A continuous and at least zero lower derivative function is increasing. Let $f:(a,b) \to \mathbb{R}$ be a continuous function on the interval $(a, b)$ such that its lower derivative is at least zero on $(a, b)$. Prove that the function $g(x)=f(x)+a\cdot x$ is increasing on $(a, b)$ for any constant $a>0$.
I need to do this homework question and have no idea where to start.
 A: Suppose the opposite is true : $g(x)=f(x) + a.x$ is non increasing.
$\exists x, z$ such that $g(x) > g(z)$.
You can take the sup on the x and the inf on the z (prove that it leads to one point y) that verify it (with x <= z) and then use the definition of lower derivative at y to find that it is absurd.
A: Let's us assume that constant $a > 0$ is different from $a$ in $(a, b)$ and thus we use symbol $k$ in place of $a$. Then we need to show that $g(x) = f(x) + kx$ is increasing on $(a, b)$.
Let $c, d$ be points in $(a, b)$ such that $c < d$. We will show that $g(c) \leq g(d)$. We are given that the lower (or left which is the correct term) derivative $f_{-}'(x) \geq 0$ for all $x \in (a, b)$. Clearly this means that $g_{-}'(x) = f_{-}'(x) + k \geq k > 0$ for all $x \in (a, b)$. Now we know that $$g_{-}'(x) = \lim_{h \to 0^{-}}\frac{g(x + h) - g(x)}{h} = \lim_{h \to 0^{+}}\frac{g(x) - g(x - h)}{h}$$ and this limit is guaranteed to be positive by the given conditions. Hence it follows that $g(x) > g(x - h)$ for all sufficiently small $h > 0$.
Now we divide the real numbers $x$ of interval $[c, d]$ into two sets $L$ and $U$ in the following fashion:
We let $x \in U$ if $g(y) \leq g(d)$ for all $y \in [x, d]$ and remaining numbers are in $L$ so that $L = [c, d] - U$. Clearly $U$ is non-empty as $g_{-}'(d) > 0$ implies the existence of an interval $[d - h, d]$ such that $g(y) \leq g(d)$ for all $y \in [d - h, d]$ so that all points of interval $[d - h, d]$ belong to $U$. Now we need to proceed from here and show that $L$ is empty. This will require the use of Dedekind theorem on sets $L$ and $U$. This would establish the fact that $g(c) \leq g(d)$. Continuing this argument a bit further and noting that $g_{-}'(x) > 0$ we can show that in fact $g(c) < g(d)$ so that that we have strict monotonicity.
