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# . . . , we have S = 2n. Hence, the function g with the

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. . . , we have S = 2n. Hence, the function g with the book

# . . . , we have S = 2n. Hence, the function g with the

DOI link for . . . , we have S = 2n. Hence, the function g with the

. . . , we have S = 2n. Hence, the function g with the book

## ABSTRACT

PROPOSITION II. 1. Let X and Y be normed spaces and B : X Y an injective linear operator, with a dense domain D(B) X. Assume that Y is an n – space and that B is an a – operator.

If H(R) X is a (continuously embedded) RKHS, and H = B(H(R)), then H is a Hilbert space that is isometrically isomorphic to H(R). Moreover, H is a RKHS with the reproducing kernel K given by K(t, s) = (Kt, Ks)H, where Kt = Bj*B* t, t T, and j* is the dual operator to the embedding operator j : H(R) X. Proof. Every element g H is of the form g = Bf, for some (unique) f H(R). The space H becomes a Hilbert space, and the operator B becomes an isometry between H and H(R), if we define the inner product in H by

(g1, g2)H = (Bf1,Bf2)H = (f1, f2)R . Since B is an a – operator, and the embedding j is continuous, it follows that the functions Kt are well defined and that Kt H, for every t T. To show that H is in fact a RKHS with the kernel K, it remains to show that the reproducing property holds. We have,

(g,Kt)H = (Bf,Bj*B* t)H = (f,j*B* t)R = f, B* t = Bf, t = g(t). If X = H(R), then the embedding j is the identity operator, and the delta

functionals in X* are the functions Rt.