Coefficients from Stone Weierstrass versus Fourier Transform

Re-reading your question, I think that I see what you are asking.

Per @Andrea Ferretti's comments, you have to be careful to distinguish between ${e^{inx}}$ and $span {e^{inx}}$. You certainly are interested the latter. Sorry if my comments were sloppy and confusing above.

So, I think that the it goes like this:

From some corollary of Stone-Weirstrauss you can show that $span {e^{inx}}$ is dense in $C(mathbb{S}^1)$ with the supremum norm. Because we know that $C(mathbb{S}^1)hookrightarrow L^2([0,1])$ has its image a dense subset of $L^2([0,1])$ and we know that if $f_n to f$ in the supremum topology on $C(mathbb{S}^1)$, then the images also converge in $L^2([0,1])$.

Thus, by this reasoning, for $fin L^2([0,1])$ we can find $f_n in span {e^{inx}}$ such that $f_n = L^2([0,1])$. Lets write
$$f_n = sum_{kin mathbb{Z}} c_k^{(n)} e^{ikx}$$
where all but finitely many of the $c_k^{(n)}$ are zero (this is because in the span of infinitely many objects we only take a finite number of them to add together)

Now, what I think you are asking is: what can we say about the coefficients $c_k^{(n)}$? The answer is that they converge to the $k$-th Fourier coefficient of $f$ as $ntoinfty$ because
$$hat f(k) = langle f, e^{ikx} rangle = lim_{ntoinfty} langle f_n ,e^{ikx}rangle = lim_{ntoinfty} c_k^{(n)}$$

In fact if $c_k^{(n)}$ are arbitrary complex numbers, defining $f_n$ as above, we see that
$$Vert f - f_n Vert_{L^2} = sum_{kin mathbb{Z}} |hat f(k) - c_k^{(n)}|^2$$
assuming convergence. Thus, if $(c_k^{(n)})_k to (hat f(k))_k$ as $ntoinfty$ in $ell^2(mathbb{Z})$ then $f_nto f$ in $L^2$, which is a pretty weak condition.

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