Do the Fourier coefficients imply the smoothness of the function?

Question

There is a conclusion that if a periodic function $f$ is $k$ times differentiable, then the $n$-th Fourier coefficient of $f$ satisfy the relation: $$a_n =O(1/|n|^k) \quad \text{as $n\rightarrow \infty$.}$$ where $O(1/|n|^k)$ means $|n|^k a_n$ is bounded.

My question is: Does the converse of the conclusion still holds, i.e. if we have $a_n =O(1/|n|^k) \quad \text{as $n\rightarrow \infty$}$, is the function $f$ k times differentiable?

Answer 1

For $k=1$ the function $f(x)=\sum_{|n| >2} \frac {\sin (nx)} {n \log n} $ is a counter-example. It can be shown that this series is uniformly convergent so $f$ is a continuous periodic function. But it is not differentiable ate $0$.

Answer 2

A counterexample to the question precisely as asked has already been given. But there are subtleties here; seems to me a simple "no" does not tell the whole story.

It's true that in a sense smoothness is equivalent to decay of the Fourier coefficients. But it's not as simple as you might hope; if S is some degree of smoothness there typically does not exist a decay condition $D$ such that S is equivalent to $D$; instead there is a $D_1$ which implies S and a $D_2$ which is implied by S.

Here for example $f\in C^1$ implies $a_n=O(1/n)$. The condition $a_n=O(1/n)$ does not imply $f\in C^1$, but the slightly stronger condition $\sum|na_n|<\infty$ does imply $f\in C^1$.

Similarly for $C^k$. Now noting that if $a_n=O(1/n^{k+2})$ then $\sum|n^ka_n|<\infty$, taking the intersection over all $k$ shows

Lemma. The periodic function $f$ is infinitely differentiable if and only if for every $k$ there exists $c_k$ with $|n^ka_n|\le c_k$ for all $n$.

Because $f\in C^k$ implies $a_n=O(1/n^k)$, while $a_n=O(1/n^{k+2})$ implies $f\in C^k$.

Similarly for the Schwarz space on the line and the Fourier transform; the above is a warmup for the proof that $f\in \mathcal S(\Bbb R)$ if and only if $\hat f\in\mathcal S(\Bbb R)$.

Answer 3

Theorem 1.9 from Lecture 9 Analysis Notes states:

Let $f\in L^2(-\pi, \pi), n\in\{0,1,2,\ldots\}$. Then $$ |\hat{f}_k|\leq\frac{C}{|k|^{n+1+\epsilon}} $$ for some $C>0, \epsilon>0$ implies $f\in C^n$.