I would like to show that $f(x)(t) = 1 - \int_0^t sx(s) ds$ where $t \in [0,\tfrac{1}{2}]$ and $X = \{x(t) \in C[0,\tfrac{1}{2}] : |x(t) - 1| \leq 1 \}$ defines a contraction. My attempt is as follows.
I've noted that $$|f(x_1)(t) - 1| = |\int_0^t sx_1(s) ds| \leq |\int_0^t 2s ds| = t^2 \leq \dfrac{1}{4}$$
but am I having some trouble seeing as to how this may apply in the proof of $f$ being a contraction mapping which follows.
$$\sup_{t\in [0,\tfrac{1}{2}]} |f(x_1)(t) - f(x_2)(t)| = |\int_0^t sx_1(s) - sx_2(s) ds| \leq \int_0^t s\sup_{s\in [0,\tfrac{1}{2}]}|x_1(s)-x_2(s)|ds = \dfrac{t^2}{2}\sup_{s\in [0,\tfrac{1}{2}]}|x_1(s)-x_2(s)|$$
$$\implies \sup_{t\in [0,\tfrac{1}{2}]} |f(x_1)(t) - f(x_2)(t)| \leq 1/8 \sup_{s\in [0,\tfrac{1}{2}]}|x_1(s)-x_2(s)|$$
Which means that $d_\infty(f(x_1)(t),f(x_2)(t)) \leq \dfrac{1}{8}d_\infty(x_1(s),x_2(s))$. Is this correct?