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: I prove the special case $p=q=2$:
(1) Continuity of $T_{a}\colon L^2\to L^2, u\mapsto T_a(u):=a u$ is equivalent to $\|a\|_{L^\infty}<\infty$. (The direction that we need is trivial. The other one is not much harder)
(2) Assume we had $\|a\|_{L^\infty}=\infty$. Take disjoint sets $A_n$, $n\in\mathbb{N}$, such that $|a|>2^{n}$ on $A_n$ and $|A_n|\leq 1$. Define $u:=\sum \frac{1}{2^n\sqrt{|A_n|}}1_{A_n}$. Then $\|u\|_{L^2}\leq \sum \frac{1}{2^n\sqrt{|A_n|}}\| 1_{A_n}\|_{L^2}\leq \sum 1/2^n=1$. But $\|au\|^2_{L^2}=\sum \int_{A^n}\frac{|a|^2}{2^{2n}|A_n|}\geq \sum 1=\infty$, in contradiction to the assumption.
A: Assumptions: sigma finite measure space $\Omega$ and $2\leq q\leq p<\infty$
I continue the idea that you have. We want to show that $T_a: L^p\to L^q$ defined by $T_a(u)(x)=a(x)u(x)$ is a closed operator and then using the Closed graph theorem, it will follow that $T_a$ is bounded as it is everywhere defined.
For every set $A$ with finite measure, $u(x)\equiv1\in L^p(A)$. From $au\in L^q(\Omega),\,\forall u\in L^p(\Omega)\Rightarrow$ we can take $u=1_{A}\Rightarrow a\in L^q(A)$. Then
$$\|v-au\|_{L^1(A)}\leq \|v-au_n\|_{L^1(A)}+\|a(u_n-u)\|_{L^1(A)}\\
\leq |A|^\frac{1}{p}\|v-au_n\|_{L^q(A)}+\|a\|_{L^r(A)}\|u_n-u\|_{L^p(A)}\to 0$$
Here we used Holder's inequality $2$ times. For the term $|A|^\frac{1}{p}\|v-au_n\|_{L^q(A)}$ we again used that the measure of $A$ is finite, and in the second term, $r$ is the Holder conjugate of $p$, i.e $r=\frac{p}{p-1}$. So, it is clear that in the case $r=\frac{p}{p-1}\leq q$ or equivalently $p\ge \frac{q}{q-1}$, which for $q\ge 2$ is $p\ge q\ge \frac{q}{q-1}$, we get $v=au$ a.e in $A,\,\forall A\subset \Omega$ such that $|A|<\infty$. And because $\Omega=\cup A_n:\,|A_n|<\infty\Rightarrow v=au$ a.e in $\Omega$.
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, $v = T(u)$.
A: 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.
