Analysis Lecture Notes PDF

Analysis Lecture notes for the Master degree in Stochastics and Data Sciences Academic Year 2024-2025 Bertrand Lods Comments: These lecture notes were originally written in for the Analysis, Course B, lectures of the academic year 2015-2016 of the Master Degree in Stochastics and Data Sciences of Turin University. Some editing and corrections have been provided for the current Academic Year 2024-2025. The notes cover the course in detail, and are therefore sufficient for the preparation of the exam. They also include material not covered during the course (Complements sections). The proofs that students should prepare and “know” for the exams are indicated with the symbol - in the left margin. These lectures notes have been written with the support from the project: “Start up della Laurea Magistrale in Stochastics and Data Science: pubblicizzazione, internazional- izzazione ed eccellenza interdisciplinare” funded by fondazione CRT. Copyright © 2019 B. Lods Licensed under the Creative Commons Attribution-NonCommercial 3.0 Unported License (the “License”). You may not use this file except in compliance with the License. You may obtain a copy of the License at http://creativecommons.org/licenses/by-nc/3.0. Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an “AS IS ” BASIS , WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND , either express or implied. See the License for the specific language governing permissions and limitations under the License. First printing, December 2017 Contents 1 Normed spaces................................................ 5 1.1 The norm and its properties 6 1.1.1 General properties................................................. 6 1.1.2 Topology of normed spaces – convergence of sequences..................... 7 1.1.3 Equivalent norms – Product spaces................................... 12 1.1.4 Continuity of functions............................................. 13 1.2 Fundamental example: Lebesgue spaces 14 1.2.1 The definition of L1 as a set of equivalence classes........................ 14 1.2.2 Lp -spaces, 1 6 p 6 ∞.............................................. 16 1.3 Another example: the space of linear applications 22 1.4 Compactness 24 1.4.1 Compactness and continuous functions................................. 26 1.5 Finite Dimensional Spaces 27 1.6 Problems 30 2 Banach spaces............................................... 33 2.1 Cauchy sequences-Banach spaces 33 2.2 Examples revisited 37 2.3 Simple consequences of completeness – Banach-Cacciopoli-Picard Theorem 47 2.4 Fundamental properties of Banach spaces 51 2.5 Problems 58 2.6 Complements 59 2.6.1 Compact operators in Banach spaces.................................. 59 2.6.2 More properties of the dual space/Hahn-Banach Theorem................... 60 3 Inner product spaces and Hilbert spaces....................... 63 3.1 Definitions and Elementary Properties. 63 3.1.1 General properties................................................ 63 3.1.2 Orthogonality.................................................... 65 3.2 Hilbert spaces and Projection Theorem 68 3.3 The dual space of a Hilbert space 75 3.4 Hilbert bases 76 3.5 Introduction to the weak topology 80 3.6 Problems 82 4 Complements about Lebesgue spaces......................... 85 4.1 Useful inequalities 85 4.1.1 Jensen’s inequality................................................ 85 4.1.2 Minkowski’s integral inequality........................................ 88 4.2 Integral depending on a parameter 91 4.3 Convolution 93 4.3.1 Application to Probability........................................... 96 4.3.2 Extension and consequences of Young’s convolution inequality................ 97 4.4 Complements : Compact subsets of Lp (RN ) 102 5 Fourier analysis............................................. 105 5.1 Fourier transform in L1 (Rd ) 106 5.1.1 Main properties – Riemann-Lebesgue Theorem.......................... 106 5.1.2 Example of computations.......................................... 113 5.1.3 Inversion formula................................................ 114 5.2 Fourier-Plancherel transform in L2 (Rd ) 119 5.2.1 An alternative Hilbertian construction of Fourier-Plancherel transform.......... 125 5.3 Fourier transform of measure 131 5.3.1 Basic definition................................................. 131 5.3.2 Application to probability theory..................................... 132 5.3.3 Central Limit Theorem............................................ 136 5.4 Laplace transform 138 5.4.1 Differentiation and integration....................................... 141 5.5 Complements 143 5.5.1 Vector differentiation and Sobolev spaces.............................. 143 5.5.2 Schwarz Class and tempered distributions.............................. 148 Bibliography................................................ 150 1. Normed spaces In all the notes, the vector spaces we are considering are R-spaces but everything extends in some straightforward way to vector spaces built over the complex field C. We recall more generally that, given a field K = R or K = C, a K-vector space X is defined as a set which is closed under finite vector addition and scalar multiplication by elements of K. In a more formal way, X is a K-vector space if there exist two operations, addition + and scalar multiplication, for which the following conditions hold for any elements x, y, z ∈ X and any scalars α, β ∈ K: 1. (Commutativity) x+y =y+x 2. (Associativity of vector addition) (x + y) + z = x + (y + z). 3. There is an element 0X ∈ X such that 0X + x = x + 0X = x ∀x ∈ X 4. For any x ∈ X, there is −x ∈ X such that x + (−x) = 0X. 5. (Associativity of scalar multiplication) α(βx) = (α β)x. 6. (Distributivity of scalar sums) (α + β)x = α x + βx. 7. (Distributivity of vector sums) α(x + y) = α x + αy. 6 Chapter 1. Normed spaces 8. (Scalar multiplication identity) 1 x = x. It is important to understand that, in the above point 6, the symbol + in the left-hand side stands for the “classical” addition in the field K whereas the same symbol + on the right-hand-side stands for the (generally more involved) addition on the space X. In the same way, the mulplication in K is denoted by α β and there is no possible confusion with the scalar multiplication α x since, in the first case, both α, β lie in K whereas, in the second case, we multiply a “vector” x ∈ X by a scalar α ∈ K and the result is a vector α x ∈ X. 1.1 The norm and its properties 1.1.1 General properties We start with the definition of seminorms and norms Definition 1.1.1 A seminorm on X is a function N : X → R such that, for any vectors x, y ∈ X and any α ∈ R, the following properties hold: 1. Nonnegativity: N (x) > 0, 2. Homogeneity: N (αx) = |α|N (x), 3. Triangle Inequality: N (x + y) 6 N (x) + N (y). A seminorm is then a norm if it also holds: 1. Uniqueness: N (x) = 0 if and only if x = 0. If N is a norm, we rather use the notation kxk = N (x). A vector space X endowed with a norm k · k is called a normed space and denoted by (X, k · k). The absolute value function |x| (x ∈ R) is a norm on the real line R. In the Example 1.1 same way, the Euclidean norm v u N uX kxk = t x2i , x = (x1 ,... , xN ) ∈ RN i=1 is a norm on RN (N > 1). Example 1.2 Given a sequence of real x = (xn )n∈N = (x1 , x2 ,...), we introduce the space of all absolutely summable sequences, ( ∞ ) X 1 X = ` (N) = x = (xn )n∈N ⊂ R such that kxk1 := |xn | < ∞. n=1 This set is a vector space and the mapping k · k1 : X → R+ defined as X k x k1 = |xn | ∀x = (xn )n ⊂ `1 (N) (1.1) n>1 is a norm