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A crash course in p-adic analysis

W. H. Schikhof

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It is a pleasure to make many acknowledgements. First to Mara

Soledad Alcano A. and Mara Eugenia Heckmann G. for typing

the manuscript. My thanks are also due to Carla Barrios R. and

Tonino Costa A., graduate students of the Pontificia Universidad

Catolica de Chile, who made a thorough revision of the typewrit-

ten text, and materially assisted in its preparation.

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INTRODUCTION

In many branches of mathematics and its applications the fields R and C are playing a

fundamental role. For quite some time one has been discussing the consequences of replacing

in those theories R or C by a non-archimedean valued field. This story started in algebraand number theory, where the p-adic fields, discovered by Hensel in 1909, could be used

successfully. Around 1930 analysis entered the picture through the work of Schnirlemann who

developed the basic theory of p-adic power series and analytic functions. From 1940 on the

Dutch mathematician A.F. Monna established the basics on p-adic Functional Analysis. At

this moment there are non-archimedean activities in all feasible disciplines such as Elementary

Calculus, Theory of Cn- and C-functions, p-adic Lie Groups, Analytic Functions in one or

several variables, Algebraic and Analytic Number Theory, Algebraic Geometry, Functional

Analysis, ...

Of course, people are asking for applications.

1. There do exist several applications within mathematics.

2. In 1987 Igor Volovich raised the question as to whether at Planck distances (1034 cm)

space should be disordered or disconnected. He suggested the use of p-adic numbers

to build more adequate quantum mechanical models. Right now serious mathematician

are working in this area.

Ultrametrics also appear speculatively in Psychology, Social Sciences, Economy (stock

market). Attempts are made to build a p-adic Probability Theory. A conference on

these applications was held, for the first time in history, in Moscow, October 2003.

Parts in the text, indicated Backgroundare not needed for the course, but might be interesting

to hear about.

November, 2003.

P. Universidad Catolica de Chile Fac. Matematicas MECESUP PUC-0103 FONDECYT 7020710 3

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Chapter 1

BASICS

1.1 Ultrametric Spaces

In this section we recall well-known facts.

1.1.1 Metrics

A metric on a set X is a map d : X X [0, ) (d = distance) such that for all x,y,z X

(i) d(x, y) = 0 if and only if x = y

(ii) d(x, y) = d(y, x)(iii) d(x, z) d(x, y) + d(y, z) (triangle inequality).

X = (X, d) is called a metric space. (If we relax (i) to just d(x, x) = 0 then we have a

semi-metric d)

For a X, r > 0 we setB(a, r) := {x X : d(x, a) r},

called the closed ball about a with radius r and

B(a, r) := {x X : d(x, a) < r}

called the open ball about a with radius r.

A subset U X is called open if for each a U there exists an r > 0 such that B(a, r) U.The collection of open sets form a topology on X which is called the topology induced by d.

Observe that an open ball is open and that a closed ball is closed.

The diameter of a non-empty set Y X is

diam Y := sup{d(x, y) : x, y Y}


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(possibly ). We set diam := 0. Y is called bounded if diam Y < .The distance between two non-empty sets Y, Z X is

d(Y, Z) := inf{d(y, z) : y Y, z Z}.

For y X we write d(y, Z) rather than d({y}, Z).A sequence x1, x2, . . . in X converges to x X, notation lim

nxn = x, if lim

nd(x, xn) = 0.

A sequence x1, x2, . . . in X is called Cauchy sequence if limm,n

d(xm, xn) = 0.

Exercise 1.A. A Cauchy sequence is bounded. Each convergent sequence is Cauchy, the

converse is not true.

Two metrics on a set X are called equivalent if they induce the same topology.

Exercise 1.B. Let d1, d2 be equivalent metrics on a set X. Prove that a sequence is convergentin (X, d1) if and only if it is convergent in (X, d2), but that a similar statement does not hold

for Cauchy sequences.

A metric space (X, d) is called complete if each Cauchy sequence converges.

Each metric space X can be embedded into its completion X# (a metric space X# that is

complete and contains X as a dense subset).

Let (X1, d1), (X2, d2) be two metric spaces. A map f : X1 X2 is called an isometry ifd2(f(x), f(y)) = d1(x, y) for all x, y X1.

1.1.2 Ultrametrics

A metric d on a set X is called ultrametric (and (X, d) is called an ultrametric space) if itsatisfies the so-called strong triangle inequality

d(x, z) max(d(x, y), d(y, z))

for all x,y ,z X. (Clearly this implies the ordinary triangle inequality). In the spirit ofabove one defines semi-ultrametrics. We have the fundamental

ISOSCELES TRIANGLE PRINCIPLE:

If d(x, y) = d(y, z) then d(x, z) = max(d(x, y), d(y, z)).

Example: Subsets of a non-archimedean valued field (see next Section) with the ultrametric(x, y) |x y|.[BACKGROUND: In fact we have this way all examples, since each ultrametric space can

isometrically be embedded into a non-archimedean valued field (Indag. Math. 46 (1984),

51-53)]

Exercise 1.C. Let (X, d) be a metric space. Then, among all semi-ultrametric that are dthere is a largest one. It is called the subdominant semi-ultrametric for d.


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The next Exercise is essential. For this course it is necessary that you have proved all facts

yourself once in your lifetime.Exercise 1.D. Let (X, d) be an ultrametric space.

(i) Let a Y X. Then diam Y = sup{d(x, a) : x Y}.

(ii) Each ball in X is both closed and open (clopen).

(iii) Each point of a ball is a center. Each ball has an empty boundary.

(iv) The radii of a ball B form the set {r R, r1 r r2}, where r1 = diam B, r2 =dist (B, X\ B) (r2 = if B = X). It may happen that r1 < r2, so that a ball mayhave infinitely many radii.

(v) Two balls are either disjoint, or one is contained in the other.

(vi) If two balls B1, B2 are disjoint, then dist (B1, B2) = d(x, y) for each x B1, y B2.

(vii) Let > 0. The relation d(x, y) < (x, y X) is an equivalence relation and inducesa partition of X into open balls of radius . A similar story holds for d(x, y) andclosed balls.

(viii) Let Y X, B ball in X, B Y = . Then, B Y is a ball in Y.

The topology induced by an ultrametric is zerodimensional, i.e. there is a base of the topology

consisting of clopen sets. Hence, an ultrametric space is totally disconnected.

If (X, d) is an ultrametric space and Y X is dense then

{d(y1, y2) : y1, y2 Y} = {d(x1, x2) : x1, x2 X}.

Thus, completion of an ultrametric space does not create new values of the ultrametric.

1.1.3 Compact Ultrametric Spaces

Exercise 1.E. Let (X, d) be a compact ultrametric space. Show that {d(x, y) : x, y X} iscountable and has only 0 as possible accumulation point.

Let (X, d) be an infinite compact ultrametric space (ifX is finite the process below breaks off).

By the above exercise the set of non-zero values of d is a sequence r1 > r2 > . . . tending to 0.

We shall make a picture of the collection of all balls in X as follows. We have X = B(a, r1)

for any a X. The relation d(x, y) r2 (or d(x, y) < r1 if you want) decomposes X intofinitely many closed balls of radius r2:


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Level 1:

Level 2:

r

hhhhhr

ee

eeer

r

r

X. . . . . .

dots represent balls of radius r2.connecting lines indicate inclusion.

Each of the closed balls of level 2 decomposes into finitely many closed balls of radius r2:

r

hhhhhr

r

ee

eeer

dd

dddr

fffffr

ee

eeer

r

r

r

r

......

r

r

iiiiir

r

. . . . . .

etc. Continuing this way we arrive at a tree.

The dots represent the balls; the infinite paths starting from X, going below represent the

elements of X.

The tree determines X up to homeomorphisms.

