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# CRITICAL VALUES LIE ON A LINE AZAT AINOULINE Received 9 September 2003

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AZAT AINOULINE Received 9 September 2003

We prove that critical values set of a diﬀerentiable map lies on a line of certain smoothness class.

1. Introduction

For those familiar with the “space-filling curves” topic, the headline of the paper is no surprise. G. Peano in 1890 constructed the first such continuous function fp: [0, 1]−−→onto [0, 1]2. Nowadays, the topic is well developed by a number of mathematicians (see [9]).

A further question is how smooth can the line be? Or how far from rectifiable is the line? In 1935, Whitney [10] published his example of aC1-function fW: [0, 1]2 onto−−→[0, 1]

not constant on a connected set of critical points. The author in [2] constructed Whitney- type examples of maps f Ck(Rn,Rm) for maximal possiblek.

Theorem1.1 [2]. For anyn,mN, there exist a map f : [0, 1]n[0, 1]m, contained in Ckfor all realk < n/m, and a connected setE[0, 1]nsuch that every partial derivative of

f of order< n/mvanishes onEand f(E)=[0, 1]m.

Theorem1.2 [2]. Letn,m,pbe nonnegative integer numbers,n > m > p; then there exists a mapf :RnRm, contained inCkfor all realk <(np)/(mp), and a connected subset Eof points of rankpsuch that f(E)contains an open set.

The first theorem holds important information that [0, 1]mcan be covered by a line of smoothness classC<1/m(i.e., we write f C<k0if f Ckfor everyk < k0). In this paper, the author determines the smoothness class of a line that can cover a critical values set of a diﬀerentiable map.

Main theorem1. LetF:Rn C−−→k·λ Rm,kN[0, 1); thenF(Cp(F)) fµf)for some C- function f :RRm, where µ=max{1/(p+ ((np)/(k+λ))), 1/m} and Σµf := {xR: any partial derivative off of order< µvanishes atx}.

Copyright©2004 Hindawi Publishing Corporation Abstract and Applied Analysis 2004:9 (2004) 757–776 2000 Mathematics Subject Classification: 58K05, 58C25 URL:http://dx.doi.org/10.1155/S1085337504311012

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This is a Sard-type theorem, and sharpness of theµcan be seen in the results, where necessary and suﬃcient conditions for the Morse-Sard theorem are establi shed, that are in [11,3] for the caseCk(R1,R1), and [1] for the caseCk(Rn,R1).

2. Notations and preliminary lemmas

Definition 2.1. Let f :RmRnbe a continuous function andλ[0, 1). It is said that f C0·λ if f satisfies aλ-H¨older condition: for every compact neighborhoodU, there existsM >0 such that

f(x)f(y)M· |xy|λ x,yU. (2.1) Definition 2.2. For kN, λ[0, 1), a function f :RmRn is a Ck·λ-function (or f Ck·λ) if f Ckand everykth partial derivative of f is aC0·λ-function. If f Cp·βfor allp+β < k+λ, f Ck·λ.

Definition 2.3. For f :Rm C−−→0·λ Rn, define partial derivatives of orderλ: f1(λ),...,fm(λ)by the formula

fi(λ)(a)=lim

t0sign(t)fa1,...,ai1,ai+t,ai+1,...,an

f(a)

|t|λ (2.2)

fora=(a1,...,am)Rm. If all partial derivatives of orderλare continuous, f Cλ. Definition 2.4. ForkR+, a function f :Rm C−−−−−→[k]·k[k] Rnis a Ck-function (or f Ck) if f C[k] and every [k]th partial derivative of f is aCk[k]-function, where [k] is the integer part ofk. Iff Ckfor everyk < k0, f C<k0.

We begin by settingK0n= {Qi0, i0N}, whereQi0 is a closed cube inRnwith side length 1 and every coordinate of any vertex ofQi0is an integer. In general, having con- structed the cubes ofKsn1, divide eachQi0,i1,i2,...,is1Ksn1into 2nclosed cubes of side 1/2s, and letKsnbe the set of all these cubes. More precisely, we will write

Ksn=

Qi0,i1,i2,...,is1,is;Qi0,i1,i2,...,is1,isQi0,i1,i2,...,is1Ksn1, 1is2n. (2.3) We also define

(i)Kn=

s+1NKsn(note thatKnis defined forRn);

(ii)S(δ)—the length of a side ofδKn.

