let $h: \mathbb{R}^{*}_{+}\to \mathbb{R}$ decreasing such that $h(u)=0,\quad u> 0$ and $f(x)=x \ h(x)$ is increasing Show that: $$h=0 \text{ over } [u,+\infty)$$
let $x\in \mathbb{R}^{*}_{+}$ we want to show that $$h(x)=0 \text{ over } [u,+\infty) $$ note that $h: \mathbb{R}^{*}_{+}\to \mathbb{R}$ decreasing means $$\forall x,u \in \mathbb{R}^{*}_{+},\text{s.t }\ x\leq u\quad h(x)\leq h(u)$$ or $h(u)=0$
Let $x\in [u,+\infty)$. Then $x\ge u$, so since $h$ is decreasing and $f$ is increasing, $h(x) \le h(u)$ and $f(x) \ge f(u)$. Since $h(u) = 0$, the inequality $h(x) \le h(u)$ gives $h(x) \le 0$. Since $f(u) = uh(u) = 0$, the inequality $f(x) \ge f(u)$ gives $xh(x) \ge 0$, which implies $h(x) \ge 0$, since $x > 0$. Since $h(x) \le 0$ and $h(x) \ge 0$, we conclude $h(x) = 0$.