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0, (M(~n-~m),%-~m)2 and MFf = Mf, f ¢ ~(MF). Mmi n C M F C M m a Theorem 3 . 3 . Its domain The operator MF O, n , m ~ ~, is selfadjoint and satisfies x" The Friedrichs extension MF is given by of }4F : {{f,Mf] ~ Mma x If[1 ] : Olm ]. Proof. i implies that Thus for M1 and Mmln. ~ M l C is bounded below, say by H = M1 , % {f Ac~-l(~)If(~) 2 =o~ is a Hilbert space with the inner product (f'g)H where ~l = ~i + i : [f'g]H and + ~l(f'g)2 ' Mmax. 2 we see that the closure of Hilbert space, call it ~[M0] , satisfies Z(MF) = Z[M0] O ~ M m a x ) = ~H O ~ ( M x ) We have seen that a seifadjoint generates a closed symmetric form and, when have ~H ~H) m X m ~[~] = ~H " = ~MI) , H, in this Hence and thus satisfying [ ' ]H C~([) M F = M I.

14. 2. Proof. 9), we have [f'f]H + c~f'f)2 ~ ( f ' f ) 2 ' ~H f ¢ ~H ' becomes a Hilbert space with the inner product (f'g)H = [f'g]H + c~f'g)2' the completeness of ~H ~H" ~H) The closure of f'g ~ ~H ; is equivalent to the fact that in this Hilbert space is just ~[H]. is a g c RH C L2(~), g J 0, whic~ is orthogonal to all 0 = for all is closed on If ~[H] J ~H f ~-~H), so that f'g)H = [f'g]H + °~f'g)2 = (Hf'g)2 + c~f'g)2' f c ~H). This implies that g e ~H) and (H + GI) g = 0. 12) shows that 0 = ((H + el)g, g)2 ~ (g'g)2, or [ ' ]H g = 0, a contradiction.

Rofe-Beketov in [ 28 ]. 2 imply the following result. 3. 13) Bc* : AD* Conversely~ C,D H* are any if d x Vm are (2vm-d) d = vm the and the H H d x vm form. 9). 11) is the adjoint of In particular, if A,B . 13). H = H* , if and only rank(A :B) = vm , ~K&* = AB*. 9). is a basis for v(A : B). Then fill1 f[2]I c ~(A:~), Now suppose f c ~(H), where 29 and hence for some unique ~ e Cd If[l] or f[1] = D*~, f[2] = C*~. then clearly we have ) D* Conversely, if ~ e cd and Af[1 ] - Bf[2 ] = (AD* - BC*)~ = 012vm-d " and only if = g[1] = B*~, g[2] , A*~ for some unique f[1] = D*~, Similarly, ~ e C2vm'd * f[2] = C ~ , g c ~(H*) if .