Hello fellow mathematicians:
I've got following excercise, but honestly not a single clue as to how to approach it. We got the Field of bounded functions on $\mathbb{R}$ and a function K, with the following definition: $$K(x)(t)=\int_{0}^{1}\frac{1}{2}e^{t-s}x(s)ds$$
Show that there is one unique fixpoint on the interval $[-1,1]$
My guess is obviously trying to show it's contracting either by looking at its derivative and/or directly. Sadly I don't got a single clue, especially since there is an $x(s)$ in the Integral, which really confuses me.
It might have to do with the following statement:
Let $x'=f(x,t),x(t_0)=x_0$ be an initial value problem. A continuous function x on the intervall $[0,\tau]$ is a solution if and only if following expression holds: $$x(t)=x_0+\int_0^tf(s,x(s))ds ~ \forall t\in [0,\tau]$$
Would be really nice, if one of you had an idea willing to be shared with me.