# Necessary and sufficient condition for a sequence of real numbers to be Fourier coefficients of some function

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 $$a_n=\int_a^b f(x) e_n(x) \, dx$$

• Iff $\sum |a_n|^2<\infty.$ But things are not so easy for other function spaces
– zhw.
Dec 16 '19 at 3:25
• Can you provide a proof ? I feel like you are supposing that the basis is complete ? Dec 16 '19 at 3:36
• I was assuming an orthonormal basis for $L^2.$
– zhw.
Dec 16 '19 at 3:37
• 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. Dec 16 '19 at 3:47
• 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 $l_2$. Dec 16 '19 at 4:46

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.

• Being convergent in $L^2$ is not enough: for $a_n$ to be Fourier coefficients, we need uniform convergence of $\sum a_ie_i$ Dec 16 '19 at 15:07
• @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. Dec 16 '19 at 15:18
• 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 $a_n = \int_a^b f(x) e_n(x) \, dx$. Dec 16 '19 at 15:37
• @W.Volante No, you do not need uniform convergence for that. Not sure where you got that idea.
– zhw.
Dec 16 '19 at 16:27
• 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. Dec 16 '19 at 18:18

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 $$\sum |a_n|^{2} <\infty$$.

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