I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. Let $1 \le q < p \le \infty$. Let $a:\Omega \to \mathbb R$ be measurable such that $au \in L^q (\Omega)$ for all $u \in L^p (\Omega)$. Then $a \in L^r (\Omega)$ with $$ r = \begin{cases} \frac{pq}{p-q} &\text{if } p < \infty,\\ q &\text{if } p=\infty. \end{cases} $$
Could you confirm if my below attempt is correct?
Proof The case $p = \infty$ follows by picking $u =1$. Let's assume $p < \infty$. We define a linear operator $$ T : L^p (\Omega) \to L^q (\Omega), u \mapsto au. $$
Let's prove that the graph of $T$ is closed. Let $u,u_n \in L^p (\Omega)$ and $f \in L^q (\Omega)$ such that $(u_n, Tu_n) \to (u, f)$ in $L^p \times L^q$. There is a sub-sequence $\varphi$ of $\mathbb N$ such that $u_{\varphi (n)} \to u$ $\mu$-a.e. Hence $Tu_{\varphi (n)} \to Tu$ $\mu$-a.e. Clearly, $(Tu_{\varphi(n)})$ is a Cauchy sequence. So $Tu_{\varphi (n)} \to Tu$ in $L^q$. Hence $Tu=f$ $\mu$-a.e. By closed graph theorem, $T$ is continuous. Then there is $C \in [0, \infty)$ such that $\|T u\|_q \le C \|u\|_p$ for all $u \in L^p$, or equivalently $$ \int |au|^q \le C^q \bigg ( \int |u|^p \bigg )^{q/p} \quad \forall u \in L^p (\Omega). $$
Notice that $|u| \in L^p (\Omega) \iff |u|^q \in L^{p/q} (\Omega)$. So $$ \int |a|^q |v|\le C^q \|v\|_{p/q} \quad \forall v \in L^{p/q} (\Omega). $$
We consider a linear map $$ F:L^{p/q} (\Omega) \to \mathbb R, v \mapsto \int |a|^q v. $$
Then $|Fv| \le C^q \|v\|_{p/q}$ for all $v \in L^{p/q} (\Omega)$. Hence $F$ is continuous. Let $k$ be the exponent conjugate of $p/q$, i.e., $\frac{1}{k} + \frac{1}{p/q}=1$. This implies $k=p/(p-q)$. We have the isometric isomorphism $L^{p/q} (\Omega)^* = L^k (\Omega)$. So there is a unique $b \in L^k (\Omega)$ such that $$ F v= \int bv \quad \forall v \in L^{p/q} (\Omega). $$
It follows that $|a|^q=b$ $\mu$-a.e. and thus $|a| \in L^{qk} (\Omega)$. Finally, $qk = pq(p-q)$. This completes the proof.
