Functional Analysis and Parabolic Equations

Functional analysis and parabolic equations

Florin A. Radu

April 12, 2017 Chapter 1

Introduction

Functional analysis can be seen as a natural extension of the real analysis to more general spaces. As an example we can think at the Heine - Borel theorem (closed and bounded is equivalent with compact) which will be now extended to any finite dimensional space. Moreover, it will be shown that the theorem is not true in an infinite dimensional space. Functional analysis is furnishing plenty of techniques of proofs, the proof being sometimes more important as the result itself. A special attention should be given to the proofs presented in this lecture (including the exercises). Another benefit in learning functional analysis is that one reaches a higher level of abstractization when dealing with mathematical objects and understands that many times is easier to prove for a general case as for a particular case. The latter is because when you deal with an explicitly described mathematical object (e.g. a space) one does not know which properties are now relevant for the specific question to be answered (in other words one has a problem of too much informa- tion). It is often much easier to answer the question when you think about the object as part of a class, means trying to identify properties which are general, valid for many objects of a certain type. As an example: imagine you have to show that the space of continuous functions on the interval [0, 1], denoted in this script C[0, 1] can not have an infinite but countable Hamel basis. The proof is in one line if one just know that not any Banach space can have such a basis (see Section 2.7 for details). There are also many practical applications of functional analysis. The most direct ones are connected with solving partial differential equations (PDEs), with a special focus on numeric. Solving PDEs is extremely important for real world applications, because the mathematical models for such applications are consist-

1 CHAPTER 1. INTRODUCTION 2 ing in the majority of the cases by PDEs. When modelling real worlds problems, the PDEs have often e.g. discontinuous coefficients (due to the heterogeneity of the medium) or are degenerate (e.g. from parabolic to elliptic in the case of vari- able saturated flow in porous media [27, 25, 26]) and do not admit a classical solution. Functional analysis provides the proper mathematical framework (e.g. spaces, norms, techniques) for defining weak solutions for PDEs and solving them numerically. The first two chapters are mainly based on the book of Cheney, [11], but con- tain also material from [14, 32, 34] is used. The second part of these notes are concerning the theory necessary for numerically solving partial differential equations, with a special focus on degenerate parabolic equations. Currently, only a short draft of the sections is outlined. Chapter 2

Banach spaces

2.1 Normed (linear) spaces

Let V be a (linear) vector space.

Definition 2.1.1 A functional k · k : V → R is a norm on V over R if it satisfies the following: i) kλxk = |λ|kxk, for all x ∈ V, λ ∈ R. ii) kx + yk ≤ kxk + kyk, for all x, y ∈ V. (the triangle inequality) iii) kxk = 0 ⇒ x = 0. If k · k satisfies only i)-ii) it is called a seminorm.

Remark 2.1.1 If k · k is a norm on a vector space V , it follows from i) and ii) that kxk ≥ 0 ∀ x ∈ V . Often is this property included in the deﬁnition of the norm, but it is not necessary.

Deﬁnition 2.1.2 Two norms k · k1, k · k2 are equivalent on the space V if there exists m, M > 0 s.t. there holds

mkxk1 ≤ kxk2 ≤ Mkxk1 ∀ x ∈ V.

We mention that there exists also the notion of quasi-norm (see e.g. Yosida, pg. 30). There are nevertheless a few different definitions in the literature. Let V a vector space and k · kV a norm on it. The norm induces a metric on V defined by d(x, y) = kx − ykV . Using this metric we can define a topology, which is called the topology given by the norm k · kV . A topological vector space

3 CHAPTER 2. BANACH SPACES 4

X is a space which has both an algeberaic structure (vector space) and a topology on it, and the addition and multiplication are continuous (as functions from Γ × X to X, where Γ is the ﬁeld of scalars).

Deﬁnition 2.1.3 A normed space (V, k·kV ) is a topological vector space V , with the topology given by the norm k · kV .

A very important inequality which holds in any normed space (V, k · kV ) is the following |kxk − kyk| ≤ kx − yk, (2.1) for any x, y ∈ V . The proof is straightforward. Examples of normed spaces In this section we present some important normed spaces. We deﬁne for 1 ≤ p < ∞ the p-norm on Rn by

n !1/p X p kxkp := |xi| (2.2) i=1

n Proposition 2.1.1 For all 1 ≤ p < ∞, k · kp is indeed a norm on R . Moreover, n (R , k · kp) is a Banach space. For p = ∞ we define n kxk∞ := max |xi| (2.3) i=1 n n Proposition 2.1.2 k · k∞ is a norm on R . Moreover, (R , k · k∞) is a Banach space. All the above spaces are finite dimensional. As examples of infinite dimensional Banach spaces we consider C[0, 1] := {f : [0, 1] → R|f continuous } with the norm kfk∞ = max |f(x)|. (2.4) x∈[0,1] and the Sobolev spaces

p L (Ω) := {f :Ω → R|f measurable and kfkp < ∞}, where Ω is an open set in Rd, (d ∈ N, d ≥ 1) and the norm Z 1/p p kfkp := |f(x)| dx , (2.5) Ω CHAPTER 2. BANACH SPACES 5 for 1 ≤ p < ∞ and kfk∞ = (ess) sup |f(x)|, (2.6) x∈[0,1] for p = ∞. We remind that the essential supremum of a measurable function f() on a domain Ω is deﬁned as the lowest number α ∈ R, for which f(x) ≤ α almost everywhere in Ω. Deﬁnition 2.1.4 A subspace E of V is closed iff V \ E is open, or, equivalently, for any sequence {xn}n∈N ⊂ E, if xn → x then x ∈ E. The closure of a subspace (set) E is denoted by E and it is the smallest set which contains E and is closed.

E := {x ∈ V |∃ {xn}n∈N ⊂ E with xn → x}. Deﬁnition 2.1.5 A topological vector space is called Hausdorff if for any two element x, y ∈ X it exists two open sets Vx and Vy such that x ∈ Vx, y ∈ Vy and Vx ∩ Vy = ∅. A topological vector space is called separable if it has a countable, dense set. It is easy to check that any metric space (therefore also every normed space) is Hausdorff.

2.1.1 Exercises Exercise 2.1.1 Prove that in any normed space the conditions kxk = 1 and kx − y yk < < 1 imply that kx − k < 2. kyk Exercise 2.1.2 Prove that in a normed space, if kx + yk = kxk + kyk, then kαx + βyk = kαxk + kβyk for all non-negative real numbers α, β.

Exercise 2.1.3 Let X be a vector space and k · k1, k · k2 two norms on it. i) Let kxkmax := max(kxk1, kxk2). Show that the function k · kmax defined by kxkmax := max(kxk1, kxk2) for all x ∈ X is a norm on X as well. Does the same hold also for kxkmin := min(kxk1, kxk2)? ii) Let α, β ∈ R be two positive numbers. Show that the application k · k defined by kxk := αkxk1 + βkxk2 for all x ∈ X is a norm on X. iii) Let p ≥ 1 and || · ||p be defined by p p 1/p ||x||p = (kxk1 + kxk2) , for all x ∈ X, is a norm on X. CHAPTER 2. BANACH SPACES 6

Exercise 2.1.4 We deﬁne for 1 ≤ p < ∞ the p-norm on Rn by

n !1/p X p kxkp := |xi| . i=1

Show that k · kp is indeed a norm.

Exercise 2.1.5 Let (X, k · k) be a normed space, {αn}n ⊂ R, {xn}n, {yn}n ⊂ X and {α}n ⊂ R, x, y ∈ X. Show that the following afﬁrmations hold true: i) If xn → x strongly (i.e. in norm) and αn → α, then there holds limn→∞(αnxn) = αx. ii) If xn → x and yn → y strongly, then xn + yn → x + y strongly. iii) If xn → x and xn → y strongly then x = y (the limit of a sequence is unique). iv) If xn → x strongly, then kxn − yk → kx − yk in R. Especially, for y = 0 it results that kxnk → kxk in R.

2.2 Completeness, Banach spaces

Let (X, k · k) be a normed space.

Deﬁnition 2.2.1 A sequence {xn}n ⊂ X is called Cauchy if ∀ > 0∃N ∈ N such that kxn − xmk ≤ , ∀n, m ≤ N.

Remark 2.2.1 Any convergent sequence is Cauchy, but not any Cauchy sequence is convergent. A counter-example is the space of all sequences in R with a ﬁnite number of nonzero elements. Nevertheless, any Cauchy sequence is bounded.

Deﬁnition 2.2.2 A topological vector space V is called complete if every Cauchy sequence is convergent to an element of V. A normed vector space which is complete is called a Banach space.

Remark 2.2.2 Not any normed space is a Banach space. A counter-example is the space of all sequences in R with a ﬁnite number of nonzero elements, with the inﬁnite norm. CHAPTER 2. BANACH SPACES 7

Theorem 2.2.1 The space C[0, 1] := {f : [0, 1] → R|f continuous } with the norm kfk∞ = maxx∈[0,1] |f(x)| is a Banach space. Proposition 2.2.1 Any complete space is closed. A subspace of a Banach space is complete iff is closed. Proposition 2.2.2 In a normed space a Cauchy sequence which has a convergent sub-sequence is convergent. P Deﬁnition 2.2.3 Let (X, k · k) be a normed space. A series xn is said to be P n≤1 absolute convergent if the series n≤1 kxnk is convergent. Proposition 2.2.3 Let (X, k · k) be a normed space. The the following two statements are equivalent: i) Any Cauchy sequence is convergent (to an element of X). ii) Any convergent absolute series is convergent (to an element of X).

2.2.1 Exercises Exercise 2.2.1 Show that in a normed space a Cauchy sequence which has a convergent sub-sequence is convergent. Exercise 2.2.2 Prove the following statements through adequate examples: (a) The derivative 1 0 D :(C [a, b], k · k∞) → (C[a, b], k · k∞), f → f ,

is not bounded (where a, b ∈ R, a < b and kfk∞ := max |f(t)|). t∈[a,b]

(b) The maximum norm k · k∞ (see above) is on C[a, b] not equivalent to any of the Lp-norms Z b 1/p p kfkp := |f(x)| dx , p ∈ [1, ∞). a

Exercise 2.2.3 Let (X, k · k) be a normed space and {xn}n a sequence in X. If xn → x ∈ X, then also the sequence {σn}n, deﬁned by x + x + . . . x σ := 1 2 k , k k for all k ∈ N?, is convergent to x. CHAPTER 2. BANACH SPACES 8 2.3 Continuous functions, contractions, Banach ﬁxed point theorem

Let (X, k · kX ) and (Y, k · kY ) be two normed space.

Deﬁnition 2.3.1 A function f : X → Y is called continuous in x∗ if for any se- ∗ ∗ quence {xn}n∈ ⊂ X, lim xn = x there holds lim f(xn) = f(x ). A function N n→∞ n→∞ is said to be continuous on X if it is continuous in any x∗ ∈ X.

Remark 2.3.1 The deﬁnition above is equivalent with ∀ > 0, ∃η > 0 such that ∗ ∗ kf(x) − f(x )kY < for all x ∈ X with kx − x kX < η.

Deﬁnition 2.3.2 A function f : X → R is often called functional, and f : X → Y , with X,Y being spaces of functions is often called operator.

Remark 2.3.2 Let (X, k·k) be a normed space. The norm k·k is Lipschitz continuous (seen as functional on X). This can be easily proved by using the inequality (2.1). If we have another norm, k · k1 on X, this norm is not necessarily continuous as a function from (X, k · k) in R. But if the space X is ﬁnite dimensional then any norm k · k1 will be (Lipschitz) continuous (because, as we will see later in Theorem 2.4.3, in a ﬁnite dimensional space all the norms are equivalent).

Deﬁnition 2.3.3 Let (X, k · kX ) and (Y, k · kY ) two normed spaces and f : X → Y . The function f is called Lipschitz continuous if there exists a constant Lf > 0 such that

kf(x) − f(y)kY ≤ Lf kx − ykX , for all x, y ∈ X. (2.7)

A Lipschitz continuous function with the Lipschitz constant Lf < 1 is called contraction.

Obviously, any Lipschitz continuous function is continuous. The converse is but not true.

Deﬁnition 2.3.4 The set f −1(A) := {x ∈ X|f(x) ∈ A} is called the preimage of A. CHAPTER 2. BANACH SPACES 9

Deﬁnition 2.3.5 Let f : X → Y be a function. We deﬁne the sets:

Ker(f) := {x ∈ X|f(x) = 0}is called the kernel off (2.8) Im(f) := {f(x) ∈ Y |x ∈ X}is called the image or range off (2.9)

Ker(f) is called kernel of f and Im(f) is called the image or range of f.

