4 typo in the formula

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 $$d(x,x^\prime) = \sum_{n=0}^\infty \frac 1 {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} .$$

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?

3 typo

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 $$d(x,x^\prime) = \sum_{n=0}^\infty \frac 1 {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} .$$

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 the 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?

2 added weaker question (in case the first one is negative)

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 $$d(x,x^\prime) = \sum_{n=0}^\infty \frac 1 {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} .$$

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 the 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