Markov’s inequality and polynomial mappings

Markov’s inequality is a certain estimate for the norm of the derivative of a polynomial in terms of the degree and the norm of this polynomial. It has many interesting applications in approximation theory, constructive function theory and in analysis (for instance, to Sobolev inequalities or Whitney-type extension problems). One of the purposes of this paper is to give a solution to an old problem, studied among others by Baran and Ple´sniak, and concerning the invariance of Markov’s inequality under polynomial mappings (polynomial images). We also address the issue of preserving Markov’s inequality when taking polynomial preimages. Lastly, we give a sufﬁcient condition for a subset of a Markov set to be a Markov set. ,


Introduction
Throughout the paper, K = R or C and R N will be treated as a subspace of C N . If ∅ = A ⊂ C N and f : A −→ C N , then we put f A := sup z∈A | f (z)|, where | | denotes the maximum norm. Moreover, N := {1, 2, 3, . . .} and N 0 := {0} ∪ N. We will also use the following notation: for each set ∅ = A ⊂ C N and each λ > 0, we put One of the most important polynomial inequalities is the following Markov's inequality (cf. [42]).

Theorem 1.1 (Markov) If P is a polynomial of one variable, then
Moreover, this inequality is optimal, because for the Chebyshev polynomials T n (n ∈ N 0 ), we have T n (1) = n 2 and T n [−1, 1] = 1.

Recall that
In fact, the above inequality for quadratic polynomials was discovered by the celebrated chemist Mendeleev. Markov's inequality and its various generalizations found many applications in approximation theory, analysis, constructive function theory, but also in other branches of science (for example, in physics or chemistry). There is now such extensive literature on Markov type inequalities that it is beyond the scope of this paper to give a complete bibliography. Let us mention only certain works which are most closely related to our paper (with emphasis on those dealing with generalizations of Markov's inequality on sets admitting cusps), for example [1][2][3][4][5][6][7]35,36,39,43,44,46,48,50,51,57,58]. We should stress here that the present paper owes a great debt particularly to Pawłucki and Pleśniak's work, because in [43] they laid the foundations for the theory of polynomial inequalities on "tame" (for example, semialgebraic) sets with cusps. From the point of view of applications, it is important that the constant (deg P) 2 in Markov's inequality grows not too fast (that is, polynomially) with respect to the degree of the polynomial P. This is the reason why the concept of a Markov set is widely investigated. Definition 1. 2 We say that a compact set ∅ = E ⊂ C N satisfies Markov's inequality (or: is a Markov set) if there exist ε, C > 0 such that, for each polynomial P ∈ C[z 1 , . . . , z N ] and each α = (α 1 , . . . , α N ) ∈ N N 0 , where D α P := ∂ |α| P ∂z α 1 1 . . . ∂z α N N and |α| := α 1 + · · · + α N .
Clearly, by iteration, it is enough to consider in the above definition multi-indices α with |α| = 1. We begin by giving some examples.
• Obviously, if ∅ = E 1 , . . . , E p ⊂ C N are compact sets satisfying Markov's inequality, then the union E 1 ∪ · · · ∪ E p satisfies Markov's inequality as well. In general, this is no longer so for the intersection E 1 ∩ · · · ∩ E p .
• It is straightforward to show that the Cartesian product of Markov sets is a Markov set. More precisely, if ∅ = E j ⊂ C N j (N j ∈ N) is a compact set satisfying Definition 1.2 with ε j , C j > 0 ( j = 1, . . . , p), then E 1 × · · · × E p ⊂ C N 1 +···+N p satisfies this definition with ε := max{ε 1 , . . . , ε p } and C := max{C 1 , . . . , C p }. • In Sect. 5, we give a sufficient condition for a subset of a Markov set to be a Markov set-see Note that a UPC set is in particular fat, that is E = IntE. In [43,44,46], some large classes of UPC sets (and hence Markov sets) are given. These classes include for example all compact, fat and semialgebraic subsets of R N (see Sect. 2 for the definition).
The following result is due to Baran and Pleśniak (cf. [3]).
Since each compact UPC set satisfies Markov's inequality, the Baran-Pleśniak theorem says that, under a certain assumption on a Markov set E ⊂ R N and under a certain assumption on a polynomial map h : R N −→ R N , the image h(E) also satisfies Markov's inequality.
Our aim is among others to show that in Theorem 1.3: • Very strong UPC assumption on the Markov set E is superfluous.
• The assumption that Jac h(ζ ) = 0 for each ζ ∈ IntE can be replaced by much weaker assumption that h : R N −→ R N and Moreover, the latter assumption is the weakest possible condition on the polynomial map h that must be assumed (see Lemma 2.3).
More precisely, we will prove the following result in Sect. It is worth noting that there is a holomorphic version of Theorem 1.3 in [3], which reads as follows. Suppose that E ⊂ C N is a compact, polynomially convex set satisfying Markov's inequality. If h : U −→ C N is a holomorphic map in a neighbourhood U of E such that h(E) is nonpluripolar and Jac h(ζ ) = 0 for each ζ ∈ E, then h(E) also satisfies Markov's inequality. (The notion of a polynomially convex set and the notion of a nonpluripolar set are defined in Sect. 3.) In connection with Theorem 1.4, the following question naturally arises.
Question 1.5 Suppose that ∅ = E ⊂ K N is a compact set satisfying Markov's inequality and g : The precise answer is not known to us. However, we will address this issue in Sects. 3 and 4. In particular, we will give some specific examples to show a variety of situations that we encounter exploring this problem. Eventually, we will give a result (Theorem 3.8) being a partial answer to Question 1.5.

