1.4 The Hahn Banach Theorem - Texas A&M Universityschlump/lecture3and4.pdfwill be covered in Chapter...

1.4. THE HAHN BANACH THEOREM 17

1.4 The Hahn Banach Theorem

Definition 1.4.1. Suppose that V is a vector space over K. A real-valuedfunction p on V , satisfying

• p(0) = 0,

• p(x + y) ! p(x) + p(y), and

• p(!x) = !p(x) for ! > 0,

is called a sublinear functional on V .Note that 0 = p(0) ! p(x) + p("x), so that p("x) # "p(x).

Theorem 1.4.2. [The analytic Hahn-Banach Theorem, real version]Suppose that p is a sublinear functional on a real vector space V , that Wis a linear subspace of V and that f is a linear functional on W satisfyingf(y) ! p(y) for all y $ W . Then there exists a linear functional g on V suchthat g(x) = f(x) for all x $ W (g extends f) and such that g(y) ! p(y) forall y $ V (control is maintained).

Theorem 1.4.3. [The analytic Hahn-Banach Theorem, general version]Suppose that p is a seminorm on a real or complex vector space V , that Wis a linear subspace of V and that f is a linear functional on W satisfying|f(x)| ! p(x) for all x $ W . Then there exists a linear functional g on Vsuch that g(x) = f(x) for all x $ W (g extends f) and such that |g(y)| ! p(y)for all y $ V (control is maintained).

Corollary 1.4.4. Let X be a normed space Y a subspace and y! $ Y !. Thenthere exists an extension x! of y! to an element in X! with %x!% = %y!%.

Proof. Put p(x) = %y!%%x%.

Corollary 1.4.5. Let X be a normed space Y a subspace and x $ X withh = dist(x, Y ) > 0. Then there exists an x! $ X!, with x!|Y & 0 andx!(x) = 1.

Proof. Consider Z = {y + ax : y $ Y and a $K}. Note that every z $ Zhas a unique representation z = y + ax, with y $ Y and a $ K. Indeed, ify1 + a1x = y2 + a2x, with y1, y2 $ Y and a1, a2 $ K, then we observe thata1 = a2, because otherwise x = (y1 " y2)/(a1 " a2) $ Y , and thus, y1 = y2.

18 CHAPTER 1. SOME BASIC BACKGROUND

We define f : Z ' K, y + ax (' a. The unique representation of eachz $ Z implies that f is linear. The functional f is also continuous. Indeed,assume zn = yn + anx ' 0, if n ' ), but infn"N |an| # " for some " > 0,then

dist(x, Y ) = dist!zn

an" yn

an, Y

"! dist

!zn

an, Y

"' 0 if n '),

which contradicts our assumption.We can therefore apply the Hahn-Banach Theorem 1.4.2 to the linear

functional f on Z and the norm p(x) = %f%Z!%x%.

Corollary 1.4.6. Let X be a normed space and x $ X. Then there is anx! $ X!, %x!% = 1, so that *x!, x+ = %x%.

Proof. Let p(x) = %x% and f(#x) = #%x%, for #x $ span(x) = {ax : a $K}.

Definition 1.4.7. [The Canonical Embedding, Reflexive spaces]For a Banach space we put X!! = (X!)! (the dual space of the dual space ofX).

Consider the map

$ : X ' X!!, with $(x) : X! ' K, *$(x), x!+ = *x!, x+, for x$X.

The map $ is well defined (i.e. $(x) $ X!! for x $ X), and since for x $ X

%$(x)%X!! = supx!"BX!

|*x!, x+| ! %x%,

it follows that %$%L(X,X!!) ! 1 . By Corollary 1.4.6 we can find for each x $ Xan element x! $ BX! with *x!, x+ = %x%, and thus %$(x)%X!!| = %x%X .

It follows therefore that $ is an isometric embedding of X into X!!. Wecall $ the canonical embedding of X into X!!.

We say that X is reflexive if $ is onto.

Remark. There Banach spaces X for which X and X!! are isometrically iso-morphic, but not via the canonical embedding. An Example by R. C. Jameswill be covered in Chapter 3.


Definition 1.4.8. [The adjoint of an operator]Assume that X and Y are Banach spaces and T : X ' Y a linear and

bounded operator. Then the operator

T ! : Y ! (' X!, y! (' y! , T,

(i.e. *T !(y!), x+ = *y! , T, x+ = *y!, T (x)+ for y! $ Y ! and x $ X)

Proposition 1.4.9. Assume X and Y are Banach spaces and T : X ' Y alinear and bounded operator. Then T ! is a bounded linear operator from Y !

to X!, and %T !% = %T%.Moreover if T is surjective T ! is an isomorphic embedding, and if T is

an isomorphic embedding T ! is surjective.

We want to formulate a geometric version of the Hahn- Banach Theorem.

