Proceedings of the Steklov Institute of Mathematics, Vol. 218, 1997, pp. 372–383. Translated from Trudy Matematicheskogo Instituta imeni V. A. Steklova, Vol. 218, 1997, pp. 374–384. Original Russian Text Copyright \begin{matrix}c \\ ◯ \end{matrix} 1997 by Skalyga. English Translation Copyright \begin{matrix}c \\ ◯ \end{matrix} by ���� �����/Interperiodica Publishing (Russia).

Estimates of the Derivatives of Polynomials on Convex Bodies

V. I. Skalyga

Received February, 1997

In this paper, we obtain the best possible estimate in Markov’s inequality for algebraic polynomials on convex bounded closed bodies in Banach spaces.

1. INTRODUCTION

Let 𝔓_{n} be the set of all algebraic polynomials of degree not higher than n ∈ ℤ_{+}, and let [a, b] ⊂ ℝ, ∥P_{n}∥_{[a,b]} = max_{x∈[a,b]} |P_{n}(x)|. The well-known Markov inequality (see [1]) states that for any P_{n} ∈ 𝔓_{n} we have

∥P′_{n}∥_{[a,b]} ≤ \frac{2n^{2}}{b − a}∥P_{n}∥_{[a,b]},  (1)

the equality being attained only for polynomials of the form

P_{n}(x) = cT_{n}(\frac{2x − a − b}{b − a}), where T_{n}(x) = cos(n arccos x).

For interior points of the interval [a, b], Bernstein’s inequality holds (see [2]); according to which we have

|P′_{n}(x)| ≤ n(\frac{∥P_{n}∥_{[a,b]}^{2} − P_{n}^{2}(x)}{(x − a)(b − x)})^{1/2}.  (2)

The inequality (2) is an identity for the polynomial T_{n}((2x − a − b)/(b − a)). The Markov and Bernstein inequalities were extended to polynomials in several variables (see [3–9] and the references therein). Note the results of [3], where it is shown that inequality (1) can be extended to the case of a ball in a Hilbert space without a change in the constant. In [4], an analog of the Markov inequality is obtained for a convex compact body in 𝔼^{m} with the constant 4n^{2}/ω (where ω is the width of the body), and a conjecture is stated that 2n^{2}/ω can be the constant. Counterexamples to this conjecture are constructed in [8].

Let X and Y be two Banach spaces over ℝ. By 𝔓_{n}(X, Y) we denote the space of all polynomials of degree not higher than n, i.e., the set of all maps of the form

P_{n}(x) = ∑_{j=0}^{n} u_{j}(x, …, x), x ∈ X,

where u_{j}: X^{j} → Y is a j-linear continuous mapping. In the sequel, ∥P′_{n}(x)∥_{L(X,Y)} is the norm of the Frechét derivative of the polynomial P_{n} at the point x in the space L(X, Y) of all bounded linear operators from X into Y. If Y = ℝ, then we use the symbol ∇ to denote the derivative. The value of the linear mapping P′_{n}(x) on the vector h is denoted by P′_{n}(x)[h].

Everywhere in the sequel, V is a convex bounded closed body in the Banach space X. The set of all bodies of this kind is denoted by W. We set

∥P_{n}∥_{V} = sup_{x∈V} ∥P_{n}(x)∥_{Y} , ∥P′_{n}∥_{V} = sup_{x∈V} ∥P′_{n}(x)∥_{L(X,Y)}.

Let h ∈ X and ∥h∥ = 1. We denote by V (h) the set

V (h) = {[y_{1}, y_{2}] : y_{1}, y_{2} ∈ V and ∃c ∈ ℝ : y_{1} − y_{2} = ch},

and by ω_{V} (h) and ω_{V} the width of the body V in the direction of h and the width of this body V, respectively, where

ω_{V} (h) = sup_{[y_{1},y_{2}]∈V (h)} ∥y_{1} − y_{2}∥, ω_{V} = inf_{∥h∥=1} ω_{V} (h).

The following estimate for P_{n} ∈ 𝔓(X, Y) and V ∈ W is proved in [6]:

∥P′_{n}∥_{V} ≤ \frac{4n^{2}}{r(V)}∥P_{n}∥_{V} ,

where r(x_{0}) = sup{r : B(x_{0}, r) ⊂ V } and r(V) = sup{r(x_{0}) : x_{0} ∈ V }.

The stronger estimate

∥P′_{n}∥_{V} ≤ \frac{4n^{2}}{ω_{V}} ∥P_{n}∥_{V}

is proved in [9] for P_{n} ∈ 𝔓_{n}(X, Y) and V ∈ W.

