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I would like to know when is a sequence of real numbers the Fourier coefficients of some $L^2([a,b])$ function for a fixed orthonormal basis $e_n$ of $L^2([a,b])$.

I know a sufficient condition: Let $a_n$ be a sequence of real numbers. If $\sum\limits_{i=1}^\infty a_i e_i(.)$ converges uniformly, and denoting $f$ its limit, then $a_n$ are the Fourier coefficients of $f$.

But is this condition also necessary ? If not, what is a sufficent and necessary condition for $a_n$ to be Fourier coefficients for the basis $e_n$ ? In other words, under what sufficent and necessary condition on $a_n$, there exists $f \in L^2([a,b])$ such that $a_n$ are the Fourier coefficients of $f$ for the basis $e_n$ ?

Edit: Being a Fourier coefficient of $f$ for the basis $e_n$ means $a_n=\int_a^b f(x) e_n(x) \, dx$

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    $\begingroup$ Iff $\sum |a_n|^2<\infty.$ But things are not so easy for other function spaces $\endgroup$
    – zhw.
    Dec 16 '19 at 3:25
  • $\begingroup$ Can you provide a proof ? I feel like you are supposing that the basis is complete ? $\endgroup$
    – W. Volante
    Dec 16 '19 at 3:36
  • $\begingroup$ I was assuming an orthonormal basis for $L^2.$ $\endgroup$
    – zhw.
    Dec 16 '19 at 3:37
  • $\begingroup$ Clearly, if $a_n$ are the Fourier coefficients of $f$, then $\sum a_i^2 < \infty$ (Bessel inequality), so it is a necessary condition. I fail to see that it is sufficient. $\endgroup$
    – W. Volante
    Dec 16 '19 at 3:47
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    $\begingroup$ Parsevals theorem. Also, note that any $f \in L^1$ has a Fourier series, so $x \mapsto {1 \over \sqrt{x}}$ has a Fourier series, but the resulting series is not in $l_2$. $\endgroup$
    – copper.hat
    Dec 16 '19 at 4:46
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The condition is also necessary because you have an orthonormal set. If $\{a_n\}\in\ell^2(\mathbb N)$, then by the orthonormality $$ \left\|\sum_{j=n}^ma_je_j\right\|^2=\sum_{j=n}^m|a_j|^2. $$ So the convergence of $\sum_j|a_j|^2$ guarantees that the sequence of partial sums of $\sum_ja_je_j$ is Cauchy, and so it is convergent to an $f$ as $L^2$ is complete.

Now you have, since the inner product is continuous, $$ \langle f,e_m\rangle=\langle \sum_na_ne_n,e_m\rangle=\lim_{k\to\infty}\langle \sum_{n=1}^ka_ne_n,e_m\rangle=a_m. $$ That the inner product is continuous is a direct consequence of the Cauchy-Schwarz inequality.

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  • $\begingroup$ Being convergent in $L^2$ is not enough: for $a_n$ to be Fourier coefficients, we need uniform convergence of $\sum a_ie_i$ $\endgroup$
    – W. Volante
    Dec 16 '19 at 15:07
  • $\begingroup$ @W.Volante, if by "uniform convergence" of $\sum a_ie_i$ you mean $L^2$ convergence of the finite partial subsums, then the $\ell^2$-ness of $\{a_i\}$ is exactly enough. If the Hilbert space is a space of functions, there are other notions of "uniform convergence", but (from corollaries of Baire Category theorem) most continuous functions' Fourier series do not converge uniformly to the function, etc. $\endgroup$ Dec 16 '19 at 15:18
  • $\begingroup$ No, what I mean is if $\sup |\sum a_i e_i - f | \rightarrow 0$, then $a_n = \int_a^b f(x) e_n(x) \, dx$ (it takes 3 lines to prove, we use orthonormality of the basis and we swap integral and series). I never saw a result that says if $\int_a^b (\sum a_i e_i(x) - f(x) )^2 \,dx \rightarrow 0$ then $a_n = \int_a^b f(x) e_n(x) \, dx$. $\endgroup$
    – W. Volante
    Dec 16 '19 at 15:37
  • $\begingroup$ @W.Volante No, you do not need uniform convergence for that. Not sure where you got that idea. $\endgroup$
    – zhw.
    Dec 16 '19 at 16:27
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    $\begingroup$ But that's trivial; it's the most basic Hilbert space stuff. If that's what you want, I'll include it in the answer. $\endgroup$ Dec 16 '19 at 18:18
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If $H$ is a Hilbert space and $(e_n)$ is an orthonormal sequence in it (not necessarily compelte) then there exists $x \in H$ such that $x =\sum a_ne_n$ (convergence in the norm of $H$) iff $\sum |a_n|^{2} <\infty$.

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  • $\begingroup$ The argument by Martin Argerami does not require completeness of the orthonormal sequence. $\endgroup$ Dec 16 '19 at 6:24

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