Fourier coefficients of smooth functions behave like Schwartz functions? [duplicate]

Let $f : \mathbb R \to \mathbb R$ be a $1$-periodic $L^1([0,1])$ function. Hence, the Fourier coefficients $$a_n = \int_0^1 f(x)\exp(-2\pi i nx) dx$$ are well definied. Do we now have the equivalence $$f \in C^\infty(\mathbb R) \quad\Leftrightarrow\quad \forall k \in \mathbb N \lim_{n \to \infty} a_n n^k = 0? $$ I think I proved the direction "$\Leftarrow$" (if I'm not mistaken, please correct me otherwise), so I'm only left with "$\Rightarrow$". Is this true as well?

Solution 1:

As mentioned in the comments, integration by parts (take $u = f(x)$, $v' = e^{-2 \pi i n x}$) reveals \begin{align} \widehat{f}(n) & = \int_{0}^{1} f(x) e^{-2 \pi i n x} \; \text{d}x = f(x) \frac{e^{-2 \pi i n x}}{- 2 \pi i n}\bigg|_{x = 0}^{1} + \frac{1}{2 \pi i n} \int_{0}^{1} f(x) e^{-2 \pi i n x} \; \text{d}x \\ & = \frac{1}{2 \pi i n} \int_{0}^{1} f'(x) e^{-2 \pi i n x} \; \text{d}x = \frac{1}{2 \pi i n} \widehat{f'}(n), \end{align} for any $n \in \mathbb Z$, where in the second to last equality we used that $f$ is 1-periodic. One can now proceed inductively to obtain that $$ \widehat{f}(n) = \frac{1}{(2 \pi i n)^k} \widehat{f^{(k)}}(n) $$ Hence if $f \in \mathcal{C}^{k}(\mathbb{R})$, then $$ \lim_{n \to \infty} n^k \hat{f}(n) = \lim_{n \to \infty} n^k \frac{1}{(2 \pi i n)^{k}}\widehat{f^{(k)}}(n) = \lim_{n \to \infty} \frac{1}{(2 \pi i)^{k}}\widehat{f^{(k)}}(n) = 0, $$ as $\widehat{} \colon L^1 \to \mathcal{C}_0(\mathbb{R})$, that is, the Fourier transformed function is a continuous function vanishing at the boundary.