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I'm trying to use the Banach fixed point theorem to show that an intergral equation has a unique solution, but can't seem to make my answer work any help would be appreciated.

Let $f:[a,b] \rightarrow \mathbb{R}$, $g:[a,b] \rightarrow \mathbb{R}$, $k:[a,b] \times [a,b]$ be continuous and $\sup_{x \in[a,b]} \int_{a}^{b} |k(x,y)|$ = $p<1$. Show there exists a unique continuous $f$ that solves:

$f(x)=g(x)+\int_{a}^{b}k(x,y)f(y)$

My Attempt

Define $T(f(x))=g(x)+\int_{a}^{b}k(x,y)f(y)$.

T is a map that takes continuous maps of the form of f to functions of the form of f, with the sup norm this is a banach space so if i prove a contraction I'm done.

$\|T(f(x))-T(s(x))\| = \|\int_{a}^{b}k(x,y)(f(y)-s(y))\|\leq p\bigg|\int_{a}^{b}(f(y)-s(y))\bigg|$

By the MVT $\exists c \in [a,b]$ s.t.

$p\bigg|\int_{a}^{b}(f(y)-s(y))\bigg|=p\bigg|(a-b)(f(c)-s(c))\bigg|$

I want to conclude that this is $\leq p \|f-s\|$ or something similar but I don't have that $(a-b)<1$ I thought maybe I could just say if $f$ is in the space so is $(a-b)f$ so I'd be done but I'm not convinced thats correct.

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$|\int_a^bk(x,y)(f(y)-s(y))dy|\leq\int_a^b|k(x,y)|sup_{y\in[a,b]}|f(y)-s(y)|dy=\|f-s\|\int_a^b|k(x,y)|dy$.

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