Hilbert-Spaces

Hilbert-Spaces - 2 22 BRUCE K DRIVER 12 Hilbert Spaces 12.1...

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222 BRUCE K. DRIVER 12. Hilbert Spaces 12.1. Hilbert Spaces Basics. De f nition 12.1. Let H be a complex vector space. An inner product on H is a function, , ·i : H × H C , such that (1) h ax + by, z i = a h x, z i + b h y,z i i.e. x h x, z i is linear. (2) h x, y i = h y,x i . (3) k x k 2 h x, x i 0 with equality k x k 2 =0 i f x . Notice that combining properties (1) and (2) that x h z,x i is anti-linear for f xed z H, i.e. h z,ax + by i a h i + ¯ b h z,y i . We will often f nd the following formula useful: k x + y k 2 = h x + + y i = k x k 2 + k y k 2 + h x, y i + h i = k x k 2 + k y k 2 +2Re h x, y i (12.1) Theorem 12.2 (Schwarz Inequality) . Let ( H, , ·i ) be an inner product space, then for all x, y H |h x, y i| k x kk y k and equality holds i f x and y are linearly dependent. Proof. If y , the result holds trivially. So assume that y 6 . First o f notice that if x = αy for some α C , then h x, y i = α k y k 2 and hence |h x, y i| = | α |k y k 2 = k x kk y k . Moreover, in this case α := h x,y i k y k 2 . Now suppose that x H is arbitrary, let z x k y k 2 h x, y i y. (So z is the “orthogonal projection” of x onto y, seeF igure28 .)Then Figure 28. The picture behind the proof. 0 k z k 2 = ° ° ° ° x h x, y i k y k 2 y ° ° ° ° 2 = k x k 2 + |h x, y i| 2 k y k 4 k y k 2 2Re h x, h x, y i k y k 2 y i = k x k 2 |h x, y i| 2 k y k 2 from which it follows that 0 k y k 2 k x k 2 |h x, y i| 2 with equality i f z or equivalently i f x = k y k 2 h x, y i
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ANALYSIS TOOLS WITH APPLICATIONS 223 Corollary 12.3. Let ( H, , ·i ) be an inner product space and k x k := p h x, x i . Then k·k is a norm on H. Moreover , ·i is continuous on H × where H is viewed as the normed space ( k·k ) . Proof. The only non-trivial thing to verify that k·k is a norm is the triangle inequality: k x + y k 2 = k x k 2 + k y k 2 +2Re h x, y i k x k 2 + k y k 2 +2 k x kk y k =( k x k + k y k ) 2 where we have made use of Schwarz’s inequality. Taking the square root of this inequality shows k x + y k k x k + k y k . For the continuity assertion: |h x, y i h x 0 ,y 0 i| = |h x x 0 i + h x 0 y 0 i| k y kk x x 0 k + k x 0 kk y y 0 k k y kk x x 0 k +( k x k + k x x 0 k ) k y y 0 k = k y kk x x 0 k + k x kk y y 0 k + k x x 0 kk y y 0 k from which it follows that , ·i is continuous. De f nition 12.4. Let ( H, , ·i ) be an inner product space, we say x, y H are orthogonal and write x y i f h x, y i =0 . More generally if A H is a set, x H is orthogonal to A and write x A i f h x, y i for all y A. Let A = { x H : x A } be the set of vectors orthogonal to A. We also say that a set S H is orthogonal if x y for all x, y S such that x 6 = y. If S further satis f es, k x k =1 for all x S, then S is said to be orthonormal. Proposition 12.5. ( H, , ·i ) be an inner product space then (1) ( Parallelogram Law) (12.2) k x + y k 2 + k x y k 2 =2 k x k 2 k y k 2 for all x, y (2) ( Pythagorean Theorem) If S H is a f nite orthonormal set, then (12.3) k X x S x k 2 = X x S k x k 2 .
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Hilbert-Spaces - 2 22 BRUCE K DRIVER 12 Hilbert Spaces 12.1...

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