Fix $\alpha$ with $0 < \alpha < \infty$. Show that the analytic function $f$ defined by $f(z) = \sum_{n=0}^\infty 2^{-n\alpha} z^{2^n}$ for $|z| < 1$ extends continuously to the unit circle, but cannot be analytically continued past the unit circle.
The problem also provides a hint concerning nowhere differentiable functions. In Stein and Shakarchi, Fourier Analysis, p. 118, it is shown that $\sum_{n=0}^\infty 2^{-n\alpha} \cos(2^n x)$ is nowhere differentiable for $0 < \alpha < 1$. On p. 174 of that book a problem with hints shows that this result is also true for $\alpha = 1$.
I see how to do the original problem using the hint for $0 < \alpha \le 1$. My question is how does one prove that $f(z)$ cannot be analytically continued for $\alpha > 1$?
Hint: $z f'(z) = \sum_{n=0}^\infty 2^{n(1-\alpha)} z^{2^n}$. If that can't be continued, then $f(z)$ can't.