How to apply Hahn-Banach to the convex hull?

Question

I am trying to understand the proof of Lemma 4.1.2 in Michel Talagrand's publication from 1995 on concentration inequalities (see below for the precise question statement):

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A bit of context: Talagrand fixes a point $x\in X$ (he uses the notation $X=\Omega$) and a subset $A\subset X$, where $X=X_1\times X_2\times\dots X_n$ is the product space of arbitrary non-empty sets $X_1,\dots, X_n$. The $\alpha_i$ and $t$ are all supposed to be positive real numbers. He defines $A_t^c$ as follows:

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My question. I understand why (4.1.4) implies (4.1.5). However, Talagrand says that "the converse follows from the Hahn-Banach theorem". How does it follow from the Hahn-Banach theorem?

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Note: The problem can be slightly reformulated by saying that we want to prove that for $t>0$ and all $M\subset\{0,1\}^n$, we have that whenever $$\text{for all }\alpha\in]0,\infty[^n, \text{ there exists a } m\in M \text{ such that } \langle \alpha, m \rangle \le t\lVert \alpha\rVert_2,$$ then

$$\min_{m \in \text{Convex hull of } M} \lVert m \rVert_2 \le t.$$

In fact, if somebody can show this, then I will be able to prove the conjecture formulated by me yesterday.

Answer 1

To solve the problem you mention at the end you can argue in this way: $$ \min_{m \in \mathrm{conv} (M)} \|m\|_{2}=\min_{m \in \mathrm{conv} (M)} \max_{\|\alpha\|_{2}\leq 1} \langle \alpha, m\rangle = \max_{\|\alpha\|_{2}\leq 1} \min_{m \in \mathrm{Conv}(M)}\langle \alpha, m\rangle \leq \max_{\|\alpha\|_{2}\leq 1, \alpha \in [0, \infty)^{n}} \min_{m \in M}\langle \alpha, m\rangle \leq t $$

The only nontrivial observation was used is min-max theorem, which says that if $X, Y$ are convex compact sets, $f(x,y)$ continuous, convex in $x$ and concave in $y$ then $\min_{x \in X} \max_{y \in Y} f(x,y) = \max_{y\in Y} \min_{x\in X} f(x,y)$. Choose $f(x,y)=\langle x, y\rangle$