Definition 1.4.10. A subset A of a vector space V is called convex if forall a, b $ A and all ! $ [0, 1] also !a + (1" !)b $ A.If A - V we define the convex hull of A by

conv(A) =# $

C : A - C - V, C convex%

=& n'

j=1

!jaj : n$N, !j$ [0, 1], ai$A, for i=1, . . . n, andn'

j=1

!j =1(

.

A subset A - V is called absorbing if for all x $ V there is an 0 < r < )so that x/r $ A. For an absorbing set A we define the Minkowski functionalby

µA : V ' [0,)), x (' inf{! > 0 : x/! $ A}.

Lemma 1.4.11. Assume C is a convex and absorbing subset of a vectorspace V . Then µC is a sublinear functional on V , and

(1.1) {v $ V : µC(v) < 1} - C - {v $ V : µC(v) < 1}.

If V is a normed linear space and if there is an " > 0 so that "BV - C,then µA is uniformly continuous (and thus bounded on BV ).

Proof. Since C is absorbing 0 $ C and µC(0) = 0. If u, v $ V and " > 0is arbitrary, we find 0 < !u < µC(u) + " and 0 < !v < µC(v) + ", so thatu/!u $ C and v/!v $ C and thus

u + v

!u + !v=

!u

!u + !v

u

!u+

!v

!u + !v

v

!v$ C,


which implies that µC(u + v) ! !u + !v ! µC(u) + µC(v) + 2", and, since," > 0 is arbitrary, µC(u + v) ! µC(u) + µC(v).

Finally for ! > 0 and u $ V

µC(!v) = inf{r > 0 : !v/r $ C} = ! inf& r

!:!u

r$ C

(= !µC(u).

To show the first inclusion in (1.1) assume v $ V with µC(v) < 1, thereis a 0 < !< 1 so that v/! $ C, and, thus,

v = !v

!+ (1" !)0 $ C.

The second inclusion is clear since for v $ C it follows that v = v! $ C.

If V is a normed linear space and "BV - C, it follows from the sublinearityof µC , for u, v $ V , that

µC(u)" µC(v) ! µC(u" v) ! %u" v%"

,

and similarly µC(v)" µC(u) ! #u$v#" .

Theorem 1.4.12. [The Geometric Hahn-BanachTheorem, general version]Let C be a non empty, closed convex subset of a Banach space X and letx0 $ X \ C.

Then there is an x! $ X! so that

supx"C

.(*x!, x+) < .(*x!, x0+).

Proof. We first assume that K = R and we also assume w.l.o.g. that 0 $ C(otherwise pass to C"x and x0"x for some x $ C). Put " := dist(x0, C) > 0and put D = {x $ X : dist(x, C) ! "/2}. From Lemma 1.4.11 it follows thatµD is a bounded sublinear functional on X, and dist(x0, D) # "/2.

On the one dimensional space Y = span(x0) define

f : Y ' R, #x0 (' #µC(x0).

Then f(y) ! µC(y) for all y $ Y (if y = #x0, with # > 0 this follows from thepositive homogeneity of µC , and if # < 0 this is clear). By Theorem 1.4.2 wecan extend f to a linear function x!, defined on all of X, with x!(x) ! µC(x)for all x $ X. Since µC is bounded on BX it follows that x! $ X!.


Moreover, since x0[1 " "4#x0# ] /$ D (otherwise dist(x0, D) ! "/4 and thus

dist(x0, C) ! "2 + dist(x0, D) < "), it follows that f(x0) = µD(x0) > 1. This

proves our claim in the case that K = R.If K = C we first choose g, by considering X to be real Banach space, and

then put f(x) = g(x)" ig(ix). It is then easily checked that f is a complexlinear bounded functional on X.

Exercises

1. Prove Proposition 1.4.9 .

2. Let X be a Banach space with norm % ·% . Show that µBX = % ·% .

3.0 Show that there is an x! $ %!% so that

a) %x!% = 1,

b) *x!, x+ = limi&% xi, for x = (xi)$c = {(&i) : limi&% &i exists}c) If x = (&i) $ %%, and &i # 0, for i$N, then *x!, x+ # 0, and

d) If x = (&i) $ %% and x' = (&2, &3, . . .) then *x!, x'+ = *x!, x+

4. Show that %1 is not isomorphic to a subspace of c0.


1.5 Finite dimensional Banach spaces

Theorem 1.5.1. [Auerbach bases]If X = (Kn, % · %) is an n-dimensional Banach space, then X has a basisx1, x2, . . . xn for which there are functionals x!1, . . . x

!n $ X!, so that

a) %xj% = %x!j% = 1 for all j = 1, 2 . . . n,

b) for all i, j = 1, 2 . . . n

*x!i , xj+ = '(i,j) =

)if i = j,

if i /= j..

We call in this case (xj, x!j) an Auerbach basis of X.

Proof. We consider the function

Det :Xn = X 1X 1X* +, -n times

' K,

(u1, u2, . . . un) (' det(u1, u2, . . . un).

