The following theorem is taken from Behrend's M-Structure and Banach-Stone Theorem, page $3.$

Theorem $0.1$: Let $X$ be Banach space and $K_1,K_2$ be nonempty convex subsets of $X^*\times \mathbb{R}$ ($X^*$ is provided with weak$^*$-topology).

Suppose that $K_1\cap K_2=\emptyset$ and that $K_1$ is closed and $K_2$ is compact.

Then there exists $x\in X, a,r\in\mathbb{R}$ such that $$p_1(x)+aa_1<r<p_2(x)+aa_2$$ for all $(p_1,a_1)\in K_1$ and $(p_2,a_2)\in K_2.$

Note that $r>0$ if $(0,0)\in K_1$ and that $a>0$ if there exist $p\in X^*$ and $a_1,a_2\in\mathbb{R}$ such that $a_1<a_2, (p,a_1)\in K_1$ and $(p,a_2)\in K_2.$

Behrend stated that the theorem above follows from the following two facts:

$(1)$ (Hahn-Banach separation theorem) Every nonempty compact convex set in a locally convex Hausdorff space can be strictly separated from every disjoint nonempty closed convex set.

$(2)$ Continuous linear functionals on $(X^*,\text{weak}^*\text{-topology})$ are just the evaluation at the points of $X.$


The following is my attempt:

Clearly $X^*\times \mathbb{R}$ is locally convex. Since $K_1$ is a closed convex set while $K_2$ is a compact convex set, by Hahn-Banach separation theorem, there exist scalars $r,s$ and a continuous linear functional $x^*$ on $X^*\times\mathbb{R}$ such that $$x^*(p_1,a_1)<r<s<x^*(p_2,a_2)$$ for all $(p_1,a_1)\in K_1$ and $(p_2,a_2)\in K_2.$ However, I do not know how to proceed from here.

Any hint would be appreciated.


From here it is important to know that the dual of $X^*\times\mathbb R$ (with $X^*$ given the weak$^*$-topology) is $X\times\mathbb R$, with the pairing $$(x,\alpha)(p,\beta)=p(x)+\alpha\beta.$$ Indeed, for each $x\in X$ and $\alpha\in\mathbb R$, the map $$(p,\beta)\mapsto p(x)+\alpha\beta$$ from $X^*\times\mathbb R$ to $\mathbb R$ is continuous and linear, so $X\times\mathbb R\subset(X^*\times\mathbb R)^*$.
If $f\in(X^*\times\mathbb R)^*$, then the map $p\mapsto f(p,0)$ is weak$^*$-continuous, hence there is some $x\in X$ such that $f(p,0)=p(x)$ for all $p\in X^*$. Since $\beta\mapsto f(0,\beta)$ is linear, there is some $\alpha\in\mathbb R$ such that $f(0,\beta)=\alpha\beta$ for all $\beta\in\mathbb R$. By linearity, we have $f(p,\beta)=p(x)+\alpha\beta$ for all $(p,\beta)\in X^*\times\mathbb R$. Thus $(X^*\times\mathbb R)^*=X\times\mathbb R$.

Thus, the linear functional you obtain from the Hahn-Banach theorem is of the form $(p,\beta)\mapsto p(x)+\alpha\beta$ for some $\alpha\in\mathbb R$, $x\in X$, and you obtain $$p_1(x)+\alpha\beta_1<r<s<p_2(x)+\alpha\beta_2$$ for all $(p_1,\beta_1)\in K_1$ and $(p_2,\beta_2)\in K_2$. From here, it should be just be routine work to check the statements in the last paragraph. If you have any trouble with that, let me know and I will edit/expand.

  • $\begingroup$ Well, last part is just simple verification using the inequality. Thanks for proving that he inequality. I understand better now. $\endgroup$ – Idonknow May 11 '18 at 16:51
  • $\begingroup$ You're welcome. Glad I could help! $\endgroup$ – Aweygan May 11 '18 at 17:20
  • $\begingroup$ By $(X^*\times\mathbb R)^*=X\times\mathbb R$, do you mean that there exists a bijection between the two sets without any structure preserving property (like homeomorphism or isometry)? Furthermore, how do you show that the map $(p,\beta)\mapsto p(x)+\alpha\beta$ is continuous? $\endgroup$ – Idonknow May 12 '18 at 10:49
  • 1
    $\begingroup$ Sorry, I got pinged then ignored it. When I say that $(X^*\times\mathbb R)^*=X\times\mathbb R$ I mean the map taking $(x,\alpha)\in X\times\mathbb R$ to the map $(p,\beta)\mapsto p(x)+\alpha\beta$ in $(X^*\times\mathbb R)^*$ is a linear homeomorphism (so it preserves topological and linear structure). $\endgroup$ – Aweygan May 12 '18 at 22:18
  • 1
    $\begingroup$ And for your second question, note that each map $p\mapsto p(x)$ and $\beta\mapsto\alpha\beta$ is continuous (in the respective topological spaces), hence the map $(p,\beta)\mapsto(p(x),\alpha\beta)$ is continuous from $X^*\times\mathbb R$ to $\mathbb R^2$, and the map $(x,y)\mapsto x+y$ is continuous on $\mathbb R^2$. Compose the last two maps. $\endgroup$ – Aweygan May 12 '18 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.