0
$\begingroup$

given $\gamma \geq 0$ if

$lim $ $inf_{r\rightarrow \infty}$ $\displaystyle\frac{M(f;r)}{r^{\gamma}} = 0$

where $M(f;r) = max \{|f(z)|: |z|=r \}.$ Then $f$ is polynom such that his degree are at least $\gamma$.

I can see that $f$ growth is not exponencial so he must be a polynomial but how can i show that his degree are at least $\gamma$?

$\endgroup$
  • 1
    $\begingroup$ I think you mean "at most $\gamma$"... by the way, I don't think it's immediately clear that $f$ is polynomial just because it has subexponential growth. It is certainly not true in real analysis, as $f(x)=\sin(x)$ shows! Anyways, Cauchy estimates will do the trick. $\endgroup$ – Wojowu Nov 25 '16 at 19:54
  • 1
    $\begingroup$ Note: in English, we call them "polynomials", not "polynoms". $\endgroup$ – Wojowu Nov 25 '16 at 19:55
  • $\begingroup$ sorry, but i can't see how to implement the theorem in this exercise... $\endgroup$ – Eduardo Silva Nov 25 '16 at 19:57
  • $\begingroup$ I have just noticed, do you assume $f$ is analytic? If not, then the statement is false. If you do, then by using Cauchy estimates and letting $r\rightarrow\infty$ shows that $n$th coefficient in power series expansion with $n\geq\gamma$ are zero. $\endgroup$ – Wojowu Nov 25 '16 at 20:00

Your Answer

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

Browse other questions tagged or ask your own question.