2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.