# Function derivative is bounded on unit disc. Show the function is bounded from [0,1) by infinity.

Let f be analytic in the unit disk. Assume there is a positive constant M such that $\int_0^{2\pi}{ |f'(re^{i\theta})| }d\theta$ $\leq$ M, $0$ $\leq$ $<$ $1$.

Prove that $\int{_{[0,1)}|f(x)|dx}$ $<$ $\infty$.

So far I have attempted to use the Cauchy representation formula for $f'(re^{i\theta})$ and to somehow reverse parameterize the first integral using the circle $x = re^i\theta$ as x goes from 0 to 1. Both of these were to no avail, am I even looking in the right direction? Any help or suggestions?

• You should rewrite the title, as it starts off assuming something that's not given, and concludes with something that doesn't make too much sense. – zhw. Apr 13 '16 at 4:21

I think you gave up too early on dear Cauchy: Fix $r\in [0,1)$ for the moment. Then $f_r(z) = f(rz)$ is holomorphic $D(0,1/r).$ Thus by Cauchy's theorem,

$$(f_r)'(z) = \frac{1}{2\pi i}\int_{|\zeta|=1}\frac{(f_r)'(\zeta)}{\zeta - z}\, d \zeta$$

for any $z\in \mathbb D.$ Now use $(f_r)'(z)=rf'(rz)$ and make a simple estimate to get

$$r|f'(rz)| \le \frac{1}{2\pi}\frac{r}{1-|z|}\int_0^{2\pi}|f'(re^{it})|\, dt\le \frac{M}{2\pi}\frac{r}{1-|z|}.$$

This should help.