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x I x I,x<b using barcode encoder for .net framework control to generate, create data matrix barcodes image in .net framework applications. Developing with Visual Studio .NET In particular, the s Data Matrix for .NET pace { f Reg(I ; X ) . f (b) = f (b )} is a closed subspace of L (I ; X ). (ii) Let I be a closed interval which is unbounded to the right. Then every Reg(I ; X ) is a closed subspace of L (I ; X ), and, for every f Reg(I ; X ), ess sup.

f (x). = sup f (x). .. x I x I Proof (i) Trivially ess supx I f (x). supx I,x<b f (x). , so only the opposi te inequality is nontrivial. To get this inequality we x > 0 and approximate f by a step function g so that . f (x) g(x). for all x I . Since the step function g has the property that ess supx I g(x). = supx I,x<b g(x). , we nd that ess supx I f (x). supx I,x<b f (x). . This being true for all > 0 we must have ess supx I f (x). supx I,x<b f (x). . (ii) The proof of (ii) is similar to the proof of (i). Corollary A.

1.7 Let X be a Banach space, and let I = [a, b] be a closed bounded interval. Then the closure in L (I ; X ) of the set of all X -valued step functions on X can be identi ed with the set { f Reg(I ; X ) .

f (b) = f (b )}. Pr gs1 datamatrix barcode for .NET oof This follows from Propositions A.

1.2 and A.1.

6. Corollary A.1.

8 Let X be a Banach space, and let I = [a, b] be a closed bounded interval. Then the closure in L (I ; X ) of the set of all X -valued step functions on X can be identi ed with the set { f Reg(I ; X ) . f (b) = f (b )}. Proof This follows from Propositions A.1.2 and A.1.6. A.2 Square root and polar decomposition A.2 The positive square root and the polar decomposition De nition A.2.1 By a .

net framework barcode data matrix positive operator in a Hilbert space X we mean a (densely de ned, closed and) self-adjoint (possibly unbounded) operator A : X D (A) X satisfying x, Ax 0 for all x D (A). The notation A 0 means that A is positive in the above sense, and the notation A B means that both A and B are self-adjoint operators on X and that A B 0 (here we require, in addition, that B B(X )). If A I for some > 0, then we write A 0 and call A uniformly positive Lemma A.

2.2 Let A be a positive operator in a Hilbert space X . (i) A has a unique positive square root A1/2 , i.

e., there is a unique positive operator A1/2 such that A = (A1/2 )2 . In particular, D (A) = x D A1/2 A1/2 x D A1/2 .

(ii) A1/2 B(X ) if and only if A B(X ). (iii) N A1/2 = N (A) and R (A) R A1/2 R (A). In particular, (iv) (v) (vi) (vii) R A1/2 = R (A), and A1/2 is injective if and only if A is injective.

R A1/2 is closed if and only if R (A) is closed. A 0 if and only if A1/2 0, or equivalently, A1/2 has a bounded inverse if and only if A has a bounded inverse. A1/2 = A if and only if A is a projection operator, or equivalently, if and only if A1/2 is a projection operator (i.

e., A1/2 = A = A2 ). An operator B B(X ) commutes with A if and only if B commutes with A1/2 (i.

e., AB = B A if and only if A1/2 B = B A1/2 )..

Proof (i), (ii), (v) , and (vii): See, for example, Kato (1980, Theorem 3.35, p. 281) or Rudin (1973, Theorems 13.

24 and 13.31). (iii): Obviously, N A1/2 N (A) (if A1/2 x = 0, then Ax = (A1/2 )2 x = 0).

Conversely, if Ax = 0, then . A1/2 x 2 = A1/2 x, A1/2 x = x, Ax = 0, so A1/2 x = 0. The fact that N A1/2 = N (A) implies that also their orthogonal complements are equal (see Lemma 9.10.

3(iii)): R A1/2 = R (A). The inclusion R (A) R A1/2 is trivial. Thus, R (A) R A1/2 R A1/2 = R (A).

(iv): If R (A) is closed, then it follows from (iii) that also R A1/2 is closed. Conversely, if R A1/2 is closed, then it follows from the closed graph theorem that A1/2 is a boundedly invertible operator from D A1/2 onto N A1/2 = 1/2 , hence A is a boundedly invertible operator R (A) and from D (A) onto D A from D (A) onto R (A). Thus, R (A) is closed.

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