Let us consider $f$ a function in $C(\mathbb{R})\cap L^{\infty}$ admitting a Fourier transform (say, either in the $L^1$ sense or tempered distribution sense if $f \notin L^1$). We assume that $supp(\hat f)\subset \mathbb{R} \setminus [-1,1]$, that is, if $\hat f$ is a function, $\hat f(\xi)=0$ whenever $|\xi|<1$. Let $u$ be an antiderivative of $f$, i.e. $u'=f$.
Prove that $u$ is bounded.
Here is how I started: assuming that $\hat f$ is a function we have $\hat u = \frac{i}{\xi}\hat f(\xi)=\frac{i(1-\chi (\xi)}{\xi}\hat f(\xi)$ where $\chi$ is a smooth function compactly supported $\subset [-1,1]$ and $=1$ around $0$. Then we write $u=\phi *f$ where $\phi = (\frac{i(1-\chi(\xi)}{\xi})^\vee$ and by Young we have $\|u\|_{L^{\infty}} \leq \|f\|_{L^{\infty}}\|\phi\|_{L^{1}}$, thus it remains to prove that $\phi \in L^1$. We have $\phi (x)=\int_{1}^{\infty}\frac{2\sin (x\xi)}{\xi}d\xi$ and by integration by part we can prove $\phi (x)=O(1)$ at $x=0$ and $\phi(x)=O(1/x^2)$ away from zero, thus the result. Here are two questions:
- What is the best way to recover the result in full generality?
- Is there another solution? I feel like there is, since in the $L^1$ case, $u(\infty)=\int_{\mathbb{R}}f(x)dx=\hat f(0)=0$, thus antiderivatives are bounded. But how can one pass to the general case?
N.B. The website tells me that the same question already appears here but the answer given by the OP is not satisfactory, I find it higlhy nonnatural.