Hilbert Space

**problem** · November 19th 2009, 04:24 AM

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 $H$ .

**HallsofIvy** · November 19th 2009, 05:28 AM

Originally Posted by problem

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.

**Jose27** · November 19th 2009, 10:42 AM

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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