SOME MORE APPLICATIONS OF THE HAHN-BANACH THEOREM

SOME MORE APPLICATIONS OF THE HAHN-BANACH THEOREM FRANCISCO JAVIER GARCÍA-PACHECO, DANIELE PUGLISI, AND GUSTI VAN ZYL Absrac We give a new proof of he fac ha equivalen norms on subspaces can be exended This new proof is based on he Hahn-Banach Exension Theorem We also give new characerizaions for an equivalen norm on a dual space o be a dual norm Finally, a new proof of a paricular case of he Hahn-Banach Separaion Theorem is provided wihou involving he Axiom of Choice Exensions of equivalen norms Exensions of equivalen norms have a lo of applicaions in he Theory of Banach Spaces For his we refer he reader o [3, Lemma 8], where he following was proved Lemma Deville, Godefroy, and Zizler, 993) Le X be a real Banach space Le Y be a subspace of X If Y is an equivalen norm on Y, hen here exiss an equivalen norm X on X such ha X Y = Y Even hough he proof of his lemma appears in [3, Lemma 8], here we will show i wih lile modificaions We recall he reader ha a subse of a real vecor space is said o be absoluely convex provided ha i is balanced and convex Proof of Lemma Observe ha we may assume ha B Y B Y Take X o be he norm on X whose uni ball is B X := co ) B X B Y I is clear ha B X is bounded, closed, absoluely convex, and conains B X, herefore i is he uni ball of an equivalen norm on X All is lef o show is ha B X Y = B Y By consrucion we have ha B X Y B Y An elemen y of B X Y is he limi of a sequence x n + ) y n ) n N where [0, ] and for all n N we have x n B X 2000 Mahemaics Subjec Classificaion Primary 46B20 Key words and phrases equivalen norm, dual norm, Hahn-Banach, Axiom of Choice

and y n B Y If = 0, hen y B Y because B Y is closed in Y Assume hen ha 0, ] Noe ha he sequence x n y )) y n n N converges o 0 This leaves us wo possibiliies: ) There exiss a subsequence y y nk such ha for all )k N k N we have ha y y nk and hence y y nk B Y In his case, he sequence y ) ) y nk + ) y nk k N converges o y and is conained in B Y, which means ha y B Y because B Y is closed in Y 2) There exiss n 0 N such ha if n n 0, hen y y n In his case, he sequence y ) ) y n + ) y n y y n n N converges o y and is conained in B Y, which means ha y B Y because B Y is closed in Y We presen here a new proof of Lemma ha relies on he Hahn- Banach Exension Theorem Noice ha his new proof is legiimae because he original proof of he Hahn-Banach Exension Theorem is no based upon exending any norm New Proof of Lemma We can assume wihou any loss of generaliy ha B Y B Y Take X o be he norm on X whose uni ball is B X := B X f [, ]) f T where T := {f X : f Y Y =, f Y = f } Observe he following: ) X is an equivalen norm on X Indeed, B X is closed, bounded, and absoluely convex I suffices o prove ha B X has nonempy inerior Since Y is an equivalen norm on Y, here mus exis a consan K > 0 such ha Y K on Y We will show ha B X 0, min {, K}) B X I is obvious ha 2

B X 0, min {, K}) B X Finally, if x B X 0, min {, K}) and f T, hen f x) f x = f Y x K f Y Y x 2) B X Y = B Y Indeed, le y B X Y and assume ha y Y > We can find f Y such ha f Y = and f y) = y Y In accordance o he Hahn-Banach Exension Theorem, we can exend f linearly and coninuously o he whole of X and preserving is -norm We will keep denoing his exension by f By consrucion, f T Now we ge he conradicion ha y Y = f y) Conversely, le y B Y If f T, hen f Y Y = so we obviously have ha f y) In order o exend dual norms, we will follow a similar process as in he original proof of [3, Lemma 8] Lemma 2 Le X be a real Banach space Le Y be a w -closed subspace of X If Y is an equivalen norm on Y, hen here exiss an equivalen dual norm X on X such ha X Y = Y Proof Assume ha B Y B Y and ake X o be he norm on X whose uni ball is B X := co ) B X B Y We can easily see ha B X is w -compac, bounded, and absoluely convex Therefore, i defines a new equivalen dual norm on X I remains o see ha B X Y = B Y Obviously, B X Y B Y An elemen z of B X Y is of he form x + ) y where [0, ], x B X and y B Y If = 0, hen z is rivially in B Y Assume hen ha 0 Since z Y, we mus have ha x Y, so x B Y B Y Therefore, z B Y The nex poin here is o adap he new proof of Lemma o exend equivalen dual norms However, we will show now ha i canno be adaped quie well Remark 3 Le X be a real Banach space Le Y be a w -closed subspace of X Le Y be an equivalen norm on Y As in he new proof of Lemma, we may assume wihou any loss of generaliy ha B Y B Y Take X o be he norm on X whose uni ball is B X := B X f T f [, ]) where T := {f X : f is w -coninuous, f Y Y =, f Y = f } 3

