Some preserver problems on algebras of smooth functions

c 2009 by Institut Mittag-Leffler. All rights reserved

Some preserver problems on algebras of smooth functions

F´elix Cabello S´anchez and Javier Cabello S´anchez

Abstract. We study bijections between algebras of smooth functions preserving certain parts of its structure. In particular, we show that multiplicative bijections are implemented by diffeomorphisms and they are automatically algebra isomorphisms. This confirms a conjecture by Mrˇcun and ˇSemrl.

1. Introduction

This paper deals with bijections of algebras of (real) smooth functions that preserve certain parts of its structure.

By a manifold we understand a Hausdorff topological spaceX locally homeo- morphic to a fixed Banach spaceEof dimension at least 1. We are not assumingX second-countable, connected or paracompact. AC^k manifold, 1≤k≤∞, is a man- ifold with an atlas whose transition maps are all k-times continuously (Fr´echet) differentiable, with the usual convention whenk=∞. We require moreover that E admits aC^k bump function (one with nonempty bounded support). This condition is automatically satisfied whenE is finite-dimensional. For the general case, please consult [3] or [8, Chapter III]. A less intimidating reference is [4].

It is worth noticing that in this setting the form of the isomorphisms between algebras of smooth functions has been elucidated only very recently, even in finite dimensions. Indeed, as observed by Weinstein (2003), the commonly known proofs that algebra isomorphisms between algebras of smooth functions are all induced by composition with aC^k diffeomorphism between the underlying manifolds strongly depend either on a second countability assumption or on paracompactness. (See [5]

Research supported by DGICYT projects MTM2004-02635 and MTM2007-6994-C02-02.

JCS was supported in part by a grant of the UEx (programa propio–acci´on 2).

for a ‘typical’ proof in infinite dimensions.) Soon afterwards, (2005) Grabowski and Mrˇcun filled the gap with two essentially different proofs appearing in [6] and [10].

Here, we generalise this result in two ways. We will show that every additive bijectionT:C^k(Y)→C^k(X) preserving the order in both directions has the form

T f(x) =a(x)f(τ(x)), f∈C^k(Y), x∈X,

wherea=T1 is strictly positive and τ: X→Y is a C^k diffeomorphism.

Also, we prove that every multiplicative isomorphism T:C^k(Y)→C^k(X) is automatically linear (actually what we shall see is thatT is just composition with a C^k diffeomorphism). This settles a conjecture by Mrˇcun and ˇSemrl who discovered the result and proved it for finitek in [11] (see [2] for related material).

Both results depend on a not very satisfying and rather incomplete, yet useful, description of the bijectionsT:C^k(Y)→C^k(X) preserving the order in both direc- tions (nothing more is assumed) that could be considered as the main result of this note (Theorem1).

The paper though self-contained is based on the ideas of Shirota’s seminal paper [12]. However the functional representation for T in Theorem 1 originates with Kaplansky’s classical [7]. See [1] for further references.

2. Order preserving bijections

Let us present our main result. The order we consider on C^k(X) is that inherited from R, sof≤g meansf(x)≤g(x) for every x∈X.

Theorem 1. Let T: C^k(Y)→C^k(X) be a bijective mapping preserving the order in both directions. Then there is a homeomorphismτ:X→Y such that

T f(x) =t(x, f(τ(x))), f∈C^k(Y), x∈X, wheret:X×R→Ris given by t(x, c)=T c(x).

The proof consists of several steps. First of all we need to introduce some relations in the space of smooth functions. This will be done in a somewhat artificial way becauseC^k(X) is not a lattice.

Given f∈C^k(X), the support of f, denoted suppf, is the closure of the set {x∈X:f(x)=0}. We write Uf for the interior of suppf. This is clearly a regular open set (one which equals the interior of its closure). Whether every regular open set can be obtained in this way is open to reflection.

SetC₊^k(X)={f∈C^k(X):f≥0}. Our immediate aim is to show that, forf, g∈ C^k₊(X), the relations Uf⊂Ug andUf⊂Ug can be expressed within the order struc- ture ofC₊^k(X). To this end, given f1, f2∈C₊^k(X), put

f1∩f2={f∈C₊^k(X) :f≤fi fori= 1,2}.

