Functions of Two Variables

with

Large Tangent Plane Sets

Zoltán Buczolich*

Eötvös Loránd University, Department of Analysis, Budapest, Múzeum krt 6-8, Hungary, H- 1088.

* Research supported by the Hungarian National Foundation for Scientific Research, Grant No. T 016094.

AMS Classification, Primary: 26B05, Secondary: 28A75, 53A05 Running head: C^{1}-FUNCTIONS AND TANGENT PLANES

Send proofs to: Zoltán Buczolich, Eötvös Loránd University, Department of Analysis, Budapest, Múzeum krt 6-8, Hungary, H-1088.

ABSTRACT

We show that there exists a C^{1} function, f, of two variables and a set E ⊆ R^{2} of zero Lebesgue measure such that using the natural three-dimensional parametrization of tangent planes z = ax+by +c to the surface z = f(x, y) the (three-dimensional) interior of the set of parameter values, (a, b, c), of tangent planes corresponding to points (x, y) in E is non-empty. From the Morse-Sard theorem it follows that there are no such C^{2} functions. We also study briefly the relationship of our example with the Denjoy-Young-Saks theorem.

INTRODUCTION

Given f(x, y), a differentiable function of two variables let, S(x_{0}, y_{0}) = (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 (x_{0}, y_{0}). We give an example of a C^{1} function, f(x, y), and a nowhere dense closed set E ⊆ R^{2} with zero two-dimensional measure such that the interior of S(E) ⊆ R^{3} is nonempty. This implies, for example, that λ_{3}(S(E)) > 0, where λ_{3} is the Lebesgue measure in R^{3}.

Since from S(x_{0}, y_{0}) = (a, b, c) it follows that f(x_{0}, y_{0})−ax_{0} −by_{0} = c and the partials of f(x, y) − ax − by vanish at (x_{0}, y_{0}) 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 C^{2} functions the Morse-Sard theorem [3], [5], [7] implies that for any (a, b) ∈ R^{2} the set of critical values of f(x, y) − ax − by is of λ_{1} measure zero, and hence cannot contain an interval. Therefore Fubini’s theorem implies that for C^{2} functions S(E) is of zero λ_{3}-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 λ_{2}-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 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, so called, 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 : I → R is a differentiable function of one variable and at x_{0} ∈ R let y = ax + b be the equation of its tangent line. Denote by S_{1} : I → R^{2} the mapping which assigns to each x_{0} ∈ I the parameters (a, b). Is it possible that S_{1}(I) has non-empty interior? Or, is it possible that λ_{2}(S_{1}(I)) > 0? Well, if f is a C^{1} 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) ∈ S_{1}(I) is of measure zero. Therefore it would be natural to ask whether S_{1}(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, Ch. IX. (3.7) Theorem, p.267] answers the question. It follows from this result that if y = a′x + b′ is a given line in the plane, π′ denotes the othogonal projection onto this line, and E′ denotes the π′ image of those points of the graph of f where the tangent of f is perpendicular to y = a′x + b′ then λ_{1}(E′) = 0. This easily implies that λ_{2}(S_{1}(I)) = 0 for differentiable functions as well.

Returning to dimension two one can consider the corresponding higher dimensional result [7, Ch. 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 P_{n} be a subset of R at every point of which the set R has a tangent plane perpendicular to n. Let T_{n} denote the orthogonal projection of P_{n} onto the line {tn : t ∈ R} and T be the union of the sets T_{n} for all unit vectors n. Is it true that T is always of λ_{3}-measure zero?” The example given in our paper provides a negative question 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 (x_{0}, y_{0}) is given by

n = (\frac{−∂_{x}f}{\sqrt{1 + (∂_{x}f)^{2} + (∂_{y}f)^{2}}} , \frac{−∂_{y}f}{\sqrt{1 + (∂_{x}f)^{2} + (∂_{y}f)^{2}}} , \frac{1}{\sqrt{1 + (∂_{x}f)^{2} + (∂_{y}f)^{2}}} ) ,