We will see later in Section 2.5 that if the function is linear then the kernel and the range are linear subspaces of X and Y , respectively.

Proposition 2.3.1 Let (X, k · kX ) and (Y, k · kY ) be two normed space, and f : X → Y be a function. The following statements are equivalent:

i) f is continuous.

ii) f(A) ⊆ f(A) for any subset A of X.

iii) The preimage of any closed set in Y is a closed set in X.

iv) The preimage of any open set in Y is a open set in X.

A very important result in functional analysis is the following theorem.

Theorem 2.3.1 (Banach ﬁxed point theorem) Let X a Banach space, U ⊆ X a subset and f : U → X. If

• U is closed

• f(·) is a contraction with Lipschitz constant L < 1

• f(U) ⊆ U

? then f(·) has an unique ﬁxed point x ∈ U. Moreover, the sequence xn := ? f(xn−1), n ≥ 1 with x0 ∈ U arbitrary, converges to x . There holds also L kx − x?k ≤ kx − x k a posteriori error estimate (2.10) k X 1 − L k k−1 X Lk kx − x?k ≤ kx − x k a priori error estimate (2.11) k X 1 − L 1 0 X CHAPTER 2. BANACH SPACES 10

Proof. We will show that the sequence deﬁned by xn := f(xn−1), n ≥ 1, ? x0 ∈ U arbitrary, converges to x ∈ U. The sequence is well deﬁned due to the assumptionf(U) ⊆ U. We show that the sequence is Cauchy, the result will then follow because the space is Banach and the set U is closed. There holds for k ≥ 2

k−1 kxk − xk−1k = kf(xk−1) − f(xk−2)k ≤ Lkxk−1 − xk−2k ≤ ... ≤ L kx1 − x0k (2.12) because f is a contraction. By using (2.12) and that L < 1, it follows for any m, n ∈ N, m ≥ n

kxm − xnk = kxm − xm−1 + xm−1 − xm−2 + ... + xn+1 − xnk

≤ kxm − xm−1k + kxm−1 − xm−2k + ... + kxn+1 − xnk m−1 m−2 n ≤ (L + L + ... + L )kx1 − x0k 1 ≤ Ln kx − x k. (2.13) 1 − L 1 0

From (2.13) we get immediately that the sequence {xn}n is Cauchy. The space is ? Banach, then {xn}n converges to x ∈ V . But xn ∈ U for all n, and the set U is closed, then x? ∈ U. We have shown the convergence, we will prove now also the estimates. We have

? ? kxk − x k = kf(xk−1) − f(x )k ? ≤ Lkxk−1 − x k ? ≤ Lkxk − xk−1k + Lkxk − x k. (2.14)

From (2.14) it immediately follows (2.10). We still have to show the a priori error estimate (2.11). There holds

? ? kxk − x k = kf(xk−1) − f(x )k ? k−1 ? ≤ Lkxk−1 − x k ≤ ... ≤ L kx1 − x k. (2.15)

On the other hand we have

? ? ? ? kx1 − x k = kf(x0) − f(x )k ≤ Lkx0 − x k ≤ Lkx1 − x k + Lkx1 − x0k.

From above we get L kx − x?k ≤ kx − x k. (2.16) 1 1 − L 1 0 CHAPTER 2. BANACH SPACES 11

The result (2.11) follows then from (2.15) and (2.16). We still have to show that the ﬁxed point is unique. Assume that f has two ﬁxed points, x and y. Then, by using that f is a contraction we can write kx−yk = kf(x) − f(y)k ≤ Lkx − yk which implies 1 ≤ L, contradiction. Q. E. D.

Corollary 2.3.1 Let X a Banach space and f : X → X a contraction with Lipschitz constant L < 1. Then f(·) has an unique ﬁxed point x? ∈ X. Moreover, ? the sequence xn := f(xn−1), n ≥ 1 with x0 ∈ X arbitrary, converges to x and the estimates (2.10) and (2.11) hold.

Corollary 2.3.2 Let X a Banach space, k ≥ 1 an integer and f : X → X such that a f k := f ◦ f ◦ . . . f is a contraction with Lipschitz constant L < 1. | {z } k−times ? Then f(·) has an unique ﬁxed point x ∈ X. Moreover, the sequence xn := ? f(xn−1), n ≥ 1 with x0 ∈ X arbitrary, converges to x .

2.3.1 Applications The Banach ﬁxed point theorem is very useful tool for solving equations numerically (because any equation can be seen as ﬁxed point problem). The theorem not only says that the equation has a unique solution (which is a qualitative, theoreti- cal result) but also furnishes an algorithm how to obtain an approximation of this solution and gives also error estimates.

2.3.2 Exercises Exercise 2.3.1 Let f :[a, b] → [a, b] be a continuous function (it is actually enough to have the Darboux property). Show that f has at least one ﬁxed point. Does this result hold true also in more dimensions?

2.4 Compactness and ﬁnite dimensional spaces

In this section we study the properties of compact sets in normed spaces. We will especially look at the validity of the Heine-Borel theorem in normed spaces and the compactness of the closed unit ball. CHAPTER 2. BANACH SPACES 12

Deﬁnition 2.4.1 A set K in a normed space X is called (sequentially) compact if any sequence xnn ⊂ K has a sub-sequence which is convergent to an element of K. A set K in a normed space X is called relatively (sequentially) compact if its closure is compact.

Definition 2.4.2 Let (X, k·k) be a normed space, x ∈ X and A ⊂ X be a subset. We define the distance of x to A the number (possibily infinite):

dist(x, A) := inf kx − yk. (2.17) y∈A

Proposition 2.4.1 (Closest point property) Let K be a compact set in a normed space X. To each point x ∈ X it exists a point in K of minimum distance to x. In other words it exists y ∈ K such that dist(x, K) = kx − yk.

Theorem 2.4.1 Let X,Y be normed spaces and f : K ⊂ X → Y be a continuous function, K being compact. Then Im(f) is compact.

Theorem 2.4.2 Let X be normed spaces and f : K ⊂ X → R be a continuous function, K being compact. Then f is bounded and it attains its inﬁmum and supremum.

Deﬁnition 2.4.3 Let f : X → Y be a function between two normed spaces. f is said to be uniformly continuous if ∀ > 0 it exists η > 0 such that

kf(x) − f(y)kY ≤ ∀x, y with kx − ykX ≤ η.

The next few results are connected with the compactness of balls in ﬁnite dimensional spaces.

k Proposition 2.4.2 Let (R , k · k2) be the k-dimensional real space with the Eu- k clidean norm. The closed ball Bk·k2 (y0, c) is compact for any y0 ∈ R and c > 0.

k Corollary 2.4.1 Let (R , k · k2) be the k-dimensional real space with the Eu- k clidean norm. The boundary of the ball Bk·k2 (y0, c) is compact for any y0 ∈ R and c > 0.

A very important result is the following:

Theorem 2.4.3 Any two norms on a ﬁnite dimensional space are equivalent. CHAPTER 2. BANACH SPACES 13

As a consequence of the results before we have:

Theorem 2.4.4 The closed ball Bk·k(y0, c) is compact in a ﬁnite dimensional space for any norm, and for any y0 and c > 0. We can give now the extension of the Heine-Borel theorem to ﬁnite dimensional spaces.

Theorem 2.4.5 A set in a ﬁnite dimensional space is compact iff it is closed and bounded.

Theorem 2.4.6 Any ﬁnite dimensional normed space is a Banach space.

Proposition 2.4.3 Any ﬁnite dimensional subspace of a normed space is closed.

To show that a space is ﬁnite dimensional iff the unit closed ball is compact we need a technical lemma.

Lemma 2.4.1 (Riesz’s lemma) Let (X, k · k) be a normed space and U be a proper, closed subspace of X. For any λ ∈ (0, 1) it exists x ∈ X such that kxk = 1 and dist(x, U) > λ.

Theorem 2.4.7 A normed space (X, k · k) is ﬁnite dimensional iff the unit closed ball is compact.

2.4.1 Exercises Exercise 2.4.1 Let (X, k · k) be a normed space. Show that any compact set in X is closed and bounded. Is also the converse always true?

Exercise 2.4.2 Let (X, k · k) be a normed space, {xn}n ⊂ X and λ ∈ (0, 1) arbitrary. i) Assume that kxn − xkk ≥ λ for any n, k ∈ N. Show that the sequence can not have a Cauchy sub-sequence. ii) Assume now that kxn − xkk ≥ λ for any n, k ∈ N, |n − k| ≤ p, with p being a (ﬁxed) given integer. Can this sequence have a Cauchy sub-sequence?

Exercise 2.4.3 Let (X, k·k) be a normed space, and U be a linear subspace of X. Prove the following properties of the distance function (d(x, U) := infu∈U kx − uk): CHAPTER 2. BANACH SPACES 14

i) dist(λx, U) = |λ|dist(x, U) for all λ ∈ R, x ∈ X. ii) dist(x − u, U) = dist(x, U) for all u ∈ U, x ∈ X.

iii) dist(x + y, U) ≤ dist(x, U) + dist(y, U) for all x, y ∈ X.

Exercise 2.4.4 Let U be a ﬁnite dimensional subspace of a normed space X. Show that for each x ∈ X it exists u ∈ U such that kx − uk = dist(x, U).

Exercise 2.4.5 Let U be a ﬁnite dimensional proper subspace of a normed space X. Show that it exists x ∈ X such that kxk = 1 = dist(x, U).

Exercise 2.4.6 Let (X, k · k) be a normed space. Are the statements below valid? i) Let U be a proper subspace of X. Then it exists x ∈ X such that kxk = 1 and dist(x, U) > λ. ii) Let U be a closed subspace of X. Then it exists x ∈ X such that kxk = 1 and dist(x, U) > λ. iii) Let U be a proper, closed subset of X. Then it exists x ∈ X such that kxk = 1 and dist(x, U) > λ.

2.5 Linear and continuous functions

This section concerns linear and continuous functions between normed spaces, the dual (or adjoint) space of a normed space, the Neumann theorem and some applications.

Deﬁnition 2.5.1 Let (X, k · kX ) and (Y, k · kY ) two normed spaces. A map (operator) f : X → Y is called: i) linear if f(αx + βy) = αf(x) + βf(y), ∀ α, β ∈ R, x, y ∈ X. ii) bounded if it is linear and there exists C > 0 s.t. kf(x)kY ≤ CkxkX , ∀ x ∈ Y .

Remark 2.5.1 For linear functions, the boundedness means to be bounded in the regular sense on the unit ball.

Proposition 2.5.1 Let f : X → Y be a linear operator. Then the kernel and the image of f are subspaces of X and Y , respectively. CHAPTER 2. BANACH SPACES 15

Proposition 2.5.2 Let (X, k · kX ) and (Y, k · kY ) two normed spaces and T : X → Y a linear operator. The following afﬁrmations are equivalent: i) T (·) is continuous in 0. ii) T (·) is continuous. iii) T (·) is bounded.

Proof. (i) =⇒ (ii) Sei x0 arbitrary. T is continuous in x0 ⇔ ∀ > 0 ∃ η > 0 such that kT x − T x0kY < ∀ x mit kx − x0kX ≤ η. Let now > 0. T is continuous in 0 =⇒ it exists η > 0 such that

kT xkY < ∀ x mit kxkX ≤ η. (2.18)

It follows that x → (x − x0) in (2.18),

kT x − T x0kY = kT (x − x0)kY < ∀ x mit kx − x0kX ≤ η, therefore T is continuous in x0. q.e.d.

(ii) =⇒ (iii) T is continuous in 0 =⇒

∃η1 > 0 so dass kT ykY < 1 ∀ y mit kykX < η1.

x Let now x 6= 0 be arbitrary and 0 < η2 < η1. We take y = η2 above kxkX (kykX = η2 < η1) =⇒

x kT (η2 )kY < 1 =⇒ kxkX η2 kT xkY < 1 =⇒ kxkX kT xkY < 1/η2 < ∞ =⇒ kxkX kT xk sup Y < ∞. x6=0 kxkX q.e.d. CHAPTER 2. BANACH SPACES 16

(iii) =⇒ (i) T is bounded =⇒

∃M > 0 so dass kT xkY < MkxkX ∀ x. Let now > 0 be arbitrary. It exists η = /M such that

kT xkY < ∀x mit kxkX ≤ η,

(because kT xkY < MkxkX < ηM = ) =⇒ T is continuous in 0. Q.E.D. Q. E. D. Proposition 2.5.3 Let f : X → Y a linear and continuous function. Then Ker(f) is a closed subspace of X.

For functionals it holds true also the converse.