Also, let us declare f⊂g when for every h∈C₊^k(X) one has f∩h={0} whenever g∩h={0}. Finally we write fg if, whenever the family {h_α}α has an upper bound inC₊^k(X) andhα⊂f for all α, there is an upper boundh∈C₊^k(X) such that h⊂g.

Step 1.1. Let f, g∈C₊^k(X). Then (a) f∩g={0} if and only if Uf∩Ug=∅.

(b) f⊂g if and only if Uf⊂Ug which holds if and only if suppf⊂suppg.

(c) If fg, thensuppf⊂Ug andUf⊂Ug.

(d) If there isu∈C₊^k(X)such that u=0 offU_g and u=1on U_f, thenfg.

Proof. (a) The ‘if’ part is clear, so assume Uf∩Ug=∅. After a moment’s reflection we see there is some point where both f and g are strictly positive. It follows thatf∩g contains some nonzero function.

(b) By the very definition, we havef⊂gif and only ifg∩h={0}impliesf∩h=

{0}. By part (a), this is equivalent to ‘Ug∩Uh=∅ implies Uf∩Uh=∅’, which is clearly equivalent toUf⊂Ug. The last equivalence is obvious.

(c) Assumefg. For eachx∈Ufpickhx∈C^k₊(X) such thathx≤1 andhx(x)=1, withhx⊂f. Now, the hypothesis gives an upper boundhfor the family{hx}xsuch thath⊂g. Clearlyh≥1 onU_f and thusU_f⊂U_h⊂U_g.

(d) Assumeu=0 offUgandu=1 onUf. Ifhα⊂f andhis an upper bound for {hα}α, then uhis also an upper bound and quite clearlyuh⊂f.

Throughout this sectionT will be as in Theorem1. Besides, we assumeT0=0.

This causes no loss of generality since f →T f−T0 preserves the order in both directions and sends 0 to 0.

Step 1.2. Given f, g, h∈C₊^k(X) one has f≤g on Uh if and only if g∩ucon- tainsf∩ufor every u⊂h.

Therefore,givenf, g, h∈C₊^k(Y), one hasf≤g on Uh if and only ifT f≤T g on UT h. The same is true if we replace ≤ by≥or =.

Proof. If f≤g onU_h andu⊂h, then it is straightforward that every function lower thanf anduis lower thang, sof∩u⊂g∩u.

As for the converse, assumef(x)>g(x) for some x∈Uh. Take somev∈C₊^k(X) such thatv≤1 andv(x)=1 and having support in the setUh∩{z:f(z)>g(z)}. Then u=vf belongs tou∩f but not tou∩g.The ‘therefore’ part is now obvious.

LetR^k(X) denote the class of those regular open sets of X arising as Uh for h∈C₊^k(X), and similarly forY. We consider in R^k(X) the (partial) order given by inclusion. It is not hard to see thatR^k(X) then becomes a lattice, but we will not use this fact.

Notice that our assumptions on the model Banach space already imply that R^k(X) is a base for the topology of X. We shall use this fact without further mention in the sequel. We consider the mapping T: R^k(X)→R^k(Y) sendingUT h

to Uh. In view of Step 1.1(b), T is a well-defined bijection preserving (the order given by) inclusion in both directions.

Step 1.3. Given f, g∈C^k(X) and U∈R^k(X) one has f≤g on T(U) if and only ifT f≤T g on U. The same is true replacing≤by≥or =.

Proof. The case f, g≥0 is contained in Step 1.2. Thus we have the following:

for eachU∈R^k(X) there isV=T(U) inR^k(Y) such that wheneverf, g∈C₊^k(Y) one hasf≤gonV if and only ifT f≤T gonU. Moreover this property characterisesV in R^k(Y).

But 0 plays no special rˆole here (that is, in the ordered set of smooth functions) and we have an analogous statement for each fixed u∈C^k(Y): there is a bijection Tu: R^k(X)→R^k(Y) preserving inclusions in both directions such that whenever f, g∈C^k(Y) satisfyf, g≥uandU∈R^k(X) one hasf≤gin V=Tu(U) if and only if T f≤T g on U. And, moreover, this property characterisesV amongst the regular open sets ofR^k(Y).

