FUNCTIONAL ANALYSIS HAHN-BANACH THEOREM. F (m 2 ) + α m 2 + x 0

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1 FUNCTIONAL ANALYSIS HAHN-BANACH THEOREM If M is a linear subspace of a normal linear space X and if F is a bounded linear functional on M then F can be extended to M + [x 0 ] without changing its norm. Proof We first suppose that X is a vector space over the real numbers. We may suppose without loss of generality F =. We may define F (x 0 ) = α and F (m + λx 0 ) = F (m) + λα m M. We must have F (m) + λα m + λx 0 i.e. F (m) + α m + x 0 (m arbitrary m λ arbitrary If m m 2 M F (m ) + α m + x 0 F (m 2 ) + α m 2 + x 0 F (m 2 ) m 2 x 0 α F (m ) + m + x 0 (I) F (m ) F (m 2 ) = F (m m 2 ) m m 2 m + x 0 + m 2 + x 0 therefore α satisfying (I). Now if X = X(C) F (x) = G(g) + ih(x) if (x) = F (ix) = ig(x) H(x) therefore H(x) = G(ix) = G(ix) + ih(x) therefore F (x) = G(x) ig(ix) and G can be extended by the first part.

2 By Zorn s lemma there will be a maximal subspace N to which M can be extended and N = X, applying the theorem. We can embed X in X as follows: Let x X and define f(λx) = λ x. Then f =. f can be extended to the whole space without changing its norm. therefore x x x(f) = f(x) = x Also x(f) = f(x) f x therefore x x. Adjoint of an operator Let T be a continuous linear transformation from X Y. The adjoint T of T is a linear transformation from Y to X defined as follows: Let f Y. We define T (f) X by (T f)x = f(t X) T (f) = sup ft (x) x = f sup T x x = F T Therefore T is continuous and T T. Now T maps X to Y. If X is regarded as a subspace of X then T is an extension of T therefore T T T therefore T = T. Weak topology Let X be a normed vector space and let X be the dual of X. We define a topology on X, called the weak topology, by taking the sets V (x) f...f nε = {y X : f i (x) f i (y) < ε i =,..., n} as a basis of neighbourhoods of the point x, where f... f n are any functionals in X and ε is any positive number. 2

3 As all the f are continuous this set will be open in the original topology and so this topology is weaker than the original one. Example Let ξ = (x n ) l 2 Let f = (y n ) f(ξ) = x n y n = (ξ, η). ξ α 0 in the weak topology ( xi α, η) 0. let ε = (, 0, 0,...) ε 2 = (0,, 0,...) etc. ε m ε n = 2 m n. But for the weak topology this sequence converges to zero as ε n η) = y n 0 as y n 2 <. Example Consider the space of all real valued functions defined on [0 ]. Consider the topology given by f α f f α (x) f(x) for each x in [0 ]. Basic neighbourhoods: Given x... x n and ε > 0 N = {g : f(x i ) g(x i ) < ε i =... n Let C be the subspace of continuous functions. Let d(x) = { x irrational 0 x rational d C for this topology, for given x... x n we can find f C such that and so f N(d). f(x i ) = d(x i ) i =,..., n But no sequence of continuous functions converges to d in this topology for, given {f n } C and f n (x) d(x) at every x. Let H n = r n {x : f r (x) 2 }. H n is closed. H n contains no rational and so is nowhere dense, therefore n=h n is of the first category. But n=h n = irrationals - of second category. Theorem Suppose X is a Banach space, than the unit sphere of X compact in the weak topology. is 3

4 Proof For each x X define C x = {z : z x } c x is a compact set therefore C = x C x is compact. x X C x = set of all functions mapping X to the complex plane with the property that F (x) x. Hence the unit sphere of X can be regarded as a subspace and so will be compact as it is closed. Theorem If X is a Banach space, X is reflexive its unit sphere is weakly compact. Proof If X is reflexive X = X the weak topology of the unit sphere of X is the same as the weak topology which is compact by th previous theorem. Closed Graph Theorem T linear X T Y Let G(T ) = {(X, T (X)} X Y. If T is continuous G(T ) is closed. The theorem states that the converse is true. Lemma Let T be a bounded linear transformation of a Banach space X into a Banach sphere Y. If the image under T of the unit sphere S = S(0, ) is dense in some sphere U r = S(0, r) about the origin of Y than it includes the whole of U r. Proof Let δ > 0. We wish to define a sequence {y n } such that y n+ y n δ n S, y n+ y < δ n+ r () We define y 0 = 0. Suppose that y 0... y n have been defined. Since y S(y, δ n+ r) (y n + δ n U r ) this is a non-empty open subset of y n + δ n U r therefore there is an element y n+ of y n + δ n T (S ) which belongs to this set, and y n+ satisfies the conditions (). 4

5 y n y as n provided δ <. x n such that x n δ n S and T (x n ) = y n+ y n. We can define x = x n provided δ <. Since T is bounded T (X) = N lim T (x n ) N = lim N+ = y N x sumδ n = for allδ δ therefore x U r( δ) T (S ) for every δ and U r = δ U r( δ T (S ) Proof of Theorem Let N(x) = x + T (x) If {x n } is a Cauchy sequence for the norm N then it is a Cauchy sequence for X and also {T (x n )} is a Cauchy sequence for T x. Therefore x n x and T x n y as n as G(T ) is closed (x, y) G(T ) therefore y = T (x). N(x x n ) = x x n + T x T x n 0 as n Therefore X is a Banach space for the new norm N. Now the identity mapping from X with norm N to (X, ) is bounded since X N(x). If S denotes the unit sphere defined by N, it follows from Baire s category theorem that S is dense in some sphere U r about the origin defined by. Thus by the lemma applied to the identity mapping U r S i.e. if X < r N(x) < i.e. N(X) r x so T (x) N(x) x and so T is continuous. r Hilbert Space A pre-hilbert Space is a real or complex vector spade in which an inner product (x, y) is defined having the following properties. (i) (x, x) > 0 unless x = 0 5

