The Hahn–Banach theorem - UCL

1 MATHEMATICS 3103 (Functional Analysis)YEAR 2012 2013, TERM 2 HANDOUT #6: THE HAHN BANACH theorem AND DUALITY OFBANACH SPACESThe Hahn Banach theoremLetXbe a normed linear space. Three weeks ago we posed the question of whether thereare enough continuous linear functionals onXto separate the points ofX. This weekwe will prove that the answer is yes (this result is a kind of analogue, for continuouslinearfunctionals on anormed linearspaceX, of Urysohn s lemma forgeneralcontinuous functionson an arbitrarymetricspaceX). We will actually prove more: namely, we will prove anextension theorem for continuous linear functionals defined on a proper linear subspace ofX(this result is a kind of analogue of the Tietze extension theorem for general continuousfunctions defined on a properclosed subsetof an arbitrarymetricspaceX): theorem (Hahn Banach theorem for normed linear spaces)1 LetXbe a real orcomplex normed linear space, letM Xbe a linear subspace, and let M be a boundedlinear functional onM.

2 Then there exists a linear functional X that extends ( M= ) and satisfiesk kX =k kM .As in the Tietze extension theorem , the important fact here is not just the existence of acontinuous extension, but the existence of a continuous extensionthat does not increase also that here (unlikein the Tietze extension theorem ) the linear subspaceMneednot be closed. That is because a boundedlinearfunctional (unlike a general continuousfunction) can always be automatically extended continuously fromMtoM(see Proposi-tion ); so it makes no difference whetherMis closed or simplicity we will prove the Hahn Banach theorem only intherealcase. The complexcase is not really much more difficult, but it involves fiddly work that would divert us frommore important proof of the Hahn Banach theorem has two parts: First, weshow that can beextended (without increasing its norm) fromMto a subspaceone dimension larger: thatis, to any subspaceM1= span{M, x1}=M+Rx1spanned byMand a vectorx1 X\ , we show that these one-dimensional extensions can be combined to provide anextension fromMto all is the first step:1 The Hahn Banach theorem was first proven in 1912 by the Austrian mathematician Eduard Helly(1884 1943).

3 It was rediscovered independently in the 1920s by the Austrian mathematician Hans Hahn(1879 1934) and the Polish mathematician Stefan Banach (1892 1945).1 Lemma (one-dimensional extension, real case)LetXbe a real normed linear space,letM Xbe a linear subspace, and let M be a bounded linear functional onM. Then,for any vectorx1 X\M, there exists a linear functional 1onM1= span{M, x1}thatextends ( 1 M= ) and satisfiesk 1kM 1=k kM . = 0 the result is trivial, so we can assume without loss of generality thatk k= 1(why?) (this assumption is made only to simplify the formulae). Now everyx M1can beuniquely represented in the formx= x1+ywith Randy M. To define 1as anextension of , it suffices to choose the value of 1(x1), call itc1: we then have 1( x1+y) = c1+ (y).

4 ( )We want to choosec1so that| 1(x)| kxkfor allx M1, k x1+yk c1+ (y) k x1+yk( )for all Randy M. This holds for = 0 by hypothesis on , and for 6= 0 it can berewritten as x1+y (y/ ) c1 x1+y (y/ )( )for all Randy M(you should check that this is correct both for >0 and for <0),or equivalently kx1+zk (z) c1 kx1+zk (z)( )for allz M. But forz1, z2 Mwe have (z2) (z1) = (z2 z1) kz2 z1k kx1+z1k+kx1+z2k( )byk k= 1 and the triangle inequality, so that kx1+z1k (z1) kx1+z2k (z2)( )for allz1, z2 M. It follows thatc supz1 M[ kx1+z1k (z1)]( )c+ infz2 M[kx1+z2k (z2)]( )are finite and satisfyc c+; so we can choose anyc1 [c , c+]. OK, what next?

