Assume $C[0,1]$ has $p$-norm, $1 \leq p \leq \infty, \ h(x)\in C[0,1]$.
Define $T_h f(t) = h(t)f(t), \ t\in [0,1]$, find $p$ such that $T_h$ is a bounded linear operator from $L^2[0,1]$ into $L^p[0,1]$
My attempt:
Need to show $\Vert T_h f(t)\Vert_p \leq a\Vert f(t)\Vert_2$, for some $a\in [0,\infty)$.
Know by generalized $H\ddot{o}lder$’s Inequality,
$\Vert h(t)f(t)\Vert_p$ $\leq \Vert h(t)\Vert_r \Vert f(t)\Vert_2$, where $\frac{1}{p}=\frac{1}{r}+\frac{1}{2}, 1\leq r\leq \infty$
Then for $\frac{2}{3}\leq p \leq 2$, it works. But I don't think this is a complete answer. And in this way, I cannot find the matrix norm.
Any help would be appreciated.
Best Answer
Partial answer: If $p \leq 2$ then $\|hf\|_p \leq M\|f\|_p \leq M \|f\|_2 $, where $M=\|h\|_{\infty}$. So $T$ is a bounded operator whenever $p \leq 2$. But this is not a necessary condition. $T$ can be bounded even when $p >2$. (Consider $h=0$).
If $\inf \{|h(x)|: 0 \leq x \leq 1\} >0$ then we can show that $p\leq 2$ is necessary and sufficient.
EDIT: As pointed out by MaoWao in his comment it is enough if $h $ is not the zero function: there is an interval on which $|h| >0$ and, if $p>2$ we can find a sequence of continuous functions $\{f_n\}$ on that interval such that $\{f_n\} \to 0$ in $L^{2}$ but not in $L^{p}$.