Is the Convex Combination of Two Pareto Optimal Allocations Also Pareto Optimal? I am studying for my qualifiers, and I ran into this question from a previous year's exam.

$\textbf{Question:}$

Consider a two-consumer two-good pure exchange economy. Both preferences are locally non-satiated and convex. Prove or disprove the following statement: if $(x_1,x_2)$ and $(\hat{x}_1,\hat{x}_2)$ are two different pareto optimal allocations, then the convex combination, $(\alpha x_1+(1-\alpha)\hat{x}_1,\alpha x_2+(1-\alpha)\hat{x}_2$ MUST also be pareto optimal for any $\alpha\in(0,1)$.

I believe the statement is true, and here is the work for my proof below.

$\textbf{My Proof:}$
By pareto optimality of $x_i$ and $\hat{x}_i$: 
$$\not\exists\; x_i^\star\; s.t.\; u_i(x_i^\star)\geq u_i(x_i)\;\forall i\; \text{and}\; u_i(x_i^\star)&gt; u_i(x_i)\;\text{for at least one }i \;\text{or}\;u_i(x_i^\star)\geq u_i(\hat{x}_i)\;\forall i\; \text{and}\; u_i(x_i^\star)&gt; u_i(\hat{x}_i)\;\text{for at least one }i$$
$$\implies\;u_i(\alpha x_i)\geq u_i(\alpha x_i^\star)\;\forall i\;\text{and}\;u_i((1-\alpha)\hat{x}_i)\geq u_i((1-\alpha)x_i^\star)\;\forall i$$
$$\implies\not\exists\;x_i^\star\;s.t.\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)\geq u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\forall i$$
$$\text{and}\;u_i(\alpha x_i^\star+(1-\alpha)x_i^\star)&gt; u_i(\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{for at least one}\;i$$
$$\implies (\alpha x_i+(1-\alpha)\hat{x}_i)\;\text{is pareto optimal}$$
$$\blacksquare$$

This proof seemed almost too easy, so I'm wondering if it is correct/rigorous.
To complement densep's answer, here is a schematic Edgeworth box illustration of what can go wrong. The points on the dashed line are convex combinations of the Pareto optimal points $(x_1,x_2)$ and $(\hat{x}_1,\hat{x}_2)$, but the marked point is not Pareto optimal.