CRITICAL VALUES LIE ON A LINE

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 a C1-function fW : [0,1]2 onto −−→ [0,1] not constant on a connected set of critical points. The author in [2] constructed Whitneytype examples of maps f ∈ Ck(Rn,Rm) for maximal possible k.


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 Whitneytype examples of maps f ∈ C k (R n ,R m ) for maximal possible k.
Theorem 1.1 [2].For any n,m ∈ N, there exist a map f : [0,1] n → [0,1] m , contained in C k for all real k < n/m, and a connected set E ⊆ [0,1] n such that every partial derivative of f of order < n/m vanishes on E and f (E) = [0,1] m .Theorem 1.2 [2].Let n, m, p be nonnegative integer numbers, n > m > p; then there exists a map f : R n → R m , contained in C k for all real k < (n − p)/(m − p), and a connected subset E of points of rank p such 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 <1/m (i.e., we write f ∈ C <k0 if f ∈ C k for every 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.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 k (R n ,R 1 ).

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•λ if f satisfies a λ-Hölder condition: for every compact neighborhood U, there exists M > 0 such that Definition 2.3.For f : R m C 0•λ − − → R n , define partial derivatives of order λ: f (λ) 1 ,..., f (λ)  m by the formula We begin by setting K n 0 = {Q i0 , i 0 ∈ N}, where Q i0 is a closed cube in R n with side length 1 and every coordinate of any vertex of Q i0 is an integer.In general, having constructed the cubes of cubes of side 1/2 s , and let K n s be the set of all these cubes.More precisely, we will write We also define We define for every n,m ∈ N a space-filling function H n,m as follows.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 n+m congruent subrectangles with sides 1/2 n , 1/2 m , each subinterval can be mapped continuously onto one of the subrectangles.
We need to demonstrate that the subsquares can be arranged so that adjacent subintervals correspond to adjacent subsquares with an edge in common, and so that the inclusion 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 subrectangles which are ordered according to the order assigned by the method used.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 Figures   Lemma 2.7.Let E 1 , E 2 be copies of R. For all ñ, m : N → N, there exists continuous function ( Proof.The proof is similar to the proof of Lemma 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  Lemma 2.5, then here we use a corresponding method VL(n(s),m(s)) on the iteration s.
Note that a continuous cubes-preserving function f n is a space-filling and measurepreserving function, that is, with the property that if α ⊆ [0,1] and for some s ∈ N, α ∈ K 1 n•s , then f n (α) = δ for some δ ∈ K n s .Theorem 2.9 (space-filling function) [2, Theorem 1].For every n ∈ N, there exists a continuous cubes-preserving function (2.4) (2.5)

Properties of D k -functions
Extension on closure property [1].
for some k > 0, and A is the closure of A, then there exists a unique function 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 and let A = range( f B); then range(F) ⊆ A.
We now define the function f : R → R m as follows: if there exists , where f i : B → R, 1 i m, are the component functions of the function f B; then for all n ∈ N, for all x ∈ (b n ,b n ), and for all i (1 i m), we define where (following [6, page 6]) g : R 1 → [0,1] is a smooth map such that (2.8) Then f is defined for all x ∈ R, continuous, smooth on R \ B and A ⊆ range( f ).To finish the proof of C <k -extension on R property, it suffices to show that It is evident for nonlimit points of B. Let B ⊆ B be the set of limit points of B.
Then for f (t) i (b), we can write it means that where ) tends to 0 as h tends to 0. Let U b be a compact neighborhood of b; then for the K required by Definition 2.1, we can take the number T k rt +1 • (diam(U b )) k−t so that, by finishing the proof of (2.9), we finish the proof of the "C <k -extension on R property." Then there exists a continuous space-filling function k,p .Proof.We consider the following: (a) functions ñ, m : N → N such that for every s ∈ N, where [ks] is the integer part of ks; where (2.25) Additionally, are some continuous space-filling cubes-preserving functions, the existence of which follows from Theorem 2.9.
We establish some properties of the functions π 1 , π 2 , which we will need to finish the proof of Lemma 2.11.

