MHB How Do Cauchy's Estimates Apply to Analytic Functions Within Unit Discs?

[Deanmark]

Let $f$ be analytic in the disc $|z| < 1$ and assume that $|f(z)| \le \dfrac{1}{1-|z|}$.

Show that $|{f}^{(n)}(0)| ≤ e(n + 1)!$.

Any ideas on how to bound $\max|f(z)|$ in the disc?

 

[Euge]

Be careful: on what domain do you take $\max \lvert f(z)\rvert$? To handle this problem, take a Cauchy estimate on the disk $\lvert z\rvert = 1 - \dfrac{1}{n+1}$. For all $\lvert z\rvert = 1 - \dfrac{1}{n+1}$, $\lvert f(z)\rvert \le n + 1$, giving the Cauchy estimate

$$\lvert f^{(n)}(0)\rvert \le \frac{n!}{\left(1 - \frac{1}{n+1}\right)^n} (n+1) = (n+1)!\left(1 - \frac{1}{n+1}\right)^{-n}$$

Now show that $\left(1 - \dfrac{1}{n+1}\right)^{-n} \le e$ to complete the argument. [Hint: Show first that $e^x > \dfrac{1}{1 - x}$ for $0 < x < 1$.]