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 *f**p*: [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 a*C*^{1}-function *f**W*: [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* *∈**C** ^{k}*(R

*,R*

^{n}*) for maximal possible*

^{m}*k.*

Theorem1.1 [2]. *For anyn,m**∈*N*, there exist a map* *f* : [0, 1]^{n}*→*[0, 1]^{m}*, contained in*
*C*^{k}*for all realk < n/m, and a connected setE**⊆*[0, 1]^{n}*such 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* :R^{n}*→*R^{m}*, contained inC*^{k}*for all realk <*(n*−**p)/(m**−**p), and a connected subset*
*Eof points of rankpsuch that* *f*(E)*contains an open set.*

The first theorem holds important information that [0, 1]* ^{m}*can be covered by a line of
smoothness class

*C*

*(i.e., we write*

^{<1/m}*f*

*∈*

*C*

^{<k}^{0}if

*f*

*∈*

*C*

*for every*

^{k}*k < k*0). 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*:R^{n C}*−−→*^{k}^{·}* ^{λ}* R

^{m}*,k*

*∈*N

*,λ*

*∈*[0, 1); then

*F(C*

*p*(F))

*⊆*

*f*(Σ

*µ*

*f*)

*for some*

*C*

^{<µ}*- function*

*f*:R

*→*R

^{m}*, where*

*µ*

*=*max

*{*1/(p+ ((n

*−*

*p)/(k*+

*λ))), 1/m*

*}*

*and*Σ

*µ*

*f*:

*=*

*{*

*x*

*∈*R

*: 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

This is a Sard-type theorem, and sharpness of the*µ*can 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 case*C** ^{k}*(R

^{1},R

^{1}), and [1] for the case

*C*

*(R*

^{k}*,R*

^{n}^{1}).

**2. Notations and preliminary lemmas**

*Definition 2.1.* Let *f* :R^{m}*→*R* ^{n}*be a continuous function and

*λ*

*∈*[0, 1). It is said that

*f*

*∈*

*C*

^{0}

^{·}*if*

^{λ}*f*satisfies a

*λ-H¨older condition: for every compact neighborhoodU, there*exists

*M >*0 such that

*f*(x)*−**f*(y)^{}*M**· |**x**−**y**|*^{λ}*∀**x,y**∈**U.* (2.1)
*Definition 2.2.* For *k**∈*N, *λ**∈*[0, 1), a function *f* :R^{m}*→*R* ^{n}* is a

*C*

^{k}

^{·}*-function (or*

^{λ}*f*

*∈*

*C*

^{k}

^{·}*) if*

^{λ}*f*

*∈*

*C*

*and every*

^{k}*kth partial derivative of*

*f*is a

*C*

^{0}

^{·}*-function. If*

^{λ}*f*

*∈*

*C*

^{p}

^{·}*for all*

^{β}*p*+

*β < k*+

*λ,*

*f*

*∈*

*C*

^{k}

^{·}

^{λ}*.*

*Definition 2.3.* For *f* :R^{m C}*−−→*^{0}^{·}* ^{λ}* R

*, define partial derivatives of order*

^{n}*λ:*

*f*1

^{(λ)},

*...,f*

*m*

^{(λ)}by the formula

*f*_{i}^{(λ)}(a)*=*lim

*t**→*0sign(t)*f*^{}*a*1,*...,a**i**−*1,a*i*+*t,a**i+1*,...,a*n*

*−**f*(a)

*|**t**|** ^{λ}* (2.2)

for*a**=*(a1,...,a*m*)*∈*R* ^{m}*. If all partial derivatives of order

*λ*are continuous,

*f*

*∈*

*C*

^{λ}*.*

*Definition 2.4.*For

*k*

*∈*R

^{+}, a function

*f*:R

^{m C}*−−−−−→*

^{[k]}

^{·}

^{k}

^{−}^{[k]}R

*is a*

^{n}*C*

*-function (or*

^{k}*f*

*∈*

*C*

*) if*

^{k}*f*

*∈*

*C*

^{[k]}and every [k]th partial derivative of

*f*is a

*C*

^{k}

^{−}^{[k]}-function, where [k] is the integer part of

*k. Iff*

*∈*

*C*

*for every*

^{k}*k < k*0,

*f*

*∈*

*C*

^{<k}^{0}

*.*

We begin by setting*K*0^{n}*= {**Q**i*0, *i*0*∈*N*}*, where*Q**i*0 is a closed cube inR* ^{n}*with side
length 1 and every coordinate of any vertex of