Following a similar way as in he new proof of Lemma, i can be shown ha X is an equivalen norm on X and B X Y B Y In order o prove ha B X Y B Y we need o verify wo hings: ) Given any y Y wih y Y >, here mus exis a w coninuous f Y wih f Y = and f y) = y Y 2) Given any w coninuous f Y, here mus exis a w coninuous linear exension o he whole of X preserving is - norm The previous remark moivaes he following quesion: Quesion 4 In he seings of Remark 3, wha condiions does Y require in order for ) and 2) o be verified? In [3, Proposiion 82] i is also proved he following resul: Proposiion 5 Deville, Godefroy, and Zizler, 993) Le X be a separable real Banach space Le Y be a subspace of X If Y is an equivalen locally uniformly round norm on Y, hen here exiss an equivalen locally uniformly round norm X on X such ha X Y = Y The nex resuls show ha we can exend equivalen norms preserving oher geomeric characerisics Theorem 6 Le X be a real Banach space Le Y be a closed subspace of X Assume ha here exiss an equivalen norm Y on Y whose uni sphere S Y conains a face C wih non-empy inerior relaive o S Y There exiss an equivalen norm X on X such ha: ) The uni sphere S X conains a face D wih non-empy inerior relaive o S X such ha D Y = C 2) The norm X Y = Y Proof The Hahn-Banach Separaion Theorem assures he exisence of an elemen f Y such ha f Y = and C = f ) B Y The Hahn-Banach Exension Theorem assures he exisence of a linear coninuous exension of f o he whole of X preserving is -norm We will keep denoing his exension by f By hypohesis, here exiss y S Y and ε > 0 such ha B Y y, ε ) f ) C Denoe F := B X y, ε) f ) where 0 < ε < ε We will consider he closed absoluely convex se aco B Y F ) This se verifies he following properies: ) aco B Y F ) has non-empy inerior Indeed, choose 0 < τ < min { 3 f, 4 ε 3 + f y ) }

We will show ha B X 0, τ) aco B Y F ) Le x B X 0, τ) Noe ha x = f x) + ) y y + 3 x f x) y)) 3 3 Observe ha y+3 x f x) y) F and ha f x) + 3 + 3, herefore x aco B Y F ) Noice ha we have proved ha aco B Y F ) is he uni ball of an equivalen norm on X ha we will denoe by X 2) aco B Y F ) Y = B Y Indeed, obviously aco B Y F ) Y B Y An elemen y of aco B Y F ) Y is he limi of a sequence y n + sx n ) n N in which + s and for all n N i is y n B Y and x n F If s = 0, hen y rivially belongs o B Y, so we may assume ha s 0 Observe ha he sequence s x n y y ) s n f y y ) s s n n N converges o 0 Since ε < ε, we evenually have ha y y s s n f y y ) B Y y, ε ) f ) C B Y s s n for large n s, which means ha y = lim n y n + s y y s s n f y y ) s s n ) B Y 3) The boundary of aco B Y F ) has a face D wih non-empy inerior relaive o ha boundary such ha D Y = C Indeed, by consrucion F is a convex se wih non-empy inerior relaive o he boundary of aco B Y F ) o see his i is sufficien o noice ha f X = sup f aco B Y F )) = ) Now ake D := f ) B X Finally, D Y = f ) B X Y = f ) B Y = C To finish his secion we make he reader noice wihou a proof ha he dual version of he previous heorem also holds Theorem 7 Le X be a real Banach space Le Y be a w -closed subspace of X Assume ha here exiss an equivalen norm Y on Y whose uni sphere S Y conains a face C wih non-empy inerior 5

