I need some help with the next exercise. I don't know how to proceed.

Let $(X,d)$ be a complete metric space and let $\phi: X\to X$ be a continuous function. Prove that if there exist $\alpha\in(0,1)$ such that $d(\phi^2(x),\phi(x))\leq \alpha d(\phi(x),x)$ for all $x\in X$ then $\phi$ has fixed point.

Clearly, mi first idea was the theorem of the fixed point of Banach. But, I was so stuck for obtain the hypothesis. My attempt:

We want to obtain $d(\phi(x),\phi(y))\leq \alpha d(x,y)$ for all $x,y\in X$. Then, in thought in the triangle inequality. We know that $$d(\phi(x),\phi(y))\leq d(\phi(x),\phi^2(x))+d(\phi^2(x),\phi(y))\leq \alpha d(\phi(x),x)+d(\phi^2(x),\phi(y))$$From here, I don't know how can I continue.

The next idea was the contradiction. If the function doesn't have fixed points, then, for all $x\in X$ we have that $\phi(x)\neq x$. Then, we have that $\phi^2(x)\neq \phi(x)$, then, $0<d(\phi^2(x),\phi(x))\leq \alpha d(\phi(x),x)$ but, then, I don't know what can I do.

I really appreciate any hint. I don't want only the answer. Is better a hint for me. It's one of my first approaches to the analysis.


1 Answer 1


Hint: \begin{align*} d(\phi^{n}(x),\phi^{m}(x))&\leq d(\phi^{n}(x),\phi^{n-1}(x))+\cdots+d(\phi^{m+1}(x),\phi^{m}(x))\\ &\leq\alpha^{n-1}d(\phi(x),x)+\cdots+\alpha^{m-1}d(\phi(x),x)\\ &=(\alpha^{n-1}+\cdots+\alpha^{m-1})d(\phi(x),x), \end{align*} where $n>m$.

If $\phi^{n}(x)\rightarrow y$, then look at $\phi^{n+1}(x)=\phi(\phi^{n}(x))$ and use the continuity of $\phi$ to get the result.


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