We know that the core of a balanced game is non-empty. The convexity ensures balancedness. However, I was wondering if a concave function too satisfies balancedness condition.
The balancedness is defined according to Bondareva–Shapley theorem as follows: Let the pair $(N, v)$ be a cooperative game in characteristic function form, where $N$ is the set of players and where the value function $v:2^{N} \to \mathbb {R}$ is defined on $N$'s power set (the set of all subsets of $N$).
The core of $(N, v)$ is non-empty if and only if for every function $\alpha :2^{N \setminus \emptyset} \to [0,1]$, where
$ \forall i\in N:\sum _{S\in 2^{N}:\;i\in S}\alpha (S)=1$ the following condition holds:
$\sum_{S\in 2^{N\setminus \emptyset}}\alpha (S)v(S)\leq v(N) $
Edit: can I show that the characteristic function $(\sum_{i} x_i - \frac{k_1}{\sum_{i} x_i}) - k_2$ is balanced? $k_1, k_2$ are non-negative constants.
@Holger I. Meinhardt, Thank you for your comments. However, without going into the details of anti-core, I was trying to prove whether the characteristic function $(\sum_{i} x_i - \frac{k_1}{\sum_{i} x_i}) - k_2$ is balanced or not. According to the balancedness condition, we know, the grand coalition set is the member of the balanced collection and results into the characteristic function being balanced. Next, I show that each of the component of the characteristic function, namely, $\sum_{i} x_i$, $\frac{-k_1}{\sum_{i} x_i}$, and $-k_2$ are balanced, and therefore, the condition $\sum_{S\in 2^{N\setminus \emptyset}}\alpha (S)v(S)\leq v(N)$ is satisfied by the grand coalition for each of the component. And, because, the algebraic sum of each of the components results into balancedness, I could easily show that the function under consideration is balanced. Is my argument correct?
Sample result, $v(\lbrace \varnothing \rbrace) = 0$, $v(\lbrace 1 \rbrace) = 0$, $v(\lbrace 2 \rbrace) = 1.50$, $v(\lbrace 3 \rbrace) = 2.67$, $v(\lbrace 12 \rbrace) = 2.67$, $v(\lbrace 13 \rbrace) = 3.75$, $v(\lbrace 23 \rbrace) = 4.60$, $v(\lbrace 123 \rbrace) = 5.83$. Note that, $v(\lbrace 123 \rbrace) + v(\lbrace 1 \rbrace) \leq v(\lbrace 12 \rbrace) + v(\lbrace 13 \rbrace)$, a concave function. The core exists, and is $x_1 = 1.230$, $x_2 = 1.930$ and $x_3 = 2.670$.
Edit:
I have modified Lemma 3.1 of H. Meinhardt, “Common pool games are convex games,” Journal of Public Economic Theory, vol. 1, no. 2, pp. 247–270, 1999. as follows:
We know that the function, $f$ is concave, if for all, $x, y, z \in \mathbb{R}$ and $x \leq z$
$$ f(x+y) - f(x) \geq f(z+y) - f(z)$$
Now, defining $x := g(S)$ and $z := g(T)$ and $y := g(\lbrace i \rbrace)$, and $S \subset T$, we get, $ g(S) \leq g(T)$. Then we get,
$$ f(g(S)+g(\lbrace i \rbrace)) - f(g(S)) \geq f(g(T)+g(\lbrace i \rbrace)) - f(g(T))$$
Now considering additivity of $g$,
$$ f(g(S \cup \lbrace i \rbrace)) - f(g(S)) \geq f(g(T \cup \lbrace i \rbrace)) - f(g(T))$$
which implies,
$$ v(S \cup \lbrace i \rbrace) - v(S) \geq v(T \cup \lbrace i \rbrace) - v(T)$$
Therefore, $v$ is concave too.
Is this proof wrong? If it is correct, then because (x - 1/x) is concave, the defined function should have been concave too. I know I am missing something. But I can't figure it out.
Another question, if the considered function $f(\sum_{i \leq S} x_i)$ is piecewise in the sense that, it is convex if $\sum_{i \leq S} x_i \leq M$ and is balanced if $\sum_{i \leq S} x_i > M$ ($M$ is an arbitrary constant). Can the piecewise function be still balanced?