Talk by Andreas Naes Aaserud for the logic seminar (290D) on 1 May 2012.

1 Hahn-Banach theorems

Theorem. (Hahn-Banach extension) Let X be a subspace of a real linear space Y and let p : Y R be a sublinear functional. Then for any linear functional f : X R satisfying f(x) p(x) forall x X, there is a linear functional f : Y R such that f |X = f and f(y) p(y) for ally Y .Proof. We will need the following

Claim. Let a Y \X be given. Then f extends to f : span(X {a}) R such that f(y) p(y)for all y span(X {a}).We refer to Folland, p. 158, for the standard proof of this fact. Next, consider Wa = {Z FDX(Y ) : a Z} for a Y . (Here, FDX(Y ) is the set of subspacesof Y that are finite-dimensional over X, i.e., are spanned by X and some finite subset of Y \X.)As {Wa}aY has FIP, there exists Z FDX(Y ) such that Y Z Y .The claim implies that, for any Z0 FDX(Y ), there exists f : Z0 R such that f |X = fand f(z) p(z) for all z Z0. By transfer, there exists f : Z R such that f |X = f andf(z) p(z) for all z Z.Thus f = stf |Y is the desired extension of f . Theorem. (Hahn-Banach separation) LetX be a real normed space, A,B X non-empty disjointconvex sets, where int(A) 6= . Then there exists f X and R such that f(A) (, ]and f(B) [,). If A is open, we may require that f(A) (, ).Proof. Put C = A B. Then C is convex with non-empty interior and 0 / C. Fix c int(C)and put K = C c.Define p : X [0,) by p(x) = inf{t R+ : x tK}.Claim. p is a sublinear functional.

p well-defined: Given x X, x/n 0 as n . As 0 int(K), x/n K for some n. Thusx nK for some n.p positive homogeneous: Let [0,) be given. If = 0 then p(x) = 0 = p(x). If > 0then

p(x) = inf{t/ R+ : x (t/)K} = p(x).

p subadditive: If x sK and y tK then x + y = (s + t)(s/(s + t)x + t/(s + t)y) (s + t)K.Thus p(x + y) s + t. It follows that p(x + y) t p(x) whenever y tK, hence thatp(x+ y) p(x) + p(y). Define a linear functional f0 : span{c} R by f0(rc) = rp(c).Claim. f0(x) p(x) for all x span{c}.For r < 0, we have that f0(rc) = rp(c) = p(rc), as p is positively homogeneous. For s > 0, weclaim that f0(sc) p(sc). Indeed, 0 = p(c+ c) p(c)+p(c) so that sp(c) (s)(p(c)) =sp(c). Thus f0(sc) = sp(c) sp(c) = p(sc). By Hahn-Banach extension, we may extend f0 to f : X R with f(x) p(x) for all x X.As p(x) = inf{t : x tK} 1 whenever x K, it follows that f(x) 1 for all x K. As0 int(K), there exists > 0 such that B(0, ) K. Hence x |f(x)| 1. It follows thatx /n |f(x)| 1/n, so if x 0 then f(x) 0. Hence f is continuous.

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Note moreover that f(c) = f(c) = f0(c) = p(c) 1. Indeed, c / K and tK K for allt (0, 1), because 0 K and K is convex.Finally, if x A and y B then x y c K so that

f(x) f(y) + 1 f(x) f(y) f(c) = f(x y c) 1,i.e., f(x) f(y). Thus = inf{f(y) : y B} works. The statement concerning open A followsfrom the Open Mapping Theorem. Corollary 1. Let X be a real normed space, C X convex, and a X with dist(a,C) > 0.Then there exists f X such that f(a) < inf{f(x) : x C}.Proof. Take B = C and A = B(a, ) such that A B = . Corollary 2. Let X be a real normed space, Y X a closed subspace with a Y and dist(a, Y ) >0. Then there exists f X such that f(a) = 1 and f(Y ) = {0}.Proof. Take B = Y and A = B(a, ) such that A B = . Then there exists f X and Rsuch that f(y) for all y Y while f(a) < . As Y is a subspace, we must have f(y) = 0 forall y Y . Finally, replace f by f/f(a). (Note: f(a) 6= 0 as otherwise > 0 so f(0) > 0.)

2 General non-standard hull

Setup. Let X be an internal linear space over R and let W be a family of internal seminormson X such that |W | < . Define an equivalence relation on X by

x W y p W : p(x y) 0.We define the monad with respect to W (of x X) by

W (x) = {y X : x W y}and the finite part of X with respect to W by

FinW (X) = {x X : suppW

st(p(x))

The triangle inequality is clear.

(iii) W is complete: Let {an}n=1 be a Cauchy sequence. Given m N choose km suchthat

n > km an akmW < 1/(2m).Extend {an} to {an}n maxj=1,...,r kmj .)So by saturation, there exists a p,mFp,m.Claim 2. n,m N : (n > km) suppW st(p(a an)) 1/m.Let n,m N be given. Suppose n > km; let p W . Thenst(p(aan)) st(p(aakm)+st(p(akman)) < 1/(2m)+anakmW < 1/(2m)+1/(2m) = 1/m.

