96 CHAPTER 4. HILBERT SPACES. Spaces of square integrable functions. Take a Cauchy sequence f n in L 2 so that. f n f m 1 (b a) f n f m 2.

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1 96 CHAPTER 4. HILBERT SPACES 4.2 Hilbert Spaces Hilbert Space. An inner product space is called a Hilbert space if it is complete as a normed space. Examples. Spaces of sequences The space l 2 of square summable sequences is complete. Proved before. The space l 0 of sequences ith vanishing tails is not complete. Counterexample. Spaces of continuous functions C([a, b]) is not complete. Counterexample. C 0 (R) (space of continuous functions ith compact support) is not complete. Counterexample. Spaces of square integrable functions. Theorem Proof: Functional analysis books. The spaces L 2 ([a, b]) and L 2 (R) are complete. Take a Cauchy sequence f n in L 2 so that as n, m. By Scharz inequality f n f m 2 0 f n f m 1 (b a) f n f m 2. So, f n is Cauchy in L 1. Since L 1 is complete it converges to some f L 1. Clearly, the subsequence f pn is Cauchy in L 2. Then there is a subsequence f pn convergent to f (almost everyhere). Therefore, f L 2. mathphyshass1.tex; October 2, 2013; 17:01; p. 93

2 4.2. HILBERT SPACES 97 Theorem The space L 2 (R) is complete. Proof. Construct a subsequence of a Cauchy sequence in L 2 (R) that converges almost everyhere by a diagonal argument. Theorem The space L 2 ([a, b], µ) is complete. Proof. Rescale functions by µ. More generally, Theorem The space L 2 (, µ, V) is complete. Sobolev Spaces Let be an open set in R n (in particular, can be the hole R n ) and V a finite-dimensional complex vector space. Let C m (, V) be the space of complex vector valued functions that have continuous partial derivatives of all orders less or equal to m. Let α = (α 1,..., α n ), α j N, be a multiindex of nonnegative integers, α i 0, and let Define α = α α n. D α f = α x α 1 1 xα n n Then f C m (V, ) iff α, α m, i = 1,..., N, x e have D α f i (x) <. The space H m (, V) is the space of complex vector valued functions such that α, α m, D α f L 2 (, V) i.e. such that α, α m, dx D α f, D α f = f. N D α f i (x) 2 dx <. i=1 mathphyshass1.tex; October 2, 2013; 17:01; p. 94

3 98 CHAPTER 4. HILBERT SPACES It is an inner product space ith the inner product ( f, g) = α, α m D α f, D α g = α, α m N i=1 D α f i D α g i The Sobolev space H m (, V) is the completion of the space H m (, V) defined above Strong and Weak Convergence Strong Convergence. A sequence (x n ) of vectors in an inner product space E is strongly convergent to a vector x E if Notation. x n x. lim n x n x = 0 Theorem Let E be an inner product space. Then for every y E the linear functional ϕ y : E C defined by ϕ y (x) = (y, x) x E is continuous (and, therefore, bounded). Proof: Use Scharz inequality. Weak Convergence. A sequence (x n ) of vectors in an inner product space E is eakly convergent to a vector x E if for any y E lim n (x n x, y) = 0 Notation. x n x Theorem A strongly convergent sequence is eakly convergent to the same limit. Proof: Use Scharz inequality. Converse is not true. Counterexample later. mathphyshass1.tex; October 2, 2013; 17:01; p. 95

4 4.2. HILBERT SPACES 99 Theorem Let (x n ) be a sequence in an inner product space E and x E. Suppose that 1. x n x and 2. x n x. Then x n x. Proof: Easy. Theorem Let S be a subset of an inner product space E such that span S is dense in E, (x n ) be a bounded sequence in E and x E. Suppose that for any y S, lim n (x n x, y) = 0. Then x n x. Proof: Let z E. Then there is y n span S such that y n z. Then converges to zero as n, m. (x n x, z) = (x n x, y m ) + (x n x, z y m ) It is possible that x n x but xn does not converge to x. But at least the seqence x n has to be bounded. Theorem Weakly convergent sequences in a Hilbert space are bounded. Proof: In functional analysis books. Let f n be a functional defined by Then f n (x) = (x n, x). f n = x n On another hand such sequence of bounded functionals must be bounded (a theorem). mathphyshass1.tex; October 2, 2013; 17:01; p. 96

5 100 CHAPTER 4. HILBERT SPACES Orthogonal and Orthonormal Systems Orthogonal and Orthonormal Systems. Let E be an inner product space. A set S of vectors in E is called an orthogonal system if any pair of distinct vectors in S is orthogonal to each other. An orthogonal system of unit vectors is an orthonormal system. Every orthogonal system can be made orthonormal. Let S be a set of vectors in E. We say that x S if x y for any y S. If x S then x span S. The orthogonal complement of S is the space S = {x E x S } Theorem Orthogonal systems are linearly independent. Proof: Easy. Orthonormal Sequence. A sequence of vectors hich is an orthonormal system is an orthonormal sequence; then Examples. (x n, x m ) = δ nm 1. Canonical basis in l 2. (e n ) i = (δ i n) 2. Fourier basis. The functions are orthonormal in L 2 ([ π, π]). 3. Legendre polynomials. f n (x) = 1 2π e inx P n (x) = 1 2 n n! n x(x 2 1) n Sho that the functions f n = are orthonormal in L 2 ([ 1, 1]). n P n mathphyshass1.tex; October 2, 2013; 17:01; p. 97

6 4.2. HILBERT SPACES Hermite polynomials. H n = ( 1) n e x2 n xe x2 Sho that the functions f n (x) = ( ) 1 2n n!π exp x2 H n (x) 1/4 2 are orthonormal in L 2 (R). Any orthogonal sequence can be alays made orthonormal. Gram-Schmidt orthonormalization process. Any sequence of linearly independent vectors can be made orthonormal. 1. Let (y n ) be a linearly independent sequence. 2. Let z n = y n y n 3. Recall that for any orthonormal sequence e n the operator P n defined by n 1 P n = I e k e k or k=1 n 1 P n x = x (e k, x)e k is the projections to the orthogonal complement of span {e 1,..., e n 1 }. 4. Then the sequence (e n ) defined by k=1 e 1 = z 1, n 1 e n = P n z n = z n (e k, z n )e k k=1 is orthonormal. mathphyshass1.tex; October 2, 2013; 17:01; p. 98