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

Question

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...

Answer 1

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.

Answer 2

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.