What is preserved by lipeomorphisms?

A lipeomorphism is a continuous surjective function $f: X \to Y$ between metric spaces such that

$$A d_X(x,y)\leq d_Y(f(x),f(y))\leq B d_X(x,y),$$

where $A,B > 0$ are constant.

I want to know what kind of metric properties are preserved by this kind of function. Is it possible to not just intuitively, but formally define which metric properties are preserved?

It is easy to prove that boundedness, Cauchy sequences and completeness are preserved by a lipeomorphism.

Is there some metric property that is not preserved by a lipeomorphism?

• many authors say bi-Lipschitz Sep 22 '15 at 4:04

Lipeomorphisms preserve completeness and boundedness. For, let $X$ and $Y$ be lipeomorphic, where $f: X \to Y$ is a lipeomorphism, and let $X$ be bounded and complete.

If $(y_l)_{l \in \mathbb N}$ is a Cauchy sequence in $Y$, then if we define $x_n := f^{-1}(y_n)$, then $(x_l)_{l \in \mathbb N}$ is a Cauchy sequence in $X$, for $d(x_n, x_k) \le \frac{1}{A} d(y_n, y_k)$. Since $X$ is complete, there exists a limit $x$ of $(x_l)_{l \in \mathbb N}$. By continuity, $y_l \to y := f(x), l \to \infty$.

If $b > 0$ is a constant such that $\forall x, y \in X : d_X(x, y) < b$, then $c := Bb$ satisfies $d_Y(z, w) < c$ for all $z, w \in Y$.

Note that in our proof for the boundedness of $Y$, we used only the boundedness of $X$ (and not the completeness). Similarly, in our proof for the completeness of $Y$, we used only the completeness of $X$, and not the boundedness. Hence, we may indeed claim that lipeomorphisms preserve boundedness and completeness, even if only one of these properties is given.

Lipeomorphisms do not preserve distance. There even are spaces which are lipeomorphic, but not isometric. As an example, take $X := [0, 1]$ and $Y := [0, 2]$, which are lipeomorphic by the function $x \mapsto 2x$, but not isometric.

You may also want to have a look at those valuable notes.

There are some more properties which are preserved. For example: We have a metric space $X$ with metric $d_S$ and a metric space $Y$ with metric $d_Y$, and a lipeomorphism $f : X \to Y$. Let $(Z, d_Z)$ be another metric space.

1) If $F \subset \mathcal C(Z; X)$ is equicontinuous, for each $z \in Z$ and $\epsilon > 0$ we may choose $\mu > 0$ such that for $d_X(x, y) < \mu$ we have $d_Y(f(x), f(y)) < \epsilon$, and then we may choose $\delta > 0$ such that for $w \in B_\delta(z)$ $\sup_{g \in F} d_X(g(z), g(w)) < \mu$ to obtain that for $w \in B_\delta(z)$ and $g \in F$ $$d_Y(f(g(z)), f(g(w))) < \epsilon.$$ Hence, $f \circ F := \{f \circ g | g \in F\} \subset \mathcal C(Z, Y)$ is equicontinuous.

Furthermore, if $F \subset \mathcal C(Y, Z)$ is equicontinuous, for each $y \in Y$ and $\epsilon > 0$ we may choose $\mu > 0$ such that $d_Y(x, y) < \mu$ implies $\sup_{g \in F} d_Z(g(x), g(y)) < \epsilon$, and then $\delta > 0$ such that $d_X(x, y) < \delta$ implies $d_Y(f(x), f(y)) < \mu$. If $d_X(x, y) < \delta$ and $g \in F$, it follows that $$d_Z(g(f(x)), g(f(y))) < \epsilon.$$

Hence, $F \circ f := \{g \circ f | g \in F\} \subset \mathcal C(X, Z)$ is equicontinuous.

Thus, in a sense we could say that equicontinuity is preserved; but note that we only used that $f$ is continuous. Hence, all homeomorphisms preserve equicontinuity in the above sense.

