My attempt was to show, $|f(z)|^2<|f(0)|^2+|f(z)-f(0)|^2\leq|f(0)|^2+(\int_\gamma|f'(z)|)^2$, where $\gamma$ is any rectifiable path connecting 0 to $z$. But now it is a line integral, not the integral over a disc is involved.
Could you tell how to apply the condition given? Or is my approach wrong?
Thanks!
Let $\epsilon >0$ It is enough to show that the family is uniformly bounded on $\{z:|z|\leq 1-\epsilon\}$. Let $|z|\leq 1-\epsilon$. Consider the disk $D_{\epsilon}$ of radius $\epsilon /2$ around $z$. It is well known that analytic functions have mean value property so $f'(z)$ is the average of its values on $D_{\epsilon}$. Using the hypothesis and Cauchy -Schwarz inequality we get an upper bound for $|f'(z)|$ on $D_{\epsilon}$. Since $|f(0)| \leq 1$ and $f(z)$ is $f(0)$ plus the integral of $f'$ on the line segment from $0$ to $z$ we have proved that the family is uniformly bounded on $D_{\epsilon}$.