where the derivatives should be evaluated at (x_{0}, y_{0}) and the parameter value t for which t · n ∈ T_{n} equals (−x_{0}∂_{x}f − y_{0}∂_{y}f + f(x_{0}, y_{0}))/\sqrt{1 + (∂_{x}f)^{2} + (∂_{y}f)^{2}} where, again, the derivatives should be taken at (x_{0}, y_{0}). Therefore choosing the open set G ⊆ R^{2} such that for any (a, b) ∈ G the set of critical values of f(x_{0}, y_{0})−ax_{0} −by_{0} contains an open interval we obtain that T_{n} also contains an interval for

n = (\frac{−a}{\sqrt{1 + a^{2} + b^{2}}} , \frac{−b}{\sqrt{1 + a^{2} + b^{2}}} , \frac{1}{\sqrt{1 + a^{2} + b^{2}}} ).

This implies that T is of positive measure.

It is also worth mentioning some related results which show that C^{1} functions in R^{2} can behave quite badly. The classical result is due to Whitney [9], showing that there are C^{1} 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. Körner [4] showed that there are functions f(x, y) such that any two points (x_{0}, y_{0}) and (x_{1}, y_{1}) 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 projection of R^{2} onto the x and y-axis is denoted by π_{x} and π_{y}, respectively (that is, for example, π_{x}(x, y) = x). The Lebesgue measure in R^{n} is denoted by λ_{n}. For A ⊆ R we also use the alternate notation |A| = λ_{1}(A).

MAIN RESULT

Theorem. There exists a C^{1}-function f : 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 (x_{0}, y_{0}) ∈ E for which the equation of the tangent plane to the surface z = f(x, y) at (x_{0}, y_{0}) 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 g_{ℓ_{0},h_{0}}(x) and g_{ℓ′_{0},h′_{0}}(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 g_{ℓ_{0},h_{0}} and g_{ℓ′_{0},h′_{0}}.

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 Q_{j}, j = 0, ..., 63. Put

T_{j} = [\frac{j}{64}, \frac{j}{64} + \frac{1}{128}] × [\frac{j}{64}, \frac{j}{64} + \frac{1}{256}] .

If j is even put Q_{j} = T_{j}. If j is odd take the above T_{j} and keeping the lower left corner of T_{j} fixed increase the slope of the horizontal side of T_{j} to d, that is, Q_{j} is bounded by the lines x = \frac{j}{64}, x = \frac{j}{64} + \frac{1}{128}, y = \frac{j}{64} + d(x − \frac{j}{64}), y = \frac{j}{64} + \frac{1}{256} + d(x − \frac{j}{64}).

Let

T̃_{j} = [\frac{j}{64}, \frac{j}{64} + \frac{1}{128}] × [\frac{j}{64}, \frac{j}{64} + \frac{1}{128}] .

From 0 < d ≤ 1/2 it follows that Q_{j} ⊆ T̃_{j} for any j = 0, ..., 63. Elementary computation yields, that if (x, y), (x′, y′) ∈ ∪_{j=0}^{63}Q_{j} ⊆ ∪_{j=0}^{63}T̃_{j} and these points belong to different Q_{j}’s (which also implies that they belong to different T̃_{j}’s) then

\frac{1}{3} ≤ \frac{y′ − y}{x′ − x} ≤ 3.  (1)

Assume that 0 ≤ d_{0} < 1/128. Let \overline{π}_{d_{0}} denote the affine projection onto the y-axis parallel with the line y = d_{0}x, that is, \overline{π}_{d_{0}}(x, y) = (0, y − d_{0}x). Put \overline{I} = \overline{π}_{d_{0}}([0, 1] × [0, 1]). Then 1 ≤ |\overline{I}| ≤ 129/128. Also put \overline{I}_{j} = \overline{π}_{d_{0}}(Q_{j}). Again an easy computation shows that the \overline{I}_{j}’s are disjoint and if we take any subinterval \overline{J} of \overline{I} with |\overline{J}| > 6/128 then there is an even j and an odd j′ such that \overline{I}_{j} ∪ \overline{I}_{j′} ⊆ \overline{J}.

