proving a mapping is contraction mapping

Question

I want to prove $T : C[0,3] \rightarrow C[0,3] $, defined by $$(T(f))(x) = 1/5 \int_0^x t(f(t)) \,dt $$ is a contraction mapping on $(C[0,3],\|\cdot \|_{sup} )$.

Here's what I've done: for any $f,g \in C[0,3] $ I have, $$\| T(f) -T(g) \|_\sup = \sup_{x\in [0,3]} \left| 1/5 \int_0^x t(f(t) - g(t))\, dt \right| = 1/5 \sup_{x\in [0,3]} \left| \int_0^x t(f(t) - g(t)) \,dt\right|$$

I'm stuck here. I gotta end up to something like $c\|f-g\|_\sup$ where $c$ is the contraction mapping constant. Any help?

Answer 1

Notice that $T$ is a linear map on $C[0,3]$.

\begin{align} \|Tf\|_\infty &= \left\|x\mapsto \frac15\int_0^x tf(t)\,dt\right\|_\infty\\ &= \sup_{x\in[0,3]}\left|\frac15\int_0^x tf(t)\,dt\right|\\ &\le \frac15 \sup_{x\in[0,3]}\int_0^x t\underbrace{|f(t)|}_{\le \|f\|_\infty}\,dt\\ &\le \frac{\|f\|_\infty}5 \sup_{x\in[0,3]}\int_0^x t\,dt\\ &= \frac{\|f\|_\infty}5\int_0^3 t\,dt\\ &= \frac{9}{10}\|f\|_\infty \end{align}

Thus, $T$ is bounded and $\|T\| \le \frac9{10}$.

Now in particular, $T$ is also a contraction:

$$\|Tf-Tg\|_\infty = \|T(f-g)\|_\infty \le \|T\|\|f-g\|_\infty \le \frac9{10}\|f-g\|_\infty$$

Answer 2

From where you are, try bringing the absolute value inside the integral (as said in the comments to your post). Then consider when the resulting integral is largest (i.e. for what $x \in [0, 3]$ is $\int_0^x (\text{something positive})$ largest)?