Thus, we consider ui $ Kn, to be column vectors and take for u1, u2, . . . un $Kn the determinant of the matrix which is formed by vectors ui, for i =1, 2, . . . n. Since (BX)n is a compact subset of Xn with respect to the prod-uct topology, and since Det is a continuous function on Xn we can choosex1, x2, . . . xn in BX so that

..Det(x1, x2, . . . xn).. = max

u1,u2,...un"BX

..Det(u1, u2, . . . un)...

By multiplying x1 by the the appropriate number # $ K, with |#| = 1, wecan assume that

Det(x1, x2, . . . xn) $ R and Det(x1, x2, . . . xn) > 0.

Define for i = 1, . . . n

x!i : X ' K, x (' Det(x1, . . . xi$1, x, xi+1, . . . , xn)

Det(x1, x2, . . . xn),

1.5. FINITE DIMENSIONAL BANACH SPACES 23

It follows that x!i is a linear functional on X (taking determinants is linearin each column), and

*x!i , xi+ = 1,

%x!i % = supx"BX

|*x!i , x+| = supx"BX

...Det(x1, . . . xi$1, x, xi+1, . . . , xn)

Det(x1, x2, . . . xn)

... = 1

(by the maximality of Det(x1, x2, . . . xn) on (BX)n),

*x!i , xj+ =Det(x1, . . . xi$1, xj, xi+1, . . . , xn)

Det(x1, x2, . . . xn)= 0 if i /= j, i, j${1, 2 . . . n}

(by linear dependence of columns)

which finishes our proof.

Corollary 1.5.2. For any two n-dimensional Banach spaces X and Y itfollows that

dBM(X, Y ) ! n2.

Remark. Corollary 1.5.2 Is not the best result one can get. Indeed from thefollowing Theorem of John (1948) it is possible to deduce that for any twon-dimensional Banach spaces X and Y it follows that

dBM(X, Y ) ! n.

Theorem 1.5.3. [John’s theorem]Let X = (Kn, % ·% ) be an n-dimensional Banach space. Then there is aninvertible matrix T so that

B#2 - T (BX) -2

nB#2 .

Theorem 1.5.4. For any Banach space X

X is finite dimensional 34 BX is compact.

Proof. The implication “4” was already noted in the remark in Section 1.2the implication “3” will follow from the following Proposition.

Proposition 1.5.5. The unit ball of every infinite dimensional Banach spaceX contains a 1-seperated sequence.


Proof. By induction we choose for each n $ N an element xn $ Bx, sothat %xj " xn% # 1, for j = 1, 2 . . . n " 1. Choose an arbitrary x1 $ SX .Assuming x1,x2,. . .xn$1 has been chosen, let F = span(x1, . . . xn$1), (thespace generated by xj, j = 1, 2 . . . , n" 1). X/F is infinite dimensional, thusthere is a z $ X so that

1 = %z%X/F = infy"F

%z + y% = infy"F,#y#(1+#z#

%z + y% = miny"F,#y#(1+#z#

%z + y%.

We can therefore choose xn = z + y so that y $ F and

%z + y% = miny"F,#y#(1+#z#

%z + y% = 1,

it follows that

1 = %xn%X/F ! %xn " xj% for all j = 1, 2 . . . n" 1.

Remark. With little bit more work (see Exercise 3) one can we find in theunit ball of each infinite dimensional Banach space X a sequence (xn) with%xm " xn% > 1, for all m /= n in N.

A much deeper result by J. Elton and E. Odell (see [EO]) says that foreach Banach space X there is a " > 0 and a sequence (xn) - BX with%xm " xn% # 1 + ", for all m /= n in N.

Definition 1.5.6. An operator T : X ' Y is called a finite rank operatorif T (X) is finite dimensional. In this case we call dim(T (X)) the rank of Tand denote it by rk(T ).

For y $ Y and x! $ X! we denote the operator

X ' Y, x (' y*x!, x+

by y 5 x!. Clearly, y 5 x! is of rank one.

Proposition 1.5.7. Assume that X and Y are Banach spaces and thatT : X ' Y is a linear bounded operator of finite rank n. Then there arex!1, x

!2 . . . , x!n $ X and y1, y2. . . yn in Y so that

T =n'

j=1

yj 5 x!j .

1.5. FINITE DIMENSIONAL BANACH SPACES 25

Exercises:

1. Prove Corollary 1.5.2 using the existence of Auerbach bases. Prove theclaim in the following remark, by using John’s Theorem.

2.0 Find two spaces X and Y which are not isometric to each other, butfor which dBM(X, Y ) = 1.

3.0 Prove that in the unit ball of each infinite dimensional Banach spaceX there is a sequence (xn) with %xm " xn% > 1, for all m /= n in N.

4. Prove Proposition 1.5.7.

5.0 For n $ N prove that dBM(%n2 , %

n1 ) = dBM(%n

2 , %n%) =

2n.

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