Why do there exist continuous functions with $\sum_n |c_n|^{2-\epsilon} = \infty$ for all $\epsilon > 0$?

Question

I came across this post on Fourier coefficients of $L^{\infty}$ functions, and @Conrad's comment reads:

"...nothing more can be said as there are continuous (hence bounded) functions with Fourier coefficients for which $\sum |c_n|^{2-\epsilon} = \infty$ for all $\epsilon > 0$."

I'm wondering why there exist continuous functions $\phi \in C(\Bbb T)$ with $\sum |c_n|^{2-\epsilon} = \infty$ for all $\epsilon > 0$ - is there a construction someone could describe (or proof of existence, at least)? Of course, the sequence $\{c_n\}_{n \in \Bbb Z}$ is just $\widehat\phi(n) = c_n$ for all $n\in \Bbb Z$.

If I had to construct such a $\phi$, I'd be inclined to use the fact that $\sum_{n=1}^\infty \frac{1}{n^p}$ diverges for $0\le p \le 1$. It'd help to find $\phi \in C(\Bbb T)$ such that $|\widehat\phi(n)| \approx \frac{1}{\sqrt{|n|}}$ for $n \ne 0$, as then $$\sum_{n} |c_n|^{2-\epsilon} \ge \sum_{n\ne 0} |\widehat\phi(n)|^{2-\epsilon} \approx \sum_{n=1}^\infty \frac{1}{n^{1-\epsilon/2}} = \infty$$ for every $\epsilon > 0$. However, we cannot find such a $\phi$, as $C(\Bbb T)\subset L^2(\Bbb T)$ and the Fourier coefficients $\{\widehat\phi(n)\}$ would not be square-summable as $\sum_{n=1}^\infty \frac 1 n$ diverges. I'm stuck here.

Thanks for your help!

Answer 1

The classical example is the Hardy-Littlewood function (where any $\log ^{1+\delta}(n)$ instead of $\log ^2 n$ in the denominator will do):

$$f(t)=\sum_{n \ge 2}\frac{e^{i n \log n}e^{int}}{\sqrt n \log^2 n}$$

Clearly its Fourier coefficients are of the size $|c_n|=\frac{1}{\sqrt n \log^2 n}$ so they satisfy $\sum |c_n|^{2-\epsilon} = \infty$

Now one has that $\sum_{k=1}^N e^{i n \log n+nt}=O(\sqrt N)$ uniformly in $t \in [0, 2\pi]$ for which we will skecth a proof below.

Assuming that, partial summation immediately shows that the partial sums $\sum_{N \le n \le M}\frac{e^{i n \log n}e^{int}}{\sqrt n \log^2 n}$ are uniformly $O(\sum_{N \le n \le M}\frac{1}{n \log^2 n}) \to 0, N,M \to \infty$ since $\sum_{n \ge 2}\frac{1}{n \log^2 n} < \infty$

Hence the partial sums $\sum_{2 \le n \le N}\frac{e^{i n \log n}e^{int}}{\sqrt n \log^2 n}$ converge uniformly by Cauchy-criterion, so $f(t)$ is continuous and clearly $\sum_{n \ge 2}\frac{e^{i n \log n}e^{int}}{\sqrt n \log^2 n}$ is its Fourier series by term by term integration (allowed by uniform continuity).

$\sum_{k=1}^N e^{i n \log n+nt}=O(\sqrt N)$ uniformly in $t \in [0, 2\pi]$ follows from the second derivative test for exponential sums as with $2\pi g_t(x)=x \log x +tx$ one has that $2\pi g_t''(x)=1/x$ so for $ M \le x \le 2M $ we have $\frac{1}{4 \pi M} \le g_t''(x) \le \frac{1}{2 \pi M}$ which means $\sum_{M \le n \le 2M} e^{i(x \log x +tx)}=O(M/\sqrt M+\sqrt M)=O(\sqrt M)$ where the implied constant is universal

(the second derivative test says that if $f \in C^2(I)$ and there is $\lambda >0, a \ge 1, \lambda \le |f''(x)| \le a\lambda, x \in I$ interval, then $\sum_{n \in I}e^{2\pi i f(n)}=O(a|I|\sqrt \lambda+1/\sqrt \lambda)$ and we apply this with $\lambda=\frac{1}{4 \pi M}, a=2, |I|=M$, so $\sqrt \lambda \sim 1/\sqrt M)$; the proof is elementary - see for example this blog page

Hence for $N$ fixed, splitting the interval $[2,N]$ diadically and noting that the series $\sum_{k \ge 1}2^{-k/2} < \infty$ (as the partial sum bound on $[N/2^{k-1}, N/2^k]$ is $O(\sqrt N/2^{k/2})$) we get the claimed result $\sum_{k=1}^N e^{i n \log n+nt}=O(\sqrt N)$ uniformly in $t \in [0, 2\pi]$

Note that the above example is fairly elementary but allows coefficients $|c_n| \sim 1/(\sqrt n \log^{1+\delta}n)$ while the only constraint on the Fourier coefficients of a continuous function is $\sum |c_n|^2 < \infty$ so there should be examples with $|c_n|\sim 1/(\sqrt n \log^{1/2+\delta}n)$ or even $|c_n| \sim 1/(\sqrt n \log^{1/2}n(\log \log)^{1/2+\delta}n)$ etc and indeed it is known that such exist by a Theorem of Salem, but the coefficients are complicated (or better put, their arguments are so as one can take them to have absolute value precisely the above, eg $|c_n|= 1/(\sqrt n \log^{1/2+\delta}n)$ etc)

Answer 2

Instead of letting $|\hat{\phi}(n)|$ be $\Theta(n^{-1/2})$, let these coefficients be $\Theta(n^{-1/2-\epsilon/4})$. Then you will have divergence of the sums in question, since $$\sum_{n\neq 0} \Big(\frac{1}{n^{1/2+\epsilon/4}}\Big)^{2-\epsilon} = \sum_{n\neq 0} \frac{1}{n^{1-(\epsilon/2)^2}} = \infty$$ whereas the sequence would still be square-summable, since $$\sum_{n\neq 0} \bigg(\frac{1}{n^{1/2+\epsilon/4}}\bigg)^{2} = \sum_{n\neq 0} \frac{1}{n^{1+\epsilon/2}} < \infty$$ Of course, you still need to do some work from here to show that such coefficients would define a continuous function, but at the very least this does not suffer from being non-square-summable. Maybe you can use this to craft your counterexamples.

---

EDIT: Just found out that I misinterpreted the question, and that we need something that works for all $\epsilon$ simultaneously! In this case, I would recommend coefficients that are $$|\hat{\phi}(n)|=\Theta\Big(\frac{1}{\sqrt{|n|} \log |n|}\Big)$$ For in this case we have that for all $\epsilon > 0$, the summation of coefficients to the $2-\epsilon$ power will diverge as desired, since: $$\sum_{|n| \geq 2} \Big(\frac{1}{\sqrt{|n|} \log |n|}\Big)^{2-\epsilon} = \sum_{|n| \geq 2} \frac{1}{n^{1-\epsilon/2}\log^{2-\epsilon} |n|} = \infty$$ while the coefficients are still square-summable: $$\sum_{|n| \geq 2} \Big(\frac{1}{\sqrt{|n|} \log |n|}\Big)^{2} = \sum_{|n| \geq 2} \frac{1}{n\log^{2} |n|} < \infty$$