Showing this is a problem from Axler's textbook on Real Analysis.
$q$ is not the dual exponent of $p$, necessarily, they both just satisfy $1 \leq p, q\leq \infty$
My idea was to use the Closed Graph Theorem:
$(f_k, hf_k) \to (f, w)$. It would suffice to show that a subsequence converges to $(f, hf)$. Because $f_k \to f$ in $p$-norm, I can take a subsequence that converges to $f$ pointwise a.e.: $f_{n_k}(x) \to f(x)$. So if I could show that $hf_{n_k} \longrightarrow hf$ in $q$-norm, I would be done. What I'd like to be able to do is something like this:
$\lim_{n \to \infty}\|hf - hf_{k_n}\|_q = \lim_{n \to \infty}\int |hf - hf_{k_n}|^q d\mu = \int \lim_{n \to \infty}|hf(x) - hf_{k_n}(x)|^qd\mu = 0$
This would show that $hf_k \longrightarrow hf$.
But I can't find a dominating sequence in order to pass the limit into the integral. It seems like I would need to claim that $f \mapsto hf$ is bounded, which is what I'm trying to prove.
Is this approach hopeless? Hints would be appreciated.
You are making things complicated. $f_n \to f$ in $L^{p}$ and $hf_n \to w$ in $L^{q}$ implies $f_k \to f$ almost everywhere and $hf_k \to w$ almost everywhere along some subsequence. Hence, $w=hf $ almost everywhere which proves that the graph is closed.