MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $L = \sum_{i,j=1}^n -\frac{\partial}{\partial x^i} (a^{ij}(x)\frac{\partial}{\partial x^j}) + \sum_{i=1}^n b^i(x) \frac{\partial}{\partial x^i} + c(x)$ be a second order elliptic operator with smooth coefficients, $\Omega$ a bounded open domain with smooth boundary in $\mathbb{R}^n$, and $f$ be a function in $L^p(\Omega)$. We say that $u \in W_0^{1,p}(\Omega)$ (one weak derivative in $L_p$ and vanishing boundary values) is a weak solution of $Lu = f$ if for all $g \in W_0^{1,q}(\Omega)$ $(q = p^*)$ we have

$\int_{\Omega} \sum_{i,j=1}^n a^{ij}(x)\frac{\partial u}{\partial x^i}\frac{\partial g}{\partial x^j} + \sum_{i=1}^n b^i(x)\frac{\partial u}{\partial x^i}g(x) + c(x)u(x)g(x) = f(x)$.

The standard result is of course that all such weak solutions $u$ actually belong to $W_0^{2,p}(\Omega)$.

I am trying to complete the following proof of this statement:

(1) First we establish an a priori estimate for strong solutions $v \in W_0^{2,p}(\Omega)$ of $Lv = f$:

$\vert\vert v\vert \vert_{W_0^{2,p}(\Omega)} \leq C(\vert\vert f\vert\vert_{L^p(\Omega)} + \vert\vert v\vert\vert_{L^p(\Omega)})$

This is non-trivial but can be established by proving the relevant estimate for the Laplacian with a Newton Potential argument and then using the freezing coefficients technique.

(2) Next we observe that if $L$ is injective on $W_0^{1,p}$, then we are done. This is because $L$ injective implies

$\vert\vert v\vert\vert_{L^p(\Omega)} \leq C\vert\vert Lv\vert\vert_{L^p(\Omega)}$

One proves this by assuming it was false and then using Rellich compactness to produce a non-zero solution to $Lv = 0$.

Having established this estimate, we consider the smooth mollifications $f_{\epsilon}$ of $f$. By $L^2$ theory we can find smooth $v_{\epsilon}$ strong solutions of $Lv_{\epsilon} = f_{\epsilon}$. Since we have

$\vert\vert v_{\epsilon} - v_{\epsilon'}\vert\vert_{W_0^{2,p}} \leq C\vert\vert f_{\epsilon}-f_{\epsilon'}\vert\vert_{L^p(\Omega)}$

The $v_{\epsilon}$ converge to some $v \in W_0^{2,p}(\Omega)$ which will solve $Lv = f$ strongly. Since strong solutions are clearly weak solutions, by the injectivity of $L$ on $W_0^{1,p}(\Omega)$ we conclude that $u = v \in W_0^{2,p}(\Omega)$ and we are done.

(3) We have no way to guarantee that $L$ is injective, for example $0$ might be an $L^2$ eigenvalue of $L$. However, if $p=2$ then we could guarantee that $L_{\lambda} = L + \lambda I$ is injective for some large $\lambda$. If we could establish this fact in the general case we would be done since $L_{\lambda}u = f + \lambda u \in L^p$ and $L_{\lambda}$ injective imply that (2) is applicable.

Question: What is the simplest way to prove that $L_{\lambda} = L + \lambda I$ is injective on $W_0^{1,p}(\Omega)$ for large $\lambda$? Do there exist weak $L^P$ maximum principles?

Of course, I would prefer that the proof not use $L^p$ regularity.

share|cite|improve this question
up vote 3 down vote accepted

I stumbled across the book Second Oder Elliptic Equations and Elliptic Systems by Yah-Ze Chen which appears to contain an answer to my question. It is available on google books here

To save time for those who are interested, here is the relevant argument:

For large $\lambda > 0$ we want to show that $L_{\lambda} = L - \lambda I$ is injective on $W_0^{1,p}$.

