Let $\Omega$ be a bounded smooth domain, $Lu = D_i \left( a^{ij} (x) D_ju \right)$, and two constants $\lambda, \Lambda > 0$. Suppose the coefficient $a$ is measurable, symmetric, and satisfies $$ a^{ij} \xi_i \xi_j \ge \lambda \vert{\xi} \vert^2 \quad \text{ and} \quad \sum_{i,j}^{} \vert{a^{ij}(x)}\vert \le \Lambda^2, $$ for all $x \in \Omega, \xi \in \mathbb{R}^n$. However, $a$ can be discontinuous and not belong to any $VMO$ or $BMO$ spaces.

Let $u \in W^{1,2}(\Omega)$ be a weak solution of $Lu = g$ for $g \in L^{q/2},~ q > n$. Denote by $\mathtt{data} = (\lambda,\Lambda,n,q)$. Then Theorem 8.15 in Gilbarg-Trudinger, Elliptic Partial Differential Equations of Second Order, says that $$ \Vert u \Vert_{L^{\infty}(\Omega)} \le C (\mathtt{data}) \left( \Vert u \Vert_{L^2(\Omega)} + \Vert g \Vert_{L^{q/2}(\Omega)} \right), $$ and Theorem 8.24 in the same book says that for any $\Omega' \subset \Omega$, $$ \Vert u \Vert_{C^{\alpha} \left( \Omega' \right)} \le K (\mathtt{data}, \text{dist}(\Omega',\Omega)) \left( \Vert u \Vert_{L^2(\Omega)} + \Vert g \Vert_{L^{q/2}(\Omega)}\right). $$

On page 214, after Theorem 8.37, the authors claim that solutions $w \in W^{1,2}_0(\Omega)$ of the eigenvalue problem $$ Lw + \sigma w = 0, $$ belong to $L^{\infty} (\Omega) \cap C^{\alpha}(\Omega)$, thanks to the above theorems 8.15 and 8.24.

My question: If $2 \le n \le 4$, then by Sobolev embedding, we have $w \in W^{1,2}_0 \hookrightarrow L^{q/2}$ for some $q > n$. Therefore, we can apply theorems 8.15 and 8.24. What about the case $n \ge 5$?

Any insights or references are appreciated! Thank you.


2 Answers 2


I recommend Luigi Orsina's Lecture Notes. They are beautifully written, and page 24 you will read Stampacchia's approach, which is (in my view) more elegant than Moser iterations and gives you the result you need in a jiffy.


I'm sorry I ignore the condition of $a_{ij}$, but there is a Moser's iteration for $-\Delta u + V(x)u=0$. For instance, see Struwe's Variational Methods Appendix B lemma B.3. And I think this iteration doesn't need the continuous of $a$.

More precisely, integral the equation $Lu+cu=0$ by $u\min\{|u|^{2s}, M^2\}\in H_{0}^{1}(\Omega)$, then we have $$\int a^{ij}u_i u_j \min\{|u|^{2s}, M^2\} +\frac{s}{2}\int _{|u|^s\leq M} a^{ij} (|u|^2)_i (|u|^2)_j|u|^{2s-2}\leq c\int |u|^2\min\{|u|^{2s},M^2\}.$$ For $s$ small enough such that $u\in L^{2+2s}(\Omega)$ we have $$\int a^{ij}u_i u_j \min\{|u|^{2s}, M^2\} \text{ is bounded uniformly with} ~M.$$ Hence let $M\to \infty$, we have $$\int |D|u|^{s+1}|^2<\infty,$$ which means, by Sobolev embedding, $u\in L^{\frac{(2s+2)n}{n-2}}(\Omega)$. By iteration, you can obtain that $u\in L^{q}(\Omega)$ for all $q>1$.


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