Geometry Hahn Banach theorem for weak* topology

We have geometric Hahn Banach theorem in standard functional analysis course, saying that there exists a separating hyperplane separates two special sets in usual topology. However this theorem can be used for some special sets in weak* topology. This is the problem 9 in the functional analysis book by Prof. Brezis.

Theorem 1 Let ${E}$ be a Banach space, ${A,B \subset E^*}$ be two nonempty disjoint convex subsets. Assume ${A}$ is open in weak* topology. Then there exist some ${x \in E, x \neq 0}$ and a constant ${\alpha}$ such that the hyperplane ${\{ f \in E^* : \langle f, x \rangle_{E^* \times E} = \alpha \}}$ separates ${A}$ and ${B}$.

Proof: Since ${A}$ is open in weak* topology, which is coarser than the norm topology for ${E^*}$, so ${A}$ is open in norm topology. By geometric Hahn-Banach theorem, there exist ${\phi \in E^{**}}$ and a constant ${\alpha}$ such that

$\displaystyle \langle \phi, f \rangle_{E^{**} \times E^*} \le \alpha \le \langle \phi, g \rangle_{E^{**} \times E^*} \quad \forall f \in A, \forall g \in B.$

Fix an element ${h \in A}$, since ${A}$ is open in weak* topology, we can find a weak* open symmetric neighborhood around 0, say ${V}$. Then we have

$\displaystyle \langle \phi, h + v \rangle_{E^{**} \times E^*} \le \alpha \quad \forall v \in V.$

Since ${C:= \alpha - \langle \phi, h \rangle_{E^{**} \times E^*} \ge 0}$ and ${V}$ is symmetric, we have

$\displaystyle \left| \langle \phi, v \rangle_{E^{**} \times E^*}\right| \le C \quad \forall v \in V.$

Therefore ${\phi}$ defines a bounded linear functional on ${E^*}$ and is continuous in weak* topology. Hence there exists an ${x \in E}$ such that ${\langle \phi, f \rangle_{E^{**} \times E^*} = \langle f, x \rangle_{E^* \times E}, \forall f \in E^*}$. $\Box$

There is a more applicable version of this theorem, in which case ${A}$ is closed and ${B}$ is compact in weak* topology. Before proving it, we need the following lemma:

Lemma 2 Assume that ${A,B \subset E^*}$ be two nonempty convex subsets, ${A}$ is closed and ${B}$ is compact in weak* topology. Then ${A-B := \{a-b : a\in A, b \in B \}}$ is convex and closed in weak* topology.

Proof: To see ${A-B}$ is convex. Since ${\theta a_1 + (1-\theta) a_2 \in A, \forall a_1,a_2 \in A, \forall \theta \in [0,1]}$; ${\theta b_1 + (1-\theta) b_2 \in B, \forall b_1,b_2 \in B, \forall \theta \in [0,1]}$. Then ${\theta (a_1-b_1) + (1-\theta) (a_2-b_2) = (\theta a_1 + (1-\theta) a_2) - (\theta b_1 + (1-\theta) b_2) \in A-B}$.

To see ${A-B}$ is weak* closed, we only need to show ${(A-B)^{\mathsf{c}}}$ is weak* open. Fixed a point ${x \in (A-B)^{\mathsf{c}}}$, since ${A}$ is weak* closed, for any ${y \in B}$, there exists a weak* open, convex neighborhood around 0, say ${V_y}$, such that

$\displaystyle (A-y) \cap (x + V_y) = \emptyset.$

Clearly we have ${B \subset \cup_{y \in B} (y + \frac{V_y}{2})}$, and since ${B}$ is weak* compact, we can find a finite coverings that cover ${B}$, i.e. ${B \subset \cup_{i \in I} (y_i + \frac{V_{y_i}}{2})}$. Define ${V:= \cap_{i \in I} \frac{V_{y_i}}{2}}$, I want to show

$\displaystyle (x+V) \cap (A-B) = \emptyset.$

Assume ${v \in V}$ such that ${(x+v) \cap (A-B) \neq \emptyset}$. Then ${(x+v) \in A-y_i - \frac{V_{y_i}}{2}}$ for some ${i}$. Since ${v \in \frac{V_{y_i}}{2}}$ and ${V}$ is convex (intersection of convex sets is convex), ${v + \frac{V_{y_i}}{2} \in V}$. This implies there exists a ${v' \in V}$ such that ${x + v' \in A - y_i}$, which leads to a contradiction. $\Box$

Theorem 3 Let ${E}$ be a Banach space, ${A,B \subset E^*}$ be two nonempty disjoint convex subsets. Assume ${A}$ is closed in weak* topology and ${B}$ is compact in weak* topology. Then there exist some ${x \in E, x \neq 0}$ and a constant ${\alpha}$ such that the hyperplane ${\{ f \in E^* : \langle f, x \rangle_{E^* \times E} = \alpha \}}$ strictly separates ${A}$ and ${B}$.i.e. there exists an ${\varepsilon>0}$ such that

$\displaystyle \langle f, x \rangle_{E^* \times E} < \alpha- \varepsilon \quad \forall f \in A$

and

$\displaystyle \langle f, x \rangle_{E^* \times E} > \alpha+ \varepsilon \quad \forall f \in B$

Proof: By the lemma we know ${A-B}$ is convex and closed in weak* topology. Since ${A}$ and ${B}$ are disjoint, ${A-B}$ does not contain 0. We can find a weak* open, convex neighborhood around 0, say ${V}$, that does not intersect with ${A-B}$. Then by Theorem 1, there exists a hyperplane ${\{ f \in E^* : \langle f, x \rangle_{E^* \times E} = \beta \}}$ that separates ${V}$ and ${A-B}$.

Since ${V}$ is convex, we can take ${v, -v \in V}$, which implies ${\beta >0}$. Therefore we have

$\displaystyle \langle f, x \rangle_{E^* \times E} \ge \langle g, x \rangle_{E^* \times E} + \beta \quad \forall f \in A, g \in B$

Define ${\alpha:= \sup_{g \in B} \langle g, x \rangle_{E^* \times E} + \frac{\beta}{2}}$, ${\varepsilon:= \frac{\beta}{4}}$. Then we have the desired separating property. $\Box$