I would like to prove the following:
Let $u$ be defined on $\overline{B_1} \subset \mathbb{R}^d$ with $d > 2$. Let $u$ be harmonic and positive, in $B_1$. Let $x_0 \in \partial B_1$ such that $u(x_0) = 0$. Prove that $\partial u_{\nu} (x_0) < 0$ where $\nu$ is the exterior normal to the boundary.
A hint was provided, which said that prove $u(x) \geq v(x) = c(1-\|x\|^{2-d})$ for $c$ sufficiently small.
I am not sure where to begin with this question. I have an amalgamation of ideas that I would like to share. First, we see that $v(x)$ is harmonic, as it is a constant plus a linear multiple of the fundamental solution. Moreover, we see that $v$ is $0$on the unit sphere. I am not sure what else I can use here. The big issue for me seems to be the singularity in $v(x)$ at $x = 0$. I am thinking the hint allows us to define $c$ to control the singularity, but this doesn't seem to make sense.
I guess I 1)cannot prove the hint, 2) cannot see how the hint implies the inequality. Thanks in advance for any help.
Something like the following could work:
Let's start with the basic idea. Since $u$ is positive and harmonic on $B_{1}$ and $u(x_{0}) = 0$, we have that $x_{0}$ is a local min so $u_{\nu}(x_{0}) \leq 0$.
Let $v(x) = \Phi(x) -\Phi(x_{0})$ where $\Phi$ is the fundamental solution to Laplace's equation. We observe that $v$ is positive on $B_{1}-B_{0.5}$ (via the form of the solution), $v=0$ on $\partial B_{1}$ (via radial symmetry), and harmonic on $B_{1}-B_{0.5}$. Moreover, we can pick $\epsilon$ small enough so that $\epsilon v \leq u$ as in the the hint. Finally, we observe $v_{\nu}(x_{0}) < 0$.
Let $w = u - \epsilon v$. We observe that $w(x_{0}) = 0$ as $v=0$ on the boundary, $\Delta w = 0$ and $w$ positive on $B_{1}-B_{0.5}$. Since $w$ is harmonic and positive, the maximum principal tells us that $w$ has a local min at $x_{0}$ so $w_{\nu}(x_{0}) \leq 0 $ so $u_{\nu}(x_{0}) \leq \epsilon v_{\nu}(x_{0}) < 0$ since $v_{\nu}(x_{0}) < 0$. Thus, $u_{\nu}(x_{0}) < 0$.