Let $T: C[0,1] \rightarrow C[0,1]$ be a linear operator defined as $$ T f(x)= \int_{0}^x f(t)dt, \forall x \in [0,1]$$ Space $C[0,1]$ is equipped with supremum norm. Prove that $T^2$ is a contraction i.e. prove there is a constant $q\in (0,1] $ such that $ d(T^2f, T^2g)\leq q \cdot d(f,g), \forall f,g \in C[0,1] $. $T^2$ is a linear operator, so I consider $T^2 $.
I tried to write $T^2$ as $T(Tf)(x) = \int_{0}^x (T f)(t) dt = \int_{0}^x \int_{0}^t f(s) ds dt $
How should I proceed now? Thanks a lot in advance.
For all $x\in [0,1]$, you have $$\left|T^2 f(x)-T^2 g(x)\right|=\left|\int_0^x\int_0^t (f(s)-g(s)) dsdt\right|\leq \int_0^x\int_0^t \|f-g\|_{C[0,1]} ds dt.$$ Therefore, $$\left|T^2 f(x)-T^2 g(x)\right|\leq \frac{x^2}{2}\|f-g\|_{C[0,1]}\leq \frac{\|f-g\|_{C[0,1]}}{2}.$$ You thus have your contraction with $q=1/2$.