In some paper, authors claim that $f(x)=2x$ has shadowing property, this means that for every $\epsilon>0$ there is a $\delta>0$ such that every $\delta$-pseudo orbit $\{x_n\}$, there is $p\in \mathbb{R}$ with $|f^n(p)- x_n|<\epsilon$ for all $n\in\mathbb{Z}$. Note that $\{x_n\}$ is $\delta$-pseudo orbit, if $|f(x_n)-x_{n+1}|<\delta$ for all $n$.
Take $g(x)= 2x+ r$ where $0<r<\delta$ and $x_n=g^n(x)$. Then it is easy to see that $\{x_n\}$ is $\delta$-pseudo orbit for $f(x)=2x$. Hence for every $x\in \mathbb{R}$, there is $y\in \mathbb{R}$ such that $|f^n(y)- g^n(x)|<\epsilon$ for all $n$. But $g^n(x)= 2^nx+ \sum_{i=0}^{n-1} 2^ir$ and $f^n(y)= 2^ny$. Since $f(x)=2x$ has shadowing property, we must have the following relation. But it is not clear for me, because I think that it does not hold for sufficiently big $n$.
$$|2^n(y-x)- \sum_{i=0}^{n-1} 2^ir|<\epsilon$$ and $|y-x|<\epsilon$.
Please help me to know it.
You haven't really given a counterexample because assuming that $f$ has the shadowing property, for every $\epsilon>0$ choose $\delta>0$ such that $\delta<\epsilon$. Now you need to fix $r$ as $0<r<\delta$ for $g(x)$ to be a $\delta$-pseudo orbit. For this $r$ and a starting point $x$, $x_n=g^n(x)$, fix $p=x+r \in \mathbb{R}$, then $$|f^n(p)-x_n|=|2^n p - 2^n x - 2^n r + r|=|2^n(p-x-r)+r|=|r|<\delta<\epsilon$$ where $\sum\limits^{n-1}_{i=0}2^i=2^n-1$ for $n>0$, and $\sum\limits^{-1}_{i=n}2^i=1-2^n$ for $n<0$ (I don't know why you didn't simplify that).