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Intersection of a closed convex set

  1. Jan 14, 2015 #1
    Let X be a real Banach Space, C be a closed convex subset of X.

    Define Lc = {f: f - a ∈ X* for some real number a and f(x) ≥ 0 for all x ∈ C} (X* is the dual space of X)

    Using a version of the Hahn - Banach Theorem to show that

    C = ∩ {x ∈ X: f(x) ≥ 0} with the index f ∈ Lc under the intersection


    Could someone help me to solve this problem, i cant see how Hahn - Banach can imply the above statement ( I used the separation version to obtain g(x)<a<g(y) for some linear functional g)
     
  2. jcsd
  3. Jan 17, 2015 #2
    You will need the following corollary of the Hahn--Banach theorem (with sublinear functional): if ##U## is a convex open set, and ##a\notin U## then there exists a non-zero linear functional ##f\in X^*## such that ##f(x)\le f(a)## for all ##x\in U##. To prove the corollary you first notice that because of translation invariance, you can assume without loss of generality that ##0\in U##.

    Then you define a sublinear functional ##q## to be the Minkowski functional of ##U##, and define ##f_0## to be the linear functional on the one-dimensional space spanned by ##a## such that ##f_0(a)=1##. Extending ##f_0## by Hahn--Banach you get a functional ##f## such that ##f(x)\le q(x)## for all ##x\in X##. Then for all ##x\in U## $$ f(x) \le q(x) \le 1 = q(a),$$ so we get the desired inequality.
    We still need to show that ##f\in X^*## (i.e. that ##f## is bounded), but that is easy: ##U## is open, so an open ball ##B_\varepsilon \subset U##, so ##q(x), q(-x)\le \frac1\varepsilon \|x\|## (##B_\varepsilon## is an open ball of radius ##\varepsilon## centered at ##0##), and therefore ##|f(x)|\le \frac1\varepsilon \|x\|##.

    To get the statement you need, take ##a \notin C##. Then for some ##\varepsilon>0## the whole open ball ##x+B_\varepsilon## does not intersect ##C##, or, equivalently, the (open) set ##U:= C+B_\varepsilon## does not contain ##a##. Applying the above corollary yo get that there exists ##f\in X^*## such that
    $$ f(x)\le f(a)$$ for all ##x\in U##, or, equivalently, $$f(x) \le f(a) + f(y) \qquad \forall x\in C, \ \forall y\in B_\varepsilon.$$ Noticing that for any non-zero ##f\in X^*## the point ##0## is an interior point of ##f(B_\varepsilon)## (in fact ##f(B_\varepsilon)## is an open set), we conclude that there exists real ##b## such that $$f(x) \le b <f(a) \qquad \forall x\in C.$$ The last statement means that ##C## contains the intersection of the half-spaces. The opposite inclusion is trivial.
     
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