on X. Proposition 1.1.1 Let I = [a, b] ⊂ R be a given interval. We introduce the space C (I) 1.1 The norm and its properties 7 as the set of all continuous functions f : I → R. Let Z b kf k1 = |f (t)|dt ∀f ∈ C (I) a and kf k∞ = sup |f (t)|, f ∈ C (I). t∈I Then, (C (I), k · k1 ) and (C (I), k · k∞ ) are two normed spaces. Proof. The proof that k · k∞ is a norm on C (I) is an easy task. To prove that k · k1 is also a norm on C (I), one first has to notice that kf k1 < ∞ for any f ∈ C (I) which is a classical result of integral calculus (the Riemann integral of a continuous function on some finite interval is well-defined and finite). Then, it is very easy to prove that k · k1 is a seminorm on C (I). To prove it is actually a norm, one has to show that, given f ∈ C (I), Z b |f (t)|dt = 0 =⇒ f (t) = 0 ∀t ∈ [a, b]. a This comes from the fact that the function t ∈ [a, b] 7→ |f (t)| is a continuous and nonnegative function and that the integral of a continuous and nonnegative result is zero if and only if the function is vanishing everywhere (Check this arguing by contradiction!). We can also introduce derivatives to build function spaces: Proposition 1.1.2 Let I = [a, b] ⊂ R be a given interval. We introduce the space C (I) as the space of all continuously differentiable functions f : I → R (i.e. f is 1 differentiable over I and its derivative f 0 is continuous over I). Let Z b Z b kf k1,1 = |f (t)|dt + |f 0 (t)|dt ∀f ∈ C (I) a a and kf k1,∞ = sup (|f (t)| + |f 0 (t)|) f ∈ C 1 (I). t∈I Then, (C 1 (I), k · k1,1 ) and (C 1 (I), k · k1,∞ ) are normed spaces Proof. The proof is left as an Exercise. Remark 1.1.1 Notice that, if one simply defines N (f ) = sup |f 0 (t)| f ∈ C 1 (I) t∈I one sees that N is a semi-norm on C 1 (I) but not a norm ! Indeed, any constant function f ≡ c is such that N (f ) = 0 so the uniqueness property is not satisfied. 1.1.2 Topology of normed spaces – convergence of sequences Let recast here the standard definitions of topological spaces in the special case of normed spaces: 8 Chapter 1. Normed spaces Definition 1.1.2 Let (X, k · k) be a normed space. For any x0 ∈ X and r > 0, the open ball centered at x0 with radius r > 0 is defined as B(x0 , r) = {x ∈ X ; kx − x0 k < r} while the closed ball centered at x0 with radius r > 0 is Bc (x0 , r) = {x ∈ X ; kx − x0 k 6 r}. A subset U ⊂ X is an open set if, for any x ∈ U there exists r > 0 such that B(x, r) ⊂ U. A subset U ⊂ X is closed if its complementary U c = X \ U is an open subset of X. One checks that ∅ and X are both open and closed subsets of (X, k · k). Example 1.3 Given x ∈ X and % > 0, the ball B(x, %) is an open set of (X, k · k) whereas Bc (x, %) is a closed set. (Check this!) Notice that there exist subsets of X which are neither open nor closed. Open and closed sets enjoy important properties whose proof is left as an Exer- cise: Proposition 1.1.3 A finite intersection of open sets of (X, k·k) is an open set of (X, k·k). Any union of open sets of (X, k · k) is an open set of (X, k · k). A finite union of closed subsets of (X, k · k) is a closed set of (X, k · k). Any intersection of closed subsets of (X, k · k) is a closed subset of (X, k · k). We recall several examples in R Example 1.4 Let X = R endowed with the natural norm | · |. Let a < b be given. One checks easily that 1. The sets (a, b), (b, +∞) and (−∞, a) are open subsets of (R, | · |); 2. the sets [a, b], [b, +∞) and (−∞, a] and {a} are closed sets of (R, | · |); 3. the sets [a, b) and (a, b] are neither open nor closed subsets of (R, | · |). Moreover, one checks that \ 1 1 [0, 1] = − ; 1+ , n>1 n n which illustrates the fact that an arbitrary (infinite) intersections of open subsets is not necessarily an open subset. In the same way, since [ 1 1 (0, 1) = ; 1− n>1 n n one sees that an arbitrary union of closed subsets is not necessarily a closed subset. We will assume the concept of sequence to be familiar. We just recall that a sequence of elements of X is a function f : N → X, i.e. f (n) ∈ X for any n ∈ N but a more useful way to see a sequence is to regard it as an ordered list of elements of X which can be written as (xn )n ⊂ X (here of course, xn = f (n), n ∈ N). A subsequence of (xn )n is then a sequence of the form (xnk )k where (nk )k ⊂ N is a strictly increasing sequence of N 1. 1 Equivalently, a subsequence of (xn )n is a sequence of the form (xϕ(n) )n where ϕ : N → N is a strictly increasing function. 1.1 The norm and its properties 9 We introduce now the notion of convergence with respect to the norm Definition 1.1.3 — Convergent sequence. Let (X, k · k) be a normed space. A sequence (xn )n ⊂ X of elements in X is said to converge to x ∈ X if lim kxn − xk = 0 n→∞ i.e., if for any ε > 0, there exists N = N (ε) ∈ N such that kxn − xk < ε for any n > N. We write then lim xn = x or xn → x. n The point x is then called the limit of (xn )n in (X, k · k). One has the following whose proof is left as an Exercise: Theorem 1.1.4 Let (X, k · k) be a normed space and let (xn )n be a convergent sequence in X. Then: 1. the limit x = limn xn is unique; 2. any subsequence of (xn )n also converges to x. 3. if (xn )n ⊂ X is a convergent sequence with limn xn = x then lim kxn k = kxk, n in particular sup kxn k < ∞. n 4. Continuity of vector addition: if (xn )n , (yn )n ⊂ X are converging sequence of X and if (αn )n ⊂ R is a convergent sequence of real numbers with lim xn = x ∈ X, lim yn = y ∈ X lim αn = α ∈ R n n n then (xn + αn yn )n ⊂ X is a convergent sequence with lim(xn + αn yn ) = x + αy. n Proof. Notice that the proof of point (iii) comes from the Reverse Triangle Inequality: |kxk − kyk| 6 kx − yk ∀x, y ∈ X. The proof is left as an Exercise. The use of sequences allows to have a more tractable characterization of closed sets Proposition 1.1.5A subset C ⊂ X is a closed subset of (X, k · k) whenever the limit of any convergent sequence (xn )n ⊂ C belongs to C. In other words, C is closed if and only if once a sequence of elements of C is converging, the limit cannot escape C. This explains the terminology... 10 Chapter 1. Normed spaces - Proof. Assume first that C is closed and let (xn )n ⊂ C be a convergent sequence with x = limn xn. Let us show that x ∈ C. We argue by contradiction assuming that x ∈ / C. Since X \ C is open, there exists r > 0 such that B(x, r) ⊂ X \ C. By definition of convergence, there exists N ∈ N such that kxn − xk < r for any n > N. In other words, xn ∈ B(x, r) for any n > N. This contradicts the fact that (xn )n ⊂ C since B(x, r) ⊂ X \ C. This proves that x ∈ C. Conversely, assume that the limit of any convergent sequence of elements in C belongs to C. We show that C is closed arguing by contradiction. Let us assume that C is not closed, i.e. X \ C is not open and, by definition, there exists x ∈ X \ C such that, for any r > 0, B(x, r) 6⊂ X \ C. This means that, for any r > 0, B(x, r) ∩ C 6= ∅. For any n > 1, we pick then xn ∈ C ∩ B(x, n1 ). This sequence is clearly a sequence of elements of C and, for any n > 1, kxn − xk < n1 so that x = limn xn. Since x ∈ / C, we just constructed a sequence of elements of C which converges to a limit not belonging to C. This is a contradiction. We can then characterize the closure of a given set: Definition 1.1.4 Let (X, k · k) be a normed space and let Y ⊂ X. The closure Y is given by n o Y = x ∈ X , ∃(xn )n ⊂ Y with x = lim xn. n Remark 1.1.2 Observe that Y ⊂ Y. It can also be proved that Y is a closed set of (X, k · k) and that Y is the smallest closed subset of (X, k · k) which contains Y , i.e. \ Y = C C closed Y ⊂C Check this as an Exercise. As