5 IfXis finite-dimensional or more generally ifMhas finite codi-mension inX, the quotient spaceX/Mis finite-dimensional then we simply need torepeat the one-dimensional extension step a finite number oftimes, and we are separable or more generally if the quotient spaceX/Mis separable thena slight refinement of this argument works: We first extend fromMtoMusing Proposi-tion Then we choose a total sequence of linearly independent vectors [x1],[x2],[x3], ..inX/M(see Problem 8(b) of Problem Set #3), and we then successively extend to a2linear functional ndefined on the spaceMn= span{M, x1, .. , xn}for eachn. Since thesearesuccessiveextensions, we have n Mn = n whenevern < n.

6 It follows that theunion of the ndefines a linear functional on the linear subspaceM = n=1Mn. But byconstructionM is dense inX, so by Proposition , can be extended (uniquely) to abounded linear functional onX(without changing its norm).Alas, ifXis nonseparable, no such simple inductive construction canwork, and we needto appeal to more powerful set-theoretic tools to show (nonconstructively) that the one-dimensional extensions can be pieced together to reach the whole spaceX. The tool weneed isZorn s (The technique of using Zorn s lemma to make nonconstructiveexistence proofs is sometimes calledZornication.) Here are the needed concepts:Definition a set.

7 Then apartial orderonSis a binary relation onSthat satisfies(a)a a(reflexivity);(b)a bandb aimplya=b(antisymmetry); and(c)a bandb cimplya c(transitivity)for alla, b, c S. The pair(S, )is called apartially ordered set(orposet). Wesometimes also refer toSalone as a partially ordered set if the relation is understoodfrom the let (S, ) be a partially ordered set. A subsetT Sis calledtotally ordered(with respect to ) if for every paira, b Twe have eithera borb a. A totally orderedsubset is also called achain. An elementu Sis said to be anupper boundfor a subsetT Sifa ufor alla T. (Note that the upper bounduneed not belong toTitself.)Finally, amaximal elementofSis an elementm Ssuch thatm ximpliesm=x.

8 (Amaximal element need not exist; and if one exists, it need notbe unique.) The usual order onRis a total order. There is no maximal The usual order onR=R { ,+ }is also a total order. Now there is a uniquemaximal element + .3. The usual partial order onRnis defined byx yif and only ifxi yifor 1 i 2 it isnota total order. There is no maximal Consider the usual partial order onR2restricted to the three-element subsetS={(0,0),(0,1),(1,0)}. Then (0,1) and (1,0) are maximal Thelexicographic orderonR2is defined byx yif and only if eitherx1< y1orelsex1=y1andx2 y2. (Think of the ordering of words in a dictionary!) This is a totalorder (why?). There is no maximal LetAbe an arbitrary set, and letP(A) be the set of all subsets ofA.

9 Then therelation of set inclusion is a partial order onP(A). (It is not a total order except in twodegenerate cases can you see what they are?) There is a unique maximal s lemma was first proved by the Polish mathematician Kazimierz kuratowski (1896 1980) in was rediscovered and applied by the German/American mathematician Max Zorn (1906 1993) in LetVbe a vector space, and letL(V) be the set of all linear subspaces ofV. Thenthe relation of set inclusion is a partial order onL(V). (It is not a total order except intwo degenerate cases can you see what they are?) There is a unique maximal elementV. We then have:Proposition (Zorn s lemma)Let(S, )be a partially ordered set in which every to-tally ordered subset has an upper bound.

10 Then(S, )contains at least one maximal s lemma is a result of set theory that can be proven usingthe axiom of choice. Moreprecisely, Zorn s lemma isequivalentto the axiom of choice in Zermelo Fraenkel (ZF) settheory. Other important statements of set theory that are equivalent to the axiom of choicein ZF set theory are the well-ordering theorem and the Hausdorff maximal principle. Weshall not enter into the details of these statements or the proof of their equivalence, whichbelong to a course in set theory or mathematical logic; rather, we shall simply take Zorn slemma as a set-theoretic result that we can use without are now ready to prove the Hahn Banach theorem :Proof of the Hahn Banach theorem (real case).