*Q*

*i*0is an integer. In general, having con- structed the cubes of

*K*

_{s}

^{n}*1, divide each*

_{−}*Q*

*i*0,i1,i2,...,i

*s*

*−*1

*∈*

*K*

_{s}

^{n}*1into 2*

_{−}*closed cubes of side 1/2*

^{n}*, and let*

^{s}*K*

_{s}*be the set of all these cubes. More precisely, we will write*

^{n}*K*_{s}^{n}*=*

*Q**i*0,i1,i2,...,i*s**−*1,i*s*;*Q**i*0,i1,i2,...,i*s**−*1,i*s**⊆**Q**i*0,i1,i2,...,i*s**−*1*∈**K*_{s}^{n}* _{−}*1, 1

*i*

*s*2

^{n}^{}

*.*(2.3) We also define

(i)*K*^{n}*=*

*s+1**∈*N*K*_{s}* ^{n}*(note that

*K*

*is defined forR*

^{n}*);*

^{n}(ii)*S(δ)—the length of a side ofδ**∈**K** ^{n}*.

Lemma 2.5. *Let* *E*1*,* *E*2 *be copies of* R*. For all* *n,m**∈*N*, there exists continuous* *H**n,m*:
[0, 1]*−−→*^{onto} [0, 1]^{2}*⊆**E*1*×**E*2*such that*

(1)*ifα**∈**K*_{(n+m)}^{1} _{·}_{s}*, thenH**n,m*(α)*=**α*^{}*×**α*^{}*, whereα*^{}*∈**K**n*^{1}*·**s**,α*^{}*∈**K**m*^{1}*·**s**,α*^{}*∈**E*1*, and*
*α*^{}*∈**E*2*,*

(2)*if* *α*^{}*×**α*^{}*⊆*[0, 1]^{2} *such that* *α*^{}*⊆**E*1, *α*^{}*⊆**E*2*,* *α*^{}*∈**K*_{n}^{1}_{·}_{s}*, and* *α*^{}*∈**K*_{m}^{1}_{·}_{s}*, then*
*H**n,m*^{−}^{1}(int(α^{}*×**α** ^{}*))

*⊆*

*α*

*∈*

*K*

_{(n+m)}

^{1}

_{·}

_{s}*.*

*Proof.* We define for every*n,m**∈*Na space-filling function*H**n,m*as follows.

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 2* ^{n+m}*congruent subintervals and [0, 1]

^{2}into 2

*congruent sub- rectangles with sides 1/2*

^{n+m}*, 1/2*

^{n}*, each subinterval can be mapped continuously onto one of the subrectangles.*

^{m}Next, each subinterval is, in turn, partitioned into 2* ^{n+m}* congruent subintervals, and
each subrectangle into 2

*congruent subrectangles with sides 1/2*

^{n+m}^{2n}, 1/2

^{2m}and the ar- gument is repeated. If this is carried on indefinitely, [0, 1] and [0, 1]

^{2}are partitioned into 2

^{(n+m)s}congruent replicas, each with sides 1/2

*, 1/2*

^{ns}*for*

^{ms}*s*

*∈*N.

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 2* ^{n+m}*equal 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.

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 define*H**n,m*for any*n,m**∈*N.

*Definition 2.6.* Every*t**∈*[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
image*H**n,m*(t) of*t.*

The function*H**n,m*satisfies the properties (1), (2) ofLemma 2.5by its definition.

