Strengthening the Comparison Theorem and Kolmogorov Inequality in the Asymmetric Case

. We obtain an strengthening the Kolmogorov comparison theorem. In particular, it gives us the opportunity to obtain such strengthening Kolmogorov inequality in the asymmetric case: for functions x ∈ L r ∞ ( R ) where k, ) r is the asymmetric perfect spline of Euler of order r and E 0 ( x ) ∞ is the best uniform approximation of the function x by constants.


Strengthening the Comparison Theorem and Kolmogorov Inequality in the Asymmetric Case
Abstract. We obtain an strengthening the Kolmogorov comparison theorem. In particular, it gives us the opportunity to obtain such strengthening Kolmogorov inequality in the asymmetric case: {|x(β) − x(α)| : x (t) = 0 ∀t ∈ (α, β)}, k, r ∈ N, k < r, α, β > 0, ϕr(· ; α, β)r is the asymmetric perfect spline of Euler of order r and E0(x)∞ is the best uniform approximation of the function x by constants. Key words: Kolmogorov comparision theorem, Kolmogorov inequality, asymmetric case, strengthening.
1. Introduction. Let G be the real line R or the unit circle T which is realized as the interval [0, 2π] with coincident endpoints. We will consider the spaces L p (G), 1 ≤ p ≤ ∞, of all measurable functions x : G → R such that For α, β > 0 and x ∈ L ∞ (G) set For r ∈ N denote by L r ∞ (G) the space of all functions x ∈ L ∞ (G) for which x (r−1) is locally absolutely continuous and x (r) ∈ L ∞ (G).
Notice that ϕ r (· ; 1, 1) is the spline of Euler of order r.
Hörmander [1] proved the following theorem. Theorem А. Let k, r ∈ N, k < r, G = R or G = T. Then for any function x ∈ L r ∞ (G) and for any α, β > 0 there is the sharp inequality where E 0 (x) ∞ is the best uniform approximation of the function x by constants. The equality in (1) is achieved for the functions The proof of Theorem A in [1] is based on the comparison theorem. In view of the importance of this theorem for further exposition, we present its formulation. For and let ϕ λ,r (t; α, β) := λ −r ϕ r (λt; α, β) for λ > 0.
Theorem В. Let r ∈ N; α, β > 0; x ∈ W r ∞;α,β (R) and the number λ is such that If the points ξ and η satisfying conditions In the symmetric case α = β Theorems A and B are due to Kolmogorov [2]. In this paper, we obtain (Theorem 1) a strengthening of Theorem B in which condition (2) At the same time, the conclusion of Theorem 1 is stronger than the conclusion of Theorem B. It is clear that |||x||| ∞ ≤ E 0 (x) ∞ , and it is easy to give examples of infinitely differentiable functions x, for which the ratio Using Theorem 1, we obtain (Theorem 2) a strengthening of inequality (1), in which the quantity E 0 (x) ∞ is replaced by a more delicate characteristic |||x||| ∞ . As an application, we obtain (Theorem 3) a strengthening of Ligun's inequality [3] and Babenko inequality [4].
Proof. For r = 1 the assertion of the lemma is obvious. Therefore, we will assume that r ≥ 2.
Let us now prove that condition (8) implies the inequality those. condition (4) of Lemma 1 is satisfied. Let's assume the opposite. Let, for example.
The following theorem is a strengthened version of inequality (1).