How does a function conveex on an open interval has both left and right hand derivatives at every point of that interval?

Question

Let $(a, b)$ be an open interval on the real line, let $f \colon (a, b) \rightarrow \mathbb{R}$ be a convex function, and let $c \in (a, b)$. Then how to show rigorously that the right- and left-hand derivatives given respectively by $$ f_+^\prime(c) \colon= \lim_{x \to c+ 0} \frac{ f(x) - f(c) }{ x-c } $$ and $$ f_-^\prime(c) \colon= \lim_{x \to c - 0} \frac{ f(x) - f(c) }{ x-c } $$ exist in $\mathbb{R}$?

Here is a post of mine on a Baby Rudin problem on the continuity of a convex function.

My Attempt:

As $f$ is convex on $(a, b)$, so we have the inequality $$ f \big( (1-\lambda) x + \lambda y \big) \leq (1- \lambda) f(x) + \lambda f(y) $$ whenever $a < x < b$, $a < y < b$, and $0 \leq \lambda \leq 1$.

First, let us take $x$ such that $c < x < b$. Let $d$ be such that $x < d < b$. Then $$ x = (1- \lambda) c + \lambda d $$ for $$ \lambda = \frac{ x-c}{d-c}. $$

What next? How to proceed from here?

Or, is there some alternative approach that will lead to the desired proofs?

I would of course like rigorous proofs using elementary enough ideas.

Answer 1

Making appropriate coordinate changes if necessary,we may assume,without loss of generality that we are checking left and right differentiability at $0$,with $f(0)=0$.

I shall demonstrate a geometric argument here.

See, a convex function is one for which given any two points $x$ and $y$ in the domain, the graph of the function lies below the line joining $(x,f(x))$ and $(y,f(y))$.

To prove right differentiability, we need to show that $\frac{f(h)}{h}$ tends to a limit as $h \rightarrow 0+$.

Define, for all $h>0$, a function $$ g(h)=\frac{f(h)}{h}. $$

We shall show that $g$ is monotone decreasing, and bounded below. Then it's infimum will be our required limit. Pick $u, v$ such that $0 < u < v$. Since $f(0)=0$ the line joining $(0,f(0))$ and $(v,f(v))$ is given by $$ y= \frac{xf(v)}{v} . $$

Now, since $0< u <v $, we have, due to convexity (graph lying below the line segment, remember?), $$ f(u) \leq \frac{uf(v)}{v} , $$ which implies that $$ \frac{f(u)}{u} \leq \frac{f(v)}{v} , $$ in other words $$ g(u) \leq g(v). $$

In a similar way, show that the function $$ p(h)=\frac{f(h)}{h} $$ for all $h<0$ is monotone increasing.

Now, due to convexity, for all $h>0$ we have $$ \frac{f(h)+f(-h)}{2} \geq 0 , $$ or $$ \frac{f(h)}{h} \geq \frac{f(-h)}{-h}. $$ This, together with the fact that $p$ is monotone increasing and $g$ is monotone decreasing, implies that $g$ is bounded below and $p$ is bounded bove(Why?). This demonstrates that both $g$ and $p$ have finite limits as $h \rightarrow 0+$ and $h\rightarrow 0-$, respectively, thus proving left and right differentiability of $f$.

Hope this helps.