Suppose we have a function $f:\mathbb{R}^d \rightarrow \mathbb{R}$ that has a infimum and attains it, i.e., there exists (not necessarily unique) $x \in \mathbb{R}^d$ such that $f(x) \leq f(y)$ for all $y \in \mathbb{R}^d$.

Let $A$ be a non-empty closed convex subset of $\mathbb{R}^d$ (not necessarily bounded - in fact in my case $A$ is not bounded).

Does $f$ restricted to the domain $A$ also attain its infimum, i.e., does there exist $a \in A$ such that $f(a) \leq f(a^{\prime})$ for all $a^{\prime} \in A$?

This is not true, for we can take $f(x) = \frac{1}{|x|}$ for $x \neq 0$ and $f(0) = 0$. Its infimum is $0$ and attains it at $x=0$. But let $A=[1,\infty)$, its infimum is $0$ but nowhere does it attain it.

What if $f$ is positive homogeneous or convex (or both)?


This is not true for convex $f$: Take $$ f(x,y)= \max( e^x + y,0), $$ which is a maximum of convex functions, hence convex. The minimum $0$ is attained at, e.g., $(x,y)=(0,-1)$.

Define $A=\{(x,0): \ x\in \mathbb R\}$. Minimizing $f$ on $A$ is equivalent to minimizing $e^x$ for $x\in \mathbb R$, where the infimum is not attained.

| cite | improve this answer | |

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.