Showing a function can be continuously extended to the unit circle Fix $0 < a < \infty.$ Define for $|z| < 1$ the function $$ f(z) = \sum_{n=0}^\infty 2^{-na} z^{2^n}$$
and show that $f$ extends continuously to the unit circle but can not be analytically continued past the unit circle. 
The second part of the problem is a whole new can of worms but I'm even having difficulty with the first part: it's easy to show that $|f|$ is bounded if $z = e^{i\theta}$ but I can't quite figure out the continuity piece. 
Let $z_0 = re^{i\phi}$ be a point in the unit disk. Then 
$$f(z) - f(z_0) = \sum_{n=0}^{\infty} 2^{-na}(e^{i\theta 2^n} - r^{2^n} e^{i\phi 2^n}) 
\\= \sum_{n=0}^\infty 2^{-na}[(1+r^{2^n})(e^{i\theta 2^n} - e^{i\phi2^n}) + 
(e^{i\phi 2^n} - r^{2^n} e^{i\theta 2^n})]$$
and I tried finding some bounds for $|f(z) - f(z_0)|$ given bounds for $|z-z_0|$ in this manner (and some other ways) but I wasn't able to find any results. 

Finding explicit bounds for $\lvert f(z) - f(z_0)\rvert$ is not nice. It's nicer if one uses the absolute and uniform convergence of the series on the closed unit disk, but then it's still nicer to use the theorem that the uniform limit of continuous functions is continuous.
To see that the series converges absolutely and uniformly on the closed unit disk, note that the geometric series
$$\sum_{n = 0}^{\infty} 2^{-na} = \frac{1}{1 - 2^{-a}}$$
converges and consists of positive terms, so for $\lvert z\rvert \leqslant 1$ we have
$$\Biggl\lvert f(z) - \sum_{n = 0}^k 2^{-na} z^{2^n}\Biggr\rvert \leqslant \sum_{n = k+1}^{\infty} 2^{-na} \lvert z\rvert^{2^n} \leqslant \sum_{n = k+1}^{\infty} 2^{-na} = \frac{2^{-ka}}{2^a-1}.$$
For the second part of the exercise, note that
$$f(z^2) = \sum_{n = 0}^{\infty} 2^{-na} (z^2)^{2^n} = \sum_{n = 0}^{\infty} 2^{-na} z^{2^{n+1}} = 2^a \sum_{m = 1}^{\infty} 2^{-ma} z^{2^m} = 2^a\bigl(f(z) - z\bigr).$$
Thus a hypothetical analytic continuation to a neighbourhood of $z_0 = e^{i \varphi_0}$ would imply an analytic continuation to a neighbourhood of $z_0^2$. Continuing the argument, we'd have an analytic continuation to a larger disk, and that contradicts the fact that the radius of convergence of the series is $1$.
