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Also, for simplicity of illustration, we assume that each source has a distinct best relay.
We shall assume that the non-self mapping S has two distinct best proximity points x ′ and x ″.
For the uniqueness, suppose that (x_{ast}) and (y_{ast}) are two distinct best proximity points of T. Let (s=d(x_{ast},y_{ast})>0).
Moreover, if S has two distinct best proximity points, then d ( A, B ) > 0 and hence the sets A and B must be disjoint.
Assume that z1(≠ z2) ∈ A are two distinct best proximity points of T A ∪ B → A ∪B in A. Thus, Tz1(≠ Tz2) ∈ B are two distinct best proximity points in B. Otherwise, Tz1 = Tz2 ⇒ T2z1 = T2z2 ⇒ z1 = z2, since z1 and z1 are best proximity points, contradicts z1 ≠z1.
Suppose, on the contrary, that x ∗ and y ∗ are two distinct best proximity points of T. Thus, we have d ( x ∗, T x ∗ ) = d ( A, B ) = d ( y ∗, T y ∗ ).
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CEO of Professional Science Editing for Scientists @ prosciediting.com