A proof of Theorem 1.4
We will need the notion of a semialgebraic set and the notion of a semialgebraic map.
All semialgebraic sets constitute the simplest polynomially bounded o-minimal structure (see [59,60] for the definition and properties of o-minimal structures). However, the knowledge of o-minimal structures is not necessary to follow the present paper. Whenever we say "a set (a map) definable in a polynomially bounded o-minimal structure", the reader who is not familiar with the basic notions of o-minimality can just think of a semialgebraic set (map).
Before going to the proof of Theorem 1.4, it is worth noting that the assumption that rank h = N is necessary in this theorem, as is seen by the following lemma. Proof By Sard's theorem, the set h(K N ) has Lebesgue measure zero.
Case 1: K = R. By the Tarski-Seidenberg theorem (cf. [8,9,40]), the set h(R N ) is semialgebraic. Therefore h(R N ) = s j=1 H j , where s ∈ N and We can clearly assume that each H j is nonempty. Put P := P 1 · . . . · P s . Note that P ≡ 0 1 and P| h(R N ) ≡ 0. Take a point a ∈ h(E). For each w ∈ R N , we have and therefore D α P(a) = 0 for some α ∈ N N 0 . Since P h(E) = 0, it follows that h(E) does not satisfy Markov's inequality.
Case 2: K = C. By Chevalley's theorem, the set h(C N ) is constructible (see [41, pp. 393-396], for the definition and details). Moreover, h(C N ) = C N 2 and h(C N ) is a complex algebraic set (see [41, p. 394]), that is the set of common zeros of some collection of complex polynomials. In particular, there exists P ∈ C[w 1 , . . . , w N ] such that P ≡ 0 and P| h(C N ) ≡ 0. Arguing as in Case 1 we see that h(E) does not satisfy Markov's inequality. 3 We will try to keep the exposition as self-contained as possible. It should be stressed, however, that our proof of Theorem 1.4 is influenced by ideas from the original proof of Theorem 1.3 by Baran and Pleśniak.
Proof of Theorem 1.4. Clearly, it suffices to consider the case K = C. Put Take an open and bounded set I ⊂ C N such that E ⊂ I and χ −1 (I ) is semialgebraic (for example, a sufficiently large open polydisk).
Put A := I \T , where Since the set T is (complex) algebraic and nowhere dense (see [41, p. 158]), it follows that χ −1 (A) ⊂ R 2N is open, semialgebraic and χ −1 (A) = χ −1 (I ). Consequently, by Corollary 6.6 in [43], χ −1 (A) is UPC. Therefore there exist υ, θ > 0 and d ∈ N such that, for each x ∈ χ −1 (A), we can choose a polynomial map S x : R −→ R 2N satisfying the following conditions: By Lemma 3.1 in [44], the maps G 0 , G 1 , . . . , G d : are bounded. Thus there exists C 1 > 0 such that, for each z ∈ A and t ∈ [0, 1], where (3) (Use the fact that T ⊂ C N \A and consider two cases: T = ∅ and T = ∅.) Take ε, C > 0 such that, for each polynomial P ∈ C[z 1 , . . . , z N ] and each α ∈ N N 0 , (see Definition 1.2). Put σ := max 1 2 , κυ , where k := deg h ≥ 1. Let w 1 , . . . , w N denote the variables in C N . We will show that, for each polynomial Obviously, the above estimate proves the required assertion that h(E) satisfies Markov's inequality. Fix therefore Q, l, a as above. First, we will show that, for each ζ ∈ C N and each j ∈ {1, . . . , N }, By Taylor's formula and (4), which completes the proof of (6). We will show moreover that, for each ζ ∈ A, To this end, take the integers (see (3)). Consider the system of equations: . . .
Now it is enough to apply Cramer's rule, Hadamard's inequality 4 and the estimates (6) and (8).
Combining this with (1) and (7), we get for each t ∈ (0, 1] the following estimate: Note that By Schur's inequality, 5 5 Schur's inequality: For each polynomial R of one variable, -see [10, p. 233], where this inequality is stated for real polynomials. If however R ∈ C[τ ] and R = which proves (♦).