Trees appear in social sciences (hierarchical structures), so no wonder why compact ultrametric

spaces enter the scene!

1.1.4 Spherical Completeness

An ultrametric space (X, d) is called spherically complete if each nested sequence of balls

B1 B2 . . . has a non-empty intersection. (It is not required that the diameters of the Bnapproach 0).

It is not difficult to see that a spherically complete ultrametric space is complete. Surprisingly,

the converse is not true, as we will see later.

Exercise 1.F. Prove that a compact ultrametric space is spherically complete.

The next exercise reveals the main reason why attention is devoted to spherical completeness.Exercise 1.G. Let X be a spherically complete ultrametric space embedded in an ultrametric

space Y. Then each y Y has a best approximation in X, i.e.

min{d(y, x) : x X}

exists.


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1.2 Non-Archimedean Valued Fields

Again, most of this section is old stuff.

1.2.1 Valued Fields

Let K be a (commutative) field. A valuation is a map | | : K [0, ) such that for all, K

(i) || = 0 if and only if = 0(ii) || = || ||

(iii) | + | || + || (triangle inequality).Obvious examples are Q, R, C with the ordinary absolute value function. The mapping

(, ) | | is a metric on K making it into a topological field (i.e. the basic arithmeticoperations are continuous).

The metric completion of K is in a natural way again a valued field.

Two valuations on a field K are called equivalent if they induce the same topology.

[BACKGROUND: e.g. G. Bachman, Introduction to p-adic numbers and valuation theory,

Academic Press, New York, 1964: Two valuations | . |1, | . |2 on K are equivalent if and onlyif there is a positive constant c such that | . |1 = | . |2c .]The following theorem essentially separates R, C from all other valued fields. It shows the

alternative character of our non-archimedean analysis.

[BACKGROUND: Bachman, see before, page 127:

Theorem Let(K, | . |) be a valued field. Then, there are only two possibilities. Either(i) K is a subfield of C and | . | is equivalent to the absolute value function, restricted to

K, or

(ii) the valuation | . | is non-archimedean (n.a) i.e. it satisfies the strong triangle inequality

| + | max(||, ||) (, K). ]

So, by excluding C and its (valued) subfields we obtain in return the strong triangle inequality.

1.2.2 Non-Archimedean Valued Fields

In this subsection (K, |.|) is a non-archimedean valued field. The expression non-archimedeanis explained by the fact that for each n N

|n 1| = |1 + 1 + . . . + 1| max(|1|, |1|, . . . , |1|) = 1


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contrasting with the Archimedean Axiom for R stating that N is unbounded.

We recall some facts that you know from a previous life.

The completion of a non-archimedean valued field is non-archimedean valued.

Let (K, |.|) be a non-archimedean valued field. Then K := { K, = 0} is amultiplicative group and so is |K| = {|| : K}; it is called the value group of K.This value group can either be dense in (0, ) or discrete.In the last case, if the valuation is not trivial, := max{r : r |K|, r < 1} exists. Anyelement K for which || = is called a uniformizing element.Furthermore, B(0, 1), the closed unit disk, is a ring and B(0, 1), the open unit disk,

is a maximal ideal in B(0, 1).

The quotient is a field:k := B(0, 1)/B(0, 1)

and is called the residue class field of K.

The p-adic valuation on Q is determined by

|n|p := p( number of factors p in n) (n N).

The completion of (Q, |.|p) is called (Qp, |.|p), the field of the p-adic numbers.Its value group is {pn : n Z}, its residue field is the field Fp of p elements.(Of course, in the above, p is a prime number)

p is a uniformizing element. The fields Q2,Q3,Q5,Q7, . . . are mutually non-isomorphic.

The closed unit disk ofQp is denoted Zp and called the ring of the p-adic integers. (This

name can be explained by the fact that Zp is the closure ofZ).

Zp is compact. More generally, each ball in Qp is compact.

Qp is therefore spherically complete, separable, locally compact.

Qp is not algebraically closed.

|.|p can be extended uniquely to the algebraic closure Qap and the completion of (Qap, |.|p) iscalled Cp, the field of the p-adic complex numbers.

Cp is no longer locally compact, but separable and algebraically closed. Its value group is

{pr : r Q},

so the valuation is dense, and the residue class field is the algebraic closure of Fp, hence

infinite.

Cp is spherically complete! This follows from:


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Theorem 1 Let (K, |.|) be a n.a. valued field. Suppose K is separable and has a densevaluation. Then K is not spherically complete.

Proof There exist r1 > r2 > . . . in |K| such that r := limn rn > 0. Let {s1, s2, . . .} be acountable dense subset. There is a closed ball B1 of radius r1 such that s1 / B1. Since B1decomposes into infinitely many closed balls of radius r2 (why infinite?) there is a closed

ball B2 B1 such that s2 / B2. Going on this way we find a nested sequence B1 B2 . . .of closed balls with radius Bn = rn, and sn / Bn for each n.Let B :=

n Bn. If B = , it would be a closed ball of radius r, hence open so the set

{s1, s2, . . .} must meet B. But on the other hand, by construction {s1, s2, . . .} B = , acontradiction.

FROM NOW ON IN THIS COURSE K = (K,|.|) IS A N.A. VALUED COMPLETE FIELD.

WE ASSUME THAT |.| IS NON-TRIVIAL I.E. THERE IS A K WITH || = 0, || = 1.Exercise 1.H. Show that { K : |1 | < 1} is a multiplicative subgroup of { K :|| = 1}.

1.2.3 Sequences and series in K

We recall a few fundamental facts.

Let a1, . . . , an K and |ai| = |aj | whenever i = j.Then THE STRONGEST WINS:

|a1 + . . . + an| = max{|ai| : 1 i n}.

Let a1, a2, . . . K. This sequence is called summable if limn

ni=1

ai exists. Then

A STUDENTS DREAM COME TRUE:

limn

an = 0 = a1, a2, . . . is summable.

If limn

an = a = 0 then |an| = |a| for large n. Thus, we have in Qp

11 p = 1 +p +p

2 + . . .

Because limn

n! = 0 in every Qp the sumi=0

n! exists in every Qp. The following problem has

been open since 1971.

PROBLEM: Cani=0

n! be rational for some prime p?


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Everybody believes that

i=0

n! is irrational for every p. Several proofs have proposed but

they all failed.

It is not even known ifi=0

n! = 0 in every Qp.Recently Vladimirov (2002) introduced

!n :=n1k=0

k!

and proposed the following related CONJECTURE:

g.c.d.(!n, n!) = 2.

Exercise 1.I. Computen=0

n (n!) in Qp.

1.2.4 Power Series

Like in the archimedean case, given a0, a1, . . . K then

R := (lim sup n

|an|)1

(where 1 := 0 , 01 := ) is called the radius of convergence of the power series

anxn.

Just like in the complex case one proves that a0, a1x, a2x

2

, . . . is summable for |x| < R, notsummable for |x| > R. The behaviour on the boundary |x| = R is much easier that in thecomplex case: a0, a1x , . . . is either summable everywhere on {x : |x| = R} or nowhere. This isbecause the region of convergence {x K : a0, a1x , . . . is summable} = {x K : lim

nanx

n =

0}.Exercise 1.J.

(i) Prove that the region of convergence ofn=0

xn is {x K : |x| < 1}.

(ii) Prove that, in Cp,n=1

xn

n exists if and only if |x| < 1.So, we can define the p-adic logarithm via the formula

logp(1 x) =n=1

xn

n(|x| < 1)

Exercise 1.K. Let n N be written in base p as

n = a0 + a1p + . . . + asps


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where ai {0, 1, . . . , p 1} and as = 0.Define sn (the sum of digits) as a0 + a1 + . . . + as. Show that the number of factors p in n!equals

n snp 1

and prove with the help of this fact that in Cp

exp x :=n=0

xn

n!

is defined for |x| < p 11p and that the series is not summable if |x| p 11p .From now on we denote

{x

Cp :

|x

|< p

1

1p

}by Ep.