In the present paper, we obtain the unimprovable estimate

∥P′_{n}∥_{V} ≤ 2n cot(\frac{π}{4n})\frac{∥P_{n}∥_{V}}{ω_{V}}

for P_{n} ∈ 𝔓_{n}(X, Y) and V ∈ W.

All the results are formulated for Y = ℝ, but this does not lead to lose of generality for the majority of estimates as Lemma 1 shows, which follows from the Hahn–Banach theorem.

2. AUXILIARY LEMMAS

Lemma 1. Let V ⊂ X, P_{n} ∈ 𝔓_{n}(X, Y), ∥P_{n}∥_{V} ≤ 1. If the inequality

∥P̃′_{n}(x)∥_{X^{∗}} ≤ C(x), x ∈ V

holds for any P̃_{n} ∈ 𝔓_{n}(X, ℝ) such that ∥P̃_{n}∥_{V} ≤ 1, then the inequality

∥P′_{n}(x)∥_{L(X,Y)} ≤ C(x), x ∈ V,

also holds for P_{n}.

The proof of this result see in [9].

Let {cZ_{n,σ}(t)} be the set of Zolotarev polynomials, where c, σ ∈ ℝ, Z_{n,σ}(t) ∈ 𝔓_{n}(ℝ, ℝ), and let Z_{n,σ}(t) be a Zolotarev polynomial of the form

t^{n} − nσt^{n−1} + ∑_{i=0}^{n−2} α_{i}t^{i}

which has the minimal norm in C[−1, 1]. For 0 ≤ σ ≤ tan^{2}(π/(2n)), Z_{n,σ}(t) is defined by the relation

Z_{n,σ}(t) = LT_{n}(\frac{t − σ}{1 + σ} ), where L = (1 + σ)^{n}2^{−(n−1)}.

For σ > tan^{2}(π/(2n)), Z_{n,σ}(t) is expressed in terms of the well-known elliptic functions [10]. For σ ≤ 0, Z_{n,σ}(t) is defined by the relation

Z_{n,σ}(t) = (−1)^{n}Z_{n,−σ}(−t).  (3)

Let ε, τ ∈ [−1, 1] and let M_{n,ε,τ} be the subset of polynomials Z_{n}(t) from the set {cZ_{n,σ}(t)} such that

Z_{n}(τ) = ε,  ∥Z_{n}(t)∥_{C[−1,1]} = 1,  Z′_{n}(τ) ≤ 0.

Let Z_{n,ε,τ} (t) be a polynomial from M_{n,ε,τ} such that Z′_{n,ε,τ} (τ) ≤ Z′_{n}(τ) for any Z_{n}(t) ∈ M_{n,ε,τ}.

Lemma 2. The polynomial Z_{n,ε,τ} (t) is a solution to the problem

P′_{n}(τ) → min,  P_{n}(τ) = ε,  τ ∈ [−1, 1],  ∥P_{n}(t)∥_{C[−1,1]} ≤ 1,  (4)

for P_{n}(t) ∈ 𝔓_{n}(ℝ, ℝ).

Proof. Problem (4) is a convex programming problem. We replace it by the equivalent problem

α_{1} → min,  max_{−2+β≤t≤β}|ε + ∑_{i=1}^{n} α_{i}t^{i}| ≤ 1,  β = 1 − τ ≥ 0.  (5)

Let |ε| < 1. It follows from the Weierstrass compactness principle that there exists a solution

P̂_{n}(t) = ε + ∑_{i=1}^{n} α̂_{i}t^{i}

to problem (5). It follows from the Kuhn–Tacker theorem (see [11]) that there exists a number λ ≥ 0 such that

0 ∈ ∂_{α}(α_{1} + λ∥ε + ∑_{i=1}^{n} α_{i}t^{i}∥_{C[−2+β,β]})|_{α=α̂},  (6)

λ(∥ε + ∑_{i=1}^{n} α̂_{i}t^{i}∥_{C[−2+β,β]} − 1) = 0,  (7)

where ∂ is the symbol of the Clarke generalized gradient (see [11]).

From (6)–(7) and the theorem of the pointwise maximum gradient [11] we have the system of equations

1 + ∑_{k=1}^{r} c_{k}ν_{k}t_{k} = 0,  ∑_{k=1}^{r} c_{k}ν_{k}t_{k}^{i} = 0, i = \overline{2, n},  (8)

where r ≤ n + 1, |P̂_{n}(t_{k})| = 1, t_{k} ∈ [−2 + β, β], ν_{k} = sign P̂_{n}(t_{k}), k = \overline{1, r}, c_{k} ≥ 0, ∑_{k=1}^{r} c_{k} = λ. The system of equations (8) implies the following equality for any P_{n}(t):

P′_{n}(0) − P_{n}(0) ∑_{k=1}^{r} c_{k}ν_{k} + ∑_{k=1}^{r} c_{k}ν_{k}P_{n}(t_{k}) = 0.  (9)

It follows from (9) that r ≥ n, since, otherwise, there would exist a polynomial P_{n}(t) having n − 1 zeros on the set of nonzero points which includes \{t_{k}\} (k = \overline{1, r}) and the double zero at zero.

