Basic embeddings and Hilberts 13th problem

Let us recall informally the concept of superposition. Suppose that there is a set of functions of several variables, including all variables considered as functions. Represent each of the functions as an element of a circuit with several entries and one exit. Then a superposition of functions of this set is a function that can be repsesented by a circuit constructed from given elements; the circuit should not contain oriented cycles. For example, a polynomial a n x n + a n-1 x n-1 + • • • + a 1 x + a 0 is a superposition of the constant functions and the functions f (x, y) = x + y, g(x, y) = xy. It is clear that any elementary function can be represented as a superposition of functions of at most two variables. Is it possible to represent each function of several arguments as a superposition of functions of at most two arguments? Since there is a 1-1 correspondence between a segment and a square, any function of three and more variables is superposition of (in general, discontinuous) functions of two variables. So the above question is only interesting for continuous functions. 3 From now on we assume all functions to be continuous, unless the contrary is explicitly specified. Hilbert's 13th problem. Can the equation x 7 + ax 3 + bx 2 + cx + 1 = 0 of degree seven be solved without using functions of three variables? This question was answered affirmatively in 1957 by Kolmogorov and Arnold. They proved that any continuous function of n variables defined on a compact subset of R n can be represented as a superposition of continuous functions of one variable and addition. For an exposition accessible to undergraduates see [Ar58]. See also [Vi04]. Ostrand extended the Kolmogorov-Arnold Theorem this theorem to arbitrary n-dimensional compacta [St89]. It is in the Kolmogorov-Arnold-Ostrand papers that the notion of basic subset 1 This is an English version of the paper in Russian under the same title. The English version has much shorter first section (which corresponds to two sections in Russian version), but contains solutions of problems 14a and 16c from the third section. Whenever possible I give references to surveys not to original papers. I would like to acknowledge V.I.Arnold, Yu.M. Burman, I.N. Shnurnikov, A.R. Safin, S.M. Voronin and M. Vyaliy for useful discussions, and M. Vyaliy for preparation of figures. 2 skopenko@mccme.ru, http://dfgm.math.msu.su/people/skopenkov/papersc.ps 3 Denote by |x, y| = |(x 1 , . . . , x n ), (y 1 , . . . , y n )| = (x 1y 1 ) the ordinary distance between points x = (x 1 , . . . , x n ) and y = (y 1 , . . . , y n ) of R n . Let K be a subset of R n . A function f : K → R is called continuous if for each point x 0 ∈ K and number ε > 0 there exists a number δ > 0 such that for each point 2 is continuous on the plane, whereas the function f (x 1 , x 2 ) equal to the integer part of x 1 + x 2 is not. 4 This subsection is not used in the sequel and so can be omitted. 1 It is interesting to compare this theorem with the Nöbeling-Menger-Pontryagin theorem on embeddability of any n-dimensional compact space into R 2n+1 and the example of an n-dimensional polyhedron non-embeddable into R 2n . Obviously, K is basically embeddable into R if and only if K is topologically embeddable into R. It follows from Theorem 1 that a compactum K is basically embeddable into R m for m > 2 if and only if dim K < m/2. Thus, the only remaining case is m = 2 (Sternfeld's problem). A subset K of R 2 is basic if for each continuous function f : K → R there exist continuous functions g, h : R → R such that f (x, y) = g(x) + h(y) for each point (x, y) ∈ K. Let us present the characterization of arcwise connected compacta basically embeddable into the plane [Sk95] (this is a partial solution of Sternfeld's problem). We formulate the criterion