Exercise 1.L. By brute force one can prove that

exp(x + y) = exp x exp y for x, y Ep andlogp(xy) = logp x + logp y for |x 1| < 1, |y 1| < 1, and

exp logp x = x (x 1 + Ep)logp exp x = x (x Ep).

More surrealistic is the following. Prove that

| exp x exp y| = |x y| (x, y Ep)| logp x logp y| = |x y| (x, y 1 + Ep).

[BACKGROUND: One can prove that the p-adic logarithm maps {x Cp : |1 x| < 1} ontoCp and that logp x = 0 if and only if x is a root of unity.]

1.2.5 Continuous Functions

Let X be a topological space. The set of all continuous functions X K is a K-vectorspace (even a K-algebra) under pointwise operations, and is denoted C(X K), or C(X)if no confusion about the scalar field is to be expected.

We define the K-valued characteristic function of a subset Y X by

Y(x) :=

1 if x Y0 if x X\ Y

It is easily seen that Y is continuous if and only if Y is clopen.

A function f : X K is locally constant if each point of X has a neighbourhood on whichf is constant. Locally constant functions are of course continuous.


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Theorem 2 Let X be a topological space, f C(X K), > 0. Then there is a locallyconstant function g : X K such that |f(x) g(x)| < for all x X.Proof(Compare Exercise 1.D (vii)) The relation |f(x)f(y)| < (x, y X) is an equivalencerelation and yields a partition of X into clopen sets: X =

iI

Ui.

Choose ai Ui for each i and define

g(x) := f(ai) whenever i I, x Ui

This g does the job.

1.2.6 Differentiability

Now we take for X a subset of K. To avoid problems, assume X is without isolated points.

A function f : X K is called differentiable at a X if

f(a) := limxa

f(x) f(a)x a

exists (you can imagine how to define limxa

). In the same spirit we define (everywhere) dif-

ferentiable function, derivative. The well-known rules for differentiation of sums, products,

quotients, compositions (chain rule) hold also in the n.a. case. The proofs are classical.

A differentiable function is continuous. Rational functions without poles in X are differen-

tiable.

To be able to define analytic functions properly we need the following exercise on double

sequences.

Exercise 1.M. For m, n N, let amn K. We say that limm+n amn = 0 if, for each > 0, the set {(m, n) N N : |amn| } is finite. Show that the following statements(), (), () below are equivalent.

() limm+n amn = 0.

() For each n, limm amn = 0; limn amn = 0 uniformly in m.

() For each m, limn amn = 0; limm amn = 0 uniformly in n.

If () () hold then

m=1

n=1

amn andn=1

m=1

amn both exist and are equal.

Lemma 3 Suppose a power series in x a converges in a neighbourhood of a K:

f(x) :=n=0

an(x a)n (x B(a, r)) .


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Then, for each b B(a, r), f can be expanded into a power series in x b on the same ball:there are b0, b1, . . . K such that

f(x) =n=0

bn(x b)n (x B(b, r) = B(a, r)).

Proof We may assume r |K|. Then limn

|an|rn = 0.Put

(x a)n = (x b + b a)n =n

k=0

n

k

(x b)k(b a)nk

and

tkn := ann

k

(x b)k

(b a)nk

.

Now clearly limk tkn = 0 for each n and since

|tkn| |an|

n

k

rkrnk |an| rn(nk

is an integer, hence

nk

1)we have that lim

ntkn = 0 uniformly in k. So, by the exercise we have

f(x) =

n=0

k=0

tkn =

k=0

n=0tkn =

k=0

bk(x b)k

where bk :=n=0

annk

(b a)nk.

Definition Let B be a closed ball in K with radius r |K|. A function f : B K iscalled analytic if there exists an a B and a0, a1, . . . K such that

() f(x) =n=0

an(x a)n (x B).

With the help of Lemma 3 one easily derives that sums and products of analytic functions on

B are analytic. Analytic functions are differentiable. If f is as in () then

f(x) =n=1

nan(x a)n1.

If f is a complex analytic function defined on some bounded domain the maximum principle

states that |f| takes its maximum on the boundary. A similar result holds also for our n.a.case; we include a proof since it is completely different from the complex proof.


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Theorem 4 (Maximum Principle) Let K = Cp and B,r,f,a,an be as in the above defi-

nition. Then |f| attains its maximum on B and, moreover,max{|f(x)| : x B} = max{|f(x)| : |x a| = r}

= max{|an|rn : n {0, 1, 2, . . .}}.

Proof We only consider the case where B = B(0, 1). (The reader can furnish the missing

details). Thus, we have

() f(x) =n=0

anxn (|x| 1).

We clearly have|f(x)

| maxn

|anx

n

| maxn

|an

|.

To prove that max{|f(x)| : |x| 1} max |an| we may assume that maxn |an| = 1 andderive a contradiction from the assumption |f(x)| < 1 for all x B(0, 1). Let be thehomomorphism B(0, 1) k (where k is the residue class field). Then from (**) we obtain

f(x) = 0 =Nn=0

anxn (x k)

for some N (since |an| < 1 for large n), for which aN = 0. Thus, a nonzero polynomial in k[x]has each element of k as a root. But k is infinite, a contradiction.

Thus we have max{|f(x)| : |x| 1} = maxn |an|. To prove that also max{|f(x)| : |x| = 1} =maxn |an| observe that |f(x)| < 1 for all |x| = 1 yields

0 =Nn=0

anxn

for all x k \ {0}. Now reason as above, using that k \ {0} is infinite as well.[BACKGROUND: Krull Valued Fields.

When looking at the requirements for a non-archimedean valuation:

(i) |x| = 0 if and only if x = 0

(ii) |xy| = |x| |y|(iii) |x + y| max(|x|, |y|)

one notices that, unlike for the archimedean triangle inequality, addition of real numbers

does not play a role; one only needs multiplication and ordering. This leads to the following

generalized concept of valued fields.

Let G be a commutative, multiplicatively written, totally ordered group such that


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Now let F := R(X1, X2, . . .) be the field of rational functions with real coefficients in countably

many variables X1, X2, . . .. It is easily seen that the requirements

|0| : = 0|| : = 1 if R \ {0}|Xn| = (1, 1, . . . , gn

n-th place

, 1, 1, . . .)

determine a Krull valuation | | : F G {0}.The completion of (F, | |) can be constructed in the natural way and leads to a completeKrull valued field L. To see a striking difference with the ordinary valuation, consider the

element X11 F. Then |X11 | = (g11 , 1, 1, . . .), so 0 < |X11 | < 1. But Xn1 does not tendto 0, as |Xn1 | = (gn1 , 1, 1, . . .) > (1, g12 , 1, . . .) > 0 for all n!We will return to this subject later on.]

1.3 Normed and Banach spaces

1.3.1 Normed Spaces

Let E be a vector space over K. A norm on E is a map : E [0, ) satisfying(i) x = 0 if and only ifx = 0,

(ii) x = ||x,(iii) x + y max(x, y)

for all x, y E, K.(Requirement (iii) is not a logical necessity but there are good reasons for assuming it, see

also Exercise 1.N).

A complete normed space is, as customary, called a Banach space.

Two norms .1 and .2 on a K-vector space E are called equivalent if the metrics (x, y) x y1 and (x, y) x y2 induce the same topology. The following Proposition can beproved as in the archimedean case.

Proposition 5 Two norms on a K-vector space E, say 1 and 2, are equivalent if andonly if there exist constants 0 < c < C such that

c 1 2 C 1.On a finite-dimensional space all norms are equivalent, and the space is a Banach space with

respect to each norm.


