Let $U=B_1(0)$ and $ f(z)=\sum_{n=1}^{\infty}2^{-n^2}z^{2^n}$. Show that $f$ has not analytic extensions to any open set $G$ with $U\subsetneq G$.

Question

Let $U=B_1(0)$ and $$f:U \rightarrow \mathbb{C},\qquad f(z)=\sum_{n=1}^{\infty}2^{-n^2}z^{2^n}.$$ Show that $f$ has not analytic extensions to any open set $G$ with $U\subsetneq G$.

Remark: Suposse that $f$ has analytic extensions to some open set $G$ with $U\subsetneq G$. Let $g$ such extension. Let $\lambda=\inf\left\{|z|:z\in \mathbb{C}\setminus G\right\}$, then we have two cases:

If $\lambda >1 $, then Taylor serie of $g$ about $z=0$ is $g(z)=\sum_{n=1}^{\infty}2^{-n^2}z^{2^n}$ whit radius of convergence $\lambda$, which is a contradiction because $\sum_{n=1}^{\infty}2^{-n^2}z^{2^ n}$ is convergent when $|z|<1$.

If $\lambda =1 $, then have problems, it is in this case where requires your help.

I would also like to know if the way I'm showing this fact is correct, and if there is another way.

Answer 1

By differentiating we can obtain a functional equation for $f$. For $|z|<1$, $$ z\cdot f'(z) = \sum_{n=0}^\infty 2^n\cdot 2^{-n^2} z^{2^n} = \sum_{n=0}^\infty 2^{-n^2+n} z^{2^n} $$ $$ \frac{z}2 \cdot \Big(z\cdot f'(z)\Big)' = \sum_{n=0}^\infty 2^{-n^2+2n-1} z^{2^n} = \frac{z}2 + \sum_{n=1}^\infty 2^{-(n-1)^2} (z^2)^{2^{n-1}} = \frac{z}2 + f(z^2). $$ Hence, $$ f(z^2) = \frac{z^2}2 f''(z) + \frac{z}2 f'(z) - \frac{z}2. \tag{*} $$

Now suppose that $f$ can be continued through some arc $A$ of the unit circle. Then (*) provides a continuation through the arc $A_1=\{z^2: z\in A\}$. (Note that $A_1$ is twice longer than $A$.) Then there exists another continuation through the arc $A_2=\{z^2: z\in A_1\}$. Repeating this we can see that $f$ can be continued through every point of the unit circle. Due to the uniqueness, these local continuations can be merged tho a joint continuation to a larger disk, but this is a contradiction because the convergence radius of the power series is $1$.

Answer 2

Note that the power series defining $f$ converges uniformly on $\overline {\mathbb D},$ so we can think of $f$ as continuous on $\overline {\mathbb D}.$ Suppose $G= \mathbb D \cup D(\zeta, r),$ where $\zeta \in \partial D,$ and $f$ extends to be analytic on $G.$ Then $G$ contains an open arc on the unit circle. Thus $f(e^{it})$ will be real analytic as a function of $t$ on some $(a,b) \subset \mathbb R.$

It turns out that $f(e^{it})$ is a smooth function on $\mathbb R$ that is "nowhere analytic". We can show this by considering points of the form $ t=\pi p/2^q,$ where $p\in \mathbb Z,q\in \mathbb N.$ These points form a dense subset of $\mathbb R.$ At each such $t,$ $\limsup |D^kf(e^{it})/k!|^{1/k} = \infty.$* Thus at such points the radius of convergence of the Taylor series is $0.$ But $f(e^{it})$ is real analytic on $(a,b),$ so we have a contradiction. Therefore $f$ has no such analytic continuation.

Note that if $1\le n_1 < n_2 < \dots,$ the above will also show $\sum_{k=1}^{\infty} z^{2^{n_k}}/2^{n_k^2}$ has no anaytic continuation to $G.$

*This is a nice exercise.