The equality r = n+1 can be attained only in the case where Chebyshev polynomial cT_{n}(t+1−β) (|c| = 1) is an extremal for the interval [−2 + β, β] satisfying the conditions

cT_{n}(1 − β) = ε, cT′_{n}(1 − β) ≤ 0

at zero.

Let r = n. Suppose that we have the numbers ν_{k} and t_{k} (k = \overline{1, n}). For system (8) to have a solution with positive c_{k} it is necessary that the elements of the first column a_{k,1} of the matrix A, which is an inverse of the matrix (ν_{k}t_{k}^{i}) (k, i = \overline{1, n}), be negative. From the Cramer rule and the properties of the Vandermonde determinant we have

a_{k,1} = (−1)^{k+1} ∏_{i≠k} ν_{i}t_{i}^{2} ∏_{\begin{matrix}n≥j>i≥1 \\ j,i≠k \end{matrix}}(t_{j} − t_{i})/( ∏_{n≥i≥1} ν_{i}t_{i} ∏_{n≥j>i≥1}(t_{j} − t_{i})).  (10)

It follows from (10) that the condition

(−1)^{k+p}ν_{k} < 0,  (11)

where p is the number of positive t_{k}, is necessary for ν_{k}.

It follows from (11) that the desired extremal should have an n-alternance. Hence, by the Chebyshev uniqueness theorem, the Zolotarev polynomial is a solution to problem P̂_{n}(t).

The problem is solved by the use of a limiting process for |ε| = 1.

Remark 1. The set M_{n,ε,1} consists of one polynomial for ε ∈ (−1, 1) and the following relation holds:

Z_{n,ε,1}(t) = −T_{n}(\frac{2t − y}{2 + y} ),  (12)

where y = 2 tan^{2}(a(−ε)/(2n)), a(ε) = arccos(ε), ε ∈ [−1, 1].

The relation

Z′_{n,ε,1}(1) = −(1 − ε^{2})^{1/2} \frac{n}{2} cot(\frac{a(−ε)}{2n})

follows from (12) and (2).

Remark 2. Note the following obvious equality for τ ∈ [−1, 1] which follows from the symmetry considerations:

max_{\begin{matrix}P_{n}(τ)=ε \\ ∥P_{n}∥_{C[−1,1]}≤1 \end{matrix}} |P′_{n}(τ)| = max\{|Z′_{n,ε,τ} (τ)|, |Z′_{n,−ε,τ} (τ)|\}.  (13)

Remark 3. Let Q_{n,ε,τ} (t) = Z_{n,ε,2τ−1}(2t − 1), where τ ∈ [0, 1]. Obviously, Q_{n,ε,τ} (t) is a Zolotarev polynomial for the interval [0, 1] which solves a problem similar to (4). This implies the inequality

|Q′_{n,ε,τ_{1}}(τ_{1})|τ_{1} ≤ |Q′_{n,ε,τ_{2}}(τ_{2})|τ_{2}  for τ_{1} < τ_{2}.  (14)

Indeed, let τ_{1}/τ_{2} = τ_{3} < 1. Consider a problem which is similar to (4) on the closed interval [0, τ_{3}] at the point τ_{1} (τ_{1} ≤ τ_{3}); assume that P_{n}(t) is a solution to this problem. Then the inequality

|P′_{n}(τ_{1})| = τ_{3}^{−1}|Q′_{n,ε,τ_{1}/τ_{3}}(\frac{τ_{1}}{τ_{3}} )| ≥ |Q′_{n,ε,τ_{1}}(τ_{1})|

holds; and this implies (14).

Note an obvious equality which follows from (3):

Q′_{n,−ε,1−τ} (1 − τ) = Q′_{n,ε,τ} (τ).  (15)

Corollary. Let U_{n}(θ): ℝ → [−1, 1] be an even trigonometric polynomial of order n. Then the inequality

|U′_{n}(θ_{0})| ≤ max{|Z′_{n,ε,c}(c)|, |Z′_{n,−ε,c}(c)|}| sin θ_{0}|

holds, where c = cos θ_{0}, ε = U_{n}(θ_{0}).

Lemma 3. The function

C_{n}(ε) = n(1 − ε^{2})^{1/2}(cot(\frac{a(ε)}{2n}) + cot(\frac{a(−ε)}{2n})),

where a(ε) = arccos ε, ε ∈ [−1, 1], is an even and decreasing function on the interval [0, 1].

