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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.

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  • $\begingroup$ What are $n, \theta(n), C(n)$? $\endgroup$ – Robert Lewis Nov 4 '18 at 17:55
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    $\begingroup$ $n$ 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
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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 $$ \int_{B_r(0)}[f-f(0)]^2\leq 4^n\left(\frac rs\right)^{n+2}\int_{B_s(0)}[f-f(0)]^2 $$

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