first for graphs and then for the general case. A conjecture on embeddability of (not necessarily arcwise connected) connected compacta into the plane can be found in [Sk95]. Compacta used in the statements are defined after the statements. A finite graph K is basically embeddable into the plane if and only if any of the following two equivalent conditions holds: (a) K does not contain subgraphs homeomorphic to S, C 1 , C 2 (fig. 1), that is, a circle, a fivepoint star, and a cross with branched endpoints; (b) K is contained in one of the graphs R n , n = 1, 2, 3, . . . (fig. 2). Let F 1 be a triod. The graph F n+1 is obtained from F n by branching its endpoints (fig. 2). The graph R n is obtained from F n by by adding a hanging edge to each non-hanging vertex. Theorem 3. [Sk95] An arcwise-connected compactum K is basically embeddable into the plane if and only if it is locally connected (i.e., is a Peano continuum) and any of the two following (equivalent) conditions hold: (1) K does not contain S 1 , C 2 , C 4 , B as subcompacta and contains only finitely many subcontinua F n , H n (fig. 1,2,3); (2) K does not contain any of the continua A sequence of sets is called a null-sequence if their diameters tend to zero. Define • H n to be the union of I with a null-sequence of triods having endpoints attached to • C 3 to be a cross with a null-sequence of arcs attached to one of its branches and converging to its center; • C 4 to be a cross with a sequence of points converging to its center; • B to be the union of the arc I and a null-sequence of arcs attached to (0; 1) by their endpoints at rational points; • F to be the union of I with a null-sequence of sets F n each having an endpoint attached to the point 1/n ∈ I; • H + (H -) to be the union of I with a null-sequence of continua H n connected to the points 1/n ∈ I by arcs that intersect H n at the points 1 ∈ I ⊂ H n (0 ∈ I ⊂ H n-1 , respectively); • h + (h -) to be obtained from a null-sequence of continua H n by pasting together the points 1 ∈ I ⊂ H n and 0 ∈ I ⊂ H n-1 (0 ∈ I ⊂ H n and 1 ∈ I ⊂ H n-1 , respectively). An embedding K ⊂ X × Y is basic if for any continuous function f : K → R there exist continuous functions g : Denote by T n an n-od, i.e., an n-pointed star. A vertex of a graph K is called horrible if its degree is greater than 4 and awful if its degree is equal to 4 and it is not an endpoint of a hanging edge. The defect of a graph K is the sum δ(K) = (degA 1 -2) + • • •+ (degA k -2), where A 1 , . . . , A k are all the horrible and awful vertices of K. Theorem 4. [Ku99] A finite graph K admits a basic embedding K ⊂ R × T n if and only if K is a tree and either δ(K) < n or δ(K) = n and K has a horrible vertex with a hanging edge. The material is presented as a sequence of problems, which is peculiar not only to Zen monasteries but also to elite mathematical education (at least in Russia). Difficult problems are marked by a star, and unsolved problems by two stars. If the statement of a problem is an assertion, then it is required to prove this assertion. Discontinuously basic subsets. 1. (a) Is it true that for any four numbers f 11 , f 12 , f 21 , f 22 there exist four numbers g 1 , g 2 , h 1 , h 2 such that f ij = g i + h j for each i, j = 1, 2? (b) Andrey Nikolaevich and Vladimir Igorevich play the 'Dare you to decompose!' game. Some cells of chessboard are marked. A. N. writes numbers in the marked cells as he wishes. V. I. looks at the written numbers and chooses (as he wishes) 16 numbers a 1 , . . . , a 8 , b 1 , . . . , b 8 as 'weights' of the columns and the lines. If each number in a marked cell turns out to be equal to the sum of weights of the line and the row (of the cell), then V. I. wins, and in the opposite case (i.e., when the number in at least one marked cell is not equal to the sum of weights of the