Lemma 2.5. Let E1, E2 be copies of R. For all n,mN, there exists continuous Hn,m: [0, 1]−−→onto [0, 1]2E1×E2such that

(1)ifαK(n+m)1 ·s, thenHn,m(α)=α×α, whereαKn1·sKm1·sE1, and αE2,

(2)if α×α[0, 1]2 such that αE1, αE2, αKn1·s, and αKm1·s, then Hn,m1(int(α×α))αK(n+m)1 ·s.

Proof. We define for everyn,mNa space-filling functionHn,mas follows.

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Figure 2.1. HU.

Figure 2.2. HD.

If the interval [0, 1] can be mapped continuously onto the square [0, 1]2, then after partitioning [0, 1] into 2n+mcongruent subintervals and [0, 1]2into 2n+mcongruent sub- rectangles with sides 1/2n, 1/2m, each subinterval can be mapped continuously onto one of the subrectangles.

Next, each subinterval is, in turn, partitioned into 2n+m congruent subintervals, and each subrectangle into 2n+mcongruent subrectangles with sides 1/22n, 1/22mand the ar- gument is repeated. If this is carried on indefinitely, [0, 1] and [0, 1]2are partitioned into 2(n+m)scongruent replicas, each with sides 1/2ns, 1/2msforsN.

We need to demonstrate that the subsquares can be arranged so that adjacent subin- tervals correspond to adjacent subsquares with an edge in common, and so that the in- clusion relationships are presented, that is, if a rectangle corresponds to an interval, then its subrectangles correspond to the subintervals of that interval.

We will use here combination of four diﬀerent methods to construct these space-filling curves. These methods are based on an idea of Peano [9]. For future use, we designate them as VL(n,m), VR(n,m), HD(n,m), HU(n,m).

If we have a rectangle, then using any of those methods gives us 2n+mequal subrectan- gles which are ordered according to the order assigned by the method used.

Figures2.1,2.2,2.3, and2.4give us a basic idea of how these four methods work.

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Figure 2.3. VL.

Figure 2.4. VR.

Note. As soon as the curves in all four methods are passing through all the subrectangles, the only essential diﬀerence among the four methods is the disposition of start and end points. That is denoted in abbreviations of the methods: V-vertical, H-horizontal, L-left, R-right, U-up, D-down.

Further, to create the next iteration curve, we will give the means of how to present each of the subrectangles from the previous iteration (see Figures2.5,2.6,2.7, and2.8).

Finally, in Figures2.9and2.10, we indicate how this process is to be carried out for the next iteration.

Now we can defineHn,mfor anyn,mN.

Definition 2.6. Everyt[0, 1] is uniquely determined by a sequence of nested closed intervals (that are generated by our successive partitioning), the lengths of which shrink to 0. With this sequence, there corresponds a unique sequence of nested closed squares, the diagonals of which shrink into a point, and which define a unique point in [0, 1]2, the imageHn,m(t) oft.

The functionHn,msatisfies the properties (1), (2) ofLemma 2.5by its definition.

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VL

HU V L

V L

VR

HU V R

V R

VL

HU V L

V L

VR

HU V R

V R

VL

HU V L

V L

VR

HU V R

V R

VL

HU V L

V L

VR

HU V R

V R

Figure 2.5. HU.

HD

VL V L

V L

HD

VR V R

V R

HD

VL V L

V L

HD

VR V R

V R

HD

VL V L

V L

HD

VR V R

V R

HD

VL V L

V L

HD

VR V R

V R

Figure 2.6. HD.

Lemma2.7. LetE1,E2be copies ofR. For alln, ˜˜ m:NN, there exists continuous function Hn, ˜˜m: [0, 1]−−→onto [0, 1]2E1×E2such that

(1)ifαK1si=1n(i)+ ˜˜ m(i) for somesN, thenH˜n, ˜m(α)=α×α, where αK1si=1n(i)˜ , αK1si

=1m(i)˜ ,

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HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD VL

VL VL

VL

Figure 2.7. VL.

