Show that $f'_k(z)$ ($f_k(z)=z^k/k$) does not converge uniformly for $|z|<1$ ($z$ a complex number).

Question

The problem reads

Show that $f'_k(z)$ ($f_k(z)=z^k/k$) does not converge uniformly for $|z|<1$ ($z$ a complex number).

Since $z^{k-1}\rightarrow 0$, for $k\rightarrow\infty$ and $|z|<1$, I don't see why this function does not converge uniformly. I am aware that for $|z|=1$, the function converges pointwise and not uniformly, but since I am taking $|z|<1$ it should converge uniformly.

Answer 1

Since $f'_k(z)\to 0$ for all $|z|<1$ pointwise, you have to prove that $\sup_{|z|<1}|f_k'(z)|\not\to 0$.

For $k\geq 2$ define $z_k:=1-\frac1{k-1}$. Since $|z_k|=1-\frac1{k-1}<1$ you get for $k\geq 2$ $$ \sup_{|z|<1}|f_k'(z)|\geq |f_k'(z_k)|=z_k^{k-1}=\left(1-\frac1{k-1}\right)^{k-1}\to e^{-1}>0. $$

Edit: I just noticed, that here is a dublicate...

Edit 2: Marks remark in detail:

Define $z_n:=1-\frac{1}{n}$ with $|z_n|=1-\frac{1}{n}<1$ then we get $$ \sup_{|z|<1}|f'_k(z)|\geq \lim_{n\to\infty}|f_k'(z_n)|=\lim_{n\to\infty}\left(1-\frac1{n}\right)^{k-1}=1^{k-1}=1. $$ On the other hand you have $$ \sup_{|z|<1}|f'_k(z)|=\sup_{|z|<1}|z^{k-1}|=\sup_{|z|<1}|z|^{k-1}\leq1^{k-1}=1. $$ Together you get $\sup_{|z|<1}|f'_k(z)|=1$ for all $k\in\mathbb{N}$.