We want so see that Tu does not depend on u. First, is easily checked that Tu=Tv if v≤u. Now, for arbitraryu, v∈C^k(Y), the functionw=√

1+u²+v² is in C^k(Y) and dominates bothuand v, so

T_u=T_w=T_v=T₀=T andTu=Tis independent ofu.

To complete the proof, take f, g∈C^k(Y) and U∈R^k(X). Set V=T(U) and u=−

1+f²+g², so thatu≤f, g. AsV=Tu(U) we havef≤g onV if and only if T f≤T g onU and we are done.

Next, consider the set-valued mapτ:X→2^Y given by τ(x) =

T(U),

where the intersection is taken over thoseU∈R^k(X) containing the point x.

Step 1.4. The setτ(x) is a singleton for everyx∈X. The map sendingxto the only point inτ(x)is a homeomorphism(still denotedτ).

Proof. Let{Un}^∞n=1be a neighbourhood base atxtaken fromR^k(X) satisfying Un+1⊂Un. We require moreover that for each n there is h∈C^k₊(X), depending on n, such that Un=Uh and h=1 on Un+1. This guarantees that if Vn=T(Un), thenV_n+1⊂V_n, by parts (c) and (d) of Step 1.1. Therefore,

τ(x) = ∞ n=1

T(U_n) = ∞ n=1

V_n= ∞ n=1

V_n.

Let us see thatτ(x) cannot be empty. Assume on the contrary thatτ(x)=∅. For eachn, letf_n∈C₊^k(X) be such thatf_n(x)=n and suppf_n⊂U_n. Takeg_n such that T gn=fn. ThenUgn⊂Vn. We claim that{gn}^∞n=1is locally finite. Indeed, picky∈Y. Asy /∈

nVn there ism∈Nsuch thaty /∈Vm and we can choose a neighbourhood ofy, sayW, that does not meetVm. As suppgn⊂Vn we see that gn vanishes on W as long asn≥m.

Therefore the sequence {g_n}^∞n=1 has an upper bound in C^k(Y), namely

_∞

n=1g_n, while{f_n}^∞n=1 lacks it. A contradiction.

We see thatτ(x) has exactly one point. Ify∈τ(x), then by the very definition ofτ, we have y∈T(U) as long asU∈R^k(X) contains x. LetS:C^k(X)→C^k(Y) be the inverse of T, S:R^k(Y)→R^k(X) the order isomorphism associated to S and σ: Y→2^X the set-valued function associated toS. Taking into account the trivial fact thatSis nothing but the inverse ofT, we get

σ(y) =

y∈V

S(V)⊂

x∈U

S(T(U)) =

x∈U

U={x}.

And sinceσ(y)=∅we see thatx=σ(y). Hence, forU∈R^k(X), we havex∈U if and only ify∈T(U). Equivalently, givenV∈R^k(Y), one hasy∈V if and only ifx∈S(V).

ButS is an order isomorphism onto R^k(Y) (which separates points ofY) and so there is at most oney satisfying that condition.

This shows that, for every x∈X, τ(x) is a singleton. That the map X→Y, sending x into the only element of τ(x), is continuous is trivial. That this map is a homeomorphism follows by symmetry: the map sending eachy into the only element ofσ(y) is the inverse ofτ:X→Y.

Step 1.5. Letf, g∈C^k(Y)andx∈X. Iff(τ(x))=g(τ(x)),thenT f(x)=T g(x).

Proof. Writey=τ(x). It clearly suffices to see that iff(y)≤g(y), thenT f(x)≤ T g(x). Or else, that ifT f(x)>T g(x), thenf(y)>g(y). So, assume on the contrary that T f(x)>T g(x), but f(y)≤g(y). As T f≥T g on some neighbourhood of x we know from Steps1.2and1.3thatf≥gon some neighbourhood ofy, whencef(y)=

g(y) andDf=Dg aty—where f−g attains a local minimum.

