Existence of strictly convex functions with same ordering as convex one

[Ed Seneca]

Existence of strictly convex functions with same "ordering" as convex one

Consider any real-valued convex function $c : R^n \rightarrow R$. I am interested in whether there exists some strictly convex function d, that satisfies d(x) > d(y) if c(x) > c(y).

That is, given a convex function, can we always find a strictly convex function that preserves strict inequalities between points?

Originally I thought this would be straightforward, but now I am not so sure.

If we restrict ourselves to considering a function $c : R \rightarrow R$ then one way we can think of attempting a solution might be to take the minima, or set of minima, and choose one arbitrarily, then divide the the function with a vertical plane and add some increasing with a strictly increasing derivative function to either side of the division. This would create a strictly convex function but not preserve the ordering. The ordering is preserved to either side of the division, but the relative ordering of both sides is now unclear.

Of course, the question does not call for the construction of such a strictly convex function, but I can see no immediate steps to a nonconstructive proof, either.

 

[fresh_42]

$d:=c+1$

 

[Infrared]

Careful @fresh_42 $d$ is required to be strictly convex, but $c$ is not.

 

[fresh_42]

I know, but the question only asked for the strict inequality. All other cases have to be properly defined first.

 

[Infrared]

I think the question requires $d$ to be strictly convex, but $d=c+1$ is not if $c$ is not.

 

[fresh_42]

I first thought about something like $d=c+\|x\|$ or $d=1+c^2$ but this will soon result in a lot of cases which needed clarification first. I would have asked the OP, but ...

 

[Infrared]

If $c$ is the zero function, neither of those candidates for $d$ are strictly convex.

If you restrict the domain to a compact set (instead of being all of $\mathbb{R}^n$) then $d=c+\varepsilon f$ would work where $f$ is any strictly convex function, if $\varepsilon>0$ is sufficiently small. I don't see how to extend this, seems like it could be tricky.

 

[fresh_42]

Yes, that's what I encountered: we likely need more conditions. E.g. how do the equilibra $d(x)=d(y)$ look like? In the end we want to transform a geometric property into an analytical one, which requires a better picture of the geometry first. I don't think this is especially interesting as specific examples usually provide more information, and convexity alone is probably not strong enough.