1
$\begingroup$

While reading the book "Elliptic Partial Differential Equations" by Han and Lin, I failed to understand the proof of the interior gradient estimate for harmonic functions. The theorem says that if $u$ is harmonic in $B_1 (\subset \mathbb{R}^n)$, then $ \sup_{B_{1/2}}{|Du|} \le c\sup_{\partial B_1}{|u|}$ for some positive constant $c=c(n).$ The proof illustrated in the book begins with choosing a cut-off function $\phi = \eta^2$ for some $\eta \in C_{0}^{1}(B_1)$ with $\eta = 1$ in $B_{1/2}.$ Directly calculating, one can show $\Delta(\eta^2|Du|^2) = 2\eta\Delta\eta|Du|^2 + 2|D\eta|^2|Du|^2 + 8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u} + 2\eta^2\sum_{i,j=1}^{n}(D_{ij}u)^2.$ What I do not know is the following inequality: $\Delta(\eta^2|Du|^2) \ge (2\eta\Delta\eta-6|D\eta|^2)|Du|^2.$ The book explains that "Holder inequality" is used here, but the only thing that I know regarding to Holder inequality is what appears in the usual real analysis book in the chapter of $L^p$ spaces (inequality for integral). I guess by the form of the equation, we need to say something about the third term: $8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u}.$ To be specific, obtaining the inequality $8\eta\sum_{i, j =1}^{n}{D_{i}\eta D_{j}u D_{ij}u} \ge -8|D\eta|^2|Du|^2$ will end the proof. Can anyone help me to see how Holder inequality could be used to prove the claimed inequality? Thanks in advance.

$\endgroup$
3
$\begingroup$

From googling, I managed to end up at these MIT notes that expand slightly on the proof (but don't use Holder's). The trick is as follows. Note that it is enough to show

$$ 8 \eta \sum_{ij} D_i \eta D_j u D_{ij} u + 2\eta^2 \sum_{ij} (D_{ij}u)^2 \ge - 8 |D\eta |^2 |Du|^2$$

Note that in each term, $\eta$ and $D_{ij}u$ appear in equal powers. So define $$ X_{ij} = \eta D_{ij} u ,\quad Y_{ij} = D_i \eta D_j u$$ Then the above inequality is equivalent to $$ \sum_{ij} 4X_{ij}Y_{ij} + 4Y_{ij}^2 + X_{ij}^2 \ge 0$$ but of course this LHS is nothing but $$ \sum_{ij} 2X_{ij}(2Y_{ij}) + (2Y_{ij})^2 + X_{ij}^2 = \sum_{ij} (2Y_{ij}+X_{ij})^2$$ which is clearly non-negative, so the original inequality that we wanted to prove is also true.

For vectors, the inequality $|a+b|^2\ge 0$ can be used to prove Cauchy-Schwarz ($\sum_i a_ib_i \le |a||b|$) so I can imagine there is a version of this proof that uses Cauchy-Schwarz...but I don't see it.

In addition, this book (A Basic Course in Partial Differential Equations) by Han expands on the proof in a different way, using the Cauchy inequality (i.e. Young's inequality for products, a.k.a. the "rob-Peter-to-pay-Paul" inequality). enter image description here

$\endgroup$
  • $\begingroup$ Thanks Khor! I've learned a lot reading your solution. $\endgroup$ – Euduardo Jan 12 at 2:57
  • $\begingroup$ @Euduardo You're welcome :) $\endgroup$ – Calvin Khor Jan 12 at 10:32

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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