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Find a function such that $f\notin C^1(\mathbb{T})$ but it's Fourier series converges to it uniformly

So (I think) if $f$ is $C^1(\mathbb{T})$ then there's a theorem says that it's Fourier series converges uniformly.

I need to find an example where $f\notin C^1(\mathbb{T})$.

How do I find an example? It's like shooting in the dark...

AlonAlon
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  • Would $f=\chi_{[-1,1]}$ be suffice? – Bombyx mori Dec 30 '14 at 16:19
  • Please remind me the meaning of $\chi$ in this context. – AlonAlon Dec 30 '14 at 16:24
  • It's the indicator function: 1 on $[-1, 1]$, 0 otherwise. I think it's fourier transform can be put in terms of the $\operatorname{sinc}$ operator, which isn't Lebesgue integrable. This is usually the goto example for a function that's Lebesgue integrable, but it's fourier transform isn't. – Robert Cardona Dec 30 '14 at 16:29
  • $$\hat{f}(n) = \frac{1}{2\pi} \int_0^{2\pi} f(x) e^{-inx} dx = \frac{1}{2\pi} \int_{-1}^{1} f(x) e^{-inx} dx = \frac{1}{2\pi} \int_{-1}^{1} 1 e^{-inx} dx = \frac{1}{2\pi} e^{-inx} |_{-1}^{1} = \frac{1}{2\pi} \left( \frac{e^{-in}-e^{in}}{-in} \right) = \frac{1}{2\pi} \frac{e^{-in}-e^{in}}{-in} = \frac{1}{2\pi}\frac{-2i\sin n}{-in} = \frac{\sin n}{\pi n}$$ – AlonAlon Dec 30 '14 at 16:58
  • Now, how do I know that the Fourier series converges to $f$? – AlonAlon Dec 30 '14 at 16:58

2 Answers2

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If you have seen that any periodic absolutely continuous function has a uniformly convergent Fourier series, then there are an abundance of examples, such as $f(x)=1-|x|$ where $\mathbb{T}=[-1,1]$. A short proof that $f$ is absolutely continuous comes from $f(x)=-\int_{-1}^x \text{sign}(y) dy$ for $x \in [-1,1]$.

Any periodic piecewise linear function will work, too. These only have finitely many points of non-differentiability. You can make nastier examples where this set is infinite, though it will always have measure zero, and will always be in a certain Borel class. (Cf. Characterization of sets of differentiability for the Borel class remark.)

Edit: also, any Lipschitz function will work.

Community
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Ian
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My earlier "example" in the comment would not work as it is not continuous at all. Let us consider a different example. I think a modification of the sawtooh wave function would work. It is continuous but not differentiable. Its Fourier series is essentially given by terms like $$ (-1)^{n}\frac{\sin[nx]}{n} $$ which converges by Dirichlet's test. However I do not have a clean proof of its uniform convergence without using some arguments like Abel summation or Dirichlet's test.

Bombyx mori
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