By Froberg R.

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Since fn(x) ---+ 0, \:Ix E X, it follows that I:i G;Jn(Xi) ---+ 0. Hence, for n sufficiently large, Now, a contradiction. It remains to be shown that if a space il has R(x, y) as its reproducing kernel, then il must be identical to the space H constructed above. Since Rx = R(x,') E il, \:Ix E X, so H ~ il. Now, for any h E ile H, by the orthogonality, h(x) = (Rx, h) = 0, \:Ix E X, so il = H. The proof is now complete. 0 30 2. Model Construction From the construction in the proof, one can see that the space 11.

The corresponding reproducing kernels are Roo(x,y) = 1, ROI(x,y) = k 1(x)k 1(y), and R 1(x,y) = k2(X)k2(y) - k4(X - y). Note that fo1 ROI (x, y)dy = fo1 Rdx,y)dy = 0, "Ix E [0,1]. Using this space on both marginal domains, one can construct a tensor product space with nine tensor sum terms. 7) on page 7, the subspaces 1iOO(l) 0 (1i01(2) EB H 1(2») and (1i01(1) EB 1i1(1») 01i00(2) span the main effects, and the subspace (H 01 (1) EB H 1(1») 0 (H0 1(2) EB H 1(2») spans the interaction. 3. 28). 13.

It is easy to compute IIf - gW a contradiction. = IIf - fgW - 0'2/(3 < IIf - fgW, D The linear subspace gc = {j : (j, g) = 0, 'Vg E 9} is called the orthogonal complement of g. By the continuity of (j, g), gc is closed. 1, it is easy to verify that Ilf - fg - fg ll 2 = (j e fg - fge,f - fge - fg) = (j - fg,f - fge) - (j - fg,fg) - (jge,f - fge) + (jge,fg) = 0, where fg E 9 and fge E 9 c are the projections of f in 9 and 9 c, respectively. Hence, there exists a unique decomposition f = fg + fge for every f E H.