# Show: $\exists n\in\mathbb{N}$ so that $T^n\colon C([a,b])\to C([a,b])$ is $q$-contractive

Consider $-\infty<a<b< +\infty, F\in C([a,b]^2\times\mathbb{R}), f\in C([a,b])$ with $$\exists L\geq 0~\forall x,y\in[a,b]~\forall u,v\in\mathbb{R}:~~\lvert F(x,y,u)-F(x,y,v)\rvert\leq L~\lvert u-v\rvert.$$ Furthermore consider the non-linear operator $T\colon C([a,b])\to C([a,b])$ given by $$T(u)(x):=\int_a^xF(x,y,u(y))\, dy+f(x)~~\text{for}~~x\in [a,b].$$ Show that it exists a $n\in\mathbb{N}$, so that $T^n\colon C([a,b])\to C([a,b])$ is $q$-contractive.

Hello!

At first I think it makes sense to add our definition of $q$-contractivity:

Let $(X,d)$ be a metric space. A function $T\colon X\to X$ is called $q$-contractive (on $X$), if it exists a $0\leq q<1$ so that $$\forall x,y\in X:~~d(T(x),T(y))\leq q~d(x,y).$$

So if I got it right, the task is to find a $n\in\mathbb{N}$ so that there exists a $0\leq q<1$ with $$\forall u,v\in C([a,b]):~~~\lVert T^nu-T^nv\rVert_{\infty}\leq \lVert u-v\rVert_{\infty}.$$

What I already got (but I do not know if it is helpful) is that for $x\in [a,b]$ it is $$\lvert T(u)(x)-T(v)(x)\rvert\leq L\lVert u-v\rVert_{\infty}\cdot (b-a),$$ so $$\lVert Tu-Tv\rVert_{\infty}\leq L\cdot (b-a)\lVert u-v\rVert_{\infty}.$$

But as I said I do not really now if this helps.

Can you help me?

First, show an estimate like $$|(T\, u - T \, v)(t)| \le \|u - v\|_\infty \, L \, (t-a).$$ Now, you can iterate this, to get a similar estimate for $T^n$: $$|(T^n\, u - T^n \, v)(t)| \le \|u - v\|_\infty \, L \, \frac{(t-a)^n}{n!}.$$ And this gives you contractivity, for $n$ large enough.
• When estimating the integrand, don't use the estimate $|u(t)-v(t)| \le \|u - v\|_\infty$, but the previous estimate for $n-1$. – gerw Jan 14 '14 at 17:20