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Consider this function $f(x) = x + \frac{1}{x}$ mapping from $[2, \infty)$ to $\mathbb{R}$. Then $[2, \infty)$ is a complete metric space. We also have $d(f(x),f(y)) < d(x,y)$. However, there is no fixed point for this function.

Since $d(f(x),f(y)) < d(x,y)$, I thought we could find $\alpha$ such that $0 < \alpha < 1$ by doing $\frac{d(f(x),f(y))}{ d(x,y) }$. The function is also mapping from a complete metric space. Why doesn't the contracting mapping theorem apply here?

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The contracting mapping theorem applies if there is a constant $q \in [0,1)$ with $d(f(x),f(y)) \le q d(x,y)$. But no such $q$ exists !

Compute $d(f(x),f(y))/d(x,y)$ for $x \ne y$ and you will get $d(f(x),f(y))/d(x,y) \to 1$ for $x,y \to \infty$.

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The contraction mapping theorem does not apply because its' assumptions are not met. The theorem requires the existence of a constant $C<1$ such that $$d(f(x), f(y) )<Cd(x,y)$$ Your example shows that $C=1$ won't do.

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  • $\begingroup$ Can't I get a $C$ by doing $d(f(x),f(y))/d(x,y)$ $\endgroup$ – user1691278 Dec 8 '17 at 6:04
  • $\begingroup$ @user1691278 Go on, find it. Can you? $\endgroup$ – Ivan Neretin Dec 8 '17 at 6:05
  • $\begingroup$ @IvanNeretin You're saying that approach doesn't yield a constant but a variable that depends on x and y? $\endgroup$ – user1691278 Dec 8 '17 at 6:08
  • $\begingroup$ @user1691278 Sure you can find $C$. You'll get $C=1$. And that is not sufficient as I tried to point out. $\endgroup$ – Thomas Dec 8 '17 at 6:09
  • $\begingroup$ @Thomas You're saying that C = 1 is the sup of $d(f(x),d(y))/d(x,y)$? $\endgroup$ – user1691278 Dec 8 '17 at 6:13

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