# Aufgaben:Problem 11

## Problem

Let \(u\) be a harmonic function on \(\mathbb{R}^3\). Assume there exists \(C>0\), independent of \(x\), such that \(|u(x)| \leq C(1+|x|)\) on \(\mathbb{R}^3\). Show that then

- \(\partial_i u\) is constant on \(\mathbb{R}^3\), \(\forall i = 1,2,3\).
- \(u\) is a linear function, i.e. \(\exists a_0, a_1, a_2, a_3 \in \mathbb{R}\) such that \(u(x) = a_0 + a_1 x_1 + a_2 x_2 + a_3 x_3.\)

*NOTE:* if you choose to use the Corollary on page 3 of ”Newtonian Potential” lecture
notes, you have to give a proof of it.

## Proof

Aforementioned **corollary** from NewtonianPotential.pdf, p.3:

Suppose \(u(x)\) is harmonic on the domain \(\Omega\) and the closed ball \(\overline{B_r(a)} \subset \Omega\), \(C_n=\frac{|S_1(0)|}{|B_1(0)|}\) and \(S_r(a) = \{x \in \mathbb{R}^n : |x-a|=r\}\), then $$\left|\frac{\partial u}{\partial x_j}(a)\right| \leq \frac{C_n}{r} \sup_{y\in S_r(a)}|u(y)|$$

**Proof** (p.3f):

\(0=\frac{\partial}{\partial x_j}(\Delta u) = \Delta \left(\frac{\partial}{\partial x_j} u\right)\) so that \(\frac{\partial}{\partial x_j}u\) is also harmonic on \(\Omega\). Apply the divergence theorem to the vector field \(V_j\) with components \((V_j)_k=u(x)\delta_{jk}\) to obtain
$$
|B_r(0)|\: \left|\frac{\partial u}{\partial x_j}(a)\right|
\overset{(*)}{=} \left| \int_{B_r(0)} \frac{\partial u}{\partial x_j} (x+a) \: \mathrm{d}x\right|
= \left| \int_{B_r(0)} (\mathrm{div}\: V_j)(x+a) \: \mathrm{d}x\right| \\
= \left| \int_{S_r(0)} \langle V_j(y+a), \frac{1}{r}y \rangle \: \mathrm{d}y\right|
= \left| \int_{S_r(0)} u(y+a)\frac{1}{r}y_j \: \mathrm{d}y\right|
\leq \int_{S_r(0)} |u(y+a)| \left| \frac{1}{r}y_j \right| \mathrm{d}y \\
\leq \int_{S_r(0)} |u(y+a)| \: \mathrm{d}y
\leq |S_r(0)| \sup_{y\in S_r(a)}|u(y)|
= |S_1(0)|r^{n-1} \sup_{y\in S_r(a)}|u(y)|
$$
where \((*)\) follows by the *Mean Value Property of Harmonic Functions* (see NewtonianPotential.pdf p. 2).

Divide by \(|B_r(0)|=|B_1(0)|r^n\) and we're done.

\(\square\)

First, note that \(C_3=4\pi / \left( \frac{4}{3}\pi \right) = 3\)

Let \(a\in\mathbb{R}^3\) and \(r>0\).

\(u\) is harmonic on all of \(\mathbb{R}^3\) and thus on \(\overline{B_r(a)}\), too.

Using the corollary: $$|\partial_i u(a)| \leq \frac{3}{r} \sup_{y\in S_r(a)}|u(y)| \leq \frac{3}{r}\sup_{y\in S_r(a)}C(1+|y|) \leq \frac{3}{r}C(1+|a|+r)=3C \left( \frac{1+|a|}{r} + 1 \right)$$ Since \(u\) is harmonic on all of \(\mathbb{R}^3\), we can choose \(r\) arbitrarily large, so let \(r \rightarrow \infty\) $$\Rightarrow |\partial_i u(a)| \leq 3C$$ We thus just showed that \(\partial_i u\) is bounded on \(\mathbb{R}^3\). As \(u\) is harmonic, so is \(\partial_i u\). Making use of Liouville's theorem (see NewtonianPotential.pdf, p.4), claim a) follows directly.

Now, let \(a_i = \partial_i u = \mathrm{const.}\). We see that for \(f(x) := a_1 x_1 + a_2 x_2 + a_3 x_3\), \(\partial_i f = a_i = \partial_i u\) and since functions with identical partial derivatives can only differ by a constant, claim b) follows as well.

\(\square\)