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.
Thanks in advance.
 A: 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 $\lvert u(x)\rvert \leqslant \sqrt[4]{n}\sqrt{\lVert x\rVert}$, and plugging that into our estimate for the derivative yields
$$\left\lvert\frac{\partial u}{\partial x_i}(x)\right\rvert \leqslant \frac{\sqrt[4]{n}C}{\sqrt{R}}.$$
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 $u \equiv 0$.)
