A Hilbert space is a complete inner product space. It is separable if there's a complete orthonormal family V1, V2,....

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A Hilbert Space Is A Complete Inner Product Space It Is Separable If There S A Complete Orthonormal Family V1 V2 1 (100.11 KiB) Viewed 41 times

A Hilbert space is a complete inner product space. It is separable if there's a complete orthonormal family V1, V2,.... All separable Hilbert spaces are the same! A model for all of them is ² (N), the space of sequences v = (a1, a2, ...) with a; C such that Σ, la converges. An operator A on a Hilbert space V means is a linear map A: V → V. We say A is bounded if there exists M20 such that ||A(v)|| ≤ M||v|| for every ve V. In that case, the operator norm ||A|| is the sup of ||A(v)|| over all |v|| 1. We say A is compact if whenever 0₁, 02,... is a sequence of vectors with |v|| ≤ 1, then A(v₁), A(v2),... contains a convergent subsequence. 1. Let A be the n x n matrix with 1s on the anti-diagonal (going from lower left to upper right), and Os everywhere else. Then A is symmetric, so the spectral theorem predicts that there is an orthonormal basis for R consisting of eigenvectors for A. Find such a basis. 2. Let V be a separable Hilbert space with complete orthonormal family V1, V2, Let A₁, A2,... be a sequence of complex numbers such that |A₂| is bounded. .... (a) Show that there is a unique bounded linear operator A satisfying Aivi, with formula Avi = A(v) = λi (v, v₁) vi. You should think of A as an infinite-dimensional diagonal matrix with A₁, A2,... along the diagonal. (b) Find the operator norm ||A||.