Show that mapping is a contraction? Show that the mapping $f:\Bbb R \to \Bbb R $, $f(x)=1-xe^x$ is a contraction. I tried everything i could think of but i cant get it to work. Witch is not much since i couldn't really find any inequality that would be helpful. Thanks. 
 A: [solution to: Show that the equation $x+xe^x-1=0$ has a unique solution in $\mathbb{R}$ using Banach's fixed point theorem]
Rearranging these equations for iterative approximations of the form $x_{n+1}=f(x_n)$, we generally want to aim for $f$ which decay or grow slowly, so as to make them contractive. Given that the largest term in our equation is an $e^x$, we'd like to divide by this if possible. According to this heuristic, we might rearrange our equation as:
$x(1+e^x)=1\implies x= \frac{1}{1+e^x}$
So, we could choose to iterate using $f(x)=\frac{1}{1+e^x}$, suspecting that it might be contractive. We differentiate using the chain rule:
$f'(x)=\frac{-e^x}{(1+e^x)^2}\implies\vert f'(x)\vert=\frac{e^x}{e^{2x}+2e^x+1}=\frac{1}{e^x+2+e^{-x}}\leq1/2$
(with a little extra work, we can tighten this to $\vert f'(x)\vert\leq 1/4$)
In any case, the bound on the derivative gives us a Lipschitz condition on $f$, as the mean value theorem tells us:
$$\vert \frac{f(x) - f(y)}{x-y}\vert=\vert f'(\xi)\vert\leq1/2\implies\vert f(x)-f(y)\vert\leq(1/2)\vert x-y\vert$$
Thus, $f$ is contractive on $\mathbb{R}$,  which is complete. We can then apply the Banach fixed point theorem to deduce that $f$ has a unique fixed point in $\mathbb{R}$, and thus that the equation $x+xe^x-1=0$ has a unique solution in $\mathbb{R}$.
A: A function is a contraction if the points in the image are closer together than in the source. In symbols: $|\mathrm f(y) - \mathrm f(x)| < |y-x|$.
Let's choose $y$ to be a tiny bit bigger than $x$, i.e. $y=x+\varepsilon$. The definition becomes
$$|\mathrm{f}(x+\varepsilon)-\mathrm{f}(x)| < |(x+\varepsilon)-x|$$
$$|\mathrm{f}(x+\varepsilon)-\mathrm{f}(x)| < |\varepsilon|$$
Since $\varepsilon \neq 0$, we can divide by $|\varepsilon|$ to give
$$\frac{|\mathrm{f}(x+\varepsilon)-\mathrm{f}(x)|}{|\varepsilon|} < 1$$
$$\left|\frac{\mathrm{f}(x+\varepsilon)-\mathrm{f}(x)}{\varepsilon}\right| < 1$$
If we let $y$ get closer and closer to $x$ we have $\varepsilon \to 0$, and so
$$\lim_{\varepsilon \to 0}\left|\frac{\mathrm{f}(x+\varepsilon)-\mathrm{f}(x)}{\varepsilon}\right| < 1$$
The thing on the right looks just like the definition of a derivative. In fact, it becomes
$$\color{red}{|\mathrm{f}'(x)| < 1}$$
All you need to do is check that the absolute value of the derivative is less than 1 for the set of $x$-values you are given.
EDIT
The OP's definition of a contraction seems to be that, there exists $0 \le q < 1$, for which $\mathrm d(\mathrm f(x),\mathrm f(y)) \le q \cdot \mathrm d(x,y)$.
With the standard Euclidean metric $\mathrm d(x,y) = |x-y|$. If we follow the same argument through, we get: There needs to be a $0 \le q < 1$ for which $\color{red}{|\mathrm f'(x)| < q}$. 
Of course, if $|\mathrm{f}'(x)| < 1$ then there exists a $0 \le q < 1$ such that $|\mathrm f'(x)| < q$. In this case $q=1$. 
