How do I evaluate the coefficients of a function, that is analytic in the unit disk with a power series $\Sigma a_n z^n$ representation that has the following property $$|f'(z)| \leq \frac{1}{1-|z|}$$ in a way such as to see that $|a_n|<e$ for all n.
What I have gotten so far is $$|a_n| = \frac{|f^{(n)}(0)|}{n!}\leq\frac{\frac{n!}{R^n}||f||_{\partial D(0,R)}}{n!}\leq\frac{||f||_{\partial D(0,R)}}{R^n}\leq e*||f||_{\partial D(0,1-1/(n+1))} = e * \sup_{|z| = 1-1/(n+1)}f(z),$$ where $R=1-1/n<1.$ But now I should be able to conclude with the help of the assumption that $||f||_{\partial D(0,1-1/n)}<1$, when it would follow that $|a_n|<e$. Can I get some help here?
We apply the Cauchy relation for $g(z)=f'(z)=\sum {b_nz^n}$ and get that:
$|b_n| \le \frac{\sup_{|z|=R}{|g(z)|}}{R^n} \le \frac{1}{(1-R)R^n}$. But with $R_n=1-\frac{1}{n+1}, n \ge 1$, $R_n^n > \frac{1}{e}, 1-R=\frac{1}{n+1}$, so $|b_n| < (n+1)e, n \ge 1$, while $|b_0| \le 1 <e$. But we have $b_n=(n+1)a_{n+1}, n \ge 0$ hence $|a_n| < e, n \ge 1$.
Note that the derivative inequality cannot determine $a_0$ as $f+c$ satisfies same for arbitrary $c$, so we must assume $|a_0| <e$ for completion