Therefore either
which establishes the estimate (5) and hence completes the proof of the theorem.

Markov's inequality and polynomial preimages
In this section, we will look at Markov's inequality from the point of view of polynomial preimages. We begin by a brief discussion of another concept, called the HCP property, which is related to Markov's inequality. For a compact set ∅ = E ⊂ C N , the following function [34,37,38,49,54,55]). It is an elementary check that E ≥ 1 in C N , E ≡ 1 in E and E ≤ K provided that ∅ = K ⊂ E and K is compact. However, except for some very special cases, no explicit expression for E is known.
We have a very simple formula (yet with nontrivial proof) connecting the function E with potential and pluripotential theory: and L(C N ) denotes the class of plurisubharmonic 6 functions φ in C N satisfying the condition (cf. Theorem 4.12 in [55] or Theorem 5.1.7 in [34]). The upper semicontinuous regularization V * E of V E is often called the pluricomplex Green function, because for a compact set E ⊂ C with positive logarithmic capacity V * E is the Green function with pole at infinity of the unbounded component of C\E.
If ∅ = E ⊂ C N is a compact set and E is continuous at every point of E, then E is continuous in C N , in other words, the set E is L-regular (cf. Proposition 6.1 in [55] or Corollary 5.1.4 in [34]). Definition 3. 1 We say that a compact set ∅ = E ⊂ C N has the HCP property if E is Hölder continuous in the following sense: there exist , μ > 0 such that We will also need the notion of a pluripolar set. Definition 3.2 (see [34]) A set A ⊂ C N is said to be pluripolar if one of the following two equivalent conditions holds: Let us add that the implication (i) ⇒ (ii) is the content of Josefson's theorem (saying that every locally pluripolar set in C N is globally pluripolar). We have moreover the following characterization of pluripolar sets in terms of the pluricomplex Green function: for each set ∅ = A ⊂ C N , (cf. Corollary 3.9 and Theorem 3.10 in [55]). Recall also that pluripolar sets have Lebesgue measure zero (cf. Corollary 2.9.10 in [34]) and countable unions of pluripolar sets are pluripolar (cf. Corollary 4.7.7 in [34]). There is a close relation between Markov's inequality and HCP property. Namely, • HCP ⇒ Markov's inequality (see [43]).
• The validity of the reverse implication still remains open (see [48], where this problem is posed by Pleśniak).
Furthermore, it is worth noting that, for each compact subset of R N , UPC ⇒ HCP (see [43]), yet the implication cannot be reversed. It should come as no surprise that the inverse image of a Markov set under a polynomial map need not be a Markov set, even if it is a compact set. Consider for example the map h : R w −→ w 2 ∈ R and the set E := [−1, 0]. Then h −1 (E) = {0} does not satisfy Markov's inequality. Clearly, the inverse image of an interval under any polynomial map h : R −→ R is a finite union of intervals (with infinite endpoints allowed) and points. Markov's inequality for sets consisting of finitely many intervals was deeply investigated by Totik in [57].
The situation is quite different if we consider the complex case (K = C). But also in this case the claim that the polynomial preimage of a Markov set is a Markov set is still far from being valid.
which is impossible.
The situation described in the above example is particular, because the set g −1 (E) is not L-regular. 8 This is no longer the case in the next example (Example 3.6).
It will be convenient to state beforehand, for easy reference, two results. The first one gives a sufficient and necessary condition for a bounded set A ⊂ R 2 definable in some polynomially bounded o-minimal structure to be UPC (cf. [44], Theorem B).