Lethbe any function inC^k(Y) such thath(y)=f(y) andDh(y)=Df(y). Then every neighbourhood ofy contains points (and so an open set) where h>f as well as points where h<g. It follows that T f(x)≤T h(x)≤T g(x), a contradiction.

This completes the proof of Theorem1.

Applications. Our first application, in the spirit of [9], shows that the behav- iour of a bijection that preserves order depends largely on the action on constant functions.

Corollary 1. Let T: C^k(Y)→C^k(X) be a bijection preserving the order in both directions.

(a) T is linear (or additive)if and only if it is linear (additive) on the set of constant functions on Y. In this case T f(x)=a(x)f(τ(x)),where a=T1 is strictly positive andτ is aC^k diffeomorphism.

(b) T is an algebra (or ring) isomorphism if and only if it sends each con- stant on Y into the same constant on X. If so, T f(x)=f(τ(x)) and τ is a C^k diffeomorphism.

It is worth noticing that a ring isomorphism between algebras of smooth func- tions must preserve order, so Corollary1implies the following result.

Corollary 2. (Grabowski [6], Mrˇcun [10]) Every ring isomorphism between C^k(Y) and C^k(X) is given by composition with a C^k diffeomorphism between the underlying manifolds.

3. Multiplicative bijections

In this section we move to the multiplicative structure of smooth functions.

The following result settles a conjecture in [11], where the casek<∞is proved.

Theorem 2. Every multiplicative bijectionC^k(Y)→C^k(X)is induced by com- position with a C^k diffeomorphism. Therefore they are all linear.

As before, we break the proof into a number of steps. From now on we assume thatT is a bijection satisfyingT(f g)=T f·T gfor allf, g∈C^k(Y). We use the same notation as in the proof of Theorem1, with the only exception that we will use sets U_f for arbitraryf. Note, however, thatU_f is the same as U_f2, so this leads to the same class of regular open sets.

First of all, we remark that T0=0 and T1=1. Also, T preserves the set of strictly positive functions (they are just the invertible squares). Moreover,T acts in a local way, meaning that, givenf,g andhinC^k(Y), one hasf=gonUhif and only ifT f=T g on UT h. And this is so because the former condition is equivalent tof h=gh. It follows that T(−1)=−1 since −1 is the only idempotent which does not agree with 1 on some nonempty open set. Next notice that U_f⊂U_g if and only if UT f⊂UT g since the former just means that gh=0 impliesf h=0 for every h∈C^k(Y).

So, as we did in Section 1, we can define an order isomorphism T:R^k(X)→ R^k(Y) takingT(UT h)=Uhforh∈C^k(Y). Now, consider the set-valued mapτ:X→ 2^Y given by

τ(x) = T(U),

where the intersection is taken over thoseU∈R^k(X) containing the point x.

Step 2.1. The setτ(x) is a singleton for everyx∈X. The map sendingxto the only point inτ(x)is a homeomorphism between X andY.

Proof. Let {Un}^∞n=1 be a neighbourhood base atxconsisting of regular open sets of R^k(X), with U_n+1⊂U_n. We require moreover that for each n there is h∈C^k(X), depending on n, such that h=1 on U_n+1 and h=0 outside U_n. This guarantees that ifVn=T(Un), thenVn+1⊂Vn, and so

τ(x) = ∞ n=1

T(U_n) = ∞ n=1

V_n= ∞ n=1

V_n.

Let us see that τ(x) is not empty. If we assume the contrary, passing to a subse- quence if necessary we find a sequence{yn}^∞n=1 withyn∈Vn\Vn+1 and a sequence of regular open sets{Wn}^∞n=1 having disjoint closures, withyn∈Wn.

Now, for evenn, leth_n∈C^k(Y) be a hat function aroundy_n(that is, one which agrees with 1 on a neighbourhood of the point) with support inWn and set

f=

neven

h_n.

As the sum is locally finite,f is inC^k(Y) and has the following property: everyVn

contains an open set wheref=1 and also an open set wheref vanishes. Therefore T f is discontinuous aty and we have reached a contradiction.

That τ(x) has exactly one point and that sending x to that point defines a homeomorphism is shown as in Step1.4.

