I am really struggling to solve the following, I don't even know how to start. I would appreciate if anyone could give me some help.

Let $$m$$ be a positive integer and $$u : \mathbb{R}^{n} \rightarrow \mathbb{R}$$ be a harmonic function. If $$u(x) = O(\left|x \right|^m)$$ when $$\left|x \right| \to \infty$$, show that $$u$$ is polynomial of degree at most $$m$$.

Thanks in advance for any help.

• So maybe give some context because this is not all trivial and will likely use quite some knowledge about harmonic functions. – hal4math Oct 6 at 20:33

Use the following analogue of the Cauchy estimates for harmonic functions: if $$u:\mathbb{R}^n\to\mathbb{R}$$ is harmonic, then for any multi-index $$\alpha$$ and $$r>0$$, $$$$\vert D^{\alpha}(0)\vert\leq \frac{C_{\alpha}\sup_{\vert x\vert=r} \vert u(x)\vert}{r^{\vert \alpha\vert}},$$$$ where $$C_{\alpha}$$ is some constant depending only on $$\alpha$$, and $$\vert \alpha\vert$$ is the order of $$\alpha$$. For any multi-index $$\alpha$$ with $$\vert \alpha\vert\geq m+1$$, your assumption shows that $$$$\vert D^{\alpha}(0)\vert\leq \frac{C'_{\alpha}r^m}{r^{m+1}}\to 0,$$$$ where $$C'_{\alpha}$$ is some other constant depending only on $$\alpha$$ as $$r\to \infty$$, so every partial derivative at $$0$$ of order at least $$m+1$$ is zero. For any other $$x\in \mathbb{R}^n$$, simply shift the function appropriately and use the same argument to show all derivatives of order $$\geq m+1$$ are zero at $$x$$.
The reason why the Cauchy estimates hold is by writing $$u$$ using the Poisson kernel and differentiating under the integral: see page 33 of these notes for a full proof.