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?


1 Answer 1


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 $X$.

  • $\begingroup$ 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

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