Proposition 2.5.4 Let f : X → R a linear functional. Then Ker(f) is closed subspace of X iff f is continuous. Examples of linear and continuous functions between two normed spaces. Example of a linear function which has a closed kernel but it is not bounded (not continuous). The next theorem states that linear functions defined on finite dimensional spaces are always continuous (this result being not true for infinite dimensional spaces).

Proposition 2.5.5 Let (X, k · kX ) and (Y, k · kY ) be two normed spaces, with X ﬁnite dimensional. Then any linear map f : X → Y is continuous. Example of a linear function which is not bounded (not continuous).

Deﬁnition 2.5.2 Let (X, k · kX ) and (Y, k · kY ) two normed spaces. We deﬁne L(X , Y ) := {f : X → Y | f(·) linear and continuous}, and ? X := L(X , R) = {f : X → R| f (·) linear and continuous} X? is called the dual space of X. We endow the spaces above with the norms

kf(x)kY kfkL(X ,Y ) := sup , and x∈X,x6=0 kxkX |f(x)| kfkX? := sup . x∈X,x6=0 kxkX CHAPTER 2. BANACH SPACES 17

Proposition 2.5.6 Let f ∈ L(X , Y ) and kfk = kfkL(X ,Y ) the operator norm. There holds i) kfk = sup kf(x)kY = sup kf(x)kY . kxkX ≤1 kxkX =1 ii) kf(x)kY ≤ kfkkxkX .

Proposition 2.5.7 i) (L(X , Y ), k · kL(X ,Y )) is a normed space. ii) If (Y, k · kY ) is a Banach space, then (L(X , Y ), k · kL(X ,Y )) is also a Banach ? space. Especially, (X , k · kX? ) is always a Banach space.

Theorem 2.5.1 (Neumann’s Theorem) Let X be a Banach space and A ∈ L(X , X ) be a linear and continuous function. If kAkL(X ,X ) < 1, then I − A is invertible and

∞ X (I − A)−1 = I + A + A2 + A3 + ... = Ak. k=0

Remark 2.5.2 Let X be a Banach space. Then A ∈ L(X , X ), kAkL(X ,X ) < 1 is equivalent to A is a linear contraction, with contraction constant being given by L = kAkL(X ,X ).

Deﬁnition 2.5.3 Let (X, k·kX ) and (Y, k·kY ) two normed spaces and f : X → Y a linear operator. f is called a compact operator if it is continuous and f(B) is relatively compact in Y for any bounded set B ⊂ X.

2.5.1 Applications of Neumann’s theorem. 2.5.2 Exercises Exercise 2.5.1 Let f : X → Y be a linear map between two normed spaces, satisfying f(x) ≤ 1, ∀x ∈ B(0, r) with r > 0 being a real number. Show that f(·) is continuous.

Exercise 2.5.2 We know that any Lipschitz continuous function is continuous. Show that any linear and continuous function between two normed spaces is Lips- chitz continuous. Give an example of a continuous function which is not Lipschitz continuous (obviously this will be not a linear function). CHAPTER 2. BANACH SPACES 18

Exercise 2.5.3 Let X be a normed space and φ : X → R a linear function. Show that we have either φ(X) = R or φ(X) = {0}, i. e. any non-trivial linear function is surjective.

Exercise 2.5.4 Let X be a normed space and φ : X → Rk a linear function. Then φ(X) is closed.

Exercise 2.5.5 Let T : X → Y be a linear function between two normed spaces. Show that T is continuous iff it brings sequences convergent to zero to bounded sequences.

Exercise 2.5.6 Let T : X → Y be a linear function between two normed spaces. Show that T is continuous iff it brings sequences convergent to zero to bounded sequences.

Exercise 2.5.7 Prove that a linear transformation is continuous iff it maps Cauchy sequences in Cauchy sequences.

Exercise 2.5.8 Let L(X , Y ) := {f : X → Y |f (·) linear and continuous} be the space of linear and continuous functions from X to Y and

kf(x)kY kfkL(X ,Y ) := sup x∈X,x6=0 kxkX for f ∈ L(X , Y ). Show that L(X , Y ) is a vector space and (L(X , Y ), k·kL(X ,Y )) is a normed space.

Exercise 2.5.9 Let X,Y, L(X , Y ), k · kL(X ,Y ) as above. Show that

kf(x)kY kf(x)kY kfkL(X ,Y ) = sup = sup . x∈X,kxk≤1,x6=0 kxkX x∈X,kxk=1 kxkX

Exercise 2.5.10 Prove that a linear functional φ (deﬁned on a vector space X) is a linear combination of the linear functionals φ1, . . . , φn if and only if Tn i=1 Ker(φi) ⊆ Ker(φ). Exercise 2.5.11 Let X be a Banach space and A ∈ L(X , X ) be a linear and continuous function.

k k k • Show that A ∈ L(X , X ) and kA kL(X ,X ) ≤ kAkL(X ,X ) for any k ∈ N, k ≥ 1. CHAPTER 2. BANACH SPACES 19

• The Neumann theorem is saying that if kAkL(X ,X ) < 1, then I − A is invertible and (I − A)−1 = I + A + A2 + A3 + ... Prove the Neumann theorem by using the Banach ﬁx point theorem. I denotes the identity application from X to X.

2.6 Zorn’s lemma, Hamel bases and the Hahn- Banach theorem.

This section is devoted to the existence of a Hamel basis for a vector space, the existence of linear functions with certain properties (Hahn-Banach theorem and corollaries) and with fundamental sets and annihilator of a set. We begin this section by introducing relations of order and equivalence on sets.

Deﬁnition 2.6.1 Let M be a set. A binary relation R ⊆ M × M is called i) reﬂexive if aRa ∀a ∈ M, i.e.(a, a) ∈ R. ii) symmetric if aRb ⇒ bRa. ii) antisymmetric if aRb and bRa implies a = b. iv) transitive if aRb and bRc implies aRc.

Deﬁnition 2.6.2 A relation ≤ is called relation of order on a set M if it is reﬂex- ive, antisymmetric and transitive.

Deﬁnition 2.6.3 A relation ∼ is called relation of equivalence on a set M if it is reﬂexive, symmetric and transitive.

Deﬁnition 2.6.4 A partially ordered set (A, ≤) is a set with a relation of order on it. Two elements of a partially ordered set are called comparable, if there holds a ≤ b or b ≤ a. In a partially ordered set not any two elements are comparable. A chain or totally ordered set is a partially ordered set in which any two elements are comparable.

Deﬁnition 2.6.5 Let (M, ≤) be a partially ordered set and A be a subset of M. An element x ∈ M is called an upper bound for A, if a ≤ x for any a ∈ A. An element m ∈ A is called maximal element of A if a ∈ A and m ≤ a implies a = m. CHAPTER 2. BANACH SPACES 20

Lemma 2.6.1 (Zorn’s lemma) Let M be a nonempty, partially ordered set. Sup- pose that every chain C ⊂ M has an upper bound. Then M has at least one maximal element.

Deﬁnition 2.6.6 (Hamel basis) Let X be a vector space. A subset H ⊂ X is called a Hamel basis for the space X, if any element of X can be written in a unique way as linear combination of a ﬁnite number of elements of H.

Assuming that the Zorn’s lemma holds true, one can prove that any nontrivial vector space has a Hamel basis. Theorem 2.6.1 Any nontrivial vector space has a Hamel basis.

Deﬁnition 2.6.7 A functional p : X → R is called sub-additive if it satisﬁes i) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X. ii) p(λx) = λp(x) for all λ ≥ 0, x ∈ X.

Deﬁnition 2.6.8 Let f, g : X → R be two functionals. We say that f is dominated by g, if f(x) ≤ g(x), ∀x ∈ X.

Theorem 2.6.2 (Hahn-Banach) Let X be a vector space and p : X → R a sub- additive function. Then any linear functional f : X0 → R,X0 ≤ X dominated by p can be extended to a linear functional on X dominated by p.

Corollary 2.6.1 Let Y ≤ X be a subspace of a normed space X and let φ a linear functional on Y satisfying |φ(y)| ≤ Mkyk for all y ∈ Y . Then the functional φ can be prolonged to a linear functional on X which satisﬁes the inequality above. In other words, a linear and continuous functional on Y can be prolonged (not uniquely) to a linear and continuous functional on X.

Corollary 2.6.2 Let Y ≤ X be a subspace of a normed space X. If w ∈ X and dist(w, Y ) > 0, then there exists a linear and continuous functional φ : X → R 1 such that φ(y) = 0 for all y ∈ Y, φ(w) = 1 and kφk = . dist(w, Y )

Corollary 2.6.3 Let X be a normed space. To each point w ∈ X, w 6= 0 it exists at least one linear and continuous functional φ such that φ(w) = 1 and 1 kφk = . kwk CHAPTER 2. BANACH SPACES 21

Corollary 2.6.4 Let X be a normed space and X? its dual. Then (∀)x ∈ X there holds

kxk = max{φ(x)|φ ∈ X?, kφk = 1} = max{|φ(x)||φ ∈ X?, kφk = 1}. (2.19)

Deﬁnition 2.6.9 A set A is called fundamental in X, if < A > = X, i.e. the linear space generated by A is dense in X.

Example. Let X = C[0, 1] be the space of continuous functions on C[0, 1]. n Then A := {un|n ∈ N, un(t) = t } is a fundamental set in X (due to Weierstrass approximation theorem).

Deﬁnition 2.6.10 The annihilator of a set A ⊂ X is deﬁned by

A⊥ := {φ ∈ X?|φ(a) = 0 ∀a ∈ A}.

Theorem 2.6.3 A subset A in a normed space X is fundamental iff its annihilator is {0}, i.e. < A > = X ⇔ A⊥ = 0.

2.6.1 Quotient spaces

Let (X, k · kX ) a normed space, F ⊆ X a linear subspace, and ” ∼ ” the relation of equivalence deﬁned through

x ∼ y ⇔ x − y ∈ F.

It is easy to prove that ∼ is an equivalence relation (reﬂexive, symmetric and transitive). The class of an element x ∈ X is denoted by

xˆ = {y ∈ X|x − y ∈ F } and X/F , the spaces of classes is called the quotient space. The seminorm

kxˆk := inf kx + ykX = inf kx + ykX = dist(x, F ), (2.20) y∈F y∈F is called the quotient seminorm. CHAPTER 2. BANACH SPACES 22

Proposition 2.6.1 i) The quotient seminorm is a norm iff F is closed. ii) If X is a Banach space and F is a closed subspace of X, then X/F is a Banach space (see Yoshida, [34] pg. 60).

Deﬁnition 2.6.11 The function p : X → X/F, p(x) :=x ˆ is called the canonical projection. dim(X/F ) := codim(F ) = the co-dimension of F in X.

Proposition 2.6.2 i) The canonical projection p : X → X/F, p(x) :=x ˆ is linear, continuous and open. ii) If F is a proper, closed subspace than kpk = 1.

Remark 2.6.1 Let X a Banach space, F a linear subspace of finite co-dimension. Then F is not necessary closed. As a counter example take a linear but discontinuous functional f (i.e. defined on X with values in R). The co-dimension of Ker(f) is one, so finite but Ker(f) is not closed, because f is not continuous.

2.6.2 Exercises Exercise 2.6.1 Let A ⊂ X be a subset of normed space X. Prove that

⊥ ⊥ A⊥ = A = hAi⊥ = hAi .

Exercise 2.6.2 Let Y ≤ X be a subspace of normed space X. Prove that

dist(x, Y ) = sup{φ(x)|φ ∈ X?, φ ⊥ Y, kφk = 1}, where φ ⊥ Y means φ(y) = 0 for all y ∈ Y.

Exercise 2.6.3 Let Y ≤ X be a subspace of normed space X. Prove that

Y ⊥ := {φ ∈ X?|φ(y) = 0 ∀y ∈ Y } is a closed subspace of X?.

Exercise 2.6.4 Let X be a normed space and let Z ≤ X? be a subspace of X?. We deﬁne Z⊥ := {x ∈ X|φ(x) = 0 ∀φ ∈ Z}. ⊥ Prove that Y ⊥ = Y for any closed subspace Y of X. CHAPTER 2. BANACH SPACES 23

Exercise 2.6.5 Show that if H is a Hamel basis of a normed space X then also h |h ∈ H is a Hamel basis of X. khk

Exercise 2.6.6 Let (X, k · kX ) be a normed space and F ⊆ X be a linear subspace. Show that the quotient seminorm, deﬁned by

kxˆk := inf kx + ykX , y∈F is indeed a seminorm. Moreover, show that if F is ﬁnite dimensional then the seminorm is a norm.

