Optimality condition of a convex function that is a.e. differentiable

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be convex, and let $A \subseteq \mathbb{R}$ be a Lebesgue-measurable set such that $\lambda(\mathbb{R}\setminus A)=0$, and such that for every $x \in A$, $f$ is differentiable at $x$ and $f'(x)$ is finite.

Suppose there is a partition of $\mathbb{R}$ into three non-empty intervals, $\mathbb{R} = I\uplus J\uplus K$, such that for every $(x,y,z) \in I\times J\times K$, $x<y<z$, and such that

for every $x \in I\cap A$, $f'(x) < 0$,
for every $y \in J\cap A$, $f'(y) = 0$,
for every $z \in K\cap A$, $f'(z) > 0$.

a. Can it be inferred that, for every triple $(x,y,z)\in I\times J\times K$, $f(x) \geq f(y) \leq f(z)$?

b. Can it be inferred that, for every triple $(x,y,z)\in I\times J\times K$, $f(x) > f(y) < f(z)$?

Take $f(x) = \operatorname{dist}(x,[0,1])$ and $I = (-\infty,0],, J = (0,1),, K = [1,+\infty)$. If $I$ has a maximal element, or $K$ has a minimal element, you don't have the strict inequality there. — Daniel Fischer, Oct 18 '16 at 17:53
@DanielFischer: Thanks. So this counter-example resolves question (b). Do you have an idea concerning question (a)? — Evan Aad, Oct 18 '16 at 17:59
We have the strict inequalities for $(x,y,z) \in \overset{\Large\circ}{I} \times J \times \overset{\Large\circ}{K}$ (as shown in copper.hat's answer), hence a follows. — Daniel Fischer, Oct 18 '16 at 18:05
Note that $f$ is 1. strictly decreasing on $I$, 2. constant on $J$, 3. strictly increasing on $K$. — Daniel Fischer, Oct 18 '16 at 18:21

copper.hat · Accepted Answer · 2021-12-24T22:12:59.223

3

If $f$ is convex and $f'(x) = 0$, then $x$ is a global minimiser of $f$, hence a. is true.

In the following, let $x$ be any global minimiser.

For b., if $f'(a)>0$, for example, then there is some nearby $a'$ such that $f(a')< f(a)$, and since $f(a') \ge f(x)$ we see that $f(a) > f(x)$. Similarly if $f'(a) <0$.

However, Daniels example above shows that there may be (at most) one point $a \in I$ at which $f(a) = f(x)$ and similarly for $J$.

Note that if $f(x') = f(x)$ then $f(y) = f(x) $ for all $y \in [x',x]$, hence there can be at most one point $a \in I$ for which $f(a) = f(x)$.

As an aside, note that $f'$ is non decreasing, whenever it exists.

Also, if $f$ is convex and defined on an open set, it is locally Lipschitz hence ae. differentiable.

Addendum: The above reasoning doesn't cover the case where $J$ is a singleton and the singleton is not contained in $A$. Let $J = \{x\}$.

Since $f$ is locally Lipschitz, it is absolutely continuous on any bounded interval, hence we have $f(a) = \int_x^a f'(t) dt$. If $ a < x$, then $f'(t) <0$ for ae. $t \in [a,x]$ and hence $f(a) > f(x)$. Similarly for $a>x$. Hence $x$ is a strict minimum.

edited Dec 24 '21 at 22:12

answered Oct 18 '16 at 17:50

copper.hat

172,524

Thanks. Why is $x$ a global minimizer of $f$ if $f'(x) = 0$? – Evan Aad Oct 18 '16 at 18:04
If $f$ is convex and differentiable at $x$ then $f(y)-f(x) \ge f'(x)(y-x)$. This is one reason convexity is so useful. – copper.hat Oct 18 '16 at 18:05
Thanks. What if $J$ consists of a single point, $J = {y^}$, and $f$ is not differentiable at $y^$? Does the conclusion still hold? – Evan Aad Oct 18 '16 at 18:12
Yes, but the reasoning needs to be refined slightly. – copper.hat Oct 18 '16 at 18:13
Would you mind refining the reasoning accordingly, please, since this case is built into the question. – Evan Aad Oct 18 '16 at 18:15
Sure; I mistakenly presumed the intervals were non trivial. – copper.hat Oct 18 '16 at 18:15

Optimality condition of a convex function that is a.e. differentiable

1 Answers1

Linked