tells us that C > 0 (Gretton et al., 2012). Hence, for almost all ( x$_{1}$, y$_{1}$ ), 〈 φ ( x$_{1}$ ) − φ ( y$_{1}$ ) , µ$_{X}$ − µ$_{Y}$ 〉$_{H}$ − 〈 µ$_{X}$ − µ$_{Y}$ , µ$_{X}$ − µ$_{Y}$ 〉$_{H}$ = 0 , where µ$_{X}$ and µ$_{Y}$ in H have the property that 〈 µ$_{X}$, f 〉$_{H}$ = E$_{Q}$f ( X ) and 〈 µ$_{Y}$ , f 〉$_{H}$ = E$_{Q}$f ( Y ) for all f ∈ H (Lemma 3 in Gretton et al., 2012). The above holds if and only if φ ( x$_{1}$ ) − φ ( y$_{1}$ ) = µ$_{X}$ − µ$_{Y}$ . Noting that µ$_{X}$ − µ$_{Y}$ does not rely on x$_{1}$, y$_{1}$ , it follows that φ ( x$_{1}$ ) − φ ( y$_{1}$ ) must not rely on x$_{1}$, y$_{1}$ for all ( x$_{1}$, y$_{1}$ ) in some Q probability one set D ⊆ Z $^{2}$. Fix a continuous function f : Z →$_{R}$ and x$_{1}$, y$_{1}$ ∈ D . For any ϵ > 0, the universality of H ensures that there exists an f$_{ϵ}$ ∈ H such that ‖ f$_{ϵ}$ − f ‖$_{∞}$ ≤ ϵ . By the triangle inequality, | f ( x$_{1}$ ) − f ( y$_{1}$ ) − f$_{ϵ}$ ( x$_{1}$ ) + f$_{ϵ}$ ( y$_{1}$ ) | ≤ 2 ϵ. Because φ ( x$_{1}$ ) − φ ( y$_{1}$ ) is constant and f ∈ H , 〈 φ ( x$_{1}$ ) − φ ( y$_{1}$ ) , f$_{ϵ}$ 〉$_{H}$ = f$_{ϵ}$ ( x$_{1}$ ) − f$_{ϵ}$ ( y$_{1}$ ) does not rely on x$_{1}$, y$_{1}$ for any ϵ . Furthermore, the fact that f$_{ϵ}$ converges to f in supremum norm ensures that | f$_{ϵ}$ ( x$_{1}$ ) − f$_{ϵ}$ ( y$_{1}$ ) | converges to a fixed quantity K (which does not rely on x$_{1}$ or y$_{1}$ ) as ϵ → 0. Applying this to the above yields that f ( x$_{1}$ ) − f ( y$_{1}$ ) = K . As f was an arbitrary continuous function and X$_{1}$ ̸≡ Y$_{1}$ , we can apply this relation to z ↦→ z and z ↦→ z 2 to show that x$_{1}$ − y$_{1}$ and x$_{1}$ + y$_{1}$ do not rely on the choice of ( x$_{1}$, y$_{1}$ ) ∈ D . Hence ( x$_{1}$ − y$_{1}$ + x$_{1}$ + y$_{1}$ ) / 2 = x$_{1}$ and ( x$_{1}$ + y$_{1}$ − y$_{1}$ + x$_{1}$ ) / 2 = y$_{1}$ do not rely on the choice of ( x$_{1}$, y$_{1}$ ) ∈ D . This can only occur if ( x$_{1}$, y$_{1}$ ) are constant over the probability 1 set D , i.e. if Q is degenerate. For the two-sample problem in Gretton et al. (2012), one can take Q to be a product distribution of the marginal distribution of X and the marginal distribution of Y . Proof of 1. We first prove sufficiency. If (i) holds, then 2 D RS = D RR + D $^{SS}$. It follows that D Ψ 1 ≡ 0 under H$_{0}$ . Now suppose (ii) holds. It is a simple matter of algebra to verify that D RR 1 ≡ D RS 1 ≡ D SS 1 ≡ 0. Hence D Ψ 1 ≡ 0, yielding the sufficiency of the stated conditions. 44