The above is the question, and here is a solution:
In the solution above, I do not quite understand:
Why the assumptions that $C[a,b]$ is complete with the sup metric guarentee that $Tf\in C[a,b]$?
Why do we say that $|(T f)(x) − (T h)(x)| ≤ \int^b_a |k(x, y)||f(y) − h(y)| dy$? I think it is $|(T f)(x) − (T h)(x)| = |\int^b_a k(x, y)[f(y) − h(y)] dy|= \int^b_a |k(x, y)||f(y) − h(y)| dy$. Is that correct?
Could someone please explain? Thanks!
$T$ is continuous since it is the composition of continuous functions: pointwise multiplication by $k(x,y)$ (which is continuous), integration, and addition.
Hence, $T \circ f:C[a,b] \to C[a,b]$ will be continuous as well.
The completeness is used to conclude that there is a fixed point, by the Banach fixed point theorem.
In the last inequality, this is just an application of the "triangle inequality," so that we bring the absolute value inside of the integral:
$$\left| \int_a^b g(x) dx\right| \leq \int_{a}^{b} |g(x)| dx.$$
your last equality should be an inequality.