Claim: Let $L^T_{\lambda}$ be the transpose of $L^{\lambda}$ with respect to the paring that defines weak solutions. Then we claim that $L^T_{\lambda}$ inective on $W_0^{2,p}$ implies that $L_{\lambda}$ is injective on $W_0^{1,p}$

Proof: Suppose that $L^T_{\lambda}$ is injective on $W_0^{2,p}$. Then, by an argument contained in the original post above, for every $f \in L^p(\Omega)$ we can find $u \in W_0^{2,p}(\Omega)$ such that $L^T_{\lambda}u = f$. Now, suppose that $L_{\lambda}v = 0$ for some $v \in W_0^{1,p}$. After an integration by parts and the definition of weak solution, we see that $\varphi \in W_0^{2,q}$ implies that

$\int_{\Omega}uL^T_{\lambda}\varphi = 0$.

Now choose $\Omega'' \subset\subset \Omega' \subset\subset \Omega$ and a bump function $\rho$ identically one in $\Omega''$ with support in $\Omega'$. $\rho\text{sgn}(u)$ is in $L^q$, and we can find $g \in W_0^{2,q} $ such that $L^T_{\lambda}g = \rho\text{sgn}(u)$. Plugging this $g$ into the above equality gives

$\int_{\Omega''}|u| = -\int_{\Omega\setminus\Omega''}\rho |u|$

Due to the arbitrariness of $\Omega''$, this implies that $\int_{\Omega} |u| = 0$ and hence $u$ is $0$ a.e.

Claim: For $\lambda$ large enough, $L_{\lambda}$ is injective on $W_0^{2,p}$.

Proof: Suppose $L_{\lambda}u = 0$ for $u \in W_0^{2,p}$. Let $\tilde{\Omega} = \Omega \times (-1,1)$, and $\tilde{\Omega'} = \Omega \times (-1/2,1/2)$. Let $(x,t)$ be the coordinates on $\Omega \times (-1,1)$. Then define $v(x,t) = \cos(\sqrt{\lambda}t)u(x)$. Let $\hat{L_{\lambda}} = L_{\lambda} + \partial_t^2$. We have $\hat{L_{\lambda}}v = 0$. The strong solution estimates give

$\vert\vert v\vert\vert_{W^{2,p}(\tilde{\Omega'})} \leq C\vert\vert v\vert\vert_{L^p(\tilde{\Omega})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $

$\vert\vert \partial_t^2v\vert\vert_{L^p(\tilde{\Omega'})} \leq C\vert\vert u\vert\vert_{L^p(\Omega)} \Rightarrow $

$\lambda\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos(\sqrt{\lambda}t|^p)^{1/p} \leq C\vert\vert u \vert\vert_{L^p(\Omega)} \Rightarrow$

$\lambda^{1 - \frac{1}{2p}}\vert\vert u\vert\vert_{L^p(\Omega)}(\int_{-1/2}^{1/2}|\cos t|^p)^{1/p} \leq C\vert\vert u\vert\vert_{L^p(\Omega)}$

Now taking $\lambda$ large enough implies that $u = 0$ almost everywhere.

share|cite|improve this answer

If you have $u\in W_0^{1,p}$ solving $L_\lambda u=0$, then by Sobolev embedding, $u$ is also in $W_0^{k,2}$ for some negative number $k$. (I've never seen this version of the Sobolev embedding theorem, but I'm assuming that one can prove it using Fourier analysis.) Once you have $u\in W_0^{k,2}$ solving $L_\lambda u=0$, where we now think of $L_\lambda:W_0^{k,2}\to W^{k-2,2}$, the $L^2$ theory takes over and tells us that $u$ is actually smooth.

share|cite|improve this answer
How are you proving $L^2$ regularity for negative derivatives? I presume this follows easily from pseudo-differential operator type arguments, but then one might as well use that straightaway for $L^p$ regularity in general. My goal was to avoid such machinery, which is presumably overkill for $L^P$ regularity (I thought), and just use the standard $L^2$ physical space theory. I apologize for not specifying this in my question. – Yakov Shlapentokh-Rothman Mar 3 '11 at 0:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.