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Math Help - Hilbert Space

  1. #1
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    Hilbert Space

    Consider the Hilbert Space H = L^2 (-1,1) of Lebesgue integrable real valued function on (-1,1) with inner product (f,g) = \int_{(-1,1)} fg  dm.
    Prove that the set of all even function in H is a closed subspace of H.
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  2. #2
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    Quote Originally Posted by problem View Post
    Consider the Hilbert Space H = L^2 (-1,1) of Lebesgue integrable real valued function on (-1,1) with inner product (f,g) = \int_{(-1,1)} fg  dm.
    Prove that the set of all even function in H is a closed subspace of H.
    Well, you know that all you have to prove is that a non-empty subset of vector space is closed under addition and scalar multiplication to prove it is a subspace, don't you? So you want to prove that the sum of two even functions is an even function and that the product of a number and an even function is an even function.
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  3. #3
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    Well, proving it's a subspace is quite easy. For the fact that it's closed you'll need the following:

    1) If (f_n) \subset L^1 (\Omega) is a sequence of integrable functions such that \lim_{n\rightarrow \infty } \int_{\Omega } \vert f_n \vert =0 then there exist a subsequence (f_{n_k} ) such that f_{n_k} \rightarrow 0 a.e. on \Omega

    Take n_k<n_{k+1} such that \int_{\Omega } \vert f_{n_k} < \frac{1}{2^k} then \int_{\Omega } \sum_{k=1}^{\infty } \vert f_{n_k} \vert \leq \sum_{k=1}^{\infty } \int_{\Omega } \vert f_{n_k} \leq 1 and so f_{n_k} \rightarrow 0 a.e. on \Omega.

    2)If (f_n) \subset L^p(\Omega ) is a sequence of integrable functions such that f_n\rightarrow f in L^p(\Omega ) then there exists a subsequence (f_{n_k}) such that f_{n_k}(x) \rightarrow f(x) a.e. on \Omega.

    Since \lim_{n\rightarrow \infty } \int_{\Omega } \vert f_n -f \vert ^p =0 by (1) we have that there exists a subsequence f_{n_k} such that \vert f_{n_k}(x)-f(x) \vert ^p \rightarrow 0 a.e. on \Omega and so the result follows. Notice that p\in [1,\infty ) but in fact the case p=\infty is easier since you don't even need to pick a subsequence (you see why?).

    Now let (f_n) \subset L^2(-1,1) be even functions such that f_n\rightarrow f in L^2(-1,1) then there exists a subsequence f_{n_k} such that f_{n_k}(x) \rightarrow f(x) a.e. on (-1,1) and so let Z:=\{ x\in (-1,1) : f_{n_k}(x) \nrightarrow f(x) \} then it's obvious \mu (Z)=0 and if x\in (-1,1) \setminus Z then f(x) = \lim_{k\rightarrow \infty } f_{n_k}(x)= \lim_{k\rightarrow \infty } f_{n_k}(-x)=f(-x). Now just define g(x)=f(x) if x\in (-1,1)\setminus Z and g(x)=0 otherwise then g=f in L^2(-1,1) and g is even.
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