Theorem 3.4 Let A ⊂ R 2 be bounded and definable in some polynomially bounded o-minimal structure (for example, semialgebraic). Then the following two statements are equivalent: • A is UPC. • A is fat and, for each a ∈ A, ρ > 0 and any connected component S of the set
Int A ∩ B(a, ρ) such that a ∈ S, there is a polynomial arc γ : The second result is a special case of the (semi)analytic accessibility criterion due to Pleśniak (cf. [47]). 9 Theorem 3.5 Let K ⊂ K N be a compact set. Suppose that there exists a polynomial mapping γ : K −→ K N such that γ ((0, 1]) ⊂ IntK . Then K is L-regular at γ (0), i.e., K is continuous at γ (0). • lim t→0 f (t) t r = 0 for each r > 0,  9 An alternative proof can also be found in [45,Corollary 2.8] and the set Note that F −1 (E) is compact. We will show that: • E satisfies Markov's inequality, does not satisfy Markov's inequality for K = R but does satisfy Markov's inequality for K = C.
To this end, put On the other hand, by Theorem 3.4, the set E 2 is UPC. 10 Therefore, there exist υ, θ > 0 and d ∈ N such that, for each x ∈ E 2 , we can choose a polynomial map S x : R −→ R 2 with degS x ≤ d satisfying the following conditions: Note that, for each x ∈ E 2 and each t ∈ [0, 1], 10 The assumption that f | [R 1 , R] is definable in a certain polynomially bounded o-minimal structure is used here to guarantee definability of E 2 in a polynomially bounded o-minimal structure and to guarantee the existence of a polynomial arc γ : (0, 1) −→ IntE such that lim t→0 γ (t) = (R, f (R)). An explicit example of such an arc is γ : where η > 0 is sufficiently small and m ∈ N is sufficiently large, which follows from the definition of a polynomially bounded o-minimal structure.
Upon combining the above estimates for the sets E 1 and E 2 , it is straightforward to show that E = E 1 ∪ E 2 is UPC and hence, by Theorem 3.1 in [43], is a Markov set.
Case 1: K = R. Note first that By Theorem 3.5, the set F −1 (E) is L-regular. 11 Suppose, to derive a contradiction, that F −1 (E) is a Markov set. In particular, there exist ε, C > 0 such that, for each polynomial P ∈ C[w 1 , w 2 ], For each n ∈ N, put Moreover, take r > ε and set Note that Combining this with (12), we get which is impossible, because the right-hand side tends to zero as n → ∞.
Case 2: K = C. Note that, for each w ∈ C 2 , By Theorem 5.3.1 in [34], for each w ∈ C 2 , Since E is UPC, it has the HCP property: there exist M 1 , μ > 0 such that, for each z ∈ E (1) , Put (1) . By (14), for each z ∈ F(K ), Take M 3 > 0 such that F| K is Lipschitz with the constant M 3 , that is for all w, w ∈ K . For each w ∈ K , we have which yields the HCP property for the set F −1 (E). Consequently, F −1 (E) is a Markov set and is L-regular.
The previous examples may suggest that a compact, L-regular set, which is the inverse image of a Markov set under a complex (i.e., holomorphic) polynomial map, is also a Markov set. This claim is however not valid.
In the same way as we handled the set E of Example 3.6, we can show that D 1 , D 2 satisfy Markov's inequality. Moreover, put Note that After repeating the argument from Case 1 of Example 3.6, we conclude that G −1 (D) is L-regular and does not satisfy Markov's inequality. On the other hand, the set D, as the Cartesian product of the Markov sets, is a Markov set.
After giving the above examples illustrating various situations which occur naturally when we consider Markov's inequality in the context of polynomial preimages, we conclude this section with the statement of the following result, to be proved in the next section.  (N , N ∈ N). Suppose that a compact set ∅ = E ⊂ C N has the HCP property, E ⊂ g(U ) and g −1 (Ê) is compact. Then

E) has the HCP property and, in particular, is a Markov set.
Recall thatÊ denotes the polynomially convex hull of E: IfÊ = E, then we say that E is polynomially convex. For example, each compact subset of R N is polynomially convex in C N (cf. Lemma 5.4.1 in [34]).

