Zack Wise

2022-04-25

How can we show that $x\left(t\right)-y\left(t\right)={\int}_{s-\epsilon}^{t}f(x\left(r\right),a)-f(y\left(r\right),a\left(r\right)):\mathrm{dr}$ implies $x\left(s\right)-y\left(s\right)=(f(x\left(s\right),a)-f(x\left(s\right),\alpha \left(s\right)))\epsilon +o\left(\epsilon \right)$

Daisy Patrick

Beginner2022-04-26Added 16 answers

Step 1

Given: $\Vert x\left(s\right)-y\left(s\right)-\epsilon \left(f\right(x\left(s\right),a)-f(x\left(s\right),\alpha \left(s\right)\left)\right)\Vert \le {\int}_{s-\epsilon}^{s}\Vert f\left(x\right(t),a)-f\left(x\right(s),a)\Vert dt+{\int}_{s-\epsilon}^{s}\Vert f\left(y\right(t),\alpha (t\left)\right)-f\left(x\right(s),\alpha (s\left)\right)\Vert dt.$

Using continuity the first term is bounded by

$\epsilon \underset{t\in [s-\epsilon ;s]}{\mathrm{sup}}\Vert f\left(x\right(t),a)-f\left(x\right(s),a)\Vert =o\left(\epsilon \right).$

Step 2

For the second term we use

$\Vert f\left(y\right(t),\alpha (t\left)\right)-f\left(x\right(s),\alpha (s\left)\right)\Vert \le \Vert f\left(y\right(t),\alpha (t\left)\right)-f\left(y\right(s),\alpha (s\left)\right)\Vert +\Vert f\left(y\right(s),\alpha (s\left)\right)-f\left(x\right(s),\alpha (s\left)\right)\Vert .$

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

$\Vert f\left(y\right(s),\alpha (s\left)\right)-f\left(x\right(s),\alpha (s\left)\right)\Vert \le C\Vert x\left(s\right)-y\left(s\right)\Vert =o\left(1\right).$

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