MTH 503: Functional Analysis

Transcription

1 MTH 53: Functional Analysis Semester 1, Dr. Prahlad Vaidyanathan

2 Contents I. Normed Linear Spaces 4 1. Review of Linear Algebra Definition and Examples Bounded Linear Operators Completeness Finite Dimensional Spaces Quotient Spaces II. Hilbert Spaces Orthogonality Riesz Representation theorem Orthonormal Bases Isomorphisms of Hilbert spaces The Stone-Weierstrass Theorem Introduction to Fourier Series III. The Hahn-Banach Theorem The Duals of l p and L p spaces The Hahn-Banach Extension theorem The Dual of Subspaces and Quotient Spaces Separability and Reflexivity IV.Operators on Banach Spaces Baire Category Theorem Principle of Uniform Boundedness Open Mapping and Closed Graph Theorems Fourier Series of L 1 functions V. Duality Weak Convergence The Hahn-Banach Separation Theorem The Weak Topology Weak Sequential Compactness The Weak- Topology Weak- Compactness

3 VI.Operators on Hilbert Spaces Adjoint of an Operator Diagonalization: The Finite Dimensional Case Compact Operators Diagonalization: The Compact Self-Adjoint Case VII.Instructor Notes 149 3

4 I. Normed Linear Spaces 1. Review of Linear Algebra Note: All vector spaces will be over k = R or C Definition: A Hamel basis for a vector space E over k is a set B E such that every element x E can be expressed uniquely as a (finite) linear combination of elements in B Theorem: For a subset B E, TFAE : (i) B is a Hamel basis for E (ii) B is a maximal linearly independent set (iii) B is a minimal spanning set. (without proof) 1.3. Zorn s Lemma: Let (F, ) be a partially ordered set such that every totally ordered subset has an upper bound. Then F has a maximal element. (without proof) 1.4. Theorem: Every vector space has a basis. Proof. Assume E {} and choose x E. independent. So consider Then A := {x} is linearly F := {B E : A B, and B is linearly independent} Then F is a partially ordered set (under inclusion). If C F is a totally ordered subset, then consider B := C C C [Check!] B F, and C B for all C C clearly holds. Hence, B is an upper bound for C. Thus, by Zorn s lemma, F has a maximal element B. By Theorem 1.2, B must be a Hamel basis for E. Note: We have proved something stronger: If A E is any linearly independent set, then a Hamel basis B of E such that A B Examples: (i) Standard basis {e i : 1 i n} for E = R n 4

5 (ii) Define c and note that it is a vector space over k (Check!). Write e i as above, and note that {e i : i N} is a basis for c. (iii) Define c. Note that {e i : i N} as above is a linearly independent set, but not a basis for c. (We will prove later that any basis of c must be uncountable) (iv) Let a, b R with a < b, then C[a, b] is a vector space. For n N, let e n (x) := x n, then {e n : n N} is a linearly independent set, but it is not a basis for C[a, b] (HW) 1.6. Theorem: If E is a vector space and B 1 and B 2 are two bases for E, then B 1 = B 2. This common number is called the dimension of E (without proof) 1.7. Definition: Let E and F be two vector spaces and T : E F a function. (i) Linear transformation or operator (ii) We write L(E, F ) for the set of all linear operators from E to F. Note that L(E, F ) is a k-vector space. (iii) If F = k, then a linear transformation T : E k is called a linear functional Examples: (i) Let E = R n, F = R m, then any m n matrix A defines a linear transformation T A : E F given by x A(x). Conversely, if T L(E, F ), then the matrix whose columns are {T (e i ) : 1 i n} defines an m n matrix A such that T = T A. Hence, there is an isomorphism of vector spaces L(E, F ) = M mn (R) given by T A A Note: Given another basis B of E, we get another isomorphism from L(E, F ) M mn (R). Thus, the isomorphism is not canonical (it depends on the choice of basis) (ii) Let E = c and define ϕ : E k given by (x n ) n N x n Note that ϕ is well-defined and linear. Thus, ϕ L(c, k) (iii) Let E = C[a, b] and define ϕ : E k given by ϕ(f) := Note that ϕ is linear and so ϕ L(C[a, b], k) (iv) Let E = F = C[, 1]. Define T : E F given by T (f)(x) := 1 x f(t)dt f(t)dt Note that T is well-defined (from Calculus) and linear. Thus, T L(E, F ). 5

6 1.9. Definition: (i) Let E, F be two k-vector spaces, then E F is a vector space under the usual component-wise operations. This is called the external direct sum and is denoted by E F. (ii) Let E be a vector space and F 1, F 2 E be two subspaces such that E = F 1 + F 2 and F 1 F 2 = {}, then there is an isomorphism F 1 F 2 = E Thus, E is called the internal direct sum of F 1 and F 2 and we once again write E = F 1 F Definition: Let E be a vector space F E a subspace (i) The quotient space E/F (End of Day 1) (ii) Note that natural the quotient map π : E E/F is a surjective linear transformation such that ker(π) = F. (iii) Furthermore, we define the codimension of F by codim(f ) := dim(e/f ) (iv) If codim(f ) = 1, then we say that F is a hyperplane of E Theorem: Let E be a finite dimensional vector space and F < E. Then codim(f ) = dim(e) dim(f ) Proof. HW (First Isomorphism Theorem) Let T : E F be a linear transformation of k- vector spaces. Then (i) ker(t ) < E and Image(T ) < F (ii) Furthermore, as vector spaces. E/ ker(t ) = Image(T ) (Rank-Nullity theorem): If T : E F is a linear transformation and E is finite dimensional, then Proof. Theorem dim(ker(t )) + dim(image(t )) = dim(e) 6

7 2. Definition and Examples 2.1. Definition: A norm on a k-vector space E is a function : E R + which satisfies the following for all x E, α k (i) x and x = iff x = (ii) αx = α x (iii) (Triangle inequality) x + y x + y The pair (E, ) is called a normed linear space Remark: If (E, ) is a normed linear space, then (i) The function d(x, y) := x y defines a metric on E. This is called the metric induced by the norm. This makes E a Hausdorff topological space. (ii) Note that a sequence (x n ) E converges to a point x E iff lim n x n x = (iii) By the triangle inequality, vector space addition is a continuous map from E E E (iv) Similarly, scalar multiplication is also a continuous map from k E E (v) By the triangle inequality, it follows that 2.3. Examples: x y x y and thus the norm function E R + is continuous. (i) k with absolute value norm (ii) k n with (a) 1-norm given by (x 1, x 2,..., x n ) 1 := n i=1 x i (b) sup norm given by (x 1, x 2,..., x n ) := sup 1 i n x i (iii) c with 1-norm or sup norm (iv) c with sup norm (v) C[a, b] with (a) 1-norm given by f 1 := b f(t) dt. Note that it is a norm because if a f 1 = and f is continuous, then f. (b) sup-norm given by f := sup x [a,b] f(x) 7

8 2.4. Definition: An inner product on a vector space E is a function, : E E k such that for all x, y, z E, α, β k, we have (i) αx + βy, z = α x, z + β y, z (ii) x, y = y, x (iii) x, x and x, x = iff x = 2.5. (Cauchy-Schwartz inequality): If E is an inner product space and x, y E, then x, y 2 x, x y, y Proof. The inequality is clearly true if x =, so if x, set Then z, x =, so z := y y, x x, x x z, z = z, y y, x = y x, x x, y y, x x, y = y, y x, x x, y 2 = y, y x, x 2.6. Corollary: If E is an inner product space, then the function x := x, x defines a norm on E. This is called the norm induced by the inner product. Proof. We only check the triangle inequality, since the other axioms are obvious. x + y 2 = x + y, x + y = x 2 + x, y + y, x + y 2 = x 2 + 2Re( x, y ) + y 2 x x, y + y 2 x x y + y 2 = ( x + y ) 2 and so x + y x + y 8

