Algebraic functions have a discrete set of singularities. Lacunary functions, e.g. $f(z)=\sum_{n=0}^\infty z^{2^n}$, have a continuum of singularities at every point of the boundary of their disk of convergence. The Ostrowski–Hadamard gap theorem gives a sufficient condition for the exponents with non-vanishing coefficients for a power series which defines a lacunary function.

Can one also construct a lacunary function using only a finite number of algebraic operations, i.e. arithmetics and solving plane algebraic equations over the complex numbers, and a finite number of analytic operations, particularly applications of the exponential function and the logarithm?


1 Answer 1

  1. You have to define what a "lacunary function" means. There are various lacunary conditions in the literature.

  2. If you mean sufficiently large lacunas so that your series have a whole circle of singularities, like in the example you give, such functions cannot be constructed as a finite composition of algebraic, exponential, and logarithmic ones.

One reason for this is the "Iversen property". Which is the following. Let $(f,z_0)$ be a germ of an analytic function. If for every curve $\gamma(t)$ starting at $z_0$, and every $\epsilon >0$, there exists a curve $\gamma'(t)$ starting at $z_0$ such that $|\gamma(t)-\gamma'(t)|<\epsilon$ for all $t$, and $(f,z_0)$ has an analytic continuation along $\gamma'$, we say that $f$ has the Iversen property.

Composition of functions with this property also has it. Algebraic, entire and log functions have it. But a function with a whole line of singularities, evidently does not have it.


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