# Prove that limit goes to infinity if a convex function's derivative > 0

I do understand what a convex function is and I can see geometrically that following statement is true:

$(1)$ If for a given convex function $f$ which is differentiable over $\mathbb{R}$, if $f'(x_0)>0$ for some $x_0$ then $\lim_{x \to \infty}{f(x)} = \infty$.

I see it like this: if the derivative $f'(x_0)>0$ then the slope is increasing all the time as long as you approach infinity from $x_0$ due to the definition of convex function:

$$f(tx_0 + (1-t)x_1 \leq tf(x_0) + (1-t)f(x_1)$$

In other words, the secant is above the curve all the time. But how do I prove the statement $(1)$?

A continuously differentiable function of one variable is convex on an interval if and only if the function lies above all of its tangents:

$f(x) \ge f(y)+f'(y)[x-y]$ for all x and y in the interval.

By substituting $y=x_0$, we obtain

$f(x) \ge f(x_0)+f'(x_0)[x-x_0]$

Thus, since $f'(x_0)>0$,

$\lim_{x\to\infty}f(x_0)+f'(x_0)[x-x_0]=\infty \le \lim_{x\to\infty}f(x)$

If $f$ is convex, $g(x,y)=\frac{f(x)-f(y)}{x-y}$ is an increasing function of $x$ and an increasing function of $y$. Thus, if we know that $f'(x_0)\gt0$, then because $$f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}g(x,x_0)\tag{1}$$ we must have for all $x\gt x_0$, $$\frac{f(x)-f(x_0)}{x-x_0}=g(x,x_0)\ge\lim_{x\to x_0}g(x,x_0)=f'(x_0)\tag{2}$$ $(2)$ implies that for $x\gt x_0$, $$f(x)\ge f'(x_0)(x-x_0)\tag{3}$$ Differentiability is only assumed at $x_0$.