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This is a Q&A; I hope simply posting a question and then answering it is the right protocol. This is stuff I thought everybody knew, but in at least two recent threads it's turned out to be somewhat mysterious. So:

Q: How do you show that there exists a continuous function on the circle whose Fourier series diverges at the origin?


Edit @TrialAndError points out that Stackexchange officially encourages asking and answering your own question.

Disintegrating By Parts
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  • This is discussed quite extensively in Wikipedia, https://en.wikipedia.org/wiki/Convergence_of_Fourier_series – Thomas Jul 15 '16 at 16:13
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    I am biased (as I was clueless in one of the two recent threads mentioned by David), but I do not think this is ground for closing the question. Namely, at least 99% of this website's content is addressed in Wikipedia or reference books in a much straightforward manner, and this particular problem at least has the merit of coming up in a lot of places. – Clement C. Jul 15 '16 at 16:28
  • @Thomas I looked there first. Yes, the general topic is discussed extensively there, but what I'm talking about here is just one sentence "follows from UBP". Twice recently I've said something about the question in more detail than that, and I've had multiple requests for further elaboration. – David C. Ullrich Jul 15 '16 at 16:37
  • @ClementC. The other clueless one in that thread was a justifiably highly respected 140,000 guy. Not that that proves anything... – David C. Ullrich Jul 15 '16 at 17:04
  • @ClementC. Is this addressed to me? I did not say anything about closing the question. – Thomas Jul 15 '16 at 17:36
  • Sorry, ignore the part about closing -- this was more generally addressed towards the downvoters, as a counterpoint. – Clement C. Jul 15 '16 at 17:39
  • @user1952009 Perhaps. Let us know when you get it. More or less everything known about this sort of thing is in Zygmund, and there's no example there anything like this simple... Hmm, doing exactly what you say is impossible, of course: The Fourier coefficients of a continuous function tend to zero. – David C. Ullrich Jul 15 '16 at 23:33
  • The downvotes for another member who regularly offers Q&A posts stopped when he starting adding something about Q&A at the bottom of the post. That's why I added something for you, and not so that you would be relieved. – Disintegrating By Parts Jul 17 '16 at 15:01
  • @TrialAndError I feel maybe I should clarify. If you detected a bit of sarcasm in my rewrite of your edit it's wasn't meant to be directed at you, just at the general idea that we have to be sooo careful about the rules, instead of just posting what we think will enhance the site. Yesterday I appreciated your support but thought maybe what you said would make more sense as a comment - given what you just said I suppose, sigh, including it in the body of the post makes sense. Thanks... – David C. Ullrich Jul 17 '16 at 15:11
  • Unfortunately, I saw that the downvotes did not stop until something was added to the body of the post; it took a long time for the member to discover how to stop that reaction. The downvoters seemed to feel it was an abuse of the site, which is the opposite of what is true. Others would immediately work to close such questions. I wanted this one to stay. :) – Disintegrating By Parts Jul 17 '16 at 15:25
  • @ DavidC.Ullrich @TrialAndError I have a question specially designed for you about analytic function on $|z| < 1$ continuous on $|z| \le 1$ whose Taylor/Fourier series diverges – reuns Sep 13 '16 at 16:57

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This is one of the reasons functional analysis is a useful thing: Giving an explicit example is not easy (there's one in Zygmund, due to Fejer), but proving the existence using a little bit of Banach-space theory is very simple.

We need the following special case of the Uniform Boundedness Principle, aka the Banach-Steinhaus Theorem:

Theorem (UBP, Special Case) Suppose $X$ is a Banach space and $S\subset X^*$. If $\sup_{\Lambda\in S}||\Lambda||=\infty$ then there exists $x\in X$ with $\sup_{\Lambda\in S}|\Lambda x|=\infty$.

Now define $\Lambda_n\in C(\Bbb T)^*$ by saying $\Lambda_n f$ is the $n$-th partial sum of the Fourier series at the origin: $$\Lambda_n f=s_n(f,0)=\frac{1}{2\pi}\int_0^{2\pi}f(t)D_n(t)\,dt,$$where $D_n$ is the Dirichlet kernel $$D_n(t)=\sum_{k=-n}^ne^{ikt}=\frac{\sin\left((n+\frac12)t\right)}{\sin\left(\frac12 t\right)}.$$Suppose we can prove two things: $$||D_n||_1\to\infty\quad(n\to\infty)$$and $$||\Lambda_n||_{C(\Bbb T)^*}=||D_n||_1.$$Then UBP says there exists $f\in C(\Bbb T)$ such that $\Lambda_n f$ is unbounded and we're done.

Showing that $||D_n||_1\to\infty$ is easy: $$\int_0^\pi|D_n(t)|\,dt\ge2\int_0^\pi\frac{\left|\sin\left((n+\frac12)t\right)\right|}{t}\,dt=2\int_0^{(n+1/2)\pi}\frac{|\sin(t)|}{t}\,dt.$$

The fact that $||\Lambda_n||_{C(\Bbb T)^*}=||D_n||_1$ is immediate from the Riesz Representation Theorem, plus the fact that the norm of an $L^1$ function is the same as its norm as a complex measure. One can also see it directly: Choose $\phi_n\in C(\Bbb T)$ so that $|\phi_n|\le 1$, and such that $\phi_n=1$ on "most" of the set where $D_n>0$ while $\phi_n=-1$ on most of the set where $D_n<0$.

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    That should be $\sin ((n+1/2)t)/\sin(t/2).$ – zhw. Jul 15 '16 at 21:12
  • You also want absolute values on the middle integral above. – zhw. Jul 18 '16 at 20:19
  • @zhw. [sigh...] Thanks. – David C. Ullrich Jul 18 '16 at 20:42
  • Unless I'm missing something, it seems Lebesgue formed a similar argument. This is proof (i) of Theorem (1.1) in Chapter VIII in Zygmund. – David Mitra Jul 23 '16 at 09:50
  • @DavidMitra Lebesgue "also" formed a similar argument? I take it Zygmund credits Lebesgue here - I know nothing abaout the history but I see no reason to conjecture that it's not simply due to Lebesgue. ??? (Hmm. Possibly there was a misunderstanding here: I didn't say Fejer came up with this argument, I mentioned Fejer as the author of the other argument in Zygmund, the "explicit" construction of an example...) – David C. Ullrich Jul 23 '16 at 12:46
  • Yes, Zygmund attributes the proof I referenced above to Lebesgue; sorry for being unclear. – David Mitra Jul 23 '16 at 12:57