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Functions of Two Variables with Large Tangent Plane Sets
journal of mathematical analysis and applications 220, 562–570 (1998) article no. AY975848
Functions of Two Variables with Large Tangent Plane Sets Zolt´an Buczolich∗ Department of Analysis, E¨ otv¨ os Lor´ and University, M´ uzeum krt 6-8, Budapest H-1088, Hungary Submitted by Brian S. Thomson Received November 6, 1995
We show that there exist a C1 function, f; of two variables and a set E ⊆ R 2 of zero Lebesgue measure such that using the natural three-dimensional parametrization of planes z = ax + by + c tangent to the surface z = f x; y, the (threedimensional) interior of the set of parameter values, a; b; c, of tangent planes corresponding to points x; y in E is nonempty. From the Morse–Sard theorem it follows that there are no such C2 functions. We also study briefly the relationship of our example with the Denjoy–Young–Saks theorem. © 1998 Academic Press AMS classifications: primary: 26B05; secondary: 28A75, 53A05.
INTRODUCTION Given f x; y, a differentiable function of two variables, let Sx0 ; y0 = a; b; c denote the parameters of the equation z = ax + by + c of the tangent plane to the surface z = f x; y at the point x0 ; y0 : We give an example of a C1 function, f x; y, and a nowhere dense closed set E ⊆ R 2 with zero two-dimensional measure such that the interior of SE ⊆ R 3 is nonempty. This implies, for example, that λ3 SE > 0, where λ3 is the Lebesgue measure in R 3 : Since from Sx0 ; y0 = a; b; c it follows that f x0 ; y0 − ax0 − by0 = c and the partials of f x; y − ax − by vanish at x0 ; y0 , our result can also be interpreted in the following way. There is an open set G ⊆ R 2 such that for any a; b ∈ G the set of critical values of f x; y − ax − by contains a nonempty interval. For C2 functions the Morse–Sard theorem [3, 5, 6] * Research supported by the Hungarian National Foundation for Scientific Research, Grant T 016094. 562 0022-247X/98 $25.00 Copyright © 1998 by Academic Press All rights of reproduction in any form reserved.
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implies that for any a; b ∈ R 2 the set of critical values of f x; y − ax − by is of λ1 measure zero, and hence cannot contain an interval. Therefore Fubini’s theorem implies that for C2 functions SE is of zero λ3 -measure for any subset E of the domain of f . The two-variable version of the gradient problem of C. E. Weil [8] is the following. Let f be a differentiable function of two variables. Is it always true that for any open set G ⊆ R 2 the set of those x; y’s for which ∇f x; y = ∂x f x; y; ∂y f x; y belongs to G is either empty, or of positive λ2 -measure? Trying to give a positive answer to the gradient problem we found the counterexample discussed in our paper. If our counterexample did not exist among all differentiable functions, that would answer the gradient problem. However, its existence shows that there might be a counterexample for the gradient problem as well. The one-dimensional positive answer to the gradient problem is the socalled Denjoy–Clarkson property of the derivative [1, 2]. It is natural to ask what is true about the question discussed in our paper when we work with functions of one variable. Thus assume that I is an open interval and f x I → R is a differentiable function of one variable and at x0 ∈ R let y = ax + b be the equation of its tangent line. Denote by S1 x I → R 2 the mapping which assigns to each x0 ∈ I the parameters a; b. Is it possible that S1 I has nonempty interior? Or is it possible that λ2 S1 I > 0? Well, if f is a C1 function then by the one-dimensional version of Sard’s theorem for any a ∈ R the critical values of f x − ax are of measure zero and hence we obtain that for any fixed a the set of b’s for which a; b ∈ S1 I is of measure zero. Therefore it would be natural to ask whether S1 I can be “larger” if we assume only the differentiability of f . However, in this case, a part of the Denjoy–Young–Saks theorem [7, Chap. IX, (3.7) Theorem, p. 267] answers the question. It follows from this result that if y = a0 x + b0 is a given line in the plane, π 0 denotes the othogonal projection onto this line, and E 0 denotes the π 0 image of those points of the graph of f where the tangent of f is perpendicular to y = a0 x + b0 then λ1 E 0 = 0. This easily implies that λ2 S1 I = 0 for differentiable functions as well. Returning to dimension 2 one can consider the corresponding higher dimensional result [7, Chap. IX, (13.11) Theorem, p. 309]. “Given a set R ⊆ R 3 , let P be a subset of R at every point of which the set R has an extreme tangent plane parallel to a fixed straight line D. Then the orthogonal projection of P on the plane perpendicular to D is of plane measure zero.” A variant of the above theorem would be the following. “Given a set R ⊆ R 3 , for any unit vector n ∈ R 3 let Pn be a subset of R at every point of which the set R has a tangent plane perpendicular to n. Let Tn denote the orthogonal projection of Pn onto the line tn x t ∈ R and let T be the union of the sets Tn for all unit vectors n. Is it true that T is always
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of λ3 -measure zero?” The example given in our paper provides a negative answer to the above question. Let R be the graph of our function f x; y. An easy computation shows that the normal vector of the tangent plane to f at x0 ; y0 is given by n=