A proof of Theorem 3.8
For the convenience of the reader we recall first the relevant notions and results from [41]. (We assume here that the maximum on the empty set is equal to −∞.) The subsequent proofs make use of the following two results. open (N , N ∈ N). Suppose that B ⊂ W is a locally analytic set such that, for some m ∈ N,

Then f (B) is a countable union of submanifolds of dimension ≤ m.
Proof See [41, p. 254].

Theorem 4.5 Every compact analytic subset of C N is finite.
Proof See [41, p. 235].
Before proceeding with the proof of Theorem 3.8, let us also state the following lemma.  (N , N ∈ N). Suppose that a set A ⊂ f (W ) is nonpluripolar. Then f −1 (A) is nonpluripolar as well.
Proof We will consider two cases.
Case 1: N < N . Obviously, rank d w f ≤ N for each w ∈ W . By Theorem 4.4, f (W ) is a countable union of submanifolds of dimension ≤ N . In particular, the set f (W ) (and hence A) is pluripolar, which is a contradiction. The case N < N cannot therefore occur.
Clearly, the set B is an analytic subset of W . As in Case 1, we show via Theorem 4.4 that f (B) is pluripolar. In particular, the set A ∩ f (W 0 ) is nonpluripolar. By the rank theorem, for each a ∈ W 0 , there exists an open set U a such that a ∈ U a ⊂ W 0 , f (U a ) is open, and there exist biholomorphic mappings ϕ a : U a −→ a × a , ψ a : f (U a ) −→ a , where a ⊂ C N , a ⊂ C N −N are open sets, such that the mapping is the natural projection. Clearly, there is a sequence a j ∈ W 0 ( j ∈ N) such that Take l ∈ N such that A ∩ f (U a l ) is nonpluripolar. Then the set ψ a l (A ∩ f (U a l )) is also nonpluripolar. Suppose, to derive a contradiction, that f −1 (A) is pluripolar. Then ϕ a l ( f −1 (A) ∩ U a l ) is also pluripolar and therefore for some plurisubharmonic function u in C N . Note that, for each y ∈ a l , Since ψ a l (A ∩ f (U a l )) is nonpluripolar, it follows that u ≡ −∞ in C N × a l , which is impossible. 12 In the proof of Theorem 3.8, we will use the notion of the relative extremal function. Suppose that ⊂ C N is an open set and A ⊂ . The relative extremal function for A in is defined as follows: where PSH( ) denotes the plurisubharmonic functions in .
Proof of Theorem 3.8. We will consider three cases.
. By the formula on p. 169 of [41], On the other hand, is compact, analytic and hence finite (cf. Theorem 4.5), in contradiction with (17). This means that the case N > N cannot occur. Case 2: N < N . It follows from Theorem 4.4 that g(U ) is a countable union of submanifolds of dimension ≤ N . In particular, g(U ) (and hence E) is pluripolar, which is a contradiction. The case N < N cannot therefore occur.
Suppose, towards a contradiction, that this is not the case and take a sequence a j ∈ U ( j ∈ N) such that dist(a j , K ) = λ and dist g(a j ),Ê → 0.
Passing to a subsequence if necessary, we can assume that a j → a ∈ K (λ) \K λ . Consequently, dist(a, K ) = λ and dist g(a),Ê = 0, which means that a / ∈ K and a ∈ K , a contradiction. Put where > 0 is of (18). For each compact set T ⊂Ê , we have and therefore the set g −1 (T ) ∩ is compact. It follows that is also a proper holomorphic map. Note moreover that g −1 (E) ⊂ K ⊂ ,Ê = g(K ) ⊂ and , are bounded. (Use the fact that ⊂Ê .) By Lemma 4.6, the set g −1 (E) is nonpluripolar and hence V * g −1 (E) ∈ L(C N ) (see (11)). In particular, (1) for some M 1 > 0. It is clear that Moreover, by Proposition 5.3.3 in [34], there exists M 2 > 0 such that Let M 3 , μ > 0 be such that, for each z ∈ E (1) , (see Definition 3.1). Obviously, for each z ∈ , where (1) , then we set M 4 := M 3 .) By Proposition 4.5.14 in [34], Consequently, for each w ∈ , (w), E)) μ and therefore where The estimate (23) implies that, for each w ∈ g −1 (E) (1) , where M > 0 is such that e M 6 t ≤ 1 + Mt for t ∈ [0, 1]. The case N = N is therefore settled, and the proof of the theorem is complete. With regard to Theorem 3.8, we have the following remark.
Remark 4.7 In Theorem 3.8, even if U = C N and g : C N −→ C N is a polynomial map, the assumption thatÊ ⊂ g(U ) and g −1 (Ê) is compact cannot be replaced by the assumption that E ⊂ g(U ) and g −1 (E) is compact and L-regular.

