Analytic function on unit disk, Schwartz lemma type inequality

Question

Let $f:\Bbb{D}\to\Bbb{C}$ be an analytic function that satisfies the inequality $|f(z)|\le\dfrac{1}{1-|z|}.$
How can I show that $|f'(0)|\le 4.$

Also, I am looking for the equality case. Is this a sharp upper bound?

Answer 1

We can show that $|f'(0)|\le 4$ using Cauchy's Integral Formula

$$f'(0)=\frac{1}{2\pi i}\oint_{|z|= r<1}\frac{f(z)}{z^2}\,dz \tag1$$

Taking the magnitude of both sides of $(1)$ reveals

$$\begin{align} |f'(0)|&=\left|\frac{1}{2\pi i}\oint_{|z|=r<1}\frac{f(z)}{z^2}\,dz\right|\\\\ &\le \frac{1}{2\pi }\oint_{|z|= r<1}\frac{|f(z)|}{|z^2|}\,|dz|\\\\ &=\frac1{2\pi}\frac{(2\pi r )\left(\frac{1}{1-r}\right)}{r^2}\\\\ &=\frac{1}{r(1-r)}\tag 2 \end{align}$$

The minimum of $\frac{1}{r(1-r)}$ occurs when $r=1/2$ in which case we have

$$|f'(0)|\le 4$$

as was to be shown!

Answer 2

With the Schwarz lemma I get :

If $r \in (0,1)$ then for $|z| < 1$ : $|f(rz)| \le \frac{1}{1-r}$ and $|f(0)| \le 1$ so that $$g(z)=\frac{f(rz)-f(0)}{1+\frac{1}{1-r}}$$ is analytic on $\mathbb{D} \to \mathbb{D}$ and $g(0) = 0$. Thus by Schwarz's lemma $$1 \ge |g'(0)| = \left|\frac{r f'(0)}{1+\frac{1}{1-r}}\right|=|f'(0)|\frac{r(1-r)}{2-r}$$ whose optimum is at $r \approx 0.6$ with $\frac{r(1-r)}{2-r} \approx 0.18$

I wonder if there is a way to improve this to $|f'(0)| \le 4$ as Mark Viola did with the Cauchy's integral formula.