line and the row) A. N. wins. Prove that V. I. can win no matter how A. N. plays if and only if there does not exist a closed route of a rook starting and turning only at marked cells (the route is not required to pass through each marked cell). Let R 2 be the plane with a fixed coordinate system. Let x(a) and y(a) be the coordinates of a point a ∈ R 2 . An ordered set (either finite or infinite) {a 1 , . . . , a n , . . . } ⊂ R 2 is called an array if for each i we have a i = a i+1 and x(a i ) = x(a i+1 ) for even i and y(a i ) = y(a i+1 ) for odd i. It is not assumed that points of an array are distinct. An array is called closed if a 1 = a 2l+1 . 2. Consider a closed array {a 1 , . . . , a n = a 1 }. A decomposition for such an array is an assignment of numbers at the projections of the points of the array on the x-axis and on the y-axis. Is it possible to put numbers f 1 , . . . , f n ∈ R, where f 1 = f n , at the points of the array so that for each decomposition there exists an f i that is not equal to the sum of the two numbers at x(a i ) and y(a i )? A criterion for a subset of the plane to be discontinuously basic. A subset of the plane is discontinuously basic if and only if it does not contain any closed arrays. 4.** Given a set of marked unit cubes in the cube 8 × 8 × 8, how can we see who wins in the 3D analogue of the 'Dare you to decompose!' game? In this analogue V. I. tries to choose 24 numbers a 1 , . . . , a 8 , b 1 , . . . , b 8 , c 1 , . . . , c 8 so that the number at the unit cube (i, j, k) would be equal to the sum a i + b j + c k of the three weights. 5.** (a) Define discontinuous basic subsets of the 3-space. Discover and prove the 3D analogue of the above criterion. (b) The same for higher-dimensional case. Solutions. (b) The statement 'only if' follows from the problem 2. Let us prove the 'if' part by induction on the number of the marked cells. If only one cell is marked then we are done. Let K be the set of centres of the marked cells. The set E(K) is defined in the following subsection after Problem 9. The set K does not contain any closed array, therefore #E(K) < #K. So by the induction hypothesis V. I. can win for E(K). Each cell from K -E(K) is the only marked cell on its line or column, thus V. I. can choose the remaining weights for K. 2. Yes, it is. If every f i is equal to the sum of two numbers at x(a i ) and y(a i ), then but this is false for some numbers f i . Set h(y) = f (0, y) and g(x) = 0. (b) Set g(x) = f (x, 0) and h(y) = f (0, y)f (0, 0). (c) The statement 'only if' follows from the problem 2. Let us prove the 'if' part. Consider a function f : K → R. Our aim is to construct functions g and h so that f (x, y) = g(x) + h(y). Two points a, b ∈ K are called equivalent if there is an array {a = a 1 , . . . , a n = b} ⊂ K. Now take an equivalence class K 1 ⊂ K. Define function g : x(K 1 ) → R and h : y(K 1 ) → R in the following way. Take any point a 1 ∈ K 1 and set g(x(a 1 )) = f (a 1 ) and h(y(a 1 )) = 0. If {a 1 , a 2 , . . . , a 2l } is an array, then set )) is already defined). Make this construction for each equivalence class. Then set g = 0 and h = 0 at all other points of R. A subset K ⊂ R 2 is called (continuously) basic if for each continuous function f : K → R there exist continuous functions g, h : R → R such that f (x, y) = g(x) + h(y) for each point (x, y) ∈ K. The Arnold problem. Which subsets of the plane are basic? [Ar58] In order to approach a solution consider some examples. 