an example, we show here that closed balls coincide with the closure of open balls: Lemma 1.1.6 Let (X, k · k) be a normed space and let x0 ∈ X, r > 0. Then, Bc (x0 , r) = B(x0 , r) where the closure is meant with respect tos k · k. Proof. Since the closed ball Bc (x0 , r) is closed and contains B(x0 , r), it contains B(x0 , r), i.e. B(x0 , r) ⊂ Bc (x0 , r) holds. Prove then the other inclusion and let x ∈ Bc (x0 , r). We have to prove that x ∈ B(x0 , r). Of course, if kx − x0 k < r, there is nothing to prove. Let us then assume that kx − x0 k = r. Since r > 0, we can define u = 1r (x − x0 ). Clearly, kuk = 1 i.e. u is the unit vector in the direction of x − x0. One has x = x0 + ru. For any n ∈ N set 1 xn = x0 + r 1 − u. n Then, kxn − x0 k = |r − nr | kuk = r(1 − n1 ) < r. Thus, xn ∈ B(x0 , r) for any r > 0 and r r kxn − xk = u = n n 1.1 The norm and its properties 11 so that limn xn = x. We thus proved that x is the limit of some sequence (xn )n contained in B(x0 , r), i.e. x ∈ B(x0 , r). We introduce the following Let (X, k · k) be a normed space and let Y ⊂ X be given. We say that Definition 1.1.5 Y is dense in X if Y =X We have the following useful characterization: Lemma 1.1.7 Let (X, k · k) be a normed space and Y ⊂ X. Then, Y is dense in X if and only if, for any non-empty open set O ⊂ X, it holds Y ∩ O = 6 ∅. Proof. The proof is left as an Exercise. Remark 1.1.3 The meaning of the above is just that, if Y is dense in X, then it meets any open subset of X or, more precisely, it meets any ball of X. Remark 1.1.4 From the definition of closure of Y , we see that Y is dense in X if and only if any element x ∈ X can be approximated by a sequence in Y , i.e. for any x ∈ X there is {xn }n ⊂ Y such that limn kxn − xk = 0. Remark 1.1.5 As well known, in (R, | · |), we have that Q is dense in R, i.e. Q = R. One also have that R \ Q = R. As explained by Lemma 1.1.7, this exactly means that, between two real numbers a < b, there exist q ∈ Q and x ∈ R \ Q such that a < q < b, a 0 such that B(y, ε) ⊂ Y }. Remark 1.1.6 Observe that Int(Y ) ⊂ Y ⊂ Y and Int(Y ) is an open subset of X. It can also be proved that Int(Y ) is the biggest open subset of (X, k · k) contained in Y , i.e. [ Int(Y ) = U U open U ⊂Y 12 Chapter 1. Normed spaces Check this as an Exercise. The link between closure and interior is made in the following Lemma 1.1.8 Let (X, k · k) be a normed space and let Y ⊂ X be given. Then X \ Y = X \ Int(Y ) i.e. the closure of the complement is the complement of the interior. Proof. The proof is left as an Exercise. We introduce now the notion of limit point Let (X, k · k) be a given normed space and let (xn )n ⊂ X be a given Definition 1.1.7 sequence in X. We say that x ∈ X is a limit point of the sequence (xn )n if it is the limit of a subsequence of (xn )n , i.e. there exists a subsequence (xϕ(n) )n such that lim kxϕ(n) − xk = 0. n Example 1.5If X = R endowed with the absolute value, the sequence ((−1)n )n has exactly two limit points −1 and 1. The sequence (n)n has no limit point in R. The link between the closure and the notion of limit points is given by the following whose proof is left as an Exercise: Proposition 1.1.9 Given Y ⊂ X. The closure Y is the set of all limit points of sequences of Y. Exercise 1.1 Given a sequence (xn )n ⊂ X, prove that x ∈ X is a limit point of (xn )n if and only if card ({k ∈ N; : xk ∈ B(x, ε)}) = ∞ ∀ε > 0, where card(A) denote the cardinal of the set of A. 1.1.3 Equivalent norms – Product spaces One can define the notion of equivalent norms Definition 1.1.8 Let X be a vector space over R. We say that two norms N1 and N2 on X are equivalent if there exist two positive constants C1 , C2 > 0 such that N1 (x) 6 C1 N2 (x) and N2 (x) 6 C2 N1 (x) ∀x ∈ X. Remark 1.1.7 Geometrically, two norms N1 , N2 are equivalent if the unit ball centered in 0 for N1 contains a (non-empty) open ball centered in 0 for N2 and viceversa. As we shall see later on, on some finite dimensional spaces, all the norms are equivalent. This is no more the case for infinite dimension spaces as illustrated by the following Example 1.6Let X = C (I) with I = [0, 1] and let k · k1 , k · k∞ be the norms defined in Proposition 1.1.1. Consider, for any k ∈ N, the function fk (t) = max{(1 − kt), 0}, t ∈ I, 1.1 The norm and its properties 13 so that (fk )k ⊂ X. Then, one has 1 kfk k1 = , kfk k∞ = 1 ∀k > 1. 2k In particular, there cannot be a positive constant C > 0 such that kfk k∞ 6 Ckfk k1 for any k. This shows that k · k1 and k · k∞ are not equivalent. Proposition 1.1.10 Let (X1 , k · k1 ) and (X2 , k · k2 ) be two normed spaces. Define X = X1 × X2 = {x = (x1 , x2 ), xi ∈ Xi , i = 1, 2} and let us define q kxk+ = kx1 k1 + kx2 k2 , kxke = kx1 k21 + kx2 k22 and kxkmax = max (kx1 k1 , kx2 k2 ) ∀x = (x1 , x2 ), ∈ X. Then, (X, k · k+ ), (X, k · ke ) and (X, k · kmax ) are equivalent normed spaces. Proof. It is easy to check that (X, k · k+ ), (X, k · ke ) and (X, k · kmax ) are normed spaces. Moreover, since kxkmax 6 kxke 6 kxk+ 6 2kxkmax ∀x ∈ X the three normeds are equivalent. 1.1.4 Continuity of functions We can define the notion of continuity of mappings between two normed spaces: Let (X, k · k1 ) and (Y, k · k2 ) be two normed spaces. A mapping Definition 1.1.9 f : X → Y is said to be continuous at x0 ∈ X if, for any ε > 0 there exists δ > 0 such that kx0 − xk1 < δ =⇒ kf (x0 ) − f (x)k2 < ε (x ∈ X). We say that f is continuous over X if f is continous at any x0 ∈ X. We have the strongest definition: Definition 1.1.10 Let (X, k · k1 ) and (Y, k · k2 ) be two normed spaces. A mapping f : X → Y is said to be uniformly continuous over X if, for any ε > 0 there exists δ > 0 such that, for any x0 , x ∈ X it holds kx0 − xk1 < δ =⇒ kf (x0 ) − f (x)k2 < ε. It is important to distinguish the two above notions. Clearly, an uniformly continuous function f : X → Y is continuous over X but the converse is not true. The difference between the two notions lies in the fact that, in the definition of uniform continuity, the choice of δ depends only on ε and not on the point x0 (the same δ will “work” for any x0 whereas, in the definition of continuity, different x0 ’s would yield different choices of δ). For instance, the mapping x 7→ x2 is continuous over R but is not uniformly √ continuous over R (Check this). On the contrary, the mapping x ∈ [0, ∞) 7→ x is uniformly continuous over [0, ∞) thanks to the inequality √ √ p | x − y| 6 |x − y| ∀x > 0, y > 0 14 Chapter 1. Normed spaces (Check the inequality and its consequence on uniform continuity). Continuity can also be characterized in terms of converging sequences: Let (X, k·k1 ) and (Y, k·k2 ) be two normed spaces and let f : X → Y Proposition 1.1.11 be a given mapping. Then, f is continuous at x ∈ X if and only if, for any sequence (xn )n ⊂ X with limn xn = x it holds limn f (xn ) = f (x) (in Y ), i.e. f is continuous at x ∈ X if and only if lim kxn − xk1 = 0 =⇒ lim kf (xn ) − f (x)k2 = 0. n n Proof. The proof is left as an Exercise. With this characterization, it is very easy to prove the following result. Let (Xi , k · ki ), i = 1, 2, 3 be three normed spaces and let f : X1 → Proposition 1.1.12 X2 and g : X2 → X3 be continuous. Then, g ◦ f : X1 → X3 is continuous where g ◦ f (x) = g(f (x)) for any x ∈ X1. 