We pause for the construction of certain real-valued functions of a single real variable which we will use later.

Step 2.2. Let 0<r<1 be fixed. There exists C^∞ smooth functionsϕ, φ:R→

[0,1) having the following properties:

(a) For everys=randN there is a nonempty open interval I lying between r ands and an integerp≥N such thatϕ(t)=t^p for all t∈I.

(b) Every neighbourhood of 0contains nonempty open sets where φ(t)=r^p for arbitrarily largep.

Proof. (a) We start with aC^∞smooth hat functionharound the origin having support in (−1,1). Set rn=r+(−¹₂)ⁿ and let Vn be the open interval of radius 1/2ⁿ⁺¹ centred at rn. Notice that Vn∩Vm is empty unless n=m. We define hn

by h_n(t)=h(2ⁿ⁺¹(t−r_n)). This is a hat function around r_n with support in V_n. Finally, forp:N→Nwe define ϕby

ϕ=

∞

n=n₀

ı^p(n)h_n,

whereıdenotes the identity onRandn0 is so thatVn⊂(0,1) for everyn≥n0. The proof will be complete if we show that a good choice of the sequence p(n)→∞makesϕsmooth.

After all, let us remark that ϕ(r)=0 and that ϕ is C^∞ smooth at all points with the only possible exception ofr. So it suffices to see that

(1) D^mϕ(t)

r−t →0, ast→r,

for eachm∈N. Indeed this implies thatD^mϕ(0)=0 for allm.

We use the following notation in the remainder of the proof. Ifuis a function whose domain containsA, we put

uA= sup

t∈A

|u(t)|.

Asϕvanishes outside theVn’s and taking into account that ϕ=ı^p(n)hn onVn and that|r−t|≥2⁻⁽ⁿ⁺¹⁾fort∈Vn we see that (1) is implied by the condition

(2) D^m(ı^p(n)hn)V_n

2⁻⁽ⁿ⁺¹⁾ →0, as n→∞. But D^m(ı^p(n)h_n)=m

l=0

_m

l

D^lı^p(n)D^m−lh_n, and it suffices to see that, whenever 0≤l≤m,

(3) 2ⁿ⁺¹D^lı^p(n)D^m−lhnVn

converges to zero asn→∞. On the other hand

D^m−lhn(z) = 2ⁿ⁺¹...2ⁿ⁺¹D^m−lh(2ⁿ⁺¹(z−rn)) (the factor 2ⁿ⁺¹ appearing m−l times), so

D^m−lhnVn= 2^(n+1)(m−l)D^m−lh[−1,1], while forp(n)>lone has

D^lı^p(n)=p(n)(p(n)−1)...(p(n)−l+1)ı^p(n)−l. Thus, if we taker<ρ<1, the sequence in (3) is dominated by

D^m−lh[−1,1]2ⁿ⁺¹2^(n+1)(m−l)p(n)^lρ^p(n)−l,

which goes to zero as long asp(n)/n→∞. So, takingp(n)=n² suffices.

Part (b) is simpler. Let hn(t)=h 2ⁿ⁺¹

t−

−¹₂n

. Given p: N→N, define φ(t)=_∞

n=1r^p(n)hn. As before we want so see thatD^mφ(t)/t→0, ast→0, ifp(n)→

∞fast enough. LetVn be the open interval of radius ₁

2

n+1

centred at

−¹₂n

. It suffices to see that

2ⁿ⁺¹D^mr^p(n)hnV_n= 2ⁿ⁺¹r^p(n)2^m(n+1)D^mh[−1,1]

goes to zero asnincreases. Again, takingp(n)=n²suffices.

Step 2.3. If 0<f≤1,then 0<T f≤1.

Proof. Assume on the contrary that 0<f≤1, but T f(z)>1 for some z∈X. Take U∈R^k(X) such that T f >R on U, with R>1. Choose y∈T(U) such that f(y)<1. We will construct a function g∈C^k(Y) having the following property:

For every neighbourhoodW ofy and everyN >0 there is an integerp≥N and an open subset ofW whereg=f^p.