9 2.7. Example: (i) k n with the Euclidean inner product. The induced norm is denoted by 2 (ii) c (iii) l 2 Proof. If (x n ), (y n ) l 2, then by the Cauchy-Schwartz inequality in k n, we have ( n n ) 1/2 ( n ) 1/2 ( ) 1/2 ( ) 1/2 x i y i x i 2 y i 2 x i 2 y i 2 i=1 i=1 i=1 This is true for each n N, so the inner product is well-defined, and the other axioms are trivial to check Definition: Fix a < b in R and < p < (i) A p integrable measurable function f : [a, b] C (ii) Let L p [a, b] be the set of all p-integrable measurable functions. Note that: If f, g L p [a, b], then f + g p [2 max{ f, g }] p 2 p [ f p + g p ] f + g L p [a, b] (I.1) and so L p [a, b] is a vector space. (iii) Define µ p : L p [a, b] R + by Then clearly, (a) µ p (f) for all f L p [a, b] (b) µ p (αf) = α µ p (f) for all α C i=1 ( b ) 1/p µ p (f) := f p (c) Note that µ p (f) = does not imply that f =. It merely implies that f a.e. (d) We are yet to prove that µ p satisfies the triangle inequality. (iv) Define N = {f L p [a, b] : µ p (f) = } = {f L p [a, b] : f a.e.} Then N is a subspace of L p [a, b] (by Equation I.1). We define the quotient space to be L p [a, b] := L p [a, b]/n Then, L p [a, b] is a vector space a i=1 9

10 (v) For f + N L p [a, b], we write f + N p := µ p (f). Note that if [f] = [g], then f g a.e., and so µ p (f) = µ p (g). Hence p : L p [a, b] R + is well-defined. It also clearly satisfies the first two axioms of a norm. (End of Day 2) Note: Henceforth, we identify two functions that are equal a.e. and merely write f p for f + N p Example: For < p < 1, the triangle inequality fails: Take f = χ (,1/2), g = χ (1/2,1) L p [, 1], then f p = g p = 2 1/p, so 1 = f + g p and f p + g p = 2 1/p + 2 1/p = 2 1 1/p < Lemma (See [Folland, Lemma 6.1]): If a, b and < λ < 1, then and equality holds iff a = b a λ b 1 λ λa + (1 λ)b Proof. If b =, there is nothing to prove, so assume b and set t = a/b, then we WTS that t λ λt + (1 λ) with equality iff t = 1. But the function f : t t λ λt satisfies f (t) = λt λ 1 λ. Since < λ < 1, f is increasing for t < 1 and decreasing for t > 1. Hence, the max value of f occurs at t = 1, when f(1) = 1 λ (Holder s inequality): Let 1 < p < and q R such that 1/p + 1/q = 1. If f, g : [a, b] C be measurable functions, then b a fg f p g q Furthermore, equality holds iff α f p β g q a.e. for some constants α, β C with αβ. Proof. If either term on the RHS is or +, there is nothing to prove. Furthermore, if the inequality holds for any pair f, g, then it also holds for all pairs αf, βg for α, β C. Therefore, replacing f by f/ f p and g by g/ g q, it suffices to assume that f p = g q = 1 1

11 So fix x [a, b] and let a = f(x) p, b = g(x) q, and λ = 1/p in Lemma 2.1, so that Integrating both sides, we get b a f(x)g(x) f(x) p p + g(x) q q b b fg p 1 f p + q 1 g q = p 1 + q 1 = 1 = f p g q a Furthermore, equality holds iff f(x) p = g(x) q a.e (Minkowski s inequality) : If 1 p < and f, g L p [a, b], then a f + g p f p + g q Thus, (L p [a, b], p ) is a normed linear space. Proof. The result is obvious if p = 1 or f + g = a.e. Otherwise, f + g p ( f + g ) f + g p 1 By Holder s inequality f + g p f p f + g p 1 q + g p f + g p 1 q [ ] 1/q = [ f p + g p ] f + g (p 1)q Now (p 1)q = p and since f, g L p [a, b], it follows by Equation I.1 that b a f + g p < Thus we may divide by [ f + g p] 1/q on both sides to obtain [ f + g p = f + g p ] 1 1/q f p + g p Definition: (i) An essentially bounded measurable function (ii) L [a, b] is the set of all essentially bounded measurable functions. (iii) For f L [a, b], define µ (f) := inf{m > : m({x [a, b] : f(x) > M}) = } 11

12 2.14. Lemma: For any f L [a, b], f µ (f) a.e. Proof. For each n N, the number µ (f) + 1/n is not a lower bound for the set A f := {M > : m({x [a, b] : f(x) > M}) = } Hence, M n A f such that M n µ (f) + 1/n. Now, {x : f(x) > µ (f)} = {x : f(x) > µ (f) + 1/n} n=1 {x : f(x) > M n } n=1 But each set {x : f(x) > M n } has measure zero, and thus m({x : f(x) > µ (f)}) = Definition/Theorem: Consider N := {f L [a, b] : µ (f) = } Then N is a subspace of L [a, b]. We define the quotient space to be and for any f + N L [a, b], we write L [a, b] := L [a, b]/n f + N := µ (f) Then is well-defined and a norm on L [a, b] (HW) Definition: For 1 p < (i) k n with p (ii) l p Note: Triangle inequality follows from Lemma 2.1 exactly as in Minkowski s inequality (iii) l (End of Day 3) 12

13 3. Bounded Linear Operators 3.1. Notation: Let E be a NLS. (i) For x E, r >, we write B(x, r) := {y E : y x < r} and B[x, r] := {y E : y x r} Note that B(x, r) is open and B[x, r] is closed. The closed unit ball is the set B[, 1] and the open unit ball is B(, 1) (ii) The unit sphere is the set {x E : x = 1} 3.2. Definition: Let E, F be normed linear spaces. A linear operator T : E F is said to be (i) continuous if it is continuous with respect to the norm topologies on E and F. (ii) bounded if M such that T (x) M x for all x E. Note: A bounded linear operator maps B[, 1] to a subset of B[, M] 3.3. Theorem: For a linear operator T : E F between normed linear spaces, TFAE: (i) T is continuous (ii) T is continuous at any one point in E (iii) T is continuous at E (iv) T is bounded (v) T is uniformly continuous Proof. (i) (ii) and (v) (i): By definition (ii) (iii): If T is continuous at a point x E, then for any ɛ >, choose δ > such that x x < δ T (x) T (x ) < ɛ So if x < δ, then let z := x + x, then z x < δ, so and so T is continuous at. T (z) T (x ) < ɛ T (x) < ɛ (iii) (iv): Suppose T is continuous at, then for ɛ = 1 >, δ > such that x < δ T (x) < 1 So for any y E, let x := δ y 2 y 13

