Comparison of Exact Constants in Inequalities for Derivatives of Functions Defined on the Real Axis and a Circle

• Дата: 2003
• Автори: Babenko, V. F., Kofanov, V. A., Pichugov, S. A., Бабенко, В. Ф., Кофанов, В. А., Пичугов, С. А.
• Формат: Стаття
• Мова: Російська; Англійська
• Опубліковано: Institute of Mathematics, NAS of Ukraine 2003
• Онлайн доступ: https://umj.imath.kiev.ua/index.php/umj/article/view/3934
• Назва журналу: Ukrains’kyi Matematychnyi Zhurnal

We investigate the relationship between the constants K(R) and K(T), where \(K\left( G \right) = K_{k,r} \left( {G;q,p,s;\alpha } \right): = \mathop {\mathop {\sup }\limits_{x \in L_{p,s}^r \left( G \right)} }\limits_{x^{(r)} \ne 0} \frac{{\left\| {x^{\left( k \right)} } \right\|_{L_q \left( G \right)} }}{{\left\| x \right\|_{L_q \left( G \right)}^\alpha \left\| {x^{\left( r \right)} } \right\|_{L_s \left( G \right)}^{1 - \alpha } }}\) is the exact constant in the Kolmogorov inequality, R is the real axis, T is a unit circle, $$L_{p,s}^r (G)$$ is the set of functions x ∈ L p(G) such that x (r) ∈ L s(G), q, p, s ∈ [1, ∞], k, r ∈ N, k < r, We prove that if $$\frac{r - k + 1/q - 1/s}{r + 1/q - 1/s} = 1 - k/r$$ thenK(R) = K(T),but if $$\frac{r - k + 1/q - 1/s}{r + 1/q - 1/s} < 1 - k/r$$ thenK(R) ≤ K(T); moreover, the last inequality can be an equality as well as a strict inequality. As a corollary, we obtain new exact Kolmogorov-type inequalities on the real axis.