Is every convex function on an open interval continuous?

Question

Let $f:(a,b)\rightarrow \mathbb{R}$.

$f$ satisfied the following property:

If $\forall x_{1},x_{0},x_{2}\in(a,b)$ and $x_{1}<x_{0}<x_{2};$then$\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}.$

My question : whether the function $f\in C((a,b))?$

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We can get $\displaystyle\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}\geq \frac{f(x_{2})-f(x_{0})}{x_{2}-x_{0}}. $Let $\displaystyle g(x)=\frac{f(x)-f(x_{0})}{x-x_{0}},$ if we can prove $g(x)$ is bound and decreasing (not strictly ) in $(x_{0}-\delta ,x_{0});$ then we have $f^{'}_{-}(x_{0}) $ exists .In a similar way ,$f^{'}_{+}(x_{0}) $ exists. $f^{'}_{-}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}-}f(x)=f(x_{0});$ $f^{'}_{+}(x_{0}) $ exists $\Rightarrow\displaystyle\lim_{x\rightarrow x_{0}+}f(x)=f(x_{0}).$ Obviously, $f$ is coutinuous at $x=x_{0}.$Further, $f\in C((a,b))!$

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But I failed to prove $g(x)$ is bound and decreasing (not strictly ) in $(x_{0}-\delta ,x_{0}).$Sometimes I doubt the conculsion that $f\in C((a,b)).$ Either make a counterexample to deny it ,or prove the conculsion is correct?

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Answer 1

You can prove $g(x)$ is decreasing over all of $(a,x_0)$. To do this, I'll use the lemma that if $\frac{a}b\ge \frac{c}d$, then $\frac{c+a}{d+b}\ge \frac{c}d$, whenever $a,b,c,d\ge0$ (this is easy to prove). Then, for any $x<y<x_0$, we have

$$ g(x)=\frac{f(x_0)-f(x)}{x_0-x}=\frac{f(x_0)-f(y)+f(y)-f(x)}{x_0-y+y-x}\ge\frac{f(x_0)-f(y)}{x_0-y}=g(y) $$ where the $\ge$ part follows from the lemma since $\frac{f(y)-f(x)}{y-x}\ge \frac{f(x_0)-f(y)}{x_0-y}$ by the convexity property. This proves $g$ is decreasing; you can also show $g$ is bounded below in this region, namely, for $x<x_0<x_2$, we always have $g(x)\ge \frac{f(x_2)-f(x_0)}{x_2-x_0}$, and then the rest of the proof goes jsut like you said.

Answer 2

Let $x < y < z$. The inequalities: \begin{eqnarray*} \frac{f(y)-f(x)}{y-x} \le \frac{f(z)- f(y)}{z-y} \\ \frac{f(y)-f(x)}{y-x} \le \frac{f(z)- f(x)}{z-x}\\ \frac{f(z)-f(x)}{z-x} \le \frac{f(z)- f(y)}{z-y} \end{eqnarray*} are equivalent.

A function $f$ is convex if for any $x<y<z$ in the domain any of the equivalent inequalities from above hold.

It is easy to see now that if $f$ is convex and $x<y$, $x'<y'$ and $x\le x'$, $y\le y'$ then \begin{eqnarray*} \frac{f(y)-f(x)}{y-x} \le \frac{f(y')- f(x')}{y'-x'} \end{eqnarray*} (the slopes are increasing). Indeed we have \begin{eqnarray*} \frac{f(y)-f(x)}{y-x} \le \frac{f(y')- f(x)}{y'-x}\le \frac{f(y')- f(x')}{y'-x'} \end{eqnarray*} Let $[c,d]$ a closed interval contained in $(a,b)$. Let $c'< c$ and $d'>d$ so that $[c,d]\subset[c',d']\subset (a,b)$. For any $x<y$ in $[a,b]$ we have \begin{eqnarray*} \frac{f(c)- f(c')}{c - c'} \le \frac{f(y)- f(x)}{y-x}\le \frac{f(d')- f(d)}{d'-d} \end{eqnarray*} Let $M = \max \{ | \frac{f(c)- f(c')}{c - c'} |, | \frac{f(d')- f(d)}{d'-d}|\}$. It follows that \begin{eqnarray*} |\frac{f(y)- f(x)}{y-x}|\le M \end{eqnarray*} for all $x<y$ in the interval $[c,d]$ so $f$ is Lipschitz on that interval.

Note that $M$ can improved to $\max\{ |f'_{-}(c)|, |f'_{+}(d)|\}$.

In any case, $f$ is Lipschitz on any closed interval and hence continuous.

$\bf{Added:}$ We can prove that a convex function defined on an open subset of $\mathbb{R}^n$ is Lipschitz on any compact subset, and so continuous. Let's sketch the argument for an open subset of $\mathbb{R}^2$. It is enough to show that is is Lipschitz on any square $S$ contained in the domain. Consider a slightly larger $S'\supset S$ also in the domain. On the boundary of $S'$ and $S$ the function is continuous, so bounded. Take two points $x$, $y$ in $S$. With the argument above we have $$\frac{|f(x)-f(y)|}{\|x-y\|} \le \frac{2A}{\delta}$$ where $A= \sup_{\partial S \cup \partial S'} |f|$ and $\delta = \operatorname{dist}(\partial S, \partial S')$.

A similar argument with induction on the dimension would show it for convex functions on open subsets of $\mathbb{R}^n$.