14 Then x < δ, and so T (x) < 1, whence T (x) = δ 2 y T (y) < 1 T (y) < 2 δ y (iv) (v): Suppose M > such that T (x) M x for all x E, then for any ɛ >, choose δ := ɛ 2M so if x y < δ, then T (x) T (y) = T (x y) M x y ɛ 2 < ɛ 3.4. Examples: (i) Let E be any inner product space, and y E be fixed. Define ϕ : E k given by x x, y Then ϕ(x) x y by the Cauchy-Schwartz inequality, and so ϕ is bounded by y and so it is continuous by Theorem 3.3 (ii) Let T : k n E be any operator where k n is endowed with the sup-norm and E is any NLS, then for any x = (x 1, x 2,..., x n ) k n, we have ( n n n ) T (x) = x i T (e i ) x i T (e i ) x T (e i ) i=1 i=1 and so T is bounded by M := n i=1 T (e i). Thus, any linear operator T : k n E is continuous. (iii) Let E = c and ϕ : E k be given by i=1 ϕ : (x n ) x n (a) If E has the 1-norm, then ϕ is continuous since ϕ(x) x 1 (b) If E has the sup-norm, let x k = (1, 1,..., 1,,,...) be the sequence where the first k terms are 1 and the rest are zero, then x k = 1 k N but ϕ(x k ) = k Hence, there does not exist M > such that ϕ(x) M x for all x E. Hence, ϕ is not continuous. 14

15 (iv) Let E = F = C[, 1] with the sup-norm, and T : E F be the integral operator T (f)(x) = Then for any x [, 1] T (f)(x) x x f(t)dt f(t) dt x f f Hence, T (f) f and so T is continuous. (v) Let E = C[, 1], and define ϕ : E k by f f() (a) If E has the sup-norm, then ϕ(f) f so ϕ is continuous. (b) If E has the 1-norm, then consider a sequence f k of non-negative continuous functions such that f k () = k and 1 f k (t)dt = 1 (A triangle of large height but area 1). Thus, f k = 1 for all k N, but ϕ(f k ) = k, so so ϕ is not continuous Definition: Let E, F be a normed linear spaces (i) B(E, F ) is set of all bounded linear operators from E to F. Note that B(E, F ) is a vector space. (ii) We write B(E) := B(E, E) (iii) We write E := B(E, k) and E is called the (continuous) dual space of E (iv) For any T B(E, F ), we write T := inf{m > : T (x) M x x E} 3.6. Lemma: For any T B(E, F ) and any x E, T (x) T x Proof. For any x E, choose a sequence M n such that T (x) M n x n N and M n T By taking limits, it follows that T (x) T x 15

16 3.7. Theorem: If E, F are normed linear spaces, then the function : B(E, F ) R + defined above is a norm on B(E, F ) Proof. (i) Clearly, T and =, so suppose T =, we WTS: T =. This follows from Lemma 3.6, since T (x) for all x E. (ii) Fix T B(E, F ) and α k, and define For any M A 1, A 1 = {M > : T (x) M x x E} A 2 = {K > : (αt )(y) K y y E} αt (x) = T (αx) M αx = M α x x E and so M α A 2, and so αt = inf A 2 M α M A 1 αt α T Now we may replacing α with 1/α, we have T = 1 α αt 1 αt αt α T α (iii) Fix S, T B(E, F ), then for any x E, we have (S+T )(x) = S(x)+T (x) S(x) + T (x) S x + T x = ( S + T ) x and so (S + T ) is bounded by M := S + T and so by definition S + T S + T (End of Day 4) 3.8. Theorem: Let T B(E, F ) then T = sup{ T (x) : x E, x = 1} Proof. Let α := sup{ T (x) : x E, x = 1}, then (i) For any x E with x = 1, T (x) T x = T α T 16

17 (ii) Set A = {M > : T (x) M x x E} For any n N, then T 1/n / A and so x n E such that T (x n ) > ( T 1/n) x n Thus, x n, so if y n := x n / x n, then y n = 1 and T (y n ) > T 1/n α > T 1/n This is true for all n N and so α T 3.9. Examples: (See Example 3.4) (i) Let E be an inner product space, y E and define ϕ : E k by x x, y. Then by Cauchy-Schwartz, ϕ(x) x y x E ϕ y Furthermore, if x = y, then y 2 = ϕ(y) ϕ y, and so ϕ = y (ii) Let E = k n with the 1-norm, F any NLS and T : E F linear. Then for x = n i=1 x ie i we have T (x) n x i T (e i ) i=1 and so T is continuous with T M := max 1 i n T (e i ). If x = e i, then x 1 = 1 and T (x) = T (e i ), and so by Theorem 3.8 T T (e i ) i T M (iii) Let E = c with the 1-norm and ϕ : E k be given by (x n ) x n, then ϕ 1. Also, for x = e 1, we have x = 1 and ϕ(x) = 1, so that ϕ = 1 (iv) Define T : L 1 [, 1] L 1 [, 1] by T (f)(x) = x (a) Note that T is well-defined because 1 1 x 1 T (f)(x) dx = f(t)dt f(t)dt x f(t) dt (b) This also proves that T is bounded with T f(t) dt = f 1 17

18 (c) To prove T = 1, we set then f n 1 = 1, and Hence, T (f n )(x) = T (f n ) 1 = 1/n x ntdt + Thus, T (f n ) 1 1, and so Thus proves that T = 1 f n = nχ [,1/n] nχ [,1/n] (t)dt = 1 1/n { 1 : x 1/n nt : x < 1/n dt = n 1 2n n = 1 1 2n T = sup{ T (f) 1 : f 1 = 1} 1 (v) Let E = F = C[, 1] with the sup norm. Fix K C([, 1] 2 ) and consider T : E F given by T (f)(x) = 1 K(x, t)f(t)dt (a) To prove that T is well-defined, given ɛ >, we use the uniform continuity of K to conclude that δ > such that Hence, x y < δ K(x, t) K(y, t) < ɛ t [, 1] T (f)(x) T (f)(y) and so T (f) C[, 1] (b) Now for any x [, 1] T (f)(x) Since the function 1 1 K(x, t) K(y, t) f(t) dt < ɛ 1 1 K(x, t) f(t) dt f K(x, t) dt x 1 is continuous on [, 1], it follows that M := sup x [,1] 1 K(x, t) dt and that T is continuous with T M. K(x, t) dt < f(t) dt 18

19 (c) To prove T = M, fix ɛ > and choose x [, 1] such that Then consider M ɛ = = K(x, t) dt = M ( K(x, t) ɛ) dt K(x, t) 2 ɛ 2 K(x, t) + ɛ dt K(x, t) K(x, t) K(x, t) + ɛ dt = T (f ɛ )(x ) where T (f ɛ ) f ɛ (t) = K(s, t) K(s, t) + ɛ Now note that f ɛ 1, and so C[, 1] M ɛ T f ɛ T This is true for all ɛ >, so M T as required. This T is called an integral operator with kernel K. 4. Completeness 4.1. Definition: (i) A complete NLS is called a Banach Space (ii) An inner product space that is complete (with respect to the norm induced by the inner product is) is called a Hilbert space 4.2. Theorem: For 1 p, E = k n with p is a Banach space Proof. Assume p = +, as the case when p < is similar. Suppose x m = (x m 1, x m 2,..., x m n ) is Cauchy, then for any 1 i n, the sequence (x m i ) k is Cauchy. Since k is complete, y i k such that x m i y i for all 1 i m. Thus, for any ɛ >, N i N such that x m i y i < ɛ m N i Let N = max{n i : 1 i n}, then for all m N and so x m y in norm. sup x m i y i < ɛ x m y < ɛ 1 i n 19

