) dξ. (1 + ξ 4 ) ξ 2 R Since

1+ξ 2 1+ξ 4

≤

2 1+ξ 2 ,

kv2 k2H s

it follows that 2 sin2 (tξ) ˆ 2 ψ(ξ) dξ (1 + ξ ) 2) ξ 2 (1 + ξ R Z 2 2 2 s−1 sin (tξ) ˆ 2 (1 + ξ ) ψ(ξ) dξ ξ2 R Z 2 2 2 s−1 sin (tξ) ˆ 2 (1 + ξ ) ψ(ξ) dξ 2 ξ |ξ| 12 and by Sobolev embedding theorem, for a constant C1 (T 0 ) depending on T 0 , i = 1, 2 and for 0 ≤ t ≤ T 0 < T , it follows that kui (t)k∞ ≤ C1 (T 0 ). Thus the Cauchy inequality implies Z [f (u1 ) − f (u2 )] ut dx ≤ kf (u1 ) − f (u2 )k kut k 2 2 R

≤ C2 (T 0 ) kuk2 kut k2 ,



Erhan Pis¸kin and Baver Okutmus¸tur

An Introduction to Sobolev Spaces

where C2 (T 0 ) is constant depending on C1 (T 0 ). Next by the Young inequality, we have   Z t

2 1
−

2 2 2 kuτ k2 dτ

−∂x2 2 ut + kuk + kut k + kuxx k + 2α 0 Z th i ≤ C2 (T 0 ) kuk2 + kuτ k2 dτ. 0

The inequality above implies 2

2

0

kuk + kut k ≤ [C2 (T ) + 2 |α|]

Z th

2

2

kuk + kuτ k

i

dτ.

(7.47)

0

By the Gronwall inequality, for 0 ≤ t ≤ T 0 , (7.47) gives kuk2 + kut k2 ≡ 0. Then for 0 ≤ t ≤ T 0 we have u ≡ 0. In other words, the problem (7.29)– (7.30) has at most one (unique) solution in X(T 0 ). It remains to show that, letting [0, T0 ) be the maximum time interval for u ∈ X(T0 ), if (7.45) is satisfied, then T0 = ∞. Suppose that (7.47) and T0 < ∞ are satisfied. Then consider the Cauchy problem vtt − vxx − vxxtt − vxxxxxx = [f (u) + αut ]xx (7.48) v(x, 0) = u(x, T 0 ), vt (x, 0) = ut (x, T 0 )

(7.49)

for T 0 ∈ [0, T0 ) . It follows by (7.45) that ku(t)kH s + kut (t)kH s−1 ≤ K

(7.50)

where K is a positive constant which does not depend on T 0 ∈ [0, T0 ) . For all T 0 ∈ [0, T0 ) there exists a constant T1 ∈ (0, T0 ) such that the problem (7.48)–(7.49) has a unique solution v(x, t) ∈ X(T1 ). Setting T 0 = T0 − T1 /2 and defining  u(x, t), t ∈ [0, T 0 ] , u˜(x, t) = v(x, t − T 0 ), t ∈ [T 0 , T0 + T1 /2] ,

Applications on Differential Equations

An Introduction to Sobolev Spaces 

implies that u˜(x, t) is a solution of the problem (7.29)–(7.30) on the interval [0, T0 + T1 /2] and by uniqueness, u˜ extends to u which breaks the maximum property of [0, T0 ) . Consequently, if (7.45) is satisfied, then T0 = ∞. Thus, the proof is done.