Step 2 Parallelograms and the mappings ϕ_{R} and ψ_{R}.

In the sequel 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 (x_{R}, y_{R}), ℓ, h, m if (x_{R}, y_{R}) is its lower left corner |π_{x}(R)| = ℓ, two of the sides of R are parallel to the y axis and are of length h, the other two sides are parallel to the line y = mx, that is, R is bounded by the lines x = x_{R}, x = x_{R} + ℓ, y = y_{R} + m(x − x_{R}), y = y_{R} + h + m(x − x_{R}). We will call m the slope of the parallelogram R.

We denote by ϕ_{R} the affine transformation which maps R onto Q, and for which ϕ_{R}(x_{R}, y_{R}) = (0, 0), that is,

ϕ_{R}(x, y) = (x − x_{R}, y − y_{R}) ( \begin{matrix}\frac{1}{ℓ} & −\frac{m}{h} \\ 0 & \frac{1}{h} \end{matrix} ) .

We denote by ψ_{R} the inverse of ϕ_{R}, that is,

ψ_{R}(x, y) = (x, y) ( \begin{matrix}ℓ & mℓ \\ 0 & h \end{matrix} ) + (x_{R}, y_{R}).

It is an easy computation to verify that lines parallel to the line y = m′x are mapped by ϕ_{R} onto lines with the slope (m′ − m)\frac{ℓ}{h} and lines parallel to the line y = m′′x are mapped by ψ_{R} onto lines with the slope m + \frac{hm′′}{ℓ} .

Step 3 Definition of the closed sets F (ℓ_{0}, h_{0}) and of the functions g_{ℓ_{0},h_{0}}.

Given ℓ_{0}, h_{0} with ℓ_{0}/h_{0} ≤ 1/128 we define a closed subset F (ℓ_{0}, h_{0}) ⊆ [0, ℓ_{0}] × [0, h_{0}] such that letting F_{x}(ℓ_{0}, h_{0}) = π_{x}(F (ℓ_{0}, h_{0})) for every x_{0} ∈ F_{x}(ℓ_{0}, h_{0}) there exists exactly one y_{0} such that (x_{0}, y_{0}) ∈ F (ℓ_{0}, h_{0}). Denoting this y_{0} by g_{ℓ_{0},h_{0}}(x_{0}) we will obtain a C^{1}-function on F_{x}(ℓ_{0}, h_{0}) (this property is verified in Step 4).

To define F (ℓ_{0}, h_{0}) put F_{0} = [0, ℓ_{0}] × [0, h_{0}].

