The Banach fixed point theorem is stated in my book (Applied Asymptotic Analysis by Miller) as

Let $\mathcal B$ be a Banach space with norm $\|\cdot\|$. Let $X$ be a nonempty bounded subset of $\mathcal B$ and suppose that $T \colon X \to X$ is a mapping that satisfies, for some $0 < \rho < 1$, the inequality $$ \|T(f) - T(g)\| \leq \rho \|f-g\| $$ for all $f$ and $g$ in $X$. Then there exists a unique element $f^\infty \in X$ such that (i) the sequence of iterates $\{T^k(f)\}_{k \geq 0}$ converges to $f^\infty$ whenever $f \in X$ and (ii) $f^\infty = T(f^\infty)$.

I'm having some trouble understanding this result.

Suppose $X_R$ is the ball of radius $R$, i.e. $$ X_R = \{f \in \mathcal B \colon \|f\| \leq R\}, $$ and suppose that $T$ is a contraction mapping on $X_R$. The theorem says that $T$ has a unique fixed point $f^\infty \in X_R$.

But isn't $T$ also a contraction mapping in every ball $X_S$ with $0 < S < R$? Does the theorem then imply that $T$ has a unique fixed point in $X_S$? It then seems to me that we must have $\|f^\infty\| = 0$, for otherwise it would be outside of some such ball.

Where am I going wrong?

  • 1
    $\begingroup$ A constant mapping is automatically a contraction. $\endgroup$ – Michael Greinecker Dec 17 '12 at 7:43

You must have that $T$ is also a contraction mapping in every ball $X_S$, i.e. that $T(X_S) \subseteq X_S$. This part of the criterion seems very implicit but is in fact very important ; it allows you to iterate $T$. That does not follow from the fact that $T$ is a contraction mapping.

Take the example where $T$ just "zooms in" in a sub-ball of your original ball, but that sub-ball closer to the side of the ball than the center (drawing a decreasing sequence of balls to see it is a good idea). Your map will be a contraction mapping, but the limit point will not be zero.

Hope that helps,

  • $\begingroup$ @Jonas : Sure. I guess I just like that word, but I agree it's confusing it use it here, I shouldn't have. Damn me. =) $\endgroup$ – Patrick Da Silva Dec 17 '12 at 7:42
  • $\begingroup$ If $\|T(f)-T(g)\| \leq \rho\|f-g\|$ for all $f,g \in X_R$, then isn't it true that $\|T(f)-T(g)\| \leq \rho\|f-g\|$ for all $f,g \in X_S \subseteq X_R$? The statement of the theorem does not seem to require that $T(X) \subseteq X$. $\endgroup$ – Antonio Vargas Dec 17 '12 at 7:43
  • 1
    $\begingroup$ @Antonio: $T:X\to X$ implies that $T(X)\subseteq X$. $\endgroup$ – Jonas Meyer Dec 17 '12 at 7:45
  • 1
    $\begingroup$ @Antonio Vargas : Yes, but you have $T : X \to X$, so to consider $T(T(x))$ for $x \in X$, you need $T(x) \in X$, which is necessary, but not necessarily true if you take $T : X_R \to X_R$ and restrict $T$ to $X_S$. $\endgroup$ – Patrick Da Silva Dec 17 '12 at 7:46
  • 1
    $\begingroup$ @PatrickDaSilva I see now. Thank you very much for the help! $\endgroup$ – Antonio Vargas Dec 17 '12 at 7:51

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.