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I'm not sure if this requires Liouville's theorem or the use of the integral formula for the Taylor coefficients but I cant get either to work.

By the formula for the Taylor coefficients for $f(z^2)$ we have that:

$$\vert a_n\vert = \Big\vert \frac{1}{2 \pi i} \int_{\vert z\vert =R} \frac{f(z^2)}{z^{n+1}} dz\Big\vert \leq \frac{1}{2 \pi} \int_{\vert z \vert =R} \frac{\vert f(z^2) \vert}{\vert z \vert^{n+1}} \vert dz \vert \leq \frac{1}{2 \pi} \int_{\vert z \vert =R} \frac{2\vert f(z) \vert}{\vert z \vert^{n+1}} \vert dz \vert $$

and then this can be bounded by

$$ \frac{1}{2 \pi } \frac{M_f(R)}{R^{n+1}} 2 \pi R = \frac{M_f(R)}{R^{n}} $$

where $M_f(R)$ is the max of $\vert f\vert $ on the circle $\vert z \vert =R$. I had wanted to take the limit as $R \to \infty$ and have that it equals $0$ unless $n=0$ but this isn't clear to me.

  • What is $|2|f(z)|$? – Paolo Leonetti Apr 23 '19 at 18:50
  • Fixed the title, thanks. –  Apr 23 '19 at 18:52
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    What have you tried? Show your work, please. – AlvinL Apr 23 '19 at 18:52
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    $f(z^2)/f(z)$ is meromorphic and bounded. Therefore, it is entire and constant. Now use $f(z^2)=Kf(z)$ to compute the coefficients of the Taylor of $f(z)$ at the origin. Separate the case $K=0$, in which $f\equiv 0$ and the case in which $K\neq0$, in which equating the corresponding coefficients of each side gives $0=a_1=a_2=...$ – user647486 Apr 23 '19 at 19:30
  • @user647486 thank you very much that was perfect –  Apr 23 '19 at 19:53

2 Answers2

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By induction, we have

$$|f(z^{2^n})|\le 2^n|f(z)|$$

Now let us fix a radius $r>1$, and let $R=r^{2^n}$. By hypotesis, we have

$$\text{max}_{C_{R}(0)}|f(z)|\le 2^n\text{max}_{C_r(0)}|f(z)|$$

Applying Cauchy integral inequalities to the circle $C_{R}$, we obtain

$$|a_m|\le \frac{\text{max}_{C_{r^{2^n}}(0)}|f(z)|}{r^{2^n}}\le \frac{2^n k}{r^{2^{n}}}$$

Taking the limit $n\to \infty$ we have the assertion.

JohnBLewis
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0

Suppose that $|f| \le 1$ on the disk $\Delta(0,2)$. Then $$|f| \le 2 \quad \text{on} \quad \Delta(0,4)$$ $$|f| \le 4 \quad \text{on} \quad \Delta(0,16)$$ $$|f| \le 8 \quad \text{on} \quad \Delta(0,256)$$ and so on.

Try to obtain an inequality of the form $|f(z)| \le \log_2|z|$ valid for large $|z|$.

Umberto P.
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