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?
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.
By simple differentiation we see the function is increasing on its domain.
$\displaystyle T(x) = x + \frac{1}{x} \Rightarrow \quad T'(x) = 1 - \frac{1}{{x^2 }} \geqslant 0$.
So, $\displaystyle x > y \Rightarrow \quad T(x) - T(y) > 0\,\& \,{ \color{blue}\frac{1}{x} - \frac{1}
{y} < 0}$.
$\displaystyle \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 $\displaystyle y>x$.