Quasilinear Utility: Pareto Optimality Implies Total Utility Maximization?
Edit: Edge cases suck; see comments. See also MWG Chapter 10 section C, D.



Suppose $(\vec x^*, \vec m^*)$ solves

$$\max \sum^I_{i=1} m_i + \phi_i(x_i)$$

but is not Pareto optimal.

$$\begin{align}
\implies \exists \ (x_i', m_i') \quad \text{s.t.} \quad &amp; u_i(x_i', m_i') \geq u_i(x_i^*, m_i^*) \quad \forall \ i = 1,\cdots,I \\
&amp; u_i(x_i', m_i') &gt; u_i(x_i^*, m_i^*) \quad \text{for some} \ i
\end{align}$$

$$\implies \sum^I_{i=1} m'_i + \phi_i(x'_i) &gt; \sum^I_{i=1} m^*_i + \phi_i(x^*_i)$$

which is a contradiction. If we have a solution to the utility maximization problem, it must be Pareto optimal.

(Note that this comes form continuous and increasing properties of $\phi(\cdot)$)



Suppose $(\vec x^*, \vec m^*)$ is a feasible Pareto optimal allocation, but does not solve

$$\max \sum^I_{i=1} m_i + \phi_i(x_i)$$

Because we treat $m_i$ as numeraire and $\phi_i(\cdot)$ is strictly increasing, we know $u_i(\cdot)$ is locally non-satiated. The Pareto allocation should be just feasible.

$$\exists \ (x_i', m_i') \quad \text{s.t.} \quad \sum^I_{i=1} m'_i + \phi_i(x'_i) &gt; \sum^I_{i=1} m^*_i + \phi_i(x^*_i)\\
\implies \boxed{ \sum^I_{i=1} \phi_i(x'_i) &gt; \sum^I_{i=1} \phi_i(x^*_i)}$$

If this is true because this alternative allocation simply gives an individual more of $x$, for all else equal, then the alternative allocation is infeasible. So we'd have a contradiction.

If this is true because in the alternative allocation, someone else is allocated more $x$ and just one other person is allocated less, then the original allocation would not be Pareto optimal. Suppose it was. If you took the original allocation and shifted $x$ in the way of the new allocation, then you would need a corresponding trade in the numeraire good, $m$, to keep whoever is losing $x$ at least at the same utility level. But trades in just the numeraire good can never change summed aggregate utility. From the original allocation, if you can trade $m$ for $x$ and make someone better off without hurting anyone, you weren't at a Pareto optimum, and if you can't trade $m$ for $x$ to make someone better off, you can't increase summed aggregate utility, which means the original allocation was a solution to the maximization problem.

This logic applies no matter how you rearrange $x$ between multiple people.

$\square$