p
−∂x f 2
2
1 + ∂x f + ∂y f
;p
−∂y f 1 + ∂x f 2 + ∂y f 2
p
;
1 1 + ∂x f 2 + ∂y f 2
! ;
where the derivatives should be evaluated at x0 ; y0 and the parameter p value t for which t · n ∈ Tn equals −x0 ∂x f − y0 ∂y f + f x0 ; y0 / 1 + ∂x f 2 + ∂y f 2 where, again, the derivatives should be taken at x0 ; y0 . Therefore, choosing the open set G ⊆ R 2 such that for any a; b ∈ G the set of critical values of f x0 ; y0 − ax0 − by0 contains an open interval, we obtain that Tn also contains an interval for −a −b 1 n= √ ;√ ;√ : 1 + a 2 + b2 1 + a 2 + b2 1 + a 2 + b2 This implies that T is of positive measure. It is also worth mentioning some related results which show that C1 functions in R 2 can behave quite badly. The classical result is due to Whitney [9]; it shows that there are C1 functions, f , of two variables and continuous arcs, γ, in the plane such that the gradient of f vanishes at each point of γ but f is not constant along γ. T. W. K¨ orner [4] showed that there are functions f x; y such that any two points x0 ; y0 and x1 ; y1 can be connected by an arc γ such that, apart from the endpoints, the gradient of f vanishes along γ: NOTATION The reals are denoted by R. The projections of R 2 onto the x- and y-axes are denoted by πx and πy , respectively (that is, for example, πx x; y = x). The Lebesgue measure in R n is denoted by λn . For A ⊆ R we also use the alternate notation A = λ1 A. MAIN RESULT Theorem. There exist a C1 function f x R 2 → R, a nowhere dense closed set E ⊆ 0; 1 × 0; 1 of zero two-dimensional measure, and a nonempty open set H ⊆ R 3 such that for any a; b; c ∈ H we can find an x0 ; y0 ∈ E for
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which the equation of the tangent plane to the surface z = f x; y at x0 ; y0 is z = ax + by + c. Proof. The proof is organized into nine steps. In Step 9 the function f x; y will be defined as the sum of two functions of one variable, gl0 ;h0 x and gl00 ;h00 y. These functions of one variable are defined by an inductive process in Steps 1–3. The properties of these functions and of related sets are studied in Steps 4–7. In Step 8 we introduce two parallel constructions, with suitably “balanced” constants, which are used for our functions gl0 ;h0 and gl00 ;h00 . Step 1 (d-Divisions of the Unit Square Q = 0; 1 × 0; 1). Assume that 0 ≤ d < 1/2: A d-division of Q will consist of 64 subparallelograms of Q; denoted by Qj , j = 0; : : : ; 63: Put 1 j j 1 j j Tj = ; + × ; + : 64 64 128 64 64 256 If j is even put Qj = Tj : If j is odd take the above Tj and keeping the lower left corner of Tj fixed increase the slope of the horizontal side of Tj to d, that is, Qj is bounded by the lines x = j/64; x = j/64 + 1/128; y = j/64 + dx − j/64; y = j/64 + 1/256 + dx − j/64: Let j j 1 j j 1 ; + × ; + : T˜j = 64 64 128 64 64 128 From 0 < d ≤ 1/2 it follows that Qj ⊆ T˜j for any j = 0; : : : ; 63: ElemenS S63 ˜ tary computation yields, that if x; y; x0 ; y 0 ∈ 63 j=0 Qj ⊆ j=0 Tj and these points belong to different Qj ’s (which also implies that they belong to different T˜j ’s) then y0 − y 1 ≤ 0 ≤ 3: 3 x −x
1
Assume that 0 ≤ d0 < 1/128: Let π d0 denote the affine projection onto the y-axis parallel with the line y = d0 x; that is, π d0 x; y = 0; y − d0 x: Put I = π d0 0; 1 × 0; 1: Then 1 ≤ I ≤ 129/128: Also put I j = π d0 Qj : Again an easy computation shows that the I j ’s are disjoint and if we take any subinterval J of I with J > 6/128 then there are an even j and an odd j 0 such that I j ∪ I j 0 ⊆ J: Step 2 (Parallelograms and the Mappings φR and ψR ). In the remainder of the paper we need to deal with parallelograms with two sides parallel to the y-axis. We will say that the parallelogram R is of the form xR ; yR ; l; h; m if xR ; yR is its lower left corner πx R = l; two of the
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sides of R are parallel to the y axis and are of length h, and the other two sides are parallel to the line y = mx; that is, R is bounded by the lines x = xR , x = xR + l; y = yR + mx − xR ; y = yR + h + mx − xR : We will call m the slope of the parallelogram R: We denote by φR the affine transformation which maps R onto Q, and for which φR xR ; yR = 0; 0, that is, 1/l −m/h : φR x; y = x − xR ; y − yR 0 1/h We denote by ψR the inverse of φR , that is, l ml + xR ; yR : ψR x; y = x; y 0 h It is an easy computation to verify that lines parallel to the line y = m0 x are mapped by φR onto lines with slope m0 − ml/h and lines parallel to the line y = m00 x are mapped by ψR onto lines with slope m + hm00 /l: Step 3 (Definition of the Closed Sets Fl0 ; h0 and of the Functions gl0 ;h0 : Given l0 ; h0 with l0 /h0 ≤ 1/128 we define a closed subset Fl0 ; h0 ⊆ 0; l0 × 0; h0 such that letting Fx l0 ; h0 = πx Fl0 ; h0 for every x0 ∈ Fx l0 ; h0 there exists exactly one y0 such that x0 ; y0 ∈ Fl0 ; h0 : Denoting this y0 by gl0 ;h0 x0 we will obtain a C1 -function on Fx l0 ; h0 (this property is verified in Step 4). To define Fl0 ; h0 put F0 = 0; l0 × 0; h0 : Assume that Fn is defined and consists of 64n disjoint parallelograms of the form x; y; l; h; k/2n with l/h ≤ 2n /128; and k ∈ 0; 1; : : : ; 2n − 1: If R is a subparallelogram of Fn of the above form then using d = l/h2n+1 < 1/128 take a d-division of Q: Assume that the corresponding S subparallelograms are Q0 ; : : : ; Q63 : Let Rj = ψR Qj and Fn+1 ∩ R = 63 j=0 Rj : Define this way Fn+1 ⊆ Fn for all subparallelograms of Fn : Obviously Fn+1 will be