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. *LetE*1*,E*2*be copies of*R*. For alln, ˜*˜ *m*:N*→*N*, there exists continuous function*
*H**n, ˜*˜*m*: [0, 1]*−−→*^{onto} [0, 1]^{2}*⊆**E*1*×**E*2*such that*

(1)*ifα**∈**K*^{}^{1}^{s}_{i}_{=}_{1}_{n(i)+ ˜}_{˜} _{m(i)}*for somes**∈*N*, thenH*˜*n, ˜**m*(α)*=**α*^{}*×**α*^{}*, where* *α*^{}*∈**K*^{}^{1}^{s}_{i}_{=}_{1}_{n(i)}_{˜} *,*
*α*^{}*∈**K*^{}^{1}^{s}_{i}

*=*1*m(i)*˜ *,*

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]^{2}*such thatα*^{}*⊆**E*1*,α*^{}*⊆**E*2*,α*^{}*∈**K*^{}^{1}^{s}_{i}_{=}_{1}_{n(i)}_{˜} *, andα*^{}*∈**K*^{}^{1}^{s}_{i}_{=}_{1}_{m(i)}_{˜} *for*
*somes**∈*N*, thenH**n, ˜*˜^{−}^{1}*m*(int(α^{}*×**α** ^{}*))

*⊆*

*α*

*∈*

*K*

^{}

^{1}

^{s}

_{i}

_{=}_{1}

_{n(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 iteration*s*in

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 *f**n*: [0, 1]*→*[0, 1]^{n}*cubes-preserving*if it has the follow-
ing properties:

(i) if*α**⊆*[0, 1] and for some*s**∈*N,*α**∈**K**n*^{1}*·**s*implies *f**n*(α)*⊆**δ*for some*δ**∈**K**s** ^{n}*,
(ii) if

*δ*

*⊆*[0, 1]

*and for some*

^{n}*s*

*∈*N,

*δ*

*∈*

*K*

_{s}*implies*

^{n}*f*

_{n}

^{−}^{1}(int(δ))

*⊆*

*α*for some

*α*

*∈*

*K*_{n}^{1}_{·}* _{s}*,

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

Note that a continuous cubes-preserving function *f**n*is a space-filling and measure-
preserving function, that is, with the property that if*α**⊆*[0, 1] and for some*s**∈*N,*α**∈*
*K*_{n}^{1}_{·}* _{s}*, then

*f*

*n*(α)

*=*

*δ*for some

*δ*

*∈*

*K*

_{s}*.*

^{n}Theorem2.9 (space-filling function) [2, Theorem 1]. *For everyn**∈*N*, there exists a con-*
*tinuous cubes-preserving function*

*f**n*: [0, 1]*−−−→*^{onto} [0, 1]^{n}*.* (2.4)
*Definition 2.10.* For*m,n**∈*N,*k**∈*R, a function*ψ*:*B**⊆*R^{m}*→*R* ^{n}*is

*D*

*-functionif there exist*

^{k}*K >*0 such that for all

*b,b*

^{}*∈*

*B*

*ψ(b)**−**ψ(b** ^{}*)

^{}

^{k}*K*

*|*

*b*

*−*

*b*

^{}*|*

*.*(2.5)

**2.1. Properties of***D*^{k}**-functions**

*Extension on closure property*[1]. If*f* :*A**⊆*R^{m D}*−→** ^{k}* R

*for some*

^{n}*k >*0, and

*A*is the closure of

*A, then there exists a unique function*

*f*:

*A*

*⊆*R

^{m C}*−→*

^{0}R

*such that*

^{n}*f*

*A*

*=*

*f*, and

*f*is a

*D*

*-function.*

^{k}*Composition property*[1]. If*g**∈**D** ^{k}*and

*f*

*∈*

*D*

*, then*

^{p}*g*

*◦*

*f*

*∈*

*D*

*.*

^{kp}*Subsets property*[1]. If*f* :*A**⊆*R^{m D}*−→** ^{k}* R

*for some*

^{m}*k >*0, then

*f*

*B*

*∈*

*D*

*for any*

^{k}*B*

*⊆*

*A.*

*C*^{<k}*-extension on*R*property.* If*F*:*B**⊆*R*−−→*^{D}^{1/k} R* ^{m}*,

*k >*0, then

*F*

*=*

*f*

*B*for some func-

tion *f* :R*−−→*^{C}* ^{<k}* R

*, with range(F)*

^{m}*⊆*

*f*(Σ

*k*

*f*).

