On Osgood theorem in Banach spaces
On Osgood theorem in Banach spaces
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Banach 空间中的 Osgood 定理
DOI：
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发表时间：
2003
期刊：
影响因子：
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通讯作者：
S. Shkarin
中科院分区：
文献类型：
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作者：
S. Shkarin
设X是实Banach空间，ω：[0，+∞)→ℝ是递增连续函数，使得ω(0)=0且ω(t+S)≤ω(T)+ω(S)对所有t，S∈[0，+∞).根据奥斯古德定理的无限维类比若∫10(ω(T))−1 dt=∞，则对于任意(t0，x0)∈ℝ×X和任意连续映射f：ℝ×X→X使得∥f(t，x)-f(t，y)∥≤ω(∥x-y∥)对所有t∈ℝ，x，y∈X，柯西问题$dox$(T)=f(t，x(T))，x(T0)=x0在t0的邻域内有唯一解.证明了如果X有一个具有无条件Schauder基的可补子空间且∫10(ω(T))−1 dt<；∞，则存在一个连续映射f：ℝ×X→X使得∥f(t，x)-f(t，y)∥≤ω(∥x-y∥)对所有(t，x，y)∈ℝ×X×X和柯西问题$dox$(T)=f(t，x(T))，x(T0)=x0在实直线的任何区间内没有解.
Let X be a real Banach space, ω : [0, +∞) → ℝ be an increasing continuous function such that ω(0) = 0 and ω(t + s) ≤ ω(t) + ω(s) for all t, s ∈ [0, +∞). According to the infinite dimensional analog of the Osgood theorem if ∫10 (ω(t))−1 dt = ∞, then for any (t0, x0) ∈ ℝ×X and any continuous map f : ℝ×X → X such that ∥f(t, x) – f(t, y)∥ ≤ ω(∥x – y∥) for all t ∈ ℝ, x, y ∈ X, the Cauchy problem $dot x$(t) = f(t, x(t)), x(t0) = x0 has a unique solution in a neighborhood of t0. We prove that if X has a complemented subspace with an unconditional Schauder basis and ∫10 (ω(t))−1 dt < ∞ then there exists a continuous map f : ℝ × X → X such that ∥f(t, x) – f(t, y)∥ ≤ ω(∥x – y∥) for all (t, x, y) ∈ ℝ × X × X and the Cauchy problem $dot x$(t) = f(t, x(t)), x(t0) = x0 has no solutions in any interval of the real line.