Suppose $f: [1, \infty) \to [1, \infty]$ defined by $f(x) = x + \frac{1}{x}$ for all $x \geq 1$. I want to prove that:

\begin{equation} |f(x)-f(y)| < |x-y| \end{equation} except when $x=y$, but $f$ does not have a fixed point.

By the Banach fixed point theorem we know that if a function $f: X \to X$ is a contraction of a complete metric space, then $f$ has a unique fixed point $p$ and the sequence of $(f, f \circ f, f\circ f\circ f, ...)$ that is the sequence of $f$ composed with itself $n$ times at index $n$ converges to p for all $x$.

But $[1, \infty]$ is not a complete metric space. So it seems like a good idea to proceed via contradiction? Where can I go from here

All help is greatly appreciated

  • 1
    $\begingroup$ $[1,\infty)$ is actually a complete space because it's closed in $\mathbb R$. I think your problem lies with your definition of contraction. $\endgroup$ – kahen Apr 19 '13 at 1:37

Use $$ f(x)-f(y) = x - y + \frac{1}{x} - \frac{1}{y} = (x-y) \left(1 - \frac{1}{x y}\right) $$ Thus, for $x\not= y$ $$ \left| f(x)-f(y) \right| = \left|x-y\right| \left(1 - \frac{1}{x y}\right) < \left|x-y\right| $$ since $1 - \frac{1}{x y} < 1$ for all $x>1$ and $y>1$

  • $\begingroup$ Then we know $f(x)$ can never have a fixed point because $x+\dfrac{1}{x} \neq x$. Correct? $\endgroup$ – CodeKingPlusPlus Apr 19 '13 at 1:50

Look at user40314's answer for the inequality. Regarding your second question, the space $[1, \infty)$ is complete although your map isn't a contraction map because for that, you would need there to be a constant $r$ outside the right hand side of the inequality for which $|r| < 1$. In your case, there is a factor of $1 - \frac{1}{xy}$ which can be made arbitrarily close to $1$ by picking large enough $x$ and $y$.


I want to extend kahen's comment.

It is in fact related to the contraction definition: a map such that $$d(f(x), f(y)) < d(x,y)$$ is called contractive, whereas one such that there exists some $0<k<1$ such that $$d(f(x), f(y)) \leq kd(x,y)$$ $\forall x, y$ is called a contraction (which is required in Banach's theorem). [Note that contraction $\implies$ contractive but that the opposite is wrong]

The given function (where $d$ is given by $d(x, y) = |x-y|$) is a good counterexample showing that one can not relax the contraction condition in Banach's theorem to contractive. [Note that there is, however, a theorem by M. EDELSTEIN giving a sufficent condition on a contractive map to admit a fixed point]

  • $\begingroup$ It is better that you add the reference concerning the two different definitions. Might be also the reference of Edelstein's theorem $\endgroup$ – user99914 Nov 21 '17 at 20:14

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