bounded convex function on $(a,b)$ is uniformly continuous?

Question

bounded convex function on $(a,b)$ is uniformly continuous?

Clearly, bounded convex function on $[a,b]$ is not uniformly continuous, for exmaple, $f(x)=x, x\in [0,1), f(x)=2, x=1$. But what about the open interval?

Answer 1

By the post here, every convex function on $(a,b)$ is continuous. One makes the following

Assertion. If $f$ is convex and bounded on $(a,b)$, then $f$ can be uniquely extended to a continuous function on $[a,b].$

Given the assertion, then it will follow that $f$ is uniformly continuous on $[a,b],$ hence on $(a,b).$ So it suffices to prove the assertion.

Proof of Assertion. Fix a $c\in (a,b).$ One proves that $\lim_{x\rightarrow a^+}f(x)$ exists, the existence of $\lim_{x\rightarrow b^-}f(x)$ being similar. One proves by contradiction. Assume that $\lim_{x\rightarrow a^+}f(x)$ does not exist. Then there exists two decreasing sequences $\{x_n\}$ and $\{x_n'\}$ with $x_n,x_n'\in (a,c)$ such that $f(x_n)$ and $f(x_n')$ have different limits when $x_n,x_n'\rightarrow a.$ Without loss of generality, let $$\lim_{n\rightarrow \infty}f(x_n)=\alpha>\beta=\lim_{n\rightarrow \infty}f(x_n').$$

Then one has that $$\frac{f(c)-\alpha}{c-a}<\frac{f(c)-\beta}{c-a}.\qquad (1)$$

By (1) and the construction of $x_n,x_n'$, there exists $N$ such that $$\frac{f(c)-f(x_n)}{c-x_n}<\frac{f(c)-f(x_m')}{c-x_m'}~{\rm for~all~}n,m>N,$$ which implies that $$f(c)-f(x_n)<\frac{f(c)-f(x_m')}{c-x_m'}(c-x_n),~{\rm for~all~}n,m>N.\qquad (2) $$

Now choose $m,n>N$ such that $$a<x_m'<x_n<c.$$ Consider the line joining $(c,f(c))$ and $(x_m',f(x_m')):$

$$L(x)=f(c)+\frac{f(c)-f(x_m')}{c-x_m'}(x-c).$$ Then it follows by convexity of $f$ on $[x_m',c]$ that $$f(x_n)\leq L(x_n)=f(c)+\frac{f(c)-f(x_m')}{c-x_m'}(x_n-c)$$ $$\Rightarrow f(c)-f(x_n)\geq \frac{f(c)-f(x_m')}{c-x_m'}(c-x_n),$$ contradicting (2). QED