We know that $\displaystyle T^{2}$ is compact, can we now deduce that $\displaystyle L^{2}(T^{2})$ is compact where $\displaystyle T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ ?

or is there some catch?

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- Aug 25th 2009, 03:48 AMMauritzvdwormCompact space of square integrable functions?
We know that $\displaystyle T^{2}$ is compact, can we now deduce that $\displaystyle L^{2}(T^{2})$ is compact where $\displaystyle T^{2}=\mathcal{R}^{2}/2\pi\mathcal{Z}^{2}$ ?

or is there some catch? - Aug 25th 2009, 04:25 AMnirax
it is an infinite dimensional vector space over R. any vector space is not compact under the most usual topologies put on it. to see this may be you should try to prove that a finite dim vector space is not compact first of all.

then any space of lower dim is closed in a space of higher dim (under almost all of the topologies that we consider) so if the higher dim space is compact, it follows that a closed sunset will also be compact. ... maybe now you can fill in the details. - Aug 25th 2009, 07:36 AMOpalg
As nirax says, a vector space is never going to be compact in any vector space topology. The most you can hope for is that the unit ball might be compact. The unit ball of the space $\displaystyle L^2(T^2)$ is not compact for the $\displaystyle L^2$-norm topology. But it is compact for the weak topology (in which a sequence of functions $\displaystyle f_n$ converges to the limit f if $\displaystyle \langle f_n-f,g\rangle\to0$ for every $\displaystyle L^2$-function g).