condition for an element to be limit of finite convex combination

Question

Let $B$ be a Banach space and let $S\subset B$ and $b\in B$.

Prove that $b$ is the limit of finite convex combination of elements of $S$ iff $\forall f\in B': f(b)\le \sup\{f(x), x\in S\}$.

I don't know where to start.

Thank you for your help.

Answer 1

I'm assuming $B$ is a real Banach space, otherwise the inequalities don't make sense. If $f\in B'$, take a look at $-f$ to get:

$$-f(b)≤ \sup \{ -f(x) \mid x\in S\} \implies f(b)≥\inf\{ f(x)\mid x\in S\}.$$

In other words you have $$\inf\{ f(x)\mid x\in S\} ≤ f(b) ≤ \sup\{f(x)\mid x\in S\}$$ for all functionals $f$. These inequalities clearly still hold if we replace $S$ by the closure of its convex hull.

Now if $b$ is not in this closed convex hull, you have two disjoint sets $A=\{b\}$ and $B$ the closed convex hull of $S$. Note that $A$ is compact and convex and apply the separation form of the Hahn Banach theorem to get a function $\lambda: B\to\Bbb R$ with $\lambda(b)<t <s< \lambda(x)$ for some $s,t\in\Bbb R$ and all $x\in B$. This contradicts the inequalities we have above, so $A$ and $B$ cannot be disjoint.