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

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