1.2 Fundamental example: Lebesgue spaces 1.2.1 The definition of L1 as a set of equivalence classes Let (S, Σ, µ) be a given measure space. We recall that L1 (S, Σ, µ) denotes the set of all measurable functions f : S → R such that Z |f |dµ < ∞. S It is clear that L1 (E, Σ, µ) is a vector space and the null element 0 of L1 (S, Σ, µ) is the function f ≡ 0, i.e. f (s) = 0 for all s ∈ S. It is tempting to define a norm on L1 (S) like Z kf k1 = |f |dµ S and, one checks easily that this defines a semi-norm on L1 (S, Σ, µ) but that Z kf k1 = 0 ⇐⇒ |f |dµ = 0 ⇐⇒ f = 0 µ − a. e. on S S In particular, f is not necessarily the null element of L1 (S, Σ, µ). Therefore, k · k1 is not a norm on L1 (S, Σ, µ). To avoid this flaw, we actually change the set of functions we wish to investigate. This is done in the following way, introducing the following equivalence relation: Definition 1.2.1 Let f, g ∈ L1 (S, Σ, µ). We will say that f is equivalent to g and write f ∼g if f − g is zero µ-a.e. on S, i.e. µ ({s ∈ S ; f (s) 6= g(s)}) = 0. It is readily seen that ∼ is an equivalence relation i.e. for any f, g, h ∈ L1 (S, Σ, µ): 1.2 Fundamental example: Lebesgue spaces 15 f ∼ f; f ∼ g if and only if g ∼ f ; f ∼ g and g ∼ h implies f ∼ h. In particular, given f ∈ L1 (S, Σ, µ), we introduce the equivalence class of f as [f ]µ = {g ∈ L1 (S, Σ, µ) ; f ∼ g}. We see from the above that the collection of all equivalence classes [f ]µ of R-valued measurable functions f ∈ L1 (S, Σ, µ) is a partition of L1 (S, Σ, µ), i.e. [ L1 (S, Σ, µ) = [f ]µ with disjoint union. f ∈L1 (S,Σ,µ) We introduce then the set L1 (S, µ) = [f ]µ , f ∈ L1 (S, Σ, µ). Notice that, strictly speaking, any element in L1 (S, µ) is a subset of L1 (S, Σ, µ), i.e. elements of L1 (S, µ) are sets. We endow L1 (S, µ) with a linear structure as follows: for any f, g ∈ L1 (S, Σ, µ) and any α ∈ R, one can define the sum: [f ]µ + [g]µ = [f + g]µ and the multiplication α[f ]µ = [αf ]µ. This makes L1 (S, µ) a vector space and the null element of L1 (S, µ) is the class µ = {g ∈ L1 (S, Σ, µ) , g ∼ 0} = {g ∈ L1 (S, Σ, µ) , g = 0 µ − a.e.}. Now, the space L1 (S, µ) can be endowed with a norm: Z k[f ]k1 = |f |dµ ∀f ∈ L1 (S, Σ, µ). S Notice that the norm is well-defined in the sense that it does not depend on the representative f chosen for the class [f ]µ : indeed, if g ∈ [f ]µ , then [f ]µ = [g]µ and k[f ]µ k1 = k[g]µ k1 (Important: try to understand that !). Of course, from now, we will abuse notations and identify equivalence classes [f ]µ with functions f and will simply write Z 1 f ∈ L (S, µ) ⇐⇒ |f |dµ S but we should always keep in mind that, doing so, we actually identify two functions which are equal µ-a.e. on S as a unique function. With this identification, we denote Z kf k1 = |f |dµ S and has the following 16 Chapter 1. Normed spaces Proposition 1.2.1 (L1 (S, µ), k · k1 ) is a normed vector space. 1.2.2 Lp -spaces, 1 6 p 6 ∞. In all the sequel, (S, Σ, µ) is a given, fixed, measure space. Having always in mind the above identification between measurable functions f and its equivalent class [f ]µ , we introduce the following Definition 1.2.2 Let p ∈ R be given with 1 < p < ∞. We set Lp (S, µ) = f : S → R , f measurable such that |f |p ∈ L1 (S, µ) and Z 1/p p kf kp = |f | dµ ∀f ∈ Lp (S, µ). S We shall check later on that, with such a definition, Lp (S, µ) is vector space and k · kp is a norm. Definition 1.2.3 We set ∞ L (S, µ) = f : S → R , f measurable such that there is C > 0 such that |f (x)| 6 C µ − a.e. on S. and kf k∞ = inf{C , |f (x)| 6 C µ − a.e. on S}. Example 1.7If S = N and µ is the counting measure, i.e. µ(A) = card(A) for any subset A ⊂ N, then one sees that Lp (S, µ) = `p (N) 16p6∞ is the spaces of sequences : X `p (N) = {x = (xn )n ; kxkpp := |xn |p < ∞}. n Remark 1.2.1 Notice that, if f ∈ L∞ (S, µ) then we have |f (x)| 6 kf k∞ µ-a.e. on S. (1.2) Indeed, there exists a sequence (Cn )n such that Cn → kf k∞ and for each n, |f (x)| 6 Cn µ-a.e. on S. Therefore, there is En ⊂ Σ with µ(En ) = 0 and |f (x)| 6 Cn ∀x ∈ S \ En. 1.2 Fundamental example: Lebesgue spaces 17 S Setting then E = n En , one has µ(E) = 0 and |f (x)| 6 Cn ∀n ∈ N , ∀x ∈ S \ E. In particular, |f (x)| 6 kf k∞ ∀x ∈ S \ E. From now on, given 1 < p < ∞, we indicate by q the conjugate exponent of p defined by the identity 1 1 + = 1. p q We begin with the following Lemma 1.2.2 — Young’s inequality. Let p > 1 and let q > 1 be its conjugate exponent. Then, for any nonnegative a, b ∈ R, it holds 1 1 ab 6 ap + bq. p q Proof. Though very simple, the result is very important for applications. The proof uses the convexity of the exponential function. Remember that a given function f : R → R is said to be convex if, for any x, y ∈ R and any θ ∈ [0, 1], f (θx + (1 − θ)y) 6 θf (x) + (1 − θ)f (y). Moreover, if f is twice differentiable and f 00 (x) > 0 for any x ∈ R, then f is convex. This in particular shows that the exponential function is convex over R. The result is obvious if a = 0 or b = 0, so assume that a > 0 and b > 0. One has ab = exp (log (ab)) = exp (log a + log b). Then, using the well-known multiplicative property of logarithm: 1 p 1 q ab = exp log (a ) + log (b ). p q 1 1 Since the exponential function is strictly increasing and convex, using p + q = 1 one gets 1 1 ap b q ab 6 exp (log (ap )) + exp (log (bq )) = + p q p q and this ends the proof. Remark 1.2.2 For p = 2, one sees that q = 2 and Young’s inequality simply states that 2ab 6 a2 + b2 which is an obvious consequence of the fact that (a − b)2 > 0. The above Young’s inequality can be seen as a generalization, for any p > 1, of this obvious estimate when p = 2. Moreover, given ε > 0, one sees that, replacing a with aε and b with b/ε, 18 Chapter 1. Normed spaces Young’s inequality yields εp p 1 ab 6 a + q bq ∀a, b, ε > 0. p qε An important consequence of the above inequality is the following Theorem 1.2.3 — Hölder inequality.Let 1 6 p 6 ∞ and 1/p + 1/q = 1. Assume that f ∈ L (S, µ) and g ∈ L (S, µ). Then, f g ∈ L1 (S, µ) and p q kf gk1 6 kf kp kgkq. (1.3) - Proof. The conclusion is obvious if p = 1 or p = ∞. We assume that 1 < p < ∞. We recall Young’s inequality: 1 1 ab 6 ap + bq ∀a, b > 0. p q Applying this to a = |f (s)| and b = |g(s)|, we get 1 1 |f (s)g(s)| 6 |f (s)|p + |g(s)|q for µ-a.e. on S p q Integrating this over S we get 1 1 kf gk1 6 kf kpp + kgkqq p q which shows that f g ∈ L1 (S, µ). To prove (1.3), we replace f by λ f (with λ > 0) in the above λp 1 λkf gk1 6 kf kpp + kgkqq ∀λ > 0, p q ie. λp−1 1 kf gk1 6 kf kpp + kgkqq ∀λ > 0. p λq q/p Choosing λ = kf k−1 p kgkq we get (1.3). Actually, by induction, one can prove the following Corollary 1.2.4 — Generalized Holder inequality. Given n ∈ N, let (p1 ,... , pn ) ∈ (R+ )n be such that n X 1 = 1. i=1 p i For any i = 1,... , n, let fi ∈ Lpi (S, µ). Then, n Y n Y 1 f= fi ∈ L (S, µ) and kf k1 6 kfi kpi. i=1 i=1 Proof. The proof is made by induction over n and is left as an Exercise. 