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?

  • $\begingroup$ 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... $\endgroup$ – DinosaurEgg May 22 at 11:01

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 $N \ge m$.

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

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