We prove this property as follows. Let*f* *B*be the*D*^{1/k}extension of the function*F*on
the closed set*B*the closure of*B, that exists and is unique by the “extension on closure*
property.” Then let*T*be a real number such that

*∀**b,b*^{}*∈**B* ^{}*f*(b)*−**f*(b* ^{}*)

^{}

^{1/k}

*T*

*|*

*b*

*−*

*b*

^{}*|*, (2.6) and let

*A*

*=*range(

*f*

*B); then range(F)*

*⊆*

*A.*

We now define the function *f* :R*→*R* ^{m}*as follows: if there exists a point

*b*

*∈*Rsuch that

*b*

*=*max

*{*

*b*

*∈*

*B*

*}*, then for all

*xb,*

*f*(x)

*=*

*f*(b), respectively, if there exists a point

*b*

*∈*Rsuch that

*b*

*=*min

*{*

*b*

*∈*

*B*

*}*; then for all

*xb,*

*f*(x)

*=*

*f*(b).

We designate*Z(B)**= {*(b,*b** ^{}*)

*⊆*R

*\*

*B;*

*b < b*

*,*

^{}*b,b*

^{}*∈*

*B*

*}*; this set is countable and we can write

*Z(B)*

*= {*(b

*n*,

*b*

^{}*);*

_{n}*n*

*∈*N

*}*, where

*b*

*n*,

*b*

^{}

_{n}*∈*

*B.*

Let *f* *=*(*f*1,...,*f**i*,*...,f**m*), where *f**i*:*B**→*R, 1*im, are the component functions of*
the function *f* *B; then for alln**∈*N, for all*x**∈*(b*n*,b^{}* _{n}*), and for all

*i*(1

*im), we*define

*f**i*(x)*=*
*f**i*

*b*^{}_{n}^{}*−**f**i*
*b**n*

*·**g*

*x**−**b**n*

*b**n*^{}*−**b**n* +*f**i*
*b**n*

, (2.7)

where (following [6, page 6])*g*:R^{1}*→*[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 all*x**∈*R, continuous, smooth onR*\**B*and*A**⊆*range(*f*). To finish
the proof of*C** ^{<k}*-extension onRproperty, it suﬃces to show that

*f**i*^{(t)}*B**=*0 *∀**i*(1*im),**∀**t**∈*

1, 2,..., [k]^{}*∪*

[k],*k*^{}*.* (2.9)
It is evident for nonlimit points of*B. LetB*^{}*⊆**B*be the set of limit points of*B.*

*Case 1.* If*k*1, then for all*b**∈**B** ^{}*and some fixed

*t*: 0

*t < k,*

*f*

_{i}^{(t)}(b)

^{}

*=*lim

*h**→*0

*f**i*(b+*h)**−**f**i*(b)^{}

*|**h**|*^{t}*.* (2.10)

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

*b*+*h**∈*

*b**n*,b^{}_{n}^{} for some*n**∈*N,

∆*n*:*=**b*+*h**−**b**n**.*

(2.11)

Then

*f**i*(b+*h)**−**f**i*(b)^{}

*|**h**|*^{t}^{}*f**i*
*b**n*

*−**f**i*(b)^{}+^{}*f**i*(b+*h)**−**f**i*
*b**n*
*b**n**−**b*^{}+∆*n**t*

^{}*f**i*
*b**n*

*−**f**i*(b)^{}

*b**n**−**b*^{}+∆*n**t*+^{}*f**i*(b+*h)**−**f**i*
*b**n*
*b**n**−**b*^{}+∆*n**t*

*T*^{k}^{}*b**n**−**b*^{}^{k}

*b**n**−**b*^{})* ^{t}* +

^{}

*f*

*i*(b+

*h)*

*−*

*f*

*i*

*b*

*n*

∆^{t}*n* *.*

(2.12)

We consider each summand of (2.12) separately:

*T*^{k}^{}*b**n**−**b*^{}^{k}

*b**n**−**b*^{}^{t}^{=}*T*^{k}^{}*b**n**−**b*^{}^{k}^{−}* ^{t}*, (2.13)
where

*k*

*−*

*t >*0;