184

An Introduction to Sobolev Spaces, 2021, 184-187

Bibliography

[1] N. S. Papageorgiou and P. Winkert, Applied Nonlinear Functional Analysis, De Gruyter, 2010. [2] R.A. Adams and J.J.F. Fournier, Sobolev Spaces, Academic Press, 2003. [3] L.S. Coste, Aspects of Sobolev-Type Inequalities, Cambridge University Press, 2002. [4] L. Diening, P. Harjulehto, P. Hasto and M. Ruzicka, Lebesgue and Sobolev Spaces with Variable Exponents, Springer, 2011. [5] L. Gross, Logarithmic Sobolev inequalities, Amer. J. Math., 97(4) (1975) 1061-1083. [6] M.A. Khamsi and W.A. Kirk, An Introduction to Metric Spaces and Fixed Point Theory, Wiley, 2001. [7] V. G. Mazja, Sobolev Spaces, Springer, 1985. [8] R.B. Ash, Real Variables with Basic Metric Space Topology, Dover Books on Mathematics, 2009. [9] E. Kreyszig, Introductory Functional Analysis with Applications, John Wiley&Sons, 1989. [10] S. Kesavan, Topics in Functional Analysis and Applications, Wiley, 1989. [11] B. Musayev and M. Alp, Fonksiyonel Analiz, Balci Yayinlari, 2000. [12] B.D. Reddy, Introductory Functional Analysis, Springer, 1998. Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

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An Introduction to Sobolev Spaces 15

[13] H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer, 2011. [14] B. Okutmus¸tur and A. Gheondea, Reproducing Kernel Hilbert Spaces. The Basics Bergman Spaces and Interpolation Problems. Saarbr¨ucken: LAP Lambert Academic Publishing, 2010. [15] H. K. Pathak, An Introduction to Nonlinear Analysis and Fixed Point Theory, Springer 2018. [16] D.V. Cruz-Uribe and A. Fiorenza, Variable Lebesgue Spaces, Springer, 2013. [17] S. Zaidman, Functional Analysis and Differential Equations in Abstract Spaces, Chapman & Hall/CRC, 1999. [18] L. Pick, A. Kufner, O. John and S. Fuˇcik, Function Spaces, Vol. 1, De Gruyter, 2013. [19] O. Kov´acˇ ik and J. R´akosnik, On Spaces Lp(x) (Ω) and W 1,p(x) (Ω) , Czechoslovak Math. J., 41(4) (1991) 592-618. [20] X. Fan and D. Zhao, On the Spaces Lp(x) (Ω) and W m,p(x) (Ω) , J. Math. Anal. Appl., 263 (2001) 424-446. [21] V. Komornik, Exact Controllability and Stabilization: The Multiplier Method, RAM: Research in Applied Mathematics, Masson-John Wiley, Paris, 1994. [22] P. Martinez, A New Method to Obtain Decay Rate Estimates for Dissipative Systems, ESAIM: Control, Optimisation and Calculus of Variations, 4 (1999) 419–444. [23] M. Nakao, Asymptotic Stability of the Bounded or Almost Periodic Solution of the Wave Equation with Nonlinear Dissipative Term, J. Math. Anal. Appl., 58 (2) (1977) 336–343.

16

An Introduction to Sobolev Spaces

Erhan Pis¸kin and Baver Okutmus¸tur

[24] M. Nakao, A Difference Inequality and Its Application to Nonlinear Evolution Equations, J. Math. Soc. Japan, 30(4) (1978) 747–762. [25] B. O. Turesson, Nonlinear Potential Theory and Weighted Sobolev Spaces, Sringer, 2000. [26] D. Lukkassen, A Short Introduction to Sobolev Spaces and Applications for Engineering Students, Lecture Notes, 2004. [27] L. Tartar, An Introduction to Sobolev Spaces and Interpolation Spaces, Springer, 2007. [28] W.P. Ziemer, Weakly Differentiable Functions, Springer, 1989. [29] S. Shkoller, Introduction to Sobolev Spaces, Lecture Notes, 2011. [30] E. Pis¸kin, Sobolev Uzayları, Sec¸kin Yayıncılık 2017 (in Turkish). [31] V.I. Burenkov, Sobolev Spaces on Domains, Teubner-Texte zur Mathematik, 1998. [32] G. Talenti, Best Constant in Sobolev Inequality, Ann. Mat. Pura Appl., 110 (1976) 353-372. [33] V.D. Radulescu and D.D. Repovs, Partial Differential Equations with Variable Exponents, CRC Press, 2015. [34] E. Pis¸kin, Existence, Decay and Blow up of Solutions for the Extensible Beam Equation with Nonlinear Damping and Source Terms, Open Math., 13 (2015) 408–420. [35] N. Polat and E. Pis¸kin, Existence and Asymptotic Behavior of Solution of Cauchy Problem for the Damped Sixth-Order Boussinesq Equation, Acta Mathematicae Applicatae Sinica, English Series, 31(3) (2015) 735-746.

