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give an example of a complete metric space $(X,d)$ and a mapping $T: X \rightarrow X$ which does not have a fixed point in X and satisfies; $$ d(T(x),T(y)) < d(x,y)$$ $\forall x,y \in X, x\neq y$

i thought first that this was impossible by the fixed point theorem, but then figured with the final constraint it must be possible in some way. however i haven't been able to find any such example. I've tried with $\mathbb {R}$ and some subsets mostly, and used the discrete metric, but can't find a map to make this work...

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    $\begingroup$ There are already some good answers here, but I will say the following for future reference. You said you used the discrete metric, but it is clear that this conjecture cannot be true for the discrete metric. The condition $ \forall x,y \in X, x \not= y, d(T(x),T(y)) < d(x,y)$ requires $ \forall x,y \in X, x \not= y$ $ T(x) = T(y)$ i.e. that $T$ is constant. Any constant $T:X \rightarrow X$ is certain to have a fixed point. $\endgroup$ – Tom Oldfield Dec 28 '12 at 7:34
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== Take the space $\,K:=[1,\infty)\,$ with the inherited euclidean topology from $\,\Bbb R\,$ , and remark that this is a complete metric space. (hint: it is a closed space)

== Define $\,f(x):=x+x^{-1}\,$ on the above space.

== For $\,x,y\in K\,\,,\,x\neq y\,$, check that

$$|f(x)-f(y)|=|x-y|\left|1-\frac{1}{xy}\right|<|x-y|$$

== Finally, prove $\,f\,$ has no fixed point (hint: suppose it does...)

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  • $\begingroup$ COncerning "$f$ has no fixed point", there's no need to assume the contrary. I'd say it's by its very definition that $f$ has no fixed point. $\endgroup$ – Hendrik Vogt Dec 28 '12 at 8:11
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Try thinking about it graphically. Think of an increasing function on $\{x\ge 0\}$ whose graph is above and asymptotic to $y=x$.

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Try $X = [0,\infty )$ , $(X,d)$ is complete metric space, where $d(x,y) = |x-y|$

$T(x) = x + e^{-x} $

$$\sup_{x\neq y} = \frac{d(T(x),T(y))}{d(x,y)} = 1$$

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  • $\begingroup$ using what distance function? i don't think it works for absolute value, or the discrete case... $\endgroup$ – pad Dec 28 '12 at 4:55
  • $\begingroup$ I changed $sup\underset{x\ne y}{}$ to $\displaystyle\sup_{x\ne y}$. That is standard usage. $\endgroup$ – Michael Hardy Dec 28 '12 at 6:12
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$X=\mathbb R$,

$T(x)=\sqrt{1+x^2}$,

$d=\text{the usual metric}$.

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