Assume $\sum_{n=0}^\infty a_nw^{2^n}$ is a power series with convergence radius $R\geq 1$
Assume $f$ is analytic in the disk $D(0,1+\delta)$ for some $\delta>0$ . Write its Taylor series: $$f(z)=\sum_{n=0}^\infty c_nz^n,z\in D(0,1+\delta).$$
If the series $\sum_{n=0}^\infty a_n\left(\frac{1}{2}z^2+\frac{1}{2}z^3\right)^{2^n}$ is convergent to $f$ in the unit circle $D(0,1)$ , that is $$f(z)=\sum_{n=0}^\infty a_n\left(\frac{1}{2}z^2+\frac{1}{2}z^3\right)^{2^n},z\in D(0,1),$$ how to show $\sum_{n=0}^\infty a_n\left(\frac{1}{2}z^2+\frac{1}{2}z^3\right)^{2^n}$ is also convergent in $D(0,1+\delta)$?
This is part of the proof of Ostrowski-Hadamard gap theorem. But I failed to find any useful information online.
I failed to prove it like this: Since these two series equal to each other in $D(0,1)$ , they must have the same coefficients, thus the same convergence radius $R_0$. By the first series we know $R_0\geq1+\delta$, so the second series is convergent in $D(0,1+\delta)$. (It turned out to be wrong, see Where is my mistake in this reasoning?).
What I know is that the $^{2^n}$ in these series is important, meaning that there are a lot of gaps in these series.