7
$\begingroup$

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...

  • 1
    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.2012-12-28

4 Answers 4

6

== 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...)

  • 0
    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.2012-12-28
4

Try thinking about it graphically. Think of an increasing function on $\{x\ge 0\}$ whose graph is above and asymptotic to $y=x$.

2

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

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

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

  • 0
    I changed $sup\underset{x\ne y}{}$ to $\displaystyle\sup_{x\ne y}$. That is standard usage.2012-12-28
1

$X=\mathbb R$,

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

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