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The following problem is posed in Greene and Krantz, page 297, Problem 11.

Let $g: \mathbb{R} \to \mathbb{C}$ be a continuous function, $\alpha \in \mathbb{R}$, and define $f: \mathbb{C} \to \mathbb{C}$ s.t. $$f(z) = \int^b_a e^{\alpha z t} g(t) dt .$$

Show that $f$ is entire and is of finite order. Determine whether or not the order of $f$ depends on $g$.

I am able to show that $f$ is entire and of finite order. Intuitively, it seems to me that assuming $\alpha \neq 0$ and $g$ not identically $0$, $f$ should always be of order 1, but I can't show this nor find a counterexample. Insights would be appreciated.

wellfedgremlin
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  • Yes, it's clear the order is at most 1. However, I don't see how you'd proceed with your suggestion - if you have something specific in mind please post it. – wellfedgremlin Mar 05 '14 at 10:22
  • Assuming that $|g(t)|\leq M$ over $[a,b]$, we have: $$|f(z)|\leq M\int_{a}^{b}|e^{\alpha z t}|,dt = M\int_{a}^{b}e^{\alpha(\Re{z}) t},dt = M(b-a)e^{\alpha(\Re z)\xi},\quad \xi\in(a,b).$$ – Jack D'Aurizio Mar 10 '14 at 18:46
  • @JackD'Aurizio: I think the more interesting part is the lower bound of the order of $f$. Do you have any suggestion? – user104254 Mar 11 '14 at 08:29
  • @user104254: For fixed $\alpha\neq 0$, take $z$ such that $\alpha z\in\mathbb{R}^+$ and let $r$ be the ray ${w\in\mathbb{C}:w=\lambda z,\lambda\in\mathbb{R}^+}$. For any $w\in r$ we can look at the exponential function as a weight-function in the integral $f(w)=\int_{a}^{b}e^{\alpha w t}g(t),dt$. Assuming $(a,b)\in\mathbb{R}^+$, we expect that most of the mass is concentrated near $t=b$, more and more concentrated as $w$ goes to $\infty$ along $r$, giving $f(w)\sim e^{\alpha w b}g(b)\cdot\int_{a}^{b}e^{\alpha w t}dt$. – Jack D'Aurizio Mar 11 '14 at 09:03
  • So, IMHO, the lower bound on the order of $f$ can be achieved through some concentration inequality. – Jack D'Aurizio Mar 11 '14 at 09:04
  • @user104254: sorry, I meant $(a,b)\subset\mathbb{R}^+$ and $f(w)\sim g(b)\cdot\int_{a}^{b}e^{\alpha wt}dt$. – Jack D'Aurizio Mar 11 '14 at 09:11
  • @JackD'Aurizio: Under your settings, I think the interesting situation is $g(b)=0$ and the behavior of $g$ around $b$ is extremely pathological. – user104254 Mar 11 '14 at 09:13
  • @user104254: despite any possible pathology, $g$ is continuous, so from $g(b)\neq 0$ we can derive a lower bound for the order of $f$, as shown in my answer below. – Jack D'Aurizio Mar 11 '14 at 09:52
  • @JackD'Aurizio: Thank you for your replies, but what I am most concerned about is the case mentioned in my last comment. – user104254 Mar 11 '14 at 10:07
  • @user104254: Non-trivial cases are, anyway, covered by the Weierstrass approximation theorem: see my updated answer. – Jack D'Aurizio Mar 11 '14 at 14:33
  • @JackD'Aurizio: I hardly think for general continuous $g$, the problem can be easily handled by a simple Weierstrass approximation argument. – user104254 Mar 11 '14 at 14:53

1 Answers1

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The following generalized statement holds. To recover your original setting, we can simply define $\mu$ as $\mu(E):=\int_{E\cap[a,b]}g(t)dt$ and replace $f(z)$ by $f(\alpha z)$.

Proposition: Let $\mu$ be a Borel complex measure on $\Bbb R$. Denote its support by ${\rm supp}(\mu)$ and its total variation by $\|\mu\|$. Suppose that $$\sigma:=\sup \{|x|:x\in {\rm supp}(\mu)\}\in (0,+\infty). \tag{1}$$ Then for $$f:\Bbb C\to \Bbb C,\quad f(z):=\int_{\Bbb R}e^{zt}d\mu (t), \tag{2}$$ $f$ is an entire function satisfying that for $x,y\in \Bbb R$, $$|f(x+iy)|\le \|\mu\|\cdot e^{\sigma|x|}, \tag{3}$$ and for $x\in \Bbb R$, $$\limsup_{|x|\to+\infty}\frac{\log |f(x)|}{|x|}=\sigma. \tag{4}$$ As a corollary, the order and type of $f$ are $1$ and $\sigma$ respectively.

Proof: Since $\mu$ has compact support, it is evident that $f$ is entire and $(3)$ holds. To prove $(4)$, without loss of generality, we may assume that $\sigma\in {\rm supp}(\mu)$(the case $-\sigma\in {\rm supp}(\mu)$ is similar), and it suffices to show that for every $\tau\in (0,\sigma)$, $$\limsup_{x\to+\infty}|f(x)e^{-\tau x}|=+\infty. \tag{5}$$ To prove $(5)$, let us define $$F:\Bbb C\to \Bbb C,\quad F(z):=\int_\tau^\sigma e^{z(t-\tau)}d\mu (t). \tag{6}$$ By definition, when $x\ge 0$, $|f(x)e^{-\tau x}-F(x)|\le \|\mu\|$, so $(5)$ is equivalent to $$\limsup_{x\to+\infty}|F(x)|=+\infty. \tag{7}$$ To begin with the proof of $(7)$, note that $$|F(x+iy)|\le \|\mu\|\cdot e^{(\sigma-\tau)\cdot\max\{x,0\}}.\tag{8}$$ In particular, the order of $F$ is no more than $1$. By reduction to absurdity, suppose that $(7)$ fails. Then from $(8)$ we know that there exists $M\ge \|\mu\|$, such that $|F(x)|\le M$ when $x\ge 0$ and $|F(x+iy)|\le M$ when $x\le 0$. Now we can apply Phragmén–Lindelöf principle in the linked page to $F$ on the first and the forth quadrants respectively with $\lambda=2$ and $\rho =1$, which implies that $|F(x+iy)|\le M$ when $x\ge 0$. Therefore, $|F|\le M$ on $\Bbb C$, so by Liouville's theorem, $F$ must be constant. On the other hand, from $(6)$ we know that the function $y\mapsto F(-iy)e^{-i\tau y}$ is the Fourier transform of the complex measure $\nu:=\mu|_{[\tau,\sigma]}$, so $F$ being constant implies that ${\rm supp}(\nu)=\{\tau\}$. However, by our assumption, $\sigma\in {\rm supp}(\nu)$. This contradiction completes the proof.

Remark: On the one hand, $(4)$ asserts that as $x\to +\infty$, $\max\limits_{t\in [-x,x]}|f(t)|\to +\infty$ exponentially fast in $x$; on the other hand, even if $\mu$ is defined by $\mu(E):=\int_{E\cap[a,b]}g(t)dt$ for some smooth function $g$, it is possible that $f(x)$ has infinitely many zeros on $[0,+\infty)$.

user104254
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