Quasilinear Utility: Pareto Optimality Implies Total Utility Maximization? I read that if we have quasilinear utility for all consumers, then any pareto optimal allocation maximizes the sum of utility levels of all consumers. That is:

$\textbf{What we know:}$
$$1)\quad u^i(m^i,x^i)=m^i+\phi^i(x^i)\; \quad \forall i=1,...,I$$
$$2)\quad\phi^i(\;)\;\text{is continous and strictly increasing (but not necessarily differentiable)}$$
$$3)\quad \text{An allocation,}\,x\, \text{satisfies}\;\neg\,\exists\,\hat{x}\; s.t. \;\hat{m}^i+\phi^i(\hat{x}^i)\geq m^i+\phi(x^i)\;\forall i$$
$$\text{and} \quad \hat{m}^i+\phi^i(\hat{x}^i)&gt; m^i+\phi(x^i)\,\text{for some}\,i$$

$\textbf{What to show:}$
$$x\;\text{solves}\;max\sum_{i=1}^Im^i+\phi^i(x^i)$$

Can anyone provide a proof of this? Any help would be greatly appreciated!

$\textbf{Edit:}\,$I don't know if this is the right path, but by the strict increasing property of $\phi(\,)$, preferences satisfy local non-satiation, which implies they satisfy the first welfare theorem. Now, If I could figure out whether all pareto optimal allocations are competitive equilibria with quasilinear utility, I may be on to something!
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.

$\square$