infimum for convex hulls and closures

Question

For $\alpha\in\mathbb{R}^{n}$ and a set $A\in\mathbb{R}^{n}$ hold: $$\inf\{\langle\alpha,a\rangle: a\in A\}=\inf\{\langle\alpha,a\rangle: a\in conv(A)\}=\inf\{\langle\alpha,a\rangle: a\in clo(conv(A))\}$$. By conv we mean convex hull and by clo closure. Why is this true and the infimums are equal?

Answer 1

Let the three sets be $S_1, S_2, S_3$ respectively. We immediately have $\inf S_1 \ge \inf S_2 \ge \inf S_3$ so it remains to check the two reverse inequalities.

Hints:

• For the first inequality, it suffices to show that for any $a' \in \text{conv}(A)$, there exists $a \in A$ such that $\langle \alpha, a \rangle \le \langle \alpha, a'\rangle$. (In plain words, going from $A$ to $\text{conv}(A)$ doesn't help make the inner product smaller.)
• For the second inequality, the idea is that any element of $\text{clo}(\text{conv}(A))$ can be approximated by an element of $\text{conv}(A)$. Since the function $a \mapsto \langle \alpha, a \rangle$ is continuous and linear, you can quantify how much this approximation affects the value of the inner product, and show that in the end it is negligible.

Solution for the second inequality

Let $B = \text{conv}(A)$. We already know $(\inf S_3)$ is a lower bound for $S_2$ so it suffices to show it is the greatest lower bound. To do this, it suffices to show that for any fixed $\epsilon > 0$ we are able to find $a \in B$ such that $\langle \alpha, a \rangle \le (\inf S_3) + \epsilon$. To find such an $a$, consider $a' \in \text{clo}(B)$ such that $\langle \alpha, a' \rangle \le (\inf S_3) + \epsilon / 2$. Since $a'$ is in the closure of $B$, there exists $a \in B$ such that $\|a-a'\| < \frac{\epsilon}{2\|\alpha\|}$. Then, using the Cauchy-Schwarz inequality, $$\langle \alpha, a\rangle = \langle \alpha, a' \rangle + \langle \alpha, a - 'a \rangle \le (\inf S_3) + \frac{\epsilon}{2} + \|\alpha\| \frac{\epsilon}{2\|\alpha\|} = (\inf S_3) + \epsilon.$$

Solution for the first inequality

Any $a' \in \text{conv}(A)$ can be written as $c_1 a_1 + \cdots + c_n a_n$ for $a_1, \ldots, a_n \in A$ and $c_1, \ldots, c_n \ge 0$ summing to $1$. WLOG let $\langle \alpha, a_1 \rangle \ge \cdots \ge \langle \alpha, a_n \rangle$. Then $$\langle \alpha, a' \rangle = c_1 \langle \alpha, a_1 \rangle + \cdots + c_n \langle \alpha, a_n \rangle \ge \langle \alpha, a_n \rangle.$$