*f**i*(b+*h)**−**f**i*
*b**n*

∆^{t}*n* *=**f**i*
*b**n*^{}

*−**f**i*
*b**n*

*b*^{}*n**−**b**n**k* *·**g*^{}∆*n**/(b*^{}*n**−**b**n*)^{}

∆*n**/(b*^{}*n**−**b**n*) ^{·}

∆^{1}*n*^{−}^{t}

*b*^{}*n**−**b**n*1*−**k*

*T** ^{k}*max(Dg)

^{}∆

*n*

*b*^{}_{n}*−**b**n*

1_{−}*t*

*b*^{}_{n}*−**b**n**k**−**t*

,

(2.14)

where*k**−**t >*0, 1*−**t >*0, and∆*n**b*^{}*n**−**b**n*.

Turning back to (2.12), we see that
*f**i*^{(t)}(b)^{}lim

*h**→*0

*T*^{k}^{}*b**n**−**b*^{}^{k}^{−}* ^{t}*+

*T*

*max(Dg)*

^{k}^{}∆

*n*

*b*^{}*n**−**b**n*

1*−**t*

*b*^{}*n**−**b**n**k*_{−}*t*

*=*0 (2.15)
because either∆*n**/(b*^{}_{n}*−**b**n*) or*b*^{}_{n}*−**b**n* tends to 0 as*h*tends to 0. Let *U**b* be a compact
neighborhood of*b; then for theM*required byDefinition 2.1, we can take the number

max^{}*T*^{k}^{}*b**n**−**b*^{}^{k}^{−}* ^{t}*+

*T*

*max(Dg)*

^{k}^{}

*b*

^{}

_{n}*−*

*b*

*n*

*k*

*−*

*t*

:*b**n*,*b*^{}* _{n}*,b

*∈*

*U*

*b*

*T*^{k}^{}diam^{}*U**b**k**−**t*1 + max(Dg)^{}*.* (2.16)

*Case 2.* If*k >*1 for every*t**∈*R, 1*t < k, we can suppose by induction that*
*f*_{i}^{(˜}^{t)}*B**≡*0, ˜*t**=*

[t] if*t**∈*N,

*t**−*1 if*t**∈*N*.* (2.17)

Then for all*b**∈**B,*

*f*_{i}^{(t)}(b)^{}*=*lim

*h**→*0

*f*_{i}^{(˜}* ^{t)}*(b+

*h)*

*−*

*f*

_{i}^{(˜}

*(b)*

^{t)}^{}

*h*^{t}^{−}^{t}^{˜} (2.18)

and using (2.11),

*f*_{i}^{(˜}* ^{t)}*(b+

*h)*

*−*

*f*

_{i}^{(˜}

*(b)*

^{t)}^{}

*h*

^{t}

^{−}^{˜}

^{t}

^{=}*f*_{i}^{(˜}* ^{t)}*(b+

*h)*

^{}

*b*

*n*

*−*

*b*

^{}+∆

*n*

*t*

*−*˜

*t*

*f*_{i}^{(˜}* ^{t)}*(b+

*h)*

*−*

*f*

_{i}^{(˜}

^{t)}^{}

*b*

*n*

∆^{t}*n*^{−}^{˜}^{t}

*=**f**i*^{(˜}* ^{t+1)}*(ξ)

*·*∆

^{1+˜}

*n*

^{t}

^{−}

^{t}(2.19)

for some*ξ**∈*(b*n*,b^{}*n*) (note that *f**i*^{(˜}* ^{t)}*(b)

*=*

*f*

*i*

^{(˜}

*(b*

^{t)}*n*)

*=*0 because

*b,b*

*n*

*∈*

*B, and also that*

*f*

*i*

*∈*

*C*

*on (b*

^{∞}*n*,b

^{}*)).*

_{n}From (2.7), it follows that
*f*_{i}^{(˜}* ^{t+1)}*(ξ)

^{}

*=*

*f*

*i*

*b**n*

*−**f**i*
*b*^{}_{n}^{}
*b*^{}_{n}*−**b**n*^{˜}*t+1* *·*

*g*^{(˜}^{t+1)}^{}*ξ**−**b**n*

*b*_{n}^{}*−**b**n*

*T*^{k}*·**b**n**−**b*^{}_{n}^{}^{k}

*b*^{}_{n}*−**b**n*˜*t+1* *·**r*^{˜}*t+1**=**T*^{k}^{}*b*^{}_{n}*−**b**n**k*

*b*_{n}^{}*−**b**n*˜*t+1* *·**r**t+1*^{˜} ,

(2.20)

where*r*^{˜}*t+1**=*max*{**g*^{(˜}* ^{t+1)}*(α);

*α*

*∈*[0, 1]

*}*.