Proof. The evenness of C_{n}(ε) is obvious; in order to prove the second assertion, we make the change of variables

x = a(ε), C_{n}(ε) = n sin x(cot(\frac{x}{2n}) + cot(\frac{π − x}{2n})).

We have C′_{n}(ε) = −f′(x)(1 − ε^{2})^{−1/2}; therefore, in order to prove the lemma, it suffices to show that f′(x) ≥ 0 for x ∈ [0, π/2].

Computing the derivative f′(x), after simple transformations, we obtain

f′(x) = \frac{sin(π/(2n))φ(x)}{2 sin^{2}(x/(2n)) sin^{2}((π − x)/(2n))},

where

φ(x) = n cos x(cos(\frac{2x − π}{2n} ) − cos(\frac{π}{2n})) + sin x sin(\frac{2x − π}{2n} ).

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

Since φ(0) = φ(π/2) = 0, in order to prove that φ(x) ≥ 0 on [0, π/2], it is sufficient to justify the following assertion.

A. φ(x) has one local extremum on (0, π/2), and this extremum is positive.

We have

φ′(x) = sin x φ_{1}(x),

where

φ_{1}(x) = −n(cos(\frac{2x − π}{2n} ) − cos(\frac{π}{2n})) + n^{−1} cos(\frac{2x − π}{2n} ).

Since φ_{1}(0) = n^{−1} cos(−π/(2n)) > 0 and φ′_{1}(x) = (1 − 1/n^{2}) sin((2x − π)/(2n)) < 0 for 0 ≤ x < π/2, asserion A is proved, and so is the lemma.

Lemma 4. Let X be a two-dimensional normed space, ABC be a triangle belonging to X, CD be its median, x ∈ [C, D], B_{1} ∈ [A, C], x ∈ [B, B_{1}], A_{1} ∈ [B, C], x ∈ [A, A_{1}], and let ∥D − x∥/∥D − C∥ = ρ. Then the relations

\frac{∥x − A∥}{∥A_{1} − A∥} = \frac{∥x − B∥}{∥B_{1} − B∥} = \frac{1 + ρ}{2}  (16)

hold.

Proof. The point x is the center of gravity of the system of particles (A, 1),(B, 1),(C, 2ρ/(1−ρ)), and the points A_{1} and B_{1} are the centers of gravity of the system of particles (B, 1), (C, 2ρ/(1−ρ)) and (A, 1), (C, 2ρ/(1 − ρ)) respectively. This implies relation (16).

3. ANALOGS OF MARKOV AND BERNSTEIN INEQUALITIES ON CONVEX BODIES IN BANACH SPACES

Let X be a Banach space, x ∈ V ∈ W, h ∈ X, ∥h∥ = 1. We introduce the following definitions: let ε ∈ [−1, 1], [y_{1}, y_{2}] ∈ V (h), [y_{1}, x] ⊂ [u_{11}, u_{12}], [y_{2}, x] ⊂ [u_{21}, u_{22}], u_{ij} ∈ ∂V (i, j = 1, 2),

\frac{∥x − u_{i2}∥}{∥u_{i1} − u_{i2}∥} = ρ_{1i}(x, y_{i}) = ρ_{1i}, \frac{∥x − y_{i}∥}{∥u_{i1} − u_{i2}∥} = ρ_{0i}(x, y_{i}) = ρ_{0i},  1 − ρ_{1i} = ρ_{i}, i = 1, 2,(17)
b_{1}(X, V, h, x, ε, n) = inf_{[y_{1},y_{2}]∈V (h)} ∥y_{1} − y_{2}∥^{−1} max_{i=1,2}{|Q′_{n,ε_{i−1},ρ_{1}}(ρ_{1})|ρ_{01} + |Q′_{n,ε_{i},ρ_{2}}(ρ_{2})|ρ_{02}}, (18)

where ε_{i} = (−1)^{i}ε, Q_{n,ε,ρ}(t) is defined in Remark 3.

Let x_{0} ∈ V, ρ_{0}(x, x_{0}) = ∥x − x_{0}∥/∥y − x_{0}∥, where y ∈ ∂V, x ∈ [x_{0}, y];

\overline{ρ}_{1} = \frac{1}{2}(1 + ρ_{0}(x, x_{0})), ω_{V} (x_{0}, h) = 2 max\{t ≥ 0 : [x_{0} − th, x_{0} + th] ⊂ V \},

b_{2}(X, V, h, x, ε, n) = inf_{x_{0}∈V} \{|Q′_{n,ε,\overline{ρ}_{1}}(\overline{ρ}_{1})| + |Q′_{n,−ε,\overline{ρ}_{1}}(\overline{ρ}_{1})|\}\overline{ρ}_{1}/ω_{V} (x_{0}, h),

B_{n}(V) = \{P_{n} : P_{n} ∈ 𝔓_{n}(X, ℝ), ∥P_{n}∥_{V} ≤ 1\}, Φ_{n}(X, V, h, x, ε) = sup_{\begin{matrix}P_{n}∈B_{n}(V) \\ P_{n}(x)=ε \end{matrix}} |∇P_{n}(x)[h]|.

