Differences

This shows you the differences between two versions of the page.

--- moser [2025/11/07 11:47] – [Moser from Agmon] spencer
+++ moser [2025/11/07 14:19] (current) – [Inhomogeneous Moser] spencer
@@ Line 14: / Line 14: @@
 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)) \\
+    (\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)) \\
@@ Line 30: / Line 30: @@
 \end{align*}
-==== Moser from Agmon ====
+===== Moser from Agmon =====
 \begin{align*}
@@ Line 50: / Line 50: @@
 Thus
 $$
- \| u\|_{L^\infty(B_R)} \le \prod_{j=1}^k S^{1/p_j} p_j^{1/p_j} (4^{j+3} R^{-2} + p_j \|V\|_\infty)^{1/p_j} \|u\|_{L^2(B_{2R})} \le C (R^{-2} + \|V\|_\infty)^{n/2} \| u\|_{L^2(B_{2R})}
+ \| u\|_{L^\infty(B_R)} \le \lim_{k \to \infty} \|u\|_{L^{p_{k+1}}(B_k)} \le \prod_{j=1}^k S^{1/p_j} p_j^{1/p_j} (4^{j+3} R^{-2} + p_j \|V\|_\infty)^{1/p_j} \|u\|_{L^2(B_{2R})} \le C (R^{-2} + \|V\|_\infty)^{n/2} \| u\|_{L^2(B_{2R})}
 $$
+===== Inhomogeneous Moser =====
+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*}
+To approximate the inhomogeneity, note that at a point $p$ with $R = \frac{|p|}{4}$, then $\| f\|_{L^\infty(B_{2R})} \le C R^{-n}$ by hypothesis, so
+\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^{2n/(n-2)}(B_k)} \le \| \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})} + (b+1)^2 C R^{-n} \|u^{b+\frac{1}{2}}\|^2_{L^2(B_{k-1})} \\
+\end{align*}
+To estimate the $\|u^{b + \frac{1}{2}}\|^2_2 = \| u\|^{2b+1}_{2b+1}$ term, we use the Holder inequality and the previous term in the iteration to bound the $L^{2b+2}$ norm:
+$$
+ \| u\|_{2b+1} \le C\| u \|_{2b+2} (R^n)^{\frac{1}{(2b+1)(2b+2)}}.
+$$
+Thus with $p_k = 2b_k + 2 = 2 \left( \frac{n}{n-2} \right)^k$, it holds that
+\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\|^{p_k}_{L^{p_{k+1}}(B_k)} &\le ((p_k-1) 4^{k+3} R^{-2} + \frac{p_k^2}{4} V^\infty)\| u\|^{p_k}_{L^{p_k}(B_{k-1})} + C p_k^2 R^{\frac{n}{p_k} - n} \| u\|^{p_k-1}_{L^{p_k}(B_{k-1})}
+\end{align*}