Assume that F_{n} is defined and consists of 64^{n} disjoint parallelograms of the form (x, y), ℓ, h, \frac{k}{2^{n}} with ℓ/h ≤ 2^{n}/128, and k ∈ \{0, 1, ..., 2^{n} − 1\}. If R is a subparallelogram of F_{n} of the above form then using d = ℓ/h2^{n+1} < 1/128 take a d-division of Q. Assume that the corresponding subparallelograms are Q_{0}, ..., Q_{63}. Let R_{j} = ψ_{R}(Q_{j}) and F_{n+1} ∩ R = ∪_{j=0}^{63}R_{j}. Define this way F_{n+1} ⊆ F_{n} for all subparallelograms of F_{n}. Obviously F_{n+1} will be closed and will consist of 64^{n+1} parallelograms. Observe that if R was the above subparallelogram belonging to F_{n} then R_{j} is a parallelogram of the form (x_{j}, y_{j}), ℓ_{j}, h_{j}, \frac{k_{j}}{2^{n+1}} with x_{j} = x + ℓj/64, y_{j} = y + (\frac{k}{2^{n}} + \frac{h}{ℓ})\frac{ℓj}{64}, ℓ_{j} = ℓ/128, h_{j} = h/256 and k_{j} = 2k if j is even and k_{j} = 2k+1 if j is odd. It also follows that ℓ_{j}/h_{j} = 2ℓ/h ≤ 2^{n+1}/128. It is easy to verify by mathematical induction that if n′ > n, R is a parallelogram belonging to F_{n} with slope k/2^{n} and R′ is a parallelogram belonging to F_{n′} ∩ R and m′ denotes the slope of R′ then k/2^{n} ≤ m′ < (k +1)/2^{n}. It is also clear that π_{x}(F_{n}) consists of 64^{n} disjoint closed intervals each of length ℓ_{0}/128^{n} that is |π_{x}(F_{n})| = ℓ_{0}/2^{n}, and if x_{0} ∈ π_{x}(F_{n}) then the vertical line x = x_{0} intersects F_{n} in a closed interval of length h_{0}/256^{n}. Set F (ℓ_{0}, h_{0}) = ∩_{n=0}^{∞}F_{n}. This clearly implies that |F_{x}(ℓ_{0}, h_{0})| = |π_{x}(F (ℓ_{0}, h_{0}))| = 0.

Step 4 Verification of the fact that the function g_{ℓ_{0},h_{0}} is C^{1}.

We need to verify that g_{ℓ_{0},h_{0}} is a C^{1} function on F_{x}(ℓ_{0}, h_{0}). To this end we need to investigate the process used during the definition of F_{n+1}. We use the notation introduced in Step 3. Assume that R is a given subparallelogram of F_{n} with slope k/2^{n} and (x, y), (x′, y′) ∈ ∪_{j=0}^{63}R_{j} are not belonging to the same subparallelogram R_{j}. Then ϕ_{R}(x, y), ϕ_{R}(x′, y′) ∈ ∪_{j=0}^{63}Q_{j} are not belonging to the same Q_{j} and by (1) the slope m′′ of the line passing through ϕ_{R}(x, y) and ϕ_{R}(x′, y′) satisfies 1/3 ≤ m′′ ≤ 3. Therefore (see also the end of Step 2) if m′ = \frac{y′−y}{x′−x}, which is the slope of the line passing through (x, y) = ψ_{R}(ϕ_{R}(x, y)) and (x′, y′) = ψ_{R}(ϕ_{R}(x′, y′)), then

\frac{h}{ℓ} \frac{1}{3} + \frac{k}{2^{n}} ≤ m′ ≤ \frac{h}{ℓ}3 + \frac{k}{2^{n}} .

By our definition of parallelograms belonging to F_{n} we have ℓ = ℓ_{0}/128^{n}, h = h_{0}/256^{n} and hence we infer

\frac{1}{3 · 2^{n}} + \frac{k}{2^{n}} ≤ m′ ≤ \frac{3}{2^{n}} + \frac{k}{2^{n}} .  (2)

For each x_{0} ∈ F_{x}(ℓ_{0}, h_{0}) there exists a unique sequence of parallelograms R_{0}, R_{1}, ... such that (x_{0}, g_{ℓ_{0},h_{0}}(x_{0})) ∈ R_{n} and R_{n} is a subparallelogram of F_{n}. Then R_{0} ⊃ R_{1} ⊃ ... and if m_{0}, m_{1}, ... denotes the slopes of R_{0}, R_{1}, ... respectively then for any n either m_{n+1} = m_{n} or m_{n+1} = m_{n} + \frac{1}{2^{n+1}} . This implies that if n′ ≥ n then

m_{n} ≤ m_{n′} ≤ m_{n} + \frac{1}{2^{n}} − \frac{1}{2^{n′}} .  (3)

