Let $f,g: \mathbb R \to \Bbb R$ be two $2\pi$ periodic functions.

(a) If $f$ is $C^{\infty}$ prove that for each $n$ there is $C_n \in \Bbb R$ such that$$\left|\hat{f}_k\right| \leq \frac{C_n}{\left|k\right|^n}\quad \forall k \in \Bbb Z \setminus \{0\}$$ where $\hat{f_k} = \frac{1}{2\pi}\int_0^{2\pi} f(t) e^{ikt} dt$ is the $k$-th Fourier coefficient of $f$.

(b) For $f \in C^{\infty}$ and $g \in L^{\infty}$ prove that $$\lim_{n\to \infty} \int_0^{2\pi} f(t) g(nt) dt = 2\pi \hat{f_0}\hat{g_0}$$

(c) Prove that (b) holds also if $f \in L^1$.

Now, I solved (a) using standard repeated integration by parts $n$ times, but what about the last two points? Can someone give me a hint? Thanks!

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    $\begingroup$ Hint: show the result for $f$ a trigonometric polynomial (where it is easy since if $m$ is its degree all except the constant $\hat{g_0}$ terms of the Fourier terms of $g((m+k)t, k \ge 1$ are orthogonal on $f$), and use that such approximate $L^1$ in norm plus the dominated convergence theorem $\endgroup$ – Conrad Sep 14 at 14:57
  • $\begingroup$ Thanks! If you want to, i would really appreciate if you could check out my answer below. In all the cases where we needed the dominated convergence theorem I used the fact that the fourier series converges in $L^2$ in norm, and $f \in L^2$ because of course is $C^{\infty}$. But now what about point(c) when $f$ is only in $L^1$? $\endgroup$ – astrobarrel Sep 15 at 9:37

So, I think I understood how to do point (b). As suggested in the comments, first prove it for a function $f$ like $$f_m(t)= \sum_{\left| k\right|\leq m} \hat{f_k} e^{ikt}$$ Therefore, if $n \geq m$ we can write $$\int_0^{2\pi}f(t)g(nt) dt= \sum_{\left| k\right|\leq m} \hat{f_k} \int_0^{2\pi} e^{ikt}g(nt)dt = \sum_{\left| k\right|\leq m} \hat{f_k} \int_0^{2\pi} \sum_{l \in \Bbb Z}e^{ikt} e^{inlt}\hat{g}_l dt$$ Now all the terms are zero except the one with $l = k = 0$ so all that stuff is equal to $$2\pi\hat{f_0}\hat{g_0}$$ as wanted. So now for each $n$ we get

$$\int_0^{2\pi}f(t)g(nt) dt = \sum_ k \hat{f_k} \int_0^{2\pi} e^{ikt}g(nt)dt = \sum_ { k\geq n } \hat{f_k} \int_0^{2\pi} e^{ikt}g(nt)dt +\hat{f_0}\hat{g_0} 2\pi $$

Letting $n$ go to infinity we see that the first term in the last member goes to zero so we get the result.

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  • $\begingroup$ the last part needs a bit more rigor since it involves a remainder (infinite) sum $R_n$ in $k \ge n$ and then letting $n \to \infty$ and it is not obvious why the sum $R_n \to 0$ (each term does by Riemann Lebesgue and the boundness of $g$ but why the infinite sum does?); here one uses that for each $f \in L^1$ there are $f_k$ trigonometric polynomials $|f_k-f|_1 \to 0$; this means in particular $\hat{f_{k,0}} \to \hat{f_0}$; but $|\int_0^{2\pi}f_k(t)g(nt)dt-\int_0^{2\pi}f(t)g(nt)|dt \le |f-f_k|_{1}||g||_{\infty} \to 0$ with $k$, uniform in $n$ so conclude $\endgroup$ – Conrad Sep 15 at 11:33
  • $\begingroup$ For the reimainder what if I just estimate $\left|R_n\right| \leq \Vert g\Vert_{\infty} \sum_{k\geq n}\left|\hat{f_k}\right|$ so that this goes to zero when $n$ goes to infinity because $\sum_k\left|\hat{f_k}\right|$ is a convergent series? (by point (a)). Does it make sense? $\endgroup$ – astrobarrel Sep 15 at 13:43
  • $\begingroup$ Ok, I see what you mean. My argument is ok for point (b) but not (c) and we need yours to take care of point (c). Thanks a lot! $\endgroup$ – astrobarrel Sep 15 at 13:50
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    $\begingroup$ yes - point b can be addressed directly, because you have good estimates but point c needs approximation in norm $\endgroup$ – Conrad Sep 15 at 14:24
  • $\begingroup$ very clear, thank you very much $\endgroup$ – astrobarrel Sep 15 at 14:25

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