5
$\begingroup$

Suppose $f$ has a power series at $0$ that converges in all of $\mathbb{C}$ and $$\int_{\mathbb{C}} |f(x+iy)|dxdy$$

Converges. Prove $f$ is identically zero. I don’t know Liouville’s theorem or any integral formulas yet, so I’m a bit stuck on this one.

A hint is given: “Use polar coordinates to show $f(0)=0$”

Edit: I am open to any suggestions, even those which use Liouville or Cauchy etc

$\endgroup$
  • $\begingroup$ Why $f(0) = 0$ makes things any easier? $\endgroup$ – xyzzyz Jan 29 '18 at 0:03
  • 1
    $\begingroup$ You probably need to tell us what you do know. $\endgroup$ – user99914 Jan 29 '18 at 0:07
  • 1
    $\begingroup$ @JohnMa I know about analytic functions and their properties but haven’t gotten to the derivative yet. I also welcome more advanced methods if it’s still comprehensible to someone with my knowledge.. for reference I’m using Marshall’s book drive.google.com/open?id=1_VQoQJgoow_nG5avmhdDhvBvVWEJirth and this is exercise $12$ in chapter 2 $\endgroup$ – mysatellite Jan 29 '18 at 0:13
  • $\begingroup$ You're going to have to use something from complex analysis; the statement doesn't hold without the assumption that $f$ is entire. Are you familiar with the maximum modulus principle? $\endgroup$ – anomaly Jan 29 '18 at 2:57
  • $\begingroup$ @anomaly Yes I know $\endgroup$ – mysatellite Jan 29 '18 at 3:00
5
$\begingroup$

Let $f(z)=\sum_0 ^{\infty} a_n z^{n}$ be the power series expansion. Write $z=re^{i\theta}$ and integrate with respect to $\theta$ from 0 to $2\pi$. Integrating term by term is permitted because of uniform convergence. You get $2\pi a_0= \int_0 ^{2\pi} f(re^{i\theta}) d\theta$. Note that $a_0 =f(0)$. Multiply both sides by r and integrate w.r.t. r. from 0 to some number R. Using the standard fact that $r dr d\theta =dxdy$ you will see that $|(\int_0 R rdr) 2\pi f(0)|$ is bounded by the given double integral. Hence $R^{2} |f(0)|$ has a bound independent of R. This implies $f(0)=0$. Now apply the result to $f(z+a)$ in place of f to conclude that $f(z+a)$ vanishes at 0 which means $f(a)=0$ for any a.

$\endgroup$
  • 2
    $\begingroup$ The integral is of the absolute value of the function. $\endgroup$ – Mariano Suárez-Álvarez Jan 29 '18 at 8:11
  • $\begingroup$ Yes, and that is what I have used. Absolute value of an integral is less than or equal to the integral of the absolute value. $\endgroup$ – Kavi Rama Murthy Jan 29 '18 at 22:34

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.