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Definition of a Contraction. Let $(X, d)$ be a metric space. Then a map $T : X → X$ is called a contraction on $X$ if there exists $q ∈ [0, 1)$ such that $d(T(x),T(y)) \le q d(x,y)$ for all $x, y$ in $X$.

My question: Does a contraction remain a contraction under an equivalent metric $d'$?

I know that Lipschitz continuity is not preserved in general under equivalent metrics, and since the two definitions are quite similar we may believe contractions are not preserved.

However, contractions have the additional requirement that they map a metric space to itself, so changing the scale of the metric will not create issues.

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The answer is still negative. For example, the map $Tx=x/2$ is a contraction on $\mathbb{R}$ with the standard metric. However, it is not a contraction under the spherical metric $$d(x, y) = |\tan^{-1}x - \tan^{-1}y|$$ (Note that for strictly monotone continuous function $f:\mathbb{R}\to\mathbb{R}$ the formula $d(x, y)=|f(x)-f(y)|$ defines an equivalent metric.) Indeed, $d(2, 1)>d(4, 2)$ because $$ \tan^{-1}2 - \tan^{-1} 1 > \tan^{-1}4 - \tan^{-1} 2 $$ as verified by a numerical computation.

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user357151 showed this isn't true for equivalent metrics in general.

However, if we restrict ourselves to metrics induced by equivalent norms, we get an interesting relation.

Consider $T:V \to V$, where $V$ is a normed vector space with two equivalent norms, say $||\cdot||_1$ and $||\cdot||_2$. Then, there exists some positive constants $a,b$ such that:

$$ a ||x||_1 \le ||x||_2 \le b ||x||_1 \text{ for all }x\in V $$

Additionally, assume that $d_1$ and $d_2$ are induced by the first and second norm respectively, and that we have

$$ d_1( T(x), T(y)) \le c d_1(x, y) $$

We have the relation

$$ \begin{align*} d_2(T(x),T(y) ) &\le b\times c \times d_1(x,y) \\ &\le \frac{b}{a} \times c \times d_2(x,y)\\ &= c_* d_2(x,y) \end{align*} $$

with $c_* = \frac{b}{a} \times c$.

My result was obtained with the help of this proof.

Hence we may suspect we can lose the contraction properties if the ratio $b/a$ is large enough. This is indeed the case.

Consider the following function $f: \mathbb{R}^2 \to \mathbb{R}^2$,

$$f(x) = (.9\max(|x_1|,|x_2|), .9 \max(|x_1|,|x_2|))$$

Then $f$ is a contraction under the metric induced by the maximum norm $d_\infty$, but not under the metric induced by the Manhattan norm $d_1$.

Indeed,

$$ \begin{align*} d_\infty( f(x), f(y) ) &= \max( |f_1(x) - f_1(y)|, |f_2(x) - f_2(y)| ) \\ &= .9|\max(|x_1|,|x_2|) - \max(|y_1|,|y_2|)| \\ &\le .9\max(|x_1 - y_1|,|x_2 - y_2|) \\ &= .9 d_\infty(x, y)\end{align*} $$ where the inequality is obtained from this relation. Note that $c = .9$.

However, $f$ is not a contraction for a metric induced by the Manhattan norm $d_1$. For example, taking $x = (1,0)$, $y = (0,0)$, we have

$$\begin{align*} d_1(x,y) &= 1 + 0 = 1 \\ d_1(f(x),f(y)) &= .9 + .9 = 1.8 \end{align*}$$

which proves that $f$ is not a contraction.

Note that for vectors of length 2, the maximum and Manhattan norms follow the following relation,

$$ ||x||_\infty \le ||x||_1 \le 2 ||x||_\infty $$

$b = 2$ and $a = 1$, and so $c_* = 1.8$.

Similarly,

Consider the following function $g: \mathbb{R}^2 \to \mathbb{R}^2$,

$$g(x) = (.7(|x_1|+|x_2|), 0)$$

Then $g$ is a contraction under the metric induced by the Manhattan norm $d_1$, but not under the metric induced by the maximum norm $d_\infty$

We have

$$ \begin{align*} d_1(g(x),g(y)) &= .7(||x_1| - |y_1|| + ||x_2| - |y_2||) \\ &\le .7(|x_1 - y_1| + |x_2 - y_2|) \\ &= .7 d_1(x,y) \end{align*} $$

but we get, with $x = (1,1)$ and $y=(0,0)$,

$$d_\infty(g(x),g(y)) = 2 \times .7 = 1.4 d_\infty(x,y)$$

Again, that's because

$$ .5||x||_1 \le ||x||_\infty \le ||x||_1 $$

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