20 (End of Day 5) 4.3. Theorem: For 1 p, l p is a Banach space. Proof. We prove this if p <. The p = + case is similar. Now suppose x k = (x k 1, x k 2,..., x k n,...) is a Cauchy sequence in l p (i) Then for any n N, note that x k n x m n x k x m p and so (x k n) k is Cauchy. Since k is complete, y n k such that x k n y n n N (ii) WTS: y = (y n ) l p : Since (x k ) is Cauchy, it is bounded, so R > such that x m p R m N. For any fixed j N, this implies ( j ) 1/p x m n p R n=1 Now let m in the finite sum to conclude that ( j ) 1/p y n p R n=1 This is true for all j N, so ( ) 1/p y n p n=1 R y l p (iii) WTS: x k y in p : For any ɛ >, N N such that Now if j N is fixed, consider x k x m p < ɛ k, m N ( j ) 1/p x k n x m n p x k x m p < ɛ n=1 Now let m in the finite sum, to obtain ( j ) 1/p x k n y n p ɛ n=1 2

21 This is true for all j N, so ( ) 1/p x k n y n p ɛ and so x k y p n= Theorem: For 1 p <, (c, p ) is dense in l p. In particular, (c, p ) is not complete. Proof. Fix x = (x n ) l p, ɛ >, then N N such that n=n x n p < ɛ Hence if y = (x 1, x 2,..., x N,,,...) c, then x y p p < ɛ 4.5. Theorem: L [a, b] is a Banach space. Proof. Let (f n ) L [a, b] be a Cauchy sequence, then define A k := {x [a, b] : f k (x) > f k } and B k,m := {x [a, b] : f k (x) f m (x) > f k f m } By Lemma 2.14, each of these sets has measure zero, so ( ) ( ) C := A k B k,m k=1 k,m=1 also has measure zero. Furthermore, on D := [a, b] \ C, each f k is bounded and uniformly Cauchy. So for any x D, f k (x) f m (x) f k f m implies that {f m (x)} k is a Cauchy sequence in k. f : D k by f(x) = lim f m(x) m We may extend f to all of [a, b] by defining f on C. Hence, we may define 21

22 Furthermore for ɛ >, N N such that f k f m < ɛ for all k, m N. Hence for any fixed x X, and k N fixed f k (x) f m (x) < ɛ f k (x) f(x) ɛ Hence, f f k is bounded on D and so f f k L [a, b]. Since f = (f f k ) + f k, it follows that f L [a, b]. Finally, the above inequality also proves that and so f k f in L [a, b]. f k f ɛ k N 4.6. Theorem: (C[a, b], ) is closed in L [a, b]. In particular, (C[a, b], ) is a Banach space. Proof. Real Analysis I Definition: (i) Let E be a NLS and (x n ) E, then we say that the series n=1 x n is convergent iff the sequence (s n ) of partial sums defined by s n := n k=1 x k converges to a point in E. In other words, s E such that for any ɛ >, N N such that ( n x k ) s < ɛ n N k=1 (ii) Let E be an NLS and (x n ) E, then we say that the series n=1 x n is absolutely convergent if x n < in R n= Theorem: An NLS E is a Banach space iff every absolutely convergent series is convergent in E. Proof. Let E be a Banach space and (x n ) E such that x n < n=1 22

23 Let s n denote the n th partial sum, then it suffices to show that (s n ) is a Cauchy sequence. So if ɛ >, N N such that Hence if n, m N with n > m, then s n s m = n k=m+1 n=n x n < ɛ x k Thus, the series n=1 x n is convergent. n k=m+1 x n k=m+1 x n < ɛ Conversely, suppose every absolutely convergent series is convergent in E, choose a Cauchy sequence (x n ) E. Since (x n ) is Cauchy, it suffices [Why?] to prove that (x n ) has a convergent subsequence. To this end, for each j N, N j N such that x k x l < 1 k, l N 2 j j By induction, we may choose N 1 < N 2 <..., and so we obtain a subsequence (x Nj ) such that x Nj+1 x Nj < 1 2 j j N Thus, By hypothesis, the series x Nj+1 x Nj < j=1 j=1 x Nj+1 x Nj converges in E. But consider the partial sum of this series s n = n x Nj+1 x Nj j=1 = x Nn+1 x N1 So if (s n ) converges, so does (x Nj ) 4.9. (Riesz-Fischer): For 1 p <, L p [a, b] is a Banach space. Proof. (See [Folland, Theorem 6.6]) If (f n ) L p [a, b] is such that M := f n p < n=1 23

24 (i) Define g k = k f n and g = n=1 f n Then by Minkowski s inequality g k p M, and so by Fatou s lemma b a g p lim inf b a n=1 g p k M p In particular, g(x) < a.e. Hence, we may write f = n=1 and this converges a.e. (we may define f otherwise). (ii) Now note that f is measurable, and f g, so f L p [a, b] by the above inequality. (iii) Now define s k = k n=1 then s k L p [a, b] and we WTS: s k f p. Note that s k f pointwise and f s k p (2g) p L 1 [a, b] so by the Dominated Convergence theorem, f s k p p = b a f n f n f s k p 4.1. Remark: Let X be a metric space (i) If A X is a set, then for any x X, define d(x, A) := inf{d(x, y) : y A} (End of Day 6) and note that the function x d(x, A) is continuous, and d(x, A) = iff x A. (ii) If A, B X are two disjoint closed sets, then define f(x) := d(x, A) d(x, A) + d(x, B) Then f : X R is continuous, and satisfies f on A and f 1 on B 24

25 4.11. Lemma: Let K [a, b] be a compact set, then g C[a, b] such that Proof. For each n N, consider g 1 on K and g < 1 on [a, b] \ K G n = {x [a, b] : d(x, K) 1/n} Then G n is closed and G n K =. Hence by Remark 4.1, g n C[a, b] such that g n 1 on K and g n on G n Now the function satisfies the required properties. g := n=1 1 2 n g n Theorem: If 1 p <, then (C[a, b], p ) is dense in L p [a, b]. In particular, (C[a, b], p ) is not complete. Proof. Let f L p [a, b], ɛ >, we WTS: g C[a, b] such that f g p < ɛ. (i) Suppose f = χ K where K [a, b] is compact: Let g C[a, b] be as in Lemma 4.1, then g n C[a, b] and g n f pointwise Furthermore, g n f p 2 p L 1 [a, b] for all n N, and hence the Dominated Convergence theorem, g n f p p = b a g n f p (ii) Suppose f = χ E where E [a, b] measurable: Then for ɛ >, K E compact such that m(e \ K) < ɛ. Hence, Now apply part (i) χ K χ E p p < ɛ (iii) Suppose f = n i=1 α iχ Ei L 1 [a, b] is a simple function: Then apply part (ii) to each E i and take a linear combination (iv) Suppose f L p [a, b]: Then choose a sequence of simple functions (s n ) such that s n s n+1 f pointwise. Since s n f p (2f) p L 1 [a, b], by Dominated convergence theorem, s n f p Now apply part (iii) to s N for N large enough. 25

