In a hyperbolic dynamical system, the Shadowing lemma states that every epsilon-pseudo-orbit is uniquely delta-shadowed by some orbit. see: https://en.wikipedia.org/wiki/Shadowing_lemma
It is not clear to me why this is specific to hyperbolic dynamical systems. What about the following general proof for (uniform) continuous functions.
Dynamical system $f:X\to X$ continuous, s.th. $d(f(x),f(y))<C\cdot d(x,y)$. Let $(x_n)$ be a epsilon-pseudo-orbit, i.e. $d(f(x_n),x_{n+1})<\epsilon$. Let's estimate $d(f^n(x_0),x_n)$ for some $n\in\mathbb N$: We have
- $d(f(x_0),x_1)<\epsilon$ as demanded per premise
- $d(f^2(x_0),x_2) \overset{\Delta\text{ inequality}}{\leq} \underbrace{d(f^2(x_0),f(x_1))}_{<C \cdot \underbrace{d(f(x_0),x_1)}_{<\epsilon}}+ \underbrace{d(f(x_1),x_2) }_{<\epsilon} \leq (C+1)\epsilon$
- $d(f^3(x_0),x_3)\overset{\Delta\text{ inequality}}{\leq} \underbrace{d(f^3(x_0),f(x_2))}_{<C\cdot \underbrace{ d(f^2(x_0),x_2)}_{\leq (C+1)\epsilon}}+ \underbrace{d(f(x_2),x_3)}_{<\epsilon} \leq (C^2+C+1)\epsilon$
- per induction ...
- $d(f^n(x_0),x_n) \leq (C^{n-1}+ \dots+1)\epsilon=\sum_{k=0}^{n-1}C^k \cdot\epsilon=\frac{1-C^n}{1-C}\epsilon \leq \delta$.
Thus, we find that the point $x_0$ induces a shadow orbit per $y_n:=f^n(x_0)$, that is arbitrary close ($\delta$ close) if $\epsilon$ is sufficiently small.
Question What is the error in this proof, why do you expend such an effort to proof the shadowing lemma for hyperbolic dynamical systems if it is true in such a wide sense?
Edit Given $\delta>0$, define $\epsilon(N) :=\underbrace{ \min_{n=0}^N \left( \frac{1-C}{1-C^n}\right) }_{= \frac{1-C}{1-C^N} } \cdot \delta$ . The limit for $N\to\infty$ (i.e. shadowing for all $n$ which is infinite orbits) exists for $C<1$ and is $1-C$ s.th. we choose $\epsilon:=(1-C)\delta$ independend of $N$. (for $C=1$ the geometric series expression is wrong but the limit does not exist either)
Your errors are, basically, errors of logic.
The shadowing lemma says that for any $\delta > 0$ there exists $\epsilon > 0$ such that for every $\epsilon$-pseudo-orbit $(x_n)$ there exists a true orbit $(y_n)$ such that $d(x_n,y_n) < \delta$.
So the outer ring of the logic of the proof should be: given $\delta>0$, produce $\epsilon>0$ as a function of $\delta$ independent of $n$. You seem to instead have produced $\delta$ as a function of $\epsilon$ depending on $n$.
Also, in the inner ring of logic, given an $\epsilon$ pseudo-orbit $(x_n)_{n \in \mathbb Z}$ you must produce a true orbit $(y_n)_{n \in \mathbb Z}$ and then prove that $d(x_n,y_n) < \delta$ for all $n$. You seem to have instead have proved $d(f^n(x_0),x_n) < \delta$, which is quite beside the point (and, again, using a $\delta$ which depends on $n$ and on $\epsilon$, which is quite beside the point of the outer ring of logic).
If you'd like to see how you can critique your own proof, here's a suggestion. Consider the identity function $f : \mathbb R \to \mathbb R$. It's an isometry, $d(f(x),f(y))=d(x,y)$ so it satisfies your hypothesis with $C=2$. Every orbit of $f$ is just a single point. Here's an unshadowable $\epsilon$-pseudo-orbit: $$x_n = n \, \epsilon \, / \, 2 $$ What does your proof do when confronted with this example?