Showing continuity of an operator from $L^p$ to $L^q$ Question: Let $1 \leq q \leq p < \infty$ and let $a(x)$ be a measurable function. Assume that $au \in L^q$ for all $u \in L^p$. Show that the map $u \to au$ is continuous.
My Approach: I have tried to use closed graph theorem to show continuity of $a$.
Let $\{u_n\} \subset L^p$ s.t. $u_n \rightarrow u$ in $L^p$ and $au_n \rightarrow v$ in $L^q$. Then we need to show $v = au$ a.e.
$$
|v-au|_q \leq |v-au_n|_q + |a(u_n - u)|_q
$$ 
The first term converges to zero but I got stuck in showing that the second term also converges to zero. Can anyone help?

This is not true. Let $b:L^p \to \mathbb{R}$ be a discontinuous linear functional (such functional always exists on infinite dimensional Banach spaces) and let $f\in L^q.$ Then the operator $a:L^p \to L^q ,$ defined by $a(f)(x) =b(x)\cdot f(x) $ satisfies assumptions of your exercise but $a$ is not continuous.
