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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)?

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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.

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