closed and will consist of 64n+1 parallelograms. Observe that if R was the above subparallelogram belonging to Fn then Rj is a parallelogram of the form xj ; yj ; lj ; hj ; kj /2n+1 with xj = x + lj/64; yj = y + k/2n + h/llj/64; lj = l/128; hj = h/256, and kj = 2k if j is even and kj = 2k + 1 if j is odd. It also follows that lj /hj = 2l/h ≤ 2n+1 /128: It is easy to verify by mathematical induction that if n0 > n, R is a parallelogram belonging to Fn with slope k/2n , R0 is a parallelogram belonging to Fn0 ∩ R; and m0 denotes the slope of R0 then k/2n ≤ m0 < k + 1/2n : It is also clear that πx Fn consists of 64n disjoint closed intervals each of length l0 /128n , that is, πx Fn = l0 /2n ; and if x0 ∈ πx Fn then the vertical line x T = x0 intersects Fn in a closed interval of length h0 /256n : Set Fl0 ; h0 = ∞ n=0 Fn : This clearly implies that Fx l0 ; h0 = πx Fl0 ; h0 = 0:
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Step 4 (Verification of the Fact that the Function gl0 ;h0 is C1 . We need to verify that gl0 ;h0 is a C1 function on Fx l0 ; h0 : To this end we need to investigate the process used during the definition of Fn+1 . We use the notation introduced in Step 3. Assume that R S is a given subparallelogram of Fn with slope k/2n and x; y; x0 ; y 0 ∈ 63 j=0 Rj are not belonging to S the same subparallelogram Rj . Then φR x; y; φR x0 ; y 0 ∈ 63 j=0 Qj are not belonging to the same Qj and by (1) the slope m00 of the line passing through φR x; y and φR x0 ; y 0 satisfies 1/3 ≤ m00 ≤ 3: Therefore (see also the end of Step 2) if m0 = y 0 − y/x0 − x; which is the slope of the line passing through x; y = ψR φR x; y and x0 ; y 0 = ψR φR x0 ; y 0 , then k k h h1 + ≤ m0 ≤ 3 + n : l 3 2n l 2 By our definition of parallelograms belonging to Fn we have l = l0 /128n ; h = h0 /256n and hence we infer that k 3 k 1 + n ≤ m0 ≤ n + n : 3 · 2n 2 2 2
2
For each x0 ∈ Fx l0 ; h0 there exists a unique sequence of parallelograms R0 ; R1 ; : : : such that x0 ; gl0 ;h0 x0 ∈ Rn and Rn is a subparallelogram of Fn : Then R0 ⊃ R1 ⊃ · · · and if m0 ; m1 ; : : : denotes the slopes of R0 ; R1 ; : : : respectively then for any n either mn+1 = mn or mn+1 = mn + 1/2n+1 . This implies that if n0 ≥ n then mn ≤ mn0 ≤ mn +
1 1 − n0 : n 2 2
3
Since πx Fn consists of 64n disjoint closed intervals there is a δn > 0 such that x0 − δn ; x0 + δn ∩ Fx l0 ; h0 ⊆ πx Rn : Assume that x1 6= x0 ; x1 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 : Then there exists n0 ≥ n such that x1 ; gl0 ;h0 x1 ∈ Rn0 but x1 ; gl0 ;h0 x1 6∈ Rn0 +1 : Then if we denote by Rn0 ;j the subparallelograms of Rn0 which belong to Rn0 ∩ Fn0 +1 ; the points x0 ; gl0 ;h0 x0 and x1 ; gl0 ;h0 x1 belong to different Rn0 ;j ’s. From our con0 0 struction it is clear that mn0 = kn0 /2n with a suitable 0 ≤ kn0 < 2n ; thus by (2) we have gl0 ;h0 x1 − gl0 ;h0 x0 3 1 ≤ n0 + mn0 : 0 + m n0 ≤ n 3·2 x1 − x 0 2 Using (3) we obtain gl ;h x1 − gl0 ;h0 x0 4 1 + mn ≤ 0 0 ≤ n + mn : n 3·2 x1 − x 0 2
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Also, from (3), it follows that mn converges to a number, say m, and we have mn ≤ m ≤ mn + 1/2n : Thus gl ;h x1 − gl0 ;h0 x0 4 −2 +m≤ 0 0 ≤ n +m n 3·2 x1 − x0 2 for any x1 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 , x1 6= x0 : This implies that gl00 ;h0 x0 = m; where m is the limit of the slopes of the parallelograms Rn : An argument similar to the above one can imply that if x00 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 ⊆ πx Rn and m0 = gl00 ;h0 x00 then mn ≤ m0 ≤ mn + 1/2n and hence m0 − m ≤ 1/2n : From this it follows that gl0 ;h0 is a C1 function on Fx l0 ; h0 : Using Whitney’s extension theorem we can choose a C1 extension of gl0 ;h0 onto R: In the sequel we will denote by gl0 ;h0 this extended function. Step 5 (Definition of the Parallelograms Rm i1 ;:::;in . In Steps 5–7 we assume that l0 ; h0 are fixed. Given m ∈ 0; 1 we want to find many points in Fx l0 ; h0 where gl00 ;h0 = m: For the above given m choose a sequence m0 ; m1 ; : : : such that 0 ≤ m − mn < 1/2n and mn is of the form k/2n with an integer 0 ≤ k < 2n : It is clear that F1 consists of 32 parallelograms of slope 0 and of 32 parallelograms of slope 1/2 and, in general, the set Fn contains 32n parallelograms of slope k/2n for any k ∈ 0; : : : ; 2n − 1. Assume that the 32 parallelograms in F1 with slope m1 are denoted by m Rm i1 , i1 = 0; : : : ; 31: If the parallelograms Ri1 ;:::;in are defined then each m Ri1 ;:::;in contains 32 parallelograms belonging to Fn+1 with slope mn+1 ; denote these parallelograms by Rm i1 ;:::;in ;in+1 ; in+1 = 0; : : : ; 31: Thus by induction we define for each n the parallelograms Rm i1 ;:::;in such that ij ∈ 0; : : : ; 31 for j = 0; : : : ; n; furthermore, for any sequence i1 ; i2 ; : : : we m , and the slope of Rm have Rm i1 ;:::;in ;in+1 ⊆ Ri1 ;:::;i i1 ;:::;in is mn : It is clear from n T∞ 0 then g Step 4 that if x0 = πx n=1 Rm i1 ;:::;in l0 ;h0 x0 = m: m Step 6 (Further Study of Rm i1 ;:::;in ’s and Definition of the Intervals Ii1 ;:::;in . m Assume that Ri1 ;:::;in is a parallelogram of the form x0 ; y0 ; l; h; mn with l = l0 /128n , h = h0 /256n : Put d 0 = 1/2n+1 , d = ld 0 /h; d00 = m − mn . Then 0 ≤ d00 < 1/2n and we also put d0 = ld00 /h: For ease of notation put φ = φRmi ;:::;i and ψ = ψRmi ;:::;i : Then Rm i1 ;:::;in ;in+1 is a parallelogram of the form n n 1 1 ψQj ; where Qj is a parallelogram belonging to Q0 ; : : : ; Q63 ; which is a d-division of Q (see Step 3). Lines parallel to y = mx are mapped by φ onto lines which are parallel to y = d0 x (see the end of Step 2). Let πm x; y = y − mx: Put Iim1 ;:::;in = πm Rm i1 ;:::;in : Recall that π d0 was defined in Step 1 and observe that ψ ◦ π d0 ◦ φx; y = x0 ; y − mx − x0 and hence πy ◦ ψ ◦ π d0 ◦ φx; y − mx0 = πm x; y: Since φRm i1 ;:::;in = Q and by Step 1 we know that 1 ≤ I = π d0 Q ≤ 129/128; we infer h ≤ ψ ◦ π d0 ◦ φRm i1 ;:::;in ≤
