# Hilbert Space

• Nov 19th 2009, 03:24 AM
problem
Hilbert Space
Consider the Hilbert Space $\displaystyle H = L^2 (-1,1)$ of Lebesgue integrable real valued function on $\displaystyle (-1,1)$ with inner product $\displaystyle (f,g) = \int_{(-1,1)} fg dm$.
Prove that the set of all even function in $\displaystyle H$ is a closed subspace of $\displaystyle H$.
• Nov 19th 2009, 04:28 AM
HallsofIvy
Quote:

Originally Posted by problem
Consider the Hilbert Space $\displaystyle H = L^2 (-1,1)$ of Lebesgue integrable real valued function on $\displaystyle (-1,1)$ with inner product $\displaystyle (f,g) = \int_{(-1,1)} fg dm$.
Prove that the set of all even function in $\displaystyle H$ is a closed subspace of $\displaystyle H$.

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.
• Nov 19th 2009, 09:42 AM
Jose27
Well, proving it's a subspace is quite easy. For the fact that it's closed you'll need the following:

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

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

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

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

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