Moser iteration is a technique that gives supremum estimates on a second order elliptic equation using an energy bound. The typical Moser iteration looks like

$$ \| u\|^2_{L^\infty(B_R)} \le C (R^{-2} + \sup |V|)^{n/2} \| u\|^2_{L^2(B_{2R})} $$ when $u$ solves

$$ (\Delta + V)u = 0. $$

To prove the Moser iteration, we first prove the Agmon identity. Let $\eta$ be some cutoff function to be chosen later $b$ any number, and consider \begin{align*} (\Delta u, \eta^2 u^{2b} u) &= (\nabla u, u^b \eta \nabla (\eta u^{b+1}) + \eta u^{b+1} \nabla (\eta u^b)) \\ &= (u^b \eta \nabla u, \nabla (\eta u^{b+1})) + (\eta u^b \nabla u, u \nabla (\eta u^b)) \\ &= \|\nabla (\eta u^{b+1})\|^2 - (u \nabla (\eta u^b), \nabla (\eta u^{b+1})) + (\eta u^b \nabla u, u \nabla (\eta u^b)) \\ &= \|\nabla (\eta u^{b+1})\|^2 + (\eta u^b \nabla u- \nabla (\eta u^{b+1}), u \nabla (\eta u^b)) \\ &= \|\nabla (\eta u^{b+1})\|^2 - \| u \nabla (\eta u^b) \|^2. \end{align*}

Therefore, \begin{align*} 0 &= \| \nabla (\eta u^{b+1}) \|^2 - \| u \nabla (\eta u^b)\|^2 + (V \eta u^{b+1}, \eta u^{b+1}). \end{align*} We can further manipulate the signed term in this expression, by passing $u$ through the derivative to get a term entirely in derivatives of $\eta$ and $\eta u^{b+1}$: \begin{align*} 0 &= (2b+1)\| \nabla (\eta u^{b+1}) \|^2 - \| u^{b+1} \nabla \eta \|^2 - 2b ( \nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) +(b+1)^2(V \eta u^{b+1}, \eta u^{b+1}). \end{align*}

\begin{align*} (2b+1) \| \nabla (\eta u^{b+1})\|^2 &= \| u^{b+1} \nabla \eta \|^2 + 2b (\nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) - (b+1)^2 (V \eta u^{b+1}, \eta u^{b+1}) \\ &\le \| u^{b+1} \eta\|^2 + 2b( \| \nabla (\eta u^{b+1})\|^2 + \| u^{b+1} \nabla \eta\|^2) + (b+1)^2 \|V\|_\infty \| \eta u^{b+1} \|^2 \\ \|\nabla (\eta u^{b+1}) \|^2 &\le ((2b+1) 4^{k+3} R^{-2} + (b+1)^2 \|V\|_\infty) \| u^{b+1}\|^2_{L^2(B_{k-1})} \end{align*} Now use a Sobolev embedding, $$ \| u^{b+1}\|^2_{L^{2n/(n-2)}(B_k)} \le \| \eta u^{b+1} \|^2_{2n/(n-2)} \le S^{-1} \| \nabla (\eta u^{b+1})\|^2. $$

Thus with $p = 2(b+1)$, $$ \| u\|_{L^{p n/(n-2)}(B_k)} \le S^{1/p} p^{1/p} (4^{k+3} R^{-2} + p \|V\|_\infty)^{1/p} \|u\|_{L^p(B_{k-1})}. $$

Now set $p_1 = 2$ and $p_k = n/(n-2) p_{k-1}$. Thus $$ \| u\|_{L^\infty(B_R)} \le \lim_{k \to \infty} \|u\|_{L^{p_{k+1}}(B_k)} \le \prod_{j=1}^k S^{1/p_j} p_j^{1/p_j} (4^{j+3} R^{-2} + p_j \|V\|_\infty)^{1/p_j} \|u\|_{L^2(B_{2R})} \le C (R^{-2} + \|V\|_\infty)^{n/2} \| u\|_{L^2(B_{2R})} $$

In the event that $u$ solves an inhomogeneous equation $(\Delta + V)u = f$, then Agmon's identity reads \begin{align*} (b+1)^2 (f \eta u^b, \eta u^{b+1}) &= (2b+1)\| \nabla (\eta u^{b+1}) \|^2 - \| u^{b+1} \nabla \eta \|^2 - 2b ( \nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) +(b+1)^2(V \eta u^{b+1}, \eta u^{b+1}). \end{align*} Thus we have \begin{align*} (2b+1) \| \nabla (\eta u^{b+1})\|^2 &= \| u^{b+1} \nabla \eta \|^2 + 2b (\nabla (\eta u^{b+1}), u^{b+1} \nabla \eta) + (b+1)^2 ((f - Vu) \eta u^b, \eta u^{b+1}) \\ S^{-1} \| u^{b+1}\|^2_{L^2(B_k)} \le \| \nabla (\eta u^{b+1})\|^2 &\le (2b+1) 4^{k+3} R^{-2} \| u^{b+1}\|^2_{L^2(B_{k-1})} + (b+1)^2 ((f - Vu) \eta u^b, \eta u^{b+1}) \\ \end{align*}