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An Introduction to Sobolev Spaces 17

[36] N. Duruk, A. Erkip and H.A. Erbay, A Higher-Order Boussinesq Equation in Locally Nonlinear Theory of One Dimensional Nonlocal Elasticity, IMA J. Appl. Math., 74 (2009) 97-106. [37] S. Wang and G. Chen, Small Amplitude Solutions of the Generalized IMBq Equation, J. Math. Anal. Appl., 274 (2002) 846-866.

An Introduction to Sobolev Spaces, 2021, 188-189

188

List of Symbols R

Real numbers

Rn

n-dimensional Euclidean space

c0

Space of all sequences of scalars converging to zero

`p

Space of all p−summable sequences (xj ) of scalars with
P∞ p 1/p the norm kxkp = |x| j=1

C(Ω)

Space of continuous functions

CB (Ω)

Space of bounded continuous functions

C m (Ω)

Space of m-th order continuously differentiable functions

C ∞ (Ω)

Space of infinitely continuously differentiable functions

C m ([0, T ]; X) Space of continuously differentiable functions from [0, T ] to X Lp (a, b; X)

Space of measurable functions from (a, b) to X

Lp (Ω)

Space of p-th order Lebesgue integrable functions

L∞ (Ω)

Space of functions defined everywhere in the bounded region Ω

W m,p (Ω)

Sobolev space

W0m,p (Ω)

Closure of the space C0∞ (Ω) on the space W m,p (Ω)

Lp (Ω)

= W 0,p (Ω)

H m (Ω)

= W m,2 (Ω)

Lp(·) (Ω)

Variable Lebesgue spaces

W m,p(·) (Ω)

Variable Sobolev spaces

LH0 (Ω)

Local log-H¨older continuity

LH∞ (Ω)

log-H¨older continuity at infinity

ρp(·)

Modular function

kukp(·)

Luxemburg norm Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

/LVWRI6\PEROV

An Introduction to Sobolev Spaces 19

Z

p

 p1

kukp = kukLp (Ω) = |u (x)| Ω Z  12 kuk2 = kuk = |u (x)|2 Ω

u

= u (x1 , x2 , · · · , xn )

∇u

∂u ∂u ∂u , , · · · , ∂x ) = (ux1 , ux2 , · · · , uxn ) = ( ∂x 1 ∂x2 n

∆u

=

∂2u ∂x21

+

∂2u ∂x22

+ ··· +

∂2u ∂x2n

= ux1 x1 + ux2 x2 + · · · + uxn xn

∆2 u

=

∂4u ∂x41

+

∂4u ∂x42

+ ··· +

∂4u ∂x4n

= ux1 x1 x1 x1 + ux2 x2 x2 x2 + · · · + uxn xn xn xn

190

An Introduction to Sobolev Spaces, 2021, 190-193

SUBJECT INDEX A Absolute value 3, 18, 39 Analysis 1, 15 functional 1, 15 real 1 Applying 89, 91, 92, 157, 171 Fourier transform 171 integration 89, 91, 92, 157

B Banach 1, 28, 35, 170 and Hilbert spaces 1 contraction mapping theorem 170 fixed point theorem 35 subspace 28 Banach space 1, 24, 25, 26, 27, 28, 31, 32, 38, 39, 40, 72, 73, 83, 84 infinite dimensional 28, 38 Bolzano-Weierstsrass theorem 15 Boundary points 10 Boussinesq equation, sixth-order 154 Brouwer fixed point theorem 37