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Examples (See also page 22)

1. Let e1, . . . , en be a base of an n-dimensional space over K. Thenni=1

iei maxi |i|is a norm.

2. Let l be the space (pointwise operations) of all bounded sequences in K. For x =

(1, 2, . . .) l, definex := sup{|n| : n N}.

With this norm l is a Banach space.

3. c0 := {(1, 2, . . .) l, limn n = 0} is a closed subspace of l, hence a Banachspace.

4. Let X be a topological space, let BC(X) be the space of all f C(X) that are bounded.It is a Banach space with respect to the supremum norm

f f := sup{|f(x)| : x X}.

5. Let M be a complete valued field containing K as a subfield. Then, M is a Banach

space over K. Thus, in particular, Cp is a Banach space over Qp. Observe that |Cp|strictly contains |Qp| showing that nonzero vectors cannot always be normalized. Thisleads to the following open problem.

PROBLEM (1960) Let (E, ) be a Banach space over K. Does there exist an equivalentnorm such that E |K|, i.e. for each nonzero x E there exists a K suchthat x = 1?There are some partial answers and reductions, but the answer is so far unknown! A

good test case would be l over Cp.

Exercise 1.N. A-NON-EXAMPLE.

(i) Prove that l2 := {(1, 2, . . .) KN :n=1

|n|2 < } is a K-vector space.

(ii) Show that x x :=

n=1

|n|2 satisfies (i), (ii) of page 17 and

x + y x + y (x, y l2)

but not the strong triangle inequality (iii) of page 17.

(iii) Compute the largest norm on l2 that is (compare Exercise 1.C)


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Exercise 1.O. Let (E, ) be a normed space. Show that E\{0} := {x : x E, x = 0}is a union of multiplicative cosets of |K| in (0, ), and that, conversely, each such set equalsE \ {0} for some normed space (E, ).Consider in a normed space (E, ) the ball with radius r > 0 about 0:

B(0, r) := {x E : x r}.Clearly x B(0, r), K, || 1 = x B(0, r), but also because of the strong triangleinequality, x, y B(0, r) = x + y B(0, r). So, in algebraic language, B(0, r) is a moduleover the ring { K : || 1}! Such submodules are called absolutely convex.[BACKGROUND: a seminorm on a K-vector space E is a map q : E [0, ) satisfyingq(x) =

|

|q(x), q(x + y)

max(q(x), q(y)) but q(x) is allowed to be 0 for nonzero x. Like

in the classical case one can define a locally convex topology on E as the weakest topologyfor which each member of a collection P of seminorms is continuous. A subbase of zeroneighbourhoods is formed by the sets {x E : q(x) < } where q P, > 0. By the aboveobservation these sets are absolutely convex. There is an extensive theory on locally convex

spaces on K.]

[BACKGROUND: Normed spaces over Krull valued fields?

Let (K, | |) be a Krull valued field with value group G. A norm on a K-vector space Eshould satisfy (i), (ii), (iii) of page 17 but it is not at once clear what a natural home for norm

values should be. Only admitting G {0} turns out to be too restrictive. This leads to theconcept of a so-called G-module; this is an ordered set X on which G acts monotonically. For

full details see p-adic Functional Analysis, Lecture Notes in Pure and Applied Mathematics

207, 233-293. Marcel Dekker. New York (1999).

Remark So far, nobody has attempted constructing a theory of locally convex spaces over

Krull valued fields!]

Let us return to the basic theory.

Proposition 6 Let E, F be normed spaces over K, let T : E F be linear. Then, if T iscontinuous at 0 then T is Lipschitz i.e. there is an M > 0 such thatT x T y Mx yfor allx, y E.

Proof There is a > 0 such that x E, x implies T x 1. Choose K, 0

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with M := ||11. Then by linearity, for x, y E, T x T y = T(x y) Mx y.[BACKGROUND: In theory of normed spaces over Krull valued fields (*) is not always possible

and there are, indeed, continuous linear maps that are not Lipschitz!]

Let L(E, F) be the space of all continuous linear operators E F. For each T L(E, F),among all M 0 such that T x Mx for all x E there is a smallest one which is calledT.We have T x Tx for all x E and the map T T is a norm on L(E, F).We write L(E) for L(E, E) and E (dual space) for L(E, K), where K is normed by thevaluation | |.Exercise 1.P. In complex functional analysis the dual space of c0 is isomorphic to l

1. In the

non-archimedean theory the situation is radically different:

For x c0, y l, say x = (1, 2, . . .), y = (1, 2, . . .), put

< x, y >:=

n=1

nn

(Show that the sum exists!). For each y l define T y c0 by

(T y)(x) :=< x, y > (x c0).

Prove that, indeed, T y c0 and that T maps l linearly and isometrically onto c0. Hence,in popular form: the dual of c0 is l

.

Like in the classical theory one can prove that L(E, F) is a Banach space as soon as F is aBanach space. Hence, E, E, E, . . . are all Banach spaces.

Here is a construction that works in both classical and n.a. context, but that you may not

have seen yet: forming of quotients.

Let E be a normed space over K, let D E be a closed subspace. We define a natural normon E/D as follows. Let : E E/D be the canonical map (assigning to each x E thecoset x + D). Set

(x) := dist(x, D) = inf{x d : d D}.

Straightforward verification shows that z = ||z, z + u max(z, u) for all z, u E/D, K. Suppose z = 0; we prove that z = 0.z has the form x + D, so z = inf{x d : d D} = 0. Hence, there are d1, d2, . . . Dwith limn x dn = 0 i.e., x = limn dn. Since D is closed, x D which implies thatx + D is the zero element of E/D.

Theorem 7 If E is a Banach space, D is a closed subspace then E/D is a Banach space.


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ProofLet z1, z2, . . . be Cauchy in E/D. Then limn(zn+1 zn) = 0, so there are v1, v2, . . .in Esuch that (v1) = z1, (vn+1) = zn+1zn (n 1), and limn vn = 0. Then, x :=

n=1

vn

exists in E and (x) =n=1

(vn) = limn zn.

The Uniform Boundedness Principle, the Banach Steinhaus Theorem, the Closed Graph The-

orem and the Open Mapping Theorem all rest on the Baire Category Theorem and some

linearity considerations and therefore remain valid in our n.a. theory.

We like to conclude this introductory chapter by signalling two striking differences with the

classical case.

1.3.2 Operators with an empty spectrum

Let E be a Banach space, T L(E). Like in the classical case one defines the spectrum of T,(T) as

{ K : T I has no inverse }(If dim E < , (T) is the collection of eigenvalues). A famous classical result is that (T)is non-empty if the scalar field is C.

Now take the Banach space M of Example 5 in page 18 and let a M\ K and putT x := xa (x M)

Then T

L(M) but for

K we have (T

I)x = (a

)x, hence T

I has an inverse

x (a )1x. Hence, (T) = .PROBLEM. Develop a reasonable spectral theory for operators on a Banach space.

(So far, only partial results have been obtained for so-called compact operators.)

1.3.3 Commutation Relations

It is well-known that in a complex Banach space E there do not exist A, B L(E) satisfying() AB BA = I

an important relation in Quantum Mechanics. Turning to the non-archimedean situation, let

E := c0. In 1996, A. Khrennikov and A. Kochubei independently found the following. Set

A(1, 2, . . .) = (2, 3, . . .)

B(1, 2, . . .) = (0, 1, 22, 33, . . .).

One verifies immediately that A, B L(E) (B is bounded since |nn| |n|!) and that (*)holds.

PROBLEM. Try to find all solutions of (*). (Last October 2003 I asked various experts about

this but nothing seems to be known apart from the above example!)