2.7 Baire theorem and uniform boundedness.

In this section we learn a few more speciﬁc properties of complete spaces (in particular of Banach spaces). We will also learn what are category I and II sets, and the principle of uniform boundedness.

Theorem 2.7.1 (Baire or category theorem) In a complete metric space, the in- tersection of a countable number of open and dense sets is a dense set.

Corollary 2.7.1 If a complete metric space is expressed as a countable union of closed sets, then at least one of the sets must have a nonempty interior.

Deﬁnition 2.7.1 A set A ⊆ X is called nowhere dense if its closure has an empty interior.

Deﬁnition 2.7.2 A set A ⊆ X is said to be of category I if it is a countable union of nowhere dense sets, i.e. A = ∪n≥1Fn, with Fn nowhere dense for each n ≥ 1.

Deﬁnition 2.7.3 A set A ⊆ X is said to be of category II if it is not of category I.

Lemma 2.7.1 If A ⊆ X is of category II, and A = ∪n≥1Fn, with Fn closed for each n, then at least one Fn has a nonempty interior (i.e. it is not nowhere dense).

Remark 2.7.1 A Banach space is always of category II, as a consequence of Corollary 2.7.1. CHAPTER 2. BANACH SPACES 24

Proposition 2.7.1 The following two statements are true: i) Any proper closed subspace of a normed space is nowhere dense (and therefore of ﬁrst category). ii) Any subspace of second category is dense.

Theorem 2.7.2 (Banach-Steinhaus Theorem) Let {Aα} be a family of continuous linear transformations deﬁned on a Banach space X and taking values in a normed space Y . In order that supα kAαk < ∞ it is necessary and sufﬁcient that the set {x ∈ X| sup kAαxk < ∞} α is of category II in X.

Theorem 2.7.3 (The principle of uniform boundedness) Let {Aα} be a family of continuous linear transformations deﬁned on a Banach space X and taking values in a normed space Y . If supα kAαxk < ∞ for all x ∈ X then supα kAαk < ∞

Proposition 2.7.2 A Banach space is either ﬁnite dimensional or it has a non- countable Hamel basis.

2.7.1 Exercises Exercise 2.7.1 Prove the following statements. i) A set is dense in a space X iff it intersects any ball. ii) Let F ⊆ X be a closed, with an empty interior set. Then show that the complement of F , i.e. X \ F is open and dense.

Exercise 2.7.2 Let A ⊆ X be a set of category II. If A ⊆ B, then show that B must be of category II as well.

Exercise 2.7.3 Let A ⊆ X be a closed set of category II. Then the interior of A is non-empty.

Exercise 2.7.4 Prove that any proper closed subspace of a normed space is nowhere dense (and therefore of category I).

Exercise 2.7.5 Prove that any subspace of second category in a normed space is dense. CHAPTER 2. BANACH SPACES 25

Exercise 2.7.6 Let X be a Banach space and O ⊆ X be an open and non-empty set. Then show that O is of category II.

Exercise 2.7.7 Show that any set with a non-empty interior is of category II in a Banach space.

Exercise 2.7.8 Show that a Banach space is either ﬁnite dimensional or it has a non-countable Hamel basis.

Exercise 2.7.9 Select a function x0 ∈ C[0, 1] and a sequence of reals {αn}n. Deﬁne recursively Z t xn+1(t) := αn + xn(s)ds 0 for all n ∈ N. Assume that for each t ∈ [0, 1], there is an n ∈ N s.t. xn(t) = 0. Prove that x0 = 0.

Exercise 2.7.10 Show that it does not exist a norm such that the space of polyno- mials on the interval [0, 1] with this norm is a Banach space.

2.8 The open (interior) mapping and closed mapping theorems.

Deﬁnition 2.8.1 Let X,Y be two normed spaces. A function f : X → Y is said to be closed (or have a closed graph) if {(x, f(x))|x ∈ X} is a closed set in X ×Y . In other words, f is closed iff xn → x, f(xn) → y, with xn, x ∈ X, y ∈ Y implies that y = f(x).

Remark 2.8.1 Obviously, any continuous map is closed. The converse is but not true (see example of the differential operator from C1[0, 1] to C[0, 1], which is closed but no continuous).

Deﬁnition 2.8.2 Let X,Y be two topological spaces. A function f : X → Y is said to be open (or interior) if maps open subsets of X to open subsets of Y .

Proposition 2.8.1 Let X,Y be normed spaces. Further, let f : X → Y be a linear and open map. Then f(·) is surjective. CHAPTER 2. BANACH SPACES 26

Proof. f(X) is a linear space because f(·) is linear. f(·) being open implies that f(X) contains a (non-empty) ball. But a linear space which contains a ball contains the whole space, therefore f(X) = Y , so f(·) must be surjective. Q. E. D.

Theorem 2.8.1 (The open mapping theorem or the interior mapping theorem) Let X,Y be two Banach spaces and f : X → Y be a closed, linear and surjective transformation. Then f is open.

Corollary 2.8.1 Let X,Y be two Banach spaces and f : X → Y be linear and bijective (so invertible). Then the inverse of f is a linear function and, if f is continuous, it is also continuous.

Corollary 2.8.2 Let X be a Banach space with two norms k · k1 and k · k2. If one of the norms dominates the other then the two norms are equivalent.

Theorem 2.8.2 (The closed graph theorem) A closed linear map from one Ba- nach space to another is continuous.

Lemma 2.8.1 Let L : X → Y be a linear map, X,Y be normed spaces. Then L is open iff (∃)c > 0 such that (∀)y ∈ Y, (∃)x ∈ X satisfying

L(x) = y and kxk ≤ ckyk.

It follows by this that any open and linear map is surjective.

Theorem 2.8.3 Let X be a Banach space and L : X → L(X) be linear, bounded and open. Then L(X) is also a Banach space.

Deﬁnition 2.8.3 Let L : X → Y be a linear and bounded map. We deﬁne the adjoint operator L? : Y ? → X? by

L?(φ)(x) := φ(L(x)), ∀x ∈ X, ∀φ ∈ Y ?.

Proposition 2.8.2 The adjoint operator deﬁned above is linear and continuous (and therefore also bounded). CHAPTER 2. BANACH SPACES 27

Proof. The linearity is easy to be checked. The boundedness follows from

kL?k = sup kL?φk = sup sup kL?φ(x)k φ∈Y ?,kφk=1k φ∈Y ?,kφk=1k x∈X,kxk=1 = sup sup kφ(L(x))k = sup kL(x)k = kLk. x∈X,kxk=1 φ∈Y ?,kφk=1k x∈X,kxk=1 We used above the representation (2.19) of a the norm of an element.

Q. E. D. In the finite dimensional case (it is enough that one of the spaces X or Y is finite dimensional) there holds (see Exrecise 2.8.3) L surjective ⇔ L? injective, L injective ⇔ L? surjective. In general we have a slightly different results. We need but first a lemma.

Lemma 2.8.2 Let L : X → Y be a linear map. Then

Ker(L?) = L(X)⊥(:= {φ ∈ Y ?|φ(L(x)) = 0 for all x ∈ X}).

Proof. There holds

Ker(L?) := {φ ∈ Y ?|L?φ = 0} = {φ ∈ Y ?|L?φ(x) = 0 for all x ∈ X} = {φ ∈ Y ?|φ(L(x)) = 0 for all x ∈ X} = L(X)⊥

Q. E. D. From Lemma 2.8.2 follows also that L surjective ⇒ L? injective. One can also show that we always have L? surjective ⇒ L injective. This is an immediate consequence of the following lemma. Lemma 2.8.3 Let L : X → Y be a linear and continuous map. There holds

L?(Y ?) ⊆ Ker(L)⊥, (2.21) ? ? L (Y )⊥ = Ker(L). (2.22) CHAPTER 2. BANACH SPACES 28

Proof. By deﬁnition we have Ker(L)⊥ = {ψ ∈ X?|ψ(x) = 0 for x ∈ Ker(L)}. Let now ψ ∈ L?(Y ?) arbitrarly. There exists φ ∈ Y ? such that ψ = φ ◦ L, which means that for any x ∈ Ker(L) there holds ψ(x) = φ(L(x)) = φ(0) = 0. This implies that L?(Y ?) ⊆ Ker(L)⊥. Recalling that Ker(L)⊥ is closed, from L?(Y ?) ⊆ Ker(L)⊥ we get that also L?(Y ?) ⊆ Ker(L)⊥. This implies (2.21). We prove now (2.22). There holds

? ? ? ? L (Y )⊥ = {x ∈ X|L φ(x) = 0 ∀ φ ∈ Y } = {x ∈ X|φ((L(x)) = 0 ∀ φ ∈ Y ?} = {x ∈ X|L(x) = 0} = Ker(L).

Q. E. D.

Remark 2.8.2 The inclusion in Lemma 2.8.3 above is normally strict. Neverthe- less, if X is reﬂexive, one can show that L?(Y ?) = Ker(L)⊥, see Exercise 2.10.1 in Section 2.10. This is also valid in tha case when L has a closed range (see the closed range theorem). Moreover, Ker(L)⊥ is always the same with the closure of the range of L? w.r.t. the weak? topology.

Now we can give the general result, which is a bit weaker as the one for the ﬁnite dimensional spaces.

Theorem 2.8.4 Let L : X → Y be a linear and bounded map.

L? injective ⇔ L(X) is dense in Y .

Proof. In other words we have to prove that the range of L is dense in Y iff L? is injective. Due to Lemma 2.8.2 Ker(L?) = L(X)⊥ which means that L? is injective iff the anihilator of L(X) is 0. But the annihilator being 0 is equivalent with L(X) is dense, due to Lemma 2.6.3. Q. E. D.

Lemma 2.8.4 Let X be a normed space, and A ⊆ X. Then there holds

⊥ (A )⊥ = hAi. (2.23)

⊥ In particular, if A ≤ X, then (A )⊥ = A. CHAPTER 2. BANACH SPACES 29

Proof. Recalling Exercise 2.6.1 there holds

⊥ A⊥ = hAi⊥ = hAi .

On the other hand, due to Exercise 2.6.4 we have

⊥ (hAi )⊥ = hAi.

⊥ It follows then that (A )⊥ = hAi. Q. E. D.

Theorem 2.8.5 Let L : X → Y be a linear and bounded map. The range of L and the kernel of L? are related by the fact

? Ker(L )⊥ = L(X). (2.24)

Proof. There holds Ker(L?) = L(X)⊥. It follows

? ⊥ Ker(L )⊥ = (L(X) )⊥ = L(X), with the last equality being a consequence of Lemma 2.8.4. Q. E. D.

Theorem 2.8.6 (The closed range theorem) Let L : X → Y be a linear and bounded map and X,Y be Banach spaces. Then the following statements are equivalent:

i) The range of L, i. e. L(X) is closed.

ii) The range of L?, i. e. L?(Y ?) is closed.

? iii) L(X) = Ker(L )⊥. iv) L?(Y ?) = Ker(L)⊥.

The next theorem is a very important result in the view of its application to Galerkin discretization methods (so numerical analysis).

Theorem 2.8.7 Let L : X → Y be a linear, bounded and injective map, X,Y be Banach spaces. The range of L is closed iff infkxk=1 kL(x)k > 0. CHAPTER 2. BANACH SPACES 30

2.8.1 Exercises Exercise 2.8.1 Let X be a normed space. Show that the following statements hold true: i) If A ≤ X then A¯ ≤ X. Is the converse true as well? ii) Let A ≤ X. Then A⊥ = A¯⊥. ⊥ ¯ iii) Let A ≤ X. Then (A )⊥ = A. Exercise 2.8.2 Let X,Y be two normed spaces and L : X → Y be a linear operator. i) Show that if there exists α > 0 such that infkxk=1 kL(x)k ≥ α, then L is injective. ii) When is also the converse true?

Exercise 2.8.3 Let X,Y be two normed spaces and L : X → Y be a linear and continuous operator. Show that if one of the spaces X,Y is ﬁnite dimensional, than L is surjective is equivalent with L? being injective, where L? denotes the adjoint operator of L.

2.9 Weak and weak? convergence.

Let X, k · k be a normed space, and X? its dual (with the usual norm). In this section we deﬁne the concept of weak and weak ? convergence.