C Cauchy 24, 123, 181 inequality 181 Schwarz inequality 123 Cauchy problem 154, 170, 182 non-linear damping term 170 Cauchy sequence 12, 13, 20, 24, 27, 32, 37, 72, 73, 98 in L1 73 in Lp 72, 98 Classical Hölder inequality 150 Closed set 10 Conductivity constant, heat 156 Cone 119, 125, 126, 131 condition 119, 125, 126, 131

finite 125 Constant 60, 61, 71, 82, 83, 116, 119, 120, 124, 138, 139, 140, 141, 178, 179, 182 T1 182 Continuous 23, 37, 38 linear operator 23 mapping 37, 38 Continuous functions 5, 13, 14, 27, 32, 35, 40, 41, 58, 80, 81, 153 real valued 27 space of 32, 41, 80 vector space of 40 Contraction mapping 35, 154, 161, 170, 179, 180

D Derivatives 87, 95, 96, 97 classical 95 mixed 96 multivariable 87 partial 97 Differential equations 55, 154, 155, 157, 159, 161, 163, 165, 167, 169, 171, 173, 175, 177, 179, 181 ordinary 55, 154, 171 Dimensional spaces, finite 27 Dirac 86, 94, 110 Delta function 86, 94 functions 110 Domain 71, 125, 126, 127, 131, 132, 138, 139, 140, 141, 143, 151, 152, 153 arbitrary 132 bounded 131, 151, 152, 153 class 132 open 125, 131, 152 smooth 153 Dominated convergence theorem 1, 73

E Embedding 75, 126, 152

Erhan Pis¸kin and Baver Okutmus¸tur (Eds.) All rights reserved-© 2021 Bentham Science Publishers

Subject Index

of Lp 75 theorems 126, 152 Equation 89, 91, 92, 154, 155, 157, 158, 160, 161, 165, 170, 171, 172, 178, 181 linear 170, 178 Equivalence 42 classes 42 relation 42 Existence 55, 154, 159, 170 of solutions 55, 154, 170 of solutions of nonlinear Timoshenko equation 154, 159

F Finite dimensional normed spaces 20, 23, 24, 26, 27 Fixed point theorems 1, 34, 35, 37, 38 Fourier transform 115 inverse 115 Fourier transformations 114, 173 Function 44, 45, 58, 59, 60, 61, 63, 64, 101, 138, 139, 170 cosine 101 distance 45 exponent 138, 139 exponential 58 initial value 170 non-increasing 60, 61, 63, 64 non-linear 170 non-negative summable 58, 59 square-integrable 44 variable exponent 139

G Green’s 69, 70 first identity 69 second identity 70 Gronwall inequality 41, 58, 59, 154, 182 differential form 58 integral form 59 Gronwall type inequality 66

H Heat 154 source density 154 transfer coefficient 154

An Introduction to Sobolev Spaces 191

Heaviside function 86, 93, 96 Heine-Borel theorem 15, 16, 23 for finite dimensional normed spaces 23 Hilbert spaces 1, 31, 32, 33, 39, 43, 45 Hölder 41, 82, 83,1 62 and Minkowski inequalities 41 and Sobolev-Poincaré inequalities 162 condition 83 exponent 82 space 41, 82, 83 Hölder inequality 49, 50, 51, 52, 53, 56, 76, 78, 122, 123, 129, 130, 133, 134, 150 generalized 150

I Indices, multiple 80 Inequalities side 121 Inequality 45, 47, 49, 51, 52, 53, 54, 55, 58, 59, 61, 62, 63, 64, 66, 67, 68, 118, 119, 120, 122, 123, 124, 130, 163, 166 integral 59 interpolation 55, 119 last 49, 53, 66, 68, 123, 130, 163, 166 Inequality results 163 Inner 1, 29 product properties 29 products and inner product spaces 1, 29 Inner product space 1, 29, 30, 31, 32, 33, 39 finite dimensional 32 Integers 97, 99, 106, 114 arbitrary nonnegative 97, 99 non-negative 106, 114 Integrable functions 34, 42, 43, 44, 113 non-negative 34 Interior points 10, 11 Interpolation inequality in Hs 134

