# Entire function of finite order with infinitely many zeros

Let $$f(z)$$ be an entire function of finite order of the form $$f(z)=\sum_{n=0}^\infty a_nz^{k_n}$$ where $$a_n\neq0$$ for all $$n$$ and $$(k_n)_{n\geq0}$$ is an increasing sequence of integers satisfying $$\limsup_n\;(k_{n+1}-k_n)=\infty.$$ Show that $$f(z)$$ has infinitely many zeros.

I know that if $$f(z)$$ has only finitely many zeros then by Hadamard's theorem $$f(z)$$ is of the form $$P(z)e^{Q(z)}$$ for some polynomials $$P$$ and $$Q$$, so a possible approach is to show that for such functions the condition $$\limsup_n\;(k_{n+1}-k_n)=\infty$$ is not possible, but I gave up after some time. Another possible way may be trying to prove that $$f(z)$$ has non-integer order, since in such a case it follows easily that $$f(z)$$ cannot have finitely many zeros. Any ideas?

• An intuitive argument, that I admittedly don't really know how to formalize in an acceptable manner, is to see that a function of the form $P(z)e^{Q(z)}$ depends on a finite number of parameters. However, if the set $k_n$ satisfies the given condition, then one would have to make an infinite number of derivatives at z=0 equal to zero, which would imply an infinite number of equations on finite parameters... doesnt imply a solution doesn't exist but at least hints at it... – DinosaurEgg May 22 at 11:01

## 1 Answer

As you said, if $$f$$ has only finitely many zeros, it is of the form $$f(z) = P(z) e^{Q(z)}$$ with polynomials $$P(z) = p_m z^m + \ldots + p_0$$ and $$Q(z) = q_n z^n + \ldots + q_0$$. (All leading coefficients are assumed to be non-zero.) Then $$P(z) f'(z) = R(z) f(z)$$ with $$R(z) = P'(z) + P(z)Q'(z) = r_N z^N + \ldots + r_0$$, where $$N=m+n-1 \ge m$$. Writing this out in coefficients, you get $$(p_m z^m + \ldots + p_0) \sum_{j=0}^\infty k_j a_j z^{k_j-1} = (r_N z^N + \ldots + r_0) \sum_{j=0}^\infty a_j z^{k_j} = \sum_{k=0}^\infty b_k z^k$$ Now pick some $$j$$ such that $$k_{j+1} - k_j > N+2$$ and consider the largest $$k \in \{ k_j, k_j +1 , \ldots, k_{j+1}-2\}$$ such that $$b_k \ne 0$$. Multiplying out terms we have $$k = m+k_j-1 = N+k_j$$, which implies that $$N = m-1$$, contradicting $$N \ge m$$.

• Thanks for your answer, but am I missing something or you've written $Rf'=Qf$ instead of $Pf'=Rf$ in the middle equation? – user246336 May 23 at 8:57
• @user246336: Oops, yes, I'll fix that later, thanks for catching my mistake. However, the main argument still works in basically the same way. – Lukas Geyer May 23 at 15:35
• @user246336: Actually, with this correction, the proof is even simpler than before. – Lukas Geyer May 23 at 20:18