Deﬁnition 2.9.1 A sequence {xn}n ⊂ X is said to be weakly convergent to ? x ∈ X if φ(xn) → φ(x) for any linear and continuous functional φ ∈ X . We write xn * x. Example (Weak convergence in the Lp spaces). Let 1 < p < ∞, and let

p L (Ω) := {f :Ω → R|f measurable and kfkp < ∞}, where Ω is an open set in Rd, (d ∈ N, d ≥ 1) and the norm Z 1/p p kfkp := |f(x)| dx . (2.25) Ω p p Let {fn}n ⊂ L (Ω) be a sequence. fn * f ∈ L (Ω) is equivalent to Z Z fnφ dx → fφ dx Ω Ω CHAPTER 2. BANACH SPACES 31

1 1 for all φ ∈ Lq(Ω). Lq(Ω) is the dual space of Lp(Ω), + = 1. p q

Proposition 2.9.1 Let {xn}n ⊂ X be a sequence which converges weakly. Then the limit is unique.

Proposition 2.9.2 The convergence in norm (also called strong convergence) implies the weak convergence. The converse is not always true.

We have just seen that a weakly convergent sequence is not necessarily strongly convergent. The next proposition is establishing that a weakly convergent sequence is at least bounded.

Proposition 2.9.3 Let {xn}n ⊂ X be a sequence which converges weakly. Then {xn}n ⊂ X is bounded. Even if the strong and weak convergence are normally not equivalent, the situation is different in ﬁnite dimensional spaces. Theorem 2.9.1 Let X be a ﬁnite dimensional normed space. Then weak and strong convergence are equivalent.

Deﬁnition 2.9.2 A subset F ⊆ X is said to be weakly sequentially closed if the weak limit of any sequence in F belongs to F .

In order to establish that actually closed and weakly sequentially closed are equivalent notions for linear subspaces, we need the following impotant result.

Lemma 2.9.1 Let {xn}n ⊆ X be a sequence which converges weakly to x ∈ X. Then x ∈ h{xn}ni, i. e. it exists a linear combination of elements of {xn}n which converges weakly to x Now we can prove that closed and weakly sequentially closed are equivalent notions for linear subspaces. Theorem 2.9.2 A subspace F ≤ X of a normed space is closed iff it is weakly sequentially closed. The next result can be used as criterion to ensure the weak convergence.

Theorem 2.9.3 Let {xn}n ⊆ X be a bounded sequence in a normed space. As- sume that it exists x ∈ X such that φ(xn) → φ(x) for all φ in a fundamental set ? of the dual space X . Then xn * x (weakly). CHAPTER 2. BANACH SPACES 32

In same special normed spaces (uniformly convex) the weak convergence is almost equivalent to strong convergence. Deﬁnition 2.9.3 A Banach space (X, k · k) is called uniformly convex if for any > 0 it exists δ > 0 such that x, y ∈ X, with kxk = kyk = 1 and kx − yk > x + y implies k k < 1 − δ. 2 The following theorem establish now that the weak convergence plus the convergence of the sequences of the norms implies the strong convergence (in uniformly convex spaces, so in particular in Hilbert spaces).

Theorem 2.9.4 Let (X, k · k) be an uniformly convex space, and {xn}n ⊆ X. Then xn → x ∈ X iff xn * x and kxnk → kxk.

2.9.1 Weak? convergence Let X be a normed space, and X? its dual. We further denote by X?? the dual of the dual space (X?? = X for reflexive spaces, in general is but a larger space than X, see Section 2.10 for more details). We can apply the weak convergence ? ? ? concept to X , so for {φn}n ⊂ X , φn * φ ∈ X if f(φn) → f(φ) for any f ∈ X??. The concept of weak? convergence concerns also convergence in the dual spaces, but it is even weaker as the weak convergence in the way defined above. ?? Let x ∈ X and we define fx ∈ X

? fx(φ) := φ(x), ∀ φ ∈ X . (2.26)

? Definition 2.9.4 We say that a sequence {φn}n ⊂ X converges weakly ? to ? φ ∈ X , if fx(φn) → fx(φ), for all x ∈ X. We denoted by fx the function defined ? in (2.26) above. We write φn * φ. Remark 2.9.1 In reflexive spaces the concept of weak and weak ? convergence will coincide.

We ﬁnish this section with a result concerning the weak ? compactness of the dual space.

? Theorem 2.9.5 Let X be a separable normed space. Then, if {φn}n ⊆ X is a bounded sequence in X?, there exists a sub-sequence that converges weakly ? to an element of X?. CHAPTER 2. BANACH SPACES 33

We ﬁnish this section with a useful result regarding weak and point-wise convergence.

Proposition 2.9.4 Let {xn}n a sequence in a Banach space X, xn * x weakly. ? ? Further, let {fn}n ⊂ X a (strong) convergent sequence, fn → f ∈ X . Then fn(xn) → f(x).

2.9.2 Exercises

?? Exercise 2.9.1 Show that the functions fx(·) deﬁned in (2.26) are indeed in X , i.e. they are linear and continuous.

Exercise 2.9.2 Let {fn}n ⊂ C[0, 1], fn * f ∈ C[0, 1]. Show that fn converges also point-wise to f.

2.10 Reﬂexive spaces.

Let X be a normed space, and X? its dual. We further denote by X?? the dual of ?? the dual space. For any point x ∈ X we deﬁned in (2.26) fx ∈ X

? fx(φ) := φ(x), ∀ φ ∈ X . We deﬁne now a function F : X → X??,

F (x) := φx.

One can easily show that F (·) is linear, injective and an isometry (which implies continuous). There holds indeed

kF (x)k = kφxk = sup kφx(f)k = sup kf(x)k = kxk, f∈X?,kfk=1 f∈X?,kfk=1 with the last equality following by one of the corollaries of the Hahn-Banach theorem.

Definition 2.10.1 A normed space X is called reflexive if X ≡ X?? or, in other words, if the map F defined above is bijective.

Remark 2.10.1 Any reﬂexive space is a Banach space, because it is the dual of a normed space. CHAPTER 2. BANACH SPACES 34

Examples.

• Any ﬁnite dimensional space is reﬂexive.

• Lp(Ω), 1 < p < +∞ is reﬂexive. L1(Ω) and L∞(Ω) are not reﬂexive.

Theorem 2.10.1 Let X be a reﬂexive space. Then any closed subspace of X is reﬂexive as well.

Theorem 2.10.2 Let X be a Banach space. Then X is reﬂexive iff X?? is reﬂexive.

Theorem 2.10.3 (Eberlein-Smulyan Theorem) Let X be a reﬂexive space and {xn}n be a bounded sequence. Then it exists a sub-sequence of {xn}n and x ∈ X such that xn * x weakly. In other words, in any reﬂexive space any closed ball is weakly sequential compact (i.e. compact w.r.t. weak topology).

Remark 2.10.2 One can actually prove that also the converse is true, i.e. a space is reﬂexive iff the unit closed ball is weakly sequential compact, cf. Yosida pg. 141.

For proving the Eberlein-Smulyan Theorem one needs the following result.

Lemma 2.10.1 If the dual space X? of a normed space X is separable, then the space itself is separable.

Proof. see Kreyszig, pg. 245. Q. E. D.

Theorem 2.10.4 (James) A Banach space X is reﬂexive iff each linear and continuous functional on X attains its supremum on the unit cloase ball of X.

2.10.1 Exercises Exercise 2.10.1 Let X,Y be normed spaces, with X being reﬂexive, and L : X → Y be a linear and continuous map. Show that there holds

L?(Y ?) = Ker(L)⊥. Chapter 3

Hilbert spaces

3.1 Deﬁnition and basic propoerties

Deﬁnition 3.1.1 Let V a vector space. h·, ·i : V × V → R is called an inner (scalar) product on V over R if: i) hx, xi ≥ 0 ∀ x ∈ V and hx, xi = 0 ⇒ x = 0. ii) hαx + βy, zi = αhx, zi + βhy, zi ∀ x, y, z ∈ V and ∀ α, β ∈ R. iii) hx, yi = hy, xi ∀ x, y ∈ V. In the case of inner product over C the property iii) must be replaced by iii)0 hx, yi = hy, xi ∀ x, y ∈ V.

Deﬁnition 3.1.2 A vector space V together with an inner product on it is called a pre-Hilbert space.

It is naturally to ask ourself if a pre-Hilbert space is also a normed space. The next proposition clariﬁes this.

Proposition 3.1.1 Let (X, h·i) be a pre-Hilbert space. The following statements hold true:

i) hx, αy + βzi = αhx, yi + βhx, zi ∀x, y, z ∈ X, α, β ∈ C.

ii) The map k · k : X → R, k · k := (hx, xi)1/2 is a norm on X. The proof of point i) is straightforward. For proving ii) one needs the Cauchy- Schwarz inequality in the form:

35 CHAPTER 3. HILBERT SPACES 36

Lemma 3.1.1 (The Cauchy-Schwarz inequality). Let (X, h·i) be a pre-Hilbert space. There holds for any x, y ∈ X |hx, yi| ≤ kxkkyk. (3.1) Proof. Q. E. D. As a consequence of Proposition 3.1.1, a pre-Hilbert space is implicitly a normed space, with the norm given by p kvkV := hv, vi. Definition 3.1.3 A pre-Hilbert space which is complete is called a Hilbert space. Every Hilbert space is obviously also a Banach space. Moreover, we will see later that every Hilbert space is reflexive (as a consequence of the Rietz-Frechet theorem), and uniformly convex (see Exercise 3.1.2). Examples of Hilbert and pre-Hilbert spaces • (Cn, h·, ·i), with the scalar product defined by n X n hx, yi = xiyi, ∀x, y ∈ C . (3.2) i=1

• (Cn, h·, ·i), with the scalar product deﬁned by n X n hx, yi = xiyi, ∀x, y ∈ R . (3.3) i=1

• L2(Ω), h·, ·i) with the scalar product deﬁned by Z hf, gi = f(x)g(x) dx, ∀f, g ∈ L2(Ω). (3.4) Ω The next two proposition concern some properties of the pre-Hilbert spaces. Proposition 3.1.2 Let (X, h·, ·i) be a pre-Hilbert space. There holds the parallelogram identity kx + yk2 + kx − yk2 = 2(kxk2 + kyk2) ∀x, y ∈ X. (3.5) and the Pythagorean law kx + yk2 = kxk2 + kyk2 ∀x, y ∈ X, hx, yi = 0. (3.6) CHAPTER 3. HILBERT SPACES 37

The proof of Proposition 3.1.2 is straightforward by using that the norm is generated by a scalar product. Actually, there holds true also the converse.

Theorem 3.1.1 Let (X, k · k) be a normed space. If the parallelogram identity holds true, i.e.

kx + yk2 + kx − yk2 = 2(kxk2 + kyk2) ∀x, y ∈ X, (3.7) then X is a pre-Hilbert space.

Proof. We deﬁne a scalar product on X by 1 hx, yi := (kx + yk2 − kx − yk2 + ikx + iyk2 − ikx − iyk2). (3.8) 4 One shows that all the properties of a scalar product are satisﬁed, so the space becomes a pre-Hilbert space. Q. E. D.

Proposition 3.1.3 Let (X, h·, ·i) be a pre-Hilbert space. The following statements hold true:

i) x = 0 iff hx, yi = 0, ∀y ∈ X.

ii) x = y iff hx, zi = hy, zi, ∀z ∈ X.

iii) kxk = sup{|hx, yi||y ∈ X, kyk = 1}.

The proof is straightforward. Please remark the similitude with the representation of the norm in Corollary 2.6.4. We will see later that the two representations are actually the same. The following two theorems concern the existence of a closest point in a set/subspace of a Hilbert space.

Theorem 3.1.2 Let (X, h·, ·i) be a Hilbert space. If K is a convex, closed, non- empty subset of X, and x ∈ X, then there exists an unique point y ∈ K, closest to X, i.e. dist(x, K) = kx − yk.

Corollary 3.1.1 If Y ≤ X is a closed, non-empty subspace of a Hilbert space X and x ∈ X, then there exists an unique point y ∈ Y , closest to X. In particularly, it is true for ﬁnite dimensional subspaces of a Hilbert space. CHAPTER 3. HILBERT SPACES 38

Before giving the next result, we need to deﬁne orthogonality in pre-Hilbert spaces.

Deﬁnition 3.1.4 Let (X, h·, ·i) be a Hilbert space. We say that x, y ∈ X are orthogonal (to each other) if hx, yi = 0 and we write x ⊥ y. If Y ⊂ X, then x orthogonal to Y means hx, yi = 0, ∀y ∈ Y. We write x ⊥ Y.

Theorem 3.1.3 Let (X, h·, ·i) be a pre-Hilbert space. Let Y ≤ X and x ∈ X, y ∈ Y . The following statements are equivalent:

i) x − y ⊥ Y .

ii) y is the closest point to x in Y.