Then for *f*_{i}^{(t)}(b), we can write
*f*_{i}^{(t)}(b)^{}*T*^{k}*·* ∆^{1+˜}*n* ^{t}^{−}^{t}*r**t+1*˜

*b*_{n}^{}*−**b**n*1+˜*t**−**k* *=**T*^{k}*r*˜*t+1**·*
∆*n*

*b*^{}_{n}*−**b**n*

^{1+˜}^{t}^{−}^{t}*·*

*b*^{}_{n}*−**b**n**k**−**t*

; (2.21) it means that

*f*_{i}^{(t)}(b)^{}lim

*h**→*0

*T*^{k}*r*^{˜}*t+1**·*
∆*n*

*b*^{}_{n}*−**b**n*

^{1+˜}^{t}^{−}^{t}*·*

*b*^{}_{n}*−**b**n**k**−**t*

, (2.22)

where*k > t,*∆*n**/(b*^{}*n**−**b**n*)1, 1 + ˜*tt. The limit is equal to 0 because either*∆*n**/(b*^{}*n**−*
*b**n*) or (b_{n}^{}*−**b**n*) tends to 0 as*h*tends to 0. Let *U**b* be a compact neighborhood of *b;*

then for the*K*required byDefinition 2.1, we can take the number*T*^{k}*r*^{˜}*t+1**·*(diam(U*b*))^{k}^{−}* ^{t}*
so that, by finishing the proof of (2.9), we finish the proof of the “C

*-extension onR property.”*

^{<k}Lemma2.11. *Letn,p**∈*N*,pn,k**∈*R*,k*1. Then there exists a continuous space-filling
*function*

*π*_{k,p}^{n}*=*
*π*1,π2

: [0, 1]*−−−→*^{onto} [0, 1]* ^{n}* (2.23)

*such thatπ*1: [0, 1]*−−−−−−→*^{D}^{(pk+n}^{−}* ^{p)/k}* [0, 1]

^{p}*andπ*2: [0, 1]

*−−−−→*

^{D}

^{pk+n}

^{−}*[0, 1]*

^{p}

^{n}

^{−}

^{p}*are component func-*

*tions ofπ*

_{k,p}

^{n}*.*

*Proof.* We consider the following:

(a) functions ˜*n, ˜m*:N*→*Nsuch that for every*s**∈*N,
*n(s)*˜ *=**p**·*

[ks]*−*

*k(s**−*1)^{},

*m(s)*˜ *=**n**−**p,* (2.24)

where [ks] is the integer part of*ks;*

(b) a function *H**n, ˜*˜*m* : [0, 1]*−−→*^{onto} [0, 1]^{2} defined in Lemma 2.7 and let *h*1,h2: [0,
1]*→*[0, 1] be the component functions of*H**n, ˜*˜*m*so that for all*t**∈*[0, 1],*H**n, ˜*˜*m*(t)*=*
(h1(t),h2(t))*∈*[0, 1]^{2};

(c) A function*π*^{n}_{p,k}*=*(π1,π2) : [0, 1]*−−→*^{onto} [0, 1]* ^{n}*, where

*π*1*=**f**p**◦**h*1, *π*2*=**f**n**−**p**◦**h*2*.* (2.25)
Additionally,

*f**p*: [0, 1]*−−−→*^{onto} [0, 1]* ^{p}*,

*f*

*n*

*−*

*p*: [0, 1]

*−−−→*

^{onto}[0, 1]

^{n}

^{−}*(2.26) are some continuous space-filling cubes-preserving functions, the existence of which follows fromTheorem 2.9.*

^{p}We establish some properties of the functions*π*1,*π*2, which we will need to finish the
proof ofLemma 2.11.

