- It's well known that if $\succsim$ satisfies independence, then it is also convex.
Since $\succsim$ satisfies independence,
$L\succsim L^{'} \iff \alpha L+(1-\alpha)L^{''}\succsim \alpha L^{'}+(1-\alpha)L^{''}$ for all $\alpha \in [0,1]$ and $ L, L^{'}, L^{''}\in \mathfrak{L} $
Convexity requires:
$L\succsim L^{''}$ and $L^{'}\succsim L^{''} \Longrightarrow \alpha L$ + $(1-\alpha)L^{'} \succsim L^{''}$ for all $\alpha \in \left[0,1\right]$ and $ L, L^{'}, L^{''}\in \mathfrak{L} $
Pick any $ L, L^{'}, L^{''}\in \mathfrak{L} $ with $L \succsim L^{''}$ and $L^{'} \succsim L^{''}$. By completeness, $L \succsim L^{'}$ or $L^{'} \succsim L$ or both. Without loss of generality, assume $L \succsim L^{'}$. Then, by independence, for all $\alpha \in [0,1]:$
$ \alpha L$ + $( 1- \alpha) L^{'} \succsim \alpha L^{'} + (1-\alpha)L^{'} = L^{'} \succsim L^{''} $, which we wanted to show.
- Is it true if $\succsim$ are convex then they satisfy independence?
On the other side, convexity does not imply independence. To see this, take a look
at Figure (see below). Triangles are indifference curves and arrows show the direction in which
utility increases. It is convex, but not independent.
