Is the characteristic function of an interval a Fourier multiplier bounded on $L^p$? Consider a Fourier multiplier, which we define as a the operator $$T_m : f \longmapsto \mathscr{F}^{-1}(m) \ast f,$$ for some $m$, where $\mathscr{F}^{-1}(\cdot)$ denotes the inverse Fourier transform. 
At the end of the lecture today, the lecturer asked the following question that I haven't been able to figure out: 
If $m : = \mathbb{1}_{[a,b]}$ for some $a,b \in \mathbb{R}$. Is $T_m$ a bounded linear operator from $L^p(\mathbb{R}^d)$ to itself for any $p \neq 2$?
 A: This operator is closely related to the Hilbert transform $H$, which is bounded on $L^p$ for $1<p<\infty$. Indeed, $T_m$ multiplies the Fourier transform of $f$ by $1_{[a,b]}(\xi)$ while  $H$ multiplies it by $i\operatorname{sgn} \xi$. So, $(I-iH)/2$ has Fourier multiplier $1_{[0,\infty)}$ and $(I+iH)/2$ has Fourier multiplier  $1_{[-\infty,0]}$. Translation in the Fourier space, i.e., modulation, is bounded on all $L^p$. Combining these, we can express  $T_m$  as:


*

*Translate Fourier transform by $-a$

*Multiply it by $1_{[0,\infty)}$

*Translate by $a-b$

*Multiply by $1_{(-\infty,0]}$

*Translate by $b$.


Therefore, $T_m$ is bounded on $L^p$ for $1<p<\infty$.
It remains to show that $T_m$ is not bounded on $L^1$ or $L^\infty$. For $L^1$, let $\phi$ be a Schwartz-class function  that is identically $1$ on $[a,b]$, and define $f = \mathcal F^{-1}(\phi)$. Then $f$ is also in the Schwartz class, in particular in $L^1$. But 
$$
T_m f = \mathcal F^{-1}(1_{[a,b]} \phi) = \mathcal F^{-1}(1_{[a,b]})
$$
is not in $L^1$, which can be seen directly (it's a form of sinc function) or by recalling that the Fourier transform of an integrable function is continuous.
For $L^\infty$ one can argue by duality. If $T_m$  was bounded on $L^\infty$, then for every $f\in L^1$, $g\in L^\infty$ we would have 
$$
\left|\int (T_mf)\bar g\right| = 
\left|\int f\overline{T_mg}\right| \le C\|f\|_1\|g\|_\infty
$$
which implies $T_m$ is bounded on $L^1$, contrary to the above.
