Let $$g(t):=EX\,1(X\ge t),$$ so that $g(t)\downarrow0$ as $t\to\infty$. Let $0=t_0<t_1<t_2<\cdots$ be such that $t_1\ge1$ and $g(t_j)<1/j^3$ for all $j\ge1$. Let $f\colon[0,\infty)\to[0,\infty)$ and $0=x_0<x_1<x_2<\cdots$ be such that $f(0)=0$ and for $j\ge0$ we have \begin{equation} x_j\ge t_j,\quad x_{j+1}\ge x_j^2, \quad x_{j+1}\ge2x_j, \end{equation} \begin{equation} f(x)=f(x_j)+(j+1)(x-x_j)\text{ for }x\in[x_j,x_{j+1}]. \end{equation} Then the function $f$ is clearly increasing and convex, and $f'(x)\to\infty$ as $x\to\infty$ (where $f'$ is the right derivative of $f$), so that $f(x)/x\to\infty$ as $x\to\infty$. The convexity of $f$ and the condition $f(0)=0$ also immediately imply that \begin{equation} f(x)y\le f(xy) \end{equation} for all real $x,y\ge1$.

It also follows that $f(x)\le(j+1)x$ for $x\in[0,x_{j+1}]$, whence \begin{equation} Ef(X)\,1(X\ge x_1)\le\sum_{j\ge1}E(j+1)X\,1(x_j\le X<x_{j+1}) \le\sum_{j\ge1}(j+1)g(x_j)\le\sum_{j\ge1}(j+1)g(t_j)\le\sum_{j\ge1}(j+1)/j^3<\infty, \end{equation} so that $Ef(X)<\infty$.

It remains to show that \begin{equation} f(xy)\le cf(x)f(y) \tag{10}\label{10} \end{equation} for some real $c>0$ and all real $x,y\ge c$.

Take any $x,y$ such that $x_1\le x\le y<\infty$. Then for some integers $j,k$ we have \begin{equation} 1\le j\le k,\quad x\in[x_j,x_{j+1}],\quad y\in[x_k,x_{k+1}]. \end{equation} So, $xy\in[x_j x_k,x_{j+1}x_{k+1}]\subseteq[x_k,x_{k+2}]$, because $x_j\ge x_1\ge t_1\ge1$ and $x_{j+1}x_{k+1}\le x_{k+1}^2\le x_{k+2}$. So, \begin{equation} f(xy)\le(k+2)xy. \end{equation} On the other hand, \begin{equation} f(y)\ge f(x_{k-1})+(k+1)(y-x_{k-1})\ge(k+1)(y-x_{k-1})\ge(k+1)y/2 \ge(k+2)y/3\ge f(xy)/(3x), \end{equation} so that \begin{equation} f(xy)\le3xf(y). \end{equation} Recalling now that $f(x)/x\to\infty$ as $x\to\infty$, we complete the proof of \eqref{10}. $\quad\Box$

Let $$g(t):=EX\,1(X\ge t),$$ so that $g(t)\downarrow0$ as $t\to\infty$. Let $0=t_0<t_1<t_2<\cdots$ be such that $t_1\ge1$ and $g(t_j)<1/j^3$ for all $j\ge1$. Let $f\colon[0,\infty)\to[0,\infty)$ and $0=x_0<x_1<x_2<\cdots$ be such that $f(0)=0$ and for $j\ge0$ we have \begin{equation} x_j\ge t_j,\quad x_{j+1}\ge x_j^2, \quad x_{j+1}\ge2x_j, \end{equation} \begin{equation} f(x)=f(x_j)+(j+1)(x-x_j)\text{ for }x\in[x_j,x_{j+1}]. \end{equation} Then the function $f$ is clearly increasing and convex, and $f'(x)\to\infty$ as $x\to\infty$ (where $f'$ is the right derivative of $f$), so that $f(x)/x\to\infty$ as $x\to\infty$. The convexity of $f$ and the condition $f(0)=0$ also immediately imply that \begin{equation} f(x)y\le f(xy) \end{equation} for all real $x,y\ge1$.

