If a holomorphic function $f$ from the open ball to the complex numbers have bounded derivative then $|\frac{f^{n}(0)}{n!}|\leq e$

Question

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 derivative of $f$ evaluated in $0$.

I already take arround four hours and don not get a good idea.

Attempt. First of all I try join the information about each coefficient of the taylor series of $f$. Since $B_1(0)$ is convex then it is simply connected and since $f$ is holomorphic we have by Cauchy integral Formula that for each $0<r<1$ and closed curve $\gamma:[0,2\pi]\to B_1(0)$ given by $\gamma(t)=re^{2\pi it}$ $$f^n(0)I(\gamma,0)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(z)}{z^{n}}dz$$ and thus

$$\left|\frac{f^{n}(0)}{n!}\right|\leq \frac{\operatorname{sup}_{z\in \gamma}|f(z)|}{r^{n}}$$

And this is all the information that I get from this.

For other side, the bound in the derivative of $f$ tell me that for $z=0$ we have that $|f^{\prime}(0)|\leq 1$. This lead me the possible intuition that I can also bound the other coefficients by $1$.

This is all the things that I have in mind, but I look myself too far of the solution. Any idea or hint is welcome.

Update

From the hints given in the comments.

Let $g:B_1(0)\to \mathbb{C}$ given by $g(z)=f^{\prime}(z)$ and notice that since $f$ is holomorphic in a simply connected set, then all their derivatives exists and also are holomorphic in $B_1(0)$. Thus we can also apply the Cauchy Integral formula to $g$ in the point $z_0=0$ to conclude that for each $n\in \mathbb{N}$ and closed curve $\gamma_n:[0,2\pi]\to \mathbb{C}$ given by $\gamma(t)=r_ne^{2\pi it}$. we have

$$\left| \frac{g^{n}(0)}{n!}\right|\leq \frac{1}{2\pi}\cdot |g(z)|\cdot 2\pi=\frac{\frac{1}{1-r_n}}{r_{n}^{n}}$$

And since $g=f^{\prime}$ we get $g^{n}(0)=f^{n+1}(0)$. Moreover, since $|g(z)|\leq \frac{1}{1-|z|}$ for all $z\in B_1(0)$ we conclude

$$ \left| \frac{f^{n+1}(0)}{n!}\right|\leq \frac{\frac{1}{1-r_n}}{r_{n}^{n}} $$

and if we multiply by $(n+1)^{-1}$ both sides, then $$ \left| \frac{f^{n+1}(0)}{(n+1)!}\right|\leq \frac{\frac{1}{1-r_n}}{r_{n}^{n}(n+1)}$$ for each $n\in \mathbb{N}$

Finally, is sufficient choose a correct sequence of radius. For this we can choose for each $n\in \mathbb{N}$ the closed curve $\gamma_n:[0,2\pi]\to \mathbb{C}$ given by $\gamma(t)=(1-\frac{1}{n+1})e^{2\pi i t}$ for which we have

$$\left| \frac{f^{n+1}(0)}{(n+1)!}\right|\leq \frac{1}{(1-\frac{1}{n+1})^{n}}$$ and taking $n\to \infty$ we have

$$\left| \frac{f^{n+1}(0)}{(n+1)!}\right|\leq e$$ as desired.

Thanks in advance to the community for their brilliant suggestions.

Answer 1

Your are very close to the solution. Applying the Cauchy inequalities to $g= f'$ gives $$ \frac{|f^{n}(0)|}{n!} = \frac 1n \cdot \frac{|g^{n-1}(0)|}{(n-1)!} \le \frac 1n \cdot \frac{\sup_{|z|=r} |g(z)|}{r^{n-1}} \le \frac{1}{n r^{n-1}(1-r)} $$ for all $r \in (0, 1)$. Now choose $r = 1-1/n$ in order to make the denominator in the fraction on the right as large as possible. It follows that $$ \frac{|f^{n}(0)|}{n!} \le \frac{1}{\left(1-\frac 1n\right)^{n-1}} < e \, . $$ For the last inequality see for example How to prove $(1-\frac{1}{n})^n \leq \frac{1}{e} \leq(1-\frac{1}{n})^{n-1}$?.

