1. ## Calculus Contraction Mapping

T(x) = X+ 1/x X=[1,INFINITY]

|T(X) - T(Y)| < |X-Y|

|(x+ 1/x) - (y+ 1/y)| < |x-y|

Can i solve the left side further?

2. Originally Posted by littlefire
T(x) = X+ 1/x X=[1,INFINITY]

|T(X) - T(Y)| < |X-Y|

|(x+ 1/x) - (y+ 1/y)| < |x-y|

Can i solve the left side further?
I am not sure what you want.
$\left| x + \frac{1}{x} - y - \frac{1}{y} \right| = \left| x - y\right| + \left| \frac{1}{x} - \frac{1}{y} \right|$
This is bounded by, $|x - y| + \frac{|x-y}{|xy|} \leq |x-y|+|x-y| = 2|x-y|$.

3. Thanks a million. Still now too sure on your answer after the bounded part.

This is my full question that I have to answer -

Let X=[1,infinity) with the metric d(x,y) = |x-y|. Define T:X->X by T(x) = x+1/x. Prove T satisfies

d(T(x),T(y)) < d(x,y) for all x,y E X

but T has no fixed points in X. Does this contradict Banach's fixed point theorem.

4. Originally Posted by littlefire
Let X=[1,infinity) with the metric d(x,y) = |x-y|. Define T:X->X by T(x) = x+1/x. Prove T satisfies
d(T(x),T(y)) < d(x,y) for all x,y E X
By simple differentiation we see the function is increasing on its domain.
$T(x) = x + \frac{1}{x} \Rightarrow \quad T'(x) = 1 - \frac{1}{{x^2 }} \geqslant 0$.
So, $x > y \Rightarrow \quad T(x) - T(y) > 0\,\& \,{ \color{blue}\frac{1}{x} - \frac{1}
{y} < 0}$
.
$\begin{array}{rcl} {\left| {T(x) - T(y)} \right|} & = & {T(x) - T(y)} \\
{} & = & {x - y +{\color{blue} \frac{1}{x} - \frac{1}{y}}} \\ {} & < & {x - y} \\
{} & = & {\left| {x - y} \right|} \\ \end{array}$
.

By symmetry this holds if $y>x$.