1.2 Fundamental example: Lebesgue spaces 19 Theorem 1.2.5 For any 1 6 p 6 ∞, Lp (S, µ) is a vector space and k · kp is a norm. Remark 1.2.3 For 1 < p < ∞, the triangular inequality kf + gkp 6 kf kp + kgkp ∀f, g ∈ Lp (S, µ) is known as Minkowski’s inequality. - Proof. The fact that, given f ∈ Lp (S, µ) one has λf ∈ Lp (S, µ) is obvious as well as the homogeneity and uniqueness property for k · kp. So, to prove that Lp (S, µ) is a vector space, we just need to prove that, given f, g ∈ Lp (S, µ), the sum f + g ∈ Lp (S, µ) while proving that k · kp is a norm amounts in proving the above Minkowski’s inequality. Let us then fix f, g ∈ Lp (S, µ). Since the function r 7→ rp is convex over R+ (Check this !) one sees that, for any x, y ∈ R it holds that 1 1 | 21 x + 12 y|p 6 |x|p + |y|p 2 2 or, equivalently, that |x + y|p 6 2p−1 (|x|p + |y|p ) ∀x, y ∈ R. (1.4) This implies in particular that, for µ-a. e. s ∈ S |f (s) + g(s)|p 6 2p−1 (|f (s)|p + |g(s)|p ) and, integrating over S this yields Z Z Z p p−1 p p |f + g| dµ 6 2 |f | dµ + |g| dµ , S S S p i.e. f + g ∈ L (S, µ) with kf + gkpp 6 2p−1 kf kpp + kgkpp. Let us now prove Minkowski’s inequality. If kf + gkp = 0 there is nothing to prove so let us assume kf + gkp 6= 0. We write Z Z p p kf + gkp = |f + g| dµ = |f + g| |f + g|p−1 dµ S S which results, since |f + g| 6 |f | + |g|, in Z Z p kf + gkp 6 |f | |f + g| dµ + |g| |f + g|p−1 dµ. p−1 (1.5) S S Now, the function ψ = |f + g|p−1 belongs to Lq (S, µ) where q is the conjugate exponent of p. Indeed, q |ψ|q = |f + g|p−1 = |f + g|p since q(p − 1) = p and therefore Z 1/q Z 1/q p q p kψkq = |ψ| dµ = |f + g| dµ = kf + gkpq < ∞. (1.6) S S 20 Chapter 1. Normed spaces This shows that |ψ|q ∈ L1 (S, µ) i.e. ψ ∈ Lq (S, µ). Since |f | ∈ Lp (S, µ) one can apply Holder’s inequality with |f | and ψ to get Z p |f | |f + g|p−1 dµ = k f ψ k1 6 kf kp kψkq = kf kp kf + gkpq S where we used (1.6). In the same way, Z p |g| |f + g|p−1 dµ 6 kgkp kf + gkpq. S Thus, turning back to (1.5), p p kf + gkpp 6 kf kp kf + gkpq + kgkp kf + gkpq p and, dividing by kf + gkpq 6= 0, we get p− pq kf + gkp 6 kf kp + kgkp p which is Minkowski’s inequality since p − q = 1. We illustrate the above results for the special case of general probability space (Ω, F, P). We have the following Proposition 1.2.6 Let (Ω, F, P) be a probability triple. Then, the following holds L∞ (Ω, F, P) ⊂ Lp (Ω, F, P) ⊂ Lq (Ω, F, P) ⊂ L1 (Ω, F, P) ∀1 6 q 6 p 6 ∞. Proof. We first prove that L∞ (Ω, F, P) ⊂ Lp (Ω, F, P) for any 1 6 p 6 ∞. Let X ∈ L∞ (Ω, F, P) be given. One has Z Z |X| dP 6 kXk∞ dP = kXk∞ P(Ω) = kXkp∞ , p p Ω Ω i.e X ∈ Lp (Ω, F, P) with kXkp 6 kXk∞. p R Letq now assume thatq X ∈ L (Ω, F, P) and let 1 6 rq < p. One has to compute Ω |X| dP. Set Z = |X| and r = p/q > 1. Clearly, Z ∈ L (Ω, F, P). Moreover Z Z Z q |X| dP = ZdP = ZY dP Ω Ω Ω 0 where Y = 1 Ω. Let r0 > 1 such that 1r + r10 = 1. Notice that Ω Y r dP = Ω Y dP = R R 0 P(Ω) = 1 so that Y ∈ Lr (Ω, F, P). According to Holder inequality ZY ∈ L1 (Ω, F, P) with Z |X|q dP = kZY k1 6 kZkr kY kr0 = kZkr. Ω 1/q This proves that X ∈ Lq (Ω, F, P) with kXkq 6 kZkr = kXkp. 1.2 Fundamental example: Lebesgue spaces 21 Remark 1.2.4 Notice that the above result remains valid if (S, Σ, µ) is a given measure space with µ(S) < ∞. In the special case in which S = N, F = P(N) and µ is the counting measure defined by X µ(A) = δk (A), A ⊂ N, k∈A using the classical identification between functions f over N and sequences x = (xn )n given by ( f : N→R n 7→ f (n) =: xn we see that ( ∞ ) X 1 1 1 L (S, µ) = L (S, µ) = ` (N) = x = (xn )n ; kxk1 := |xn | < ∞ n=1 i.e. `1 (N) is nothing that some L1 (S, µ) for some special choice of S and µ. Notice that the counting measure µ is indeed such that Z X∞ ∞ X |f (n)|dµ(n) = |f (n)| = |xn | N n=1 n=1 with the identification above. This allows in particular to define and prove the follow- ing Proposition 1.2.7 Let p > 1 be given. We define the set ( ∞ ) X `p (N) = x = (xn )n∈N ⊂ R such that |xn |p < ∞. n=1 Then, `p (N) is a vector space. Moreover, if ∞ ! p1 X kxkp := |xn |p ∀x = (xn )n ∈ `p (N) n=1 then, (`p (N), k · kp ) is a normed space. Exercise 1.2Let X = C ([0, 1]) be the space of all continuous functions f : [0, 1] → R. For any p > 1, define Z 1 p1 kf kp = |f (t)|p dt , f ∈ X. 0 22 Chapter 1. Normed spaces Use Young’s inequality to prove that (X, k · kp ) is a normed space. Show also that, if f, g ∈ X, then 1 1 kf gk1 6 kf kp kgkq + = 1. p q 1.3 Another example: the space of linear applications In normed space, there is a very useful characterization of continuous linear applications: recall that, given X, Y two vector spaces, a mapping L : X → Y is said to be linear if L(λx + y) = λL(x) + L(y) ∀x, y ∈ X, λ ∈ R. One has the following Proposition 1.3.1 Let (X, k·kX ) and (Y, k·kY ) be two normed spaces and let L : X → Y be a linear application. The following are equivalent: 1. L is continuous on X; 2. L is continuous at x = 0; 3. there is a positive constant C > 0 such that kL(x)kY 6 CkxkX ∀x ∈ X. - Proof. It is clear that i) =⇒ ii). Assume now that L is continuous at 0 and let us apply the definition of continuity with ε = 1. There exists a positive constant δ > 0 such that kx − 0X kX 6 δ =⇒ kL(x) − L(0Y )kY < ε = 1. Since L is linear, L(0X ) = 0Y so that it holds kxkX 6 δ =⇒ kL(x)kY < 1. Let now x ∈ X \ {0X } be given. Set y = kxkδ X x. By the homogeneity of the norm, kykX = δ so that kL(y)kY < 1. Using the linearity of L and again the homogeneity of the norm, this reads 1 kL(x)kY < kxk δ 1 and proves iii) with C = δ. Let us prove now that iii) =⇒ i). By linearity, for any x, y ∈ X, it holds kL(x) − L(y)kX = kL(x − y)kX 6 Ckx − ykY and this clearly implies the continuity of L at any x ∈ X (actually, L is uniformly continuous on X). Definition 1.3.1If (X, k · kX ) and (Y, k · kY ) are two normed spaces, we denote by L (X, Y ) the space of continuous linear applications from X to Y. If Y = X, we simply denote L (X) = L (X, X). We also define kLkL (X,Y ) = kLkop = sup kL(x)kY ∀L ∈ L (X, Y ). kxkX 61 1.3 Another example: the space of linear applications 23 Remark 1.3.1 According to Proposition 1.3.1, kLkL (X,Y ) is finite for any L ∈ L (X, Y ). With this, one check easily as an Exercise that k · kL (X,Y ) is a norm on L (X, Y ). Moreover, one can check (Do it!) that, by definition, kL(x)kY 6 kLkop kxkX ∀x ∈ X. In particular, kLkop = inf {C > 0, kL(x)kY 6 CkxkX ∀x ∈ X} = min {C > 0, kL(x)kY 6 CkxkX ∀x ∈ X}. Proposition 1.3.2If (X, k · kX ) and (Y, k · kY ) are two vector spaces and X is of finite dimension, prove that any linear application L : X → Y is continuous. Proof. The proof is a relatively simple Exercise. Remark 1.3.2 If dim(X) = n and dim(Y ) = p, the space L(X, Y ) can be identified with the space Mn×p (R) of matrices with n lines and p rows. (Explain why). Example 1.8 Let X = C ([0, 1]) = C ([0, 1], R) be endowed with the sup-norm kf k∞ = sup |f (t)| ∀f ∈ X. t∈[0,1] Introduce the linear mapping L : X → X defined by Z x L(f )(x) = f (t)dt ∀x ∈ [0, 1], f ∈ X. 0 One checks easily that L is well-defined (i.e. L(f ) ∈ X for any f ∈ X) and linear. Moreover, for any x ∈ [0, 1], it holds Z x Z x |L(f )(x)| 6 |f (t)|dt 6 kf k∞ dt = xkf k∞ 6 kf k∞. 0 0 This means that kL(f )k∞ 6 kf k∞ for any f ∈ X. Therefore, L is continuous (Prop. 1.3.1) and kLkL (X) 6 1. Exercise 1.3Prove that, with the notations of the previous example, one exactly has kLkL (X) = 1. (Hint: Exhibit f ∈ X with kf k∞ = 1 = kL(f )k∞ ). Exercise 1.4 Let us again consider X = C ([0, 1]) and endowed it now with the norm k · k1 defined in Proposition 1.1.1. Prove that the mapping L : X → R defined by L(f ) = f (0) is linear but not continuous (where R is endowed with its usual norm), i.e. L∈ / L (X, R). What happens if we endow X with k · k∞ ? 