(19)

Theorem 1. Let X be a Banach space, x ∈ V ∈ W, h ∈ X, ∥h∥ = 1, and let C_{n}(ε) be the function defined in Lemma 3. Then the following inequalities hold:

Φ_{n}(X, V, h, x, ε) ≤ b_{1}(X, V, h, x, ε, n) ≤ \frac{C_{n}(ε)}{ω_{V} (h)},  (20)
F_{n}(X, V, h, x) = sup_{P_{n}∈B_{n}(V)} |∇P_{n}(x)[h]| ≤ b_{2}(X, V, h, x, 0, n) ≤ \frac{C_{n}(0)}{ω_{V} (h)},  (21)
Ψ_{n}(X, V) = sup_{\begin{matrix}P_{n}∈B_{n}(V) \\ x∈V \end{matrix}} ∥∇P_{n}(x)∥_{X^{∗}} ≤ \frac{C_{n}(0)}{ω_{V}}.  (22)

Proof. Let us prove inequalities (20). Let x ∈ V, [y_{1}, y_{2}] ∈ V (h), P_{n}(x) = ε ∈ [−1, 1]. Then the relation

|∇P_{n}(x)[y_{1} − y_{2}]| = |∇P_{n}(x)[y_{1} − x] + ∇P_{n}(x)[x − y_{2}]|  (23)

holds. Let the terms on the right-hand side of (23) have different signs, and let the modulus of the sums is not exceeded, for example, |∇P_{n}(x)[y_{1} − x]|. Then the inequality

|∇P_{n}(x)[h]| ≤ \frac{|∇P_{n}(x)[y_{1} − x]|}{∥y_{1} − y_{2}∥}.  (24)

holds. This inequality, Lemma 2, and definitions (17) imply the inequality

|∇P_{n}(x)[h]| ≤ ∥y_{1} − y_{2}∥^{−1} max{|Q′_{n,ε,ρ_{1}}(ρ_{1})|, |Q′_{n,−ε,ρ_{1}}(ρ_{1})|}ρ_{01}.  (25)

If the terms on the right-hand side of (23) are of the same sign, then the values ∇P_{n}(x)[x − y_{1}] and ∇P_{n}(x)[x−y_{2}] have different signs, and we obtain from Lemma 2 and definition (17) the inequality

|∇P_{n}(x)[h]| ≤ ∥y_{1} − y_{2}∥^{−1} max_{i=1,2}{|Q′_{n,ε_{i−1},ρ_{1}}(ρ_{1})|ρ_{01} + |Q′_{n,ε_{i},ρ_{2}}(ρ_{2})|ρ_{02}},  (26)

where ε_{i} = (−1)^{i−1}ε.

The inequality

Φ_{n}(X, V, h, x, ε) ≤ b_{1}(X, V, h, x, ε, n)

follows from (25), (26), and the definition of b_{1}(X, V, h, x, ε, n). Since

C_{n}(ε) = |Q′_{n,ε,1}(1)| + |Q′_{n,−ε,1}(1)|,

the inequality

b_{1}(X, V, h, x, ε, n) ≤ \frac{C_{n}(ε)}{ω_{V} (h)}

follows from (25), (26), and (14). Inequalities (20) are proved. Let us prove inequality (21).

Let P_{n}(x) ∈ B_{n}(V), x ∈ V, [y_{1}, y_{2}] ∈ V (h) and let x_{0} be the midpoint of the interval [y_{1}, y_{2}]. Let y ∈ ∂V and y − x_{0} = (x − x_{0})/ρ_{0}(x, x_{0}) (x ∈ [x_{0}, y]). We introduce the change of variables

U_{n}(t_{1}, t_{2}) = P_{n}(x_{0} + t_{1}(y − x_{0}) + t_{2}(y_{1} − x_{0})).  (27)

U_{n}(t_{1}, t_{2}) is an nth degree polynomial in two variables, and we shall consider the vector t = (t_{1}, t_{2}) in the rectangular coordinate system. Relation (27) implies the equality

U′_{n,t_{2}}(ρ_{0}(x, x_{0}), 0) = ∇P_{n}(x)[y_{1} − x_{0}]

and the condition

|U_{n}(t_{1}, t_{2})| ≤ 1  for t_{1} + |t_{2}| ≤ 1 and t_{1} ≥ 0.  (28)