It also follows that $f(x)\le(j+1)x$ for $x\in[0,x_{j+1}]$, whence \begin{equation} Ef(X)\,1(X\ge x_1)\le\sum_{j\ge1}E(j+1)X\,1(x_j\le X<x_{j+1}) \le\sum_{j\ge1}(j+1)g(x_j)\le\sum_{j\ge1}(j+1)g(t_j)\le\sum_{j\ge1}(j+1)/j^3<\infty, \end{equation} so that $Ef(X)<\infty$.

It remains to show that \begin{equation} f(xy)\le cf(x)f(y) \tag{10}\label{10} \end{equation} for some real $c>0$ and all real $x,y\ge c$.

Take any $x,y$ such that $x_1\le x\le y<\infty$. Then for some integers $j,k$ we have \begin{equation} 1\le j\le k,\quad x\in[x_j,x_{j+1}],\quad y\in[x_k,x_{k+1}]. \end{equation} So, $xy\in[x_j x_k,x_{j+1}x_{k+1}]\subseteq[x_k,x_{k+2}]$, because $x_j\ge x_1\ge t_1\ge1$ and $x_{j+1}x_{k+1}\le x_{k+1}^2\le x_{k+2}$. So, \begin{equation} f(xy)\le(k+2)xy. \end{equation} On the other hand, noting that $x_{k-1}\le x_k/2\le y/2$, we have \begin{equation} f(y)\ge f(x_{k-1})+(k+1)(y-x_{k-1})\ge(k+1)(y-x_{k-1})\ge(k+1)y/2 \ge(k+2)y/3\ge f(xy)/(3x), \end{equation} so that \begin{equation} f(xy)\le3xf(y). \end{equation} Recalling now that $f(x)/x\to\infty$ as $x\to\infty$, we complete the proof of \eqref{10}. $\quad\Box$

Let $$g(t):=EX\,1(X\ge t),$$ so that $g(t)\downarrow0$ as $t\to\infty$. Let $0=t_0<t_1<t_2<\cdots$ be such that $t_1\ge1$ and $g(t_j)<1/j^3$ for all $j\ge1$. Let $f\colon[0,\infty)\to[0,\infty)$ and $0=x_0<x_1<x_2<\cdots$ be such that $f(0)=0$ and for $j\ge0$ we have \begin{equation} x_j\ge t_j,\quad x_{j+1}\ge x_j^2, \quad x_{j+1}\ge2x_j, \end{equation} \begin{equation} f(x)=f(x_j)+(j+1)(x-x_j)\text{ for }x\in[x_j,x_{j+1}]. \end{equation} Then the function $f$ is clearly increasing and convex, and $f'(x)\to\infty$ as $x\to\infty$, so that $f(x)/x\to\infty$ as $x\to\infty$. The convexity of $f$ and the condition $f(0)=0$ also immediately imply that \begin{equation} f(x)y\le f(xy) \end{equation} for all real $x,y\ge0$.

It also follows that $f(x)\le(j+1)x$ for $x\in[x_j,x_{j+1}]$, whence \begin{equation} Ef(X)\,1(X\ge x_1)\le\sum_{j\ge1}E(j+1)X\,1(x_j\le X<x_{j+1}) \le\sum_{j\ge1}(j+1)g(x_j)\le\sum_{j\ge1}(j+1)g(t_j)\le\sum_{j\ge1}(j+1)/j^3<\infty, \end{equation} so that $Ef(X)<\infty$.

It to show that \begin{equation} f(xy)\le cf(x)f(y) \tag{10}\label{10} \end{equation} for some real $c>0$ and all real $x,y\ge c$.

