Show that functions are affinely dependent

Question

Let $u(x),v(x)$ be continuous bounded functions on $\mathbb{R}$ such that for any Borel probability measures $\mathbb{P}_{1},\mathbb{P}_2$ on $\mathbb{R}$ $$ \int u(x) \, \mathbb{P}_1(dx) \leqslant \int u(x) \, \mathbb{P}_{2}(dx) \;\;\; \left(\text{def: } \mathbb{P}_1 \preccurlyeq_{u} \mathbb{P}_2 \right) $$ holds if and only if $$ \int v(x) \, \mathbb{P}_{1}(dx) \leqslant \int v(x) \, \mathbb{P}_{2}(dx) \;\;\; (\mathbb{P}_1 \preccurlyeq_{v} \mathbb{P}_2) $$ How to show that in this case $u(x) = av(x) + b$, $a,b = \mathrm{const}$, $a>0$?

If we take $P_1 = \delta_{x_1}$, $P_2 = \delta_{x_2}$ we can obtain that $u(x_1) \leqslant u(x_2)$ holds if and only if $v(x_1) \leqslant v(x_2)$, so $u(x) = f(v(x))$, where $f(\cdot)$ is some increasing function. But what to do next?

Answer 1

Assume first that $u(0)=v(0)=0$ and $u(1)=v(1)=1$. Choose $x$ not in $\{0,1\}$, and $\mathbb P_i=a_i\delta_1+b_i\delta_x+(1-a_i-b_i)\delta_0$ for some nonnegative $(a_i,b_i)$ such that $a_i+b_i\leqslant1$.

Then $\mathbb P_1\preccurlyeq_{u}\mathbb P_2$ if and only if $a_1+b_1u(x)\leqslant a_2+b_2u(x)$, likewise for $\mathbb P_1\preccurlyeq_v\mathbb P_2$. This means that, for every nonnegative $(a_i,b_i)$ such that $a_i+b_i\leqslant1$, $a_1+b_1u(x)\leqslant a_2+b_2u(x)$ if and only if $a_1+b_1v(x)\leqslant a_2+b_2v(x)$.

But, if $u(x)\ne v(x)$, there exists such $(a_i,b_i)$ such that $a_1+b_1u(x)\lt a_2+b_2u(x)$ and $a_1+b_1v(x)\gt a_2+b_2v(x)$. Hence $u(x)=v(x)$ for every $x$ not in $\{0,1\}$, that is, $u=v$.

The general case follows.