Let the polynomial

Û_{n}(t) = ∑_{i+j≤n} α̂_{ij}t_{1}^{i}t_{2}^{j}

has a maximum of the modulus of the derivative U′_{n,t_{2}}(t) at the point (ρ_{0}(x, x_{0}), 0) under condition (28). But then this maximum is also attained on the polynomial

\overline{U}_{n}(t) = \frac{Û_{n}(t_{1}, t_{2}) − Û_{n}(t_{1}, −t_{2})}{2}

which satisfies condition (28) and has only odd degrees with respect to t_{2}. By Lemmas 2 and 4, this implies the inequality

|Û′_{n,t_{2}}(ρ_{0}, 0)| = |∑_{i=0}^{n−1} α̂_{i1}ρ_{0}^{i}| ≤ 2|Q′_{n,0,\overline{ρ}_{1}}(\overline{ρ}_{1})|\overline{ρ}_{1},  (29)

where \overline{ρ}_{1} = (1 + ρ_{0}(x, x_{0}))/2. Now, (21) follows from (29) and (19). Inequality (22) follows from (21).

Theorem 2. The following assertions hold:

(a) if X = ℝ^{2} with a certain norm and if V is a triangle ABC with x = A, then the following equalities hold:

Φ_{n}(X, V, h, x, ε) = \{ \begin{matrix}b_{1}(X, V, h, x, ε, n) & for each h (∥h∥ = 1) and ε ∈ [−1, 1] \\ \frac{C_{n}(ε)}{ω_{V} (h)} & for each ε ∈ [−1, 1] and h = \frac{B − C}{∥B − C∥}; \end{matrix}

(b) for any Banach space X ≠ ℝ and two numbers δ ∈ (0, 1) and ε ∈ [−1, 1], there exist V_{δ} ∈ W, x_{0} ∈ V_{δ}, h (∥h∥ = 1), and P_{n}(x, δ, ε) ∈ B_{n}(V_{δ}) such that P_{n}(x_{0}, δ, ε) = ε,

Φ_{n}(X, V_{δ}, h, x_{0}, ε) = |∇P_{n}(x_{0}, δ, ε)[h]| = \frac{C_{n}(ε)}{ω_{V_{δ}}(h)} ≥ \frac{(1 − δ)C_{n}(ε)}{ω_{V_{δ}}};

(c) there exist X, V ∈ W, h ∈ X, ∥h∥ = 1, and x ∈ V such that

Φ_{n}(X, V, h, x, ε) = \frac{C_{n}(ε)}{ω_{V}}

for any ε ∈ [−1, 1].

PROCEEDINGS OF THE STEKLOV INSTITUTE OF MATHEMATICS  Vol. 218  1997

Proof. Let us prove assertion (a). First, we set X = ℝ^{2} and assume that X is endowed with the Euclidean norm, i.e., X = ℓ_{2}^{(2)}. Let V be the triangle ABC with the vertices C = (0, 1), B = (0, −1), and A = (1, 0). Let the point A_{1} ∈ [B, C], the point x_{0} ∈ [A, A_{1}], and let ρ_{1} = ∥x_{0} − A∥/∥A_{1} − A∥, h = (A_{1} − A)/∥A_{1} − A∥.

Suppose that \overline{x} is a projection of the point x onto the straight line containing the points A and A_{1}, which is parallel to the line containing the points B and C. Consider the polynomials

\overline{P}_{n}(x) = Q_{n,ε,ρ_{1}}(\frac{\overline{x}}{∥A_{1} − A∥}),  P̃_{n}(x) = Q_{n,−ε,ρ_{1}}(\frac{\overline{x}}{∥A_{1} − A∥}).

The definitions of Φ_{n} and b_{1}, and relations (13) and (15) imply the inequalities

D = max{|Q′_{n,ε,ρ_{1}}(ρ_{1})|, |Q′_{n,−ε,ρ_{1}}(ρ_{1})|}
≤ ∥A_{1} − A∥Φ_{n}(X, V, h, x_{0}, ε) ≤ ∥A_{1} − A∥b_{1}(X, V, h, x_{0}, ε, n)
≤ max{|Q′_{n,ε,ρ_{1}}(ρ_{1})|ρ_{1} + |Q′_{n,−ε,(1−ρ_{1})}(1 − ρ_{1})|(1 − ρ_{1}),
|Q′_{n,−ε,ρ_{1}}(ρ_{1})|ρ_{1} + |Q′_{n,ε,(1−ρ_{1})}(1 − ρ_{1})|(1 − ρ_{1})} = D.