6. (a) A closed array is not basic. A sequence of points {a 1 , . . . , a n , . . . } ⊂ R 2 converges to a point a ∈ R 2 if for each ε > 0 there exists an integer N such that for each i > N we have |a i , a| < ε. 7. (a) If a subset of the plane is basic, then it is discontinuously basic. (b) A completed array is the union of a point a 0 ∈ R 2 with an infinite array {a 1 , . . . , a n , . . . } ⊂ R 2 of distinct points which converges to the point a 0 . Prove that any completed array is not basic. (Note that it is discontinuously basic). (c) Let [a, b] be the rectilinear arc which connects points a and b. Prove that the cross K = . Consider the set of points (m i,2l , m i,2l ) and (m i,2l , m i,2l-2 ), where i varies from 1 to ∞ and l = 1, 2, 3, . . . , 2 i-1 . Prove that this subset of the plane does not contain any infinite arrays but contains arbitrary long arrays. (e) The union of the set from the previous problem and the point (2, 2) is not basic. A subset K ⊂ R 2 of the plane is closed, if for each sequence a i ∈ K converging to a point a this point belongs to K. A subset K ⊂ R 2 of the plane is closed if and only if for each point a ∈ K there exists ε > 0 such that if for a point b of the plane we have |a, b| ≤ ε, then b does not belong to K. The Sternfeld criterion for being a basic subset. A closed bounded subset K ⊂ R 2 of the plane is basic if and only if K does not contain arbitrary long arrays. The criterion is false without the assumption that K closed. (b) The criterion is false without the assumption that K bounded. (c)** Find a criterion of being a basic subset for closed (but unbounded) subsets of the plane. Suppose that K is a subset of R 2 . For every point v ∈ K consider the pair of lines passing through v and parallel to the x-axis and the y-axis. If one of these two lines intersects K only at point v, we colour v in white. Define E(K) as the set of noncoloured points of K: 10. A subset K of the plane does not contain arbitrary long arrays if and only if E n (K) = ∅ for some n. 12. (a)* Give an elementary proof that if K is a closed bounded subset of R 2 and E(K) = ∅, then K is basic [Mi09]. Hint. It can be proven that for piecewise-linear maps f there is a decomposition f (x, y) = g(x) + h(y) with |g| + |h| < 5|f |. (b)* Prove the 'if' part of the criterion without using the functional spaces as below. Hint. Same as above with |g| + |h| < C n |f |, where C n depends only on that n for which The set of 4 points (0, 0, 0); (1, 1, 0); (0, 1, 1); (1, 0, 1) is basic. (But E n (K) = ∅ for each n, see below.) (c)* Define E(K) analogously to the above, only instead of lines use planes orthogonal to the axes: Solutions. But this is false for some functions f . Cf. problem 2. (b),(c) Analogously to problems 3a,3b. (d) Take h(y) = 0 and g(x) = f (x, y). If the subset is not discontinuously basic, then it contains a closed array. Hence the statement follows by extension of f on the subset and using problem 6a. (b) Define function f by f (a n ) = (-1) n n . Suppose that f (x, y) = g(x) + h(y) for some g and h. Then Since lim l→∞ h(y 2l ) exists and equals to h(y(a 0 )), it follows that 2l i=1 (-1) i f (a i ) converges when l → ∞, which is a contradiction. (c) The cross contains a completed array Define a function f on this array using problem 7.b and then extend it (e.g. piecewise linearly) to the cross. Then there are no functions g and h such that f (x, y) = g(x) + h(y). (d) For every i the set (m i,2l , m i,2l l=1 is an array of 2 i points. (e) Define a function f by If f (x, y) = g(x) + h(y) for some g and h, then for every i using array of points (m i,2l , m i,2l ) and (m i,2l , m i,2l-2 ), where l = 1, 2, 3, . . . This contradicts to the continuity of h. Let us prove the 'only if' part. Let K be a closed subset of the plane. Suppose that for some point a = (x, y) ∈ K and for each ε = 1 n > 0 there exists a point a n ∈ K (at least one) such that |a, a n | ≤ 1 n . The sequence of points a n ∈ K converges to the point a, thus a ∈ K. Contradiction. Now let us prove the 'if' part. Suppose that a sequence a n converges to a point a and the point a = (x, y) is not in K. There exists ε > 0 such that for every point a n ∈ K the