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**Example text**

The solution is the Boolean sum P ⊕ Q, provided that P ∗ (X ∗ ) = F, Q∗ (X ∗ ) = G, and QP Q = QP. Many theorems of this type are given in Gordon and Cheney [1978]. To illustrate these methods, let us find a function which interpolates a given function on the perimeter of the unit square in the st-plane. Lagrange interpolation with two nodes, 0 and 1, is given by (P x)(s) = x(0)(1 − s) + x(1)s. Here x is any element of C(S), and S = {s : 0 ≤ s ≤ 1}. Similarly, in C(T ), with T = {t : 0 ≤ t ≤ 1} we have (Qy)(t) = y(0)(1 − t) + y(1)t.

If P : X → → U and Q : X → → V are projections defined on a linear space X, we construct their Boolean sums by the equations (9) P ⊕ Q = P + Q − P Q, Q ⊕ P = Q + P − QP. These are (in general) different operators and are (in general) not projections. However, we have this result: THEOREM. In order that P ⊕ Q be a projection of X onto U + V it is necessary and sufficient that P QP = QP. Proof. Obviously P ⊕ Q maps X into U + V. If v ∈ V then (10) (P ⊕ Q)v = P v + Qv − P Qv = P v + v − P v = v. If P QP = QP then for any x, (11) (P ⊕ Q)P x = P 2 x + QP x − P QP x = P x.

So far, this is purely algebraic. If a crossnorm α has been prescribed on X ⊗ Y, then X ⊗α Y (being the completion of the normed space) is a Banach space, and we would like to extend A ⊗ B to be a continuous linear map of X ⊗α Y into X ⊗α Y. The existence of such an extension depends upon the boundedness of A ⊗ B on the dense set X ⊗ Y. , m (8) α i=1 m Axi ⊗ Byi ≤ A B α i=1 xi ⊗ yi for all operators A : X → X, all operators B : Y → Y, and all elements of X ⊗ Y. If the crossnorm α has the property (8), it is said to be a uniform crossnorm.