Let us consider a game with two actors named A and B. Let we have the following pay-offs depending on the coalitions:
- (empty coalition) = $0$, (A) = $0$, (B) = $200$, (AB) = $300$.
The calculation shows that the Shapley value is (50, 250) and that the core consists of vectors $(x,300-x)$ for $0<x<100$.
I claim that any vector in the core satisfies the four conditions of the Shapley value:
efficiency: $x+(300-x)=300$,
symmetry: trivial, because there is no equivalent actors,
linearity: trivial, because there is only one game (more precisely, no other game is presented),
zero player: trivial, because there is no zero player.
So, any vector in the core is Shapley value. But Shapley's theorem says that the Shapley values is unique. Where is my mistake? Have I got a misunderstanding of the linearity condition?
Note first that linearity is not additivity. Additivity states that adding the solution of two games together produces the solution of the sum of these games. Stated more formally, it means that if we consider two arbitrary games $z$ and $w$, then
$$(*)\quad \phi(z+w) = \phi(z) + \phi(w) $$
must hold.
In terms of your example, this means that we can decompose, for instance, your game $v$ into a game $z=\{50,250,300\}$, and a game $w=\{-50,-50,0\}$. Adding both games reproduces the example $v=\{0,200,300\}$. Moreover, game $z$ has Shapley value $(50,250)$ whereas game $w$ has Shapley value $(0,0)$. However, you are right when you say that property $(*)$ will also be satisfied for the core when considering this particular decomposition of game $v$. But this property is not satisfied in general for the core. Finally, one ought to be very careful while concluding from an isolated case to its general.
Hence, uniqueness of the Shapley value is determined by the four axioms you have mentioned, but this is not the case for the core, which is not a value.