24 Chapter 1. Normed spaces Exercise 1.5 Let X = C 1 ([0, 1]) be endowed with the sup-norm kf k∞ = sup |f (t)|, f ∈ X. t∈[0,1] Let T : f ∈ X 7→ T (f ) = f 0 (0) ∈ R. Considering the sequence of functions sin(n2 x) fn (x) = , x ∈ [0, 1], n ∈ N, n / L (X, R). prove that T ∈ 1.4 Compactness We introduce the concept of compact subset of a normed space: Definition 1.4.1Let (X, k · k) be a normed space and let K ⊂ X. We say that K is compact if every sequence (xn )n ⊂ K contains a subsequence which converges to some x ∈ K. Remark 1.4.1 Of course, if K is compact then K is closed. One has the following first property Lemma 1.4.1 If K is a compact subset of a normed space (X, k · k), then K is closed and there exists M > 0 such that supx∈K kxk 6 M , i.e. K is bounded. Proof. The fact that a compact set is closed is clear from the definition. Let us now prove that K is bounded. Assume the contrary, i.e. sup kxk = ∞. x∈K This means that there is a sequence (xn )n ⊂ K such that limn kxn k = ∞. Being K compact, there is a subsequence (xϕ(n) )n of (xn )n which converges to some x ∈ K. We know that then supn kxϕ(n) k < ∞ which contradicts the previous convergence to infinity. Proposition 1.4.2Let (X, k · k) be a normed space and let K be a compact subset of X. If A ⊂ K is a closed subset then A is compact. Proof. The proof is left as an easy Exercise. On R, it is easy to describe a large class of compact sets: Lemma 1.4.3 Let R be endowed with the absolute value, | · |. Any interval [a, b] ⊂ R is compact. Proof. Without loss of generality, we only prove the compactness of I = [0, 1]. Let (xn )n be a given sequence in I. Let us define two sequences (an )n and (bn )n inductively as 1.4 Compactness 25 follows: set a0 = 0 and b0 = 1 and assume a1 6... 6 an 6 bn... 6 b1 to be constructed. Then, one sets an + b n an + b n an+1 = an and bn+1 = if k ∈ N ; xk ∈ an , is infinite 2 2 otherwise, one sets an+1 = an +b 2 n and bn+1 = bn. With such a construction, the sequence (an )n is nondecreasing and the sequence (bn )n is nonincreasing. Moreover, for any n ∈ N, |an − bn | = 2−n and {k ∈ N ; xk ∈ [an , bn ]} is infinite. One sees that then that the two sequence are convergent to some common limit x (Check all these details). For any ε > 0 there exists n ∈ N such that x − ε 6 an 6 bn 6 x + ε. In particular, {k ∈ N ; xk ∈ (x − ε, x + ε)} is infinite so that x is a limit point of (xn )n (see Exercise 1.1). This shows that I is compact. Let (X1 , k · k1 ) and (X2 , k · k2 ) be two Proposition 1.4.4 — Product of compact spaces. compact normed spaces and let X = X1 × X2. Then, (X, k · k+ ) and (X, k · kmax ) are compact normed spaces where k · k+ and k · kmax have been defined in Proposition 1.1.10. - Proof. Let (ξn )n ⊂ X be a given sequence. It means that there are two sequences (xn )n ⊂ X1 and (yn )n ⊂ X2 such that ξn = (xn , yn ) for any n ∈ N. Since (X1 , k · k1 ) is compact, there is a subsequence (xϕ(n) )n ⊂ X1 which converges to x ∈ X1. Consider then the sequence (yϕ(n) )n ⊂ X2. Since (X2 , k · k2 ) is compact, there is a subsequence (yφ(ϕ(n)) )n of (yϕ(n) )n which converges to y ∈ X2. Then, the subsequence (xφ(ϕ(n)) )n of (xϕ(n) )n is still convergent in X1 to x and one sees easily that the subsequence (ξφ(ϕ(n)) )n = (xφ(ϕ(n)) , yφ(ϕ(n)) )n ⊂ X1 × X2 is converging in X to x = (x, y). This proves that X is compact. Remark 1.4.2Of course, the above result readily extends to any finite product of compact normed spaces. Corollary 1.4.5 — Heine-Borel Theorem. A subset K of RN (where RN is endowed with, say, the usual Euclidean normed) is compact if and only if it is closed and bounded. - Proof. The fact that a compact subset of RN is closed and bounded is a well-know property true in any normed space (see Lemma 1.4.1). Conversely, let K ⊂ RN be closed and bounded. Being bounded, there exists R > 0 such that K ⊂ [−R, R]N. Since K is closed, from the previous Proposition, it suffices to prove that [−R, R]N is a compact subset of RN. This is a simple consequence of the fact that [−R, R] is a compact subset of R (Lemma 1.4.3) together with Corollary 1.4.4. 26 Chapter 1. Normed spaces Remark 1.4.3 The previous Corollary can be reformulated as follows: every bounded sequence of RN has a convergent subsequence. 1.4.1 Compactness and continuous functions Continuous functions over compact normed spaces enjoy nice and useful properties. We first state the following Let (X, k · kX ) and (Y, k · kY ) and let f : X → Y be continuous. If Proposition 1.4.6 K ⊂ X is a compact subset of X then f (K) is a compact subset of Y. - Proof. Let (yn )n be a sequence of f (K). It means that there is a sequence (xn )n ⊂ K such that yn = f (xn ) for any n. Being K compact, there is a subsequence (xϕ(n) )n of (xn )n which converges to x ∈ K, i.e. limn kxϕ(n) − xkX = 0. Being f continuous, one has limn kf (xϕ(n) ) − f (x)kY = 0, i.e. limn kyϕ(n) − f (x)kY = 0. Since y = f (x) ∈ f (K), we get that any sequence of f (K) has a subsequence which converges to a limit in f (K), i.e. f (K) is compact. As a consequence Let (X, k · k) be a normed space and let K ⊂ X be compact. Let Theorem 1.4.7 f : K → R be continous. Then, f assumes its maximum and minimum on K. - Proof. According to Proposition 1.4.6, f (K) is a compact subset of R and, from Heine- Borel Theorem, f (K) is bounded and closed. This shows that f is bounded. The fact that it reaches is minimum and maximum value is a simple consequence of the fact that f (K) is closed. Indeed, let M = sup{f (x) , x ∈ K}. By definition, there is a sequence (xn )n ⊂ K such that limn f (xn ) = M. The sequence (f (xn ))n lies in f (K) which is compact and os is closed. Thus, its limit M also lies in f (K), i.e. there is x ∈ K such that f (x) = M. This shows that M is a maximum value of f. One proceeds in the same way with the minimum value. One also has Let (X, k · kX ) and (Y, k · kY ) be two normed spaces Theorem 1.4.8 — Heine Theorem. and let K ⊂ X be compact. Assume that f : K → Y is continuous. Then, f is uniformly continuous on K. Proof. Suppose f is not uniformly continuous. This means that there exists ε0 > 0 such that ∀δ > 0, ∃x, y ∈ K, with kx − ykX < δ and kf (x) − f (y)kY > ε0. Now choosing δ = n1 , n ∈ N this allows to build two sequences (xn )n and (yn )n such that 1 kxn − yn kX < and kf (xn ) − f (yn )kY > ε0 ∀n ∈ N. n Since K is compact, then we can extract a subsequence of xn , that we call (xϕ(n) )n , which converges to some x0 ∈ K. It follows that (yϕ(n) )n converges also to x0 (Explain this). 1.5 Finite Dimensional Spaces 27 Because f is continuous we get that limn→∞ f (xϕ(n) ) = f (x0 ) = limn→∞ f (yϕ(n) ) in Y , i.e. lim kf (xϕ(n) ) − f (yϕ(n) )kY = 0 n→∞ but this contradicts the fact that kf (xϕ(n) ) − f (yϕ(n) )kY > 0 for each n ∈ N+. 1.5 Finite Dimensional Spaces We aim here to characterize the compact subsets of finite dimensional normed spaces. We begin with the following Proposition 1.5.1 Let (X, k·k) be an finite dimensional normed vector space, dimX = d and let {e1 ,... , ed } be a basis for X. Then, there are positive constants C0 , C1 > 0 such that d X d X d X C0 |xi | 6 xi e i 6 C 1 |xi | ∀ (x1 ,... , xd ) ∈ Rd. i=1 i=1 i=1 Proof. We aim to prove that there are positive constants C0 , C1 > 0 such that C0 kxk1 6 kxk 6 C1 kxk1 , ∀x ∈ X where x is written on the basis {e1 ,... , ed } as x = di=1 xi ei and where we denote by P kxk1 = di=1 |xi | the norm of the vector x = (x1 ,... , xd ) of Rd. First, using the triangle P inequality, one has d X d X d X kxk = k xi e i k 6 kxi ei k = |xi |kei k. i=1 i=1 i=1 Setting C1 = max16j6d kej k, one has C1 < ∞ since we have a finite number of vectors ej and clearly Xd kxk 6 C1 |xi | = C1 kxk1. i=1 Let us prove the converse inequality. We denote by S1 the closed unit sphere of (Rd , k · k1 ), i.e. S1 = x = (x1 ,... , xd ) ∈ Rd ; kxk1 = 1. Since S1 is closed and bounded in (Rd , k · k1 ), it is compact. Consider the mapping f : Rd → R+ given by d X f (x) = xi e 1 , ∀x = (x1 ,... , xd ). i=1 Using the continuity of the norm k · k (as an application from X to R) we see easily that f is continuous over Rd (i.e. if {xn }n converges in Rd to x then limn f (xn ) = f (x) in R). Being S1 compact, Heine-Borel Theorem (Corollary 1.4.5) asserts that inf f (x) = min f (x) = C0 > 0 x∈S1 x∈S1 28 Chapter 1. Normed spaces i.e. there exists α = (α1 ,... , αd ) ∈ S1 such that C0 = f (α). Let us prove that C0 > 0. Otherwise, k