Answer 2

Corollary:
Given holomorphic $\color{purple}{f:B_1(0)\to\mathbb{C}}$ such that: $$\forall z\in dom(f):\text{ }|f'(z)|\leq \frac{1}{1-|z|},$$ we have: $$\forall n\in \mathbb{N}:\text{ }\bigg|\frac{f^{(n)}(0)}{n!}\bigg|\leq e.$$

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Proof:
Applying C.I.F., we have that $f$ is infinitely differentiable, satisfying: $$\color{blue}{\forall n\in \mathbb{N}:\forall a\in dom(f):\bigg[f^{(n)}(a) = \frac{n!}{2\pi i}\oint_{\gamma}\frac{f(z)}{(z-a)^{n+1}}dz\bigg]},$$ where for a radius $r\in (0,1)$, the integral can be taken over any $\color{purple}{\gamma(t):= re^{2\pi i t}}$, with $t\in [0,1]$.

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Since $f'$ is holomorphic over the same domain, we also have: $$\color{black}{\forall n\in \mathbb{N}:\forall a\in dom(f'):\bigg[(f')^{(n)}(a) = \frac{n!}{2\pi i}\oint_{\gamma}\frac{(f')(z)}{(z-a)^{n+1}}dz\bigg]}.$$ Noting that $(f')^{(n-1)} := f^{(n)}$, we get: $$\forall n\in \mathbb{N}:\forall a\in dom(f):\text{ } \bigg[f^{(n)}(a) = \frac{(n-1)!}{2\pi i}\oint_{\gamma}\frac{f'(z)}{(z-a)^{n}}dz\bigg].$$

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Now, take $a=0$ and apply $\color{red}{``\text{properties of norms"}}$ to get: $$\forall n\in \mathbb{N}: \big|f^{(n)}(0)\big| = \bigg|\frac{(n-1)!}{2\pi i}\oint_{\gamma}\frac{f'(z)}{z^{n}}dz\bigg|$$ $$\leq \frac{(n-1)!}{2\pi}\oint_{\gamma}\bigg|\frac{f'(z)dz}{z^n}\bigg|$$ $$\leq \frac{(n-1)!}{2\pi}\oint_{\gamma}\frac{|f'(z)||dz|}{|z|^n}.$$ From here, we can apply the hypothesis, so the above is $$\leq \frac{(n-1)!}{2\pi}\oint_{\gamma}\frac{|dz|}{(1-|z|)|z|^n}.\text{ }(\star)$$

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What's left is to evaluate the integral, so note: $$|\gamma(t)| = |re^{2\pi i t}| \equiv r\text{ }\text{ }\text{ and }\text{ }\text{ }|d(\gamma(t))| = |2\pi i\cdot r e^{2\pi i t}dt|\equiv 2\pi r\text{ }dt.$$ Then: $$(\star) = \frac{(n-1)!}{2\pi}\int_0^1 \frac{2\pi r}{(1-r)r^n}dt$$ $$= (n-1)!\int_0^1 \frac{1}{(1-r)r^{n-1}}dt$$ $$= \frac{(n-1)!}{(1-r)r^{n-1}}.$$

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Hence we've arrived at: $$\color{green}{\forall n\in \mathbb{N}:\text{ }\bigg|\frac{f^{(n)}(0)}{n!}\bigg|\leq \frac{1}{n(1-r)r^{n-1}}},$$ where $r\in (0,1)$ is free. To minimize, we can maximize the denominator: $$\frac{d}{dr}(n(1-r)r^{n-1}) = n(-1)r^{n-1}+n(1-r)(n-1)r^{n-2} = 0$$ $$\implies (n-1)(1-r)r^{n-2} = r^{n-1}$$ $$\implies \frac{1}{r}-1 = \frac{1}{n-1}$$ $$\implies r = \frac{n-1}{n} = 1-\frac{1}{n}.$$ We've extremized it, but one needs to check further that this is a maximum ($\color{red}{\text{sign of second derivative}}$). Naively plugging this in yields the "mimimum" bound: $$\frac{1}{(1-\frac{1}{n})^{n-1}}.$$ This of course varies with $n$, but all expressions for $\big|\frac{f^{(n)}(0)}{n!}\big|$ will be bounded above by $\sup\limits_{n\in\mathbb{N}}1/(1-\frac{1}{n})^{n-1}$. Which proves the result (via trivia): $$\forall n\in \mathbb{N}:\text{ }\bigg|\frac{f^{(n)}(0)}{n!}\bigg|\leq e.\text{ }\text{ }\text{ }\square$$

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Everyone involved in the comment discussions should be credited here, just tried to summarize with more detail. I noted in red places to further be cautious.