 Solution

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.