Assume $E$ has finite measure, $\{f_n\}$ is a sequence of functions in $L^p(E)$, and $f$ is in $L^p(E)$, $1<p<\infty$. Suppose $\lim_{n\to\infty}\int_Ef_ng=\int_Efg$ for all $g\in L^q(E)$. Show that there is a constant $M$ such that $\|f_n\|_p\le M$ for all $n$.
Solution so far: We know that $L^p(E)^*$ is a normed linear space. We also know that every function $h\mapsto \int_E gh$ is given by some $T\in L^p(E)^*$, and we may call $\mathcal F\subseteq L^p(E)^*$ the set of all of these $T$. It is probably possible to prove that $\sup_{T\in \mathcal F}|T(f)|<\infty$ for every $f\in L^p(E)$. If so we can apply the Weak Principle of Uniform Boundedness.
With this we would have that $\sup_{T\in \mathcal F}\|T\|_* < \infty$.
We could also say by the Riesz Representation theorem that every $T\in L^p(E)^*$ is of the form $h\mapsto \int_E gh$ for some choice of $g$. So in effect $\mathcal F = L^p(E)^*$.
This is about where I'm stuck. I'm not perfectly certain how to prove that the condition for uniform boundedness holds, but if I could, I'm not certain what to do with the result. Is it that the sup over all the star-norms is in some sense the star-norm over the star-norms, and because $(L^p(E)^{*})^{*} = L^p(E)$? If so, then I take it that we have shown that each $\|f\|_p$ is some $\sup_{T\in \mathcal F}\|T\|_*$ which is finite ... this can't be the right idea because it wouldn't imply uniform boundedness over the set.
Am I on a very wrong track?
In this setting the collection of linear functionals (operators) $T_n: L^q \mapsto \mathbb{R}$ defined by $T_n(g) = \int f_n g $ satisfies $\sup_{n\ge 1} |T_n(g)| < \infty$ for all $g$. The reason for this is that $T_n(g)$ is a convergent sequence in $\mathbb{R}$ for each fixed $g$, and is hence bounded. The uniform boundedness principle then implies that $$ \sup_{n\ge 1, \|g\|_q = 1}|T_n(g)| = \sup_{n\ge 1} \|T_n\|_{op} < \infty. $$ Now what we wish to check is that $\|T_n\|_{op}= \|f_n\|_p$. By Holders inequality, $|T_n(g)| \le \|f_n\|_p \|g\|_q$, so $\|T_n\|_{op}\le \|f_n\|_p$. Further, define whenever $\|f_n\|_p >0$ (otherwise $f_n = 0 $ a.e. and $\|T_n\|_{op}=0=\|f_n\|_p$) $g_n(x) = \|f_n\|_p^{p-1}|f_n(x)|^p/f_n(x)$, if $f_n(x)\ne 0$, and $g_n(x)=0$ otherwise. Check that $\|g_n\|_q=1$, and $|T_n(g_n)|=\|f_n\|_p$. These imply $\|T_n\|_{op}= \|f_n\|_p$, hence $\sup_{n\ge 1} \|f_n\|_p < \infty.$