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Let $f$ be a convex function on a convex domain $\Omega$, such that, $f:2^{\Omega} \rightarrow \mathbb R _+$ and $g$ a convex non-decreasing function defined on $[0, + \infty]$ such that $g(0) = 0$. prove that the composition of $g \circ f$ is convex on $\Omega$.

My attempt:

Since $f$ is convex from Shapley’s definition of convexity,

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$$f(S\cup T) + f(S\cap T) \geq f(S) + f(T), S, T \subset \Omega$$

Let. $S\cap T = \varnothing$, we get, $f(S\cup T) \geq f(S) + f(T)$.

Given. g is non-decreasing,

$$g(f(S\cup T)) \geq g(f(S) + f(T))$$

Now, we know that every convex function defined in $[0, +\infty]$ with $f(0) = 0$ is is superadditive.

With superadditivity property of $g$,

$$g(f(S) + f(T)) \geq g(f(S)) + g(f(T))$$

Finally,

$$g(f(S\cup T)) \geq g(f(S)) + g(f(T))$$

Therefore, $g \circ f$ is convex.

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The only dubious point is the second inequality where you claim that

$$g(f(S) + f(T)) \geq g(f(S)) + g(f(T))$$

by convexity. The property

$$g(x+y) \ge g(x) + g(y)$$

is called superadditivity. It is known that a convex function $g(x)$ with $g(0)\le 0$ is superadditive; see https://en.wikipedia.org/wiki/Convex_function or https://math.stackexchange.com/posts/507611/edit.

Thus, if you add the assumption that $g(0) \le 0$, your claim holds. (In cooperative game theory, it is usually assumed $g(0)=0$.)

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