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129h/128: Using that πy and the translation by −mx0 are not changing vertical distances we obtain that h0 129h0 ≤ Iim1 ;:::;in ≤ : n 256 128 · 256n Step 7 (The J Subinterval Property of the Iim1 ;:::;in ’s). By our construction the parallelograms Rm i1 ;:::;in ;in+1 are either of the form ψQ2in+1 in+1 = 0; : : : ; 31 when mn ≤ m < mn + 1/2n+1 , or of the form ψQ2in+1 +1 in+1 = 0; : : : ; 31 when mn + 1/2n+1 ≤ m < mn + 1/2n : In the first case put J in+1 = π d0 Q2in+1 ; in the second put J in+1 = π d0 Q2in+1 +1 : The remark at the end of Step 1 implies that if J is a subinterval of I = π d0 Q = π d0 ◦ φRm i1 ;:::;in with J > 6/128 then there is an in+1 ∈ 0; : : : ; 31 such that J in+1 ⊆ J: Therefore an easy computation yields that if J is a subinterval of Iim1 ;:::;in = n πy ◦ ψ ◦ π d0 ◦ φRm i1 ;:::;in − mx0 and J > 6h/128 = 6h0 /128 · 256 then m there is an in+1 ∈ 0; : : : ; 31 such that Ii1 ;:::;in ;in+1 ⊆ J: This last property will be referred to as the J subinterval property. Step 8 (Application of the J Subinterval Property for Two Parallel Constructions). Assume that 0 ≤ m < 1; 0 ≤ m0 < 1, l0 = 1; h0 = 128; l00 = 1, h00 = 2048 = 16h0 : Put R = 0; 1 × 0; 128; R0 = 0; 1 × 0; 2048; 0 I m = πm R; I m = πm0 R0 : Then 0; 128 ⊆ I m ⊆ −1; 128; 0; 2048 ⊆ 0 0 I m ⊆ −1; 2048: If c ∈ 128; 2047 then J0 = c − I m ⊆ 0; 2048 ⊆ I m and J0 ≥ 128 > 6h00 /128: Therefore by the J subinterval property there exists 0 i10 ∈ 0; : : : ; 31 such that Iim0 ⊆ J0 : Then using Step 6 we obtain 1
6= 0
h0 6h0 129h00 0 < 8 = 0 ≤ Iim0 ≤ : 1 128 256 128 · 256 0
Put J10 = c − Iim0 : Since Iim0 ⊆ J0 = c − I m we have J10 ⊆ I m : From J10 = 1 1 0 Iim0 > 6h0 /128 it follows that there exists i1 such that Iim1 ⊆ J10 : Then 1
6h01 129h0 6h00 6 · 16h0 h = = < 0 ≤ Iim1 ≤ : 128 128 · 256 128 · 256 256 128 · 256 0
0
Put J1 = c − Iim1 : Since Iim1 ⊆ J10 = c − Iim0 we have J1 ⊆ Iim0 : From J1 = 1 1 0 Iim1 > 6h01 /128 it follows that there exists i20 such that Iim0 ;i0 ⊆ J1 : Then 1
2
h0 129h01 6h0 h00 6h1 0 = < = 1 ≤ Iim0 ;i0 ≤ : 1 2 128 128 · 256 256 · 256 256 128 · 256 0
Put J20 = c − Iim0 ;i0 : : : 1 2 From the above steps it should be clear that by mathematical induction we can choose sequences i1 ; i2 ; : : : ; and i10 ; i20 ; : : : such that Iim1 ;:::;in ⊆
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0
c − Iim0 ;:::;i0 , and Iim0 ;:::;i0 ;i0 1
0
n
n
1
n+1
⊆ c − Iim1 ;:::;in for any n. Since Iim1 ;:::;in → 0;
and Iim0 ;:::;i0 → 0 as n → ∞ there are unique c0 and c00 such that T1 n m T ∞ m0 0 0 c0 = ∞ n=1 Ii1 ;:::;in and c0 = n=1 Ii10 ;:::;in0 : It is also clear that c0 + c0 = c: T∞ T∞ m m0 0 0 Let x0 ; y0 = n=1 Ri1 ;:::;in and x0 ; y0 = n=1 Ri0 ;:::;i0 : Then y0 − mx0 = n 1 πm x0 ; y0 = c0 and y00 − m0 x00 = πm0 x00 ; y00 = c00 : Using the properties of the functions gl0 ;h0 and gl00 ;h00 we have y0 = gl0 ;h0 x0 ; y00 = gl00 ;h00 x00 ; gl00 ;h0 x0 = m; and gl00 ;h0 x00 = m0 : 0
0
Step 9 (Definition of f x; y. Put f x; y = gl0 ;h0 x + gl00 ;h00 y with l0 ; h0 ; l00 ; h00 from Step 8. Then by the argument of Step 8 for any m; m0 ∈ 0; 1 and c ∈ 128; 2047 there exist x0 ∈ Fx l0 ; h0 and x00 ∈ Fx l00 ; h00 such that letting a = m = gl00 ;h0 x0 = ∂x f x0 ; x00 , and b = m0 = gl00 ;h0 x00 = ∂y f x0 ; x00 we have c = c0 + c00 = gl0 ;h0 x0 − ax0 + gl00 ;h00 x00 − 0 0 bx00 = f x0 ; x00 − ax0 − bx00 for any a; b; c ∈ 0; 1 × 0; 1 × 128; 2047: Since it is easy to see that the set E = Fx l0 ; h0 × Fx l00 ; h00 is a nowhere dense closed set and λ2 E = 0 our proof is complete. Questions and Remarks. It is not difficult to verify that the set E in our theorem is of Hausdorff dimension less than 2. Since f is a C1 function and for a fixed a; b; c ∈ H the set f x; y − ax − by x x; y ∈ E contains an interval it is not difficult to see that the set E should be at least of Hausdorff dimension 1. By changing some constants in our construction for any > 0 one can obtain an example when the Hausdorff dimension of E is less than 1 + . However, it is not clear to us whether there exists an example where E is of Hausdorff dimension 1. REFERENCES 1. J. A. Clarkson, A property of derivatives, Bull. Amer. Math. Soc. 53 (1947), 124–125. 2. A. Denjoy, Sur l’integration des coefficients differentiels d’ordre sup´erieur, Fund. Math. 25 (1935), 273–326. 3. M. W. Hirsch, “Differential Topology,” Graduate Texts in Mathematics, Vol. 33, SpringerVerlag, Berlin/New York, 1976. 4. T. W. K¨ orner, A dense arcwise connected set of critical points—Molehills out of mountains, J. London Math. Soc. (2) 38 (1988), 442–452. 5. A. P. Morse, The behavior of a function on its critical set, Ann. of Math. (2) 40 (1939), 62–70. 6. A. Sard, The measure of the critical values of differentiable maps, Bull. Amer. Math. Soc. 48 (1942), 883–890. 7. S. Saks, “Theory of the Integral,” Dover, New York, 1964. 8. Queries section, Real Analysis Exchange 16 (1990–1991), 373. 9. H. Whitney, A function not constant on a connected set of critical points, Duke Math. J. 1 (1935), 514–517.
