# Let $u \in C^3(\mathbb{R}^n)$ and harmonic such that $u(x) = o(|x|)$ when $|x| \rightarrow \infty$. Show that $\nabla u(0) = 0$ and u is constant

Let $$u \in C^3(\mathbb{R}^n)$$ and harmonic such that $$u(x) = o(|x|)$$ when $$|x| \rightarrow \infty$$.
1. Show that $$\nabla u(0) = 0$$
2. Show that u is a constant function.

To show that $$\nabla u(0) = 0$$, I tried to show that $$\frac{\partial u}{\partial x_i}(0) = 0, \forall i$$, but came up with nothing.
For example, I know that the partial derivatives are also harmonic functions, so using the mean value theorem i get that:

$$\frac{\partial u}{\partial x_i} = \frac{1}{\operatorname{Vol}(r \cdot B^n)}\int_{r \cdot B^n}\frac{\partial u}{\partial x_i}(x) dx$$

and then I tried to use the divergence theorem on the vector field $$F = (0,\ldots,0,u,0,\ldots,0)$$ where the $$u$$ is in the i-th coordinate, so I get: (sorry about the strange formatting of that vector)

$$= \frac{1}{\operatorname{Vol}(r \cdot B^n)} \int_{r \cdot S^{n-1}} \left< \left(\matrix{0\\ \vdots\\u\\ \vdots \\ 0}\right),\frac{x}{r}\right> dS =\frac{1}{\operatorname{Vol}(r \cdot B^n)} \int_{r \cdot S^{n-1}}u \cdot \frac{x_i}{r}dS$$

and if $$u \cdot x_i$$ was harmonic, I would've finished here by using the mean value theorem again because obviously $$u \cdot x_i = 0$$ at $$x=0$$, but it isn't harmonic for sure.

Any help will be very appreciated.
Do note that this is a question from a past exam in an advanced multivariate calculus class, so we don't have so much claims about harmonic functions and that's why I thought I could find a solution with the divergence theorem.
I did find this answer, which somehow answers both questions together, but seems like a complete overkill, and probably this is not close to the solution the professor thought of when he wrote the questions and the solution to the second question somehow involves the first question.

If you take the Poisson integral formula and differentiate under the integral sign it follows that there exists $$C$$ such that $$|\nabla u(0)|\le C\sup_{|x|=1}|u(x)|.$$
Now for $$r>0$$ let $$u_r(x)=u(rx)$$. Then $$|\nabla u_r(0)|=r|\nabla u(0)|$$, so $$r|\nabla u(0)|\le C\sup_{|x|=r}|u(x)|=o(r);$$that can't happen unless $$\nabla u(0)=0$$.
Now if you let $$v(x)=u(x+p)$$ this shows that $$\nabla u(p)=\nabla v(0)=0.$$
Details added on request. The Poisson kernel can be found for example here; for $$r=1$$ it's $$P(x,\zeta)=\frac{1-|x|^2}{\omega_{n-1}|x-\zeta|^n}.$$So above if $$|x|<1$$ we have $$u(x)=\int_{|\zeta|=1}P(x,\zeta)u(\zeta)\,d\sigma(\zeta).$$Since $$P(x,\zeta)$$ depends smoothly on $$x$$ it follows that $$\nabla u(0)=\int_{|\zeta|=1}\nabla_x P(0,\zeta)u(\zeta)\,d\sigma(\zeta),$$hence $$|\nabla u(0)|\le c\max_{|x|=1|}|u(x)|,$$with $$c=\int_{|\zeta|=1}|\nabla_x P(0,\zeta)|\,d\sigma(\zeta).$$