2.1 The construction

In this section, we will resolve Question 1.1. Let us write $h(x)=\exp (f(x)),$ where $f(x)$ is convex. Since $h\in L^1(\mathbb {R}^+)$ , it must be that $\lim _{x\rightarrow \infty }f(x)=-\infty $ . The remaining condition on h implies $\lim _{n\rightarrow \infty }f(n)/n=0$ , or equivalently (by the convexity of f) $\lim _{x\rightarrow \infty }f'(x)=0$ provided $f'(x)$ exists. In fact, the function f we construct will be piecewise linear and hence $f'(x)$ will be undefined on a discrete set. We will choose sequences $\{a_n\}_{n=0}^{\infty }$ , $\{m_n\}_{n=0}^{\infty }$ , and $\{b_n\}_{n=0}^{\infty }$ so that

(2.1) $$ \begin{align} f(x)=m_nx+b_n,\qquad x\in[a_n,a_{n+1}], \end{align} $$

and f is continuous.

To begin our construction, let $\{m_n\}_{n=0}^{\infty }$ be a fixed sequence of negative real numbers that monotonically increases to $0$ . With this fixed sequence in hand, we will construct the sequence $\{a_n\}_{n=0}^{\infty }$ inductively, and the sequence $\{b_n\}_{n=0}^{\infty }$ will then be defined implicitly in order to make f continuous.

We begin our construction of the sequence $\{a_n\}_{n\geq 0}$ by defining $a_0=0$ and we also define $b_0=0$ . Now, choose $a_1$ large enough so that

• •

  $$\begin{align*}-\frac{e^{m_0a_1}}{m_1}<\frac{1}{2} \end{align*}$$

• •

  $$\begin{align*}a_1>1=a_0+e^{-b_0}. \end{align*}$$

Now, set $b_1=m_0a_1-m_1a_1$ and observe that $m_1a_1+b_1=m_0a_1$ .

Now, let us assume that $\{a_j\}_{j=0}^n$ and $\{b_j\}_{j=0}^n$ have already been defined. We will now show how one can choose $a_{n+1}$ and then $b_{n+1}$ to complete the construction. Indeed, as above, we will choose $a_{n+1}$ large enough so that

$$ \begin{align*} -\frac{e^{m_na_{n+1}+b_n}}{m_{n+1}}<\frac{1}{2^{n+1}}, \end{align*} $$

$$ \begin{align*} a_{n+1}>\max\left\{a_n+e^{(1-t)b_n}:t\in[0,n]\right\}. \end{align*} $$

Then define

$$\begin{align*}b_{n+1}=m_na_{n+1}+b_n-m_{n+1}a_{n+1} \end{align*}$$

and observe that

(2.2) $$ \begin{align} m_{n+1}a_{n+1}+b_{n+1}=m_na_{n+1}+b_n. \end{align} $$

Proceeding inductively, we arrive at two sequences $\{a_n\}_{n=0}^{\infty }$ and $\{b_n\}_{n=0}^{\infty }$ . It is clear from our construction that $a_{n+1}>a_n+1$ (since $1\in [0,n]$ ) and so each $a_n$ is positive (except $a_0$ ) and $\lim _{n\rightarrow \infty }a_n=\infty $ . Therefore, this procedure defines f on all of $(0,\infty )$ if we define f by (2.1). Notice also that $a_{n}>0$ and $m_n>m_{n-1}$ inductively implies that each $b_n<0$ when $n>0$ .

Now, let us check that this function has the desired properties. First of all, since $m_n\rightarrow 0$ monotonically, it is clear that $f(n)/n\rightarrow 0$ and also that f is convex. Now we calculate

$$ \begin{align*} \int_0^{\infty}h(x)dx&=\int_0^{\infty}e^{f(x)}dx=\sum_{n=0}^{\infty}\int_{a_n}^{a_{n+1}}e^{m_nx+b_n}dx\\ &\leq\sum_{n=0}^{\infty}\int_{a_n}^{\infty}e^{m_nx+b_n}dx\\ &=\sum_{n=0}^{\infty}-\frac{e^{m_na_n+b_n}}{m_n}\\ &=-\frac{1}{m_0}+\sum_{n=1}^{\infty}-\frac{e^{m_{n-1}a_n+b_{n-1}}}{m_n}\qquad\qquad\mbox{(we use}\ (2.2)\ \mbox{here)} \\&<-\frac{1}{m_0}+\sum_{n=1}^{\infty}2^{-n}, \end{align*} $$

which is clearly finite. Therefore, $h\in L^1(\mathbb {R}^+)$ as desired.

Finally, fix $r\in (0,\infty )$ and choose $N\in \mathbb {N}$ so that $r<N$ . Notice that

$$\begin{align*}\frac{h^r(x)}{h(rx)}=e^{(r-1)b_n},\qquad x\in[a_n,a_{n+1}]. \end{align*}$$

Therefore,

$$ \begin{align*} \int_0^{\infty}\frac{h^r(x)}{h(rx)}dx&=\sum_{n=0}^{\infty}e^{(r-1)b_n}(a_{n+1}-a_n)>\sum_{n=N}^{\infty}e^{(r-1)b_n}(a_{n+1}-a_n)>\sum_{n=N}^{\infty}1, \end{align*} $$

by construction, so $\frac {h^r(x)}{h(rx)}\not \in L^1(\mathbb {R}^+)$ . This completely answers Question 1.1.

2.2 The conjecture

In this section, we will prove Theorem 1.4. The log convexity of h implies that h is either monotone increasing on $(A,\infty )$ for some $A\geq 0$ or monotone decreasing on $(0,\infty )$ . In the latter case, we have $h(x)\geq h(2x)$ and so

$$\begin{align*}\int_0^{\infty}h(x)dx\leq\int_0^{\infty}\frac{h^2(x)}{h(2x)}dx<\infty, \end{align*}$$

so h is integrable.

If h is monotone increasing on $(A,\infty )$ , then $\log (h(x))$ is also increasing on $(A,\infty )$ . Note that the convexity implies that we can choose A so that $\log (h(x))$ is strictly monotone increasing on $(A,\infty )$ , for otherwise h would be constant on some interval $[B,\infty )$ . This would contradict the assumption that $h^2(x)/h(2x)$ is integrable on $(0,\infty )$ . Since $\log (h(x))$ is convex, it must be that there is some constant $c>0$ so that

$$ \begin{align*} \log(h(n+1))-\log(h(n))\geq c,\qquad n>A. \end{align*} $$

This implies $h(n+1)/h(n)\geq e^c$ , which means h cannot satisfy (1.1). Therefore, this case cannot occur, and we have proven Theorem 1.4.