VR

VR VR

VR HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD HU

HD

HU

HD

Figure 2.8. VR.

(2)ifα×α[0, 1]2such thatαE1E2K1si=1n(i)˜ , andαK1si=1m(i)˜ for somesN, thenHn, ˜˜1m(int(α×α))αK1si=1n(i)+ ˜˜ m(i).

Proof. The proof is similar to the proof ofLemma 2.5, with the only diﬀerence that if we used, for instance, a method VL(n,m) to decompose a subrectangle on an iterationsin

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Figure 2.9. Next iteration when started with method VL.

Figure 2.10. Next iteration when started with method HD.

Lemma 2.5, then here we use a corresponding method VL(n(s),m(s)) on the iterations.

Definition 2.8[2]. Call a function fn: [0, 1][0, 1]ncubes-preservingif it has the follow- ing properties:

(i) ifα[0, 1] and for somesN,αKn1·simplies fn(α)δfor someδKsn, (ii) ifδ[0, 1]n and for somesN,δKsnimplies fn1(int(δ))αfor someα

Kn1·s,

where int(δ) is the set of interior points ofδ.

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Note that a continuous cubes-preserving function fnis a space-filling and measure- preserving function, that is, with the property that ifα[0, 1] and for somesN,α Kn1·s, then fn(α)=δfor someδKsn.

Theorem2.9 (space-filling function) [2, Theorem 1]. For everynN, there exists a con- tinuous cubes-preserving function

fn: [0, 1]−−−→onto [0, 1]n. (2.4) Definition 2.10. Form,nN,kR, a functionψ:BRmRnisDk-functionif there existK >0 such that for allb,bB

ψ(b)ψ(b)kK|bb|. (2.5)

2.1. Properties ofDk-functions

Extension on closure property[1]. Iff :ARm D−→k Rnfor somek >0, andAis the closure ofA, then there exists a unique function f :ARm C−→0 Rnsuch that f A= f, and f is aDk-function.

Composition property[1]. IfgDkand f Dp, thengf Dkp.

Subsets property[1]. Iff :ARm D−→k Rmfor somek >0, thenf BDkfor anyBA.

C<k-extension onRproperty. IfF:BR−−→D1/k Rm,k >0, thenF= f Bfor some func-

tion f :R−−→C<k Rm, with range(F)fkf).

We prove this property as follows. Letf Bbe theD1/kextension of the functionFon the closed setBthe closure ofB, that exists and is unique by the “extension on closure property.” Then letTbe a real number such that

b,bB f(b)f(b)1/kT|bb|, (2.6) and letA=range(f B); then range(F)A.

We now define the function f :RRmas follows: if there exists a pointbRsuch thatb=max{bB}, then for allxb, f(x)= f(b), respectively, if there exists a point bRsuch thatb=min{bB}; then for allxb, f(x)= f(b).

We designateZ(B)= {(b,b)R\B; b < b,b,bB}; this set is countable and we can writeZ(B)= {(bn,bn); nN}, wherebn,bnB.

Let f =(f1,...,fi,...,fm), where fi:BR, 1im, are the component functions of the function f B; then for allnN, for allx(bn,bn), and for alli(1im), we define

fi(x)= fi

bnfi bn

·g

xbn

bnbn +fi bn

, (2.7)

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where (following [6, page 6])g:R1[0, 1] is a smooth map such that g(−∞, 0]=0,

g[1,)=1, g(x)>0 for 0< x <1.

(2.8)

Then f is defined for allxR, continuous, smooth onR\BandArange(f). To finish the proof ofC<k-extension onRproperty, it suﬃces to show that

fi(t)B=0 i(1im),t

1, 2,..., [k]

[k],k. (2.9) It is evident for nonlimit points ofB. LetBBbe the set of limit points ofB.

Case 1. Ifk1, then for allbBand some fixedt: 0t < k, fi(t)(b)=lim

h0

fi(b+h)fi(b)

|h|t . (2.10)

Note that we may suppose without loss of generality that h >0,

b+h

bn,bn for somenN,

n:=b+hbn.