(I) If*α**∈**K*_{p[ks]+(n}^{1} _{−}* _{p)s}*for some

*s*

*∈*N, then

*f*

*p*

*h*1(α)^{}*=*

*π*1(α)^{}*∈**K*_{[ks]}* ^{p}* ,

*f*

*n*

*−*

*p*

*h*2(α)^{}*=*

*π*2(α)^{}*∈**K**s*^{n}^{−}^{p}*.* (2.27)
We prove this property as follows. As *p[ks] + (n**−**p)s**=*_{s}

*i**=*1*n(i) + ˜*˜ *m(i), by prop-*
erty (1) ofLemma 2.7, one has that*H**n, ˜*˜*m*(α)*=**α*^{}*×**α** ^{}*, where

*h*1(α)

*=*

*α*

^{}*∈*

*K*

^{}

^{1}

^{s}

_{i}

_{=}_{1}

_{n(i)}_{˜}

*=*

*K*

^{1}

*and*

_{p[ks]}*h*2(α)

*=*

*α*

^{}*∈*

*K*

^{}

^{1}

^{s}

_{i}

_{=}_{1}

_{m(i)}_{˜}

*=*

*K*

_{(n}

^{1}

_{−}*. Then according toDefinition 2.8of cubes- preserving functions*

_{p)s}*f*

*p*,

*f*

*n*

*−*

*p*, we can see that

*f**p**h*1(α)^{}*∈**K*_{[ks]}* ^{p}* ,

*f*

*n*

*−*

*p*

*h*2(α)

^{}

*∈*

*K*

*s*

^{n}

^{−}

^{p}*.*(2.28) (II) If

*α*

*∈*

*K*

_{p[ks]+(n}^{1}

_{−}*for some*

_{p)s}*s*

*∈*N, then

*|**α**| =**S*^{}*π*1(α)^{}^{(p[ks]+(n}^{−}^{p)s)/[ks]}*=**S*^{}*π*2(α)^{}^{(p[ks]+(n}^{−}* ^{p)s)/s}*, (2.29)
where

*S(π*1(α)),

*S(π*2(α)) are lengths of sides of cubes

*π*1(α),

*π*2(α), respectively.

We prove this property as follows. If*α**∈**K*_{p[ks]+(n}^{1} _{−}* _{p)s}*for some

*s*

*∈*N, then by property (2.27), it means that

*S*^{}*π*1(α)^{}*=* 1

2^{[ks]}, *S*^{}*π*2(α)^{}*=* 1

2^{s}*.* (2.30)

On the other hand,*α**∈**K*_{p[ks]+(n}^{1} _{−}* _{p)s}*so that

*|**α**| =* 1
2^{p[ks]+(n}^{−}* ^{p)s}*,

*|**α**| =*

*S*^{}*π*1(α)^{}^{(p[ks]+(n}^{−}^{p)s)/[ks]}

*=*

*S*^{}*π*2(α)^{}^{(p[ks]+(n}^{−}^{p)s)/s}*.*

(2.31)

(III) To prove that*π*1*∈**D*^{(pk+n}^{−}* ^{p)/k}*,

*π*2

*∈*

*D*

^{pk+n}

^{−}*, it suﬃces to show that there exists*

^{p}*K >*0 such that for all

*a,b*

*∈*[0, 1],

*a < b,*

diam^{}*π*1

[a,b]^{}^{p+(n}^{−}^{p)/k}*< K(b**−**a)>*^{}diam^{}*π*2

[a,b]^{}^{pk+n}^{−}^{p}*.* (2.32)
We prove this property as follows. If [a,b]*⊆*[0, 1], then there exists*s*1*∈*N,*s*1*s*0,
such that

1

2^{p[k(s}^{1}^{+1)]+(n}^{−}^{p)(s}^{1}^{+1)}^{}*b**−**a* 1

2^{p[ks}^{1}^{]+(n}^{−}^{p)s}^{1}; (2.33)
then [a,*b]**⊆**α*^{}*∪**α** ^{}*for some