distance |a, a n | > ε. This is a contradiction. Let us prove the 'only if' part. Suppose that E n (K) = ∅ for each n. For each n take a point a 0 ∈ E n (K). Then there exist points a -1 , a 1 ∈ E n-1 (K) such that x(a -1 ) = x(a 0 ) and y(a 1 ) = y(a 0 ). Analogously there exist points a -2 , a 2 ∈ E n-2 (K) such that {a -2 , a -1 , a 0 , a 1 , a 2 } is an array. Analogously we construct an array of 2n + 1 points in K, which is a contradiction. Let us prove the 'if' part. Suppose that K contains an array of 2n + 1 points {a -n , . . . , a 0 , . . . , a n }. Then there is an array of 2n -1 points {a -n+1 , . . . , a n-1 } in E(K). Analogously a 0 ∈ E n (K). Thus if E n (K) = ∅, then K does not contain an array of 2n + 1 points. and 2l(1) = -f (0, 0, 0)f (1, 1, 0) + f (1, 0, 1) + f (0, 1, 1). Proof of the criterion for being a basic subset. Let K be a closed bounded subset of the plane. It is known that each continuous function Beginning of the proof of the 'only if ' part of the criterion. Assume to the contrary that K contains arbitrary long arrays and is basic. Choosing subsequences we may assume that points of each array are distinct. Therefore for each n there is an array {a n 1 , . . . , a n 2n+5 } of 2n + 5 distinct points in K. Then there exists continuous function (Indeed, first define such a continuous function f : R 2 → R. Denote s = min i n for each n. For this it suffices to prove that Lemma. Let m ≥ 4, • K = {a 1 , . . . , a 2m+5 } be an array of 2m + 5 distinct points, Proof. We may assume that a 1 a 2 Ox. Then Therefore g(x(a 1 ))g(x(a 2m+4 )) ≥ (2m + 4) -3 > 2m. This implies the required inequality. Completion of the proof of the 'only if ' part of the criterion. We have Apply the Lemma to This is possible because f (x, y) = g(x) + h(y) and (since . By Lemma we obtain Proof of the criterion. 5 The proof is based on a reformulation of the property of being a basic subset in terms of bounded linear operators in Banach functional spaces. Denote by C(X) the space of continuous functions on X with the norm |f | = sup{|f (x)| : x ∈ X}. In this proof denote by pr x (a) and pr y (a) the projections of a point a ∈ K on the coordinate axes. For K ⊂ I 2 := [0; 1] × [0; 1] define a map (linear superposition operator) Clearly, the subset K ⊂ I 2 is basic if and only if φ is surjective, or equivalently, epimorphic. Denote by C * (X) the space of bounded linear functions C(X) → R with the norm |µ| = sup{|µ(f )| : f ∈ C(X), |f | = 1}. For a subset K ⊂ I 2 define a map (dual linear superposition operator) Since |φ * µ| ≤ 2|µ|, it follows that φ * is bounded. By duality, φ is epimorphic if and only if φ * is monomorphic. 6It is clear that φ * is monomorphic if and only if (*) there exists ε > 0 such that |φ * µ| > ε|µ| for each unzero µ ∈ C * (K). We leave as an excercise the proof that (*) implies the abcense of arbitrarily large arrows. (This proves the 'only if' part of the criterion, for which we already have an elementary proof.) So it remains to prove that E n (K) = ∅ implies the condition (*). We present the proof for n ∈ {1, 2}. The proof for arbitrary n is analogous. We use the following non-trivial fact: C * (X) is the space of σ-additive regular real valued Borel measures on X (in the sequel we call them simply 'measures'). We have φ * µ = (µ x , µ y ), where µ x (U) = µ(pr -1 x U) and µ y (U) = µ(pr -1 y U) for each Borel set U ⊂ I. If µ = µ +µ -is the decomposition of a measure µ into its positive and negative parts, then |µ| = μ(X), where μ = µ + + µ -is the absolute value of µ. Let D x (D y ) be the set of points of K which are not shadowed by some other point of K in x-(y-) direction. Take any measure µ on K of the norm 1. If n = 1, then Therefore without loss of generality, μ(D x ) ≥ 1/2. Since the projection onto the x-axis is injective over D x , it follows that |µ x | ≥ 