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I think the answer to this question must be well known. Is it possible to characterize those functions $f \colon \mathbb{R} \to \mathbb{R}_+$ which are of the form $f(x) = |g(x)|^2, x \in \mathbb{R},$ for some entire function $g \colon \mathbb{C} \to \mathbb{C}$. As a simple counterexample let $f(x) = e^{-1/x^2}$.

Edit: Alexandre Eremenkos answer allows to reformulate the question.

Which entire functions $f \colon \mathbb{C} \to \mathbb{C}$ are nonnegative on $\mathbb{R}$?

In his proof he has (essentially) given a characterization with the help of the Weierstrass factorization theorem. Are there other (more direct) characterizations? I know, it's vague.

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    $\begingroup$ $e^{-1/x^2}$ is certainly not of this form, since it vanishes too quickly around $x=0$. $\endgroup$
    – Wojowu
    Jun 24 '20 at 12:03
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    $\begingroup$ It was meant as a counterexample. I've edited my post. $\endgroup$ Jun 24 '20 at 12:06
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    $\begingroup$ It is not clear what "chatacterize" means. One characterization is this: there are exactly those positive functions on the real line which are entire. If this characterization answers your question, I can write a proof. $\endgroup$ Jun 24 '20 at 12:12
  • $\begingroup$ This would answer my question partially. If you have a reference this would suffice. Of course a short sentence of the form "An entire function $f$, defined on $\mathbb{C}$ is not negative on $\mathbb{R}$ iff ..." would be better. $\endgroup$ Jun 24 '20 at 12:15
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These $f$ are exactly those non-negative functions on the real line which are entire (=represented by their Taylor series on the whole real line). For example, $f(x)=(\arctan x)^2$ is not in your class since the Taylor series at $0$ has finite radius of convergence. Neither $f(x)=e^{-1/x^2}$ is in your class since the Taylor series at zero does not converge to the function).

Proof. Suppose that $g$ is an entire function. Define $g^*(z)=\overline{g(\overline{z})}$ which is also entire. Then on the real line $f(z)=|g(z)|^2=g(z)g^*(z)$, so your function $f(x)$ is non-negative on the real line and entire (as a product of entire functions).

Conversely. Let $f$ be an entire function which is non-negative on the real line. Then all real roots are of even multiplicities, and the rest are symmetric with respect to the real line. Let $X$ be the divisor in the plane which consists of those roots which lie in the open upper half-plane with their multiplicities, and real roots with half of their multiplicities. We have the Weierstrass factorization $f=P e^h$ where $P$ is the canonical product, and $h$ is entire, both $P$ and $h$ real on the real line. Let $P_1$ be the canonical product over $X$, then $P=P_1P_1^*$, and set $g=P_1e^{h/2}$. Then on the real line $$|g(x)|^2=|P_1(x)|^2|e^{h(x)}|=P(x)e^{h(x)}=f(x).$$

Remark. If $f$ has infinitely many non-real zeros, then there are infinitely many different $g$'s which give such a representation: the zeros can be split between $P_1$ and $P_1^*$ in many ways: if $Y$ is the divisor of zeros of $f$, then any $X$ such that $Y=X+\overline{X}$ will do the job.

Remark 2. How to determine that a function of a real variable is in fact entire. A criterion is that $|f^{(n)}(x)|^{1/n}/n\to 0$ uniformly on compact subsets of the real line. This follows from the Taylor formula with remainder combined with Stirling's formula.

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  • $\begingroup$ What is the definition of an entire function $\mathbf{R}\to\mathbf{C}$? I'm only used to the definition of entire function $\mathbf{C}\to\mathbf{C}$. $\endgroup$
    – YCor
    Jun 24 '20 at 12:38
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    $\begingroup$ I wrote in the first sentence in parentheses. $\endgroup$ Jun 24 '20 at 12:40
  • $\begingroup$ Ah indeed, sorry. $\endgroup$
    – YCor
    Jun 24 '20 at 12:42
  • $\begingroup$ @Ivan Meir: Yes. With essentially the same proof. Weierstrass representation can be generalized to any simply connected neighborhood of the real line. $\endgroup$ Jun 24 '20 at 13:02
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    $\begingroup$ After thinking a little bit about your answer I think its just what I wanted. Thank you. $\endgroup$ Jun 24 '20 at 18:27

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