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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?

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    $\begingroup$ A constant mapping is automatically a contraction. $\endgroup$ Dec 17, 2012 at 7:43

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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,

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  • $\begingroup$ @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. =) $\endgroup$ Dec 17, 2012 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 $T(X) \subseteq X$. $\endgroup$ Dec 17, 2012 at 7:43
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    $\begingroup$ @Antonio: $T:X\to X$ implies that $T(X)\subseteq X$. $\endgroup$ Dec 17, 2012 at 7:45
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    $\begingroup$ @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 $X_S$. $\endgroup$ Dec 17, 2012 at 7:46
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    $\begingroup$ @PatrickDaSilva I see now. Thank you very much for the help! $\endgroup$ Dec 17, 2012 at 7:51

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