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 $0 < x < 1$.]

 

[math771]

One way to bound $\max|f(z)|$ in the disc is to use the Cauchy integral formula. Since $f$ is analytic in the disc $|z| < 1$, we can write $f(z) = \frac{1}{2\pi i} \int_{|w| = r} \frac{f(w)}{w-z} dw$, where $r < 1$ is the radius of the disc. Then, using the triangle inequality, we have $|f(z)| \le \frac{1}{2\pi} \int_{|w| = r} \frac{|f(w)|}{|w-z|} |dw|$. Since $|w-z| \ge |w|-|z|$, we can further bound this by $\frac{1}{2\pi} \int_{|w| = r} \frac{|f(w)|}{|w| - |z|} |dw|$.

Now, using the given inequality $|f(z)| \le \frac{1}{1-|z|}$, we have $\frac{|f(w)|}{|w| - |z|} \le \frac{1}{1-|z|}$. Therefore, we can write $|f(z)| \le \frac{1}{2\pi} \int_{|w| = r} \frac{1}{1-|z|} |dw| = \frac{1}{1-|z|}$.

Now, using the Cauchy integral formula for the derivative, we have $f^{(n)}(0) = \frac{n!}{2\pi i} \int_{|w| = r} \frac{f(w)}{w^{n+1}} dw$. Therefore, we can bound $|f^{(n)}(0)|$ by $\frac{n!}{2\pi} \int_{|w| = r} \frac{|f(w)|}{|w|^{n+1}} |dw|$. Using the same argument as before, we have $\frac{|f(w)|}{|w|^{n+1}} \le \frac{1}{|w|^{n+1}(1-|z|)}$. Thus, we can write $|f^{(n)}(0)| \le \frac{n!}{2