Prove that $f(z) = \sum\limits_{k = 1}^\infty \frac{z^{2^k}}{2^k}$ is continuous in the closed unit disc and holomorphic inside it.

Question

I have started off by assuming that there is a disc of radius $r$ for which $|z|<r$ for $r \in (0,1)$ and $z \in D_r$.

This implies that $|z|^{2^k} < r^{2^k}$

And after that, I don't know where to go

This problem has given me nightmares

Answer 1

This function is holomorphic in the open unit disc and it is continuous in the closed unit disc.

1. $f(z)$ is holomorphic in $D$. This is obtained simply by using the root test for power series. We have $f(z)=\sum_{n=0}^\infty a_nz^n$, where $$ a_n=\left\{ \begin{array}{cc} 2^{-k} & n=2^k, \\ 0 & \text{otherwise}. \end{array} \right. $$ Thus $$ \lvert a_n\rvert^{1/n}=\left\{ \begin{array}{cc} \lvert 2^{-k}\rvert^{1/2^k} & n=2^k, \\ 0 & \text{otherwise}. \end{array} \right. $$ So, $\lvert a_n\rvert^{1/n}$ contains infinitely many terms of the sequence $n^{1/n}$ and infinitely many terms of the zero sequence, which implies that $$ \limsup_{n\to\infty}\lvert a_n\rvert^{1/n}=1, $$ and thus the radius of convergence of $f$ is equal to $1$, and finally, $f$ is analytic in the open unit disc, and not in any larger disc.

2. $f(z)$ is continuous in $\overline D$. It suffices to shoe that the sequence of continuous functions $$ f_n(z)=\sum_{k=0}^n \frac{z^{2^k}}{2^k}, $$ converges uniformly in $\overline D$. But $$ \lvert f_m(z)-f_n(z)\rvert =\Big\lvert\sum_{n<k\le m}\frac{z^{2^k}}{2^k}\Big\rvert \le \sum_{n<k\le m} 2^{-k} <2^{-n}, $$ which implies that $\{f_n\}$ is uniformly Cauchy in $\overline D$, and hence uniformly convergent.

Answer 2

First check that the given function is convergent in closed unit disc, that is it converges for the points $|z|=1$. So this defines a holomorphic function in open unit disc and also has a finite value at the boundary points ($|z|=1$). So it is continuous by power series representation (continuous from the left hand side as this is domain of definition.)