A function $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ with $\varphi(x)=x,$ $\|\varphi(y)-x\|\leq K\|y-x\|^\alpha$ for $\alpha>1, K>0$ If we have a function $\varphi:\mathbb{R}^n\to\mathbb{R}^n$ with $\varphi(x)=x,$ $\|\varphi(y)-x\|\leq K\|y-x\|^\alpha$ for $K>0,$ and we define $\varphi^1:=\varphi, \varphi^2:=\varphi\circ\varphi,\dots,\varphi^n:=\varphi\circ\cdots\circ\varphi,$ then if $\alpha>1$ it is claimed (according to a problem in some lecture notes) that there exists some open set $X$ with $x\in X$ for which $$\lim\limits_{n\to\infty}\varphi^n(y)=x$$ for all $y\in X.$ My attempt at a proof was:
Let $h=x-y$ for distinct $x$ and $y$ (otherwise it is trivial). Then since $\alpha>1$ we can write
$$\frac{\|\varphi^n(x+h)-\varphi(x)\|}{\|h\|}\le\frac{\|\varphi(x+h)-\varphi(x)\|}{\|h\|}\le K\|h\|^{\alpha-1}\to 0$$ as $\|h\|\to 0,$ so I'm thinking there exists some $\delta>0$ such that if $\|h\|<\delta$ then $\varphi^n(x+h)=\varphi^n(y)\to\varphi(x)=x;$ ie. choose $X=B_\delta(x).$ 
Could somebody just confirm/correct the above? I feel like it's not a very formal proof and it just seems too simple to be correct to be honest. Thanks ahead.
 A: First make some observations:
$$\|\varphi(y)-x\|\leq K\|y-x\|^{\alpha}$$
So :
$$\|\varphi(y)-x\|^{\alpha}\leq K^{\alpha}\|y-x\|^{\alpha^2}$$
Next (using inequality above):
$$\|\varphi^2(y)-x\| \leq K\|\varphi(y)-x\|^\alpha \leq K \cdot K^{\alpha}\|y-x\|^{\alpha^2}$$
So:
$$\|\varphi^2(y)-x\|  \leq K^{\alpha} \cdot K^{\alpha^2}\|y-x\|^{\alpha^3}$$
Next do similar thing:
$$\|\varphi^3(y)-x\|  \leq K \cdot \|\varphi^2(y)-x\|^{\alpha}\leq K \cdot K^{\alpha} \cdot K^{\alpha^2}\|y-x\|^{\alpha^3}$$
Now you probably see the patern, so you can prove folwing lemma by induction:

$\textbf{Lemma}$ For $n \in \mathbb{N}$ 
  $$\|\varphi^n(y)-x\| \leq K^{\frac{\alpha^n-1}{\alpha -1}}\|y-x\|^{\alpha^n}=K^{-\frac{1}{\alpha -1}}\left( K^{\frac{1}{\alpha-1}} \| y-x\|\right)^{\alpha^n}$$

Now it's easy to complete the proof, because if $\left( K \| y-x\|\right) < 1$ (so $y$ is inside ball with radius $K^{\alpha-1}$) then because $\alpha>1$ and for $n \to \infty$ you have $\alpha^n \to \infty$:
$$K^{-\frac{1}{\alpha -1}}\left( K^{\frac{1}{\alpha-1}} \| y-x\|\right)^{\alpha^n} \to 0$$
when $n \to \infty$.So by lemma $$\|\varphi^n(y)-x\| \to 0$$ when $n \to \infty$ for all $y$ in ball with origin $x$ and radius $K^{\alpha-1}$.