di=1 αi ei k = 0 i.e. P Pd i=1 αi ei = 0 (the null element of X). Being the family {e1 ,... , ed } a basis of X, the vectors ei are linearly independent Pd and this means that αi = 0 for any i = 1,... , d. This is impossible since kαk1 = i=1 |αi | = 1. Therefore C0 > 0 and, by definition, for any vector x = (x1 ,... , xd ) ∈ S1 , one has d X xi ei > C0. i=1 This proves the desired inequality on the sphere S1. To deduce the inequality over Rd , we just consider, for any x ∈ Rd non zero the normalized vector 1 y = (y1 ,... , yd ) = (x1 ,... , xd ) ∈ S1 kxk1 Pd so that i=1 yi ei > C0 which reads d 1 X xi ei > C0 kxk1 i=1 and the proof is achieved. The above Proposition actually asserts that, if dim(X) = d, any norm of k·k is “related” to the k · k1 norm of Rd. This easily translates in the following Proposition 1.5.2 If X is a finite dimensional vector space, all norms over X are equivalent. Proof. As before, assume dim(X) = d and let {e1 ,... , ed } be a basis of Rd. Let x ∈ X be written on the basis {e1 ,... , ed } as x = di=1 xi ei and denote by kxk1 = di=1 |xi | P P the norm of the vector x = (x1 ,... , xd ) of Rd. Let N1 and N2 be two norms on X. From the previous proposition (applied to (X, N1 )) there are positive constants C0 , C1 > 0 such that C0 kxk1 6 N1 (x) 6 C1 kxk1 ∀x ∈ X whereas there are two positive constants C00 , C10 > 0 such that C00 kxk1 6 N2 (x) 6 C10 kxk1 ∀x ∈ X. It is easy to see then that C1 C10 N1 (x) 6 N2 (x) and N2 (x) 6 N2 (x) C00 C0 which proves the result. 1.5 Finite Dimensional Spaces 29 The above Proposition also allows to identify – in a continuous way – a finite dimension space (X, k · k) and the space Rd (d being the dimension of X). Indeed, introducing a basis {e1 ,... , ed } a basis of X, the mapping Φ : X → Rd which, to some x = di=1 xi ei ∈ X associates Φ(x) = x = (x1 ,... , xd ), we see that Φ P is a bijection from X to Rd which is continuous whose inverse is also continuous. This results in the following whose proof is a simple Exercise. If (X, k · k) is a finite dimensional vector space and K ⊂ X is closed Corollary 1.5.3 and bounded then K is compact. Again, this is very specific to finite dimensional spaces and, as we shall see, this actually characterizes finite dimensional spaces. Indeed, in infinite dimensional normed spaces, the closed unit ball cannot be compact. This shows that, in infinite dimensional spaces, the compact subsets do not coincide with closed and bounded subsets !! We first state the following technical lemma: Lemma 1.5.4 — Riesz Lemma. Let (X, k · k) be a normed vector space and let Y be a closed subspace of X (i.e. Y is closed in X and Y is a linear subspace of X). If Y 6= X then, for any ε ∈ (0, 1), there exists x ∈ X with kxk = 1 such that inf kx − yk > 1 − ε. y∈Y Remark 1.5.1 The Lemma asserts that, if Y 6= X is a closed subspace, then, for any ε ∈ (0, 1), there is some unit vector x ∈ X such that dist(x, Y ) > 1 − ε. - Proof. Let z ∈ X \ Y. Since Y is closed and z ∈ / Y , one has α = dist(z, Y ) = inf kz − yk > 0 y∈Y α (Explain why!). Pick ε ∈ (0, 1). There exists y ∈ Y such that ky − zk 6 1−ε (otherwise, α we would get dist(z, Y ) > 1−ε > α !). Notice that, y 6= z so that r := ky − zk > 0. Set 1 x := (z − y). r Clearly, kxk = 1. Let y ∈ Y be given. One can write 1 kx − yk = kz − y − r yk r and, since Y is a linear subspace, y + ry ∈ Y so that kz − y − ryk > α. Therefore kx − yk > αr > 1 − ε by assumption on r = kz − yk. Since this is true for any y ∈ Y , this proves the result. We also need the following Lemma 1.5.5 If (X, k · k) is a normed space, any linear subspace of finite dimension is closed. 30 Chapter 1. Normed spaces Proof. Let V ⊂ X be of finite dimension, say dim(V ) = d and let {e1 ,... , ed } be a basis of V. Let {xn } be a sequence in V which converges to some x ∈ X. In particular, supn kxn k < ∞. From Proposition 1.5.1, if we introduce the components vectors xn ∈ Rd (n) (n) such that xn = (x1 ,... , xd ) we see that the sequence {xn }n is bounded in (Rd , k · k1 ). In particular, it admits a subsequence {xϕ(n) }n converges to some x = (x1 ,... , xd ) in (Rd , k · k1 ). Using again Proposition 1.5.1, we deduce that {xϕ(n) }n converges to the element y = di=1 xi ei. Since it also converges to x we get P d X x= xi ei i=1 and in particular x ∈ V. This proves that V is closed. We can prove the following fundamental result Theorem 1.5.6 — Riesz Theorem. A normed space (X, k · k) is finite dimensional if and only if the closed unit ball Bc (0, 1) = {x ∈ X ; kxk 6 1} of X is compact. - Proof. We already saw that in finite dimensional spaces the closed and bounded subsets are compact. Let us assume that Bc (0, 1) is compact. Argue by contradiction assuming that X is infinite dimensional. Let us then pick e0 ∈ X with ke0 k = 1 and let F0 = Span(e0 ) = {te0 , t ∈ R}. According to the previous Lemma, F0 is a closed linear subspace of X (since it is of dimension 1). Being X infinite dimensional, F0 6= X. According to Riesz Lemma, there exists e1 ∈ / F0 such that ke1 k = 1 and minx∈F0 kx − e1 k > 1/2. Set then F1 := Span(e0 , e1 ). One constructs inductively a sequence 2 (en )n ⊂ X such that ken k = 1 and such that, the finite dimensional space Fn = Span(e0 ,... , en ) satisfies 1 en ∈ / Fn−1 , and inf kx − en k >. x∈Fn−1 2 In particular, (en )n ⊂ Bc (0, 1) which is compact so that a subsequence of (en )n should converge. But, for any n > m ∈ N, ken − em k > 21 (since en ∈ / Fm ). This is a contradiction. Remark 1.5.2 Of course, in the above statement, the choice of the unit ball is arbitrary. You can check easily (Do it !) that the closed unit ball is compact if and only if every closed ball is compact which, again, is equivalent to the fact that any closed and bounded subset of X is compact. 1.6 Problems Exercise 1.6 Let (E, k · kE ) and (F, k · kF ) two normed spaces and L : E → F a linear mapping such that (L(xn ))n is bounded in F for any sequence (xn )n ⊂ E which converges to 0 ∈ E. Prove that L is continuous. 2 Notice that this is an infinite sequence since the space X is of infinite dimension so that, for any n ∈ N, Fn 6= X 1.6 Problems 31 Exercise 1.7Let (X, k · k) be a normed space. Let Y ⊂ X be a linear subspace with Y 6= X. Prove that Int(Y ) = ∅. Exercise 1.8 Compute the norm of the following linear applications in L (X): For X = `∞ (N) with the usual norm, consider the shift-mapping defined by S : x = (xn )n 7→ S(x) where Sn+1 (x) = xn and S0 (x) = 0. For X = C ([0, 1]) endowed with k · k∞ , consider T f (x) = f (x)g(x) where g ∈ X is given. Exercise 1.9 Compute the nom of the following linear applicationsR 1 in L (X, R): For X = C ([0, 1]) endowed with k · k∞ , consider T (f ) = 0 f (x)g(x) dx where g ∈ X is a given mapping with g(1/2) = 0 and g(x) 6= 0 for any x 6= 1/2. For X = `2 (N) with the usual norm, consider the mapping ∞ X u(x) = an xn n=1 for some (an )n ∈ X. For X = `1 (N) with the usual norm, consider the mapping ∞ X u(x) = an x n n=1 for some (an )n ∈ `∞ (N). For X the set of all real convergent sequences x = (xn )n endowed with the norm kxk∞ = supn |xn |, consider the mapping u(x) = limk xk ∈ R. Preamble We admit in the first two exercises the change of variables formula (polar coordinate) in RN : if F (x) = Φ(kxk) depends only on the norm kxk (we say F is a radial function), then 3 Z Z ∞ N −1 F (x)dx = |S | Φ(%)%N −1 d% RN 0 N −1 where |S | is the surface area the unit sphere SN −1 = {y ∈ RN ; kyk = 1} given by 2π N/2 |SN −1 | = Γ(N/2) where Γ is the Gamma function. 