Functions of Two Variables with Large Tangent Plane Sets Zolt´an Buczolich∗ Department of Analysis, E¨ otv¨ os Lor´ and University, M´ uzeum krt 6-8, Budapest H-1088, Hungary Submitted by Brian S. Thomson Received November 6, 1995
We show that there exist a C1 function, f; of two variables and a set E ⊆ R 2 of zero Lebesgue measure such that using the natural three-dimensional parametrization of planes z = ax + by + c tangent to the surface z = f x; y, the (threedimensional) interior of the set of parameter values, a; b; c, of tangent planes corresponding to points x; y in E is nonempty. From the Morse–Sard theorem it follows that there are no such C2 functions. We also study briefly the relationship of our example with the Denjoy–Young–Saks theorem. © 1998 Academic Press AMS classifications: primary: 26B05; secondary: 28A75, 53A05.
INTRODUCTION Given f x; y, a differentiable function of two variables, let Sx0 ; y0 = a; b; c denote the parameters of the equation z = ax + by + c of the tangent plane to the surface z = f x; y at the point x0 ; y0 : We give an example of a C1 function, f x; y, and a nowhere dense closed set E ⊆ R 2 with zero two-dimensional measure such that the interior of SE ⊆ R 3 is nonempty. This implies, for example, that λ3 SE > 0, where λ3 is the Lebesgue measure in R 3 : Since from Sx0 ; y0 = a; b; c it follows that f x0 ; y0 − ax0 − by0 = c and the partials of f x; y − ax − by vanish at x0 ; y0 , our result can also be interpreted in the following way. There is an open set G ⊆ R 2 such that for any a; b ∈ G the set of critical values of f x; y − ax − by contains a nonempty interval. For C2 functions the Morse–Sard theorem [3, 5, 6] * Research supported by the Hungarian National Foundation for Scientific Research, Grant T 016094. 562 0022-247X/98 $25.00 Copyright © 1998 by Academic Press All rights of reproduction in any form reserved.
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implies that for any a; b ∈ R 2 the set of critical values of f x; y − ax − by is of λ1 measure zero, and hence cannot contain an interval. Therefore Fubini’s theorem implies that for C2 functions SE is of zero λ3 -measure for any subset E of the domain of f . The two-variable version of the gradient problem of C. E. Weil [8] is the following. Let f be a differentiable function of two variables. Is it always true that for any open set G ⊆ R 2 the set of those x; y’s for which ∇f x; y = ∂x f x; y; ∂y f x; y belongs to G is either empty, or of positive λ2 -measure? Trying to give a positive answer to the gradient problem we found the counterexample discussed in our paper. If our counterexample did not exist among all differentiable functions, that would answer the gradient problem. However, its existence shows that there might be a counterexample for the gradient problem as well. The one-dimensional positive answer to the gradient problem is the socalled Denjoy–Clarkson property of the derivative [1, 2]. It is natural to ask what is true about the question discussed in our paper when we work with functions of one variable. Thus assume that I is an open interval and f x I → R is a differentiable function of one variable and at x0 ∈ R let y = ax + b be the equation of its tangent line. Denote by S1 x I → R 2 the mapping which assigns to each x0 ∈ I the parameters a; b. Is it possible that S1 I has nonempty interior? Or is it possible that λ2 S1 I > 0? Well, if f is a C1 function then by the one-dimensional version of Sard’s theorem for any a ∈ R the critical values of f x − ax are of measure zero and hence we obtain that for any fixed a the set of b’s for which a; b ∈ S1 I is of measure zero. Therefore it would be natural to ask whether S1 I can be “larger” if we assume only the differentiability of f . However, in this case, a part of the Denjoy–Young–Saks theorem [7, Chap. IX, (3.7) Theorem, p. 267] answers the question. It follows from this result that if y = a0 x + b0 is a given line in the plane, π 0 denotes the othogonal projection onto this line, and E 0 denotes the π 0 image of those points of the graph of f where the tangent of f is perpendicular to y = a0 x + b0 then λ1 E 0 = 0. This easily implies that λ2 S1 I = 0 for differentiable functions as well. Returning to dimension 2 one can consider the corresponding higher dimensional result [7, Chap. IX, (13.11) Theorem, p. 309]. “Given a set R ⊆ R 3 , let P be a subset of R at every point of which the set R has an extreme tangent plane parallel to a fixed straight line D. Then the orthogonal projection of P on the plane perpendicular to D is of plane measure zero.” A variant of the above theorem would be the following. “Given a set R ⊆ R 3 , for any unit vector n ∈ R 3 let Pn be a subset of R at every point of which the set R has a tangent plane perpendicular to n. Let Tn denote the orthogonal projection of Pn onto the line tn x t ∈ R and let T be the union of the sets Tn for all unit vectors n. Is it true that T is always
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of λ3 -measure zero?” The example given in our paper provides a negative answer to the above question. Let R be the graph of our function f x; y. An easy computation shows that the normal vector of the tangent plane to f at x0 ; y0 is given by n=
p
−∂x f 2
2
1 + ∂x f + ∂y f
;p
−∂y f 1 + ∂x f 2 + ∂y f 2
p
;
1 1 + ∂x f 2 + ∂y f 2
! ;