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1.3.4 Addendum

On page 18 I should have included the following example of a space of analytic functions. Let

K = Cp (more generally, let K have infinite residue class field), let r |K| and consider thespace Hr of all analytic functions B(0, r) K (see page 14).From the maximum principle (Theorem 4) it follows that for f Hr, f(x) =

n=0

anxn (|x| r):

f := max{|f(x)| : |x| r} = maxn

|an|rn

(so the sequence a0, a1, . . . is uniquely determined by f) and the space Hr is (linearly) iso-

metrically isomorphic to {(a0, a1, . . .) : |an|rn 0}, with the max norm.For r = 1 we see that H1

= c0. (Show that also Hr

= c0 if r

|K

|.) So, Hr is a Banach

space.

COROLLARY. The uniform limit of a sequence of analytic functions on B(0, r) (r |K|) isagain analytic.

Exercise 1.Q. Show that on Hr is multiplicative i.e. for f, g Hr

f g = fg

Hence, the ring Hr has no zero divisors!


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Chapter 2

ORTHOGONALITY

2.1 Definition of orthogonality

The finest examples of classical Banach spaces are Hilbert spaces, so one may look for a non-

archimedean pendant.

An inner product ( , ) in a space over C satisfies(i) x (x, y) is linear in x, for each y,

(ii) (x, y) = (y, x),

(iii) (x, x) > 0 whenever x = 0.In the non-archimedean case we replace complex conjugation by a field automorphism for which = and || = || for all K. (We allow to be the identity). As K is notordered, we replace (iii) by (x, x) = 0 whenever x = 0. So, on a K- vector space E, let us saythat a map ( , ) : E E K is an inner product if

(i) x (x, y) is linear for each y E

(ii) (x, y) = (y, x) for all x, y E

(iii) (x, x) = 0 whenever x = 0, x E.

Exercise 2.A. Requirement (iii) seems to be a kind of arbitrary relaxation of (iii), born out

of emergency. But it is not such a wild generalization as one may think. In fact, prove the

following. In a vector space over C, any form ( , ) satisfying (i), (ii) and (iii) is either aninner product or else ( , ) is an inner product!For E as above and x E we put

x =

|(x, x)|.


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(Notice the valuation signs, compare with Exercise 2.A.)

From this point we could proceed with the theory like in classical theory. However, we willnever reach the goal we had in mind, according to the following theorem which we state

without proof.

Theorem 8 Let( , ) be an inner product on a Banach space E over K such that |(x, x)| =x2 for allx E. Suppose that each closed subspace has an orthogonal complement. Thendim E < .

[BACKGROUND: Theorem 8 is not the last word that is to be said about the subject. The

picture changes completely if we allow Krull valuations on the scalar field. In fact, let L be

the completion of the Krull valued field R(X1, X2, . . .) as constructed on pages 1617. Put

X0 = 1. Let E be the L- vector space of all sequences (0, 1, . . .) LN for which i=0 2i Xiconverges in L. For x, y E, say x = (1, 2, . . .) and y = (1, 2, . . .) put

(x, y) :=i=0

iiXi.

H. Keller showed in 1980 that ( , ) is an inner product in the sense of (i), (ii), (iii) of page 23(where = for all L), but, what is more, he proved has every closed subspace D ofEhasan orthogonal complement, i.e. D + D = E when D = {x E : (x, y) = 0 for all y D}.The study those non-classical Hilbert spaces is in full progress.]

So, for our theory to develop, inner products do not seem appropriate. But we do have apowerful non-archimedean concept, valid in any normed space:

Definition Let x, y be elements of a normed space E over K. We say that the vector x is

(norm-)orthogonal to y if the distance of x to the space Ky is precisely x:

x = min{x y : K}

(so the minimum is attained for = 0. Draw a picture). We denote this by x y.

Before continuing we state

VAN ROOIJS PRINCIPLE:

If x, y E, x y x, then x y y.

The proof is obvious.

It can be used for

Proposition 9 If x y then y x.


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Proof We only have to prove that y x y for all K. This is clear for = 0. If = 0 we have y x = || x 1y, which by x y is ||x = x.By van Rooijs Principle, y x y.It is the symmetry that makes important. It is revealed in the following alternative defini-tion.

Proposition 10 x y if and only if for each , K

x + y = max(x, y).

Proof Left to the reader.

[BACKGROUND: The relation

can in the same way be introduced in archimedean normed

spaces, so one may wonder why you have never met it in the courses you took. The reason is

that in classical theory is not symmetric. Essentially the only spaces where is symmetricare inner product spaces and there x y is equivalent to (x, y) = 0.]The following two Exercises show that one has to be careful by not automatically assuming

form orthogonality properties to hold for our norm orthogonality.

Exercise 2.B. Let K2 be normed by

(1, 2) max(|1|, |2|).

Prove that (1, 0) (0, 1) (not surprising), but also that (1, 1) (1, 0) and (1, 1) (0, 1).Exercise 2.C. Let E be a normed space. Prove that (not orthogonal) is an equivalencerelation on E\ {0}. Thus, if x y and x z then y z!Inspired by Proposition 10 we now define:

Definition A sequence e1, e2, . . . of non-zero vectors in a normed space E is called orthogonal

if for each n N, 1, . . . , n Kni=1

iei

= max{iei : 1 i n}(i.e. each ei is orthogonal to every vector in the linear span of

{ej : j

= i

}). The sequence is

orthonormal if, in addition, en = 1.The difference with classical orthogonality is demonstrated again in the following

Theorem 11 (Perturbation theorem) Let e1, e2, . . . be an orthogonal sequence in a nor-

med space E and let f1, f2 . . . E be such thatfn en < en for all n. Then f1, f2, . . . isorthogonal.


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Proof Choose 1, . . . , n

K. Set x =n

i=1

iei, y =n

i=1

ifi .

By assumption we have fi = ei for all i. Hence, yx = ni=1

i(fiei) maxi

|i|fiei < (if not all i = 0) < maxi |i|ei = x. It follows that y = x = maxi |i|ei =maxi |i|fi.

2.2 Orthogonal bases

In this section E is a Banach space over K. It is not hard to see that, for an orthogonal

sequence e1, e2, . . . and 1, 2, . . .

K with

nen

0 we have

i=1

iei

= maxi iei.

Definition e1, e2, . . . is called an orthogonal base for E ifx E has an expansion x =i=1

iei

for some suitable i K.Clearly such an expansion is unique.

Proposition 12 If e1, e2, . . . is orthogonal and [e1, e2, . . .] (K-linear span) is dense in E then

e1, e2, . . . is an orthogonal base.Proof Let K, 0 < || < 1. Then there exists 1, 2, . . . K such that || nen 1for all n N. For (1, 2, . . .) c0, put T(1, 2, . . .) =

n=1

nnen.

One verifies immediately that T is a linear map c0 E, that, for x = (1, 2, . . .) c0:

||x = || maxn

|n| T x x,

so that T is a homeomorphism, hence T c0 is complete. But also T c0 contains all finite linear

combinations of e1, e2, . . ., hence T c0 is dense. It follows that T c0 = E, which is what we

wanted to prove.

The most famous example of an orthogonal base is the so-called Mahler base ofC(Zp Cp):For x Zp and n {0, 1, 2, . . .} we set

en(x) :=

x

n

:=

x(x 1) . . . (x n + 1)n!

Then we have


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1. en is a polynomial of degree n,

2. en(m) = 0 if n > m, en(n) = 1,

3. en(1) = (1)n,

4. en = 1 for all n,

5. e0, e1, . . . is an orthonormal sequence.

(Proof. Let 0, 1, . . . , n K. Then by taking the value at 0, we get

0e0 + . . . + nen |0e0(0)| = |0|e0.

By van Rooijs Principle,

0e0 + . . . + nen 1e1 + . . . + nen.

