3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So maybe give some context because this is not all trivial and will likely use quite some knowledge about harmonic functions. $\endgroup$ – hal4math Oct 6 at 20:33
1
$\begingroup$

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 $r>0$, \begin{equation} \vert D^{\alpha}(0)\vert\leq \frac{C_{\alpha}\sup_{\vert x\vert=r} \vert u(x)\vert}{r^{\vert \alpha\vert}}, \end{equation} 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 \begin{equation} \vert D^{\alpha}(0)\vert\leq \frac{C'_{\alpha}r^m}{r^{m+1}}\to 0, \end{equation} 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.