I found this problem Proposition but I am completely stuck. Let $u$ be an harmonic function satisfying $$ \int_{B_1(0)}|\nabla u|^2 \mathrm{d}x \leq 1, $$ where $B_1(0)$ is the unitary ball in $\mathbb{R}^n$. Then $$ \frac{1}{r^n}\int_{B_r(0)}|\nabla u - \nabla u(0)|^2\mathrm{d}x \leq C(n)r^2 \leq \frac{1}{2} $$ for $r\leq \theta(n)$. Obviously I understand how to prove the final bound $1/2$, but I don't get how to obtain the quadratic decay bound.

  • $\begingroup$ What are $n, \theta(n), C(n)$? $\endgroup$ – Robert Lewis Nov 4 '18 at 17:55
  • 1
    $\begingroup$ $n$ is the dimension of the space and $\theta(n), C(n)$ are dimensional constant. $\endgroup$ – Gio712 Nov 5 '18 at 11:13
  • $\begingroup$ Thanks for the clarification. $\endgroup$ – Robert Lewis Nov 5 '18 at 16:07

This follows from the scalar version of the estimate:

Theorem: Let $0<r\leq s$. There are constant $C(n)$ such that $$ \int_{B_r(0)} [f-f(0)]^2 \leq C(n)\left(\frac{r}{s}\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2 $$ for all harmonic function $f$ on $B_s(0)$. The constant $C(n)$ can be chosen to be $4^{n+2}$.

noting that, since $u$ is harmonic, $D_iu$ is harmonic too.

The proof relies on Caccioppolli inequality and the estimate

Lemma: If $f$ is harmonic on $B_{2r}(0)$, then $$ \sup_{B_r(0)} f^2\leq2^n-\!\!\!\!\!\!\!\int_{B_{2r}(0)} f^2 $$ Proof of Lemma: For $y\in B_r(0)$, Cauchy-Schwarz with the mean-value property of harmonic function gives $[f(y)]^2=\left[-\!\!\!\!\!\!\int_{B_r(y)} f\right]^2\leq-\!\!\!\!\!\!\int_{B_r(y)} f^2-\!\!\!\!\!\!\int_{B_r(y)} 1=-\!\!\!\!\!\!\int_{B_r(y)} f^2$. Now expand the volume of integration to $B_{2r}(0)$.

Proof of Theorem: If $4r\geq s$, then we have simply $$ \int_{B_r(0)}[f-f(0)]^2\leq\left(\frac{4r}s\right)^{n+2}\int_{B_r(0)}[f-f(0)]^2\leq 4^{n+2}\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2. $$ Otherwise, we have $r\leq s\leq 4r$ and we need to do a little more work. Separately bounding each component of $\nabla f$ using Lemma, we have $$ \sup_{B_r(0)}\lvert\nabla f\rvert^2\leq2^n-\!\!\!\!\!\!\!\int_{B_{2r}(0)} \lvert\nabla f\rvert^2 $$ and so $$ \begin{align*} -\!\!\!\!\!\!\int_{B_r(0)}[f-f(0)]^2 &\leq -\!\!\!\!\!\!\int_{B_r(0)}\left(r\sup_{B_r(0)}\lvert\nabla f\rvert\right)^2\\ &=r^2\sup_{B_r(0)}\lvert\nabla f\rvert^2\\ &\leq r^2\sup_{B_{s/4}(0)}\lvert\nabla f\rvert^2\\ &\leq r^2 2^n-\!\!\!\!\!\!\!\int_{B_{s/2}(0)}\lvert\nabla f\rvert^2\\ &\leq r^2 2^n \frac{2^n}{s^2}-\!\!\!\!\!\!\!\int_{B_s(0)}[f-f(0)]^2 \end{align*} $$ where the last inequality is by Caccioppolli. Rearranging, we obtain $$ \int_{B_r(0)}[f-f(0)]^2\leq 4^n\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2 $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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