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,

$$f(S\cup T) + f(S\cap T) \geq f(S) + f(T), ~~~~\forall S, T \subseteq \Omega$$

$$f(S\cup T) \geq f(S) + f(T), ~~~~\forall S, T \subseteq \Omega$$

• For the usual definition of superadditivity $S$ and $T$ should be restricted to disjoint sets. And a more conventional name for convexity of $f$ is supermodularity (equivalent to the concavity of its Lovász extension) – Dap Nov 10 '17 at 13:31
• Can you share any reference? – Eval Nov 10 '17 at 13:33
• Well, Wikipedia. Superadditivity is a common assumption in coalition games, so a reference would be Owen's "Game Theory". Supermodularity is sometimes studied in non-cooperative games, and is discussed in Fudenberg and Tirole's "Game Theory". – Dap Nov 10 '17 at 13:56

No. Define $\Omega = \{1\}$ and $f(\{1\}) = 1$. It is easy to check that $f$ is convex but not superadditive.
• $f(\lbrace 1 \rbrace \cup \lbrace \varnothing \rbrace) \geq f(\lbrace 1 \rbrace) + f(\lbrace \varnothing \rbrace)$, because, $f(\lbrace \varnothing \rbrace) = 0$. – Eval Nov 10 '17 at 13:20
• @Eval $2^\Omega$ only contains the sets $\varnothing$ and $\{1\}$, so you can just check out all combinations and find that $f$ is convex but not superadditive. For example $$f(\{1\}\cup\{1\})=f(\{1\})=1\not\ge2=1+1=f(\{1\})+f(\{1\}),$$hence not superadditive. – M. Winter Nov 10 '17 at 13:20