Fixing n > k1 in claim 2 shows that a an FinW (X) so that a FinW (X). Moreover, claim 2shows that a anW 0 for n. Proposition. W (0) is closed in (FinW (X), W ) and XW = FinW (X)/W (0) as Banachspaces.

Proof. Trivial. See also Proposition 2.6 concerning norm non-standard hulls.

3 Weak non-standard hull

Let X be a real normed space. We will apply the notions from the previous section to X.

Consider the weak topology on X generated by the semi-norms pf (x) = |f(x)| for f X . For thisfamily W of seminorms,

x W y f X : f(x) f(y).

Proposition. Let C X be a set. Then(i) C weakly closed C is norm-closed;(ii) C is norm-closed and convex C is weakly closed.Proof. (i) Trivial.

(ii) Let c C, X 3 a W c. We want to show that a C. (By Proposition 1.14(ii), this willsuffice.) Suppose a / C. By the Hahn-Banach separation theorem (cf. corollary 1), there existsf X such that f(a) < infxC f(x). By transfer, f(a) < infxC f(x). In particular, f(a) < f(c)so that f(a) 6 f(c). As a W c, this is a contradiction.

Setup. We put now W = {pf : f BX} and call XW

the weak non-standard hull of X.

Define an infinite-valued seminorm st X : X [0,] by setting stxX = if xX /Fin(R). On the quotient space X/W (0), we define the quotient seminorm

x+ W (0)q = inf{stx+ yX : y W (0)} [0,]for x X. The finite part of X/W (0) is defined as

Finq(X/W (0)) = {x+ W (0) : x X x+ W (0)q

and forms a subspace of X/W (0).

We will use

Proposition. Let X be a normed space and f X . Then there exists a BX such thatf f(a) [0,).Proof. Consider the internal sets En = {x BX : f |f(x)|+ 1/n}, n N. As {En}nN hasthe FIP, there exists a0 BX such that f |f(a0)|. As transfer yields x BX : |f(x)| f,it follows that f |f(a0)|. Choose next F such that f(a0) = |f(a0)| and put a = a0.Then a BX does the trick. Theorem. We have that

(i) x FinW (X) : xW = x+ W (0)q;

(ii) XW = Finq(X/W (0)) as Banach spaces.

Proof. (i) Fix a FinW (X). Then, for y W (0), we have that

aW = supfBX

st|f(a)| = supfBX

st|f(a+ y)| sta+ yX .

Hence aW infyW (0) sta+ yX = a+ W (0)q.We next prove the reverse inequality. Without loss of generality, we may assume that aW = 1.Assuming this, we wish to show that a+ W (0)q 1.Write, for f X , rf = st(f(a)).Claim. For A SX finite and n,m N, there exists c (1 + 1/m)BX such that

f A : |f(c) rf | 1/n.

Suppose for contradiction that the claim fails for A,n,m. Then

x X :fA

(|f(x) rf | 1/n) x / (1 + 1/m)BX .

Put A = (1+1/m)BX and B = {x X : fA|f(x)rf | 1/n}. Then AB = by the previousline. Moreover, A and B are convex and A is open. Hence the Hahn-Banach separation theoremimplies that there exist f X and R such that f(A) (, ) and f(B) [,).Hence, for any x, y X, if x < 1 and fA[|f(y) rf | 1/n], we have that

f(x) > 1, a contradiction. Put Ef,n,m = {x X : x (1 + 1/m)BX |f(x) rf | 1/n} for f SX and n,m N. The claim implies that {Ef,n,m}f,n,m has FIP. Thus we get by saturation that there exists

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c f,n,mEf,n,m. This c satisfies c 1 and f(x) rf = stf(a) f(a) for all f X , i.e.,(c a) W (0). It follows that

a+ W (0)q = infyW (0)

sta+ yX sta+ (c a)X = stc 1,

showing what we want.

(ii) Note that XW Finq(X/W (0)) as sets and, by part (i), also as Banach spaces. [In

other words, the map x x + W (0) is a linear isometric embedding.] To see that we haveequality [surjectivity], let x X with x+ W (0)q

Hence the corollary to Hellys theorem implies that there exists a X such that

sta FX X : (a) F ().

In particular, a Fin(X) FinW (X) and, for each X , pi(a) = st(a) = F (), i.e.,pi(a) = F . Moreover,

aW = supBX

st|(a)| = supBX

|pi(a)| = supBX

|F ()| = FX ,

showing what we want. Theorem. Let X be a normed space and let K,C X be disjoint non-empty sets such that Kis weakly compact and C is norm-closed and convex. Then dist(K,C) > 0.

