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I'm not able to prove this theorem:

Let a map $f:M \rightarrow M$ be a contraction. Then $\forall \varepsilon > 0$ there is a $\delta > 0$ such that every $\delta$-pseudo-orbit is $\varepsilon$-shadowed by some orbit of $f$.

So, $\exists L \in (0,1)$ such that $d(f(x), f(y)) \leq L \cdot d(x,y)$. I denoted $\delta$-pseudo-orbit as $(x_n)$ and true orbit as $(y_n)$:

We know:

$x_{n+1} = f(x_n) + \delta_n$

$y_{n+1} = f^n(y_0)$

$d(x_{n+1}, y_{n+1}) < \delta$

We must prove that:

$d(x_n, y_n) < \varepsilon$

I don't know what direction to take. Tried to express $x_n$ in such way so it depends on $x_0$, but was unable to do so. Any help?

Egor N
  • 105

1 Answers1

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Given $\epsilon>0$, let $\delta:=(1-L)\epsilon$. Given a $\delta$-pseudo-orbit $(x_n)_{n\ge 0}$, let $(y_n)_{n\ge 0}$ be the true orbit with initial condition $y_0=x_0$. Let us show that $(x_n)$ is $\epsilon$-shadowed by $(y_n)$.

For every $n\ge 0$, denote $d_n:=d(x_n,y_n)$. By definition, $d_0=0$ and for every $n\ge 0$,

$$d_{n+1}=d(x_{n+1}, f(y_n))\le d(f(x_n), f(y_n))+d(x_{n+1}, f(x_n))\le L\cdot d_n+\delta.$$ Or equivalently, $$d_{n+1}-\epsilon\le L\cdot (d_n-\epsilon).$$ Then by induction, for every $n\ge 0$, $$d_n-\epsilon\le L^n\cdot (d_0-\epsilon)\Longrightarrow d_n\le \epsilon,$$ which completes the proof.

user104254
  • 1,937