# How to prove a harmonic function is a polynomial?

Given $u(x)$ be entire harmonic function on $\mathbb R^n$ , and satisfy $u(x)\geq-C \left(1+|x|\right)^m$, where $C$ and $m$ are constants. ($m$ is an integer) Prove $u(x)$ is a polynomial of degree less than or equal to $m$.

I tried Taylor series and tried to show the derivatives after $m$ vanish. but it seems unsuccessful.

• Use the Poisson integral to show that the derivatives of higher order vanish. – Daniel Fischer Sep 29 '13 at 16:26
• Check the result I gave in my answer. – Mhenni Benghorbal Sep 29 '13 at 16:37

Subtracting $u(0)$ from $u$, we may assume $u(0)=0$. Since the average of $u$ on every sphere $S_r=\{x:|x|=r\}$ is zero, it follows that the average of $|u|$ is controlled by the negative part $u^-=\max(0,-u)$: $$\frac{1}{|S_r|}\int_{S_r }|u| = \frac{1}{|S_r|}\int_{S_r }(u^+ +u^-) =\frac{2}{|S_r|}\int_{S_r } u^- \le 2C(1+r)^m \tag{1}$$
Harmonic functions satisfy the following interior regularity estimate: For every multiindex $\alpha$, $$|D^{(\alpha)}u(0)|\le \frac{C}{|S_r|\,r^{|\alpha|}} \int_{S_r }|u| \tag{2}$$ This is probably somewhere in Chapter 2 in the book by Gilbarg-Trudinger, but is not hard to prove anyway. By rescaling, (2) reduces to the case $r=1$. Then write $u$ as the Poisson integral, differentiate the kernel accordingly, and use the inequality of the sort $$|D^{(\alpha)}u(0)|\le C\int_{S_1 }|u P^{(\alpha)} | \le C\int_{S_1 }|u |$$ From (1) and (2), the desired conclusion $D^{(\alpha)}u(0)=0$ for $|\alpha|>m$ follows by letting $R\to \infty$.
• @mononono Well, isn't it that we can adjust the constant $C$ and make this happen? We couldn't actually add any constant to $u$ and push it up or down? – ares Sep 27 '18 at 22:40