Unique Solution for IVP of Continuous Functions with Global Lipschitz Condition

[Eulogy]

Homework Statement

For the space of continuous functions C[0,T] suppose we have the metric ρ(x,y) =sup [itex]_{t\in [0,T]}[/itex]e[itex]^{-Lt}[/itex][itex]\left|x(t)-y(t)\right|[/itex] for T>0, L≥0.

Consider the IVP problem given by

x'(t) = f(t,x(t)) for t >0,
x(0) = x[itex]_{0}[/itex]

Where f: ℝ×ℝ→ℝ is continuous and globally Lipschitz continuous with
respect to x.

Find an integral operator such that the operator is a contraction on (C[0,T],ρ) and hence deduce the IVP has a unique solution on C[itex]^{1}[/itex][0,T]

The Attempt at a Solution

I was able to show that the metric space (C[0,T],ρ) is complete, but I'm having problems finding an integral operator that is a contraction on the space. I've tried the operator
(Tx)(t) = [itex]x_{0}[/itex] + [itex]\int^{t}_{0}f(s,(x(s))ds [/itex]but I was not able to get a contraction. Any help would be much appreciated!

 

[Ratpigeon]

try finding T|x-y| and then use the lipschitz continuity?

 

[Eulogy]

I have tried this but was unable to show it was a contraction. I'm not to sure if I have the wrong integral operator for this particular question or if I'm trying to show a contraction in the wrong way.

 

[sunjin09]

I'm not sure if your operator is contraction, but it does not seem to be a fixed point iteration operator at all ...