How to show that $u-Ku=f$? Let $X \subset C(\mathbb R;\mathbb R)$ be the space of all continuous functions $u: \mathbb R \to \mathbb R$ where 

$$\lim_{x \to \pm \infty} u(x)=0$$

provided with the $\sup$-norm. Let $k \in L^1(\mathbb R)$, $u \in X$ and 

$$(Ku)(x) := \int_\mathbb R k(x-y)u(y)\,dy, \,\,\,x \in \mathbb R.$$

Furthermore $$\int_\mathbb R |k(s)|\, ds &lt;1$$

How can I show that for every $f \in X$ there exists exactly one $u \in X$ such that $$u-Ku=f\,\,?
$$
Fix $f\in X$ and consider the mapping $\varphi:X\rightarrow X$ defined by
$$\varphi(u)=Ku+f$$
We apply the Banach-fixed point theorem on $\varphi$. To do so we need to check three conditions:


$X$ is complete (enough to show that is closed in the Banach space of bounded continuous functions).
The functional $K$ (and hence $\varphi$) maps $X$ to $X$.
We have $||\varphi(u)-\varphi(v)||_{\infty}\leq C||u-v||_{\infty}$ for every $u,v\in X$ where $C&lt;1$.


If these conditions hold, then $\varphi$ has a unique fixed point in $X$ that is, there exists a unique $u\in X$ such that
$$Ku+f=u\Leftrightarrow u-Ku=f$$
$K$ is a bounded linear operator with

$||K||=\int_\mathbb R |k(s)|\, ds &lt;1$.

Hence (Neumann seies !) $I-K$ is invertible.