6 (ii) (x, y) = (y, x) (iii) (x + y, z) = (x, z) + (y, z) (iv) λx, y) = λ(x, y) A pre-hilbert space can be normed by defining x = (x, x) 2. A Hilbert space is a pre-hilbert space which is complete for this norm. A Banach space is a Hilbert space x + y 2 + x y 2 = 2 x y 2 Schwarz inequality (x, y) x Y Proof (λx y, λx y) = λ 2 x 2 2Rλ(x, y) + y 2 2Rλ(x, y) λ 2 x 2 + y 2 Choose λ so that λ = Y X Hence the result. and argλ = arg(x, y). 2 Y (x, y) 2 y 2 X Minkowski Inequality x + y x + y Proof x + y 2 = (x + y, x + y) = X 2 + 2R(x, y) + y 2 x 2 + Y (x, y) ( x + Y ) 2 using Schwarz. Theorem A closed convex subset C of a Hilbert space contains a unique element of smallest norm. Proof Let d = inf{ x : x C}. Then {x n } C such that x n d. Since C is convex xn+xm 2 C therefore x n + x m 2d. 6

7 x n x m 2 = 2 x n x m 2 x n + x m 2 if n and m are sufficiently large. 2{ x n 2 d 2 } + 2{ x m d 2 } < ε Hence the sequence is a Cauchy sequence and has a limit point x which belongs to C as C is closed, and x = d. If y C and y = d then x + y 2d = x + y and so y = λx where λ > 0 y = λ x λ = therefore x = y. Theorem Let M be a closed subspace of a Hilbert space H. Then any x = x + x 2 where x M and x 2 perpendicular M (i.e. (x 2, y) = 0 for all y M). Proof Suppose x M. Let x 2 be the element in the closed convex set x+m which is closest to 0. Put x = x x 2 M. If y M then for any scalor λ Since 2Rλ(x 2, y) + λ 2 y 2 0 Put λ = (x 2 y) y 2. x 2 + λy 2 x 2 2 Then (x 2 y) 2 y 2 0 therefore (x 2 y) = 0. Suppose x = x + x 2 = x + x 2 therefore x x = x 2 x 2 = 0. Hence uniqueness. If M is closed H = M + M If M is closed and x M x = x + x 2 x M x 2 M (x x 2 ) = (x x 2 ) + (x 2 x 2 ) Therefore (x 2 x 2 ) = 0 therefore x 2 = 0 therefore x M. Theorem Suppose H is any Hilbert Space and let f X. Then there is an element y H such that f(x) = (x, y) for every x H. 7

8 Proof Let M= null space of f. y 0 M such that if x H x = m + λy 0 m M f(x) = λf(y 0 ) (x, y 0 ) = λ y 0 2 f(x) = (x, y 0) y 0 2 f(y 0) = ( x, f(y ) 0) y 0 y 2 0 Write y = f(y 0) y 0 2 y 0. If M is any closed subspace and x H x = x + x 2 x M x 2 M x = Proj M x If T (x) = x T is a linear operator from H to itself, and T =. (T x, y) = (x y) = (x y ) Therefore T is self-adjoint. T T = T (x, T y) = (x y ) = (x y ) Theorem If M,..., M n are n mutually perpendicular closed subspaces of a Hilbert space H and if x H and x i,..., x n are the projections of x on M,... M n respectively, then xi 2 leq x 2 Proof Put M = M + +M n x = x x n + y, y M, then x 2 = x 2 + y 2. Theorem Let {M α } be a family, possibly uncountable, of pairwise orthogonal closed subspaces of H, and let M be the closure of their direct sum. 8

9 If x α = proj Mα x x H then x α = 0 except for a countable set of indices α n. xαn is convergent and its sum is the projection of x on M. Proof r x βi x 2 i= Hence for any n the number of indices satisfying x α is finite n therefore the number of indices satisfying x α > 0 is countable. N x αn 2 x 2 for each N therefore x αn 2 < +. N If y n = x αn n y n y m 2 x αi 2 < ε m+ if m is sufficiently large. Therefore {y n } is a Cauchy sequence which tends to a limit y = in M, as M is closed. x αn It remains to prove that x y M. It is sufficient to prove that w β + w β w βr x y where w βi M β M. as the class of all such vectors is everywhere dense in If β as an α n (x y, w β ) = (x w β ) (x β w β ) = (x β w β ) (x β w β ) = 0 If β is not an α n then w β x and y and so to x y. Orthonormal vectors A set N of vectors in a Hilbert space H is said to be orthonormal if x = for every x in N, and (x, y) = 0 for all y in N x. 9

10 An orthonormal set N of vectors is conplete if N = {0}. Let M x be the -dimensional subspace generated by x in N. If y H proj Mx y = (y x).x = (y, x).x x as x = (y, x) = 0 except for a sequence {x n } subsetn and for this sequence y = (y x n )x n y 2 = (y x n ) 2 This condition of completeness is equivalent to: (i) for any y in H y = x N(y x)x (ii) for any y in H y 2 = z (y x) 2. x N 0