$\inf\{c^\intercal x \mid x \in M\}$ achieved $\Rightarrow$ $\inf\{c^\intercal x \mid x \in X\}$ achieved for closed, convex $X \subseteq M$?

Question

Let $c \in \mathbb{R}^n$, $M \subseteq \mathbb{R}^n$ such that there exists $x' \in M$ with $f(x') = \inf_{x\in M} \{c^\intercal x \}$. Let $X \subseteq M$ be a closed and convex subset.

Does it follow that $\inf_{x \in X} \{ c^\intercal x\}$ is attained by some $x^* \in X$?

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Thoughts: I know that this doesn't hold if we only assume that $X$ is closed. Counterexample: $\inf\{x_3 \mid \underbrace{x \in \mathbb{R}^3, x_3 \geq (x_1 - 1)^2 e^{-x_2^2}}_{M} \}$ and $X = \{x \in \mathbb{R}^3 \mid x_3 \geq (x_1 - 1)^2 e^{-x_2^2}, x_1 = 0 \}$ (see: here). But I'm hoping that this holds for a convex $X$.

Answer 1

Counterexample in $\mathbb R^2$. Let $M$ be the closed right half-plane and let $c^\intercal x$ be the first component $x_1$ (i.e. $c^\intercal=(1,0)$). This attains a minimum of $0$ on $M$. Now let $X=\{x\in M:x_1x_2\geq1\}$, i.e., the part of $M$ on or above the hyperbola $x_1x_2=1$. The infimum of $c^\intercal x$ on $X$ is $0$, but it is not attained, even though $X$ is closed and convex.