Let $g: \Bbb R \to \Bbb R$ be convex, and $\lim_{x\to\infty} \frac{g(x)}{x}=0$. Prove $g(x)$ is monotone decreasing

Question

Let $g: \Bbb R \to \Bbb R$ be convex, and $\lim_{x\to\infty} \frac{g(x)}{x}=0$. Prove $g(x)$ is monotone decreasing.

My idea: define a function $f(x)=\frac{g(x)-g(0)}{x-0}$. We notice that $\lim_{x\to\infty} f(x)=0$.

Using a theorem about convex functions we can also see that $f(x)$ is monotone increasing for $x>0$, implying that $f(x)<0$. I'm not sure how to prove that $g(x)$ is monotone increasing from this.

Answer 1

Assume not. Then there exists $x_1 < x_2$ such that $g(x_1) < g(x_2)$. Let $h = x_2 - x_1$. Examine the functional value of $g$ at $x_2 + h, x_2 + 2h, \cdots$ and see if you can get a contradiction with $\lim_{x \to \infty} g(x)/x = 0$.