I've tried to solve the following problem from Ahlfors' complex analysis text:
If a function element $(f,\Omega)$ has no direct analytic continuations other than the ones obtained by restricting $f$ to a smaller region, then the boundary of $\Omega$ is called a natural boundary for $f$. Prove that the series $\sum_{n=0}^\infty z^{n!}$ has the unit circle as a natural boundary. Hint: Show that the function tends to infinity on every radius whose argument is a rational multiple of $\pi$.
My attempt:
Let $q \in \mathbb Q$ be a rational number, with reduced form $a/b$ where $a \in \mathbb Z,b \in \mathbb N$. For $r<1$ we have $$f\left(r e^{i q \pi}\right)=\sum_{n=0}^\infty r^{n!} e^{i q n! \pi}= \sum_{n=0}^{b+1} r^{n!} e^{i q n! \pi}+\sum_{n=b+2}^\infty r^{n!} e^{i \frac{a}{b} n! \pi} $$
In the last sum we have $e^{i \frac{a}{b} n! \pi}=1$ for $n \geq b+2$, since the exponent is an even multiple of $\pi i$. Thus using the reverse triangle inequality $$\left|f \left( r e^{i q \pi} \right) \right|=\left|\sum_{n=0}^{b+1} r^{n!} e^{i q n! \pi}+\sum_{n=b+2}^\infty r^{n!} \right| \geq \left| \sum_{n=b+2}^\infty r^{n!}- \left| \sum_{n=0}^{b+1} r^{n!} e^{i q n! \pi} \right| \right|.$$
Taking $r$ sufficiently close to $1$ the series $\sum_{n=b+2}^\infty r^{n!}$ can be made arbitrarily large. Indeed, if $r \geq \sqrt[N!]{1/2}$ with $N$ large the series contains at least $N-b-1$ terms which are $\geq 1/2$. It follows that for $r$ sufficiently close to $1$ the sum $\sum_{n=b+2}^\infty r^{n!}$ can be made larger than $$M:= \max_{r \in [0,1]} \left| \sum_{n=0}^{b+1} r^{n!} e^{i q n! \pi} \right|<\infty ,$$ making the external absolute value redundant. For all such sufficiently close numbers $r$ we have $$\left| f \left( r e^{i q \pi} \right) \right| \geq \sum_{n=b+2}^\infty r^{n!}-M $$ which tends to $\infty$ as $r \to 1^-$.
Since $\left\{ e^{i q \pi} : q \in \mathbb Q \right\}$ is a dense subset of the unit circle the above prevents the existence of direct analytic continuation by function elements $(g,\Omega')$, unless $\Omega' \subseteq \Omega$. This is true since if $\Omega'$ leaves $\Omega$ it must contain a boundary point whose corresponding radius has an argument which is a rational multiple of $\pi$. Since $f=g$ on $\Omega \cap \Omega'$ it follows that $g$ has infinity as the radial limit towards that boundary point - which contradicts the analyticity of $g$ there.
Is the above correct? If not please help me fix it. Thanks!
The solution is correct, but you could save some labor.
When $z$ is on the radius $(0,1)$, for every $N$ we have $$ f(z)\ge \sum_{n=0}^N z^{n!}\to N+1,\quad z\to 1^-$$ hence $f(z)\to +\infty$ as $z\to 1^{-}$.
For every rational number $a/b$, the function $f(e^{2\pi i a/b}z)-f(z)$ is a polynomial, because $(e^{2\pi i a/b})^{n!} =1$ when $n\ge b$.
From 1 and 2, $|f(e^{2\pi i a/b}z)|\to \infty$ as $z\to 1^-$, which is the same as saying that $|f|\to\infty$ along the radius from $0$ to $e^{2\pi ia/b}$.