Consider the following functional equation: $$V(x)=\max_{y\in [0,f(x)]}[u(f(x)-y)+\beta V(y)]$$ where $u$ is continuous, strictly increasing, and strictly concave; the function $f$ is continuous and strictly increasing, and $\beta\in (0,1)$. Let $g(x)$ be the decision rule associated with this functional equation: $$V(x)=u[f(x)-g(x)]+\beta V[g(x)]$$ How can I show that $g$ is non-decreasing (if $x'>x$, then $g(x')\geq g(x)$)?
Assume that $f$ and $u$ are not necessarily differentiable.
Suppose $x'>x$. We'll show that $g(x')\geq g(x)$. Suppose that is not the case and we have $x'>x$ but $g(x')<g(x)$. Since $0\leq g(x') < g(x) \leq f(x)< f(x')$, $g(x')$ is feasible in case of $x$, and $g(x)$ is feasible in case of $x'$, consequently both of the following inequalities are true: \begin{eqnarray*}V(x')=u(f(x')-g(x'))+\beta V(g(x')) &\geq & u(f(x')-g(x))+\beta V(g(x)) \\ V(x)=u(f(x)-g(x))+\beta V(g(x))&\geq & u(f(x)-g(x'))+\beta V(g(x'))\end{eqnarray*}
Adding them we get the following inequality:
$u(f(x')-g(x'))+\beta V(g(x'))+u(f(x)-g(x))+\beta V(g(x))\geq u(f(x')-g(x))+\beta V(g(x))+u(f(x)-g(x'))+\beta V(g(x'))$
which yields $u(f(x')-g(x'))-u(f(x)-g(x'))\geq u(f(x')-g(x))-u(f(x)-g(x))$
which, by strict concavity of $u$, implies that $g(x')\geq g(x)$ contradicting that $g(x') < g(x)$.
I'm going to assume that everything is smooth and that the optimal solution $g(x)$ is interior (in $]0, f(x)[$)
First you can show that the function $V$ is concave as the Bellman operator maps concave functions to concave functions.
Now assume that the solution $g(x)$ is interior. Then the first order condition (with respect to $y$) gives: $$ -u'(f(x) - g(x)) + \beta V'(g(x)) = 0. $$ Taking the derivative with respect to $x$ gives: $$ -u''(f(x) - g(x))f'(x) + u''(f(x) - g(x))g'(x) + \beta V''(g(x)) g'(x) = 0, $$ So: $$ g'(x) = \frac{u''(f(x) - g(x))f'(x)}{u''(f(x) - g(x)) + \beta V''(g(x))}. $$ Both numerator and denominator on the right side are negative, so $g'(x) > 0$.
