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The proof of the existence of a continuous function $f\in C(\Bbb T)$ with divergent Fourier series using the Banach-Steinhaus theorem is well-known. It uses the unboundedness of the Dirichlet kernel $D_n$ - note that $\|D_n\|_1 \sim \log n$. For example, see this answer.

I'm trying to extend the argument to a countable set of points. Let $\{x_k\}_{k=1}^\infty$ be a countable subset of $\Bbb T = \Bbb R/\Bbb Z$. Define the functionals $$\Lambda_{N,k}(f) = S_N(f;x_k) = \sum_{|n| \le N} \widehat{f}(n) e^{2\pi in x_k}$$ Then, by the Banach-Steinhaus theorem, if $$\sup_{N,k} \|\Lambda_{N,k}\| = \infty,$$ then there exists $f\in C(\Bbb T)$ such that $$\sup_{N,k} |\Lambda_{N,k} (f)| = \infty.$$ As the supremum is over $N$ and $k$, this doesn't seem to help us show that a function $f\in C(\Bbb T)$ exists such that its Fourier series diverges at every $x_k$, i.e., for every $k\in \Bbb N$, $$\sup_{N} |\Lambda_{N,k}(f)| = \infty.$$ Is there a way to fix this argument, i.e., use the Banach-Steinhaus theorem to arrive at the result? That's what I'd be most interested in. If not, it'd be nice to see other ways to show that such an $f$ exists. Thank you!

stoic-santiago
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    I suggest you read the corresponding section in Rudin’s RCA. The point is that the origin $x=0$ in the other answer is not special. So, for each $x\in\Bbb{R}$, you can find a dense $G_{\delta}$ set $\Phi_x$ of continuous functions on the circle such that for each $f\in\Phi_x$, $\sup\limits_{n\geq 1}|S_n(f,x)|=\infty$. Now, take a countable collection of points ${x_k}$, and let $\Phi=\bigcap\Phi_{x_k}$; by Baire’s theorem this is a dense $G_{\delta}$ set, and so for eevry function $f\in \Phi$ and every $k$, we have the divergence $\sup\limits_{n\geq 1}|S_n(f;x_k)|=\infty$. – peek-a-boo Dec 13 '23 at 23:19
  • Thanks! That's a neat argument. Would it be possible for you to point me to the specific section in RCA (on how to construct $\Phi_x$)? @peek-a-boo – stoic-santiago Dec 13 '23 at 23:49
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    It’s page 102 (Chapter 5, “Fourier Series of Continuous Functions”). Anyway, the $\Phi_x$ is just what Banach-Steinhaus gives you (you don’t ‘construct’ it). – peek-a-boo Dec 14 '23 at 01:03

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