Take any $x,y$ such that $x_1\le x\le y<\infty$. Then for some integers $j,k$ we have \begin{equation} 1\le j\le k,\quad x\in[x_j,x_{j+1}],\quad y\in[x_k,x_{k+1}]. \end{equation} So, $xy\in[x_j x_k,x_{j+1}x_{k+1}]\subseteq[x_k,x_{k+2}]$, because $x_j\ge x_1\ge t_1\ge1$ and $x_{j+1}x_{k+1}\le x_{k+1}^2\le x_{k+2}$. So, \begin{equation} f(xy)\le(k+2)xy. \end{equation} On the other hand, \begin{equation} f(y)\ge f(x_{k-1})+(k+1)(y-x_{k-1})\ge(k+1)(y-x_{k-1})\ge(k+1)y/2 \ge(k+2)y/3\ge f(xy)/(3x), \end{equation} so that \begin{equation} f(xy)\le3xf(y). \end{equation} Recalling now that $f(x)/x\to\infty$ as $x\to\infty$, we complete the proof of \eqref{10}. $\quad\Box$

Let $$g(t):=EX\,1(X\ge t),$$ so that $g(t)\downarrow0$ as $t\to\infty$. Let $0=t_0<t_1<t_2<\cdots$ be such that $t_1\ge1$ and $g(t_j)<1/j^3$ for all $j\ge1$. Let $f\colon[0,\infty)\to[0,\infty)$ and $0=x_0<x_1<x_2<\cdots$ be such that $f(0)=0$ and for $j\ge0$ we have \begin{equation} x_j\ge t_j,\quad x_{j+1}\ge x_j^2, \quad x_{j+1}\ge2x_j, \end{equation} \begin{equation} f(x)=f(x_j)+(j+1)(x-x_j)\text{ for }x\in[x_j,x_{j+1}]. \end{equation} Then the function $f$ is clearly increasing and convex, and $f'(x)\to\infty$ as $x\to\infty$ (where $f'$ is the right derivative of $f$), so that $f(x)/x\to\infty$ as $x\to\infty$. The convexity of $f$ and the condition $f(0)=0$ also immediately imply that \begin{equation} f(x)y\le f(xy) \end{equation} for all real $x,y\ge1$.

It also follows that $f(x)\le(j+1)x$ for $x\in[0,x_{j+1}]$, whence \begin{equation} Ef(X)\,1(X\ge x_1)\le\sum_{j\ge1}E(j+1)X\,1(x_j\le X<x_{j+1}) \le\sum_{j\ge1}(j+1)g(x_j)\le\sum_{j\ge1}(j+1)g(t_j)\le\sum_{j\ge1}(j+1)/j^3<\infty, \end{equation} so that $Ef(X)<\infty$.

It remains to show that \begin{equation} f(xy)\le cf(x)f(y) \tag{10}\label{10} \end{equation} for some real $c>0$ and all real $x,y\ge c$.

Take any $x,y$ such that $x_1\le x\le y<\infty$. Then for some integers $j,k$ we have \begin{equation} 1\le j\le k,\quad x\in[x_j,x_{j+1}],\quad y\in[x_k,x_{k+1}]. \end{equation} So, $xy\in[x_j x_k,x_{j+1}x_{k+1}]\subseteq[x_k,x_{k+2}]$, because $x_j\ge x_1\ge t_1\ge1$ and $x_{j+1}x_{k+1}\le x_{k+1}^2\le x_{k+2}$. So, \begin{equation} f(xy)\le(k+2)xy. \end{equation} On the other hand, \begin{equation} f(y)\ge f(x_{k-1})+(k+1)(y-x_{k-1})\ge(k+1)(y-x_{k-1})\ge(k+1)y/2 \ge(k+2)y/3\ge f(xy)/(3x), \end{equation} so that \begin{equation} f(xy)\le3xf(y). \end{equation} Recalling now that $f(x)/x\to\infty$ as $x\to\infty$, we complete the proof of \eqref{10}. $\quad\Box$