In $C^0([a, b])=\{f:[a, b]\longrightarrow \mathbb R: f\ \textrm{continuous}\}$ consider the metric $$d_\infty(f, g)=\sup_{x\in [a, b]}|f(x)-g(x)|.$$Let $T:C^0([a, b])\longrightarrow C^0([a, b])$ given by $$(Tf)(x)=g(x)+\int_{a}^b k(x, y)f(y)\ dy,$$ where $k:[a, b]\times [a, b]\longrightarrow \mathbb R$ and $g:[a, b]\longrightarrow \mathbb R$ are continuous functions. Suppose $$\sup_{x\in [a, b]}\int_{a}^b |k(x, y)|\ dy<1.$$ Is it true that $T$ is a contraction?
Recall $T:X\longrightarrow X$ is a contraction in the metric space $(X, d)$ if there is $c<1$ such that $d(Tx, Ty)\leq c d(x, y)$.
$$d_\infty(Tf, Th)= \sup_x \left| \int_a^b k(x,y) (f(y)-h(y)) dy \right| \\ \le \sup_x \left( \sup_y |f(y)-h(y)| \int_a^b |k(x,y)| dy \right) = \sup_x \int_a^b |k(x,y)| dy \times d_\infty(f, h)$$