By substituting the value x = 1 we get:

1e1 + . . . + nen |1e1(1)| = |1|e1.

Continuing this way we obtain

0e0 + . . . + nen max{iei : 0 i n}= max

{|i|

: 0

i

n}

).

To conclude the proof that e0, e1, . . . is an orthonormal base some work has to be done. First,

we need

Exercise 2.D. (Continuation of Exercise 1.K)

(i) Show that for j {1, 2, . . . , pn}

pn

j

= pn.(ii) Show that spj = sj for each j N (here again sm is the sum of digits of n in base p)

(iii) Use (i) and (ii) for

pn

j

= |pn||j| (j {1, 2, . . . , pn}).To get an idea for the proof we imagine that f C(Zp Cp) has an expansion. Then whatwould be the candidates for the coefficients? So suppose f(x) = a0

x0

+ a1

x1

+ . . .

By taking x = 0 we find f(0) = a0.

By taking x = 1 we find f(1) = a0 + a1.


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By taking x = 2 we find f(2) = a0 + 2a1 + a2

...So, a0 = f(0), a1 = f(1) f(0), a2 = f(2) 2f(1) + f(0).In general we proceed as follows: Let E : C(Zp Cp) C(Zp Cp) be the shiftoperator

(Ef)(x) = f(x + 1) (x Zp).Then

(Ef)(x) =n=0

an

x + 1

n

.

By using

x + 1n

=x

n

+ x

n 1 (n 1)we find

(Ef)(x) =n=0

an

x

n

+

n=1

an

x

n 1

= f(x) +n=0

an+1

x

n

so that

(E

I)f(x) =

n=0

an+1xn

(I =identity operator). By applying E I k times

(E I)kf(x) =n=0

an+k

x

n

and by putting x = 0:

(E I)kf(0) = ak.

Because (E I)n =n

k=0

n

k

(1)nkEk we have

(*) an =n

k=0

n

k

(1)nkf(k)

and

(E I)kf(x) =n

k=0

n

k

(1)nkf(x + k).


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Since

0 = (1 1)n =n

k=0

nk

(1)nk =

nk=0

nk

(1)nkf(x)

we get

(**) (E I)nf(x) =n

k=1

n

k

(1)nk(f(x + k) f(x)).

We will prove that for f C(Zp Cp), (E I)nf 0.Since f (EI)f (EI)2f . . . it suffices to show that (EI)pnf 0.Let > 0. By uniform continuity off there is an l such that |k| < pl implies |f(x+k)f(x)| < for all x.

Choose n so large that pn < pl. Let k {1, . . . , pn}.If |k| < pl then

pn

k

(1)pnk(f(x + k) f(x))

< .If |k| > pl then

pn

k

(1)pnk(f(x + k) f(x))

pn

k

f |pn||k| f (by Exercise2.D.) < pn+lf < f. Thus, we have proved that (E I)nf 0, using ().From this fact and () it follows that an 0 for every f C(Zp Cp). This means that

g : x n=0

an

x

n

is a well-defined continuous function on Zp.Inductively, from (EI)nf(0) = (EI)ng(0) for each n, we find f(0) = g(0), f(1) = g(1), . . ..Since N is dense in Zp we have f = g. So we have proved:

Theorem 13 (Mahler) The functions e0, e1, . . . given by

en(x) =

x

n

form an orthonormal base of C(Zp Cp). If f C(Zp Cp) has the expansion

f =

n=0 anen

then

an =n

k=0

n

k

(1)nkf(k).

Corollary 14 (Weierstrass, p-adic version) Continuous functions on Zp can uniformly

be approximated by polynomial functions.


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[BACKGROUND: One can prove the following generalization.

STONEWEIERSTRASS: Let X be a compact topological space, let H be a subalgebra ofC(X K) that contains the constants and separates the points of X. Then H is uniformlydense in C(X K).]Exercise 2.E. (Another application: the indefinite sum).

In classical analysis a function of the form x xa f(t)dt is often called indefinite integralof f. Show that in the p-adic case we have an indefinite sum for an f C(Zp Cp):There is an F C(Zp Cp) such that

F(n) = f(0) + f(1) + . . . + f(n 1) (n 1)F(0) = 0.

Also prove the next

Corollary 15 (No p-adic Haar integral) Let C(Zp Cp) have the property that(f1) = (f) (f C(Zp Cp), where f1 is the shift f1(x) = f(x + 1). Prove that = 0.

Exercise 2.F. Let Cp, || < p1

1p . Then exp x is defined for all x Zp. Show that itsMahler expansion is

n=0

(exp 1)n

x

n

.

[BACKGROUND: One can prove that if a Banach space E has an orthogonal base then so hasevery closed subspace D. However it is not always true that D has an orthogonal complement

i.e. a closed subspace S with D + S = E and D S. It depends on K and E.]In the next Chapter we will use the following version of approximate orthogonality.

Definition Let t (0, 1]. A sequence e1, e2, . . . in a normed space is called t-orthogonal if forall n N, 1, . . . , n K

ni=1

iei

t max{iei : i {1, . . . , n}}.In the same spirit we have the notion of t-orthogonal base of a Banach space.

1- orthogonal = orthogonal.If e1, e2, . . . is a t-orthogonal base of E and we introduce a new norm by

n=0

nen

= maxn

nen

then e1, e2, . . . is an orthogonal base of E with respect to , and the norms and are equivalent.


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Thus, the fact that E has a t-orthogonal base for some t (0, 1] means nothing else but: thereis an equivalent norm for which E has an orthogonal base.Exercise 2.G. Verify the t-version of Van Rooijs Principle: If t (0, 1] and x y txthen x y ty.Prove also a t-version of Proposition 12.


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Chapter 3

COUNTABILITY

3.1 Introduction

A topological space is called separable if it has a countable dense subset. If we have a metric

space that is separable then any subset is also separable.

In classical Functional Analysis some role is played by the separable Banach spaces.

In our n.a. case separability is not the right notion to work with. In fact, if K happens to be

non-separable then each onedimensional space has the same property.

To overcome this problem one has linearizedthe definition of separability as follows.

Definition A normed space E is called of countable type if it has a countable subset whose

linear hull is dense.

It is an easy exercise to show that if K itself is separable then a normed space over K is of

countable type if and only if it is separable.

It has been an open question for quite some time as to whether each separable Banach space

over C has a Schauder base e1, e2, . . . i.e. each vector has a unique expansionn=1

nen, where

n C. Only in 1974 Enflo proved that the answer was negative.Further, in the complex case there are many non-isomorphic separable Banach spaces e.g. lp

for 1 p < . In the non-archimedean case the situation is much simpler as we will shownow.

Lemma 16 Every finite-dimensional space has, for each t (0, 1), a t-orthogonal base.ProofLet Ehave dimension n, let x1, . . . , xn be an algebraic base for E. Choose t1, . . . , tn1 (0, 1) such that their product t1t2 . . . tn1 t (e.g. ti = n1

t).

Set e1 = x1, D1 := [e1] = [x1] ([] indicates linear span). Now D1 is closed, x2 / D sodist(x2, D2) > 0.


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leading to

1e1 + . . . + nen tntn1n1en1.Inductively we arrive at

1e1 + . . . + nen tnnen1e1 + . . . + nen tntn1n1en1

...

1e1 + . . . + nen tntn1 . . . t11e1.

Now tn, tntn1, . . . are all t and we obtain

1e1 + . . . + nen t max{iei : 1 i n},

which was to be shown.

Remark. One can prove that, if K is spherically complete, all distances in the above proof

are attained and hence that each finite-dimensional space has an orthogonal base! But, ifK is

not spherically complete one can construct a two-dimensional space without orthogonal base.