.32)
We prove this property as follows.If [a,b] ⊆ [0,1], then there exists s 1 ∈ N, s 1 s 0 , such that From property (2.29), we can see that (2.35) and using the fact that Considering inequality (2.38), we may suppose that diam(π 2 ([a,b])) 1; also using and after the routine arithmetic transformation, we find that there exists n 2 ∈ N, which does not depend on s 1 , such that (2.40) Now we look at inequality (2.37).Knowing that (b − a) 1/2 p[k(s1+1)]+(n−p)(s1+1) , inequality (2.37) can be transformed into (2.42) The existence of such n 1 only depends on whether the expression ( and by the definition of π n k,p : π n k,p (t) = y.From ( 1) and ( 2), it follows that π n k,p is a continuous space-filling function.
Proof.Similar to the proof of the C k+β+ inverse function theorem in [7].
Proof.Similar to the proof of Zygmund preimage theorem in [8].
Lemmas 2.14 and 2.15 are generalized Morse vanishing lemma and Morse theorem; see Morse [5], and for more general version of the lemmas, see also Norton [7,8] and Moreira [4].
Lemma 2.14.Let k, n be nonnegative integers, λ ∈ [0,1), and with the following property: every f ∈ C k•λ (R n ,R) vanishing on A satisfies for each i and some K i 0, Proof.Fix λ.The proof is by double induction on n and k.Let n,k stand for the statement of the lemma for R n and C k•λ .We will prove 0,k for all k, n,0 for all n, and n − 1,k and n,k − 1 imply n,k .Define vanishing on A satisfies D y g ≡ 0 on A , (2.47) We prove the result separately for A * * and A * .
On A * * .Since f vanishes on A, D y f = (D yj f ) p< j n ≡ 0 on A * * , where y = (y p+1 ,..., y j ,..., y n ) so that for each j (p < j n), if any, D yj f vanishes on A * * , and , ψ i as in the statement, and or let K i = √ n − p max p< j n K i j , then x 0 , y , x 0 , y 0 ∈ V i × B i , Now by the mean value theorem, for some θ ∈ B i lying on a line segment betweeny and y 0 (2.50) We recall that ψ i (x, y) = (x, ψi (x, y)) so that Dψ i (x, y) is presented in the following matrix consisting of n rows, where the last n − p rows constitute the Jacobian matrix for the function ψi (x, y): (2.53) Now using (2.52), (2.53), (2.49) in (2.50), we have f ψ i x 0 , y − f ψ i x 0 , y 0 D y f ψ i x 0 ,θ • 0,D ψi x 0 ,θ • 0, y − y 0 D y f ψ i x 0 ,θ − D y f ψ i x 0 , y 0 Ki 0, y − y 0 where D y f ψ i x 0 , y 0 = 0 because ψ i x 0 , y 0 ∈ A * * , and Ki is a Lipschitz constant of the C 1 -function ψi on the bounded cube V i × B i , that we may suppose to exist) (2.54) On A * .If (x 0 , y 0 ) ∈ A * , there is g as above, and by Lemma 2.13, there is ε > 0 such that as in the statement, and A ∩ B ε (x 0 , y 0 ) ⊆ g −1 (0).Taking a countable subcovering of A * by these balls, we reduce the proof in this case to a case with smaller n.Lemma 2.15.Let k, n be nonnegative integers, λ ∈ [0,1), and A ⊆ R n = R n−p × R p for some p n. Then there are sets A 1 ,A 2 ,... ⊆ A such that A = ∞ i=1 A i , where for each i = 1,2,..., there is a function (2.46) for each i, and some K i 0.

45) holds with the following property: every
Proof.The same as in the case "On A * * " of Lemma 2.14 if we make there the following corrections: (1) delete " f vanishes on A," (2) replace " n,k − 1 hypothesis" with "Lemma 2.14," (3) replace A * * with A, (4) replace A * * i with A i .

Proof of the main theorem
It follows from [2, Theorem 1] and Theorems 3.2, 3.3, and 3.4.
If x 0 ∈ Cp (F), we can consider with accuracy to within a change of coordinates of class We have x = (z, y) ∈ Cp (F) if and only if D y G(z, y) = 0.By applying the results of Lemma 2.15 to a set It is not difficult to see that the proof of the theorem is reducible to a proof of the following statement: for each i, and some K i 0, Since every component function F j : R n → R (1 j m) of the function F = (F 1 ,...,F j , ...,F m ) satisfies F j ∈ C k•λ , D y F j ≡ 0 on A, then by Lemma 2.15, F j satisfies (2.46), and then it is not difficult to see that F itself satisfies (2.46), it means that for each i; and for some K i 0, F ψ i x 0 , y − F ψ x 0 , y 0 K i y − y 0 k+λ ∀ x 0 , y ∈ V i × B i , x 0 , y 0 ∈ A i .

Figures 2. 1
, 2.2, 2.3, and 2.4 give us a basic idea of how these four methods work.
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 the M required by Definition 2.1, we can take the number max T k b n − b .41) Note that [ks 1 ]/(p[ks 1 ] + (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).

( a )
Proof of 0,k for all k is trivial.(b) Proof of n,0 for all n follows directly from the definition of f ∈ C 0•λ .(c) Induction step: we assume n − 1,k and n,k − 1 , and we prove n,k .