Taikov 情形下有限区间上的 Landau-Kolmogorov 问题
偏微分方程分析
2021-07-06 v1 经典分析与常微分方程
泛函分析
摘要
我们在区间 I = [ − 1 , 1 ] \mathbb{I} = [-1,1] I = [ − 1 , 1 ] ∥ f ∥ 2 ⩽ δ \|f\|_2 \leqslant \delta ∥ f ∥ 2 ⩽ δ ∥ f ( r ) ∥ 2 ⩽ 1 \left\|f^{(r)}\right\|_2\leqslant 1 f ( r ) 2 ⩽ 1 ∣ f ( k ) ( t ) ∣ → sup \left|f^{(k)}(t)\right|\to\sup f ( k ) ( t ) → sup t ∈ I t\in\mathbb{I} t ∈ I δ > 0 \delta > 0 δ > 0 r = 1 r = 1 r = 1 r = 2 r = 2 r = 2 sup t ∈ I ∣ f ( k ) ( t ) ∣ = ∣ f ( k ) ( − 1 ) ∣ \sup\limits_{t\in \mathbb{I}}\left|f^{(k)}(t)\right| = \left|f^{(k)}(-1)\right| t ∈ I sup f ( k ) ( t ) = f ( k ) ( − 1 ) I \mathbb{I} I ( − 1 ) r u ( 2 r ) + λ u = − λ f (-1)^ru^{(2r)} + \lambda u = -\lambda f ( − 1 ) r u ( 2 r ) + λ u = − λ f u ( s ) ( − 1 ) = u ( s ) ( 1 ) = 0 u^{(s)}(-1) = u^{(s)}(1) = 0 u ( s ) ( − 1 ) = u ( s ) ( 1 ) = 0 s = 0 , 1 , … , r − 1 s = 0,1,\ldots,r-1 s = 0 , 1 , … , r − 1 λ \lambda λ f f f k ∈ { r − 2 , r − 1 } k \in \{r-2, r-1\} k ∈ { r − 2 , r − 1 } ∥ f ( k ) ∥ ∞ ⩽ A ∥ f ∥ 2 + B ∥ f ( r ) ∥ 2 \left\|f^{(k)}\right\|_\infty \leqslant A\|f\|_2 + B\left\|f^{(r)}\right\|_2 f ( k ) ∞ ⩽ A ∥ f ∥ 2 + B f ( r ) 2 A > 0 A > 0 A > 0 B = B ( A ) B = B(A) B = B ( A )
引用
@article{arxiv.2107.01698,
title = {The Landau-Kolmogorov Problem on a Finite Interval in the Taikov Case},
author = {Dmytro Skorokhodov},
journal= {arXiv preprint arXiv:2107.01698},
year = {2021}
}