Consider a continuous and convex function $F(x):[0,1]\longrightarrow\mathbb{R}$. I am wondering if

  1. $F(x)$ is continuously differentiable in $[0,1]$

  2. $F(x)$ is of bounded variation in $[0,1]$

  3. $F(x)$ is absolute continuous in $[0,1]$.

The second one is correct, due to this post Proving a convex function is of bounded variation.

However, the remaining two became mysterious to me. Royden's chapter 6 answers them if we have an open interval.

Corollary 17: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is Lipschitz, and therefore absolutely continuous on each closed, bounded subinterval $[c,d]$ and $(a,b)$

Theorem 18: Let $\varphi$ be a convex function on $(a,b)$. Then $\varphi$ is differentiable except at a countable number of points.

By the Theorem 18, it is hard to believe that $F(x)$ will become differentiable in $[0,1]$. But I cannot find a counterexample. That is, a convex function that is continuous on $[0,1]$ but is not differentiable.

The Corollary 17 gives us pretty nice result, but seems like it does not apply to the closed interval. Is it possible to say that if we have $F(x)$ on $[0,1]$ is convex, then it will be convex on $(-\epsilon, 1+\epsilon)$? and then we can use Corollary 17 to conclude that it is absolutely continuous on $[0,1]\subset (-\epsilon, 1+\epsilon)$.

Thank you!

  • 1
    $\begingroup$ As a counterexample for 1, you can use $F(x)=|x - \frac{1}{2}|$. It is not differentiable at $x=\frac{1}{2}$. $\endgroup$ – JPMarciano Nov 26 '20 at 22:56
  • $\begingroup$ @JPMarciano oh right. I thought $|x|$, but only god knows why i did not think about $|x-a|$ thank you!! $\endgroup$ – JacobsonRadical Nov 26 '20 at 23:01
  • 1
    $\begingroup$ 3) is true. $F$ is the indefinite integral of its right-hand derivative. $\endgroup$ – Kavi Rama Murthy Nov 26 '20 at 23:23
  • $\begingroup$ @KaviRamaMurthy would you mind elaborating it? $\endgroup$ – JacobsonRadical Nov 26 '20 at 23:24
  • 1
    $\begingroup$ @JackM but I don't have tools to prove $F(x)$ is absolutely continuous on $(0,1)$. Royden's theorem only tells us if it is convex on $(a,b)$ then it will be absolutely continuous on every compact subintervals $[c,d]$ of it. Not from $[a,b]$ to $(c,d)$ $\endgroup$ – JacobsonRadical Nov 27 '20 at 6:53

Given real numbers $a<b$, let us show that a continuous and convex function $F(x):[a,b]\longrightarrow\mathbb{R}$ is absolutely continuous. Since $F$ is a continuous function on a compact set $[a,b]$ it attains its minimum at some point $c\in [a,b]$. Convexity of $F$ implies that $F$ is non-increasing on $[a,c]$ and non-decreasing on $[c,b]$. So it suffices to consider a case when $F$ is monotonic on $[a,b]$.

Let $\varepsilon>0$ be any number. Since the function $F$ is continuous at $a$ and $b$, there exist $0<\delta'<|b-a|$ such that if $x,y\in [a,b]$ and $|x-a|\le\delta’$, $|y-b|\le\delta’$ then $|F(a)-F(x)|\le\varepsilon/3$ and $|F(b)-F(y)|\le\varepsilon/3$. The monotonicity of $F$ implies that for any family $(x_n,y_n)$ of disjoint open intervals contained in $[a,a+\delta’]\cup [b-\delta’,b]$ we have $\sum |F(y_n)-F(x_n)|\le 2\varepsilon/3$.

By Corollary 17, $F$ is absolutely continuous on $(a+\delta’, b-\delta’)$, so there exists a real number $\delta\le \delta’$ such that for any finite family $(x_n,y_n)$ of disjoint open intervals contained in $(a+\delta’, b-\delta’)$ of the total length at most $\delta$ we have $\sum |F(y_n)-F(x_n)|\le \varepsilon/3$.

The above easily implies that any finite family $(x_n,y_n)$ of disjoint open intervals contained in $[a, b]$ of the total length at most $\delta$ we have $\sum |F(y_n)-F(x_n)|\le \varepsilon$.

  • $\begingroup$ Hi. Sorry for the late reply. Why does Corollary 17 imply the absolute continuity in that open interval? The corollary gives us the absolute continuity in the compact sub intervals, right? $\endgroup$ – JacobsonRadical Nov 29 '20 at 17:36
  • $\begingroup$ I think perhaps you mean it is absolutly continuous on $[a+\delta', b-\delta']$? $\endgroup$ – JacobsonRadical Nov 29 '20 at 19:30
  • $\begingroup$ and why you can choose $\delta$ to be such that $\delta\leq \delta'$? $\endgroup$ – JacobsonRadical Nov 29 '20 at 22:23
  • 1
    $\begingroup$ Okay. All set. Brilliant treatment around the boundary points! $\endgroup$ – JacobsonRadical Nov 30 '20 at 5:23
  • 1
    $\begingroup$ Thank you so much! $\endgroup$ – JacobsonRadical Nov 30 '20 at 13:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.