@JonasTeuwen Thanks! Actually, now I'm not sure anymore whether I still have a question. I was going to ask you why the Dirichlet kernel is interesting.
Wait, before you say anything, let me add something:
From my notes I gather two things we can do with it:
(1) We can take an $f$ in $C(\mathbb R / \mathbb Z)$. If $D_n$ denotes the $n$-th Dirichlet kernel of $f$ then we know that $T_n (f) = \int_0^1 f(x) D_n(x) dx $ is a continuous linear functional.
@JonasTeuwen Well, I'm not comfy with these yet. The only time I heard of this before was in Pseudo-differential operators class and I've not figured out yet what I want to do with them.
Yes, sure. Just explaining. You wanted that, right? Wanted to start with a general remark.
So often you have this abstract operator $T$ which can concretely given (perhaps for a subclass of functions) as integration against a so-called integral kernel.
Hey, Dirichlet kernel is sum over some characters (the ones centered at zero) and by magic, if I convolute $f$ with it I get the partial Fourier series of $f$ over those characters.
Ah not magic!
The sum is finite so I just swap sum and integral.
Wait no, let me do it on paper, one sec.
@JonasTeuwen Sorry but this is already not clear to me. The definition of convolution of two functions is over all of $\mathbb R$: $f \ast g = \int_{-\infty}^\infty f(x) g(y-x) dx $ (ugh I can never remember this formula : /). But the convolution of the Dirichlet kernel is on $\int_{-\pi}^\pi$.
Versus the group operation on $\mathbb T$: $arg(e^{ia}) + arg(e^{ib}) = a + b$.
@JonasTeuwen I suspect the consequence is that on $\mathbb T$ my characters are $e^{2 \pi i k x}$ whereas on $S^1$ my characters are $e^{ikx}$. Is this correct?
How is this possible? That should be finite since the space has finite measure and hence every $L^2$ function is also $L^1$. By assumption $f$ is $L^2$ and by Parseval, $\|f\|_2 = \|\hat{f}\|_2$...
@JonasTeuwen Hey, I think I have it. These two lemmas are used to show that there exists $f \in C(\mathbb T)$ such that $\hat{f}(e^{ix0}) = \hat{f}(1) = \infty$.