26 (v) Suppose f L p [a, b] is real-valued, then write it as f = f + f and apply part (iv) to each of f + and f. (vi) Suppose f L p [a, b] is complex valued, then apply part (v) to the real and imaginary parts of f Theorem: Let E {} be any NLS. (i) If F is complete, then B(E, F ) is complete. (ii) In particular, E is a Banach space. Proof. Suppose (T n ) B(E, F ) is a Cauchy sequence, then for any x E, the inequality T n (x) T m (x) T n T m x implies that (T n (x)) F is a Cauchy sequence. Hence we may define T : E F by T (x) = lim T n (x) and it is clear that T is linear. Furthermore, since (T n ) is Cauchy, M > such that T n M for all n N and hence T (x) M x x E and so T B(E, F ). We now WTS: T n T. To this end, choose ɛ > and N N such that Then for any x E, and n N fixed T n T m < ɛ n, m N T (x) T n (x) = lim m T m(x) T n (x) lim m T m T m x ɛ x and so T T n ɛ for all n N Definition: A topological space E is said to be separable if it has a countable dense subset Remark: Let E be an NLS (i) If E has a dense, separable subspace, then E is separable. (ii) If E contains an uncountable family of disjoint open sets, then E is not separable Examples: (i) (k n, p ) is separable since Q n R n and (Q Q) n C n are dense. 26

27 (ii) c is separable since we may choose sequences with only rational entries. This would give a subset n=1q n which is countable and dense in c (iii) By Theorem 4.4, and Example (ii), l p is separable if 1 p < (iv) l is not separable Proof. For each subset A N, choose χ A l. Then if A B, then χ A χ B = 1 Thus, {B(χ A ; 1/3) : A N} forms an uncountable family of disjoint open sets. (v) (C[a, b], p ) is separable since polynomials with rational coefficients form a dense subset of C[a, b] (Note: They are dense in ), but since f g p f g, it follows that they are dense in p ) as well. (vi) By Theorem 4.11 and Example (v), L p [a, b] is separable for 1 p < (vii) L [a, b] is not separable. Proof. For each t [a, b], consider f t = χ [a,t], then if s t f s f t = 1 and so once again we obtain an uncountable family of disjoint open sets. 5. Finite Dimensional Spaces 5.1. Definition: Let E be a vector space and 1 and 2 be two norms on E. We say that these norms are equivalent (In symbols, 1 2 ) if they generate the same metric topologies on E. Note that this is an equivalence relation on the class of norms on E. (End of Day 7) 5.2. Theorem: Two norms 1 and 2 are equivalent iff α, β > such that α x 1 x 2 β x 1 x E ( ) Proof. Let τ 1 and τ 2 be the topologies generated by 1 and 2 respectively. (i) Suppose 1 2, then B 1 = {x E : x 1 < 1} τ 1 B 1 τ 2 In particular, since B 1, δ > such that B 2 = {x E : x 2 < δ} B 1 27

28 Now for any y E, consider z := δ y 2 y 2 Then z 2 < δ, so z 1 < 1, and hence By symmetry, β such that δ 2 y 1 < y 2 y E y 2 < β y 1 y E as well. (ii) Now suppose ( ) holds. Then choose U τ 1. We WTS: U τ 2. So for any x U, r > such that B 1 (x, r) = {y E : y x 1 < r} U Let V := B 2 (x, αr), then for any y V, y x 1 y x 1 α and so V U. This is true for every x U, and so U τ 2.Hence, τ 1 τ 2. By symmetry, τ 2 τ 1 as well. < r 5.3. Remark/Examples: (i) Let E = k n, and consider 1 and on E. Note that for any x k n, and so 1 x x 1 n x (ii) Let E = c and consider 1 and on E. Then x x 1 for all x E, but if x k = (1, 1,..., 1,,,...), then and so 1. x k 1 = k and x k = 1 (iii) Suppose E is a vector space with two equivalent norms 1 and 2. If E is complete wrt 1, then it is complete wrt 2. [Check!] (iv) Hence, if E = C[a, b] then for any 1 p <, p Lemma: Let E = (k n, 1 ), then every closed, bounded subset of E is compact. (HW) 28

29 5.5. Theorem: Any two norms on a finite dimensional vector space are equivalent. Proof. Let (E, E ) be a finite dimensional NLS with basis {e 1, e 2,..., e n }. For any x = n i=1 x ie i E, define x 1 := n x i i=1 and note that 1 is a norm on E. Since the equivalence of norms is an equivalence relation, it suffices to show that E 1 (i) If D := max{ e j }, then x E D x 1. (ii) We WTS: C > such that x E C x 1 for all x E. Dividing both sides by x 1, it suffices to prove that C > such that x E C x E such that x 1 = 1 Suppose not, then α k = (α k 1, α k 2,..., α k n) k n such that α k 1 = 1 and α k i e i E < 1/k Now, (α k ) k n is a bounded in 1, so by Lemma 5.4, there is a subsequence (α n k ) and α k n such that α n k α 1 By continuity of the norm, it follows that α 1 = 1 and n α i e i E = i=1 n α i e i = i=1 But this contradicts the fact that the {e i } are linearly independent Corollary: Let E be finite dimensional, and F any NLS. Any linear operator T : E F is continuous. Proof. Define a norm on E by x := x + T (x) It is easy to see that this is a norm. Thus by Theorem 5.5, M > such that T (x) x M x x E and so T is continuous by Theorem

30 5.7. Definition: A linear operator T : E F is said to be a (i) topological isomorphism if T is an isomorphism of vector spaces and a homeomorphism (ie. T and T 1 are both continuous). (ii) isometric isomorphism is a topological isomorphism that is isometric. We write E = F if they are isometrically isomorphic Corollary: Let E be a finite dimensional NLS with dim(e) = n, then there is a topological isomorphism T : (k n, 1 ) E. Proof. Choose any bijection T : k n E and apply Corollary 5.6 to both T and T Corollary: Let E be a finite dimensional NLS, then E is a Banach space. Proof. Let T : (k n, 1 ) E be as in Corollary 5.8. Note that (k n, 1 ) is a Banach space. Thus, if (x n ) E is Cauchy, then (T 1 (x n )) k n is Cauchy. Thus T 1 (x n ) x k n, whence x n T (x) in E Corollary: Let E be an NLS and F < E be a finite dimensional subspace, then F is closed in E Definition: A topological space E is said to be locally compact if every point in E has an open neighbourhood with compact closure (Riesz Lemma): Let E be an NLS and F < E be a proper closed subspace. Then for any < t < 1, x t E such that x t = 1 and d(x t, F ) t Proof. Since F < E is a closed proper subspace, x E \ F, whence d := d(x, F ) > For < t < 1, d/t > d so y F such that Now x t := x y x y satisfies d(x t, F ) = x y d/t 1 x y d(x, F ) t (End of Day 8) Theorem: Let E be an NLS, then TFAE: (i) E is finite dimensional (ii) Every closed bounded set in E is compact. 3

31 (iii) E is locally compact. (iv) B[, 1] is compact (v) B[, 1] = {x E : x = 1} is compact. Proof. (i) (ii): If dim(e) = n, let T : (k n, 1 ) E be as in Corollary 5.8. Let B E be a closed bounded set, then T 1 (B) is closed and bounded in k n (Why?). Hence, T 1 (B) is compact by Lemma 5.4. Thus, B = T (T 1 (B)) is compact since T is continuous. (ii) (iii): If every closed, bounded set in E is compact, then in particular, B[x, 1] is compact for all x E. Since it follows that E is locally compact. B[x, 1] = B(x, 1) (iii) (iv): If E is locally compact, then open U such that U and U is compact. Hence r > such that B(, r) U B[, r] U and so B[, r] is compact. But B[, r] = rb[, 1] and the map x rx is a homeomorphism, so B[, 1] is compact. (iv) (v) : Obvious since B[, 1] is closed subset of B[, 1]. (v) (i): Suppose B[, 1] is compact and E is infinite dimensional, then choose x 1 E with x 1 = 1, and let F 1 = span{x 1 } Then F 1 < E is a closed proper subspace, and hence x 2 E such that x 2 = 1 and d(x 2, F 1 ) 1/2 Thus proceeding, we get a sequence {x n } such that x n = 1 and d(x n, F n 1 ) 1/2 where F n 1 = span{x 1, x 2,..., x n 1 }. In particular, x n x m 1/2 n m Hence, (x n ) B[, 1] cannot have a convergent subsequence [Why?]. 31

