I need to solve the equation $ x + f(x) = d$, where $f: \mathbb{R}\to \mathbb{R}_+$ is decreasing and convex, $d$ is a positive constant. Suppose the equation has a solution $x^* > 0$. Does the following method converge? Given $x^{(1)}$ close to $x^*$ and $\epsilon > 0$
while $\Delta > \epsilon$
$ \quad x^{(i)} = d-f(x^{(i-1)})$
$ \quad\Delta = d-f(x^{(i)})-x^{(i)}$
I have no idea whether or not it is true. Can anyone give me some hints? Thank you in advance!
If $d-f(x^{(1)})>x^{(1)},$ then $x^{(i)}$ will be increasing, otherwise decreasing (proof by induction, as $d-f(x)$ is increasing). Due to convexity, there can be no more than two solutions. $d-f(x)\le d$ is bounded, so we must have $d-f(x)<x$ for big enough $x$, and thus for all $x>x^*,$ if $x^*$ is the biggest solution. So if we start above $x^*,$ there is monotone convergence towards it. Immediately below $x^*,$ there must be $d-f(x)>x,$ so we have monotone convergence, again. If we start below a second, smaller solution, we have $d-f(x)<x,$ again, so we are moving downwards, away from that solution.