Rough Cutoff of Fourier Transform Let $I=[-\frac 1 4,\frac 1 4]$. Is it always true that we have
$$
\|\mathcal F(f1_I)\|_\infty\lesssim \|\mathcal F(f)\|_\infty,
$$
for all smooth functions $f$ supported on $[-\frac 1 2,\frac 1 2]$?
I am asking this question because the left hand side can be thought of as a rough projection of $f$ in the frequency space. The inequality $\|\mathcal F(f1_I)\|_1\lesssim \|\mathcal F(f)\|_1$ is easily seen to be false by taking $f=1$ on $[-\frac 1 4,\frac 1 4]$ and smoothly decreasing to $0$.
 A: Since $\mathcal F(f1_I)$ is a multiplier operator, it is bounded from $L^\infty$ to $L^\infty$ if and only if it is bounded from $L^1$ to $L^1$. Since it is not bounded from $L^1$ to $L^1$ as the example shows, the $L^\infty$ boundedness also fails.
A: *

*The map $L^1\to L^1,f \mapsto f1_{[-1/4,1/4]}$ is bounded of norm $1$, the map $L^1\to L^\infty,f \mapsto \mathcal{F}(f)$ is bounded of norm $1$, the map $\mathcal{F}(L^1)\subset L^\infty \to L^1,  \mathcal{F}(f) \mapsto f$ is unbounded, it doesn't mean the map $\mathcal{F}(L^1)\subset L^\infty \to L^\infty,  \mathcal{F}(f) \mapsto  \mathcal{F}(f 1_{[-1/4,1/4]})$ is unbounded.

*The Fourier transform of $-pv(\frac1{ x})$ is $\frac1{-i \pi}sign(\xi)$ and the Fourier transform of $e^{-\pi x^2}$ is $e^{-\pi \xi^2}$
thus the Fourier transform of $pv(\frac1{ x-1/4}) \ast n e^{-\pi (nx)^2}$ is 
$\frac1{i \pi}sign(\xi) e^{-\pi \xi^2/n^2} e^{2i \pi \xi/4}$ which is bounded,
on the other hand $-(pv(\frac1{ x-1/4}) \ast n e^{-\pi (nx)^2})1_{[-1/4,1/4]}$ is $\ge 0$ and of unbounded $L^1$ norm as $n \to \infty$ so that its Fourier transform is as large as wanted at $\xi=0$.
With $\phi \in C^\infty_c([-1/2,1/2]),\phi \ge 0$ and $ \phi=1$ on $[-1/4,1/4]$ we get
$$f_n =  ( pv(\frac1{ x-1/4}) \ast n e^{-\pi (nx)^2})\phi, \qquad \frac{\|\mathcal{F}(f_n 1_{[-1/4,1/4]})\|_\infty}{\|\mathcal{F}(f_n)\|_\infty} \to \infty$$