32 6. Quotient Spaces Notation: Throughout this section, E is an NLS, F < E and π : E E/F is the natural quotient map x x + F Theorem: Define p : E R + by Then (i) p defines a semi-norm on E. p(x) := d(x, F ) = inf{ x y : y F } (ii) If F is closed, p induces a norm on E/F given by Proof. By HW 1.3, it suffices to prove (i) x + F := p(x) (i) Clearly, p(x) and p(x) = iff x F. (ii) Now if α k, then consider d(αx, F ) = inf{ αx y : y F } If α =, then αx = F, so p(αx) =. If α, then the map y αy is a bijection on F, and hence p(αx) = inf{ αx αy : y F } = α p(x) (iii) Finally, if x 1, x 2 E, then for any y 1, y 2 F, we have Hence, (x 1 + x 2 ) (y 1 + y 2 ) x 1 y 1 + x 2 y 2 p(x 1 + x 2 ) x 1 y 1 + x 2 y 2 Taking infimums independently, we get that p(x 1 + x 2 ) p(x 1 ) + p(x 2 ) Theorem: If F is closed, then π is continuous and π = 1. Proof. Note that for x E, π(x) = x + F = d(x, F ) x since F. Hence π is continuous and π 1. Furthermore, by Riesz Lemma, for each < t < 1, x t E such that x t = 1 and Hence, π 1. π(x t ) t 32

33 6.3. Lemma: If F is closed, then, for any x E, r > π(b E (x, r)) = B E/F (π(x), r) Proof. (i) If z B E (x, r) then x z < r. Since F, this implies that π(x) π(z) = π(x z) x z < r (ii) Now suppose z + F B E/F (x + F, r), then π(z) π(x) = π(z x) < r So y F such that z x y < r. So if w := z y, then w B E (x, r) and z + F = π(z) = π(z y) = π(w) 6.4. Theorem: If F is closed, then (i) A set U E/F is open iff π 1 (U) X is open (ii) π is an open map. Proof. (i) If U E/F is open, then π 1 (U) E is open since π is continuous by 6.2. Conversely, suppose π 1 (U) E is open. Choose π(x) U, then it follows that By Lemma 6.3 x π 1 (U) r > such that B E (x, r) π 1 (U) B E/F (π(x), r) π(π 1 (U)) = U since π is surjective, and so U is open. (ii) If U E is open, then consider V = π(u) and note that π 1 (V ) = U + F = {u + y : u U, y F } For each y F, the map u u + y is a homeomorphism, and hence π 1 (V ) = U + F = y F(U + y) is an open set. So by part (i), V E/F is open. (End of Day 9) 6.5. Corollary: If F is closed and W < E finite dimensional, then W + F is closed in E. 33

34 Proof. Since dim(π(w )) dim(w ) < it follows from Corollary 5.1 that π(w ) < E/F is closed. Since π is continuous, is closed in E. W + F = π 1 (π(w )) 6.6. Theorem: If E is a Banach space, then E/F is a Banach space. Proof. By Theorem 4.8, it suffices to prove that every absolutely convergent series is convergent. So suppose (x n ) E such that We WTS that the sequence is convergent in E/F. M := π(x n ) < n=1 k s k := π(x n ) n=1 For each n N, y n F such that π(x n ) + 1/2 n > x n + y n Hence, By Theorem 4.8, the sequence x n + y n < n=1 k t k := (x n + y n ) n=1 converges to a point t E. But since π is continuous s k = π(t k ) π(t) E/F 6.7. (First Isomorphism Theorem): Let T : E H be a bounded linear operator and F := ker(t ). Then (i) F is a closed subspace of E (ii) There exists a unique injective bounded operator T : E/F H such that T π = T. Furthermore, T = T 34

35 Proof. Since F = T 1 ({}), F is closed. Let T : E/F H be the injective map given by T (x + M) := T (x) Then T is well-defined and linear. Furthermore, for any x E, y M, we have T (x + M) = T (x) = T (x y) T x y This is true for all y M, and hence T (x + M) T x + M, whence T is bounded and T T However, and so T T T (x) = T (x + M) T x + M T x 35

36 II. Hilbert Spaces 1. Orthogonality 1.1. Definition/Remark: Let H be an Hilbert space. (i) We say that two elements x, y H are orthogonal if x, y =. If this happens, we write x y (ii) For two subsets A, B H, we write A B if x y for all x A, y B (iii) For any set A H, write A = {x H : x y y A} If A = {x}, then we simply write x := {x} (iv) For each y H, the linear functional ϕ y : H k given by x x, y is continuous. Hence, is a closed subspace of H (v) For any A H A = y A ker(ϕ y ) A A = {} A (A ) 1.2. Theorem: Let x, y H, then (i) (Polar Identity): x + y 2 = x 2 + 2Re x, y + y 2 (ii) (Pythagoras Theorem): If x y, then x + y 2 = x 2 + y 2 (iii) (Parallelogram law): x + y 2 + x y 2 = 2( x 2 + y 2 ) Proof. Expand x + y 2 and x y 2 and use linearity/conjugate-linearity of the inner product Example: For 1 p 36

37 (i) Let E = l p, and Then x = e 1 + e 2 and y = e 1 e 2 x + y p = 2e 1 p = 2 and x y p = 2e 2 p = 2 { 2 1/p : 1 p < x p = y p = 1 : p = Hence, the parallelogram law holds iff 8 = 4 2 2/p p = 2 (ii) Let E = C[, 1] and Then f(t) = t and g(t) = 1 t f + g p = 1 and f g p = f p = g p = And once again, parallelogram law holds iff p = 2 { 1 (1+p) 1/p : 1 p < 1 : p = 1.4. Definition: A subset A H is said to be convex if, for any x, y A, the line 1.5. Examples: (i) Open/Closed unit ball (ii) Any subspace [x, y] := {tx + (1 t)y : t 1} A (iii) If A is convex, then so is A + x for any x H (End of Day 1) 1.6. (Best Approximation Property) Let A H be a non-empty, closed, convex set, and x H, then!x A such that x x = d(x, A) = inf{ y x : y A} This x A is called the best approximation of x in A Proof. Since d(x, A) = d(, A x), replacing A by A x, we may assume WLOG that x =. 37

