Q:

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? 

A:

Define the operator $T(u) = au , u\in L^p$ and let $u_n \to u$ in $L^p$ and $T(u_n) \to v$ in $L^q$. Using closed graph theorem, suffices to show that $T(u) = v$.

$u_n \to u$ in $L^p$ and $T(u_n) \to v$ in $L^q$ implies that there exist a common subsequence $u_{n_k} \to u$ a.e.  and $T(u_{n_k}) \to v$ a.e. 
$u_{n_k} \to u$ a.e.  implies $au_{n_k}\to au$ a.e.  which means $T(u_{n_k}) \to T(u)$. This together with $T(u_{n_k}) \to v$ implies, $v = T(u)$.