This implies the relation

Φ_{n}(X, V, h, x_{0}, ε) = b_{1}(X, V, h, x_{0}, ε, n) = \frac{D}{∥A_{1} − A∥}.  (30)

Let the point B_{1} ∈ [C, A], h = (B_{1}−B)/∥B_{1}−B∥, ρ = ∥A−B_{1}∥/∥A−B∥, and let the inequality

|Q′_{n,ε,1}(1)|ρ + |Q′_{n,−ε,1}(1)| ≥ |Q′_{n,−ε,1}(1)|ρ + |Q′_{n,ε,1}(1)|

hold for the given ε ∈ [−1, 1]. (In the case of the inverse inequality, the proof is similar). Consider the polynomial U_{n}(x_{1}, x_{2}), where

U_{n}(x_{1}, x_{2}) = J_{n}(x_{1}) + J_{n−1}(x_{1})x_{2},  (31)

J_{n} ∈ 𝔓_{n}(ℝ, ℝ), J_{n−1} ∈ 𝔓_{n−1}(ℝ, ℝ), and the relation

J_{n}(1 − x_{2}) + J_{n−1}(1 − x_{2})x_{2} = Q_{n,ε,1}(1 − x_{2}) = Z_{n,ε,1}(1 − 2x_{2})  (32)

holds for any x_{2} ∈ [0, 1] and the relation

J_{n}(1 + x_{2}) + J_{n−1}(1 + x_{2})x_{2} = −Q_{n,−ε,1}(1 + x_{2}) = −Z_{n,−ε,1}(2x_{2} + 1)  (33)

holds for any x_{2} ∈ [−1, 0].

Let a_{i} (i = \overline{0, n}) and b_{i} (i = \overline{0, n − 1}) be coefficients of the polynomials J_{n} and J_{n−1} respectively. Then (32) and (33) imply

(−1)^{i}a_{i} + (−1)^{i−1}b_{i−1} = c_{i},

a_{i} + b_{i−1} = d_{i},

a_{0} = ε − ∑_{i=1}^{n} a_{i},

for a_{i} and b_{i}, where c_{i} (d_{i}) are a linear combination of coefficients of the polynomial Q_{n,ε,1} (Q_{n,−ε,1}), a_{j}, and b_{j−1} (j = \overline{i + 1, n}). Therefore the polynomial U_{n}(x_{1}, x_{2}) of the form (31) satisfying identities (32) and (33) does exist. Let us show that |U_{n}(x_{1}, x_{2})| ≤ 1 for (x_{1}, x_{2}) ∈ V. From conditions (31)–(33) we have

U_{n}(0, 1) = J_{n}(0) + J_{n−1}(0) = (−1)^{n−1}, U_{n}(0, −1) = J_{n}(0) − J_{n−1}(0) = (−1)^{n}.

This implies that J_{n}(0) = 0, J_{n−1}(0) = (−1)^{n−1}, and |U_{n}(x)| ≤ 1 on the interval [B, C]. Hence, taking into account conditions (32) and (33), we find that the inequality

|U_{n}(x)| ≤ 1,  x ∈ ∂V,  (34)

holds on the boundary of V.

Suppose that the inequality

|U_{n}(x_{0})| > 1

holds at a certain point x_{0} ∈ V \ ∂V.

Then (34) implies the existence of a point x_{∗} ∈ V \ ∂V such that

|U_{n}(x_{∗})| = max_{x∈V} |U_{n}(x)| > 1.  (35)

It follows from (35) that U′_{n,x_{2}}(x_{∗}) = J_{n−1}(x_{1∗}) = 0 and |J_{n}(x_{1∗})| > 1; but this contradicts (34). Consequently, U_{n}(x) ∈ B_{n}(V).

The following relations hold:

|∇U_{n}(1, 0)[B_{1} − B]| = |∇U_{n}(1, 0)[B_{1} − A] − ∇U_{n}(1, 0)[B − A]|
= |Q′_{n,ε,1}(1)|ρ + |Q′_{n,−ε,1}(1)| = d.  (36)

The inequalities

d ≤ ω_{V} (h)Φ_{n}(X, V, h, A, ε) ≤ ω_{V} (h)b_{1}(X, V, h, A, ε, n) ≤ d

follow from (36). This implies the relations

Φ_{n}(X, V, h, A, ε) = b_{1}(X, V, h, A, ε, n) = \frac{d}{ω_{V} (h)}.  (37)

Note that d = C_{n}(ε) for ρ = 1 (when B_{1} = C).

