numerical-methods, Fixed point theorem. I am just looking to apply a result, so can someone confirm the following for me. 
Let say I have a equation of the form below:
$V_1(x) = a + bV_0(x)$,
where in theory $V_1(x) = V_0(x)$.
I have an algorithm as follow:
Start with initial guess $V_0(x)=0$ and proceed to obtain $V_1(x)=a$.
Using '$a$' as the new guess, and repeat this until say the difference is less than $0.1\%$.
What sort of regularity conditions or things to look out for to ensure I obtained the true solution? 
Thank you community once again.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
You have a sort of recurrence likes $\ds{\xi_{n + 1} = a + b\xi_{n}}$ which can be rewritten, for $\ds{b \not= 1}$, as
\begin{align}
\xi_{n + 1} - {a \over 1 - b} & =
b\pars{\xi_{n} - {a \over 1 - b}} =
b^{2}\pars{\xi_{n - 1} - {a \over 1 - b}} = \cdots =
b^{n}\pars{\xi_{1} - {a \over 1 - b}}
\end{align}

$$
\mbox{If}\ \verts{b} < 1\,,\ \mbox{it converges to}\
\color{#f00}{{a \over 1 - b}}
$$
