What is the Baire category argument here? (divergence of many Fourier series at one point)

Question

I came across this PDF file by Paul Garrett. In it, he shows via the standard application of the Uniform Boundedness Principle that there exists a continuous function $f\in C^0(\mathbb T)$ in the unit ball $B$ of $C^0(\mathbb T)$, whose Fourier series diverges at the origin.

(In one sentence, evaluation at say $x=0$ of the $N$th partial Fourier series is a linear functional, and this collection of functionals do not have a uniform norm bound.)

But curiously, he goes on to say that the collection of such $f$ is a countable intersection of open dense subsets of $B$, and I've not seen this before, or I've forgotten :) (I presume $v$ is a typo in the PDF.)

Question: What is this collection of open dense subsets?

Naturally, once the above question is solved, the Baire Category theorem gives that (as $B$ is a complete metric space), this collection of functions with diverging Fourier series is dense in $B$.

"What have you tried", I already hear you say, well I still feel that Baire Category applications are the result of a magical trick...the only obvious collection of functions I can think of are the bandlimited functions, but these (as in: the span of the first $N$ complex exponentials) are not dense.

A hint will be enough.

Answer 1

Okay... partial answer should now be a complete answer - please check!

For those who didn't read the PDF, we define $\lambda_N f=\sum_{n=-N}^N \langle f,e^{inx}\rangle_{L^2(S^1)}$ to be a partial sum of the Fourier series of $f$ evaluated at $0$.

I'm fairly certain that Garrett means that $\{f|\sup_N |\lambda_N f|=\infty\}=\cap_{M=1}^{\infty} U_M$, where $U_M=\{f|\sup_N |\lambda_N f|>M\}.$ To see that a given $U_M$ is open, let $f\in U_M$ and suitable $N_0$ and $\varepsilon$ such that $|\lambda_{N_0} f|\geq M+\varepsilon$. Then, since each Fourier Coefficient is a contraction, we get for any continuous $g:S^1\to \mathbb{C}$ that

$$ |\lambda_{N_0}g|\geq |\lambda_{N_0}f|-\sum_{n=-N_0}^{N_0} ||f-g||_{\infty}, $$ which is strictly larger than $M$ for $||f-g||_{\infty}<\frac{\varepsilon}{2N_0+1}$.

Now, the tricky part is establishing density of $U_M,$ and here, I'm not quite finished. So here's my idea:

We can try to approximate trigonometric polynomials via $U_M$ functions since the trigonometric polynomials are dense in $C(S^1)$ by Stone-Weierstrass.

Let $f$ be some function lying in $\cap_{M=1}^{\infty} U_M$ (such functions exist by the results in the PDF). Then, for any trigonometric polynomial $p(x)=\sum_{n=-N}^N c_n e^{inx}$ and any $k\in \mathbb{N}$, we claim that $f_k=\frac{f}{k}+g\in U_M$. Indeed, there exists some $N_0$such that $|\lambda_{N_0} f|> k(\sum_{n=-N}^N |c_n|+M),$ implying that $$|\lambda_{N_0} f_k|\geq \frac{1}{k}|\lambda_{N_0} f|-|\lambda_{N_0} p|> M$$ However, $f/k$ clearly tends to $0$ uniformly, so $f_k\to g$ uniformly. This establishes density.

Of course, all we showed was that $\cap_{M=1}^{\infty} U_M$ is itself dense, rather than showing that each individual $U_M$ was, but I suppose the proof works fine.