[It can be done with the knowledge built up so far. In fact, let B(1, r1) B(2, r2) . . . beballs in K with empty intersection. Define for (1, 2) K2

(1, 2) = limj

|1 2j |

(show first that this limit exists!). Verify that is a norm in K2 and that (1, 2) |K|for all (1, 2) K2. Show that (1, 2) (1, 0) = 1 = 2 = 0. Finally, show that twoarbitrary non-zero vectors in K2 are not orthogonal.]

Theorem 17 Let t (0, 1). Then each Banach space of countable type has a t-orthogonalbase.

Proof Let E be an infinite-dimensional Banach space of countable type.

Choose t1, t2, . . . (0, 1) such thati=1

ti t (show that this can be done). Let x1, x2, . . . be

such that [x1, x2, . . .] is dense in E. We may assume that x1, x2, . . . are linearly independent.By continuing the induction of the proof of Lemma 16 to infinity we obtain e1, e2, . . . E suchthat [e1, e2, . . .] = [x1, x2, . . .] is dense in E and such that e1, . . . , en is t1 . . . tn- orthogonal for

each n. Then it follows that e1, . . . , en is t-orthogonal for each n i.e. e1, e2, . . . is t-orthogonal.

Now apply Exercise 2.G.

Corollary 18 Each infinite-dimensional Banach space of countable type is linearly homeo-

morphic to c0.


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A projection is an operator P L(E) with P2 = P. If P = 0 then P 1. P is anorthoprojection ifP 1. Then Im P Ker P and Im P Ker P = E.Corollary 19 Let E be a Banach space of countable type, let D E be a closed subspace.Then D and E/D are of countable type. For each > 0 there is a projection P with P E = D

and P 1 + . Let 0 < t < t < 1. Then each t-orthogonal base of D can be extended to at-orthogonal base of E.

Proof By Theorem 7 the space E/D is Banach. That it is of countable type is obvious. By

Theorem 17 it has a

t/t-orthogonal base g1, g2, . . .. Let : E E/D be the canonicalmap. By definition of the quotient norm there are f1, f2, . . . E with (fi) = gi and fi

(t/t)gi for each i. It is easily seen that the mapQ :

i=1

igi i=1

ifi (igi 0)

is in L(E/D,E) and that Q is the identity on E/D, and that Q (t)1t. Then putP := I Q ; one verifies that P (t)1, that P is a projection onto D. Finally, ife1, e2, . . . is a t-orthogonal base of D then e1, f1, e2, f2, . . . is easily seen to be a t

-orthogonal

base of E.

Corollary 20 ((1 + )-HAHN-BANACH) LetE be a Banach space of countable type, let

D be a subspace, let f

D, let > 0. Then f can be extended to an f

E such that

f (1 + )f.Proof We may assume that D is closed. Then choose f := f P, where P is a projectiononto D with norm 1 + .[BACKGROUND: This is probably the strangest proof of the Hahn-Banach you ever encoun-

tered! The classical proof for real scalars uses the fact that a collection of closed bounded

intervals with the finite intersection property has a non-empty intersection. If K is spheri-

cally complete one can use this proof, with obvious modifications, to arrive at a Hahn-Banach

Theorem, the one you are used to (E arbitrary, = 0. See next Chapter). But ifK is not

spherically complete then taking = 0 in Corollary 20 leads to a falsity. In fact one can prove

that in K2

, normed like in Remark, page 34, the linear function f : (, 0) cannot beextended to an f (K2) with f = 1].


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Chapter 4

DUALITY

Theorem 21 (HAHN-BANACH THEOREM) Let K be spherically complete, let E be

a normed space over K, letD be a subspace and f D. Thenf can be extended to an f Esuch that f = f.Proof Like in the classical case, through Zorns Lemma it suffices to prove it for the case

E = D + Ka where a E\ D. By linearity, if we fix := f(a) then f is determined:

a + d + f(d) ( K, d D).

The requirement f f means

(*) | + f(d)| fa + d ( K, d D).

We notice that (*) is true for = 0. Then, to have (*) for = 0 it suffices to have (*) for = 1:

| + f(d)| fa + d (d D).Thus

B(f(d), fa + d) (d D).To be able to choose this way we need that the balls have a non-empty intersection. By

spherical completeness it suffices to prove that each two of them have a non-empty intersection,

i.e. we have to see if two centersf(d1), f(d2) have distance the maximum offa +d1, fa + d2. But that is true:

|f(d1) f(d2)| fd1 d2 fd1 + a a d2 max(fa + d1, fa + d2).

For each normed space E over K we have, as in the complex case, the canonical map jE :

E E given byjE(x)(f) = f(x) (f E, x E).


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Exercise 4.A. (Has nothing to do with non-archimedean aspects). Prove that jE is continu-

ous, linear and that jE 1. Also, let T L(E, F), where F is a second normed space, letT : F E be defined as T(f) := f T (f F). T is called the adjoint of F. Show thatthe diagram:

E E FT

c cE FE

T

jE jF

is commutative.

Exercise 4.B. Apply Corollary 20 and Theorem 21 for one-dimensional subspaces D to provethe following: Let E be a Banach space of countable type or let K be spherically complete.

Then the canonical map jE : E E is isometrical.A normed space E is called reflexive if jE is an isometrical bijection E E. Reflexivespaces are complete.

One proves easily that finite - dimensional spaces are reflexive. We first study reflexivity for

spherically complete K.

Lemma 22 Let K be spherically complete, let E be a reflexive Banach space over K. Then

every closed subspace is also reflexive.

Proof Let D be a closed subspace, let i : D E be the inclusion and : E E/D thequotient map.

By Exercise 4.A. we have the commutative diagram:

D E Ei E E/D

cc cD EE

iE

(E/D)

jD jE jE/D

Observe that

i = 0, hence

i = 0.

By Exercise 4.B., we only have to prove that jD is surjective. Let D. Then i() E.By reflexivity of E there is an x E such that jE(x) = i(). Applying at both sides wefind 0 = i() = jE(x) = jE/D(x). By injectivity of jE/D we have (x) = 0 i.e. x Di.e. x = i(d) for some d D. So i jD(d) = i() i.e. jD(d) Ker i. Now observethat by the Hahn- Banach Theorem i is surjective so that i is injective and it follows that

= jD(d): SURJECTIVITY !


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(B(x1, r1 )) = B(y1, r

1 ). Then we can choose an x2 B(x1, r1 ) with (x2) = y2 and form

B(x2, r2 ), whose image under is exactly B(y2, r2 ), etc. By spherical completeness of Ethere is an x

nB(xn, r

n ). Then (x)

n

B(yn, rn ).

Lemma 26 (ANTI-HAHN BANACH THEOREM) LetE be a Banach space over K.

If E is spherically complete and K is not then E = {0}.Proof Let f E, f = 0. Decompose f as follows.

(On E/Ker f the canonical norm. is the unique map making thediagram conmute.)

E E Kf

ee

ef1

E/Ker f

!

By Lemma 25, the space E/Ker f is spherically complete. But is a linear homeomorphism

between two one-dimensional spaces viz. E/Ker f and K. Then there is a constant c > 0

such that |(x)| = cx for all x E/Ker f. Hence, maps balls onto balls and so K mustbe spherically complete as well. Contradiction.

Proposition 27 For any K, the space l/c0 is spherically complete.

Proof Let : l l/c0 be the quotient map. For x = (1, 2, . . .) l we have

(x) = limn

sup(|n|, |n+1|, . . .)

(verify this!) Like in the proof of Lemma 25 we can find, for given balls B(y1, r1 )

B(y2, r2 ) . . . in l/c0, balls B(x1, r1 ) B(x2, r2 ) . . . in l such that (xn) = yn,

(B(xn, rn )) = B(yn, r

n ) for each n. Now write

x1 = (x11, x12, . . .)

x2 = (x21, x22, . . .)

...

and take the diagonal sequencea := (x11, x22, . . .).

