How can we show that&nbsp;x(t)-y(t)=∫s-εtf(x(r),a)-f(y(r),a(r)) dr&nbsp;implies&nbsp;x(s)-y(s)=(f(x(s),a)-f(x(s),α(s)))ε+o(ε)

Step 1Given:&nbsp;‖x(s)-y(s)-ε(f(x(s),a)-f(x(s),α(s)))‖≤∫s-εs‖f(x(t),a)--f(x(s),a)‖dt+∫s-εs‖f(y(t),α(t))-f(x(s),α(s))‖dt.Using continuity the first term is bounded byεsupt∈[s-ε;s]‖f(x(t),a)-f(x(s),a)‖=o(ε).Step 2For the second term we use‖f(y(t),α(t))-f(x(s),α(s))‖≤‖f(y(t),α(t))-f(y(s),α(s))‖++‖f(y(s),α(s))-f(x(s),α(s))‖.We only need to show that the expression above is o(1). For the first term this follows from continuity and for the second term we use that f is differentiable to get‖f(y(s),α(s))-f(x(s),α(s))‖≤C‖x(s)-y(s)‖=o(1).

Key to "How can we show thatx(t)-y(t)=∫s-εtf(x(r),a)-f(y(r),a(r)) drimpliesx(s)-y(s)=(f(x(s),a)-f(x(s),α(s)))ε+o(ε)"

Tony Mccarthy

Answered question

2022-03-25

How can we show that $x (t) - y (t) = \int_{s - ε}^{t} f (x (r), a) - f (y (r), a (r)) d r$ implies $x (s) - y (s) = (f (x (s), a) - f (x (s), α (s))) ε + o (ε)$

Answer & Explanation

Janessa Foster

Beginner2022-03-26Added 12 answers

Step 1
Given: $‖ x (s) - y (s) - ε (f (x (s), a) - f (x (s), α (s))) ‖ \leq \int_{s - ε}^{s} ‖ f (x (t), a) -$

$- f (x (s), a) ‖ d t + \int_{s - ε}^{s} ‖ f (y (t), α (t)) - f (x (s), α (s)) ‖ d t .$
Using continuity the first term is bounded by
$ε \sup_{t \in [s - ε; s]} ‖ f (x (t), a) - f (x (s), a) ‖ = o (ε) .$
Step 2
For the second term we use
$‖ f (y (t), α (t)) - f (x (s), α (s)) ‖ \leq ‖ f (y (t), α (t)) - f (y (s), α (s)) ‖ +$

$+ ‖ f (y (s), α (s)) - f (x (s), α (s)) ‖ .$
We only need to show that the expression above is o(1). For the first term this follows from continuity and for the second term we use that f is differentiable to get
$‖ f (y (s), α (s)) - f (x (s), α (s)) ‖ \leq C ‖ x (s) - y (s) ‖ = o (1) .$

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