Introduction to Analysis: Convexity A friend and I were trying to figure out this problem from our assignment.

Prove that on an open $I$, a geometrically convex function $f(x)$ is continuous. 

To better assist the audience, it is best I give you the definition of a geometrically convex function from the book, "Introduction to Analysis," by Arthur Mattuck.

Let $f(x)$ be defined on any type of interval I. For any subinterval $[a,b] \subset I$, we let $P:(a,f(a))$ and $Q:(b,f(b))$ be the two points of the graph lying over the endpoints of the interval. We say $f(x)$ is geometrically convex on $I$ if the graph on f(x) lies on or below the chord PQ, for all $[a,b]\subset I$ An equivalent analytic formulation of this is $\frac{f(x)-f(a)}{x-a}\leq\frac{f(b)-f(x)}{b-x}$, for all $a<x<b$ in $I.$ 

How would we go about pursuing this problem? I would figure a way to approach it would be to show $\lim_{\bigtriangleup x\rightarrow0^-}\frac{\bigtriangleup y}{\bigtriangleup x}$ exists at each point of $I$, and some how deduce $\lim_{\bigtriangleup x\rightarrow0^-}\bigtriangleup y=0$. Is this correct? If so, how would I proceed?
 A: Given any point $b$ in $I$, choose a point $x_0<b$ in $I$. Choose $x_1$ as the average point of $x_0$ and $b$, so $\frac{f(b)-f(x_0)}{b-x_0} \le \frac{f(b)-f(x_1)}{b-x_1}$ by geometrically convex. Choose $x_2$ as the average point of $x_1$ and $b$, and so on. Then we get an increasing sequence $\{\frac{f(b)-f(x_n)}{b-x_n} \}$, i.e. $\{ \frac{\Delta y_n}{\Delta x_n}\}$.
The sequence is bounded above. For if not, choose a point $c>b$ in $I$(open), and there is a point $x_n<b$ such that slope of $x_nb >$ slope of $x_nc$, contradicting to geometrically convex.
So the sequence has a limit, i.e. $\lim_{\Delta x\to 0^-}\frac{\Delta y}{\Delta x}$ exists. So, $$\lim_{\Delta x\to 0^-}\Delta y =\lim_{\Delta x\to 0^-}\frac{\Delta y}{\Delta x} \cdot \Delta x=\lim_{\Delta x\to 0^-}\frac{\Delta y}{\Delta x}\cdot \lim_{\Delta x \to 0^-}=0.$$
You can show that $\lim_{\Delta x\to 0^+}\Delta y=0$ by the same way.
A: If $f$ is convexe then it is differentiable at the right and at the left on any interior point, hence continuous on all the open set.
Just use the inequality written in the OP post by noticing the first term is increasing with $x$ and it is bounded above by the second term, hence it converges when $x \to a^-$, hence $f$ is differentiable at the left of $a$. The same for the right.