3 More generally, if F ∈ L1 (RN ), denoting any x ∈ RN , x 6= 0 as x = % σ, with % = kxk and kσk = 1, we have Z Z ∞ Z F (x)dx = %N −1 d% F (% σ)dσ RN 0 SN −1 N −1 where dσ denotes the surface Lebesgue integral over S. 32 Chapter 1. Normed spaces Exercise 1.10 Let 1 6 p < ∞ be given. 1. For which values of α does the mapping 1 x ∈ RN 7→ (1 + kxk2 )α belong to Lp (Rn ) ? 2. For which values of β does the mapping 1 − kxk2 x ∈ RN 7→ e 2 kxkβ belong to Lp (Rn ) ? 3. Let now 1 6 q < p 6 +∞. Using the previous two points, find a mapping f belonging to Lq (Rn ) but not belonging to Lp (Rn ) and a mapping g ∈ Lp (Rn ) / Lq (Rn ). with g ∈ Exercise 1.11 Let α > 0 and β > 0. Set −1 f (x) = (1 + kxkα )−1 1 + | log kxk|β , ∀x ∈ RN. Under what conditions on α, β does f belong to Lp (RN )? Exercise 1.12 (Generalized Holder’s inequality) Let f1 ,... , fn be measurable func- tions over (S, Σ, µ) such that fi ∈ Lpi (S, µ) ∀i = 1,... , n Pn 1 where 1 6 pi 6 ∞ and i=1 pi = 1. Set n Y f= fi. i=1 1. Show by induction over n that f ∈ L1 (S, µ) with n Y kf k1 6 kfi kpi. i=1 2. Deduce that if f ∈ Lp (S, µ) ∩ Lq (S, µ) with 1 6 p 6 q 6 ∞, then for any f ∈ Lr (S, µ) for any p 6 r 6 q with 1 α 1−α kf kr 6 kf kαp kf k1−α q where α ∈ [0, 1] is such that = +. r p q 2. Banach spaces 2.1 Cauchy sequences-Banach spaces An important related concept is the one of Cauchy sequence: Let (X, k · k) be a normed space. A sequence (xn )n ⊂ X of elements Definition 2.1.1 in X is a Cauchy sequence if, for any ε > 0, there exists N = N (ε) ∈ N such that kxn − xm k < ε ∀n, m > N. Remark 2.1.1 One checks easily that every Cauchy sequence (xn )n ⊂ X is bounded. Indeed, by definition, for any ε > 0, there is N ∈ N such that kxn − xm k < ε ∀n, m > N. For, say, ε = 1 and m = N we see that, for any n > N , kxn k = kxn − xN + xN k 6 kxn − xN k + kxN k 6 1 + kxN k. Thus, with C1 = 1 + kxN k, sup kxn k 6 C1. n>N Setting now C2 := max (kx1 k,... , kxN −1 k), one sees that C2 is finite since it is the maximum of only a finite number of real numbers. By definition, kxn k 6 C2 ∀n < N. Therefore, sup kxn k 6 C = max(C1 , C2 ) < ∞ n i.e. (xn )n is bounded. 34 Chapter 2. Banach spaces The triangle inequality readily implies the following Lemma 2.1.1 Let (X, k · k) be a normed space. Any convergent sequence is a Cauchy sequence. - Proof. Let (xn )n ⊂ X be a convergent sequence in X and let x be its limit. By definition, given ε > 0, there is N ∈ N such that kxn − xk < ε for any n > N. Pick then n, m > N , one has from the triangle inequality kxn − xm k 6 kxn − xk + kxm − xk < 2ε. Since ε > 0 is arbitrary, we proved that the sequence is a Cauchy sequence (since we can rename 2ε as ε !). It is very important to understand that the converse result is not true in general: there exists Cauchy sequence which cannot converge. Example 2.1 Consider the set of rational numbers Q endowed with the norm | · |. There exist Cauchy sequences of rational numbers that do not converge to a rational number. For example, let (xn )n be the following rational numbers based on the decimal expansion of π: x1 = 3, x2 = 3.1, x3 = 3.14,.... Then, (xn )n is a Cauchy sequence in Q however, (xn )n is not a convergent sequence in Q since there is no rational q ∈ Q with xn → q (the limit is π ∈ / Q). - Example 2.2We introduce here the set c00 (N) of all sequences of real numbers that have only finitely many nonzero components, i.e. c00 = {x = (xn )n ⊂ X , ∃N ∈ N , xk = 0 ∀k > N }. It is clear that c00 (N) is a subset of the space `1 (N) introduced in Example 1.2. In particular, it is a norm space with respect to the norm induced by the one in `1 (N) and defined in (1.1). Now, for any n ∈ N, let x(n) be the sequence given by x(n) = (2−1 , 2−2 ,... , 2−n , 0, 0,...) (n) (n) (n) i.e. x(n) = (xk )k with xk = 2−k if k 6 n and xk = 0 if k > n. Clearly x(n) ∈ c00 (N) for any n ∈ N. Therefore, the family x(n) n is a sequence of elements of c00 (N) (pay attention that this is a sequence whose elements are again sequences !!). Moreover, if m < n, then X∞ Xn (n) (m) (n) (m) kx − x k1 = |xk − xk | = 2−k. k=1 k=m+1 (n) One sees then that (Explain this) (x )n is a Cauchy sequence in (c00 (N), k·k1 ). However, it does not converge in (c00 (N), k · k1 ). Indeed, assume the contrary, i.e. there exists x = (xk )k ∈ c00 (N) such that lim x(n) = x. It is not difficult to check that, necessarily, for any k ∈ N, it must hold (n) lim xk = xk n→∞ and therefore that xk = 2−k for any k ∈ N. However, one sees that the sequence x = (2−k )k does not belong to c00 (N). This proves the claim. One has the following 2.1 Cauchy sequences-Banach spaces 35 Lemma 2.1.2 Let (X, k · k) be a given normed space and let (xn )n be a Cauchy sequence in X. If (xn )n admits a limit point then it is convergent. Proof. Let x be a limit point of (xn )n and let ε > 0. Let us pick N0 ∈ N such that kxn −xm k < ε/2 for any n, m > N0. Let now N1 > N0 be given such that kxN1 , xk < ε/2 (the existence of such N1 is clear since x is a limit point). Then, for any n > N1 , one has kxn − xk 6 kxn − xN1 k + kxN1 − xk < ε which proves that the sequence converges to x. We introduce now the notion of complete normed space Definition 2.1.2A normed space (X, k · k) is said to be complete if any Cauchy sequence is convergent in X. A complete normed space (X, k · k) is called a Banach space. Example 2.3 The examples we already deal with showed that Q endowed with the norm | · | is not a complete space and c00 (N) endowed with the `1 (N)-norm is not a complete space. Exercise 2.1 Prove that, introducing now the supremum norm kxk∞ := sup |xn |, x = (xn )n ∈ c00 , n then again, (c00 , k · k∞ ) is not a Banach space. The most fundamental example of complete normed space is the set of real num- bers Theorem 2.1.3 If X = R is endowed with the usual norm | · | for any x, y ∈ R, then (R, | · |) is a complete normed space. In other words, any Cauchy sequence in R is convergent. Proof. Let (xn )n ⊂ R be a Cauchy sequence in R. One checks easily that (xn )n is a bounded sequence in R. Therefore, a well-known consequence of this is that it admits a subsequence which converge. In other words, the Cauchy sequence (xn )n has a limit point and we conclude with Lemma 2.1.2. We establish here a characterisation of Banach spaces in terms of convergent series. We start with the following: Definition 2.1.3 Let (X, k·k) be a normed space and let {xn }n ⊂ X be a given sequence. We define the sequence (sN )N ⊂ X of partial sums N X sN := xn , N ∈ N. n=1 P The series n xn is said to be convergent if there exists x ∈ X such that ∞ X lim ksN − xk = 0, we write then x = xn. N →∞ n=1 36 Chapter 2. Banach spaces P The series n xn is said to be absolutely convergent if X kxn k < ∞. n It is a well-known property of (R, | · |) that series which are absolutely convergent in (R, | · |) are actually convergent. The next result actually shows that it is a general feature (actually a characterisation) of Banach spaces: Theorem 2.1.4 Let (X, k · k) be a normed space. The following are equivalent: 1. (X, k · k) is a Banach space; 2. Every absolutely convergent series is convergent in (X, k · k), i.e. for any {xn }n ⊂ X, ∞ N ! X X kxn k < ∞ =⇒ xn converges in (X, k · k). n=1 n=1 N - P Proof. (1 =⇒ 2) Let {xn }n be a sequence such that n kxn k < ∞ and let n X Sn = xk , k ∈ N. k=1 We need to prove that (SN )N converges in (X, k · k). Since, by assumption, (X, k · k) is a Banach space, it is enough to show that (Sn )n is a Cauchy sequence. Thus, for m > n > 1, we compute Xm X n Xm kSm − Sn k = xk − xk = xk. k=1 k=1 k=n+1 Thanks to the triangle inequality, we deduce that m X ∞ X

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