*α*

*,α*

^{}

^{}*∈*

*K*

^{1}

_{p[ks}_{1}

_{]+(n}

_{−}

_{p)s}_{1},

*α*

^{}*∩*

*α*

^{}*= ∅*, and also

*b**−**a* ^{|}*α*^{}*|*

2^{p([k(s}^{1}^{+1)]}^{−}^{[ks}^{1}^{])+(n}^{−}^{p)}*>* ^{|}*α*^{}*|*

2^{2kp+n}*.* (2.34)

From property (2.29), we can see that

*|**α*^{}*| =**S*^{}*π*1(α* ^{}*)

^{}

^{(p[ks}

^{1}

^{]+(n}

^{−}

^{p)s}^{1}

^{)/[ks}

^{1}

^{]}

*=*

*S*

^{}

*π*2(α

*)*

^{}^{}

^{(p[ks}

^{1}

^{]+(n}

^{−}

^{p)s}^{1}

^{)/s}

^{1}(2.35) and using the fact that

diam^{}*π*1(α^{}*∪**α** ^{}*)

^{}2

^{}

*pS*

^{}

*π*1(α

*)*

^{}^{},

diam^{}*π*2(α^{}*∪**α** ^{}*)

^{}2

^{}

*n*

*−*

*pS*

^{}

*π*2(α

*)*

^{}^{}, (2.36) we get

diam^{}*π*1(α^{}*∪**α** ^{}*)

^{}2

^{}

*p*

^{}2

^{p([k(s}^{1}

^{+1)]}

^{−}^{[ks}

^{1}

^{])+n}

^{−}

^{p}*·*(b

*−*

*a)*

^{}

^{[ks}

^{1}

^{]/(p[ks}

^{1}

^{]+(n}

^{−}

^{p)s}^{1}

^{)}, (2.37) diam

^{}

*π*2(α

^{}*∪*

*α*

*)*

^{}^{}2

^{}

*n*

*−*

*p*

^{}2

^{p([k(s}^{1}

^{+1)]}

^{−}^{[ks}

^{1}

^{])+n}

^{−}

^{p}*·*(b

*−*

*a)*

^{}

^{s}^{1}

^{/(p[ks}^{1}

^{]+(n}

^{−}

^{p)s}^{1}

^{)}

*.*(2.38) Considering inequality (2.38), we may suppose that diam(π2([a,b]))1; also using

[a,b]*⊆**α*^{}*∪**α** ^{}*,

*k*

^{}

*s*1+ 1

^{}

*−*

*ks*1

*<*2k+ 1, (2.39)

and after the routine arithmetic transformation, we find that there exists*n*2*∈*N, which
does not depend on*s*1, such that

diam^{}*π*2

[a,b]^{}^{pk+n}^{−}^{p}*n*2(b*−**a).* (2.40)
Now we look at inequality (2.37). Knowing that (b*−**a)*1/2^{p[k(s}^{1}^{+1)]+(n}^{−}^{p)(s}^{1}^{+1)}, inequal-
ity (2.37) can be transformed into

diam^{}*π*1

[a,*b]*^{}2^{}*p*^{}2^{2kp+n}*·*(b*−**a)*^{}^{1/(p+(n}^{−}^{p)/k)}

*×*

2^{2kp+n}*·* 1

2^{p[k(s}^{1}^{+1)]+(n}^{−}^{p)(s}^{1}^{+1)}

[ks1]/(p[ks1]+(n*−**p)s*1)* _{−}*1/(p+(n

*−*

*p)/k)*

*.*
(2.41)
Note that [ks1]/(p[ks1] + (n*−**p)s*1)*−*1/(p+ (n*−**p)/k)*0 and there exists a number
*n*1*∈*N, which does not depend on*s*1, such that

diam^{}*π*1

[a,b]^{}^{p+(n}^{−}^{p)/k}*n*1(b*−**a).* (2.42)

The existence of such*n*1only depends on whether the expression

*−*

*p*^{}*k*^{}*s*1+ 1^{}+ (n*−**p)*^{}*s*1+ 1^{}

*ks*1

*p*^{}*ks*1

+ (n*−**p)s*1*−* 1

*p*+ (n*−**p)/k* (2.43)