# Problem involving Fourier series

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!

• 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 – Conrad Sep 14 at 14:57
• 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 $L^1$? – 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.
• 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 – Conrad Sep 15 at 11:33
• 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? – astrobarrel Sep 15 at 13:43