We continue by deﬁning the orthogonal complement of Y ≤ X by Y ⊥ := {x ∈ X|hx, yi = 0 ∀y ∈ Y }. We recall the deﬁnition ?? of a annihilator of a set, and we will see that Y ⊥ is indeed the annihilator of Y in Hilbert spaces.

Theorem 3.1.4 Let (X, h·, ·i) be a Hilbert space and Y ≤ X be a closed subspace. Then X = Y + Y ⊥. (3.9)

3.1.1 Exercises

Exercise 3.1.1 Let (X, h·i) be a Hilbert space, and {xn}n ⊆ X. Show that xn → x ∈ X iff xn * x and kxnk → kxk.

Exercise 3.1.2 A Banach space (X, k · k) is called uniformly convex if for any > 0 it exists δ > 0 such that x, y ∈ X, with kxk = kyk = 1 and kx − yk > x + y implies k k < 1 − δ. Show that any Hilbert space is uniformly convex. 2

Exercise 3.1.3 We deﬁne for 1 ≤ p < ∞ the p-norm on Rn by

n !1/p X p kxkp := |xi| i=1

n Show that the space (R , k · kp) is a Hilbert space iff p = 2. Nevertheless, the n space (R , k · kp) is Banach for any p ≥ 1. CHAPTER 3. HILBERT SPACES 39

Exercise 3.1.4 Let again 1 ≤ p < ∞ be a real norm and consider the space Lp(Ω) with the norm Z 1/p p kfkp := |f(x)| dx Ω

Show that the space (Lp(Ω), k · kp) is a Hilbert space iff p = 2. Again, the spaces are but Banach for any p ≥ 1.

Exercise 3.1.5 Let X, k · k be a complex Hilbert space. Let A : X → X be a linear and continuous map. Show that if Ax ⊥ x for all x ∈ X, then A = 0. Show that this is not true for real Hilber spaces.

3.2 Orthogonality

In this section we deﬁne orthogonal systems, orthogonal and orthonormal basis and show that any pre-Hilbert space admits an orthonormal basis.

Deﬁnition 3.2.1 A set V of vectors in a pre-Hilbert space X is called orthogonal if hx, yi = 0 ∀x, y ∈ V . V is called orthonormal if it is orthogonal and kuk = 1, ∀u ∈ V.

Theorem 3.2.1 (Generalized Pythagorean law) Let (X, h·, ·i) be a pre-Hilbert space. If {x1, . . . , xn} is orthogonal, then

n n X 2 X 2 k xik = kxik . (3.10) i=1 i=1

The proof is straightforward. There exists also a more general Pythagorean law.

Theorem 3.2.2 (Generalized Pythagorean law) Let (X, h·, ·i) be a Hilbert space P P 2 and {xj}j∈J be orthogonal. Then the series j xj converges iff j |xj| < ∞.

n n X 2 X 2 k xik = kxik . (3.11) i=1 i=1 The set J above might be non-countable. We define a non-countable sum as P P J xj = S if ∀ > 0 it exists a finite set J ⊆ J such that | j∈F xj − S| ≤ ∀F finite, with J ⊆ F ⊆ J. CHAPTER 3. HILBERT SPACES 40

Theorem 3.2.3 If {y1, . . . , yn} is an orthonormal set in a pre-Hilbert space, and Pn if Y := hy1, . . . , yni, then the closest point to x ∈ X in Y is i=1hx, yiiyi.

Definition 3.2.2 (orthogonal projection) In the same condition as above, we call Pn y = i=1hx, yiiyi the orthogonal projection of x ∈ X to hy1, . . . , yni. The Pn operator p : X → hy1, . . . , yni, defined by p(x) := i=1hx, yiiyi is called the orthogonal projector. The scalars hx, yii are called generalised Fourier coefficients of x.

Pn We remark that for x ∈ hy1, . . . , yni holds obviously x = i=1hx, yiiyi.

Theorem 3.2.4 (Bessel’s inequality) If {ui}i∈I is an orthonormal set in a pre- Hilbert space X, then for every x ∈ X there holds

X 2 hx, uii ≤ kxk . (3.12) i∈I

Corollary 3.2.1 If {ui}i∈I is an orthonormal system in a pre-Hilbert space X, then limn→∞hx, uni = 0.

Corollary 3.2.2 Let {ui}i∈I be an orthonormal system in a pre-Hilbert space X. Then for each x ∈ X, at most a countable number of the generalised Fourier coefﬁcients hx, uii are non-zero.

Deﬁnition 3.2.3 Let X be a pre-Hilber space. An orthonormal basis for X is a maximal orthonormal system (in the sense that it can not be properly contained in other orthonormal system).

Theorem 3.2.5 Every non-trivial pre-Hilbert space has an orthonormal basis.

Theorem 3.2.6 (The orthonormal basis theorem) For an orthonormal system {ui}i∈I in a Hilber space X the following statements are equivalent:

i) {ui}i∈I is an orthonormal basis.

ii) If x ∈ X and x ⊥ ui, ∀i ∈ I ⇒ x = 0. P iii) For any x ∈ X holds true x = i∈I hx, uiiui. P iv) For all x, y ∈ X there holds hx, yi = i∈I hx, uiihy, uii. CHAPTER 3. HILBERT SPACES 41

2 P v) kxk = i∈I hx, uii (Parceval’s inequality).

Further, the next theorem concerns the properties of the projector operator.

Theorem 3.2.7 (The orthogonal projection theorem) Let X be a Hilbert space, Y ≤ X a closed subspace and {yi}i∈I an orthogonal basis of Y . The orthogonal Pn projector p(·): H → Y , deﬁned through p(x) := i=1hx, yiiyi, for all x ∈ X has the following properties

i) It is well-deﬁned, i.e. p(x) exists and it is unique for any x ∈ X.

ii) If Y 6= 0, then kpk = 1.

iii) It is surjective.

iv) It is linear.

v) x − p(x) ⊥ Y ∀x ∈ X.

vi) p is Hermitian, i.e. hP x, vi = hx, P vi ∀x, v ∈ X.

vii) p is idempotent, i.e. p ◦ p = p.

viii) p(y) = y for all y ∈ Y .

ix) There holds kxk2 = kp(x)k2 + kx − p(x)k2 for all x ∈ X.

There is a famous algorithm to construct an orthonormal sequence of vectors by starting with a linear independent system.

Theorem 3.2.8 (The Gram-Schmidt construction) Let {vn}n be a linear indepen- v1 dent sequence of vectors in a pre-Hilbert space. Having set u1 := deﬁne kv1k recursively for n ≥ 2

v − Pn−1hv , u iu u := n i=1 n i i . (3.13) n Pn−1 kvn − i=1 hvn, uiiuik CHAPTER 3. HILBERT SPACES 42

3.2.1 Exercises

Exercise 3.2.1 Let (X, h·i) be a Hilbert space, x, y ∈ X and λ0 ∈ Γ such that hy − λ0x, xi = 0. Then there holds

dist(y, hxi) = ky − λ0xk. (3.14)

Generalization.

Exercise 3.2.2 Let (X, h·i) be a Hilbert space, z, v ∈ V such that z − v ⊥ V . Show that dist(z, V ) = kz − vk. (3.15)

3.3 Linear functional and operators in Hilbert spaces

Theorem 3.3.1 (Riesz-Frechet representation theorem) Let X be a Hilbert space. Then for each linear and continuous functional φ : X ∈ C it exists a unique vφ ∈ X such that φ(x) = hx, vφi ∀x ∈ X. CHAPTER 3. HILBERT SPACES 43 3.4 Spectral theory

Theorem 3.4.1

Proposition 3.4.1

Theorem 3.4.2

Corollary 3.4.1

Deﬁnition 3.4.1

Remark 3.4.1

Lemma 3.4.1 44 CHAPTER 4. PARABOLIC EQUATIONS: WEAK FORMULATIONS, NUMERICAL APPROXIMATIONS AND ANALYSIS.45

Chapter 4

Parabolic equations: weak formulations, numerical approximations and analysis.

4.1 Preliminaries (Chapter 5 in [12]).

4.1.1 Sobolev spaces. 4.1.2 Inequalities. 4.1.3 Bochner spaces. 4.1.4 Compactness arguments. Papers: [33]. 4.2 Linear parabolic equations (Chapter 7 in [12] and Chapter 6 in [12]).

4.2.1 Weak formulation (continuous variational fromulation), regularity of solution. 4.2.2 Rothe method, backward-Euler - Galerkin FE approximations/MFEM. 4.2.3 Stability and error estimates (Galerkin FE and MFEM). 4.3 Non-linear parabolic equations (different sources)

4.3.1 Weak solutions, regularity. 4.3.2 Fixed point theorems (Chapter 9 in [12]). 4.3.3 Stability and error estimates (Galerkin FE and MFEM). Papers: [4, 3, 31]. 4.3.4 Linearization methods. 4.4 Degenerate parabolic equations (different sources).

4.4.1 Weak formulation (continuous variational fromulation), regularity of solution. Papers: [2, 30]. 4.4.2 Numerical approximations (Galerkin FE and MFEM). 4.4.3 Stability and error estimates, Kirchhoff’s transformation and Green’s operator. Papers: [20, 21, 27, 30, 25]. 4.4.4 Linearization methods. Papers: [22, 19, 28] 4.5 Coupled elliptic and parabolic equations (different sources).

4.5.1 Reactive transport in porous media [15, 16, 18, 26, 29]. 4.6 Poromechanics (Biot’s model). Papers: [17, 8, 5, 6]. Appendix A

Inequalities

In this section we give a collection of the most used inequalities in analysis of partial differential equations. We start with probably the most known inequality, the mean inequality. It can be proved by induction or by using Jensen’s inequality (see below).

Theorem A.1.1 (Mean inequality) Let n ∈ N and ai ∈ R, i = 1, . . . , n be strictly positive numbers. There holds Pn n n i=1 ai ≤ Πk=1ai ≤ . (A.1) Pn 1 n i=1 ai

Theorem A.1.2 (Cauchy-Schwarz inequality) Let n ∈ N and ai, bi ∈ R, i = 1, . . . , n. There holds

n !2 n ! n ! X X 2 X 2 aibi ≤ ai bi . (A.2) i=1 i=1 i=1

Theorem A.1.3 (Jensen inequality) Let f : R → R. Then i) if f is convex, i.e. f(ax + (1 − a)y) ≤ af(x) + (1 − a)f(y) for all a ∈ [0, 1] and x, y ∈ R, there holds n n X X f( λixi) ≤ λif(xi), (A.3) i=1 i=1 Pn for all λi ∈ [0, 1], i=1 λi = 1 and xi ∈ R, i ∈ {1,...,N}.

46 APPENDIX A. INEQUALITIES 47

ii) if f is concave, i.e. f(ax+(1−a)y) ≥ af(x)+(1−a)f(y) for all a ∈ [0, 1] and x, y ∈ R, there holds

n n X X f( λixi) ≥ λif(xi), (A.4) i=1 i=1 Pn for all λi ∈ [0, 1], i=1 λi = 1 and xi ∈ R, i ∈ {1,...,N}.

Proof. We consider the case f convex, the other being absolutely similar. We prove the result by induction. For n = 2, the inequality (A.4) is just the deﬁnition of convexity. Suppose now the inequality (A.4) holds for n − 1 and prove it for n. Pn Let λi ∈ [0, 1], i=1 λi = 1 and xi ∈ R, i ∈ {1,...,N}. By using the hypothesis of induction we have

n n−2 X X λn−1xn−1 λnxn f( λ x ) = f( λ x + (λ + λ )( + )) i i i i n−1 n λ + λ λ + λ i=1 i=1 n−1 n n−1 n n−2 X λn−1xn−1 λnxn ≤ λ f(x ) + (λ + λ )f( + ) i i n−1 n λ + λ λ + λ i=1 n−1 n n−1 n n−2 X ≤ λif(xi) + λn−1f(xn−1) + λnf(xn). i=1 Q. E. D.