Since π_{x}(F_{n}) consists of 64^{n} disjoint closed intervals there is a δ_{n} > 0 such that (x_{0}−δ_{n}, x_{0}+ δ_{n}) ∩ F_{x}(ℓ_{0}, h_{0}) ⊆ π_{x}(R_{n}). Assume that x_{1} ≠ x_{0}, x_{1} ∈ (x_{0} − δ_{n}, x_{0} + δ_{n}) ∩ F_{x}(ℓ_{0}, h_{0}). Then there exists n′ ≥ n such that (x_{1}, g_{ℓ_{0},h_{0}}(x_{1})) ∈ R_{n′} but (x_{1}, g_{ℓ_{0},h_{0}}(x_{1})) ∉ R_{n′+1}. Then denoting by R_{n′,j} the subparallelograms of R_{n′} which belong to R_{n′} ∩ F_{n′+1} the points (x_{0}, g_{ℓ_{0},h_{0}}(x_{0})) and (x_{1}, g_{ℓ_{0},h_{0}}(x_{1})) belong to different R_{n′,j}’s. From our construction it is clear that m_{n′} = k_{n′}/2^{n′} with a suitable 0 ≤ k_{n′} < 2^{n′} thus by (2) we have

\frac{1}{3 · 2^{n′}} + m_{n′} ≤ \frac{g_{ℓ_{0},h_{0}}(x_{1}) − g_{ℓ_{0},h_{0}}(x_{0})}{x_{1} − x_{0}} ≤ \frac{3}{2^{n′}} + m_{n′}.

Using (3) we obtain

\frac{1}{3 · 2^{n}} + m_{n} ≤ \frac{g_{ℓ_{0},h_{0}}(x_{1}) − g_{ℓ_{0},h_{0}}(x_{0})}{x_{1} − x_{0}} ≤ \frac{4}{2^{n}} + m_{n}.

Also from (3) it follows that m_{n} converges to a number, say m and we have m_{n} ≤ m ≤ m_{n} + 1/2^{n}. Thus

\frac{−2}{3 · 2^{n}} + m ≤ \frac{g_{ℓ_{0},h_{0}}(x_{1}) − g_{ℓ_{0},h_{0}}(x_{0})}{x_{1} − x_{0}} ≤ \frac{4}{2^{n}} + m

for any x_{1} ∈ (x_{0} −δ_{n}, x_{0} +δ_{n})∩F_{x}(ℓ_{0}, h_{0}), x_{1} ≠ x_{0}. This implies that g′_{ℓ_{0},h_{0}}(x_{0}) = m where m is the limit of the slopes of the parallelograms R_{n}. An argument, similar to the above one, can imply that if x′_{0} ∈ (x_{0} − δ_{n}, x_{0} + δ_{n}) ∩ F_{x}(ℓ_{0}, h_{0}) ⊆ π_{x}(R_{n}) and m′ = g′_{ℓ_{0},h_{0}}(x′_{0}) then m_{n} ≤ m′ ≤ m_{n} + 1/2^{n} and hence |m′ − m| ≤ 1/2^{n}. From this it follows that g_{ℓ_{0},h_{0}} is a C^{1} function on F_{x}(ℓ_{0}, h_{0}). Using Whitney’s extension theorem we can choose a C^{1} extension of g_{ℓ_{0},h_{0}} onto R. In the sequel we will denote by g_{ℓ_{0},h_{0}} this extended function.

Step 5 Definition of the parallelograms R_{i_{1},...,i_{n}}^{m}.