38 (i) Existence: By definition, (y n ) A such that WTS: (y n ) is Cauchy. d := d(, A) = lim n y n By the parallelogram law, 2 y n y m 2 = 1 2 ( y n 2 + y m 2 ) y n + y m 2 Since A is convex, (yn+ym) A, and so 2 y n + y m 2 For ɛ >, choose N N such that Hence, for n, m N y n y m 2 2 d 2 y n 2 < d 2 + ɛ n N and so y n y m < 2 ɛ for all n, m N. 2 < 1 2 (2d2 + 2ɛ) d 2 = ɛ 2 Thus, (y n ) is Cauchy, and hence convergent in H. Since A is closed, x A such that y n x whence by Remark I.2.2 d = lim n y n = x (ii) Uniqueness: Suppose x, x 1 A such that x = x 1 = d then by convexity, (x + x 1 )/2 A, and hence d 1 2 (x + x 1 ) 1 2 ( x + x 1 ) d and so 1(x 2 + x 1 ) = d. The parallelogram law then implies 2 2 d 2 = x + x 1 2 = d 2 x x 1 2 and so x = x 1 38

39 1.7. Theorem: Let M < H be a closed subspace and x H. Then x M is the best approximation of x in M iff x x M Proof. (i) Suppose x is the best approximation of x to M: We WTS: x x, y = for all y M. It suffices to prove this when y = 1, so fix y M with y = 1 and let α := x x, y Then z := x + αy M, so x x 2 x z 2 = (x x ) αy 2 = x x 2 + αy 2 2Re x x, αy = x x 2 + α 2 y 2 2 α 2 = x x 2 α 2 Hence, α 2 =, whence x x y. (ii) Conversely, suppose x x M: We WTS: x x = d(x, M). In other words, we WTS: x x x y y M But then for any y M, we have (x y) M, so by Pythagoras theorem x y 2 = (x x ) + (x y) 2 = x x 2 + x y 2 x x Definition: Let M < H be a closed subspace. For x X, let P M (x) M denote the best approximation of x in M. ie. x P M (x) = d(x, M) x P M (x) M The map P M : H M H is called the orthogonal projection of H onto M Theorem: Let P : H M be the orthogonal projection onto a closed subspace M < H. Then (i) P is a linear transformation (ii) P is bounded and P 1. If M {}, then P = 1 (iii) P 2 := P P = P (iv) ker(p ) = M and Image(P ) = M Proof. (i) Let x 1, x 2 H and α k, then set z = x 1 + αx 2 and z = P (x 1 ) + αp (x 2 ) 39

40 Then we WTS: P (z) = z. For any y M z z, y = x 1 P (x 1 ), y + α x 2 P (x 2 ), y = Hence, z z M, so by Theorem 1.7, P (z) = z (ii) For any x H, x = (x P (x)) + P (x) and (x P (x)) P (x). Hence by Pythagoras theorem x 2 = x P (x) 2 + P (x) 2 P (x) 2 Hence, P is continuous and P 1. (iii) Note that if y M then P (y) = y. Hence, if x M then y = P (x) M, so P (P (x)) = P (x) x M (iv) If P (x) =, then x = x P (x) M. Hence ker(p ) M. Conversely, if x M, then x M and so x P (x) = x whence P (x) = Theorem: Let M < H be a closed subspace, then (M ) = M Proof. (i) If x M, then for any y M, x, y =. Hence, x (M ) M (M ) (ii) Conversely, if x (M ), then let x = P M (x) M, then x M and x x M, so x, x x = and x, x x = x x 2 = x x, x x = whence x = x M Theorem: Let M < H be a closed subspace, and P = P M. Since M is a closed subspace, let Q = P M. Then (i) P Q = QP = (ii) P + Q = I where I : H H denotes the identity map. Proof. (i) If x H, then Q(x) M. By Theorem 1.9, ker(p ) = M, and hence P Q(x) = Similarly, P (x) M and M (M ) = ker(q) and hence QP (x) = 4

41 (ii) If x H, then x = (x P (x))+p (x) = Q(x)+(x Q(x)) (x P (x)) Q(x) = (x Q(x)) P (x) But x P (x), Q(x) M, x Q(x) (M ) = M by Theorem 1.1, and P (x) M and hence (x P (x)) Q(x) M and (x Q(x)) P (x) M and since M M = {}, it follows that x P (x) Q(x) = x = (P + Q)(x) Corollary: Let M < H be a closed subspace, then H = M M Proof. Let P : H M and Q : H M be the orthogonal projections onto M and M respectively. Then, by Theorem 1.11, H = (P + Q)(H) = P (H) + Q(H) = M + M Furthermore, M M = {}, and so H = M M Corollary: If M < H is any subspace, then M is dense in H iff M = {} Proof. HW (End of Day 11) 2. Riesz Representation theorem 2.1. Definition: If y H, define ϕ y : H k by x x, y Then ϕ y H and ϕ y = y. Hence, we get a map : H H given by y ϕ y which is an isometry. Note that (αy) = α (y) and (y 1 + y 2 ) = (y 1 ) + (y 2 ) We say that is a conjugate-linear isometry (Riesz Representation Theorem): For any ϕ H,!y H such that ϕ(x) = x, y x H Hence, : H H is a conjugate-linear isometric isomorphism. 41

42 Proof. Fix ϕ H, then M := ker(ϕ) is a closed subspace of H. Since ϕ, M H, and so by II.1.13, M {} Choose y M such that ϕ(y ) = 1. Now for any x H with α := ϕ(x), note that ϕ(x αy ) = x αy M Hence, Hence if y = y / y 2, then As for uniqueness, if ϕ y = ϕ z, then and so y = z. = x αy, y = x, y ϕ(x) y 2 ϕ(x) = x, y = ϕ y (x) ϕ y z = y z = ϕ y z = 2.3. Remark: If H is a Hilbert space, then H is a Hilbert space under the inner product (ϕ y, ϕ z ) := z, y 2.4. Theorem: Let E be an NLS, F a Banach space and E < E be dense. If T B(E, F ), then!t B(E, F ) such that T E = T and T = T T is called the norm-preserving extension of T. Proof. (i) For any x E, choose a sequence (x n ) E such that x n x. Then (x n ) is Cauchy, so (T (x n )) F is Cauchy. Since F is complete, y F such that T (x n ) y. Define T : E F by T (x) := lim n T (x n ) (ii) WTS: T is well-defined So suppose (z n ) E is another sequence such that z n x, then T (z n ) T (x n ) T z n x n and hence, lim T (z n ) = lim T (x n ) 42

43 (iii) T is linear: If x n x and y n y, then Since T is linear, it follows that x n + y n x + y T (x + y) = lim T (x n + y n ) = lim T (x n ) + lim T (y n ) = T (x) + T (y) and similarly, T (αx) = αt (x) for all α k. (iv) T is bounded: If x n x, then x n x, and so T (x) = lim n T (x n ) T lim n x n = T x and so T is bounded with T T. However, T is an extension of T, and so T T (v) Uniqueness: If T 1, T 2 B(E, F ) are two bounded linear operators such that T 1 E = T = T 2 E then for any x E, choose (x n ) E such that x n x, then T 1 (x) = lim n T 1 (x n ) = lim n T (x n ) = lim n T 2 (x n ) = T 2 (x) 2.5. Corollary: If E is a NLS, and E < E is dense in E, then the map E E given by ϕ ϕ E is an isometric isomorphism of Banach spaces. Proof. The map S : ϕ ϕ E is clearly well-defined and linear. Furthermore, by Theorem 2.4, for any ψ E,!ϕ E such that ϕ E = ψ Hence, S is surjective. Also, since ϕ = ψ, it follows that S is isometric, and hence injective Corollary: Let M < H be a subspace of H (not necessarily closed) and let ϕ : M k be a bounded linear functional. Then ψ H such that ψ M = ϕ and ψ = ϕ (We say that ψ is a norm-preserving extension of ϕ. It may not be unique - see HW 4) 43

