Question about relationship between $\ell^ p$ spaces and Fourier coefficients. let $f\in L^2(\mathbb{T})$, and let $f(x)=\sum_{n\in \mathbb{Z}}B_nsin(nx)$ be the associated Fourier sine series (which of course is an orthonormal basis for for $L^2(\mathbb{T})$). 
Consider a sequence $a_n \in \ell^2({\mathbb{Z}})$ where all terms are zero except for multiples of $3$. Is it true that there exists a unique $f\in L^2(\mathbb{T})$, up to a set of measure zero, that has the fourier coefficients $B_n=a_n$? Would this mean that $\ell^2(\mathbb{Z})$ contains all possible combinations of Fourier coefficients (at least for functions in $L^2(\mathbb{T})$)?
I think it is true because of course, such an $a_n$ exists as an appropriate linear combination of basis elements in $\ell^2({\mathbb{Z}})$. Also, you can map these to an appropriate linear combination of basis elements in $L^2(\mathbb{T})$, so such a function with those fourier coefficients must exist. 
I am unsure how to show the following analogous result for $\ell^2(\mathbb{Z})\cap \ell^p(\mathbb{Z})$, $p\geq5$. Here is the precise statement.
let $a_n\in \ell^2(\mathbb{Z})\cap \ell^p(\mathbb{Z})$, $p\geq5$, where $a_n$ are the Fourier coefficients for the Fourier series $f(x)=\sum_{n\in \mathbb{Z}}B_nsin(nx)$, 
where $f\in L^2(\mathbb{T})\cap L^p(\mathbb{T}), p\geq 5$. 
Is it still true that there exists a sequence $a_n\in \ell^2(\mathbb{Z})\cap \ell^p(\mathbb{Z})$, $p\geq5$ which is zero except for multiples of $3$, prime numbers or any other possible sequence, where $a_n$ corresponds exactly to the Fourier coefficients $B_n$ for $f\in L^2(\mathbb{T})\cap L^p(\mathbb{T}), p\geq 5$. How would I prove this? 
 A: I only know the answer to the first part of your question, I'm afraid. Hopefully an expert will answer the rest.
It's certainly true that if $a_n \in \mathcal l^2(\mathbb Z)$, then $\sum_{n\in \mathbb N} a_n e^{2\pi inx}$ defines a valid element in $L^2(\mathbb T)$, and vice versa.
[I'm going to use complex exponentials as my basis for $L^2(\mathbb T)$ but I don't think it particularly matters if you use sines instead...]
Here is the argument: If $a_n \in \mathcal l^2(\mathbb Z)$, then $$\sum_{n\in \mathbb N} |a_n |^2 < \infty,$$so the sequence of partial sums, $$ n \mapsto \sum_{m = 1}^n |a_m|^2, $$is a Cauchy sequence.
But for any $n > n'$, the orthogonality of $e^{2\pi imx}$ implies that
$$ || \sum_{m=1}^{n} a_me^{2\pi imx} - \sum_{m=1}^{n'}a_me^{2\pi imx} ||^2 \leq \sum_{m=1}^{n} |a_m |^2 - \sum_{m=1}^{n'} |a_m |^2. $$
So the sequence $$n \mapsto \sum_{m=1}^{n} a_me^{2\pi imx} $$ is a Cauchy sequence in $L^2(\mathbb T)$.
Hence the limit $$ \sum_{m=1}^\infty a_m e^{2\pi i mx} = \lim_{n \to \infty}\sum_{m=1}^{n} a_me^{2\pi imx} $$ exists in $L^2(\mathbb T)$ by completeness.
The reverse argument is identical.
I can't say anything useful about the $l^2(\mathbb Z) \cap l^p (\mathbb Z)$ conjecture, I'm afraid. Of course, the isomorphism of vector spaces $$\{ a_n \} \mapsto \sum_{n} a_n e^{2\pi i n x}$$ does not map the $l^p(\mathbb Z)$ norm to the $L^p(\mathbb T^2)$ norm unless $p = 2$, which is why it's tricky.