2) If $F \subset \mathcal C(Z; X)$ is uniformly equicontinuous, for each $\epsilon > 0$ we may choose $\mu > 0$ such that for $d_X(x, y) < \mu$ we have $d_Y(f(x), f(y)) < \epsilon$, and then we may choose $\delta > 0$ such that for $w \in B_\delta(z)$ $\sup_{g \in F} d_X(g(z), g(w)) < \mu$ to obtain that for $w \in B_\delta(z)$ and $g \in F$ $$d_Y(f(g(z)), f(g(w))) < \epsilon.$$ Hence, $f \circ F := \{f \circ g | g \in F\} \subset \mathcal C(Z, Y)$ is uniformly equicontinuous.

Furthermore, if $F \subset \mathcal C(Y, Z)$ is uniformly equicontinuous, for each $\epsilon > 0$ we may choose $\mu > 0$ such that $d_Y(x, y) < \mu$ implies $\sup_{g \in F} d_Z(g(x), g(y)) < \epsilon$, and then $\delta > 0$ such that $d_X(x, y) < \delta$ implies $d_Y(f(x), f(y)) < \mu$. If $d_X(x, y) < \delta$ and $g \in F$, it follows that $$d_Z(g(f(x)), g(f(y))) < \epsilon.$$

Hence, $F \circ f := \{g \circ f | g \in F\} \subset \mathcal C(X, Z)$ is uniformly equicontinuous.

Thus, in a sense we could say that uniform equicontinuity is preserved; but note that we only used that $f$ is uniformly continuous. Hence, all uniformly continuous homeomorphisms preserve equicontinuity in the above sense.

3) If $F \subset \mathcal C(Z; X)$ is Lipschitz equicontinuous, then we may choose a constant $L >0$ such that $d_X(g(z), g(w)) \le L d_Z(z, w)$ for all $g \in F$. By the Lipschitz continuity of $f$ follows $$\forall g \in F : d_Y(f(g(z)), f(g(w))) \le C d_X(g(z), g(w)) \le C L d_Z(z, w),$$ where $C := 1/A$. Hence, $f \circ F := \{f \circ g | g \in F\} \subset \mathcal C(Z, Y)$ is Lipschitz equicontinuous.

Furthermore, if $F \subset \mathcal C(Y, Z)$ is Lipschitz equicontinuous, there exists a constant $b > 0$ such that for all $g \in F$, $d_Z(g(z), g(w)) \le b d_Y(z, w)$, and it follows that $$\forall g \in F: d(g(f(x)), g(f(y))) \le b d_Y(f(x), f(y)) \le bB d_X(x, y).$$

Hence, $F \circ f := \{g \circ f | g \in F\} \subset \mathcal C(X, Z)$ is equicontinuous.

For this preservation to hold, we actually used that $f$ is Lipschitz continuous.

Further, all topological properties are preserved, since a lipeomorphism is a homeomorphism (inserting $x=f^{-1}(z)$ and $y = f^{-1}(w)$ into the first of your inequalities yields $d(f^{-1}(z), f^{-1}(w)) \le \frac{1}{A} d(z, w)$).

Further, if you transform your group to a metric space, lipeomorphisms preserve the quasi-isometric invariants.

There are more properties which are not necessarily preserved.

1) Isomorphisms to other spaces are not necessarily preserved.

For, if $f$ is no isomorphism (which is possible), then we may choose two points $x, y \in X$ such that $d_X(x, y) \neq d_Y(f(x), f(y))$, and if $g: Y \to Z$ is an isomorphism, then $g \circ f : X \to Z$ is not an isomorphism, and if $h: Z \to X$ is an isomorphism, then $f \circ h: Z \to Y$ is not an isomorphism.

2) Optimal Lipschitz constant is not necessarily preserved.

This follows from the previous point, since isomorphisms are exactly the lipeomorphisms with optimal Lipschitz constants $1$.

Metric equivalence is not preserved by lipeomorphisms, in fact not even isometries suffice.

I think I now see a bit better where your question was directed, but I think the best formal statement may be the one given by the definition; this I conjecture by analogy to homeomorphisms, where there is a huge quantity of properties which are preserved (namely the topological properties), and the other properties are not preserved.

• Great answer! +1 Sep 23 '15 at 20:47
• Also preserve uniform continuity.
– YTS
Sep 24 '15 at 3:58
• Even uniform Lipschitz continuity. Let me just add that. Sep 24 '15 at 4:13