44 Proof. Let ψ M, then by Corollary 2.5, ϕ : M k linear such that ϕ M = ψ and ϕ = ψ Since M < H, it is a Hilbert space, so by the Riesz Representation theorem, y M such that ϕ (x) = x, y x M Now simply define ϕ : H k by ϕ(x) = x, y x H Then clearly, ϕ is an extension of ϕ and hence of ψ. Furthermore, ϕ = y = ϕ = ψ 2.7. Remark: If M < H, then we have just proved that the restriction map S : H M given by ϕ ϕ M is surjective. Furthermore, it is injective iff M = H. (HW) 3. Orthonormal Bases 3.1. Definition: Let H be a Hilbert space and A H, (i) A is said to be orthogonal if x y for all distinct x, y A (ii) A is said to be orthonormal if it is orthogonal and x = 1 for all x A. (iii) A maximal orthonormal set is called an orthonormal basis (ONB) of H. Note: An ONB may not be a vector space basis for H (End of Day 12) 3.2. Lemma: Every orthogonal set is linearly independent. Proof. If A is orthonormal and {x 1, x 2,..., x n } A satisfy n α i x i = i=1 Then for any fixed j N, n α i x i, x j = α j = i=1 44

45 3.3. Lemma: Let A H be an orthonormal set, then TFAE: (i) A is an ONB (ii) A = {} (iii) span(a) is dense in H Proof. Let A H orthonormal (i) (ii): Suppose A is an ONB, and A {}, then choose x A such that x = 1, then A {x} is an orthonormal set. This contradicts the maximality of A (ii) (iii): Suppose A = {}, then span(a) A span(a) = {} So by Corollary II.1.13, span(a) is dense in H (iii) (i): Suppose span(a), then we WTS: A is a maximal orthonormal set. Suppose x A, then x span(a). By continuity of the inner product (by Cauchy-Schwartz), x span(a). Hence, x H and so x =. Thus, there is no x H of norm 1 such that x A. Hence, A is a maximal orthonormal set Theorem: If A H is any orthonormal set, then there is an ONB B H that contains A. Proof. HW (Use Zorn s Lemma) 3.5. Examples: (i) Let H = (k n, 2 ), then the standard basis is an orthonormal basis. (ii) Let H = l 2, then {e n : n N} is an ONB Proof. Clearly, A = {e n : n N} is orthonormal. Furthermore, if x = (x n ) A, then x n = x, e n = for all n, and hence A = {} (iii) Let H = L 2 [, 2π] and k = C. For n Z, define e n (t) = 1 2π e int Then e n, e m = 1 2π e i(n m)t dt 2π If n = m, then clearly, this is 1. Otherwise, we get e n, e m = 1 2π We will prove later that it forms an ONB. e i(n m)t 2π n m = 45

46 3.6. (Gram-Schmidt Orthogonalization): Let {x 1, x 2,..., x n } H be linearly independent, then define {u 1, u 2,..., u n } inductively by u 1 := x 1 and u j = x j j 1 i=1 Then {u 1, u 2,..., u n } is an orthogonal set and x j, u i u i, u i u i ( ) span({u 1, u 2,..., u n }) = span({x 1, x 2,..., x n }) Proof. We proceed by induction on n, since this is clearly true if n = 1. If n > 1, suppose {u 1, u 2,..., u n 1 } is an orthogonal set such that span({u 1, u 2,..., u n 1 }) = span({x 1, x 2,..., x n 1 }) Then if u n is given by ( ), then clearly, u n span({x 1, x 2,..., x n }) and u n since {x 1, x 2,..., x n } is linearly independent. Also, u n, u j = j < n and so {u 1, u 2,..., u n } is orthogonal. independent, and hence In particular, {u 1, u 2,..., u n } is linearly span({u 1, u 2,..., u n }) = span({x 1, x 2,..., x n }) since both spaces have the same dimension Corollary: If H is a Hilbert space and {x n : n N} is a linearly independent set, then an orthonormal set {e n : n N} such that, for all n N span({e 1, e 2,..., e n }) = span({x 1, x 2,..., x n }) 3.8. Example: Let H = L 2 [ 1, 1] and x n (t) = t n, then e n = [ ] 1/2 1 (2n + 1) P n (x) where P n (x) = 1 ( ) n d (t 2 1) n 2 2 n n! dt The polynomials P n are called Legendre polynomials. And since span({e n }) = span({x n }) it follows by Weierstrass approximation Theorem and Theorem I.4.12 that span({e n }) = H and hence {e n } form an ONB for H Theorem: Let {e 1, e 2,..., e n } be an orthonormal set in H and M = {e 1, e 2,..., e n }. Then, for any x H n P M (x) = x, e k e k k=1 46

47 Proof. Let n x = x, e k e k k=1 then for any 1 j n, it follows that x x, e j = Hence, x x M and so P M (x) = x by Theorem 1.7 (End of Day 13) 3.1. (Bessel s inequality): If {e n : n N} is an orthonormal set and x H, then x, e n 2 x 2 Proof. For each n N, write n=1 x n = x n x, e i e i i=1 Then x n e i for all 1 i n, so by Pythagoras theorem, x 2 = x n 2 n 2 + x, e i e i i=1 n = x n 2 + x, e i 2 i=1 n x, e i 2 This is true for all n N and hence we get the result. i= (Riemann-Lebesgue Lemma): Let {e n : n N} be an orthonormal set, and x H, then lim n x, e n = Corollary: Let A be an orthonormal set in H and x H, then is a countable set. {e A : x, e } 47

48 Proof. For each n N, define A n = {e A : x, e 1/n} If {e k } N k=1 A n, then by Bessel s inequality, N n = N 2 k=1 and hence A n must be finite. But then 1 n N x, e 2 k 2 x 2 k=1 and so this set must be countable. {e A : x, e } = n=1a n Remark: Let {x α : α A} be a (possibly uncountable) set in an NLS E. We want to make sense of an expression of the form ( ) α A If A is countable, we can do this by enumerating A = N and writing x α n=1 However, if A is uncountable, we do the following: Define For each F F, we define a partial sum x n F := {F A : F is finite} s F := α F x α We say that the expression ( ) exists if s H with the property that for all ɛ >, F F finite such that s F s < ɛ F F, F F In other words, F is a partially ordered set under inclusion and {s F : F F} forms a net. Our requirement is that this net be convergent in E Corollary: Let A be an orthonormal set in H and x H, then x, e 2 x 2 e A 48

49 Proof. Clearly, the sum is the same as the sum over the set B = {e A : x, e } By Corollary 3.12, this set is countable, and so it reduces to a countable sum. Now the result follows from Bessel s inequality Lemma: Let A be an orthonormal set in H, then for any x H, x, e e converges in H. Proof. By Corollary 3.12, the set e A {e A : x, e } is countable. Denote this set by {e n : n N}, then so define x, e e = e A x n = x, e n e n n=1 n x, e i e i i=1 and it now suffices to prove that (x n ) is Cauchy. However, by Bessel s inequality, x, e n 2 x 2 < n=1 Hence if ɛ >, N N such that and hence if n > m, we have x n x m 2 = n=n x, e n 2 < ɛ n i=m+1 x, e i e i 2 = by Pythagoras theorem. Hence, if n, m N, then x n x m 2 and so (x n ) is Cauchy as required. n i=m+1 i=n x, e i 2 < ɛ x, e i 2 49