Then, if τ(x)=y, the function T g is unbounded near xsince each neighbourhood ofxcontains points whereT g >R^N for allN.

We constructg using the functions of Step2.2 as follows. Suppose f is non- constant on every neighbourhood ofy. Theng=ϕ¨f, where ϕis as part (a), with r=f(y).

Iff is constant on a neighbourhood of y, let u∈C^k(Y) be such that u(y)=0 andDu(y)=0. Then takeg=φ¨u, with φas in part (b) andr=f(y).

An immediate consequence is that if 0<f≤g, withf, g∈C^k(Y), then 0<T f≤ T g. Indeed the hypothesis means that 0<f /g≤1, so 0<T(f /g)=T f /T g≤1. Hence T defines a bijection betweenC^k(Y,(0,∞)) andC^k(X,(0,∞)) preserving the order in both directions. AsRis order-diffeomorphic with (0,∞) we can apply Theorem1 to get the following consequence.

Step 2.4. There is a strictly positive functionp:X→Rsuch that

(4) T f(x) =|f(τ(x))|^p(x)sign(f(τ(x))) for everyf∈C^k(Y).

Proof. First supposef >0. We know from Theorem1that T f(x) =t(x, f(τ(x))),

wheret(x, c)=T c(x), withc>0. Actually only Step1.5is needed here. Next notice that for fixed x the map c →T c(x) is multiplicative and increasing on (0,∞), so T c(x)=c^p, with p>0 depending on x. This proves (4) when f >0. As T acts locally, the formula is true wheneverf(τ(x))>0, even iff assumes negative values.

On the other hand, T(−f)=−T f, so (4) holds true for all f∈C^k(Y) and every x∈X providedf(τ(x))=0. In particular, we see thatf(τ(x))=0 impliesT f(x)=0, that is, T f(x)=0 implies f(τ(x))=0. Applying the same reasoning to the inverse ofT we see that, in fact, T f(x)=0 if and only iff(τ(x))=0. Hence (4) is true for allf∈C^k(Y) and allx∈X.

Step 2.5. One has p(x)=1 for allx,and τ is aC^k diffeomorphism.

Proof. First, p is in C^k(X), since 2^p=T2 is. It follows that f¨τ belongs to C^k(X) forf∈C^k(Y) andf >0, henceτ isC^k smooth and, by symmetry, it is aC^k diffeomorphism.

Finally, let us check p(x)=1 for all x∈X. By symmetry one only has to see thatp(x)≥1 for all xin X. Asτ is a C^k diffeomorphism we infer from (4) that

Lf(x) =|f(x)|^p(x)sign(f(x))

defines a multiplicative automorphism ofC^k(X). Supposep(x)<1. Pickf∈C^k(X) with f(x)=0 and Df(x)=0. We claim that Lf=sign(f)|f|^p is not differentiable at x. Indeed, let u: R→X be a smooth curve passing through x at t=0, with Df(x)u(0)>0. ThenLf¨uhas no derivative at 0 since, taking 0<c<Df(x)(u(0)) andp(x)<p₀<1, one has

lim

t→0⁺

Lf(u(t))−Lf(u(0))

t = lim

t→0⁺

(f(u(t)))^p(u(t))

t ≥ lim

t→0⁺

(ct)^p0 t =∞. This completes the proof.

Concluding remarks

We close with a couple of questions arising from the content of this note.

(a) In Theorem 1, what can be said about τ? We do not know if τ must be differentiable even in the caseY=X=R. It is apparent that the main obstruction is ‘decoupling’ the actions oft(·,·) andτ.

(b) Does Corollary2 remain true if the ‘model’ Banach space E fails to have a bump? We do not know the answer even ifY andX are in fact Banach spaces.

See [5] for some affirmative results.

Acknowledgements. It is a pleasure to thank the referee for many remarks that greatly improved the article.

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F´elix Cabello S´anchez

Departamento de Matem´aticas Universidad de Extremadura ES-06071 Badajoz

Spain

fcabello@unex.es

Javier Cabello S´anchez Departamento de Matem´aticas Universidad de Extremadura ES-06071 Badajoz

Spain coco@unex.es

Received September 15, 2008 published online May 26, 2009