The problem is the following. Let $B_1(0)=\lbrace z\in \mathbb{C}: |z|<1\rbrace$ and consider a holomorphic function $f:B_1(0)\to \mathbb{C}$ such that $|f^{\prime}(z)|\leq \frac{1}{1-|z|}$ for all $z\in B_1(0)$. Show that $$|\frac{f^{n}(0)}{n!}|\leq e$$
where $f^n(0)$ denote the $n$-th derivativ...
@redwhisker I use your suggestion and feel like the solution is so closer. But at this point I do not know how use the fact about our $\gamma$ this is $0<r<1$.
plug in $r_n=1-1/(n+1)$ in $\left| \frac{f^{n+1}(0)}{(n+1)!}\right|\leq \frac{\frac{1}{1-r}}{r^{n+1}(n+1)}$ which is valid for all $r<1$ and see what happens
Since the perimeter of the circular arc $\gamma_n$ is $2\pi r_n$, shouldn’t you have the following instead? $$\left| \frac{g^{n}(0)}{n!}\right|\leq \frac{1}{2\pi}\cdot |g(z)|\cdot 2\pi r_n=\frac{\frac{1}{1-r_n}}{r_{n}^{n}}.$$
Yes, you are right i made a mistake. But the reasoning is the same (we only need modify the chosen radious). I will update the post with the correct assumptions
@user1551 thanks a lot for you clarification. Now the post is fixed and indeed is not necessary change the radius of the sequence of closed curves $\gamma_n$.
@BrienNavarro: You cannot take $n \to \infty$ only on the right-hand side of $\left| \frac{f^{n+1}(0)}{(n+1)!}\right|\leq \frac{1}{(1-\frac{1}{n+1})^{n}}$.
I tried this independently last night with $f'$ in the formula and ended up with an infinite limit in the end (details omitted). So I have a question, where does this problem come from?
@Kevin the problem is from the the notes of my Complex Analysis Teacher. In the notes he suggest that the problem follows as consequence of Cauchy Integral Formula. But after of spend more than 5 hours I really get frustrated and create the open question here.
While you cannot take $n$ to the limit (because $n$ is fixed), it suffices to prove that $e^{-1}\le(1-\frac{1}{n+1})^n$ for every positive integer $n$.
Discussion on question by Brien Navar…
Discussion on question by Brien Navar…