I am working the topic of Lipschitz condition. Let f(x,y) be a continuous function for $0 \leq x \leq T = const < \infty, y \in R^{1}$, satisfying the Lipschitz condition with respect to y with a constant $K > 0$ such that: $$|f(x,y_{1})-f(x,y_{2})| \leq K.|y_{1}-y_{2}|,\forall x \in [0,T]$$.
Now, I define X to be a set of all continuous functions on $[0,T]$. Define a mapping T from X to X as follows: for any continuous function y = y(x) on $[0,T]$, T(y) is a continuous function on $[0,T]$ with values:$$(T(y))(x):= y_{0} + \int_{0}^{x} f(t,y(t)) dt, \text{where } y_{0} = const \in R^{1}$$
Now, I define the distance between $y_{1}$ and $y_{2}$ in X as:$$d(y_{1},y_{2}): = \int_{0}^{T} e^{-2Kx}|y_{1}(x)-y_{2}(x)|dx$$
I would like to show that: $$d(T(y_{1}),T(y_{2})) \leq \frac12. d(y_{1},y_{2}) $$
$\underline{\text{My approach:}}$
Straight up from my definition of distance, I apply for $d(T(y_{1}),T(y_{2}))$. Assume I have done that correctly, eventually, I will have $d(T(y_{1}),T(y_{2})) \leq K. d(y_{1},y_{2})$.
My question is: can I choose K to be $\cfrac12$?
Thanks in advance!
No, you can not. Somewhere in your calculation you will get to the integration $$ \int_0^TKe^{-2Kt}dt=\frac12(1-e^{-2KT})<\frac12, $$ this is how you get your contraction factor.