# Why is this function constant?

I'm trying to prove that an entire harmonic function $u$ that satisfies $|u| \leq\sqrt{\sum_i |x_i|}$ is constant. I think that I have to use Liouville theorem for harmonic functions, however I don't know exactly how to bound it.

• Except in (real) dimension $1$, that estimate doesn't imply $u$ is constant. Consider $u(x) = x_1^2 - x_2^2$. $\lvert u(x)\rvert = \lvert x_1^2 - x_2^2\rvert \leqslant x_1^2 + x_2^2 \leqslant (\lvert x_1\rvert + \lvert x_2\rvert)^2$. – Daniel Fischer Jun 30 '13 at 23:48
• @DanielFischer that is a (nice) answer, which should be posted as such. There is nothing to add to it. – ˈjuː.zɚ79365 Jul 1 '13 at 9:34
• @ˈjuː.zɚ79365 I sort-of expected that the OP was just missing some additional constraints that would imply constancy, and wanted to give him/her time to add those. OP wasn't seen since the comment, so I'll wait for another couple of hours, I think. – Daniel Fischer Jul 1 '13 at 9:54
• @DanielFischer I had this question in a test last month. I think that I didn´t write the correct inequality, however I´m not finding the paper where it was written. I will correct it soon (when I find the paper). – user40276 Jul 2 '13 at 13:46
• @DanielFischer I've edited properly. This is the right inequality. – user40276 Jul 2 '13 at 22:12

A harmonic function is represented by its Poisson integral,

$$u(x) = \frac{1}{R\omega_{n-1}}\int_{S_R} u(y)\frac{R^2 - \lVert x\rVert^2}{\lVert x - y\rVert^n}\, d\sigma(y),$$

for $\lVert x\rVert < R$, where $n$ is the dimension of the space, $S_R = \{ \zeta \colon \lVert\zeta\rVert = R\}$, $\omega_{n-1}$ is the ($n-1$-dimensional) volume of the unit sphere $S_1$, and $d\sigma$ the measure on the sphere.

The Poisson kernel $P_R(x,y) = \frac{1}{R\omega_{n-1}}\frac{R^2 - \lVert x\rVert^2}{\lVert x-y\rVert^n}$ is (for $x \neq y$) smooth, so we can differentiate under the integral,

$$\frac{\partial u}{\partial x_i}(x) = \frac{1}{R\omega_{n-1}}\int_{S_R} u(y)\frac{\partial}{\partial x_i}\frac{R^2 - \lVert x\rVert^2}{\lVert x - y\rVert^n}\, d\sigma(y).$$

The partial derivative

$$\frac{\partial}{\partial x_i}\frac{R^2 - \lVert x\rVert^2}{\lVert x - y\rVert^n} = -\frac{2x_i}{\lVert x-y\rVert^n} - \frac{n(R^2 - \lVert x\rVert^2)(x_i-y_i)}{\lVert x-y\rVert^{n+2}}$$

is, if we constrain $\lVert x\rVert < \frac{R}{2}$, so that $\lVert x-y\rVert > \frac{R}{2}$, bounded by a constant multiple of $\frac{1}{R^{n-1}}$ (the constant depends on $n$, but not on $R$).

Thus from the differentiated Poisson integral, we obtain the estimate

$$\left\lvert\frac{\partial u}{\partial x_i}(x)\right\rvert \leqslant \frac{C}{R^n\omega_{n-1}} \int_{S_R} \lvert u(y)\rvert \,d\sigma(y) \leqslant \frac{C}{R} \max_{y \in S_R} \lvert u(y)\rvert.$$

Now, the condition $\lvert u(x)\rvert \leqslant \sqrt{\sum_i \lvert x_i\rvert}$ gives a bound $\lvert u(x)\rvert \leqslant \sqrt{n}\sqrt{\lVert x\rVert}$, and plugging that into our estimate for the derivative yields

$$\left\lvert\frac{\partial u}{\partial x_i}(x)\right\rvert \leqslant \frac{\sqrt{n}C}{\sqrt{R}}.$$

Passing to the limit $R \to \infty$, we obtain $\frac{\partial u}{\partial x_i}(x) = 0$.

That means $u$ is constant. (If we interpret the growth condition as only constraining large enough $\lVert x\rVert$, the constant can be $\neq 0$, if we interpret it globally, it forces $u \equiv 0$.)