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I have a naive question. The Fourier series give an injection between continuous function of period $T$ and the set of real valued sequences. But, don't we expect the set of continuous function of periode $T$ to be much bigger than the set of sequences? By this I mean that since "$card(\mathbb{N})<card(\mathbb{R})$" we could expect sets of function from $\mathbb{R}$ to $\mathbb{R}$ to be much larger than the set of functions from $\mathbb{N}$ to $\mathbb{R}$. Can anyone comment on this?

Thanks

Eric Wofsey
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Chevallier
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    A continuous function, periodic or not, is determined by its values on the rationals. So there are only continuum many of them. Yes, from $\mathbb{R}$ to $\mathbb{R}$, no continuity restriction, there are many more. – André Nicolas Jul 21 '15 at 00:03
  • The set of real-valued sequences has the same size as the set of reals (This is in essence because because $||\mathbb R|=2^{\aleph_0}$ and $\aleph_0\times\aleph_0=\aleph_0$). As long as the sequence determines the Fourier series, this shows that the set of Fourier series and the set of reals have the same size. This certainly holds if we only look at continuous functions, but it also holds in significantly larger generality. – Andrés E. Caicedo Jul 21 '15 at 00:04

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It is true that there are more functions $\mathbb{R}\to\mathbb{R}$ than there are functions $\mathbb{N}\to\mathbb{R}$. However, the key restriction here is that we are only considering continuous functions $\mathbb{R}\to\mathbb{R}$. Very few (relatively speaking) functions $\mathbb{R}\to\mathbb{R}$ are continuous, so few that in fact they can inject into the set of functions $\mathbb{N}\to\mathbb{R}$. A more elementary way to see this than Fourier analysis is to just notice that a continuous function $\mathbb{R}\to\mathbb{R}$ is uniquely determined by its restriction to $\mathbb{Q}$. So restriction gives an injection from the set of continuous functions $\mathbb{R}\to\mathbb{R}$ to the set of functions $\mathbb{Q}\to\mathbb{R}$, which can be identified with the set of functions $\mathbb{N}\to\mathbb{R}$ since $\mathbb{Q}$ is countable.

For some related discussion (and in particular, proofs that all of these sets actually have the same cardinality as $\mathbb{R}$ itself), see Cardinality of set of real continuous functions.

Eric Wofsey
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