Proof that $\lVert u\rVert_{L^p(M)}\leq C\lVert\Delta u\rVert_{L^p(M)}$ for each $u\in W_0^{2,p}(M)$ Let $(M,g)$ be a compact Riemannian manifold with nontrivial boundary and $W_0^{2,p}(M)$ the elements of $W^{2,p}(M)$ that vanish on $\partial M$. I'm trying to understand a proof that there exists a constant $C$ such that
$$\lVert u\rVert_{L^p(M)}\leq C\lVert\Delta u\rVert_{L^p(M)}$$
for each $u\in W_0^{2,p}(M)$. Here is the proof:

Suppose there does not exist such a constant. Then we can find a sequence $u_i$ with $\lVert u_i\rVert_{L^p}=1$ and $\Delta u_i\to 0$ in $L^p$. By the elliptic estimate
$$\lVert u\rVert_{W^{k+2,p}}\ \leq K(\lVert Lu\rVert_{W^{k,p}}+\lVert u\rVert_{L^p})$$
(existence of the constant $K$), it follows that $u_i$ is uniformly bounded in $W^{2,p}$. By the Rellich-Kondrachov compactness theorem, there exists a subsequence of $u_i$ converging to some function $u$ weakly in $W^{2,p}$ and strongly in $L^p$. Then for any compactly supported smooth test function $\psi$,
$$\langle\Delta u,\psi\rangle=\langle u,\Delta\psi\rangle=\lim_{i\to\infty}\langle u_i,\Delta\psi\rangle=\lim_{i\to\infty}\langle\Delta u_i,\psi\rangle=0,$$
and therefore $\Delta u=0$. By elliptic regularity, $u$ must be smooth. Since it vanishes on the boundary, the maximum principle implies that $u$ is identically zero. But this is a contradiction since we must have $\lVert u\rVert_{L^p}=1$.

I feel confused about many of its details:

*

*The uniform boundedness here seems strange to me. In my introductory analysis course, a uniform bound is a constant $M$ such that $|u_i(x)|\leq M$ for every $x$ and for every $i$, but in the proof, it doesn't seem to work this way.

*I have looked up the Rellich-Kondrachov compactness theorem in my PDE book and Wikipedia, and it doesn't refer to subsequences. How could the author grab such a subsequence?

*Where does the equality $\langle\Delta u,\psi\rangle=\langle u,\Delta\psi\rangle$ come from? Is there an inner product on $L^p$? I only know $L^2$ can be equipped with an inner product, which makes it a Hilbert space.

Any suggestion is welcome. Thank you.
 A: *

*I don't know why, but many people say "uniformly bounded" when they really just mean "bounded". In this case it just means that there is a constant $M>0$ such that $\lVert u_i\rVert_{W^{2,p}}\leq M$ for all $i\in\mathbb N$.

*The Rellich-Kondrachov compactness theorem tells you that the embedding $W^{2,p}(M)\hookrightarrow L^p(M)$ is compact. This means that bounded subsets of $W^{2,p}$ are compact in $L^p(M)$, and a subset of a metric space is compact if and only if every sequence has a convergent subsequence. That gives you strong convergence in $L^p(M)$. To additionally get weak convergence in $W^{2,p}(M)$, note that $W^{2,p}(M)$ is separable and reflexive (this holds only if $1<p<\infty$). Then we know from the Banach-Alaoglu theorem that every bounded sequence in $W^{2,p}(M)$ has a weakly convergent subsequence in $W^{2,p}(M)$.

*The symbol $\langle\cdot\,,\cdot\rangle$ does not denote an inner product here, but a dual pairing between say $L^1_{\mathrm{loc}}(M)$ and $C_c^\infty(M)$. In other words,
$$
\langle v,\phi\rangle=\int v(x) \phi(x)\,dx.
$$
The identity $\langle \Delta u,\psi\rangle=\langle u,\Delta \psi\rangle$ is just the definition of the distributional Laplacian.