relaive o S Y There exiss an equivalen dual norm X on X such ha: ) The uni sphere S X conains a face D wih non-empy inerior relaive o S X such ha D Y = C 2) The norm X Y = Y 2 Equivalen dual norms The following characerizaion of equivalen dual norms can be found in [3, Fac 54]: Fac 2 Deville, Godefroy, and Zizler, 993) Le X be a real Banach space Le be an equivalen norm on X The following condiions are equivalen: ) The norm is a dual norm 2) The Banach-Aloaglu Theorem is verified, in oher words, B is w -closed 3) The norm : X, w ) R is lower semi-coninuous Our conribuion o his opic consiss of showing ha an equivalen norm on a dual space is a dual norm if and only if he Goldsine Theorem is verified This will be bring ineresing consequences on he exension of dual norms Remark 22 Le X be a real Banach space Le be an equivalen norm on X Consider he equivalen norm on X given by := X We have he following basic facs: B = B X is a dual norm on X if and only if = X = X By aking ino consideraion Remark 22 we are capable of carrying ou he desired characerizaions Theorem 23 Le X be a real Banach space Le be an equivalen norm on X The following condiions are equivalen: ) The norm is a dual norm 2) The Goldsine Theorem is verified, in oher words, cl w B X ) = B 6

Proof Le := X For every x X we have ha x = sup { x x) : x B } = sup { x x ) : x B } = sup { x x ) : x B X } = sup { x x ) : x B } Goldsine Theorem) = x The following characerizaion of equivalen dual norms solves an open problem on he exension of equivalen norms Corollary 24 Le X be a real Banach space Le be an equivalen norm on X Consider he equivalen norm on X given by := X The following condiions are equivalen: ) The norm is a dual norm 2) = Proof Assume ha = We will show ha he Goldsine Theorem is verified by, ha is, cl w B X ) = B Indeed, by bearing in mind Remark 22 we have ha cl w B X ) ) = cl w B = B = B in virue of he fac ha he bidual norm verifies he Goldsine Theorem Now i only suffices o apply he previous heorem An informally open quesion among Funcional Analyss is o deermine wheher wo equivalen dual norms on a dual space are equal provided ha hey coincide on a w -dense subspace The nex example shows ha his is no rue in virue of Corollary 24 Example 25 Le X be a real Banach space Le be an equivalen norm on X which is no a dual norm Consider he equivalen norm on X given by := X In accordance o Corollary 24 we have ha boh and are equivalen dual norms on X which coincide on he w -dense subspace X bu which are no equal 3 The Hahn-Banach Theorem This secion is devoed o prove an equivalen form of he Hahn- Banach Separaion Theorem over he Zermelo-Fraenkel Axiomaic Sysem ZF) plus he Axiom of Dependen Choices DC) I is well known 7

ha DC is a weaker form of he Axiom of Choice AC) and is acually equivalen o he Baire Caegory Theorem see []) I is also known ha DC sricly implies he Axiom of Counable Choice AC ω ) The lieraure involving AC and Hahn-Banach Theorem shows, among oher hings, ha: ) The Ulrafiler Lemma, which is sricly weaker han AC, implies he Hahn-Banach Theorem, alhough he converse is no he case 2) The Hahn-Banach Theorem can in fac be proved using even weaker hypoheses han he Ulrafiler Lemma see [5] and [4]) 3) For separable Banach spaces, Brown and Simpson proved ha he Hahn-Banach Theorem follows from WKL 0, a weak subsysem of second-order arihmeic see [2]) Over ZF + DC he following hree resuls hold noice ha he exisence of converging sequences is guaraneed by AC ω which is implied by DC as we menioned a he beginning of his secion) Lemma 3 Le X be a real Banach space Le c S X and 0 < r < such ha B X c, r) S X is convex Then [B X c, r) S X ] c is absoluely convex Proof Le x [B X c, r) S X ] c All we need o show is ha x [B X c, r) S X ] c Noe ha in order o show his i is sufficien o prove ha x + c Indeed, since x + c B X c, r) S X, we have ha x + c B X c, r), and since c = 2 x + c) + x + c), 2 we necessarily have ha x + c = Le us show hen ha x + c mus be in B X Assume no Le Y := span {x, c} Le y [c, x + c] be he closes o x+c ha sill remains on S Y Noice ha y c < r We can find z S Y sufficienly close o y so ha z c < r and z is no on he segmen [c, y] Since B Y c, r) S Y is convex, z mus lie on he segmen y, x + c), which conradics he fac ha y is he closes in [c, x + c] o x + c ha sill remains on S Y Theorem 32 Le X be a real Banach space Le C S X be a convex se wih non-empy inerior relaive o S X There exiss a unique f S X such ha C f ) Proof Le c in SX C) and 0 < r < such ha B X c, r) S X C We will show he following: ) span C c) = span [B X c, r) S X ] c) Indeed, le x C There exiss 0, ) sufficienly small such ha x+ ) c 8

