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Let $X$ be a normed space over $\mathbb{R}$ and $K\subset X$ a nonempty, closed and convex subset. Show that $$K=\{x\in X:\phi(x)\leq \sup\limits_{y\in K}\phi(y)\quad\forall\phi\in X'\}$$ Here $X'$ is the dual space to $X$.

My idea so far, was to use the separation theorem for convex sets: https://en.wikipedia.org/wiki/Hyperplane_separation_theorem

Suppose

  • $\exists x_0\in X\backslash K$ where $X\backslash K$ is open so there $\exists\epsilon>0$ such that $B_\epsilon(x_0)\subset X\backslash K$ and $B_\epsilon(x_0)\cap K=\emptyset$

  • $\exists\phi\in X',\alpha\in \mathbb{R}$ such that $\phi(x)\leq\alpha\leq\phi(y)\quad \forall x\in K,y\in B_\epsilon(x_0)$

Now I'm lost ...

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You're on the right track, but I would recommend using the other version of the hyperplane separation theorem.

Suppose $x_0\notin K$. Then the set $\{x_0\}$ is compact, convex, and disjoint from $K$. Thus, the Hahn-Banach separation theorem (the second bullet point) furnishes some $\phi\in X'$, $s,t\in\mathbb R$ such that $$ \phi(y)

I'm sure you can (or already have) handled the other inclusion.

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    very comprehensive answer, thank you. For $\supset K$: K is closed and convex and therefore always a true subset of $K=\{x\in X:\phi(x)\leq \sup\limits_{y\in K}\phi(y)\quad\forall\phi\in X'\}$, but there is probably a cleaner way to express this? – 2017-01-09
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    @JonathanKrill That's basically what I had in mind. Closedness and convexity aren't required though, given any subset $C\subset X$, any $x\in C$, and any $\phi\in X'$ we have $\phi(x)\leq\sup\limits_{y\in C}\phi(y)$. – 2017-01-09