Show that $T$ is or isn't a strict contraction but have fixed points

Question

Define $T: C[0,1] \to C[0,1]$ by

$$(Tf)(x) = \int^x_0 f(t)dt.$$

Show that $T$ is not a strict contraction while $T^2$ is. What is the fixed point of $T$?

Answer 1

We need to first specify a metric on $C[0,1]$, but I think that we want the metric $d(f,g)=max_{x \in [0,1]} \{ |f(x)-g(x)| \}$.

Then to show that $T$ is a weak contraction but not a strict contraction:

$d(Tf,Tg)=max_{x \in [0,1]}\{ |\int_{0}^{x}f(t)dt-\int_{0}^{x}g(t)dt|\} = max_{x \in [0,1]} \{ | \int_{0}^{x}f(t)-g(t)dt | \} $

Using the inequality for Riemann integrable functions $|\int_{a}^{b}f(x)dx| \le \int_{a}^{b}|f(x)|dx$ we get:

$max_{x \in [0,1]} \{ | \int_{0}^{x}f(t)-g(t)dt | \} \le max_{x\in[0,1]} \{ \int_{0}^{x}|f(t)-g(t)|dt\} = \int_{0}^{1}|f(t)-g(t)|dt$

where the final equality is since we're now integrating a non-negative function. Since $d(f,g)$ is the maximal size of the difference between $f$ and $g$ we have:

$\int_{0}^{1}|f(t)-g(t)|dt \le \int_{0}^{1}d(f,g)dt =d(f,g)$

So that $d(T(f),T(g) \le d(f,g)$ and we have at least a weak contraction. Now to show that there is no constant $ c \in [0,1)$ such that $d(T(f),T(g)) \le c \cdot d(f,g)$ for all functions $f,g \in C[0,1]$ we need to come up with a pair of functions $f,g$ in our space where $c=1$. For this take $f(x)=1$ and $g(x)=0$ so that $d(T(f),T(g))=d(f,g)=1$.

To show that $T \circ T=T^2$ is a strict contraction use the following:

$|(T^2f)(x)-(T^2g)(x)|= |\int_0^x \int_0^u f(t)dtdu-\int_0^x \int_0^ug(t)dtdu| \le \int_0^x \int_0^u|f(t)-g(t)|dtdu $

and then from here:

$\int_0^x \int_0^u|f(t)-g(t)|dtdu \le \int_0^xd(f,g) \cdot udu=d(f,g) \frac{x^2}{2} \le \frac{1}{2}d(f,g)$

so we have a strict contraction with constant $c=\frac{1}{2}$.