Is minimizer containment preserved after adding the same function?

Question

Suppose I have three convex real-valued functions $f$, $g$, and $h$, which satisfy $0\leq h\leq g$ and

$$\textrm{argmin}\, g\subset \textrm{argmin} \, h.$$

Is it true that $\textrm{argmin} (f+g)\subset\textrm{argmin} (f+h)$? We can assume all argminima are nonempty.

I have tried to prove it by combining various inequalities based on being a minimizer and the inequality between $h$ and $g$ -- we easily get cancellation of the function values of $f$. However, I'm not currently seeing a path to proving the final claim. I'm beginning to wonder if it is false!

Answer 1

This is not true. Here are some hints which should enable you to find a counterexample:

• The minimizers of $f + g$ and $f + h$ are characterized by the derivatives (or, in the non-differentiable case, by subgradients).
• Your assumptions give you (almost) no relation between the derivatives of $g$ and $h$.