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One of the conditions of contraction principle is $d(fx,fy)<cd(x,y)$ for all $x,y$ where $c<1$. Now I am finding an example that when the condition is replaced by $d(fx,fy)<d(x,y)$, the theorem fails, i.e. we can not find a fixed point.

I am sure this is true.But I can not find an example.

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@derpy The typography was corrupted because of bad use of < characters outside math mode. See edited question. – Jeppe Stig Nielsen Apr 23 '14 at 14:16
up vote 2 down vote accepted

The hypotheses to Banach-Caccioppoli fixed point theorem are the following: suppose $ (X,d) $ is a complete metric space (that is, all $ d $-Cauchy sequences converge in $ X $), $ X \ne \varnothing $, and let a function $ f : X \to X $ be provided such that $$ d(f(x),f(y))\le Cd(x,y) \qquad (*)$$ for all $ x,y \in X $, where $ C \in [0,1) $ is a constant which doesn't depend on $ (x,y) $ (such $ f $ is called a contraction of $ X $); then $ f $ admits a fixed point, that is, there is an $ x_0 \in X $ with $ f(x_0) = x_0 $.

The hypothesis $ (*) $ may not in general be replaced by the weaker hypothesis $$ d(f(x),f(y)) < d(x,y) \qquad (**) $$ for all $ x,y \in X $, as shown in the following example.

Let $ X = [1,+\infty) $ and $ d(x,y) = \left|x-y\right| $; notice that $ (X,d) $ is a closed subspace of the Banach space $ (\mathbb R,d_\mathbb R) $, hence complete. Let $ f : X \to X $ be given by $$ f(x) = x + \frac{1}{x}.$$

Notice that $$ \left|f(x) - f(y)\right| = \left|x - y - \frac{x - y}{xy}\right| = \left|(x - y)\left(1 - \frac{1}{xy}\right)\right| < \left|x - y\right| $$ whenever $ x,y \in X $. Still, you can verify that $ f $ doesn't fix any point in $ X $. The failure can be appointed to $ f $ fixing the "point at infinity"; actually, hypothesis $ (**) $ is sufficient if $ X $ happens to be compact.

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