Let $F,H:\mathbb{C}\to\mathbb{C}$ be entire functions of mean exponential type and of completely regular growth. Assume further that the indicator diagrams $I_F$ and $I_H$ are on the imaginary axis and separated, e.g., $I_F=\imath[a,b]$ and $I_H=\imath[b+1,c]$.

Question: Does $F+H \in L^2(\mathbb{R})$ imply $F\in L^2(\mathbb{R} )$?

In other words, can $F$ and $H$ interfere on $\mathbb{R}$ in a way that $F+H\in L^2(\mathbb{R}^2)$ without $F$ or $H$ being $L^2(\mathbb{R})$?

Discussion: Motivation behind this question comes from real analysis. If we assume in addition that $F,H$ are polynomially bounded on $\mathbb{R}$ then they are Fourier transforms of compactly supported distributions $u_F,u_H$ on $\mathbb{R}$, and $\mathrm{supp}\,u_F=\imath I_F$, $\mathrm{supp}\,u_H=\imath I_H$ (Paley-Wiener, Plancherel-Polya). Qualitatively, the behaviour at infinity of $F$ and $H$ reflect the local regularity of $u_F$ and $u_H$. If $F+ H\in L^2(\mathbb{R})$ then $u_F+u_L\in L^2(\mathbb{R})$, i.e., the sum of the two distributions is locally integrable. However, since $\mathrm{supp}\,u_F\cap\mathrm{supp}\,u_H=\emptyset$, adding $u_H$ to $u_F$ cannot change the local regularity of $u_F$, and we expect $u_F$ to be locally integrable as well, which entails $F\in L^2(\mathbb{R})$. In the above question we do not assume that $F$ and $H$ are polynomially bounded on $\mathbb{R}$, but we let $I_F$ and $I_H$ be at a positive distance. In this case $F$ and $H$ can be thought of as Fourier transforms of analytical functionals, but the latter objects do not have the notion of compact support, and the analogy doesn't work.

Thank you.

Edit: For an entire function $F$ of mean exponential type, the indicator function can be defined as $$ h_F(\theta)=\limsup_{r\to\infty}\frac{\log|f(re^{\imath\theta})|}{r} $$ and indicates the rate of growth in direction $\theta$. It can be shown that there exists a compact convex set $I_T\subset\mathbb{C}$ (called the indicator diagram of $F$) such that $h_F$ is exactly the support function of $I_F$, $$ h_F(\theta)=\sup_{z\in I_F}\mathrm{Re}\left[ze^{-\imath\theta}\right]. $$ Intuitively, the thickness of the projection of the set $I_F$ on the ray $\theta$ shows the rate of growth of $F$ along that ray.

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    $\begingroup$ It would be helpful to recall what an "indicator diagram" is... even if you are anticipating that "experts in entire functions" would know this... $\endgroup$ Oct 4, 2019 at 23:46

1 Answer 1


The proof that you indicated for polynomially bounded case can be extended almost literally to the general case, if you use the correct generalization of Fourier-Laplace transform.

Indeed, suppose that $F+G\in L^2(R)$. Then by the Paley-Wiener, this sum is a Fourier transform of some $L^2$ function, say $h$ supported on $[a,c]$ since the indicator diagram of $F+G$ belongs to $[a,c]$. On the other hand, $F+G$ is the Borel transform of an analytic function in $C\cup\{\infty\}\backslash([a,b]\cup[b+1,c])$. By the uniqueness theorem of the Borel transform, this analytic function "coincides" with $g$ in the sense of hyperfunctions. It follows that both $F$ and $G$ are in $L^2$.

Ref. MR1026013 Kaneko, A. Introduction to hyperfunctions. Kluwer, Dordrecht, 1988.

On my opinion, this is the best introduction for beginners. Two other good books (more advanced) are

Mitsuo Morimoto, An introduction to Sato's hyperfunctions. AMS, Providence, RI, 1993. and

L. Hormander, Analysis of linear differential operators with partial derivatives, volume I, chapter 9.

  • $\begingroup$ Thank you for the answer and for the reference. I will read on hyperfunctions to understand the arguments. $\endgroup$
    – Bedovlat
    Oct 5, 2019 at 1:03
  • $\begingroup$ Excellent and very eye-opening (at least for me) answer. And a wonderful reference, which, unlike many other expositions of hyperfunctions, does not torment the innocent analyst with homological algebra. Thank you! $\endgroup$
    – Bedovlat
    Oct 11, 2019 at 2:47

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