1/2, thus the required assertion holds for ε = 1 2 . If n = 2, then In the case when μ(E(K)) < 3/4 we have μ(D x ∪ D y ) > 1/4 and without loss of generality μ(D x ) > 1/8. Then as for n = 1 we have |µ x | > 1/8, thus (*) holds for ε = 1 8 . In the case when μ(E(K)) ≥ 3/4 we have μ(K -E(K)) ≤ 1/4. By the case n = 1 above without loss of generality μx (pr Smoothly basic subsets of the plane. Let K be a subset of the plane R 2 . A function f : K → R is called differentiable if for each point z 0 ∈ K there exist a vector a ∈ R 2 and infinitesimal function α : R 2 → R such that for each point z Here the dot denotes scalar product of vectors a =: (f x , f y ) and zz 0 =: (x, y), i.e. a • (zz 0 ) = xf x + yf y . A function α : R 2 → R is infinitesimal, if for each number ε > 0 there exists a number δ > 0 such that for each point (x, y) ∈ R 2 if x 2 + y 2 < δ, then |α(x, y)| < ε. Let V be the graph of the function y = |x|, where x ∈ [-1; 1]. A function f : V → R is differentiable if and only if f (x, |x|) is differentiable on the segments [-1; 0] and [0; 1]. A subset K ⊂ R 2 of the plane is called differentiably basic if for each differentiable function f : K → R there exist differentiable functions g : R → R and h : R → R such that f (x, y) = g(x) + h(y) for each point (x, y) ∈ K. (c) The broken line whose consecutive vertices are (-2, 0), (-1, 1), (0, 0), (1, 1) and (2, 0) is not differentiably basic. (Note that it is continuously basic). (d) The completed array {([ (Note that it is also not continuously basic.) (e) The completed array {(2 )} ∞ n=1 ∪ {(0, 0)} is differentiably basic. (Note that it is not continuously basic.) (f) (I. Shnurnikov) The cross K = [(-1, -2), (1, 2)] ∪ [(-1, 1), (1, -1)] is not differentiably basic. (This assertion and Conjecture 15a imply that the property of being differentably basic is not hereditary.) (g) If a graph is basically embeddable in the plane, then it is differentiably basically embeddable in the plane. (This is non-trivial because the plane contains graphs which are basic but not differentaibly basic and vice versa.) [RZ06] 15.** Conjectures. (a) (I. Shnurnikov) A completed array {a n } ∞ n=1 ∪{(0, 0)} is differentiably basic if and only if the sequence 2 ; 1 2 ] of the plane is not differentiably basic. Hint. One can try to prove this analogously to 14f. Cf. [Vo81,Vo82]. (c) A piecewise-linear graph in R 2 is differentiably basic if and only if it does not contain arbitrary long arrays and for each two singular points a and b we have x(a) = x(b) and y(a) = y(b). A point a ∈ K is singular if the intersection of K with each disk centered at a is not a rectilinear arc. It would be interesting to find a criterion of being differentiably basic for closed bounded subsets of the plane. Apparently a simple-to-state criterion (analogous to the Sternfeld criterion) does not exist. Another interesting question: is there a continuous map [0; 1] → R 2 whose image is differentiably basic but not basic? 16. Let r ≥ 0 be an integer and K ∈ R 2 a subset. A function f : K → R is called r times differentiable if for each point z 0 ∈ K there exist a polynomial f (z) = f (x, y) of degree at most r of 2 variables x and y and an infinitesimal function α : R 2 → R such that f (z) = f (zz 0 ) + α(zz 0 )|z, z 0 | r for each point z ∈ K. (This definition differs from the one generally accepted.) (a) Functions differentiable zero times are exactly continuous functions, and functions differentiable one time are exactly differentiable functions. (b) For each positive integer r define the property of being an r times differentiably basic subset of the plane R 2 . (c) For each integer k ≥ 0 there is a subset of the plane which is r times differentiably basic for r = 0, 1 . . . k but is not r times differentiably basic for each r > k. (d)** Find a criterion for graphs in R 2 to be r times differentiably basic. 