In Steps 5-7 we assume that ℓ_{0}, h_{0} are fixed. Given m ∈ [0, 1) we want to find many points in F_{x}(ℓ_{0}, h_{0}) where g′_{ℓ_{0},h_{0}} = m. For the above given m choose a sequence m_{0}, m_{1}, ... such that 0 ≤ m − m_{n} < 1/2^{n} and m_{n} is of the form k/2^{n} with an integer 0 ≤ k < 2^{n}. It is clear that F_{1} consists of 32 parallelograms of slope 0 and of 32 parallelograms of slope 1/2 and, in general, the set F_{n} contains 32^{n} parallelograms of slope k/2^{n} for any k ∈ {0, ..., 2^{n} − 1}. Assume that the 32 parallelograms in F_{1} with slope m_{1} are denoted by R_{i_{1}}^{m}, (i_{1} = 0, ..., 31). If the parallelograms R_{i_{1},...,i_{n}}^{m} are defined then each R_{i_{1},...,i_{n}}^{m} contains 32 parallelograms belonging to F_{n+1} with slope m_{n+1}, denote these parallelograms by R_{i_{1},...,i_{n},i_{n+1}}^{m}, (i_{n+1} = 0, ..., 31). Thus by induction we define for each n the parallelograms R_{i_{1},...,i_{n}}^{m} such that i_{j} ∈ {0, ..., 31} for j = 0, ..., n, furthermore for any sequence i_{1}, i_{2}, ... we have R_{i_{1},...,i_{n},i_{n+1}}^{m} ⊆ R_{i_{1},...,i_{n}}^{m}, and the slope of R_{i_{1},...,i_{n}}^{m} is m_{n}. It is clear from Step 4 that if x_{0} = π_{x}(∩_{n=1}^{∞}R_{i_{1},...,i_{n}}^{m}) then g′_{ℓ_{0},h_{0}}(x_{0}) = m.

Step 6 Further study of R_{i_{1},...,i_{n}}^{m}’s and definition of the intervals I_{i_{1},...,i_{n}}^{m}.

Assume that R_{i_{1},...,i_{n}}^{m} is a parallelogram of the from (x_{0}, y_{0}), ℓ, h, m_{n} with ℓ = ℓ_{0}/128^{n}, h = h_{0}/256^{n}. Put d′ = 1/2^{n+1}, d = ℓd′/h, d′_{0} = m − m_{n}. Then 0 ≤ d′_{0} < 1/2^{n} and we also put d_{0} = ℓd′_{0}/h. For ease of notation put ϕ = ϕ_{R_{i_{1},...,in}^{m}} and ψ = ψ_{R_{i_{1},...,in}^{m}} . Then R_{i_{1},...,i_{n},i_{n+1}}^{m} is a parallelogram of the form ψ(Q_{j}) where Q_{j} is a parallelogram belonging to Q_{0}, ..., Q_{63} 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 = d_{0}x, (see the end of Step 2). Let π_{m}(x, y) = y − mx. Put I_{i_{1},...,i_{n}}^{m} = π_{m}(R_{i_{1},...,i_{n}}^{m}). Recall that \overline{π}_{d_{0}} was defined in Step 1 and observe that ψ ∘ \overline{π}_{d_{0}} ∘ ϕ(x, y) = (x_{0}, y − m(x − x_{0})) and hence (π_{y} ∘ ψ ∘ \overline{π}_{d_{0}} ∘ ϕ(x, y)) − mx_{0} = π_{m}(x, y). Since ϕ(R_{i_{1},...,i_{n}}^{m}) = Q and by Step 1 we know that 1 ≤ |\overline{I}| = |\overline{π}_{d_{0}}(Q)| ≤ 129/128, we infer h ≤ |ψ ∘ \overline{π}_{d_{0}} ∘ ϕ(R_{i_{1},...,i_{n}}^{m})| ≤ 129h/128. Using that π_{y} and the translation by −mx_{0} are not changing vertical distances we obtain that

\frac{h_{0}}{256^{n}} ≤ |I_{i_{1},...,i_{n}}^{m}| ≤ \frac{129h_{0}}{128 · 256^{n}} .

Step 7 The J subinterval property of the I_{i_{1},...,i_{n}}^{m}’s.