Clearly |xnn| xn for all n, so a supn xn < and we see that a l. Weclaim that (a) B(yn, rn ) for each n (which will finish the proof), i.e., we have to show


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(a) (xn) < rn. Now(a) (xn) = lim

ksup(|xkk xnk|, |xk+1,k+1 xn,k+1|, . . .)

sup(|xn+1,n+1 xn,n+1|, |xn+2,n+2 xn,n+2|, . . .) sup(xn+1 xn, xn+2 xn) sup(xn+1 xn, rn+1, rn+2, . . .)< rn

and we are done.

Corollary 28 Let K be not spherically complete. Then (l/c0) = {0}. If f (l) and

f = 0 on c0 then f is identically zero.

ProofCombine Lemma 26 and Proposition 27. For the second statement, observe that f = 0

on c0 implies that (with : l l/c0 the canonical map)

(x) f(x) (x l)defines an element of (l/c0)

which has to be 0, so f = 0.

Proposition 29 LetK be not spherically complete. Then

f : (1, 2, . . .)

n=1n ((1, 2, . . .) c0)

cannot be extended to an element of (l).

ProofWith an ingenious trick. Suppose we could extend the given functional to a h (l).Define the shift operator : l l by the formula

(1, 2, . . .) = (0, 1, 2, . . .).

Then maps c0 into c0 and f = f, so h is also an extension of f. By Corollary 28 wehave h = h. Now take any x l, x = (1, 2, . . .). Set

s = (1, 1 + 2, . . .)

Then s l

and s s = x. Then h(x) = (h h)(s) = 0, a contradiction.The idea of the proof of Proposition 29 can be generalized:

Proposition 30 LetK be not spherically complete. Let (a1, a2, . . .) l. Then

f : (1, 2, . . .) n=1

nan ((1, 2, . . .) c0)

can be extended to an element of (l) if and only if (a1, a2, . . .) c0.


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Proof We prove that if (a1, a2, . . .) / c0 then f cannot be extended.There is an > 0 such that X := {n N : |an| } is infinite. Let n1, n2, . . . be anenumeration of X. Define a continuous linear map T : l l by

T(1, 2, . . .) = (b1, b2, . . .),

where

bn =

0 if n / Xa1ni i if n = ni.

We have for (1, 2, . . .) c0

(

) (f

T)(1, 2, . . .) = f(b1, b2, . . .) = nX anbn = i=1 anibni = i=1 i.

If f had an extension f (l) then f T would be an extension of f T. But this isimpossible by the previous Proposition and ().

Theorem 31 (c0 and l are reflexive) Let K be not spherically complete. For each x c0

define fx (l) by

fx(y) =< x, y >=n=1

nn (y l)

(x = (1, 2, . . .), y = (1, 2, . . .)). Then x fx is an isometrical isomorphism. c0 (l).Proof By Proposition 30, x fx is surjective. Obviously it is linear. We have for x c0,y l:

|fx(y)| = | < x, y > | = nn max |n||n|

xy

so fx x. But also, if we put y = en (n-th unit vector) we get

fx = enfx |fx(en)| = |n|

so, fx maxn |n| = x.[BACKGROUND: The conclusion of Lemma 22 is not true if K is not spherically complete,

as one can construct non-reflexive subspaces of l. The class of reflexive spaces over non-

spherically complete K is, so far, not yet well-described.]


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Chapter 5

COMPACTOIDS

This little chapter treats a third important notion (next to orthogonalityand of countable

type), namely that of a compactoid set. We do not have much time to explain it in the

lectures, so the only purpose of this chapter is to give you an idea about these notions, its

impact on the theory but without details and without proofs.

A subset X of a Banach space E (over K or over R or C) is called precompact if, for any > 0,

X can be covered by finitely many -balls. In other words, X is precompact if for every > 0

there is a finite set F E such that

X B(0, ) + F

One proves by classical means that X is precompact if and only its closure X in E is compact.

(For this, it is necessary that E is complete).

In classical theory convex compact sets play an important role in Functional Analysis

(e.g. Alaoglu-Bourbaki Theorem, Krein-Milman Theorem, Choquet Theory). How are the

prospects in the non-archimedean case? Recall that a subset Cof a K-vector space is absolutely

convex if 0 C and for x, y C, , K, || 1 || 1 we have x + y C. So, if K isnot locally compact the only absolutely convex compact subset of a Banach space is {0}. Thisdifficulty is similar to the separability problem on page 32, and we will solve it in a similar

way; this time by convexifying the definition of precompactness.

Definition A subset X of a Banach space E over K is called compactoid if

for each > 0 there is a finite set F E such that

X B(0, ) + aco F,

where aco stands for absolutely convex hull, the smallest absolutely convex set containnig

the given set. In our case ifF = {a1, . . . , an} then aco F = {1a1 +. . .+nan : i K, |i| 1for each i}.


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One may hope now that absolutely convex complete compactoids will take over the role played

by absolutely convex compact sets in classical analysis. The following results show that thisis the case to a certain degree.

5.1 List of properties of compactoids

Compactoids are bounded.

The absolutely convex hull of a compactoid is a compactoid.

The closure of a compactoid is a compactoid.

The continuous linear image of a compactoid is a compactoid.

If K is locally compact then a subset of a K-Banach space is a compactoid if and only if it is

precompact.Those were easy properties. Now a few less trivial ones (we do not give proofs).

Let E be a K-Banach space with norm . X E is a compactoid and only if there exists a sequence x1, x2, . . . tending to 0 such

that X aco {x1, x2, . . .}. If is a norm weaker than and A is a closed absolutely convex compactoid in E,

then the topologies induced by and coincide on A. Let A E be a complete compactoid, absolutely convex. Let K, || > 1 if the

valuation is dense. = 1 otherwise. Let t

(0, 1) ifK is not spherically complete. Let

t (0, 1] otherwise. Then there exists a t-orthogonal sequence e1, e2, . . . in A such thatA aco {e1, e2, . . .} A

Let X E be bounded. Then X is a compactoid if and only if for every t (0, 1] eacht-orthogonal sequence in X tends to 0.

Let X E. Then X is a compactoid if and only if for each sequence x1, x2, . . . in Xlimn

n

Vol(x1, x2, . . . , xn) = 0

(Here, Vol(x1

, . . . , xn

) :=

x1

dist(x2

, [x1

]) dist(x3

, [x1

, x2

]) . . . dist(xn

, [x1

, . . . xn1

])).

(Arzela-Ascoli) Let F C(X) where X compact.Then Fis bounded and equicontinuous if and only if F is a compactoid.

If A E is an absolutely convex closed compactoid then (T A)e := ||>1

(T A) is closed

(where T L(E)).(Not always T A is closed)


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If T L(E) and A is a complete absolutely convex compactoid and z1, z2, . . . TA, zn 0 then for each K, || > 1 there exist x1, x2, . . . A with T xn = zn foreach n and xn 0 (The map T|A is almost open).

An A L(E) is called compact operator if it maps the unit ball into a compactoid.One can derive a theory a la Riesz for compact operators:

non-zero elements of the spectrum are eigenvalues.

0 is the only possible accumulation point of the spectrum.

Each eigenspace of a = 0 is finite-dimensional.

Alternative of Fredholm....

One may think of alternative ways of convexifying notions of compactness. The following

was introduced by Springer.

Let A E be convex. We call A c-compact if the following holds. A is closed and for eachcollection of closed convex sets {Ci : i I} in A such that Ci,1 . . . Ci,n = for anyi1, . . . , in I we have

i

Ci = (This translates a well-known property of closed subset of acompact set).

However:

If K is not spherically complete each c-compact set is either empty or a singleton set.

But:

If K is spherically complete and A E is convex and bounded then

A is a complete compactoid A is c-compact.

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