# A separation theorem in locally convex spaces

1. Aug 31, 2014

### DavideGenoa

Dear friends, my book (an Italian language translation of Kolmogorov-Fomin's Элементы теории функций и функционального анализа) proves the following separation theorem: let $A$ and $B$ be convex sets of a normed space and let $A$ have a non-empty algebraic interior $J(A)$ such that $J(A)\cap B =\emptyset$. Then there is a non-null continuous linear functional separating $A$ and $B$.
Then, with some translation errors or misprints, it says that the theorem can be generalized to locally convex topological linear spaces (Kolmogorov and Fomin don't require them to be $T_1$). If $A$ is open I easily see that it applies, but I wonder whether it can be generalized as it is stated and hope I can find a proof of it if it can...
$\infty$ thanks for any help!

2. Aug 31, 2014

### micromass

So you see how it is true with $A$ open? Well, then just apply the theorem to $J(A)$ and $B$ to find a separating functional $f:V\rightarrow \mathbb{R}$ such that $f(a)\leq c\leq f(b)$ for $a\in A$ and $b\in B$.

Then we apply the following lemma:
Let $f:V\rightarrow \mathbb{R}$ be any nontrivial functional, let $A$ be any subset of a locally convex vector space $V$ such that $J(A)\neq \emptyset$. Let $c\in \mathbb{R}$. Then the following are equivalent:

1) $f(a)\leq c$ for each $a\in A$
2) $f(a)\leq c$ for each $a\in J(A)$
3) $f(a)<c$ for each $a\in J(A)$

$(1)\Rightarrow (2)$ is trivial.

$(2)\Rightarrow (3)$ Assume by contradiction that there is an $a\in J(A)$ such that $f(a) = c$. Since $f$ is nontrivial, we can find a $v\in V$ such that $f(v)>0$. Clearly, there exists some $t>0$ such that $a+tv\in J(A)$. But then $f(a+tv)>f(a)=c$, contradicting $(2)$.

$(3)\Rightarrow (1)$ Assume by contradiction that there is some $x\in A$ with $f(x)> c$. Since $J(A)$ is nonempty, we can take some $a\in J(A)$. We can thus find a continuous seminorm $p$ such that the open ball $B_p(a,\varepsilon)$ is contained in $A$. But then we can easily check that $B_p((1-t)a + tx, (1-t)\varepsilon)\subseteq A$ for each $t\in [0,1)$. So we see that the segment $[a,x)$ is a subset of $J(A)$. But this segment must intersect $f^{-1}(c)$. So there must exist some $z\in J(A)$ such that $f(z) = c$. This is a contradiction.

3. Aug 31, 2014

### DavideGenoa

Well, so it works. I think $T_2$ is needed, as you use the seminorm in $(3)\Rightarrow (1)$: correct?
Moreover, I don't know why $J(A)$ is open: could you help me to understand it?
$\aleph_1$ thanks!

4. Aug 31, 2014

### micromass

Oh, sorry. I missed that you were working with the algebraic interior. The above post is for the topological interior (and it doesn't use Hausdorff).

Here are some generalizations of the OP:

Lemma:
let A be any subset of a vector space V such that $J(A)\neq \emptyset$. Furthermore, let $f:V\rightarrow \mathbb{R}$ be a nontrivial functional and let $c\in \mathbb{R}$. Then the following are equivalent:

1) $f(a)\leq c$ for each $a\in A$
2) $f(a)\leq c$ for each $a\in J(A)$
3) $f(a)<c$ for each $a\in J(A)$

$(1)\Rightarrow (2)$ is trivial.

$(2)\Rightarrow (3)$ Assume by contradiction that there is an $a\in J(A)$ such that $f(a)=c$. Since $f$ is nontrivial, we can find a $v\in V$ such that $f(v)>0$. By definition of the algebraic interior, there exists some $t>0$ such that $a+tv\in A$. But then $a + \frac{t}{2}v\in J(A)$. Hence $f(a+ \frac{t}{2}v)>c$, which is a contradiction.

(3)⇒(1) Assume by contradiction that there is some $x\in A$ with $f(x)>c$. Since $J(A)$ is nonempty, we can take some $a\in J(A)$. By convexity of $A$, we see that the segment $[a,x)$ is a subset of $J(A)$. But this segment must intersect $f^{-1}(c)$. So there must exist some $z\in J(A)$ such that $f(z)=c$. This is a contradiction.

Lemma
Let $L$ be a subspace of a vector space $V$ and let $A$ be a convex, absorbing set. Let $f:L\rightarrow \mathbb{R}$ be a linear functional such that there exist an $m\in \mathbb{R}$ such that $f(y)\leq m$ for each $y\in A\cap L$. Then there exists an extension $f:V\rightarrow \mathbb{R}$ such that $f(y)\leq m$ for $y\in A$.
Furthermore, if $V$ is a topological vector space and if $A$ is a neighborhood of $0$, then $f$ is continuous.

Proof:
Let $p_A$ be the Minkowski functional of $A$. Then $p_A$ is sublinear and
$$\{x\in V~\vert~p_A(x)<1\}\subseteq A\subseteq \{x\in V~\vert~p_A(x)\leq 1\}$$
It follows that $l(y)\leq mp_A(y)$ for each $y\in L$. Indeed, if $p_A(y)<\alpha$, then $\frac{1}{\alpha}y\in A$, and thus $f(\frac{1}{\alpha}y)\leq m$. Hence $f(y)\leq m\alpha$.
Now, by the algebraic Hahn-Banach theorem, we can find an extension $f:V\rightarrow \mathbb{R}$ such that $f(y)\leq mp_A(y)$ for each $y\in V$. Thus $f(y)\leq m$ for each $y\in A$.
The continuity statement follows immediately from the fact that a linear functional on a topological vector space is continuous if and only if it is bounded above on a neighborhood of $0$.

