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?

  • $\begingroup$ 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. $\endgroup$ – Jonas Lenz Mar 9 at 17:37
  • $\begingroup$ Yes, I forgot to mention that we suppose the sequence is well-defined. $\endgroup$ – 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.


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