By our construction the parallelograms R_{i_{1},...,i_{n},i_{n+1}}^{m} are either of the form ψ(Q_{2i_{n+1}}), (i_{n+1} = 0, ..., 31) when m_{n} ≤ m < m_{n}+ \frac{1}{2^{n+1}} , or of the form ψ(Q_{2i_{n+1}+1}), (i_{n+1} = 0, ..., 31) when m_{n}+ \frac{1}{2^{n+1}} ≤ m < m_{n}+ \frac{1}{2^{n}} . In the first case put \overline{J}_{i_{n+1}} = \overline{π}_{d_{0}}(Q_{2i_{n+1}}), in the second put \overline{J}_{i_{n+1}} = \overline{π}_{d_{0}}(Q_{2i_{n+1}+1}). The remark at the end of Step 1 implies that if \overline{J} is a subinterval of \overline{I} = \overline{π}_{d_{0}}(Q) = \overline{π}_{d_{0}} ∘ ϕ(R_{i_{1},...,i_{n}}^{m}) with |\overline{J}| > 6/128 then there is an i_{n+1} ∈ \{0, ..., 31\} such that \overline{J}_{i_{n+1}} ⊆ \overline{J}. Therefore an easy computation yields that if J is a subinterval of I_{i_{1},...,i_{n}}^{m} = (π_{y} ∘ ψ ∘ \overline{π}_{d_{0}} ∘ ϕ(R_{i_{1},...,i_{n}}^{m})) − mx_{0} and |J| > 6h/128 = 6h_{0}/(128 · 256^{n}) then there is an i_{n+1} ∈ {0, ..., 31} such that I_{i_{1},...,i_{n},i_{n+1}}^{m} ⊆ 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 ≤ m′ < 1, ℓ_{0} = 1, h_{0} = 128, ℓ′_{0} = 1, h′_{0} = 2048 = 16h_{0}. Put R = [0, 1] × [0, 128], R′ = [0, 1] × [0, 2048], I^{m} = π_{m}(R), I^{m′} = π_{m′}(R′). Then [0, 128] ⊆ I^{m} ⊆ [−1, 128], [0, 2048] ⊆ I^{m′} ⊆ [−1, 2048]. If c ∈ [128, 2047] then J_{0} = c − I^{m} ⊆ [0, 2048] ⊆ I^{m′} and |J_{0}| ≥ 128 > 6h′_{0}/128 therefore by the J subinterval property there exists i′_{1} ∈ {0, ..., 31} such that I_{i′_{1}}^{m′} ⊆ J_{0}. Then using Step 6 we obtain

6 = \frac{6h_{0}}{128} < 8 = \frac{h′_{0}}{256} ≤ |I_{i′_{1}}^{m′}| ≤ \frac{129h′_{0}}{128 · 256}.

Put J′_{1} = c − I_{i′_{1}}^{m′}. Since I_{i′_{1}}^{m′} ⊆ J_{0} = c − I^{m} we have J′_{1} ⊆ I^{m}. From |J′_{1}| = |I_{i′_{1}}^{m′}| > 6h_{0}/128 it follows that there exists i_{1} such that I_{i_{1}}^{m} ⊆ J′_{1}. Then

\frac{6h′_{1}}{128} = \frac{6h′_{0}}{128 · 256} = \frac{6 · 16h_{0}}{128 · 256} < \frac{h_{0}}{256} ≤ |I_{i_{1}}^{m}| ≤ \frac{129h_{0}}{128 · 256}.

Put J_{1} = c − I_{i_{1}}^{m}. Since I_{i_{1}}^{m} ⊆ J′_{1} = c − I_{i′_{1}}^{m′} we have J_{1} ⊆ I_{i′_{1}}^{m′}. From |J_{1}| = |I_{i_{1}}^{m}| > 6h′_{1}/128 it follows that there exists i′_{2} such that I_{i′_{1},i′_{2}}^{m′} ⊆ J_{1}. Then

\frac{6h_{1}}{128} = \frac{6h_{0}}{128 · 256} < \frac{h′_{0}}{256 · 256} = \frac{h′_{1}}{256} ≤ |I_{i′_{1},i′_{2}}^{m′}| ≤ \frac{129h′_{1}}{128 · 256}.