(2.11)

Then

fi(b+h)fi(b)

|h|t fi bn

fi(b)+fi(b+h)fi bn bnb+∆nt

fi bn

fi(b)

bnb+∆nt+fi(b+h)fi bn bnb+∆nt

Tkbnbk

bnb)t +fi(b+h)fi bn

tn .

(2.12)

We consider each summand of (2.12) separately:

Tkbnbk

bnbt =Tkbnbkt, (2.13) wherekt >0;

fi(b+h)fi bn

tn =fi bn

fi bn

bnbnk ·gn/(bnbn)

n/(bnbn) ·

1nt

bnbn1k

Tkmax(Dg)n

bnbn

1t

bnbnkt

,

(2.14)

wherekt >0, 1t >0, and∆nbnbn.

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Turning back to (2.12), we see that fi(t)(b)lim

h0

Tkbnbkt+Tkmax(Dg)n

bnbn

1t

bnbnkt

=0 (2.15) because either∆n/(bnbn) orbnbn tends to 0 ashtends to 0. Let Ub be a compact neighborhood ofb; then for theMrequired byDefinition 2.1, we can take the number

maxTkbnbkt+Tkmax(Dg)bnbnkt

:bn,bn,bUb

TkdiamUbkt1 + max(Dg). (2.16)

Case 2. Ifk >1 for everytR, 1t < k, we can suppose by induction that fit)B0, ˜t=

[t] iftN,

t1 iftN. (2.17)

Then for allbB,

fi(t)(b)=lim

h0

fit)(b+h)fit)(b)

htt˜ (2.18)

and using (2.11),

fit)(b+h)fit)(b) ht˜t =

fit)(b+h) bnb+∆nt˜t

fit)(b+h)fit)bn

tn˜t

=fit+1)(ξ)·1+˜n tt

(2.19)

for someξ(bn,bn) (note that fit)(b)= fit)(bn)=0 because b,bnB, and also that fiCon (bn,bn)).

From (2.7), it follows that fit+1)(ξ)=fi

bn

fi bn bnbn˜t+1 ·

gt+1)ξbn

bnbn

Tk·bnbnk

bnbn˜t+1 ·r˜t+1=Tkbnbnk

bnbn˜t+1 ·rt+1˜ ,

(2.20)

wherer˜t+1=max{gt+1)(α); α[0, 1]}.

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Then for fi(t)(b), we can write fi(t)(b)Tk·1+˜n ttrt+1˜

bnbn1+˜tk =Tkr˜t+1· n

bnbn

1+˜tt·

bnbnkt

; (2.21) it means that

fi(t)(b)lim

h0

Tkr˜t+1· n

bnbn

1+˜tt·

bnbnkt

, (2.22)

wherek > t,n/(bnbn)1, 1 + ˜tt. The limit is equal to 0 because eithern/(bn bn) or (bnbn) tends to 0 ashtends to 0. Let Ub be a compact neighborhood of b;

then for theKrequired byDefinition 2.1, we can take the numberTkr˜t+1·(diam(Ub))kt so that, by finishing the proof of (2.9), we finish the proof of the “C<k-extension onR property.”

Lemma2.11. Letn,pN,pn,kR,k1. Then there exists a continuous space-filling function

πk,pn = π12

: [0, 1]−−−→onto [0, 1]n (2.23)

such thatπ1: [0, 1]−−−−−−→D(pk+np)/k [0, 1]p andπ2: [0, 1]−−−−→Dpk+np [0, 1]np are component func- tions ofπk,pn .

Proof. We consider the following:

(a) functions ˜n, ˜m:NNsuch that for everysN, n(s)˜ =p·

[ks]

k(s1),

m(s)˜ =np, (2.24)

where [ks] is the integer part ofks;

(b) a function Hn, ˜˜m : [0, 1]−−→onto [0, 1]2 defined in Lemma 2.7 and let h1,h2: [0, 1][0, 1] be the component functions ofHn, ˜˜mso that for allt[0, 1],Hn, ˜˜m(t)= (h1(t),h2(t))[0, 1]2;

(c) A functionπnp,k=12) : [0, 1]−−→onto [0, 1]n, where

fp: [0, 1]−−−→onto [0, 1]p, fnp: [0, 1]−−−→onto [0, 1]np (2.26) are some continuous space-filling cubes-preserving functions, the existence of which follows fromTheorem 2.9.

