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A separation theorem in locally convex spaces

  1. Aug 31, 2014 #1
    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. jcsd
  3. Aug 31, 2014 #2

    micromass

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    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.
     
  4. Aug 31, 2014 #3
    Well, so it works. I think [itex]T_2[/itex] is needed, as you use the seminorm in [itex](3)\Rightarrow (1)[/itex]: correct?
    Moreover, I don't know why [itex]J(A)[/itex] is open: could you help me to understand it?
    [itex]\aleph_1[/itex] thanks!
     
  5. Aug 31, 2014 #4

    micromass

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    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
    [tex]\{x\in V~\vert~p_A(x)<1\}\subseteq A\subseteq \{x\in V~\vert~p_A(x)\leq 1\}[/tex]
    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.
     
  6. Sep 1, 2014 #5
    Wow, what a detailed answer! Thank you so much!!! As to the lemma Let [itex]L[/itex] be a subspace... I understand that [itex]\alpha>p_A(y)\Rightarrow\alpha^{-1}y\in A[/itex], but it isn't clear to me why [itex]f(p_A(y)^{-1} y)\leq m[/itex]. I suppose that [itex]p_A(y)^{-1}y\in A[/itex] but I don't understand why, since [itex]p_A[/itex] represents an infimum... [itex]\aleph_1[/itex] thanks!
     
  7. Sep 1, 2014 #6

    micromass

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    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.
     
  8. Sep 1, 2014 #7
    [itex]\aleph_2[/itex] thanks! I had to increase cardinality :cool:
    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 [itex]\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 [/itex], while I supposed you used Kadison-Ringrose's theorem 1.2.6, I now think the seminorm you talked about is the Minkowski functional [itex]P_{A-a}[/itex] for [itex]A-a[/itex]: have I understood?
     
  9. Sep 1, 2014 #8

    micromass

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    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

    [tex]B_p(a,\varepsilon),~p\in \mathcal{P},~a\in V,~\varepsilon>0[/tex]

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

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

    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.
     
  10. Sep 1, 2014 #9
    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!
     
  11. Sep 1, 2014 #10

    micromass

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    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.
     
  12. Sep 2, 2014 #11
    Wonderfully interesting, I heartily thank you!!! Written a pencil note in the book.
     
  13. Sep 2, 2014 #12

    micromass

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