# Proof of interior gradient estimate for Laplace's equation

I have a question about the proof of the estimate $$|\nabla u(x_0)| \leq \frac{n}{R} \max_{\bar{B}_R(x_0)} |u|$$ where $$u$$ is assumed to be harmonic.

Since $$u_{x_i}$$ is harmonic, by the mean value property and integration by parts, $$u_{x_i}(x_0) = \frac{r}{\omega_n R^n}\int_{B_R(x_0)} u_{x_i}(y) dy = \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu_i dS_y.$$ Taking the absolute value, we obtain $$|u_{x_i}(x_0)| \leq \frac{n}{\omega_n R^n} \int_{\partial B_R(x_0)} |u(y)| dS_y \leq \frac{n}{R}\max_{\bar{B}_R(x_0)} |u|.$$ I understand the preceding steps. What I don't understand is how this obviously proves the desired result. This is my attempt at obtaining the desired result: \begin{align*} |\nabla u(x_0)|^2 &= u_{x_1}^2(x_0) + \cdots + u_{x_n}^2(x_0) \\ &\leq \underbrace{\frac{n^2}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2 + \cdots + \frac{n^2}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2}_{\text{n times}}\\ &=\frac{n^3}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2. \end{align*} Taking the square root, $$|\nabla u(x_0)| \leq \left(\frac{n^3}{R^2}(\max_{\bar{B}_R(x_0)} |u|)^2\right)^{1/2} = \frac{n^{3/2}}{R}\max_{\bar{B}_R(x_0)}.$$

I'm not sure where my logic is wrong and I am aware this must be something simple...

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## 1 Answer

The line $$u_{x_i}(x_0) =\frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu_i \,dS_y$$ would be the components of the equation $$\nabla u(x_0) = \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} u(y) \nu\, dS_y$$ And now you do the approximation at this level: \begin{align}|\nabla u(x_0)| &= \frac{n}{\omega_n R^n}\left|\int_{\partial B_R(x_0)} u(y) \nu\, dS_y\right| \\ &\le \frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} |u(y) \nu|\, dS_y \\ &=\frac{n}{\omega_n R^n}\int_{\partial B_R(x_0)} |u(y) |\, dS_y \\ &\le \frac{n}{R}\|u\|_{L^\infty(\overline{B_R(x_0)})}. \end{align}

• Thank you for the help! Aug 9, 2020 at 17:41