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We establish some properties of the functionsπ1,π2, which we will need to finish the proof ofLemma 2.11.

(I) IfαKp[ks]+(n1 p)sfor somesN, then fp

h1(α)=

π1(α)K[ks]p , fnp

h2(α)=

π2(α)Ksnp. (2.27) We prove this property as follows. As p[ks] + (np)s=s

i=1n(i) + ˜˜ m(i), by prop- erty (1) ofLemma 2.7, one has thatHn, ˜˜m(α)=α×α, whereh1(α)=αK1si=1n(i)˜ = K1p[ks] andh2(α)=αK1si=1m(i)˜ =K(n1p)s. Then according toDefinition 2.8of cubes- preserving functions fp, fnp, we can see that

fph1(α)K[ks]p , fnph2(α)Ksnp. (2.28) (II) IfαKp[ks]+(n1 p)sfor somesN, then

|α| =Sπ1(α)(p[ks]+(np)s)/[ks]=Sπ2(α)(p[ks]+(np)s)/s, (2.29) whereS(π1(α)),S(π2(α)) are lengths of sides of cubesπ1(α),π2(α), respectively.

We prove this property as follows. IfαKp[ks]+(n1 p)sfor somesN, then by property (2.27), it means that

Sπ1(α)= 1

2[ks], Sπ2(α)= 1

2s. (2.30)

On the other hand,αKp[ks]+(n1 p)sso that

|α| = 1 2p[ks]+(np)s,

|α| =

Sπ1(α)(p[ks]+(np)s)/[ks]

=

Sπ2(α)(p[ks]+(np)s)/s.

(2.31)

(III) To prove thatπ1D(pk+np)/k,π2Dpk+np, it suﬃces to show that there exists K >0 such that for alla,b[0, 1],a < b,

diamπ1

[a,b]p+(np)/k< K(ba)>diamπ2

[a,b]pk+np. (2.32) We prove this property as follows. If [a,b][0, 1], then there existss1N,s1s0, such that

1

2p[k(s1+1)]+(np)(s1+1)ba 1

2p[ks1]+(np)s1; (2.33) then [a,b]ααfor someαK1p[ks1]+(np)s1,αα= ∅, and also

ba |α|

2p([k(s1+1)][ks1])+(np) > |α|

22kp+n. (2.34)

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From property (2.29), we can see that

|α| =Sπ1)(p[ks1]+(np)s1)/[ks1]=Sπ2)(p[ks1]+(np)s1)/s1 (2.35) and using the fact that

diamπ1α)2pSπ1),

diamπ2α)2npSπ2), (2.36) we get

diamπ1α)2p2p([k(s1+1)][ks1])+np·(ba)[ks1]/(p[ks1]+(np)s1), (2.37) diamπ2α)2np2p([k(s1+1)][ks1])+np·(ba)s1/(p[ks1]+(np)s1). (2.38) Considering inequality (2.38), we may suppose that diam(π2([a,b]))1; also using

[a,b]αα, ks1+ 1

ks1

<2k+ 1, (2.39)

and after the routine arithmetic transformation, we find that there existsn2N, which does not depend ons1, such that

diamπ2

[a,b]pk+npn2(ba). (2.40) Now we look at inequality (2.37). Knowing that (ba)1/2p[k(s1+1)]+(np)(s1+1), inequal- ity (2.37) can be transformed into

diamπ1

[a,b]2p22kp+n·(ba)1/(p+(np)/k)

×

22kp+n· 1

2p[k(s1+1)]+(np)(s1+1)

[ks1]/(p[ks1]+(np)s1)1/(p+(np)/k)

. (2.41) Note that [ks1]/(p[ks1] + (np)s1)1/(p+ (np)/k)0 and there exists a number n1N, which does not depend ons1, such that

diamπ1

[a,b]p+(np)/kn1(ba). (2.42)

The existence of suchn1only depends on whether the expression

pks1+ 1+ (np)s1+ 1

ks1

pks1

+ (np)s1 1

p+ (np)/k (2.43)

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