where the derivatives should be evaluated at x0 ; y0 and the parameter p value t for which t · n ∈ Tn equals −x0 ∂x f − y0 ∂y f + f x0 ; y0 / 1 + ∂x f 2 + ∂y f 2 where, again, the derivatives should be taken at x0 ; y0 . Therefore, choosing the open set G ⊆ R 2 such that for any a; b ∈ G the set of critical values of f x0 ; y0 − ax0 − by0 contains an open interval, we obtain that Tn also contains an interval for −a −b 1 n= √ ;√ ;√ : 1 + a 2 + b2 1 + a 2 + b2 1 + a 2 + b2 This implies that T is of positive measure. It is also worth mentioning some related results which show that C1 functions in R 2 can behave quite badly. The classical result is due to Whitney [9]; it shows that there are C1 functions, f , of two variables and continuous arcs, γ, in the plane such that the gradient of f vanishes at each point of γ but f is not constant along γ. T. W. K¨ orner [4] showed that there are functions f x; y such that any two points x0 ; y0 and x1 ; y1 can be connected by an arc γ such that, apart from the endpoints, the gradient of f vanishes along γ: NOTATION The reals are denoted by R. The projections of R 2 onto the x- and y-axes are denoted by πx and πy , respectively (that is, for example, πx x; y = x). The Lebesgue measure in R n is denoted by λn . For A ⊆ R we also use the alternate notation A = λ1 A. MAIN RESULT Theorem. There exist a C1 function f x R 2 → R, a nowhere dense closed set E ⊆ 0; 1 × 0; 1 of zero two-dimensional measure, and a nonempty open set H ⊆ R 3 such that for any a; b; c ∈ H we can find an x0 ; y0 ∈ E for
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which the equation of the tangent plane to the surface z = f x; y at x0 ; y0 is z = ax + by + c. Proof. The proof is organized into nine steps. In Step 9 the function f x; y will be defined as the sum of two functions of one variable, gl0 ;h0 x and gl00 ;h00 y. These functions of one variable are defined by an inductive process in Steps 1–3. The properties of these functions and of related sets are studied in Steps 4–7. In Step 8 we introduce two parallel constructions, with suitably “balanced” constants, which are used for our functions gl0 ;h0 and gl00 ;h00 . Step 1 (d-Divisions of the Unit Square Q = 0; 1 × 0; 1). Assume that 0 ≤ d < 1/2: A d-division of Q will consist of 64 subparallelograms of Q; denoted by Qj , j = 0; : : : ; 63: Put 1 j j 1 j j Tj = ; + × ; + : 64 64 128 64 64 256 If j is even put Qj = Tj : If j is odd take the above Tj and keeping the lower left corner of Tj fixed increase the slope of the horizontal side of Tj to d, that is, Qj is bounded by the lines x = j/64; x = j/64 + 1/128; y = j/64 + dx − j/64; y = j/64 + 1/256 + dx − j/64: Let j j 1 j j 1 ; + × ; + : T˜j = 64 64 128 64 64 128 From 0 < d ≤ 1/2 it follows that Qj ⊆ T˜j for any j = 0; : : : ; 63: ElemenS S63 ˜ tary computation yields, that if x; y; x0 ; y 0 ∈ 63 j=0 Qj ⊆ j=0 Tj and these points belong to different Qj ’s (which also implies that they belong to different T˜j ’s) then y0 − y 1 ≤ 0 ≤ 3: 3 x −x
1
Assume that 0 ≤ d0 < 1/128: Let π d0 denote the affine projection onto the y-axis parallel with the line y = d0 x; that is, π d0 x; y = 0; y − d0 x: Put I = π d0 0; 1 × 0; 1: Then 1 ≤ I ≤ 129/128: Also put I j = π d0 Qj : Again an easy computation shows that the I j ’s are disjoint and if we take any subinterval J of I with J > 6/128 then there are an even j and an odd j 0 such that I j ∪ I j 0 ⊆ J: Step 2 (Parallelograms and the Mappings φR and ψR ). In the remainder of the paper we need to deal with parallelograms with two sides parallel to the y-axis. We will say that the parallelogram R is of the form xR ; yR ; l; h; m if xR ; yR is its lower left corner πx R = l; two of the
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sides of R are parallel to the y axis and are of length h, and the other two sides are parallel to the line y = mx; that is, R is bounded by the lines x = xR , x = xR + l; y = yR + mx − xR ; y = yR + h + mx − xR : We will call m the slope of the parallelogram R: We denote by φR the affine transformation which maps R onto Q, and for which φR xR ; yR = 0; 0, that is, 1/l −m/h : φR x; y = x − xR ; y − yR 0 1/h We denote by ψR the inverse of φR , that is, l ml + xR ; yR : ψR x; y = x; y 0 h It is an easy computation to verify that lines parallel to the line y = m0 x are mapped by φR onto lines with slope m0 − ml/h and lines parallel to the line y = m00 x are mapped by ψR onto lines with slope m + hm00 /l: Step 3 (Definition of the Closed Sets Fl0 ; h0 and of the Functions gl0 ;h0 : Given l0 ; h0 with l0 /h0 ≤ 1/128 we define a closed subset Fl0 ; h0 ⊆ 0; l0 × 0; h0 such that letting Fx l0 ; h0 = πx Fl0 ; h0 for every x0 ∈ Fx l0 ; h0 there exists exactly one y0 such that x0 ; y0 ∈ Fl0 ; h0 : Denoting this y0 by gl0 ;h0 x0 we will obtain a C1 -function on Fx l0 ; h0 (this property is verified in Step 4). To define Fl0 ; h0 put F0 = 0; l0 × 0; h0 : Assume that Fn is defined and consists of 64n disjoint parallelograms of the form x; y; l; h; k/2n with l/h ≤ 2n /128; and k ∈ 0; 1; : : : ; 2n − 1: If R is a subparallelogram of Fn of the above form then using d = l/h2n+1 < 1/128 take a d-division of Q: Assume that the corresponding S subparallelograms are Q0 ; : : : ; Q63 : Let Rj = ψR Qj and Fn+1 ∩ R = 63 j=0 Rj : Define this way Fn+1 ⊆ Fn for all subparallelograms of Fn : Obviously Fn+1 will be closed and will consist of 64n+1 parallelograms. Observe that if R was the above subparallelogram belonging to Fn then Rj is a parallelogram of the form xj ; yj ; lj ; hj ; kj /2n+1 with xj = x + lj/64; yj = y + k/2n + h/llj/64; lj = l/128; hj = h/256, and kj = 2k if j is even and kj = 2k + 1 if j is odd. It also follows that lj /hj = 2l/h ≤ 2n+1 /128: It is easy to verify by mathematical induction that if n0 > n, R is a parallelogram belonging to Fn with slope k/2n , R0 is a parallelogram belonging to Fn0 ∩ R; and m0 denotes the slope of R0 then k/2n ≤ m0 < k + 1/2n : It is also clear that πx Fn consists of 64n disjoint closed intervals each of length l0 /128n , that is, πx Fn = l0 /2n ; and if x0 ∈ πx Fn then the vertical line x T = x0 intersects Fn in a closed interval of length h0 /256n : Set Fl0 ; h0 = ∞ n=0 Fn : This clearly implies that Fx l0 ; h0 = πx Fl0 ; h0 = 0:
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Step 4 (Verification of the Fact that the Function gl0 ;h0 is C1 . We need to verify that gl0 ;h0 is a C1 function on Fx l0 ; h0 : To this end we need to investigate the process used during the definition of Fn+1 . We use the notation introduced in Step 3. Assume that R S is a given subparallelogram of Fn with slope k/2n and x; y; x0 ; y 0 ∈ 63 j=0 Rj are not belonging to S the same subparallelogram Rj . Then φR x; y; φR x0 ; y 0 ∈ 63 j=0 Qj are not belonging to the same Qj and by (1) the slope m00 of the line passing through φR x; y and φR x0 ; y 0 satisfies 1/3 ≤ m00 ≤ 3: Therefore (see also the end of Step 2) if m0 = y 0 − y/x0 − x; which is the slope of the line passing through x; y = ψR φR x; y and x0 ; y 0 = ψR φR x0 ; y 0 , then k k h h1 + ≤ m0 ≤ 3 + n : l 3 2n l 2 By our definition of parallelograms belonging to Fn we have l = l0 /128n ; h = h0 /256n and hence we infer that k 3 k 1 + n ≤ m0 ≤ n + n : 3 · 2n 2 2 2
2
For each x0 ∈ Fx l0 ; h0 there exists a unique sequence of parallelograms R0 ; R1 ; : : : such that x0 ; gl0 ;h0 x0 ∈ Rn and Rn is a subparallelogram of Fn : Then R0 ⊃ R1 ⊃ · · · and if m0 ; m1 ; : : : denotes the slopes of R0 ; R1 ; : : : respectively then for any n either mn+1 = mn or mn+1 = mn + 1/2n+1 . This implies that if n0 ≥ n then mn ≤ mn0 ≤ mn +
1 1 − n0 : n 2 2
3
Since πx Fn consists of 64n disjoint closed intervals there is a δn > 0 such that x0 − δn ; x0 + δn ∩ Fx l0 ; h0 ⊆ πx Rn : Assume that x1 6= x0 ; x1 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 : Then there exists n0 ≥ n such that x1 ; gl0 ;h0 x1 ∈ Rn0 but x1 ; gl0 ;h0 x1 6∈ Rn0 +1 : Then if we denote by Rn0 ;j the subparallelograms of Rn0 which belong to Rn0 ∩ Fn0 +1 ; the points x0 ; gl0 ;h0 x0 and x1 ; gl0 ;h0 x1 belong to different Rn0 ;j ’s. From our con0 0 struction it is clear that mn0 = kn0 /2n with a suitable 0 ≤ kn0 < 2n ; thus by (2) we have gl0 ;h0 x1 − gl0 ;h0 x0 3 1 ≤ n0 + mn0 : 0 + m n0 ≤ n 3·2 x1 − x 0 2 Using (3) we obtain gl ;h x1 − gl0 ;h0 x0 4 1 + mn ≤ 0 0 ≤ n + mn : n 3·2 x1 − x 0 2
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Also, from (3), it follows that mn converges to a number, say m, and we have mn ≤ m ≤ mn + 1/2n : Thus gl ;h x1 − gl0 ;h0 x0 4 −2 +m≤ 0 0 ≤ n +m n 3·2 x1 − x0 2 for any x1 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 , x1 6= x0 : This implies that gl00 ;h0 x0 = m; where m is the limit of the slopes of the parallelograms Rn : An argument similar to the above one can imply that if x00 ∈ x0 − δn ; x0 + δn ∩ Fx l0 ; h0 ⊆ πx Rn and m0 = gl00 ;h0 x00 then mn ≤ m0 ≤ mn + 1/2n and hence m0 − m ≤ 1/2n : From this it follows that gl0 ;h0 is a C1 function on Fx l0 ; h0 : Using Whitney’s extension theorem we can choose a C1 extension of gl0 ;h0 onto R: In the sequel we will denote by gl0 ;h0 this extended function. Step 5 (Definition of the Parallelograms Rm i1 ;:::;in . In Steps 5–7 we assume that l0 ; h0 are fixed. Given m ∈ 0; 1 we want to find many points in Fx l0 ; h0 where gl00 ;h0 = m: For the above given m choose a sequence m0 ; m1 ; : : : such that 0 ≤ m − mn < 1/2n and mn is of the form k/2n with an integer 0 ≤ k < 2n : It is clear that F1 consists of 32 parallelograms of slope 0 and of 32 parallelograms of slope 1/2 and, in general, the set Fn contains 32n parallelograms of slope k/2n for any k ∈ 0; : : : ; 2n − 1. Assume that the 32 parallelograms in F1 with slope m1 are denoted by m Rm i1 , i1 = 0; : : : ; 31: If the parallelograms Ri1 ;:::;in are defined then each m Ri1 ;:::;in contains 32 parallelograms belonging to Fn+1 with slope mn+1 ; denote these parallelograms by Rm i1 ;:::;in ;in+1 ; in+1 = 0; : : : ; 31: Thus by induction we define for each n the parallelograms Rm i1 ;:::;in such that ij ∈ 0; : : : ; 31 for j = 0; : : : ; n; furthermore, for any sequence i1 ; i2 ; : : : we m , and the slope of Rm have Rm i1 ;:::;in ;in+1 ⊆ Ri1 ;:::;i i1 ;:::;in is mn : It is clear from n T∞ 0 then g Step 4 that if x0 = πx n=1 Rm i1 ;:::;in l0 ;h0 x0 = m: m Step 6 (Further Study of Rm i1 ;:::;in ’s and Definition of the Intervals Ii1 ;:::;in . m Assume that Ri1 ;:::;in is a parallelogram of the form x0 ; y0 ; l; h; mn with l = l0 /128n , h = h0 /256n : Put d 0 = 1/2n+1 , d = ld 0 /h; d00 = m − mn . Then 0 ≤ d00 < 1/2n and we also put d0 = ld00 /h: For ease of notation put φ = φRmi ;:::;i and ψ = ψRmi ;:::;i : Then Rm i1 ;:::;in ;in+1 is a parallelogram of the form n n 1 1 ψQj ; where Qj is a parallelogram belonging to Q0 ; : : : ; Q63 ; which is a d-division of Q (see Step 3). Lines parallel to y = mx are mapped by φ onto lines which are parallel to y = d0 x (see the end of Step 2). Let πm x; y = y − mx: Put Iim1 ;:::;in = πm Rm i1 ;:::;in : Recall that π d0 was defined in Step 1 and observe that ψ ◦ π d0 ◦ φx; y = x0 ; y − mx − x0 and hence πy ◦ ψ ◦ π d0 ◦ φx; y − mx0 = πm x; y: Since φRm i1 ;:::;in = Q and by Step 1 we know that 1 ≤ I = π d0 Q ≤ 129/128; we infer h ≤ ψ ◦ π d0 ◦ φRm i1 ;:::;in ≤