Theorem A.1.4 (Jensen integral inequality) Let f, p :[a, b] → R, with f con- R b tinuous and p ≥ 0, a p(x)dx = 1. If φ : R → R is differentiable and convex, then there holds Z b Z b φ( p(x)f(x)dx) ≤ p(x)φ(f(x))dx. (A.5) a a 1 1 Theorem A.1.5 (Young inequality) Let a, b ∈ and p, q > 0, + = 1. There R p q holds |a|p |b|q |ab| ≤ + . (A.6) p q APPENDIX A. INEQUALITIES 48

1 1 Proof. The function ln is concave, therefore there holds for p, q > 0, + = 1 p q the Jensen inequality

ap bq 1 1 ln + ≥ ln ap + ln bq, p q p q and the result (A.6) follows straightforward. Q. E. D. A very used variant of Young’s inequality is given by

Corollary A.1.1 Let a, b ∈ R. Then for any > 0 there holds |a|2 |b|2 |ab| ≤ + . (A.7) 2 2

Corollary A.1.2 (Generalized Young inequality) Let λi ∈ [0, 1], i ∈ Pn 1 {1,...,N} with i=1 = 1 and ai, i ∈ {1, . . . , n} be positive numbers. There λi holds n n Y X aλi a ≤ i . (A.8) i λ i=1 i=1 i Proof. It follows immediately by applying the Jensen inequality for the ln function. Q. E. D. 1 1 Theorem A.1.6 (Holder¨ inequality) Let again p, q > 0, + = 1 and x , . . . x p q 1 n be real numbers. Then there holds

n n !1/p n !1/q X X p X q |xiyi| ≤ |xi| |yi| . (A.9) i=1 i=1 i=1

Proof. We will use Young’s inequality. There holds for any i ∈ {1, . . . , n}

|x | |y | 1 |x |p 1 |y |q i i ≤ i + i . n 1/p n 1/q Pn p Pn q P p P q p |xi| q |yi| ( i=1 |xi| ) ( i=1 |yi| ) i=1 i=1 Summing up the above result from i = 1 to n gives the Holder¨ inequality. Q. E. D. A combination between Young’s inequality and Holder¨ inequality would read as APPENDIX A. INEQUALITIES 49

1 1 Corollary A.1.3 Let a, b ∈ , > 0 and p, q > 1 be s.t. + = 1. Then, R p q

p q |a| − q |b| |ab| ≤ + p . (A.10) p q

1 1 Theorem A.1.7 (Holder¨ integral inequality) Let again p, q > 0, + = 1 and p q Ω be a domain in Rd, f, g :Ω → R with |f|p and |g|q integrable two functions on Ω. Then we have: Z Z 1/p Z 1/q |fg|dx ≤ |f|pdx |g|qdx . (A.11) Ω Ω Ω For p = q = 2, the inequality above becomes

Z Z 1/2 Z 1/2 |fg|dx ≤ |f|2dx |g|2dx , (A.12) Ω Ω Ω and it is called the Cauchy-Schwarz integral inequality.

Proof. The proof is based on Young’s inequality. We have

|f| |g| 1 |f|p 1 |g|q ≤ R + R . R p 1/p R q 1/q p |f|pdx q |g|qdx Ω |f| dx Ω |g| dx Ω Ω By intergrating now the above over Ω we get the result. Q. E. D.

Theorem A.1.8 (Generalized Holder¨ inequality) Let pi > 0, i ∈ {1, . . . , n}, 1 Pn d pi with i=1 = 1 and Ω be a domain in R . Further, let fi :Ω → R with |fi| pi be integrable functions on Ω. Then we have:

Z Z 1/pi n n p Πi=1|fi|dx ≤ Πi=1 |fi| dx . (A.13) Ω Ω Proof. Induction. Q. E. D. APPENDIX A. INEQUALITIES 50

Theorem A.1.9 (Minkowski integral inequality) Let now p ≥ 1, Ω be a domain in Rd and f, g :Ω → R with |f|p, |g|p integrable, two functions on Ω. Then we have: Z 1/p Z 1/p Z 1/p |f + g|pdx ≤ |f|pdx + |g|pdx . (A.14) Ω Ω Ω Proof. The proof is based on Holder’s¨ inequality. For p = 1 the inequality is just the triangle inequality, so we can assume that p > 1. The conjugate number p of p > 1 is . There holds p − 1 Z Z |f + g|pdx = |f + g||f + g|p−1dx Ω Ω Z ≤ (|f| + |g|)|f + g|p−1dx Ω Z Z ≤ |f||f + g|p−1dx + |g||f + g|p−1dx Ω Ω Z Z Z Z ≤ ( |f|pdx)1/p( |f + g|p)dx)(p−1)/p + ( |g|pdx)1/p( |f + g|pdx)(p−1)/p. Ω Ω Ω Ω The Minkowski integral inequality is then an immediate consequence of the above result. Q. E. D. There is also a Minkowski inequality for real numbers. Its proof is very similar with the proof of Minkowski’s intergral inquality, i.e. based on the triangle and Holder¨ inequlities.

Theorem A.1.10 (Minkovski inequality) Let again p ≥ 1 and x1, . . . xn real numbers. Then there holds n !1/p n !1/p n !1/p X p X p X p |xi + yi| ≤ |xi| + |yi| . (A.15) i=1 i=1 i=1 The Gronwall inequalities are very usefull for proving of uniqueness for differential or partial differential equations. They are also very useful for converence analysis of numerical methods for PDEs (for obtaining error estimates). Theorem A.1.11 (Gronwall integral inequality) Let b(·) : [0,T ] → R be a non- decreasing function and let h(·) : [0,T ] → R be a positive function. Let now f(·) be a function on [0,T ] which satisﬁes a.e. t ∈ [0,T ] the integral inequality Z t f(t) ≤ b(t) + h(s)f(s)ds. (A.16) 0 APPENDIX A. INEQUALITIES 51

Then there holds R t h(s)ds f(t) ≤ b(t)e 0 . (A.17) for a.e. t ∈ [0,T ].

− R t h(s)ds Proof. From (A.22), by multiplying with h(t)e 0 , h ≥ 0 we get Z t − R t h(s)ds − R t h(s)ds − R t h(s)ds h(t)e 0 f(t) − h(t)e 0 h(s)f(s)ds ≤ h(t)e 0 b(t) 0 Z t 0 − R t h(s)ds − R t h(s)ds ⇒ h(s)f(s)dse 0 ≤ b(t)h(t)e 0 0 Z t Z t − R t h(s)ds − R s h(u)du ⇒ h(s)f(s)dse 0 ≤ b(s)h(s)e 0 ds 0 0 Z t Z t R t h(u)du ⇒ h(s)f(s)ds ≤ b(s)h(s)e s ds. 0 0 (A.18) Using (A.18) in (A.22) we get Z t R t h(u)du f(t) ≤ b(t) + b(s)h(s)e s ds. 0 By using b(s) ≤ b(t), the above becomes Z t R t h(u)du f(t) ≤ b(t) 1 + h(s)e s ds 0 Z t 0 R t h(u)du ≤ b(t) 1 + −e s ds 0 R t h(u)du ≤ b(t) 1 − 1 + e 0 . (A.19) Q. E. D. Corollary A.1.4 Let f(·) a positive function on [0,T ] which satisﬁes a.e. t ∈ [0,T ] the integral inequality Z t f(t) ≤ C1 f(s)ds + C2 (A.20) 0 for constants C1,C2 ≥ 0. Then there holds

C1t f(t) ≤ C2e (A.21) for a.e. t ∈ [0,T ]. APPENDIX A. INEQUALITIES 52

A more general variant of the continuous Gronwall lemma is the following.

Theorem A.1.12 Let F1,F2 : [0,T ] → R be non-negative functions, γ ∈ (0, 1) and C ≥ 0. Let now f(·) be a function on [0,T ] which satisﬁes a.e. t ∈ [0,T ] the integral inequality

Z t Z t 1−γ f(t) ≤ C + F1(s)f(s) ds + F2(s)f(s)ds. (A.22) 0 0 Then there holds

t 1/γ Z R t γ F2(s)ds f(t) ≤ C + γ F1(s)ds e 0 . (A.23) 0 for a.e. t ∈ [0,T ].

Lemma A.1.1 (Discrete Gronwall Lemma) Let an, λn positive numbers for all integers n ∈ N and B ≥ 0. If λn < 1 ∀ n ∈ N and

n X an ≤ B + λkak ∀ n ∈ N, (A.24) k=0 then there holds n X λk 1 − λk k=0 an ≤ Be ∀ n ∈ N. (A.25)

For proving the discrete Gronwall lemma we need an auxiliary result, whose proof is immediately by induction.

Lemma A.1.2 Let λn ∈ [0, 1) for all integers n ∈ N. Then we have

λ0 λ0 λn λ0 λn 1−λ 1−λ +...+ 1−λ 1−λ +...+ 1−λ 1 + λ0e 0 + ... + λne 0 n ≤ e 0 n . (A.26)

We can now give a proof for the discrete Gronwall lemma. (A.24) Proof. We do this by induction. For n = 0 we have a0 ≤ B + λ0a0 ⇒ a0 ≤ B/(1 − λ0) and because

λ λ 1 e 1−λ ≥ + 1 = , (A.27) 1 − λ 1 − λ APPENDIX A. INEQUALITIES 53 for λ ∈ [0, 1) (we used ex ≥ x + 1 ∀ x ≥ 0), we get (A.25). We assume now that (A.25) holds for n ≤ k n X λi 1 − λ i=0 i an ≤ Be , (A.28) and we prove it for n = k + 1. We have

k (A.24) X λi B a ≤ a + k+1 1 − λ i 1 − λ i=0 k+1 k+1

i X λj k (A.28) X λi 1 − λj B ≤ e j=0 B + 1 − λ 1 − λ i=0 k+1 k+1

k X λi (A.26) B 1 − λ ≤ e i=0 i 1 − λk+1

k+1 X λi (A.27) 1 − λ ≤ Be i=0 i ⇒ (A.25).

Q. E. D. Another variant of Gronwall discrete lemma is (see [7]).

Lemma A.1.3 (Discrete Gronwall Lemma 2) Let {xn}n, {an}n, {bn}n, {cn}n be sequences of non-negative real numbers and let the sequence {an}n be non- decreasing. If there holds

x0 + c0 ≤ a0 n−1 X xn + cn ≤ an + bkxk k=0 for any k ≥ 1, then we have for any n ≥ 1.

n−1 xn + cn ≤ anΠk=0(1 + bk). (A.29) APPENDIX A. INEQUALITIES 54

The proof of the above variant of discrete Gronwall lemma is immediately by induction. We give now also a generalization of the triangle integral inequality for the case of functions valued in a Banach space X, namely the Bochner inequality (for the proof see Yoshida [34], Chapter V).

Proposition A.1.1 (Bochner inequality) Let f :[t1, t2] → X an integrable function and X a Banach space. There holds

Z t2 Z t2

f(t) dt ≤ kfkX dt. (A.30) t1 X t1

0 Lemma A.1.4 Assuming that b ≥ > 0 (which particulary implies that b(·) is monotone increasing) there holds

n X (b (xj) − b (xj−1)) xj ≥ −C|x0|2 + |xn|2, (A.31) 2 j=1 for any real sequence xj, j ∈ {1, . . . , n}.

0 Proof. Since b ≥ , one has Z x Z x 0 0 2 ((b(x) − b(y))x ≥ sb ds and sb(s) ds ≥ x , y 0 2 for any reals x and y. Furthermore,

n n Z xj X j j−1 j X 0 (b(x ) − b(x ))x ≥ sb(s) ds j−1 j=1 j=1 x

Z xn Z x0 0 0 0 2 n 2 = sb(s) ds − sb(s) ds ≥ −C|x | + |x | , 0 0 2 where the constant C is half of the Lipschitz constant of b(·). Q. E. D. The following lemma is very useful for the proof of convergence for numerical schemes for nonlinear parabolic partial differential equations (when both the term with the time derivative and the second order term in space are nonlinear, as e.g. in the case of the Richards equation). The lemma which was ﬁrst ﬁrst presented in [3]. The proof, which is elementary, can be found also in [4], p. 1685. APPENDIX A. INEQUALITIES 55

Lemma A.1.5 Let un, vn be real numbers, n ∈ {1,...N}. Suppose that |un − n−1 0 00 u | ≤ Cuτ and Θ: R → R is such that 0 ≤ Θ ≤ CΘ0 < ∞ and |Θ | ≤ CΘ00 . Then Θ(un) − Θ(vn) − Θ(un−1) + Θ(vn−1) (un − vn) τ n n−1 R u Θ(µ) − Θ(vn)dµ − R u Θ(µ) − Θ(vn−1)dµ = vn vn−1 − E, τ where E ≤ C{(un − vn)2 + (un−1 − vn−1)2 + τ 2}, for some C depending on Cu,CΘ0 and CΘ00 . Appendix B

Matrix norm

Proposition B.1.2 Let (E, k · kE), (V, k · kV ) be two normed spaces and k · k a norm on L(E,V). We say that k · k is compatible with k · kE, k · kV if there holds

kT xkV ≤ kT kkxkE ∀T ∈ L(E, V ). (B.1)

Deﬁnition B.1.1 (Matrix norm) A matrix norm is a function k k : Rn,n → R,A → kAk with the following properties i) kλAk = |λ|kAk ∀λ ∈ R,A ∈ Rn,n. ii) kA + Bk ≤ kAk + kBk ∀A, B ∈ Rn,n. iii) kAk = 0 ⇒ A = 0n. iv) kABk ≤ kAkkBk ∀A, B ∈ Rn,n (submultiplicity).

