# Density of Lacunary Functions

I'm curious whether lacunary functions (functions in the complex plane that are holomorphic in some open ball about the origin, but cannot be analytically extended past that ball) are typical or the exception.

Are lacunary functions dense in the space of functions that are holomorphic on the open unit ball?

I'm going to prove density in compact-open topology, i.e. that for any $$f$$ holomorphic on the disk there is a sequence of lacunary functions convergent to $$f$$ uniformly on compact sets.
Take any function $$g$$ lacunary in the unit disk satisfying $$g(0)=0$$. Then it's easy to see that for $$g_n(x)=g(x^n)$$, $$g_n(x)$$ is lacunary again and $$g_n$$ converges to $$0$$ uniformly on compacts. Let $$f_n$$ be the $$n$$-th Taylor polynomial of $$f$$, then $$f_n$$ converges to $$f$$ uniformly on compacts. It follows $$f_n+g_n$$ is a sequence of lacunary functions convergent to $$f$$ uniformly on compacts.
• It is perhaps even simpler to consider $f((1+\frac{1}{n})z) + \frac{1}{n} g(z)$ instead. – Mateusz Kwaśnicki Jan 8 at 21:00
• @MateuszKwaśnicki That's true! I was thinking of $g(z)/n$ originally, but for some reason thought it doesn't work because $g$ is unbounded... of course, it doesn't matter on compact subsets. – Wojowu Jan 8 at 21:02
• @DylanAltschuler: I do not know if there is a reasonable measure structure on the class $\mathcal{L}$ of lacunary functions, but one can ask a more natural question: is this class residual? The answer seems to be "yes": the class $\mathcal{H}_{\alpha,r}$ of holomorphic functions in the unit disk $D$ that extend to $D \cup D(e^{i \alpha},r)$ for a fixed $\alpha$ and $r > 0$ is nowhere dense, and the complement of $\mathcal{L}$ is the union of $\mathcal{H}_{\alpha,r}$ over all rational $\alpha$ and $r$. – Mateusz Kwaśnicki Jan 9 at 9:02