K Komornik inequalities 41, 66 classical 66

L Lebesgue 1, 25, 33, 34, 42, 73, 85, 97, 114, 138, 140, 141 classical 138 dominated convergence theorem 33, 34

Erhan Pis¸kin and Baver Okutmus¸tur

192 An Introduction to Sobolev Spaces

measure 42 space 141 theorem 33 weighted 97, 114 Linear differential equation 155 Lipschitz 35, 82, 119, 125, 126, 137, 138, 152, 159, 162 and cone conditions 126 condition 125, 126 constant 159, 162 continuous function 138, 152 Logarithmic Sobolev 119, 131 inequalities 119 inequality 131 Luxemburg norm 138, 141, 145, 146 definition of 145, 146

M

Normed spaces 1, 16, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 39, 40, 43, 119 special 29 Notion variable exponent Lebesgue 138

O Order 41, 81, 87, 94, 96, 97, 99, 100, 102, 103, 107, 164, 170 derivatives 100, 102, 103 distribution derivatives upto 97 m-th 97 negative 97, 107 of differentiation 94, 96 real 97, 107

P

Measurable functions 33, 34, 42, 71, 83, 114, 138, 140, 151 non-negative 33 non-negative extended real-valued 33 real-valued 34 Measure space 33, 34, 41 finite 41 Metric 3, 16, 24, 38, 39 absolute 3 conditions 3, 17 discrete 38, 39 natural 24 normed 16 properties 4 space, compact 15 Minkowski Inequality 4, 41, 51, 175 Monotone convergence theorems 33, 34, 72 Multi indices 87

Parallelogram 30, 31 rule 31 Plancherel theorem 97, 115, 116 Poincaré inequality 119, 129, 153 Polar equivalence properties 30 Pre-Hilbert space 29 Problem 38, 154, 155, 156, 157, 158, 159, 160, 179, 180, 181, 182, 183 boundary value 154, 155, 158, 159 Properties 2, 3, 17, 18, 24, 28, 30, 32, 37, 57, 71, 75, 94, 95, 97, 119, 125, 146 commutativity 57 completeness 24 contraction 37 convex 146 embedded 119 fundamental 24, 97 integrability 95 Pythagorean formula 30

N

R

Nakao inequalities 41 Non-expansive mapping 35 Nonlinear Timoshenko equation 154, 159 Nonreflexive spaces 26 Norm 1, 16, 19, 43, 143 definition of 43, 143 equivalent 1, 19 semi 16

Real numbers 10, 12, 15, 16, 39, 40, 48, 106, 117, 118 arbitrary 39 non-negative 16, 106 Reverse 41, 53, 54 Hölder inequality 53, 54 inequalities 41 Minkowski inequality 54 Riesz-Fischer theorem 41

Subject Index

Rn space 32

An Introduction to Sobolev Spaces 193

real 29

S

W

Scalars converging, sequences of 26, 28 Schauder fixed point theorem 38 Schauder’s theorem 1 Schwartz space 97, 114 Separable Space 12 Sixth order boussinesq equation 170 Sobolev 113, 119, 120 Gagliardo-Nirenberg inequality 119, 120 Sobolev Poincaré 119, 127, 154, 162 inequalities 127, 154, 162

Weighted Sobolev space 114

T Theorem 69, 70, 158 divergence 69, 70 uniqueness 158 Thermal conductivity coefficient 154 Timoshenko equation 154 Topological concepts 1 Transformation 65, 125, 160, 164 non-linear 160 Triangle inequality 2, 4, 5, 18, 36, 43, 142 generalized 2, 36 property 18 Triangular inequality 13 Tychonoff fixed point theorem 38

U Unit ball property 138, 148 Use embedding theorems 125

V Variable exponent 138, 139, 140, 141, 143, 145, 147, 149, 150, 151, 153 Lebesgue spaces 138, 140, 141, 150 Sobolev space 138, 151, 152 Spaces 139, 141, 143, 145, 147, 149, 151, 153 Vector space 16, 19, 24, 27, 29, 39, 43 complete normed 43 finite dimensional 19, 27 linear 24 normed 16