# Question about Banach's fixed point theorem

Let $$(x_n) _{n\ge 1}$$ be a sequence and $$f:\mathbb{R} \to \mathbb{R}$$ a contraction. I know that if $$x_{n+1} =f(x_n)$$ then $$(x_n) _{n\ge 1}$$ converges to $$f$$'s unique fixed point by Banach' s fixed point theorem. What if $$x_n=f(x_{n+1})$$? Can we somehow extend the theorem?

• Just as a note, if $f$ is not injective, i.e. constant, one is already in big trouble as $x_{n+1}$ might not be well-defined. – Jonas Lenz Mar 9 at 17:37
• Yes, I forgot to mention that we suppose the sequence is well-defined. – MathEnthusiast Mar 9 at 17:43

No I do not believe so. Consider $$f$$ defined by $$x\mapsto \frac{x}{2}$$ and $$x_0=1$$. As $$f$$ is a bijection we must then have $$x_n=2^n$$ for each $$n\in \mathbb N$$. Quite cleary $$(x_n)$$ does not converge to the fixed point of $$f$$, which is $$0$$.

Because $$f$$ is a retraction any sequence satisfying this new property must tend to infinity (I'm assuming strict retraction), as long as your initial element of the sequence isn't a fixed point. If $$f$$ is invertible then we would require the inverse to be a retraction for something similar to hold.