If $f:2^{\Omega} \rightarrow \mathbb R _+$ is convex and nondecreasing, and is defined over the set $\Omega$ and $f(\varnothing) = 0$, is $f$ superadditive?
The convexity of $f$ is defined as,
$$f(S\cup T) + f(S\cap T) \geq f(S) + f(T), ~~~~\forall S, T \subseteq \Omega$$
Superadditivity is defined by,
$$f(S\cup T) \geq f(S) + f(T), ~~~~\forall S, T \subseteq \Omega$$
No. Define $\Omega = \{1\}$ and $f(\{1\}) = 1$. It is easy to check that $f$ is convex but not superadditive.