Put J′_{2} = c − I_{i′_{1},i′_{2}}^{m′}...

From the above steps it should be clear that by mathematical induction we can choose sequences i_{1}, i_{2}, ..., and i′_{1}, i′_{2}, ... such that I_{i_{1},...,i_{n}}^{m} ⊆ c − I_{i′_{1},...,i′_{n}}^{m′}, and I_{i′_{1},...,i′_{n},i′_{n+1}}^{m′} ⊆ c − I_{i_{1},...,i_{n}}^{m} for any n. Since |I_{i_{1},...,i_{n}}^{m}| → 0, and |I_{i′_{1},...,i′_{n}}^{m′}| → 0 as n → ∞ there are unique c_{0} and c′_{0} such that c_{0} = ∩_{n=1}^{∞}I_{i_{1},...,i_{n}}^{m} and c′_{0} = ∩_{n=1}^{∞}I_{i′_{1},...,i′_{n}}^{m′}. It is also clear that c_{0} + c′_{0} = c. Let (x_{0}, y_{0}) = ∩_{n=1}^{∞}R_{i_{1},...,i_{n}}^{m} and (x′_{0}, y′_{0}) = ∩_{n=1}^{∞}R_{i′_{1},...,i′_{n}}^{m′}. Then y_{0} −mx_{0} = π_{m}(x_{0}, y_{0}) = c_{0} and y′_{0} − m′x′_{0} = π_{m′}(x′_{0}, y′_{0}) = c′_{0}. Using the properties of the functions g_{ℓ_{0},h_{0}} and g_{ℓ′_{0},h′_{0}} we have y_{0} = g_{ℓ_{0},h_{0}}(x_{0}), y′_{0} = g_{ℓ′_{0},h′_{0}}(x′_{0}), g′_{ℓ_{0},h_{0}}(x_{0}) = m, and g′_{ℓ′_{0},h′_{0}}(x′_{0}) = m′.

Step 9. Definition of f(x, y).

Put f(x, y) = g_{ℓ_{0},h_{0}}(x) + g_{ℓ′_{0},h′_{0}}(y). Then by the argument of Step 8 for any m, m′ ∈ [0, 1) and c ∈ [128, 2047] there exist x_{0} ∈ F_{x}(ℓ_{0}, h_{0}) and x′_{0} ∈ F_{x}(ℓ′_{0}, h′_{0}) such that letting a = m = g′_{ℓ_{0},h_{0}}(x_{0}) = ∂_{x}f(x_{0}, x′_{0}), and b = m′ = g′_{ℓ′_{0},h′_{0}}(x′_{0}) = ∂_{y}f(x_{0}, x′_{0}) we have c = c_{0} + c′_{0} = g_{ℓ_{0},h_{0}}(x_{0}) − ax_{0} + g_{ℓ′_{0},h′_{0}}(x′_{0}) − bx′_{0} = f(x_{0}, x′_{0}) − ax_{0} − bx′_{0} for any (a, b, c) ∈ [0, 1) × [0, 1) × [128, 2047]. Since it is easy to see that the set E = F_{x}(ℓ_{0}, h_{0}) × F_{x}(ℓ′_{0}, h′_{0}) is a nowhere dense closed set and λ_{2}(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 two. Since f is a C^{1} function and for a fixed (a, b, c) ∈ H the set {f(x, y) − ax − by : (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 ϵ > 0 one can obtain an example when the Hausdorff dimension of E is less than 1 + ϵ. However it is not clear for us whether there exists an example when E is of Hausdorff dimension 1.

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