Hyperplanes and convex sets in Hilbert space

Question

The simplest Hahn-Banach extension theorem in Hilbert space $X$ avoids the use of the axiom of choice by virtue of the Riesz representation theorem. But what about the version of the theorem where the sought-for linear functional is required to remain below a given sub-linear or convex function? Also, to which extent can we separate 2 disjoint convex sets by a hyperplane without Zorn? Can we assert that any hyperplane $H\subset X$ has a translate that is tangent to a given bounded closed convex subset $C\subset X$?

Answer 1

Let $C$ be convex, closed, non-empty. Then $x_0\not\in C$ can be separated by $C$ by a closed hyperplane, i.e., there is $f\in X^*$ such that $$ f(u ) < f(x_0) \quad \forall u\in C.$$

Proof: uses that $\inf_{u\in C} \|x_0 - u\|$ has a unique solution $u^*$. Then take hyperplance perpendicular to $x_0-u^*$.

Let $C_1,C_2$ be disjoint sets that are convex, closed, non-empty, $C_1$ compact. Then $C_1,C_2$ can be separated by by a closed hyperplane, i.e., there is $f\in X^*$ such that $$ \sup_{x\in C_1}f(x) < \inf_{y\in C_2}f(y).$$

Proof: Define $C:=C_1-C_2$, which is closed due to compactness of $C_1$. Take $\tilde x\in C$. Then $-\tilde x\not\in C$ as $0\not\in C$. Now $C$ and $\tilde x$ can be separated, which yields a separation of $C_1$ and $C_2$.

In this thread https://mathoverflow.net/questions/37551/a-counter-example-to-hahn-banach-separation-theorem-of-convex-sets/37564 is a counterexample showing non-separation of two convex sets in $L^2(\mu)$ (both sets are neither compact nor open).

I am not aware of a proof of extension of linear functionals below sublinear functionals. This would also imply the non-trivial separability of an open convex set $C$ from a point $x_0\not\in C$.

Answer 2

About the question concerning the construction of a bounded linear sub-differential below a convex (lower-semicontinuous) function $f$: let us give a construction of a hyperplane tangent to a closed convex set $C \subsetneq X$ at some arbitrary point $x \in \partial C$ by using the sequential Banach-Alaoglu theorem (which does not require the axiom of choice for its proof). As a corollary, letting $C$ be the epigraph of the function $f$, we deduce a proof for the original statement.

Let $x_n \to x$ be a convergent sequence in $X\setminus C$. Because $X$ is a Hilbert-space, there is a unique $y_n\in C$ that is "closest" to $x_n$ in the sense that $dist(x_n,C)=dist(x_n,y_n)=\|x_n-y_n\|$. Now, denote $\varphi_n := (y_n-x_n)/\|y_n-x_n\|$. Since the sequence $(\varphi_n)_n$ is contained within the closed unit ball of the separable Hilbert space $\overline{span_{n\in \mathbb{N}}(\varphi_n)}\subseteq X$ (which is its own dual as stipulated by Riesz' theorem), the sequential Banach-Alaoglu theorem concludes that $(\varphi_n)_n$ must have a subsequence that weakly converges to some $\varphi \in X$. This means however that $\forall z \in C$, (in the following we replace the sequence $(\varphi_n)_n$ by the earlier-mentioned convergent subsequence) $$\langle \varphi,z-x\rangle = \lim_{n \to \infty}\langle \varphi_n,z-x\rangle = \lim_{n \to \infty}\langle \varphi_n,z-x_n\rangle$$ Now, minimality of the distance between $y_n$ and $x_n$ and the convexity of $C$ implies that $\langle y_n-x_n,z-x_n\rangle \geq 0$ so that $\langle \varphi_n,z-x_n\rangle \geq 0$ and therefore $\langle \varphi,z-x\rangle\geq 0$. So the hyperplane determined by the kernel of the bounded functional $z \mapsto \langle \varphi,z-x\rangle$ is tangent to $C$ at $x$.

Let me now shift to the question of finding a translate of a given hyperplane $H \subseteq X$ so that it is tangent to some bounded convex set $C \subseteq X$. If $\varphi \in X$ represents the normal of the plane $H$, then $H$ is the kernel of the bounded functional $z \mapsto \langle \varphi,z\rangle$ and the translate $z_0+H$ is the kernel of the functional $z \mapsto \langle \varphi,z-z_0\rangle$. Now consider the set $$I=\{t \in \mathbb{R}:\,(t\varphi + H )\cap C\neq \emptyset\}$$ Then convexity of $C$ implies that $I$ is also convex and therefore an interval. Let $t_n \overset{>}{\to} \inf I$ and let $(x_n)_n$ be a sequence such that $x_n \in (t_n\varphi + H)\cap C$.(*) That sequence is bounded and contained within the (self-dual) separable Hilbert-space $\overline{span_{n \in \mathbb{N}}(x_n)}$. Therefore $(x_n)_n$ has a subsequence that converges weakly to some $x\in X$. In fact $x \in C$ because, as already discussed in DAW's answer, for every point $x'\in X\setminus C$ there is a bounded functional that vanishes at $x'$ while being uniformly bounded below by some $\varepsilon>0$ on $C$ (hence $x_n$ cannot possibly converge weakly to $x'$). Then the plane $x+H=(\inf I) \varphi+H$ intersects with $C$ in the point $x$ while for every $t<\inf I$, $(t\varphi+H)\cap C=\emptyset$: this one way of saying that the plane $x+H$ is tangent to $C$ at $x$.

(*) A priori one needs the axiom of countable choice to select the $x_n\in t_n \varphi + H$. That can be circumvented by e.g. noting that $t_n \varphi + H$ is closed and convex so that we can fix $\xi \in X\setminus C$ and agree to define, for all $n \in \mathbb{N}$, $x_n$ as the unique element in $t_n \varphi + H$ that is closest to $\xi$.