B X c, r) S X Then x c = [x + ) c] c) span [B X c, r) S X ] c) 2) X span C c) Indeed, assume ha X = span C c), hen X = span [B X c, r) S X ] c) By Lemma 3, we have ha [B X c, r) S X ] c is a barrel of X Since X is complee, we are allowed o apply he Baire Caegory Theorem which is equivalen o DC as we menioned a he beginning of his secion) Therefore, [B X c, r) S X ] c has nonempy inerior, which means ha B X c, r) S X has non-empy inerior, and his is no possible 3) X = span C c) + Rc Indeed, le x X There exiss α R \ {0} small enough such ha Then x = c αx c αx C ) c αx c αx α c αx c + span C c) + Rc c αx c α 4) B spanc c) 0, r/2) C c Indeed, le y B spanc c) 0, r/2) Then c y c y c = c y ) c y c y = c y ) c y) c y y c y c y + y Now, = c c y + y y = c y c c y) + y = 2 y r c y c y c ) + c y ) c, 9

which shows he conradicion ha c span C c) unless c y = 0 Therefore y = c y c y c C c 5) dis rc, C c) r Indeed, if x C, hen 2 2 r 2 c x c) = + r ) c x 2 + r ) c x 2 = r 2 6) span C c) is closed If i is no closed, hen i mus be dense, since i is of codimension Therefore, here exiss a sequence y n ) n N span C c) ha converges o c Then ) ryn B spanc c) 0, r/2) C c 2y n n N converges o rc, which is impossible because dis rc, C c) 2 2 r 2 7) Since span C c) is closed and of codimension, i mus be he kernel of a unique coninuous linear funcional f S X such ha f c) > 0 All is lef o prove is ha f c) = Assume no There exiss x S X such ha f c) < f x) We can find 0, ) small enough so ha x + ) c x + ) c C However, his is no possible because f x) + ) f c) > f c) x + ) c f c) Theorem 33 The following saemens are equivalen for a real Banach space X: ) There exiss a coninuous linear funcional f X \ {0} 2) There exiss a closed bounded absoluely convex subse B of X whose boundary conains a convex subse C wih non-empy inerior relaive o he boundary of B Proof If here exiss such B, hen one can call on Theorem 32 o assure he exisence of a coninuous linear funcional f X \{0} Conversely, 0

assume he exisence of a coninuous linear funcional f X \ {0} I is sufficien o ake B := B X f [ 2, 2] and C := BX f 2) References Blair, C: The Baire caegory heorem implies he principle of dependen choices Bull Acad Polon Sci Sr Sci Mah Asronom Phys 25 977), no 0, 933 934 2 D K Brown and S G Simpson: Which se exisence axioms are needed o prove he separable Hahn-Banach heorem? Annals of Pure and Applied Logic 3 986), 23-44 3 R Deville, G Godefroy and V Zizler: Smoohness and renormings in Banach spaces Piman Monographs and Surveys in Pure and Applied Mahemaics 64, New York, 993 4 M Foreman and F Wehrung: The Hahn-Banach heorem implies he exisence of a non-lebesgue measurable se Fundamena Mahemaicae 38 99), 3-9 5 D Pincus: The srengh of HahnBanach s Theorem Vicoria Symposium on Non-sandard Analysis, Lecure noes in Mah 369, Springer 974, 203-248 Deparmen of Mahemaics, Universiy of Cadiz, Puero Real 50, Spain EU) E-mail address: garciapacheco@ucaes Deparmen of Mahemaics, Universiy of Caania, Caania 9525, Ialy EU) E-mail address: dpuglisi@dmiunici Deparmen of Mahemaics and Applied Mahemaics, Universiy of Preoria, 0002 Preoria, Souh Africa E-mail address: gusivanzyl@upacza