The limit of integral 5 Let's consider the following equation:
$u(x) = 1 + \int_0^x k(x-s)u(s)\hbox{d}s$, where $k \ge 0$, is continuous and $\int_0^{\infty}k(s)\hbox{d}s < 1$ for $x\in[0,\infty)$.
I want to show, that 
\begin{align}
 \lim_{x\to\infty}u(x) = \frac{1}{1-\int_0^{\infty}k(s)\hbox{d}s}.
\end{align}
I have shown already, that $u$ exists and is unique using Banach fixed point theorem. I also have shown, that 
\begin{align}
|u(x)| \le \frac{1}{1-\int_0^{\infty}k(s)\hbox{d}s} \ \ \ \hbox{for every } x\in[0,\infty).
\end{align}
My idea is to show, that
\begin{align}
\inf_{x\ge n}u(x) \ge \frac{1}{1-\int_0^{n}k(s)\hbox{d}s}.
\end{align}
Is this a good idea to deal with this problem?
 A: I try a solution.
1) First note that we can use the Banach fixed point theorem with as starting function $u_0(x)=0$, and $\displaystyle u_{n+1}(x)=1+\int_0^x k(x-s)u_n(s)ds$. Then $u_n(x)\geq 0$ for all $n$ and $x$, hence $u(x)\geq 0$, and in addition, as $u_1(x)=1$, we can show by an easy induction that $u_{n+1}(x)\geq u_n(x)$; it follows that $u(x)\geq u_n(x)$ for all $n$ and $x$.
2) Now we prove by induction that for $n\geq 2$, we have
$$u_{n}(x)\geq 1+\sum_{m=1}^{n-1}(\int_0^{x/2^m} k(t)dt)^m$$
For $n=2$, this is true.
Suppose that this is true for $n$.  Then
$$u_{n+1}(x)\geq 1+\int_0^xk(x-s)[1+\sum_{m=1}^{n-1}(\int_0^{s/2^m} k(t)dt)^m]ds$$
The first two terms are $1$ and $\displaystyle \int_0^xk(x-s)ds=\int_0^xk(u)du$. Now
$$\int_0^xk(x-s)(\int_0^{s/2^m} k(t)dt)^mds\geq \int_{x/2}^xk(x-s)(\int_0^{s/2^m} k(t)dt)^m ds\geq  \int_{x/2}^xk(x-s)(\int_0^{x/2^{m+1}} k(t)dt)^m ds$$
The last expression is equal to:
$$(\int_{x/2}^xk(x-s)ds)(\int_0^{x/2^{m+1}} k(t)dt)^m=(\int_{0}^{x/2}k(t)dt)(\int_0^{x/2^{m+1}} k(t)dt)^m$$
and this is $\displaystyle \geq (\int_0^{x/2^{m+1}} k(t)dt)^{m+1}$, and the property is proved.
Now as $u(x)\geq u_n(x)$ we get
$$u(x)\geq 1+\sum_{m=1}^{n-1}(\int_0^{x/2^m} k(t)dt)^{m}$$
Now put $\displaystyle I=\int_0^{+\infty}k(t)dt$. We fix $n$, and let $x\to +\infty$, and we get:
$${\rm liminf}_{x\to+\infty}(u(x))\geq 1+I+\cdots+I^{n-1}$$
Now if $n\to +\infty$, this gives $\displaystyle {\rm liminf}_{x\to+\infty}(u(x))\geq \frac{1}{1-I}$. As you have shown that $\displaystyle u(x)\leq \frac{1}{1-I}$, we are done. 
