1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $f$.

$\endgroup$
  • $\begingroup$ Thanks! Could you develop the answer a bit more? I haven't been able to proceed $\endgroup$ – Eduardo Alejandro Silva Múller Apr 3 '17 at 17:29
  • $\begingroup$ What part don't you understand? $\endgroup$ – Robert Israel Apr 3 '17 at 18:03
  • $\begingroup$ i don't see how to relate g_n to the McLaurin series of f $\endgroup$ – Eduardo Alejandro Silva Múller Apr 3 '17 at 18:18
  • $\begingroup$ 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 $0?$ $\endgroup$ – zhw. Apr 3 '17 at 18:53
0
$\begingroup$

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 $f\equiv 0.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.