It follows from i)-iii) that also kAk ≥ 0, ∀A ∈ Rn,n. Let us remember, that kT xkV on L(E,V) we deﬁned the operator norm kT k := supx∈E,x6=0 . It is well kxkE known, that L(Rn , Rm ) = Rm,n, and of course L(Rn , Rn ) = Rn,n. Moreover, each matrix A ∈ Rn,n can be identiﬁed with a operator T ∈ L(Rn , Rn ). It is now natural to ask: is the operator norm also a matrix norm?

Proposition B.1.3 Let (X, k kX ), (Y, k kY )and(Z, k kZ ) be three normed spaces and T ∈ L(X , Y ), S ∈ L(Y , Z ) two linear and bounded operators. Then there holds kS ◦ T kL(X ,Z ) ≤ kSkL(Y ,Z )kT kL(X ,Y ) (B.2)

56 APPENDIX B. APPENDIX 57

Proof. Let x ∈ X. There holds

kS ◦ T xkZ ≤ kSkL(Y ,Z )kT xkY ≤ kSkL(Y ,Z )kT kL(X ,Y )kxkX (B.3) It follows

kS ◦ T xkZ sup ≤ kSkL(Y ,Z )kT kL(X ,Y ) ⇒ (B.2). x∈X,x6=0 kxkX Q. E. D.

Corollary B.1.5 The operator norm is submultiplicative, and therefore it is a matrix norm.

Using the above results one can deﬁne some matrix norms:

kAxk1 kAk1 = sup (column sum norm) (B.4) x∈Rn,x6=0 kxk1

kAxk∞ kAk∞ = sup (row sum norm) (B.5) x∈Rn,x6=0 kxk∞

kAxk2 kAk2 = sup (euclidean norm) (B.6) x∈Rn,x6=0 kxk2

kAxkp kAkp = sup , p ≥ 1. (B.7) x∈Rn,x6=0 kxkp Deﬁnition B.1.2 When the matrix norm is deﬁned by a vector norm (as above), we say that the matrix norm is induced by the vector norm.

Proposition B.1.4 The following afﬁrmations hold true

n X kAk1 = max |aij| (column sum norm) (B.8) j=1,...,n i=1 n X kAk∞ = max |aij| (row sum norm) (B.9) i=1,...,n j=1 p kAk2 ≤ kAk1kAk∞ (B.10)

kABkp ≤ kAkpkBkp. (B.11) APPENDIX B. APPENDIX 58

Not all matrix norms are induced by a vector norm. For example the Frobenius norm, deﬁned by n !1/2 X 2 kAkF := aij (B.12) i,j=1 n is not induced by any vector norm on R for n ≥ 2. Indeed, let√ us consider the identity matrix In. The Frobenius norm of In is kInkF = n. If the Frobenius norm would be induced by a vector norm kk, then we would have kInxk kI k = sup n = 1, which is true only for n = 1. n F x∈R ,x6=0 kxk We introduce now the spectral radius of a matrix ρ(A) := max |λ| (B.13) λ eigenvalue of A and the spectrum of A

σ(A) := {λ ∈ C|λ eigenvalue of A} (B.14) 0 1 Remark B.1.1 The spectral radius is not a norm. Let e.g. A = , we 0 0 have ρ(A) = 0 but obviously A 6= 02.

Lemma B.1.6 Let (E, k · kE), (V, k · kV ) be two normed spaces with E ﬁnite dimensional. Then, for the operator norm there holds

kT k = max kT xkV . (B.15) x∈E,kxkE =1

Therefore it exists x ∈ E with kxkE = 1 such that kT k = kT xkV .

Proof. Let B(0,1) := {x ∈ EkxkE = 1} be the unit ball in E. B(0,1) is bounded and closed, E ﬁnite dimensional therefore B(0,1) is compact, which implies sup = max . x∈B x∈B(0,1) (0,1) Q. E. D.

Proposition B.1.5 Let A ∈ Rn,n. Then there holds

p T kAk2 = ρ(A A). (B.16) Proof. Q. E. D. APPENDIX B. APPENDIX 59

Remark B.1.2 One can deﬁne more generally, for A ∈ Rn,m the kAk as a operator norm from Rm to Rn. The submultiplicativity condition makes but not much sense in this case (you can not multiply matrices in Rn,m, if m 6= n). The proposition above remains valid for A ∈ Rn,m.

The next proposition is very important when studying the convergence of iterative solvers for linear systems.

Proposition B.1.6 Let A ∈ Rn,n. Then

ρ(A) ≤ kAk (B.17) for any matrix norm. Moreover, for all > 0 it exists a matrix norm induced by a vector norm which satisﬁes

kAk ≤ ρ(A) + . (B.18)

Proof. We prove ﬁrst (B.17). Let λ ∈ C be the biggest eigenvalue of A and v ∈ Cn a eigenvector associated to λ. Let B ∈ Cn,n be the matrix deﬁned by B := vvT . There holds

kABk = kAvvT k = k(Av)vT k = kλvvT k = |λ|kBk. (B.19)

By using the submultiplicativity of the matrix norm (which we actually deﬁned for Rn,n but the deﬁnition can be extended to Cn,n without problems), i.e. kABk ≤ kAkkBk we obtain from (B.19)

|λ|kBk = kABk ≤ kAkkBk ⇒

|λ| ≤ kAk ⇒ ρ(A) ≤ kAk. For the second inequality (B.18) we refer to Skript Num I + II, Hoffmann University Hamburg. Q. E. D.

Deﬁnition B.1.3 We say that a matrix norm k·k is compatible with a vector norm n n,n k · kRn if there holds for all x ∈ R and A ∈ R

kAxkRn ≤ kAkkxkRn . (B.20) APPENDIX B. APPENDIX 60

B.1.1 Exercises

Exercise B.1.1 Die p-Normen auf dem Rn sind deﬁniert durch

n !1/p X p kxkp := |xi| , p ∈ [1, ∞) und kxk∞ := max |xi|. i=1,...,n i=1 Zeigen Sie fur¨ die zugeordneten Matrixnormen

kAxkp m,n kAkp := sup ,A ∈ R x6=0 kxkp die folgenden Aussagen:

m n X X m,n (a) kAk1 = max |aij|, kAk∞ = max |aij|,A ∈ . j=1,...,n i=1,...,m R i=1 j=1

p m,n (b) kAk2 ≤ kAk1kAk∞,A ∈ R .

n,n (c) kABkp ≤ kAkpkBkp fur¨ 1 ≤ p ≤ ∞, A, B ∈ R . Bibliography

[1]R.A.A DAMS, Sobolev Spaces, Academic Press, New York, 1975.

[2] H. W. Alt and S. Luckhaus, Quasilinear elliptic-parabolic differential equations, Math. Z., 183 (1983), pp. 311-341.

[3] T. Arbogast, An error analysis for Galerkin approximations to an equation of mixed elliptic-parabolic type, Technical Report TR90-33, Department of Computational and Applied Mathematics, Rice University, Houston, TX, 1990.

[4] T. Arbogast, M. Obeyesekere and M. F. Wheeler, Numerical methods for the simulation of ﬂow in root-soil systems, SIAM J. Num. Anal. 30 (1993), pp. 1677-1702.

[5] M. Bause, F. A. Radu and U. Koecher, Space-time ﬁnite element approximation of the Biot poroelasticity system with iterative coupling., arXiv:1611.06335, 2016.

[6] M. Borregales, F.A. Radu, K. Kumar and J.M. Nordbotten, Robust iterative schemes for non-linear poromechanics, arXiv:1702.00328, 2017.

[7] M. Balazsova and M. Feistauer, On the stability of the ALE space-time discontinuous Galerkin method for nonlinear convection-diffusion problems in time- dependent domains. Appl. Math. 60 (2015), no. 5, pp. 501–526.

[8] J. Both, M. Borregales, F.A. Radu, K. Kumar and J.M. Nordbotten, Robust ﬁxed stress splitting for Biot’s equations in heterogeneous media, Applied Mathematics Letters 68 (2017), pp. 101-108.

[9] F. Brezzi and M. Fortin, Mixed and Hybrid Finite Element Methods, Springer Verlag, New York, 1991.

61 BIBLIOGRAPHY 62

[10] Z. Chen, Finite Element Methods and Their Applications, Springer Verlag, 2005.

[11] W. CHENEY, Analysis for Applied Mathematics, Springer Verlag, 2001.

[12] L. C. Evans, Partial differential equations, American mathematical society, 2010 (ISBN 978-0-8218-4974-3).

[13] P. Knabner and L. Angermann, Numerical methods for elliptic and parabolic partial differential eqations, Springer-Verlag, 2003 (ISBN 0-387-95449-X).

[14] P. KNABNER, Numerik I, Lecture Notes, University of Erlangen-Nuremberg, Germany, 2002.

[15] K. Kumar, I.S. Pop and F.A. Radu, Convergence analysis of mixed numerical schemes for reactive ﬂow in a porous medium, SIAM J. Num. Anal. 51 (2013), pp. 2283-2308.

[16] K. Kumar, I.S. Pop and F.A. Radu, Convergence analysis for a conformal discretization of a model for precipitation and dissolution in porous media, Numerische Mathematik 127 (2014), pp. 715-749.

[17] A. Mikelic and M.F. Wheeler, Convergence of iterative coupling for coupled ﬂow and geomechanics, Comput. Geosci. 17 (3) (2013), pp. 455-461.

[18] J.L. Musuuza, F.A. Radu and S. Attinger, The stability of density-driven ﬂows in saturated heterogeneous porous media, Advances in Water Resources 34 (2011), pp. 1464-1482.

[19] F. List and F.A. Radu, A study on iterative methods for Richards’ equation, Computational Geosciences 20 (2016), pp. 341-353.

[20] R. H. Nochetto and C. Verdi, Approximation of degenerate parabolic problems using numerical integration, SIAM J. Numer. Anal. 25 (1988), pp. 784– 814.

[21] I. S. Pop, Error estimates for a time discretization method for the Richards’ equation, Computational Geosciences 6 (2002), pp. 141-160.

[22] I. S. Pop, F.A. Radu and P. Knabner, Mixed ﬁnite elements for the Richards’ equation: linearization procedure, J. Comput. and Appl. Math. 168 (2004), pp. 365-373. BIBLIOGRAPHY 63

[23]A.Q UARTERONI AND A.VALLI, Numerical approximations of partial differential equations, Springer-Verlag, 1994.

[24]A.Q UARTERONI,R.SACCOAND F. SALERI, Numerical mathematics, Springer-Verlag, New York, 2000. [25] F. A. Radu, Mixed finite element discretization of Richards’ equation: error analysis and application to realistic infiltration problems, PhD Thesis, Univer- sity of Erlangen, Germany (2004). [26] F. A. Radu, Convergent mass conservative schemes for reactive solute transport and flow in variably saturated porous media, Habilitation Thesis, Univer- sity of Erlangen, Germany (2013). [27] F. A. Radu, I. S. Pop and P. Knabner, Order of convergence estimates for an Euler implicit, mixed finite element discretization of Richards’ equation, SIAM J. Numer. Anal. 42 (2004), pp. 1452-1478. [28] F.A. Radu, I.S. Pop and P. Knabner, On the convergence of the Newton method for the mixed finite element discretization of a class of degenerate parabolic equation, In Numerical Mathematics and Advanced Applications. A. Bermudez de Castro et al. (editors), Springer, 1194-1200, 2006. [29] F. A. Radu, I. S. Pop and S. Attinger, Analysis of an Euler implicit - mixed finite element scheme for reactive solute transport in porous media, Numer. Meth. PDE’s 26 (2009), pp. 320-344. [30] F. A. Radu, I. S. Pop and P. Knabner, Error estimates for a mixed finite element discretization of some degenerate parabolic equations, Numer. Math. 109 (2008), pp. 285-311. [31] F. A. Radu and W. Wang, Convergence analysis for a mixed finite element scheme for flow in strictly unsaturated porous media. Nonlinear Analysis Serie B: Real World Applications 15, 2014, pp.266-275. [32] B. Rynne and M. Youngson, Linear functional analysis, Springer-Verlag, 2000. [33] J. Simon, Compact sets in the space Lp(0,T, ; B), Annali di matematica pura et aplicata (1987), pp. 65-96.

[34]Y OSHIDA, Functional Analysis.