# What's the connection between Banach/Hilbert spaces and tools like power series, Fourier transforms etc.

I've learned, abstractly, about Banach and Hilbert spaces, and more concretely about $l^p$ and $L^p$ spaces. I also understand these ideas have something to do with a variety of tools that are useful in physics and engineering, like Power series, Fourier transforms, etc. But no one has really ever spelled out the connection to me.

This is probably a dumb question, but what's the connection?

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In Banach spaces, if you have a multiplicative norm ($\|ab\|\le\|a\|\|b\|$), then you can use power series. And a power series that would converge for $|x|< C$ would converge absolutely for $\|a\|<C$ and since the space is complete, you have that absolute convergence implies convergence. And example you be the exponential of matrices with the norm $\|M\|=\max\limits_{i,j}|m_{i,j}|$ –  xavierm02 May 11 '13 at 12:07
For example, $L^2$ is a very natural domain for the fourier transform, because it's an isomorphism on that space. In other words, the fourier transform is linear, the fourier transform of every function in $L^2$ is also in $L^2$, every function in $L^2$ is the fourier-transform of some other function in $L^2$, and the $L^2$-norm of a function and its fourier transform is always the same. Oh, and it's one-to-one also.
So basically, the Fourier transform induces a linear bijection $L^2 \rightarrow L^2$. For which other $p$ does the Fourier transform induce a bijection $L^p \rightarrow L^p$? How about $p=\infty$? –  goblin May 11 '13 at 14:24
@user18921 Linear and isometric bijection, to be precise. It's not an isomorphism on any other $L^p$ space I think, though it's for example defined for all $L^1$ functions (but the result isn't necessarily in $L^1$). –  fgp May 11 '13 at 14:31
@user18921 Another space which is closed under forward and reverse fourier transform is the schwarzian space $\mathcal{S}$ of rapidly decreasing functions. There's no metrik though which makes this space complete, though there are locally convex topologies which do. By using those, you can extend the fourier transform to the dual space of $\mathcal{S}$, which is the space of tempered distributions. That space too is closed under forward and reverse fourier transform, but and it's also not a metric space. –  fgp May 11 '13 at 14:32