# Liouville theorem - function bounded by Re(z)

Given an entire function $f:\mathbb{C}\to \mathbb{C}$ such that

$\exists \ m \in (-1,0)$ such that $\forall z \in \mathbb{C}$ with $Re(z)\neq 0$: $|f(z)|\le |Re(z)|^m$

show that $f$ is constant.

I'm almost sure that i need to construct an entire function that is bounded by this inequality and then use Liouville's Theorem, but i've been having trouble doing it. I would appreciate any help with this.

Let $g_n(z) = \int_0^{2\pi} e^{in\phi} f(e^{-i\phi} z)\; d\phi$, which is again entire.
If $z = r e^{i\theta}$ we have $$|g_n(r e^{i\theta})| \le r^m \int_0^{2\pi} |\cos(\phi-\theta)|^m \; d\phi = C r^m$$ for some constant $C$. In particular $g_n(z)$ is bounded, and thus constant.
Now relate $g_n$ to the Maclaurin series of $f$.

• Thanks! Could you develop the answer a bit more? I haven't been able to proceed – Eduardo Alejandro Silva Múller Apr 3 '17 at 17:29
• What part don't you understand? – Robert Israel Apr 3 '17 at 18:03
• i don't see how to relate g_n to the McLaurin series of f – Eduardo Alejandro Silva Múller Apr 3 '17 at 18:18
• Robert, why not go to $$\frac{f^{(n)}(0)}{n!} = \frac{1}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^{n+1}}\,dz,$$ estimate, and let $r\to \infty$ to get all Taylor coefficients at $0$ equal to $0?$ – zhw. Apr 3 '17 at 18:53

For large $r$ we have

$$\tag 1 I(r) = \int_0^{2\pi}|f(re^{it})|\, dt \le r^m\int_0^{2\pi} |\cos t|^m \, dt.$$

Now $|\cos t|^m \in L^1[0,2\pi]$ (the singularities at $\pi/2, 3\pi/2$ are like those of $|t|^m$ at $t=0$). Thus $I_r\to 0$ as $r\to \infty.$ However $|f|$ is subharmonic, hence $I(r)$ is a nondecreasing function of $r.$ It follows that $I(r)\equiv 0,$ which implies $f\equiv 0.$