Lemma
Let $C$ be a convex set in a vector space $V$. If $J(C)\neq \emptyset$ and $x_0\notin J(C)$, then $\{x_0\}$ and $C$ can be separated by a hyperplane.
If $V$ is a topological vector space and if $C$ has nonempty topological interior, then we can take the hyperplane closed.

Proof:
By translating, we can suppose WLOG that $0\in J(C)$. Then $C$ is absorbing and $x_0\neq 0$. Consider the subspace $L=\mathbb{R}x_0$. The algebraic interior of the interval $L\cap C$ contains $0$ but not $x_0$ (otherwise, $x_0$ would belong to $J(C)$. Thus the linear functional $f:L\rightarrow \mathbb{R}: tx_0\rightarrow t$ satisfies that $f(y)\leq f(x_0)$ whenever $y\in C\cap L$. By the previous lemma, there exist a linear extension $f:V\rightarrow \mathbb{R}$ such that $f(y)\leq f(x_0)$ for each $y\in C$.

Finally, the theorem:
Let $A$ and $B$ be two nonempty convex sets in a vector space $V$. If $J(A)\neq \emptyset$ and $B\cap J(A) = \emptyset$, then $A$ and $B$ can be separated by a hyperplane.
Moreover if $V$ is a topological vector space, if $A$ has nonempty topological interior, then we can take the hyperplane to be closed.

Proof:
The set $J(A)$ is convex, nonempty and disjoint from $B$. The set $C = J(A) - B$ is then convex, has nonempty algebraic interior and does not contain $0$. By previous lemma, there exists a nontrivial linear functional $f$ on $V$ such that $f(c)\leq 0$ for each $c\in C$. It follows that $f(a)\leq f(b)$ for each $a\in J(A)$ and $b\in B$. The theorem now follows by application of the first lemma in this post.

5. Sep 1, 2014

### DavideGenoa

Wow, what a detailed answer! Thank you so much!!! As to the lemma Let $L$ be a subspace... I understand that $\alpha>p_A(y)\Rightarrow\alpha^{-1}y\in A$, but it isn't clear to me why $f(p_A(y)^{-1} y)\leq m$. I suppose that $p_A(y)^{-1}y\in A$ but I don't understand why, since $p_A$ represents an infimum... $\aleph_1$ thanks!

6. Sep 1, 2014

### micromass

Is your problem that you don't how from $p_A(y)<\alpha~\Rightarrow~f(y)<m\alpha$ follows that $f(y)\leq mp_A(y)$?

To prove it, assume that $p_A(y) < \frac{1}{m} f(y)$. Choose a number $\alpha>0$ such that $p_A(y)<\alpha<\frac{1}{m}f(y)$. Then since $p_A(y)<\alpha$, it follows immediately from the hypothesis that $f(y)<m\alpha$. And thus $\alpha<\frac{1}{m}f(y)<\alpha$. This is a contraduction.

7. Sep 1, 2014

### DavideGenoa

$\aleph_2$ thanks! I had to increase cardinality
So, I see that the theorem my book stated holds for any vector space. Since I'm interested in understanding your first proof that in a locally convex space $\forall a\in A\quad f(a)\leq c\iff\forall a\in \text{Int}(A)\quad f(a)\leq c\iff\forall a\in \text{Int}(A)\quad f(a)< c$, while I supposed you used Kadison-Ringrose's theorem 1.2.6, I now think the seminorm you talked about is the Minkowski functional $P_{A-a}$ for $A-a$: have I understood?

8. Sep 1, 2014

### micromass

A locally convex space is a vector space equipped with a collection of seminorms $\mathcal{P}$ which generate the topology. This means that the collection of balls

$$B_p(a,\varepsilon),~p\in \mathcal{P},~a\in V,~\varepsilon>0$$

forms a subbasis for the topology. Or equivalently, the sets

$$V(a:p_1,...,p_n:\varepsilon) = \{x\in V~\vert~p_i(a-x)<\varepsilon, i\in \{1,...,n\}\}$$

form a basis for each $a\in V$, $p_1,...,p_n\in \mathcal{P}$ and $\varepsilon>0$. A basis means that for each open set $G$ and $a\in G$, there exists some basis open set $V(a:p_1,...,p_n:\varepsilon)\subseteq G$. If I then define $p=\max\{p_1,...,p_n\}$ then this is also a continuous seminorm and we have that $B_p(a,\varepsilon)\subseteq G$. This is exactly what I used in my proof.

9. Sep 1, 2014

### DavideGenoa

I didn't know that: I only knew theorem 1.2.6 of Kadison-Ringrose's Fundamentals of the Theory of Operator Algebras, which is valid for Hausdorff spaces. Thank you so much again!

10. Sep 1, 2014

### micromass

Theorem 1.2.6 remains valid for non-Hausdorff spaces if remove the separation condition (which is actually equivalent with Hausdorff). So locally convex spaces can (even in the non-Hausdorff case) be described in two equivalent ways: by the existence of a base of convex set or equivalently, by the existence of a collection of seminorms generating the topology. The proof of this assertion should be exactly the same as the proof of 1.2.6.

11. Sep 2, 2014

### DavideGenoa

Wonderfully interesting, I heartily thank you!!! Written a pencil note in the book.

12. Sep 2, 2014

### micromass

I wonder when I'll get $c$ amount of thanks :tongue: