I'm trying to solve this problem:
Fix $0<\alpha<\infty$. Show that 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.
This problem also appears here: Showing a function can be continuously extended to the unit circle, and has a great solution. But I solved using different approach, and it would be appreciated if you check my proof is correct. (If you have no time, it's fine to verify only part 2.)
First part, proving that $f$ can be extended continuously to the unit circle:
We can define the value of $f$ on unit circle, by $f(z)=\sum_{n=0}^{\infty}{2^{-n\alpha}}z^{2^n}$, since the sum absolutely converges.
It is enough to prove that $f$ is continuous on the unit circle. Let $z$ be any point on the unit circle, $\epsilon >0$ be any real. There exists large enough $N$ such that $\sum_{n=N+1}^{\infty}{2^{-n\alpha}} < \frac{\epsilon}{4}$, so that $|f(z)-f(t)|\le \sum_{n=0}^{N}{2^{-n\alpha}|z^{2^n}-t^{2^n}|} + \frac{\epsilon}{2} $. Since $z, z^2 , z^4 , \cdots , z^{2^N}$ are continuous, there exists $\delta>0$ such that if $|z-t|<\delta$, $\sum_{n=0}^{N}{2^{-n\alpha}|z^{2^n}-t^{2^n}|} < \frac{\epsilon}{2}$. Thus we have for every $t$ with $|z-t|<\delta$, $|f(z)-f(t)|<\epsilon$. So, $f$ is continuous on the unit circle, which completes the proof.
Second part, $f$ cannot be analytically continued past the unit circle.
If $f$ can be analytically continued, there exists a point $z$ in the unit circle and neighborhood of $z$ (Call $U$) and analytic function $g$ on $U$ such that $f=g$ on $\mathbb{D}\cap U$. ($\mathbb{D}$ denotes the open unit disc.)
Now, we can differentiate $g$ and multiply by $z$ to get a new analytic function $g_1$. Let $g_{n+1}$ be a function obtained by differentiating $g_n$ and multiplying by $z$. Then, $g_n$ is analytic. Also, since $g(z)=\sum_{n=0}^{\infty}{2^{-n\alpha}}z^{2^n}$ for $z \in \mathbb{D}\cap U$, we have $g_m (z) = \sum_{n=0}^{\infty}{2^{n(m-\alpha)}}z^{2^n}$. Let $N$ be large enough integer so that $N>\alpha$. Then $g_N (z)=\sum_{n=0}^{\infty}{2^{n(N-\alpha)}}z^{2^n}$ for $z \in \mathbb{D} \cap U$, and $g_N$ is analytic on $U$. So, $g_N$ is continuous on $U$.
But, $g_N (z)=\sum_{n=0}^{\infty}{2^{n(N-\alpha)}}z^{2^n}$ is nowhere continuous on unit circle, since $g_N (r e^{2\pi i \frac{a}{2^b}})=\sum_{n=0}^{b-1}{2^{n(N-\alpha)}}r^{2^n}e^{2\pi i \frac{a}{2^{b-n}}} + \sum_{n=0}^{\infty} {2^{n(N-\alpha)}}r^{2^n}$ diverges as $r \to 1$. So, the proof is completed.
Any advices, whether it is big or small (including typos, ways to make question more readable, and grammatical error), are really appreciated.