1

currently I am trying to solve the following exercise where I need to show that the following family is normal: $$ \mathcal{F} = \{f \colon \mathbb D \to \mathbb C : f \text{ is analytic and } \int_{\mathbb D} |f(x+iy)| dx dy \leq 1\}. $$

My try: By the theory of complex analysis it is enough to show that $\mathcal{F}$ is locally uniformly bounded. Thus take arbitary $f \in \mathcal{F}$ and any $z_0 \in \mathbb D$, take $r < 1-|z_0|$ then since $f$ is analytic it satisfies the mean value property hence $$ |f(z_0)| \leq \frac{1}{2\pi} \int^{2\pi}_0 |f(z_0+re^{i\theta})| d\theta, $$ now integrate everything with respect to $r$ yields that $$ r |f(z_0)| \leq \frac{1}{2\pi} \int^{r}_0 \int^{2\pi}_{0} |f(z_0+re^{i\theta})| d\theta \, ds, $$ now using the co-area formula yields one has $$ r |f(z_0)| \leq \frac{1}{2\pi} \int_{C_r(z_0)} |f(x+iy)| dx dy \leq \frac{1}{2\pi} \int_{\mathbb D} |f(x+iy)| dx dy \leq \frac{1}{2\pi}, $$ but rearranging yields a uniform bound for $f$ since $r$ is only dependent on $z_0$. Since $z_0 \in \mathbb D$ was arbitary we just showed that $\mathcal{F}$ is locally uniformly bounded and hence normal.

Question: Is the abuse of the co-area formula correct here, intuitively this seems to make sense since we integrate over all smaller balls inside $D_{r}(z_0)$ but maybe one can write it more precise?

a.s. graduate student
  • 840

1 Answers1

0

There is a small error since $$ \frac{1}{2\pi} \int_{B_r(z_0)} |f(x+iy)| dx dy = \frac{1}{2\pi} \int^{r}_0 \int^{2\pi}_{0} |f(z_0+re^{i\theta})| s \, d\theta \, ds \, , $$ the factor $s$ (which is the Jacobian determinant of the tranformation from cartesian to polar coordinates) is missing in your calculation.

But that is easily fixed: Multiplying $$ |f(z_0)| \leq \frac{1}{2\pi} \int^{2\pi}_0 |f(z_0+re^{i\theta})| d\theta \, , $$ with $s$ and integrating with respect to $s$ from $0$ to $r$ gives $$ \begin{align} \frac 12 r^2 |f(z_0)| &\leq \frac{1}{2\pi} \int^{r}_0 \int^{2\pi}_{0} |f(z_0+re^{i\theta})| s \, d\theta \, ds \\ &= \frac{1}{2\pi} \int_{B_r(z_0)} |f(x+iy)| dx dy \\ &\le \frac{1}{2\pi} \int_{\Bbb D} |f(x+iy)| dx dy \\ &\le \frac{1}{2\pi} \end{align} $$ and therefore $$ |f(z_0)| \le \frac{1}{\pi r^2} \, . $$

Martin R
  • 128,226
  • I have used $B_r(z_0)$ to denote the ball (disk) with center $z_0$ and radius $r$, and not $C_r(z_0)$ which I associate with a circle. – Martin R Nov 09 '23 at 15:55
  • you are correct that is the better notation to not use $C_r(z_0)$ and the rest is now also clear now. The results still should hold true if one has higher power in the inequality with respect to the funtion, i.e. $\int_{\mathbb D} |f(x+iy)|^{p} dx dy \leq 1$ right? – a.s. graduate student Nov 10 '23 at 09:08
  • @a.s.graduatestudent: Yes, because $|f(z_0)|^p \leq \frac{1}{2\pi} \int^{2\pi}_0 |f(z_0+re^{i\theta})|^p d\theta $ holds for all $p > 0$. For integers $p$ one can use that $f^p$ is holomorphic. For the general case one needs that $|f(z)|^p$ is subharmonic for all $p > 0$, see for example https://math.stackexchange.com/q/3406949/42969. – Martin R Nov 10 '23 at 09:36
  • Very cool thank you a lot for the reference ! – a.s. graduate student Nov 10 '23 at 10:07