Which of the following inequalities hold?

Question

Let $f(z) = \large\sum_\limits{n=0}^{\infty}\normalsize a_n z^n\:$ be an entire function and let . Which of the following inequalities holds?

$1.\sum _{n=0}^{\infty }\left|a_n\right|^2r^{2n}\le \frac{1}{2\pi }\int _0^{2\pi }\left|f\left(re^{i\theta }\right)\right|^2d\theta \ $

$2.\:\:\sup_\limits{|z|=r}\:|f(z)|^2\leq \large\sum_\limits{n=0}^{\infty}\normalsize |a_n|^2\:r^{2n}$

$3. \sum _{n=0}^{\infty }\left|a_n\right|^2r^{2n}\le \sup _{\left|z\right|=r}\left|f\left(z\right)\right|^2$

$4.\:\:\sup_\limits{|z|=r}\:|f(z)|^2\leq\large\frac{1}{2\pi}\Large\int _{\normalsize 0}^{\normalsize 2\pi}\normalsize|f(re^{i\theta})|^2d \theta$

I applied Parseval Inequality, I got (1) and (3) are true. How do I check the other two? I am not able to prove it.

Answer 1

1) is actually an equality and 1) implies 3). $\frac 1 {2\pi} \int_0^{2\pi} |f(re^{i\theta})|^{2} d\theta \leq \sup \{|f(z)|^{2}: |z|=r\}$ and the inequality is strict unless $|f|$ is a constant on $\{z:|z|=r\}$. So 4) implies $|f|$ is a constant on $\{z:|z|=r\}$. Take any entire function which does not have this property ( e.g., $f(z)=e^{z}$) to see that 4) is not true in general. 2) is equivalent to 4).

Answer 2

The right-hand sides of $(2)$ and $(4)$ are equal (Parseval's identity) so that they are simultaneously true or false. If $(4)$ holds then $$ \sup_{|z|=r} |f(z)|^2\leq\frac{1}{2\pi}\int _{0}^{2\pi}|f(re^{i\theta})|^2 \le \sup_{|z|=r} |f(z)|^2 $$ so that $|f|$ is constant on the circle with radius $r$. For an entire function this implies that $f$ is of the form $f(z) = c z^k$. For all other entire functions we have “$>$” in $(2)$ and $(4)$.

Proof of the last statement (sketch): Consider the function $$ h(z) = f(rz) \overline{f(r/\overline z)} \, . $$ Then show that

• $h$ is holomorphic in $\Bbb C \setminus \{ 0 \}$.
• $h$ is constant on the unit circle, and therefore constant everywhere.
• $f$ has a pole or a removable singularity at $z = \infty$, so that it is a polynomial.
• $f(z) = 0$ is only possible for $z=0$.