Contract curve and Pareto frontier Consider an exchange economy with two agents.
Each agent $i \in \{1,2\}$ derives utility $u^i(x_1,x_2) \in \mathbb R$ by consuming $(x_1,x_2) \in \mathbb R_+^2$. 
Let $u_j^i(x_1,x_2) = \partial u^i(x_1,x_2) / \partial x_j$ for $(i,j) \in \{1,2\}^2$.
The set of efficient consumption bundles is given by
\begin{align}
C = \left\{(x_1,x_2) \in \mathbb R_+^2 ~ \bigg| ~ \frac{u_1^1(x_1,x_2)}{u_2^1(x_1,x_2)} = \frac{u_1^2(x_1,x_2)}{u_2^2(x_1,x_2)}\right\}
\end{align}
Suppose the function $x_2^c :\mathbb R_+ \to \mathbb R_+$ solves 
\begin{align}
\frac{u_1^1(x_1,x_2^c(x_1))}{u_2^1(x_1,x_2^c(x_1))} = \frac{u_1^2(x_1,x_2^c(x_1))}{u_2^2(x_1,x_2^c(x_1))}
\end{align}
The contract curve is the graph $\mathcal G(x_2^c) = \{(x_1,x_2^c(x_1)) \mid x_1 \in \mathbb R_+\}$.
Let $\overline u^i(x_1) = u^i(x_1,x_2^c(x_1))$.
The Pareto frontier is now given by the following utility allocations
\begin{align}
P = \{(\overline u^1(x_1), \overline u^2(x_1)) \mid x_1 \in \mathbb R_+\} \subset \mathbb R^2.
\end{align}

Suppose that the Parento frontier is downward sloping w.r.t. $x_1$ such that
\begin{align}
\frac{\overline u_1^2(x_1)}{\overline u_1^1(x_1)} &lt; 0.
\end{align}


  Q: What conditions must be met such that the Pareto frontier is concave?


I'm especially wondering if the following condition is sufficient: 
\begin{align}
\frac{\overline u_{11}^2(x_1)}{\overline u_{11}^1(x_1)} &lt; 0.
\end{align}
Let us assume all these functions exist and are differentiable. Furthermore let us denote the inverse of the function $\overline u^1(x_1)$ by $x_1\left(\overline u^1\right)$.

Note that for any invertible function $f$ we have
$$
\left.\frac{\text{d} f(x)}{\text{d} x}\right|_{x=x_0} =
\frac{1}{\left.\frac{\text{d} f^{-1}(y)}{\text{d} y}\right|_{y=f(x_0)}}. 
$$
In our case this means (with slightly simplified notation)
$$
\frac{\text{d} x_1(\overline u^1)}{\text{d} \overline u^1} =
\frac{1}{\frac{\text{d} \overline u^1(x_1)}{\text{d} x_1}}.
$$
We will further simplify notation by writing
$$
\overline u{^i}' =  \frac{\text{d} \overline u^i(x_1)}{\text{d} x_1}.
$$
We will denote second derivatives similarly by affixing $''$.



The question is under what conditions
$$
\overline u^2\left(x_1\left(\overline u^1\right)\right)
$$ 
is concave in $\overline u^1$. This of course, depends on the sign of
$$
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2}.
$$
Differentiating
\begin{align*}
\frac{\text{d} \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}^1} &amp; =
\frac{\text{d} x_1(\overline u^1)}{\text{d} \overline u^1}
\frac{\text{d} \overline u^2(x_1)}{\text{d} x_1}
=
\frac{1}{\overline u{^1}' }\overline u{^2}' 
=
\frac{\overline u{^2}'}{\overline u{^1}' }.
\end{align*}
Going one step further
\begin{align*}
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} &amp; =
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''} 
}
{\left(\overline u{^1}'\right)^2}.
\end{align*}
This implies that the Pareto-frontier is concave when
\begin{align*}
\frac{\text{d}^2 \overline u^2\left(x_1\left(\overline u^1\right)\right)}{\text{d} \overline {u}{^1}^2} &amp; \leq 0 \\
\\
\frac{
\frac{1}{\overline u{^1}' }\overline u{^2}^{''} \overline u{^1}'
-
\overline u{^2}'\frac{1}{\overline u{^1}' }\overline u{^1}^{''} 
}
{\left(\overline u{^1}'\right)^2} &amp; \leq 0 \\
\\
\overline u{^2}^{''}
 &amp; \leq \frac{\overline u{^2}'}{\overline u{^1}' }\overline u{^1}^{''}
\end{align*}
for all $x_1$.