7
$\begingroup$

Lipschitz maps are defined over metric space as maps $f:(X,d_X) \to (Y,d_Y)$ such that $$ d\left( f(x),f(x^\prime) \right)_Y \le k d(x,x^\prime)_X \ \forall x,x^\prime \in X, $$ where $k$ is a positive constant. We usually say that $f$ is a contraction if $k<1$.

It is well know that a different equivalent metric on $X$ does not preserve contractions, i.e. a map can be a contraction with respect to a metric but not with respect to an equivalent one.

In the Banach space setting, where the spaces $X$ and $Y$ are endowed with a norm defining the topology, there is a somehow "canonical" distance given by $$ d(x,x^\prime) = \lVert x-x^\prime \rVert .$$ With this distance, Lipschitz maps can be characterized as maps satisfying, for some $k>0$ $$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_Y \le k {\left \lVert x-x^\prime \right \rVert}_X \ \forall x, x^\prime \in X. $$ It is obvious that, if $f$ satisfies the above relation, then is a $k$-lipschitz map.


In the Fréchet space setting, the topology is defined by a countable family of semi-norms $({\lVert\cdot\rVert}_n)$. The classical example of metric inducing the same topology is given by $$ d(x,x^\prime) = \sum_{n=0}^\infty {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} . $$

In analogy with the Banach case, I would like to characterize (at least some) Lipschtiz maps between Fréchet spaces as maps satisfying $$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_n \le k {\left \lVert x-x^\prime \right \rVert}_n \ \forall x, x^\prime \in X,\ \forall n \in \mathbb{N}. $$ Again, maps satisfying the last equation are Lipschitz maps with respect to the metric defined above, but the Lipschitz constant is not $k$ anymore, and in particular contraction with respect to the semi-norms (i.e. maps satisfying the last equation with $k<1$) are not contraction with respect to the metric.

Are there equivalent distances on $X$ and $Y$ such that every contraction with respect to the semi-norms is a contraction with respect with the new distance? If this is not possibile for every contraction, is it possible for a specific one?

$\endgroup$

1 Answer 1

5
$\begingroup$

Use $\sum 2^{-n}(\|x-y\|_n \wedge 1)$ for the distance on $Y$ and $\sum 2^{-n}(\|x-y\|_n \wedge 2)$ for the distance on $X$.

$\endgroup$
5
  • $\begingroup$ I understand the idea, but I would say that if I take $ \sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1)$ on $X$, I need $\sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1/k)$ on $Y$. Anyway... what about maps from $X$ into himself? Are maps $f:X\toX$ satisfying $\left\lVert f(x) -f(y)\right\rVert \le k \left \lVert x-y \right \rVert$, with $k<1$, distance-contraction (and thus have a unique fixed point)? maybe this is a different question... $\endgroup$ Aug 7, 2011 at 20:03
  • $\begingroup$ I meant $\sum 2^{-n} (\left\lVert x-y \right\rVert \wedge k)$ on $Y$. $\endgroup$ Aug 7, 2011 at 21:29
  • $\begingroup$ As for having the same distance on $X$ and $Y$ when $X=Y$, that is not possible when e.g. $X$ is the countable product of lines. For this space $I/2$ is not a contraction under any compatible metric. $\endgroup$ Aug 7, 2011 at 23:17
  • $\begingroup$ ...but it still has a unique fixed point, right? $\endgroup$ Aug 8, 2011 at 8:39
  • $\begingroup$ Sure; zero. I is the identity operator. $\endgroup$ Aug 8, 2011 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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