= Solution
Interpret a beneficial trade as weakly improving both participants and strictly improving at least one; otherwise the identity trade would always remain available. Put $g_j=\nabla U_j(y_j)$. For a globally <strictly increasing concave function>, every component of its gradient is strictly positive: <concavity> makes a coordinate derivative nonincreasing along that coordinate, and monotonicity makes it nonnegative. If it vanished at a finite point, it would vanish farther along the coordinate, contradicting strict increase on the whole real line.
To construct a <pairwise improvement from nonparallel utility gradients>, suppose two positive gradients $g,h$ are not proportional. Strict <Cauchy-Schwarz inequality> gives
$$
\frac{g\cdot h}{\|h\|^2}<\frac{\|g\|^2}{g\cdot h}.
$$
Choose $t$ between these bounds and put $\delta=g-th$. Then $g\cdot\delta>0$ and $h\cdot\delta<0$. For sufficiently small $\varepsilon>0$, the feasible transfer $y_i\mapsto y_i+\varepsilon\delta$, $y_j\mapsto y_j-\varepsilon\delta$ strictly improves both utilities by <differentiability>. This contradicts the absence of beneficial pairwise trades.
Hence all gradients are proportional. Choose $v=g_1$; positivity makes each proportionality scalar positive:
$$
\boxed{\nabla U_j(y_j)=\lambda_jv,\qquad\lambda_j>0.}
$$
Now take $\boxed{a_j=1/\lambda_j}$. The <concave supporting-tangent inequality> gives, for any other allocation $(z_j)$ with the same total endowment,
$$
\begin{aligned}
\sum_ja_jU_j(z_j)
&\leq\sum_ja_jU_j(y_j)+\sum_ja_j\nabla U_j(y_j)\cdot(z_j-y_j)\\
&=\sum_ja_jU_j(y_j)+v\cdot\sum_j(z_j-y_j)
=\sum_ja_jU_j(y_j).
\end{aligned}
$$
Thus the final allocations globally maximize the social planner's weighted objective. They need not be unique when utilities are not strictly concave, but the required positive supporting weights exist.
Back to article page
