I think this condition might be too weak and the conjecture is not true under this definition.
If Φ1⊆Φ2, then we have
Ey∼ξminμ∈Φ2Ex∼μu(x,y)≤Ey∼ξminμ∈Φ1Ex∼μu(x,y)
(because a minimum over a larger set is smaller).
Thus, Φ2 can only be the unique argmax if Φ1=Φ2.
Consider the example ^D={[0,x]:x∈[0,1]}.
Then ^D is closed.
And Θ∗=[0,1] satisfies
Θ∗=Φ∨Ψ⟹Φ⊆Ψ∨Ψ⊆Φ.
But per the above it cannot be a unique maximizer.
Maybe the issue can be fixed if we strengthen the condition so that
Φ∗ has to be also minimal with respect to ⊆.
I think this condition might be too weak and the conjecture is not true under this definition.
If Φ1⊆Φ2, then we have Ey∼ξminμ∈Φ2Ex∼μu(x,y)≤Ey∼ξminμ∈Φ1Ex∼μu(x,y) (because a minimum over a larger set is smaller). Thus, Φ2 can only be the unique argmax if Φ1=Φ2.
Consider the example ^D={[0,x]:x∈[0,1]}. Then ^D is closed. And Θ∗=[0,1] satisfies Θ∗=Φ∨Ψ⟹Φ⊆Ψ∨Ψ⊆Φ. But per the above it cannot be a unique maximizer.
Maybe the issue can be fixed if we strengthen the condition so that Φ∗ has to be also minimal with respect to ⊆.
You’re absolutely right, good job! I fixed the OP.