13. (a), (b), (c) Analogously to problems 6(a), 3(a) and 3(b). 14. (a) Take a differentiable function f : V → R. Since f is differentiable at (0, 0), it follows that there exist a, b ∈ R such that f (x, |x|) = f (0, 0) + ax + b|x| + α(x), where α(x) = o( x 2 + |x| 2 ) when x → 0. Take h(y) := by and g(x) := f (0, 0) + ax + α(x). Clearly, h is differentiable and g is differentiable outside 0. Since α(x) = o(x) when x → 0, it follows that g is differentiable also at 0. (b) See 16c for k = 0. (c) Suppose the broken line is differentiably basic. The function f (x, y) = xy is differentiable. We have f (x, y) = g(x) + h(y), where both g and h are differentiable. Then Hence h ′ (1) = -1. A contradiction. (d) Suppose that this completed array is differentiably basic. Set a n = ([ n+1 2 ] -1/2 , [ n 2 ] -1/2 ), f (a n ) := (-1) n n , n = 2, 3, . . . . If f (x, y) = g(x) + h(y) for some functions g(x) and h(y), then the series f (a 2 )f (a 3 ) + f (a 4 ) -. . . converges to g(1)g(0) (analogously to Problem 7b). This is a contradiction because the series 1 2 + 1 3 + 1 4 + . . . diverges. (e) Without loss of generality assume that f (0, 0) = 0, then take g(0) = 0 and h(0) = 0. Set where the right-hand sides are sums of alternating series. Now g(x) and h(y) may be extended to differentiable functions R → R. (f) Define w(0) = w(4 -i + 4 -3i ) = w(4 -i ) = 0 and w(4 -i + 4 -3i-1 ) = 2 3i for i = 1, 2, 3, . . . . Extend piecewice-linearly to obtain a function w : [0; 1] → R. For x ∈ [0; 1] define W (x) as the area under the graph of w on [0; x]. (This is well-defined because this area is finite.) Define f (x, -x) = W (x) for x ∈ [0; 1] and f (x, y) = 0 on the rest of the cross. Clearly, f is differentiable outside (0, 0). It is easy to check that f is differentiable at (0, 0). o([x 2 + x 2r ] r/2 ) = o 1 (x r ), we have f (x, |x| k ) = a 00 + a 01 |x| k + a 10 x + • • • + a r0 x r + o 2 (x r ). Take h(y) = a 01 y and g(x) = f (x, |x| k )h(|x| k ). Clearly, h is r times differentiable and g is r times differentiable outside 0. We also have g(x) = a 00 + a 10 x + • • • + a r0 x r + o 2 (x r ) when x → 0. So g is r times differentiable also at 0. Next we prove that V = V 1 is not r times differentiably basic for each 1 < r. Define a differentiable function f : V → R by f (x, y) = xy, where y = |x|. If V is r times differentiably basic for some r ≥ 2, then there are r times differentiable functions g, h : R → R such that f (x, |x|) = x|x| = g(x) + h(|x|). Hence g(x)g(-x) = 2x 2 for x ∈ [0; 1]. But this is impossible because g is 2 times differentiable, hence for x → +0 g(x) = g(0) + ax + bx 2 + o(x 2 ) and g(-x) = g(0)ax + bx 2 + o(x 2 ). At last we prove that V k is not r times differentiably basic for k odd and each k < r. Define a differentiable function f : V k → R by f (x, y) = xy, where y = |x| k . If V is r times differentiably basic for some r > k, then there are r times differentiable functions g, h : R → R such that f (x, |x| k ) = x|x| k = g(x) + h(|x| k ). Hence g(x)g(-x) = 2x k+1 for each x ∈ [0; 1]. But this is impossible for k odd because g is (k + 1) times differentiable, hence for x → +0 g(x) = g 0 + g 1 x + • • • + g k+1 x k+1 + o(x k+1 ) and g(-x) = g 0g 1 x + • • • + g k+1 x k+1 + o(x k+1 ). This proof is not elementary, is not used in the sequel and could be omitted. We remark that φ * can be injective but not monomorphic. In other words not only some linear relation on im φ can force it to be strictly less than C(K). If an embedding K ⊂ R 2 is basic, then we can prove that φ * is monomorphic without use of φ as follows. Define a linear operator Ψ :C * (I) ⊕ C * (I) → C * (K) by Ψ(µ x , µ y )(f ) = µ x (g) + µ y (h),where g, h ∈ C(I) are such that g(0) = 0 and f (x, y) = g(x) + h(y) for (x, y) ∈ K. Clearly, Ψφ * = id and Ψ is bounded, hence φ * is monomorphic.