Proof. Suppose for contradiction that dist(K,C) = 0. Then, for any n N, there exist a Kand c C such that a c < 1/n. Apply saturation to the sets En = {(a, c) K C :a c < 1/n} to find a K and c C such that a c. As K is weakly compact, there existsb K such that b W a. As b / C, there exists r > 0 such that A = B(b, r) is disjoint from C.The Hahn-Banach separation theorem implies that there exist f X and R such that

f(b) < x C : f(x) .

By transfer, f(c) . In particular, f(b) 6 f(c).On the other hand, b W a implies that f(b) f(a) so that f(a) 6 f(c), contradicting the factthat a c.

4 Weak* topology

Consider the weak* topology on X generated by the semi-norms pa() = |(a)| for a BX . Forthis family W of seminorms,

W x X : (x) (x).

Note for the next proof that (X) = (X ), where (X) is the space of internally continuouslinear functionals X R.Proposition. Let X be a normed space and equip (X ) with the family W = {pa : a BX} ofseminorms. Then the map pi : X (X )W given by 7 is an isometric isomorphism.Proof. Note that pi is linear and that

pi() = = supaBX

st|(a)| = X .

Let next (X )W be given and put = st |X. Then is a linear functional on X and|(x)| = st|(x)| for all x BX so that is also continuous. Finally, (x) (x) for eachx X. Hence ( ) W (0) and = = pi(). Theorem. (Alaoglu) Let X be a normed space. Then BX is weak*ly compact.

Proof. We will repeatedly use the fact that R = R (under x 7 x). Moreover, we use the usualnon-standard characterization of compactness.

Let (BX) = B(X) be given. We must show that there exists BX such that W.

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Consider the map : (X) (X) defined by ()x = (x).Claim. is a well-defined linear isometric isomorphism.

Suppose = , i.e., . Then (x) (y) for all x, y Fin(X) with x y, so ()is uniquely determined. Moreover, |(x)| |(x)| and x

X xX for all x X. Hence() = so that is a well-defined linear isometry.It remains to show that is onto. Indeed, let (X) be given. Put = |X under the usualidentification. Then the non-standard extension of , which we will denote by the same symbol

as usual, belongs to Fin((X)). Hence (X). Finally, as (x) = (x) for all x X, transferimplies that (x) = (x) for all x X. Thus (x) (x) = (x). As (x) and (x) bothbelong to R, we get that () = , showing surjectivity. Under this identification, put = |X. Then BX . Given x X, put = (x) R = R.Then (x) = (x) = (x) under our identifications. Thus W and we are done.

5 Riesz Representation Theorem

In the following we will use the identification ba(N) = (`) given by 7 , where (f) =N f d for f `(N). By transfer, each ((`)) is of the form for some ba(N).

Note also that we are here working with complex scalars, unlike in previous parts of the talk.

Moreover, we use without proof the following theorem on general Loeb measures:

Theorem. Let (,B, ) be an internal measure space, where is a finite internal complex measureand B need only be an algebra of sets. Then st extends uniquely to a -additive measure onB, where B is the -algebra generated by B.Theorem. Let be a compact topological space. Then the map : M() C() given by 7 , where M() is the space of complex regular Borel measures on and (f) =

f d,

is a linear isometric isomorphism.

Proof. We need only show that is onto so fix C().Step 1. We claim that there exists a hyperfinite setH such that H and st(H) = .Consider, for , the set W = {H Pfin() : H}. As {W} has FIP, there existsa hyperfinite set H such that H . As is compact, st = ; hence st(H) = .Step 2. We identify internally H with [0, N ] N for some N N. Thus we may embed`(H) `(N) internally. For f C(), we put f = f |H `(H) `(N). ForX FD(C()) with basis {f1, . . . , fn}, put X = span(f1, . . . , fn) `(N). For f C(),x H, we have f(x) = f(x) f(st(x)) so that stf(x) = f(st(x)). Moreover,

{f1, . . . , fn} C() lin. indep. {f1, . . . , fn} `(N) lin. indep.

by transfer.

Step 3. If X = span{f1, . . . , fn} C(), with f1, . . . , fn lin. indep., then we define an internallinear functional X : X C by fi 7 (fi). This functional is independent of the choice of basis.By Hahn-Banach extension, X extends to X (`(N)) with X = X. Note that, forf X, X(f) = X(f) = (f).Step 4. Apply saturation to the family of sets EX,n = { (`(N)) : | | < 1/n f X : (f) = (f)} for X FD(C()) and n N to find (`(N)) with and(f) = (f) for all f C().

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Step 5. By the remark before the proof, we may choose ba(N) such that = . Let bethe restriction of to the internal subsets of H. As

||() ||(N) = = R,

is a finite internal complex measure. Thus st extends to a unique -additive measure L().Put = L() st1, defined on the Borel subsets of . Then M() and, for each f C(),we get

(f) = (f) =

Nf d =

H

f d =

H

st f dL() =H

f st dL() =

f d.

This completes the proof.

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Hahn-Banach theoremsGeneral non-standard hullWeak non-standard hullWeak* topologyRiesz Representation Theorem