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In the definition of contraction mapping, it requires:

$(1)$ the function $f$ must map the domain $A$ to $A$, where $A\subseteq\mathbb{R}$.

$(2)$ There exists a constant $0<c<1$, such that $\forall x,y\in A, |f(x)-f(y)|\le c|x-y|$

My question is, why it must require $(1)$, and is there a counter-example such that $f: A\to B$, where $A\neq B$, but it is not a contraction mapping? To make this counter-example more strong, assume both $A$ and $B$ are compact sets.

In this post, it shows $\cos(x)$ is a contraction mapping on $[0, \pi]$, but clearly it is not a $A\to A$ mapping.

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  • $\begingroup$ You could let $B$ be a single point and $f$ be a constant map from any metric space $A$ to $B$. Then $f$ satisfies the conditions of the contraction mapping theorem but has no fixed points if the point in $B$ is not in $A$. $\endgroup$
    – Zarrax
    Commented Aug 10, 2022 at 4:49
  • $\begingroup$ $f(x)=\frac 1 2 x+2$ from $[0,1] \to [2,3]$. $\endgroup$ Commented Aug 10, 2022 at 4:54
  • $\begingroup$ What do you mean by a counterexample? You gave a definition. Do you mean the contraction mapping theorem, which says a contraction from a set to itself has a fixed point? $\endgroup$
    – Alan
    Commented Aug 10, 2022 at 5:21
  • $\begingroup$ Thank you! So the only motivation to require $f: A\to A$ is to make sure the fixed point $f(x_p)$ is reachable, right? In other words, if $x_p\in A\cap B,$ then we can allow $f: A\to B,~A\neq B$, right? @Zarrax $\endgroup$
    – MathFail
    Commented Aug 10, 2022 at 14:57
  • $\begingroup$ Thank you for this example! @geetha290krm $\endgroup$
    – MathFail
    Commented Aug 10, 2022 at 14:57

1 Answer 1

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The general idea of a contraction mapping is that it shrinks the distance between any two points. It is easy to generalize the idea, let $f: (A,d_a) \rightarrow (B,d_b)$, f is a contraction if $d_b(fx, fy) \leq \alpha d_a(x,y)$ for some $\alpha < 1$.

Since you mention compactness, it is possible that you are discussing this in the context of fixed point theorem. But if $(A,d_a), (B,d_b)$ share no points, then it is impossible that there will be any $x$ such that $f(x) = x$.

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  • $\begingroup$ Thank you for the explanation! So the only motivation to require $f: A\to A$ is to make sure the fixed point $f(x_p)$ is reachable, right? In other words, if $x_p\in A\cap B,$ then we can allow $f: A\to B,~A\neq B$, right? $\endgroup$
    – MathFail
    Commented Aug 10, 2022 at 14:56
  • $\begingroup$ @MathFail we can allow that it be a contraction mapping, but that doesn't mean that $x_p$ is your desired fix point. $\endgroup$
    – emesupap
    Commented Aug 10, 2022 at 17:35
  • $\begingroup$ The fixed point should be unique, right? What do you mean the "desired one"? $\endgroup$
    – MathFail
    Commented Aug 10, 2022 at 19:13
  • $\begingroup$ @MathFail I haven't worked it out, so take it with a grain of salt, but if f: A $\rightarrow$ B, then we can't necessarily iterate it on itself. Typically fix point theorem is proved by iterating f repeatedly. Hence just because A, B share a point does not mean that we can prove fixpoint theorem. $\endgroup$
    – emesupap
    Commented Aug 10, 2022 at 20:08
  • $\begingroup$ OK, thank you!. $\endgroup$
    – MathFail
    Commented Aug 10, 2022 at 21:59

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