Let now X = ℝ^{2} with a given norm, and let ABC be an arbitrary triangle in X. The relation corresponding to (30) can be proved by analogy in this case, and the relations corresponding to (37) follow from the change of coordinates

x = (x_{1}, x_{2}) = M + t_{1}(A − M) + t_{2}(C − M),
P̂_{n}(x_{1}, x_{2}) = P̂_{n}(M + t_{1}(A − M) + t_{2}(C − M)) = U_{n}(t_{1}, t_{2}),

where M is the midpoint of the interval [B, C] and U_{n}(t_{1}, t_{2}) is the polynomial satisfying conditions (31)–(33).

Let us prove item (b). Let X be an arbitrary Banach space, V_{δ} be a convex body which is a triangular “prism” whose bases are the equal triangles ABC and A_{1}B_{1}C_{1} belonging to the parallel planes Π_{1} and Π_{2} and three lateral faces Γ_{3}, Γ_{4}, Γ_{5} are parts of the hyperplanes Π_{3}, Π_{4}, and Π_{5}, where Π_{i} = H × a_{i} (i = 3, 4, 5), H is a subspace of codimension 2, and a_{i} (i = 3, 4, 5) are the straight lines containing the intervals [A, B], [B, C], and [C, A] respectively. Let the point C be chosen such that ∥C − B∥ = 1 and the inequality

d_{1} = inf_{x∈Π_{3}} ∥C − x∥ ≥ 1 − \frac{δ}{2}  (38)

holds. This can be done by the Riesz lemma on an “almost perpendicular”.

Assume that the relation

inf_{\begin{matrix}x∈Π_{1} \\ y∈Π_{2} \end{matrix}} ∥x − y∥ = 2  (39)

holds and the point A is chosen such that the following inequalities hold true:

d_{2} = inf_{x∈Γ_{4}} ∥A − x∥ ≥ 2,  (40)
d_{3} = inf_{x∈Π_{5}} ∥B − x∥ ≥ 1 − δ.  (41)

This can be done since d_{2} → ∞, and d_{3} → d_{1} as ∥A∥ → ∞. By (38)–(41) we have the inequality

ω_{V} ≥ 1 − δ.  (42)

Let \overline{x} be the projection of a point x ∈ X onto the plane Π_{1} parallel to the subspace H. Let P̂_{n}(x_{1}, x_{2}) be a polynomial in two variables constructed in item (a) for the triangle ABC, and let h = B − C, P_{n}(x) = P̂_{n}(\overline{x}). Then, by (37) and (42), the inequality

|∇P_{n}(A)[h]| = Φ_{n}(X, V_{δ}, h, A, ε) = \frac{C_{n}(ε)}{ω_{V} (h)} ≥ \frac{(1 − δ)C_{n}(ε)}{ω_{V}}

holds for any ε ∈ [−1, 1].

(c) We can consider by way of example X = ℝ^{2} with a norm such that there exists a triangle V in which the length of one side is ω_{V}. Suppose that we have a triangle ABC whose vertices are A = (1, 0), B = (0, 1), C = (0, −1).

Example 1. Let X = ℓ_{∞}^{(2)}, where ∥x∥_{ℓ_{∞}^{(2)}} = max{|x_{1}|, |x_{2}|}. Then ∥A − B∥_{ℓ_{∞}^{(2)}} = ω_{V} = 1 and

the relation

Φ_{n}(X, V, A − B, C, ε) = \frac{C_{n}(ε)}{ω_{V}}

holds.

Example 2. Let X = \overline{ℓ}_{1}^{(2)}, where ∥x∥_{\overline{ℓ}{(2) \atop 1}} = 2|x_{1}| + |x_{2}|. Then ∥B − C∥_{\overline{ℓ}{(2) \atop 1}} = ω_{V} = 2 and the

relation

Φ_{n}(X, V, \frac{1}{2}(B − C), A, ε) = \frac{C_{n}(ε)}{ω_{V}}

holds.

Remark 4. Let X and Y be two Banach spaces, and let B_{n}(V) = {P_{n} : P_{n} ∈ 𝔓_{n}(X, Y), ∥P_{n}∥_{V} ≤ 1}. Then the following inequalities for x ∈ V ∈ W and the normed h ∈ X follow from Theorem 1 and Lemma 1:

F_{n}(X, V, h, x) = sup_{P_{n}∈B_{n}(V)} |P′_{n}(x)[h]| ≤ b_{2}(X, V, h, x, 0, n) ≤ \frac{C_{n}(0)}{ω_{V} (h)}.

Items (b) and (c) of Theorem 2 also hold for an arbitrary Y if we replace the function Φ_{n} by F_{n} and ε by 0. To this end, it suffices to consider the polynomial U_{n}(t_{1}, t_{2})y, where y is a normed vector in Y, instead of the polynomial U_{n}(t_{1}, t_{2}).

ACKNOWLEDGMENTS

This work was supported by the Russian Foundation for Fundamental Research (project 96-01-00767).

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Translated by S. Vakhromeev