Proof
Set Take a compact set K ⊂ R 2 ⊂ C 2 such that: • K does not satisfy Markov's inequality.
Suppose, towards a contradiction, that g −1 (E) is a Markov set. Since is the Shilov boundary of D 2 (see Example 3.3), it follows that D 2 ∪ K is a Markov set as well. Put : C 2 (w 1 , w 2 ) −→ w 2 ∈ C. Note that (D 2 ) = D ⊂ C and (K ) ⊂ (1, 2] ⊂ R ⊂ C. Consequently, the sets (D 2 ) and (K ) are disjoint and polynomially convex. Clearly, D 2 and K are also polynomially convex. Therefore, by Kallin's separation lemma (cf. [33, p. 302]), we obtain the polynomial convexity of the set D 2 ∪ K . On account of Corollary 5.2, we get a contradiction, because K is not a Markov set.

Subsets of Markov sets
In this section, we will prove the following result announced in Introduction. Proof Choose λ > 0 such that K (λ) ∩ (E\K ) = ∅. In particular, we have Take moreover a compact and polynomially convex set Z ⊂ C N such that E ⊂ IntZ and Z ⊂ K λ ∪ C N \K (λ) -see the proof of Lemma 2.7.4 in [32]. Define g : By Theorem 8.5(1) in [55], there exist M > 0, ρ ∈ (0, 1) with the following property: for each μ ∈ N, we can choose a polynomial R μ ∈ C[z 1 , . . . , z N ] with deg R μ ≤ μ and such that We can clearly assume that M ≥ 1. By (11), V K (and hence K ) is bounded on each compact subset of C N . Thus we may choose k ∈ N such that 1 − Mρ k > 0 and Mρ k sup Moreover, let ε, C > 0 be of Definition 1.2 for the set E. Therefore, for each polynomial P ∈ C[z 1 , . . . , z N ] and each α ∈ N N 0 , Take also C 1 > 0 such that and put We will show that, for each polynomial Q ∈ C[z 1 , . . . , z N ] with deg Q ≤ n (n ∈ N) and each a ∈ K (C1(k+1) −ε n −ε ) , we have |Q(a)| ≤ e N 1 + Mρ k 1 − Mρ k Q K .

Corollary 5.2
Assume that E 1 , . . . , E p ⊂ C N ( p ∈ N) are compact, nonpluripolar and pairwise disjoint sets such that E := E 1 ∪ · · · ∪ E p is polynomially convex. Let ε > 0. Then the following two statements are equivalent: 1. E satisfies Markov's inequality with the exponent ε.
2. For each j ≤ p, the set E j satisfies Markov's inequality with the exponent ε.
We conclude this section with the following example concerning Corollary 5.2. Note that E 1 does not satisfy Markov's inequality. Indeed, suppose otherwise and take ε, C of Definition 1.2. Then, for the polynomials P n (z) := z(1 − z) n (n ∈ N), we have 1 = |P n (0)| ≤ P n E 1 ≤ C(n + 1) ε P n E 1 ≤ C(n + 1) ε 2 3 n+1 , which is impossible. By Bernstein's theorem (see Example 3.3) and the maximum principle, for each complex polynomial Q of one variable, Hence, E satisfies Markov's inequality. Moreover, E is not polynomially convex. In Corollary 5.2, the assumption that E is polynomially convex is therefore relevant even if N = 1.