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I've been tasked with the following;

Let $u : \mathbb{R}^n \to \mathbb{R}$ be harmonic and $R \ge 0$, $\beta \in \mathbb{N}$. Suppose that; $$\limsup_{|x| \to \infty}\frac{|u(x)|}{|x|^{\beta}}<\infty$$ Then show that $u$ is a polynomial of degree $<\beta$.

In lectures, we've been given a proof of Liouville's Theorem, which states that there are no non-trivial bounded harmonic functions on $\mathbb{R}^n$. This proof utilised a specific gradient estimate, but the question immediately prior to this involved proving the following gradient estimate; $$|D^{\alpha}u(x_0)| \le \frac{n^m e^{m-1} m!}{R^m}\max_{B_R(x_0)}|u|$$ Which I'm sure that I have to use within this question, but I'm slightly confused as to where it will come into play.

Using the given supposition, I've basically set; $$\frac{|u(x)|}{|x|^{\beta}} < C$$ for some constant $C$, which implies; $$|u(x)| < C |x|^{\beta}$$

This closely matches the general set up for generalising Liouville's Theorem for complex numbers, but I'm unsure how to incorporate the given gradient estimate at this point. With the initial statement including $R = 0$, I'll have a singularity if I employ the gradient estimate at this point, but again, I'm not 100% sure how to compensate for this.

Any general hints would be fantastic, as I'm just looking to get the ball rolling on this problem.

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I assume that in your inequality $m=|\alpha|$.

From $|u(x)| < C\,|x|^{\beta}$ it follows that $$ \max_{B_R(x_0)}|u|\le C\,R^\beta. $$ Then $$|D^{\alpha}u(x_0)| \le \frac{n^m e^{m-1} m!}{R^{m-\beta}}.$$ Taking limits as $R\to\infty$ we get $$ D^{\alpha}u(x_0)=0,\quad |\alpha|\ge m. $$ This implies that $u$ is a polynomial.

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  • $\begingroup$ I'm following to a point. I had the $|D^{\alpha}u(x)|$ term bounded by a constant divided by $R^{m-\beta}$. So clearly, as $R \to \infty$, the LHS becomes equal to zero. However, that only holds true for $m > \beta$. Is this even a relevant point, or should I just note that as long as $|\alpha| \ge m$, I acquire the desired result?? $\endgroup$ – Jack Mar 26 '15 at 3:42
  • $\begingroup$ All the derivatives of order $>\beta$ are $0$. This implies that $u$ is a polynomial of degree $\le\beta$. $\endgroup$ – Julián Aguirre Mar 26 '15 at 10:04

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