Boundedness of multiplication operator on $L^p$ spaces.

Question

I am asking myself when a $L^p\to L^q$ multiplication operator is continuous. The following should be true:

Let $a:[0,1]\to\mathbb{C}$ be a measurable function. Let $T_a: L^p([0,1])\to L^q([0,1])$, with $p,q\in[1,\infty]$, $p<q$, be the operator of pointwise multiplication by $a$, i.e., $(T_af)(x):=a(x)f(x)$. Then $T_a$ is continuous if and only if $a=0$ (almost everywhere).

Any suggestion for a rigorous proof?

Answer 1

If $a(x)$ is not the zero function, there is an $\epsilon > 0$ and a set $E$ of positive measure such that $|a(x)| > \epsilon$ on $E$. Use characteristic functions of subsets of $E$ to show that there cannot be such boundedness.