Finding a specific bound using Cauchy's integral formula

Question

Let $f\in\mathcal{H}(\mathbb{D})$ and $\alpha>0$ a real value so that there exists $c>0$ such that for any $|z|<1$, $(1-|z|)^\alpha|f(z)|\leq c$. I have to show that $|f^{(n)}(0)|\leq cn!(e/\alpha)^\alpha(n+\alpha)^\alpha$.

By now I have applied Cauchy's integral formula in order to get that, for any $0

The problem is that once I substitute that value of $r$ in the function, I do not obtain the bound I am supposed to get and also I do not know how to bound this $$|f^{(n)}(0)|\leq cn!(n+\alpha)^\alpha\frac{(n+\alpha)^n}{n^{n+\alpha}}.$$

Answer 1

There is an error in your last inequality. Substituting $r=\frac{n}{\alpha+n}$ in $\frac{cn!}{r^n(1-r)^\alpha}$ gives $$ |f^{(n)}(0)|\le \frac{c n! (n+\alpha)^n (n+\alpha)^\alpha}{n^n \alpha^\alpha} = \frac{c n! (n+\alpha)^\alpha}{ \alpha^\alpha} \bigl( 1+\frac{\alpha}{n} \bigr)^n $$ and it remains to show that $$ \bigl( 1+\frac{\alpha}{n} \bigr)^n = \left( \bigl( 1+\frac{\alpha}{n} \bigr)^\frac{n}{\alpha} \right)^\alpha < e^\alpha $$ which follows from $$ \bigl( 1 + \frac 1x \bigr)^x < e \text{ for } x > 0 \, . $$