The properties of convex function on the closed unit interval $[0,1]$.

Question

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

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

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

$F(x)$ 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!

As a counterexample for 1, you can use $F(x)=|x - \frac{1}{2}|$. It is not differentiable at $x=\frac{1}{2}$. — JPMarciano, Nov 26 '20 at 22:56
@JPMarciano oh right. I thought $|x|$, but only god knows why i did not think about $|x-a|$ thank you!! — JacobsonRadical, Nov 26 '20 at 23:01
Surely if a function is absolutely continuous on $(0,1)$ and continuous on $[0, 1]$, it is automatically a.c. on $[0,1]$? — Jack M, Nov 26 '20 at 23:57
@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 $(c,d)$ — JacobsonRadical, Nov 27 '20 at 06:53
@JacobsonRadical Should "and $(a,b)$" read "of $(a,b)$" in your statement of Corollary 17? — Jack M, Nov 27 '20 at 11:22
It seems to me that the key question here is about extensions of convex functions. Can a continuous convex function on a closed interval be extended to a neighborhood of that interval? Can a continuous convex function on an open interval be extended to its closure? — Jack M, Nov 27 '20 at 11:52
@JackM yes. If we can fix this, we can use the theorem and corollary from Royden. But perhaps there is another way around. — JacobsonRadical, Nov 27 '20 at 20:40
@MartinR In fact I did not find a convincing proof from this post and the post it linked to. — JacobsonRadical, Nov 27 '20 at 23:00
@JackM I guess we can construct a continuous convex function $f$ on a closed interval $[0,1]$ which cannot be extended to a continuous convex function on a neighborhood of that interval as follows. Let $a_0=0$ and $(a_n)$ be an increasing sequence of elements of $(0,1)$ tending to $1$. Let $f$ be a continuous function on $[0,1]$ such that $f$ is linear on each segment $[a_n,a_{n+1}]$, but tangents of slopes of the graph of $f$ on these segments increase and tend to infinity. — Alex Ravsky, Nov 29 '20 at 05:39
@JackM Any unbounded (continuous convex) function on an open interval (for instance, $1/x$ on $(0,1)$) cannot be extended to a continuous function on its closure. — Alex Ravsky, Nov 29 '20 at 05:42

score 3 · Accepted Answer · answered Nov 29 '20 at 06:21

3

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 $[a,b]$.

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 $\sum |F(y_n)-F(x_n)|\le \varepsilon$.

answered Nov 29 '20 at 06:21

Alex Ravsky

90,434

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? – JacobsonRadical Nov 29 '20 at 17:36
I think perhaps you mean it is absolutly continuous on $[a+\delta', b-\delta']$? – JacobsonRadical Nov 29 '20 at 19:30
and why you can choose $\delta$ to be such that $\delta\leq \delta'$? – JacobsonRadical Nov 29 '20 at 22:23
@JacobsonRadical 1. $F$ is absolutely continuous on $[a+\delta’, b-\delta’]$ and so on its subset $(a+\delta’, b-\delta’)$. 2. The number $\delta$ can be diminished to be not biger than $\delta’$, keeping the required condition. – Alex Ravsky Nov 30 '20 at 04:54
1

Okay. All set. Brilliant treatment around the boundary points! – JacobsonRadical Nov 30 '20 at 05:23
oh by the way, the choice of $\delta'$. In fact since it is continuous around the boundary points, for $\epsilon>0$, we should have $\delta_{1}$ and $\delta_{2}$ from the definition of continuity, and you take $\delta':=\min{\delta_{1},\delta_{2}}$. Right? just double check – JacobsonRadical Nov 30 '20 at 05:24
@JacobsonRadical Yes. In fact, the function $F$ is a continuous function on a compact set, so it is uniformly continuous, so given $\varepsilon$ we can pick $\delta’>0$ such that if $x,y\in [a,b]$ and $|x-y|\le\delta’$ then $|F(x)-F(y)|\le\varepsilon/3$. – Alex Ravsky Nov 30 '20 at 08:52
1

Thank you so much! – JacobsonRadical Nov 30 '20 at 13:31

The properties of convex function on the closed unit interval $[0,1]$.

1 Answers1