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129h/128: Using that πy and the translation by −mx0 are not changing vertical distances we obtain that h0 129h0 ≤ Iim1 ;:::;in ≤ : n 256 128 · 256n Step 7 (The J Subinterval Property of the Iim1 ;:::;in ’s). By our construction the parallelograms Rm i1 ;:::;in ;in+1 are either of the form ψQ2in+1 in+1 = 0; : : : ; 31 when mn ≤ m < mn + 1/2n+1 , or of the form ψQ2in+1 +1 in+1 = 0; : : : ; 31 when mn + 1/2n+1 ≤ m < mn + 1/2n : In the first case put J in+1 = π d0 Q2in+1 ; in the second put J in+1 = π d0 Q2in+1 +1 : The remark at the end of Step 1 implies that if J is a subinterval of I = π d0 Q = π d0 ◦ φRm i1 ;:::;in with J > 6/128 then there is an in+1 ∈ 0; : : : ; 31 such that J in+1 ⊆ J: Therefore an easy computation yields that if J is a subinterval of Iim1 ;:::;in = n πy ◦ ψ ◦ π d0 ◦ φRm i1 ;:::;in − mx0 and J > 6h/128 = 6h0 /128 · 256 then m there is an in+1 ∈ 0; : : : ; 31 such that Ii1 ;:::;in ;in+1 ⊆ J: This last property will be referred to as the J subinterval property. Step 8 (Application of the J Subinterval Property for Two Parallel Constructions). Assume that 0 ≤ m < 1; 0 ≤ m0 < 1, l0 = 1; h0 = 128; l00 = 1, h00 = 2048 = 16h0 : Put R = 0; 1 × 0; 128; R0 = 0; 1 × 0; 2048; 0 I m = πm R; I m = πm0 R0 : Then 0; 128 ⊆ I m ⊆ −1; 128; 0; 2048 ⊆ 0 0 I m ⊆ −1; 2048: If c ∈ 128; 2047 then J0 = c − I m ⊆ 0; 2048 ⊆ I m and J0 ≥ 128 > 6h00 /128: Therefore by the J subinterval property there exists 0 i10 ∈ 0; : : : ; 31 such that Iim0 ⊆ J0 : Then using Step 6 we obtain 1
6= 0
h0 6h0 129h00 0 < 8 = 0 ≤ Iim0 ≤ : 1 128 256 128 · 256 0
Put J10 = c − Iim0 : Since Iim0 ⊆ J0 = c − I m we have J10 ⊆ I m : From J10 = 1 1 0 Iim0 > 6h0 /128 it follows that there exists i1 such that Iim1 ⊆ J10 : Then 1
6h01 129h0 6h00 6 · 16h0 h = = < 0 ≤ Iim1 ≤ : 128 128 · 256 128 · 256 256 128 · 256 0
0
Put J1 = c − Iim1 : Since Iim1 ⊆ J10 = c − Iim0 we have J1 ⊆ Iim0 : From J1 = 1 1 0 Iim1 > 6h01 /128 it follows that there exists i20 such that Iim0 ;i0 ⊆ J1 : Then 1
2
h0 129h01 6h0 h00 6h1 0 = < = 1 ≤ Iim0 ;i0 ≤ : 1 2 128 128 · 256 256 · 256 256 128 · 256 0
Put J20 = c − Iim0 ;i0 : : : 1 2 From the above steps it should be clear that by mathematical induction we can choose sequences i1 ; i2 ; : : : ; and i10 ; i20 ; : : : such that Iim1 ;:::;in ⊆
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0
c − Iim0 ;:::;i0 , and Iim0 ;:::;i0 ;i0 1
0
n
n
1
n+1
⊆ c − Iim1 ;:::;in for any n. Since Iim1 ;:::;in → 0;
and Iim0 ;:::;i0 → 0 as n → ∞ there are unique c0 and c00 such that T1 n m T ∞ m0 0 0 c0 = ∞ n=1 Ii1 ;:::;in and c0 = n=1 Ii10 ;:::;in0 : It is also clear that c0 + c0 = c: T∞ T∞ m m0 0 0 Let x0 ; y0 = n=1 Ri1 ;:::;in and x0 ; y0 = n=1 Ri0 ;:::;i0 : Then y0 − mx0 = n 1 πm x0 ; y0 = c0 and y00 − m0 x00 = πm0 x00 ; y00 = c00 : Using the properties of the functions gl0 ;h0 and gl00 ;h00 we have y0 = gl0 ;h0 x0 ; y00 = gl00 ;h00 x00 ; gl00 ;h0 x0 = m; and gl00 ;h0 x00 = m0 : 0
0
Step 9 (Definition of f x; y. Put f x; y = gl0 ;h0 x + gl00 ;h00 y with l0 ; h0 ; l00 ; h00 from Step 8. Then by the argument of Step 8 for any m; m0 ∈ 0; 1 and c ∈ 128; 2047 there exist x0 ∈ Fx l0 ; h0 and x00 ∈ Fx l00 ; h00 such that letting a = m = gl00 ;h0 x0 = ∂x f x0 ; x00 , and b = m0 = gl00 ;h0 x00 = ∂y f x0 ; x00 we have c = c0 + c00 = gl0 ;h0 x0 − ax0 + gl00 ;h00 x00 − 0 0 bx00 = f x0 ; x00 − ax0 − bx00 for any a; b; c ∈ 0; 1 × 0; 1 × 128; 2047: Since it is easy to see that the set E = Fx l0 ; h0 × Fx l00 ; h00 is a nowhere dense closed set and λ2 E = 0 our proof is complete. Questions and Remarks. It is not difficult to verify that the set E in our theorem is of Hausdorff dimension less than 2. Since f is a C1 function and for a fixed a; b; c ∈ H the set f x; y − ax − by x x; y ∈ E contains an interval it is not difficult to see that the set E should be at least of Hausdorff dimension 1. By changing some constants in our construction for any > 0 one can obtain an example when the Hausdorff dimension of E is less than 1 + . However, it is not clear to us whether there exists an example where E is of Hausdorff dimension 1. REFERENCES 1. J. A. Clarkson, A property of derivatives, Bull. Amer. Math. Soc. 53 (1947), 124–125. 2. A. Denjoy, Sur l’integration des coefficients differentiels d’ordre sup´erieur, Fund. Math. 25 (1935), 273–326. 3. M. W. Hirsch, “Differential Topology,” Graduate Texts in Mathematics, Vol. 33, SpringerVerlag, Berlin/New York, 1976. 4. T. W. K¨ orner, A dense arcwise connected set of critical points—Molehills out of mountains, J. London Math. Soc. (2) 38 (1988), 442–452. 5. A. P. Morse, The behavior of a function on its critical set, Ann. of Math. (2) 40 (1939), 62–70. 6. A. Sard, The measure of the critical values of differentiable maps, Bull. Amer. Math. Soc. 48 (1942), 883–890. 7. S. Saks, “Theory of the Integral,” Dover, New York, 1964. 8. Queries section, Real Analysis Exchange 16 (1990–1991), 373. 9. H. Whitney, A function not constant on a connected set of critical points, Duke Math. J. 1 (1935), 514–517.