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Unformatted text preview: ANALYSIS TOOLS WITH APPLICATIONS 533 29. Unbounded operators and quadratic forms 29.1. Unbounded operator basics. De f nition 29.1. If X and Y are Banach spaces and D is a subspace of X , then a linear transformation T from D into Y is called a linear transformation (or operator) from X to Y with domain D. We will sometimes wr If D is dense in X, T is said to be densely de f ned . Notation 29.2. If S and T are operators from X to Y with domains D ( S ) and D ( T ) and if D ( S ) ⊂ D ( T ) and Sx = T x for x ∈ D ( S ), then we say T is an extension of S and write S ⊂ T. We note that X × Y is a Banach space in the norm kh x, y ik = p k x k 2 + k y k 2 . If H and K are Hilbert spaces, then H × K and K × H become Hilbert spaces by de f ning ( h x, y i , h x , y i ) H × K := ( x, x ) H + ( y, y ) K and ( h y, x i , h y , x i ) K × H := ( x, x ) H + ( y, y ) K . De f nition 29.3. If T is an operator from X to Y with domain D , the graph of T is Γ ( T ) := {h x, Dx i : x ∈ D ( T ) } ⊂ H × K. Note that Γ ( T ) is a subspace of X × Y . De f nition 29.4. An operator T : X → Y is closed if Γ ( T ) is closed in X × Y. Remark 29.5 . It is easy to see that T is closed i f for all sequences x n ∈ D such that there exists x ∈ X and y ∈ Y such that x n → x and T x n → y implies that x ∈ D and T x = y. Let H be a Hilbert space with inner product ( · , · ) and norm k v k := p ( v, v ) . As usual we will write H ∗ for the continuous dual of H and H ∗ for the continuous conjugate linear functionals on H. Our convention will be that ( · , v ) ∈ H ∗ is linear while ( v, · ) ∈ H ∗ is conjugate linear for all v ∈ H. Lemma 29.6. Suppose that T : H → K is a densely de f ned operator between two Hilbert spaces H and K. Then (1) T ∗ is always a closed but not necessarily densely de f ned operator. (2) If T is closable, then ¯ T ∗ = T ∗ . (3) T is closable i f T ∗ : K → H is densely de f ned. (4) If T is closable then ¯ T = T ∗∗ . Proof. Suppose { v n } ⊂ D ( T ) is a sequence such that v n → 0 in H and T v n → k in K as n → ∞ . Then for l ∈ D ( T ∗ ) , by passing to the limit in the equality, ( T v n , l ) = ( v n , T ∗ l ) we learn ( k, l ) = (0 , T ∗ l ) = 0 . Hence if T ∗ is densely de f ned, this implies k = 0 and hence T is closable. This proves one direction in item 3. To prove the other direction and the remaining items of the Lemma it will be useful to express the graph of T ∗ in terms of the graph of T. We do this now. Recall that k ∈ D ( T ∗ ) and T ∗ k = h i f ( k, T x ) K = ( h, x ) H for all x ∈ D ( T ) . This last condition may be written as ( k, y ) K − ( h, x ) H = 0 for all h x, y i ∈ Γ ( T ) . 534 BRUCE K. DRIVER † Let V : H × K → K × H be the unitary map de f ned by V h x, y i = h − y, x i . With this notation, we have h k, h i ∈ Γ ( T ∗ ) i f h k, h i ⊥ V Γ ( T ) , i.e....
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This note was uploaded on 10/11/2010 for the course MATH 11 taught by Professor Smith during the Three '10 term at ADFA.
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