# Series of Fourier coefficients

Let $\mathscr B=\{\phi_n\}_{n\in\mathbb N}$ be an orthonormal basis of the real Hilbert space $L^2([0,1])$, and given $f,g \in L^2([0,1])$ let $\langle f,g\rangle$ denote the usual scalar product between $f$ and $g$: $$\langle f,g\rangle = \int_{[0,1]}\hspace{-12pt}f(x)g(x)\,{\rm d}x$$ For $f\in L^2([0,1])$ let $$\widehat f_n = \langle f,\phi_n\rangle$$ denote the $n$-th Fourier coefficient of $f$ with respect to the basis $\mathscr B$. Then we know that $$\|f\|_{L^2}^2 = \langle f,f\rangle = \sum_{n\in\mathbb N}\widehat f_n{}^2$$ My question is: can we say anything about the series $$\sum_{n\in\mathbb N} \widehat f_n \quad\text{?}$$ EDIT: as @GEdgar and @Zarrax point out, $f=\sum_n \frac{1}{n}\phi_n\in L^2(0,1)$ makes the above series diverge. So I wonder:

(1) Can we find a function $f:\mathbb N\rightarrow\mathbb R^+$ (e.g. $f(n)=n^\alpha$) s.t. $$\lim_{n\to\infty}\frac{1}{f(n)} \sum_{k=0}^n\widehat f_n$$ converges and can be explicitly computed?

or/and

(2) Can we find a sequence $a_n>0$ (not depending on $f$) s.t. $$\sum_{n\in\mathbb N} a_n\widehat f_n$$ converges and can be explicitly computed?

It would be nice to know the result even for just a specific $\mathscr B$.

I need it because a similar sum came out in a problem I am working on, and I'd like to modify things to something 'nicer'.

• You know only that $\hat{f_n}$ is square-summable, not that it is summable. And you can get an arbitrary square-summable sequence. For example $\hat{f_n} = 1/n$. – GEdgar Mar 1 '13 at 18:59
• @GEdgar: Thanks, I'll edit my question. – AndreasT Mar 1 '13 at 19:03

If the question can be usefully tweaked, then there are possibilities. For example, for the specific orthonormal basis $\varphi_n(x)=e^{2\pi inx}$, and for continuous functions $f$, the "Fejer kernel" weighted form $\sum_{|n|\le N} {N-|n|+1\over N+1} \hat{f}(n)e^{2\pi i n x}$ of the partial sums of the Fourier series of $f$ does converge pointwise to $f$ (while the more obvious partial sums, corresponding to the "Dirichlet kernel", typically do not converge to $f$, even for continuous $f$).

I do not know of a general prescription for analogues of this idea... but if the orthonormal basis arises in a reasonable eigenvalue problem (Sturm-Liouville?), so that the eigenfunctions themselves are well-controlled, then it is reasonable to hope that there is such a "mollification" scheme.

Edit: yes, "my" Fejer kernel was not accurate... maybe fixed now. The length of the interval, the measure, other stuff can affect the normalization, but my dropping the absolute value on the "n" was bad.

• Thank you! Just a clarification: the Féjer's kernel is $$F_N(x) = \frac{1}{N+1}\sum_{k=0}^n\sum_{|j|\leq k} e^{2\pi ijx} = \frac{1}{N+1}\left(\frac{\sin[(N+1)\pi x]}{\sin \pi x}\right)^2$$ right? And it is the convolution $F_N*f$ that converges to $f$ pointwise, right? I computed $$F_N*f(x) = \sum_{n=0}^N \frac{N+1-|n|}{N+1}\widehat f_n e^{2\pi inx}$$ and my expression differs from yours for the abs. value $|n|$ and the $(N+1)$ in the fraction. Be that as it may, did I understand correctly? – AndreasT Mar 1 '13 at 21:06
• Besides, I believe my problem involves the eigenfunctions of the $[-\Delta]$ (Laplace) operator, so I am glad you considered that particular basis. I am thinking about something related to the $[-\Delta]^{\frac12}$ operator, i.e., since $[-\Delta]e^{2\pi in x}=(2\pi n)^2e^{2\pi inx}$, $$[-\Delta]^{\frac12}f(x) ~:=~ 2\pi\sum_{n\in\mathbb Z}n\widehat f_n e^{2\pi inx}$$ So this too (maybe?) kinda answers my question but I'd be happier to find something in the form (1) rather than (2) (regarding the numbered list in the question). But in $x=0,1$.. can $\sum_n n\widehat f_n$ converge? – AndreasT Mar 1 '13 at 21:18
• @AndreasT: again, for fairly general $f$, even with the special basis, that Fourier series will usually not converge pointwise, but easily converges in a distributional sense, or even in Levi-Sobolev spaces, and so on. So the operational equation may be as to whether you really, really need pointwise convergence, or some other notion may suffice. – paul garrett Mar 1 '13 at 21:33

The mapping $f(x) \rightarrow \{\hat{f}(n)\}_{n \in Z}$ is actually surjective, so you can choose any $\{\hat{f}(n)\}_{n \in Z}$ that is square summable but not summable and it is guaranteed to be the set of Fourier coefficients of some $L^2$ function.

Choosing the $a_n$ also is not going to work.. since you require $|a_n| > \epsilon > 0$ for infinitely many $n$, you can choose such a sequence $\{\hat{f}(n)\}_{n \in Z}$ which is nonzero only for $n$ for which $|a_n| > \epsilon$, and arrange it so that $a_n \hat{f}(n)$ is positive for these $n$. Then the same phenomenon happens.

• Thank you, you are right. I am changing the question right away; besides I misinterpreted the meaning of 'not asymptotically null' which I thought meant "$\forall N$ there exists $n>N$ s.t. $a_n\neq 0$". – AndreasT Mar 1 '13 at 19:10