Pareto efficiency and maximization of sum of utilities

Finding Pareto efficient allocations via social planner's problems is a special case of scalarization in convex optimization.

Suppose agents' utility functions are $u_i:mathbb{R}^n rightarrow mathbb{R}$, for $i = 1, 2, cdots, m$, and the feasible allocations are given by $g(x) geq 0$ for some $g:mathbb{R}^n rightarrow mathbb{R}^p$. The general result is as follows.

Sufficiency under general conditions

If an allocation $x$ solves the social planner's problem with strictly positive social weights, then it is Pareto. This is true with no assumptions on utility functions and feasibility constraint. In other words, if $$ x in argmax_{g(x') geq 0} sum_{i = 1}^m lambda_i u_i(x'), mbox{ for some } lambda in mathbb{R}^m_{++}, quad (*) $$ then it is Pareto optimal.

This is easy to see. If $y$ Pareto-improves $x$, then $$ sum_{i = 1}^m lambda_i u_i(y) > sum_{i = 1}^m lambda_i u_i(x) $$ for any $lambda in mathbb{R}^m_{++}$. It is also easy to see that zero entry in $lambda$ cannot be allowed---it would make the statement false.

The converse is not true in general---this sufficient condition is not necessary. There are Pareto allocations that do not solve $(*)$. This is because the set of achievable utility values $$ mathcal{U} = { (u_1 (x), cdots, u_m(x)); g(x) geq 0 } $$ is in general not a convex subset of $mathbb{R}^m$. The social planner's weights $lambda$ corresponds to certain supporting hyperplanes of $mathcal{U}$. If $mathcal{U}$ is not convex, its Pareto frontier cannot be recovered by varying $lambda$.

However, there is a partial converse.

Necessity under concavity

Assume $u_i$'s and $g$ are concave. If an allocation $x$ is Pareto, then it solves a social planner's problem with nonnegative social weights. In other words, if $x$ is Pareto then $$ x in argmax_{g(x') geq 0} sum_{i = 1}^m lambda_i u_i(x') mbox{ for some } lambda in mathbb{R}^m_{+}. quad (**) $$ This is true because of convex sets and separating hyperplanes.

This is a partial converse because it's necessary but not sufficient. There are allocations that solve $(**)$ but not Pareto.

Comment

Even under concavity, $(**)$ is a only a partial converse of $(*)$. Notice the gap between strictly positive and nonnegative social weights.

To recover the Pareto frontier, the general procedure is to first find solutions to $(*)$. Then check the solutions to $(**)$ case-by-case for Pareto optimality. Alternatively, one can also take the closure (the limit points) of solutions to $(*)$.