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Unformatted text preview: FUNCTIONAL ANALYSIS LECTURE NOTES: ADJOINTS IN HILBERT SPACES CHRISTOPHER HEIL 1. Adjoints in Hilbert Spaces Recall that the dot product on R n is given by x · y = x T y , while the dot product on C n is x · y = x T ¯ y . Example 1.1. Let A be an m × n real matrix. Then x mapsto→ Ax defines a linear map of R n into R m , and its transpose A T satisfies ∀ x ∈ R n , ∀ y ∈ R m , Ax · y = ( Ax ) T y = x T A T y = x · ( A T y ) . Similarly, if A is an m × n complex matrix, then its Hermitian or adjoint matrix A H = A T satisfies ∀ x ∈ C n , ∀ y ∈ C m , Ax · y = ( Ax ) T ¯ y = x T A T ¯ y = x · ( A H y ) . Theorem 1.2 (Adjoint) . Let H and K be Hilbert spaces, and let A : H → K be a bounded, linear map. Then there exists a unique bounded linear map A ∗ : K → H such that ∀ x ∈ H, ∀ y ∈ K, ( Ax,y ) = ( x,A ∗ y ) . Proof. Fix y ∈ K . Then Lx = ( Ax,y ) is a bounded linear functional on H . By the Riesz Representation Theorem, there exists a unique vector h ∈ H such that ( Ax,y ) = Lx = ( x,h ) . Define A ∗ y = h . Verify that this map A ∗ is linear (exercise). To see that it is bounded, observe that bardbl A ∗ y bardbl = bardbl h bardbl = sup bardbl x bardbl =1 |( x,h )| = sup bardbl x bardbl =1 |( Ax,y )| ≤ sup bardbl x bardbl =1 bardbl Ax bardblbardbl y bardbl ≤ sup bardbl x bardbl =1 bardbl A bardblbardbl x bardblbardbl y bardbl = bardbl A bardblbardbl y bardbl . These notes closely follow and expand on the text by John B. Conway, “A Course in Functional Analysis,” Second Edition, Springer, 1990. c circlecopyrt 2007 by Christopher Heil. 1 2 ADJOINTS IN HILBERT SPACES We conclude that A ∗ is bounded, and that bardbl A ∗ bardbl ≤ bardbl A bardbl . Finally, we must show that A ∗ is unique. Suppose that B ∈ B ( K,H ) also satisfied ( Ax,y ) = ( x,By ) for all x ∈ H and y ∈ K . Then for each fixed y we would have that ( x,By − A ∗ y ) = 0 for every x , which implies By − A ∗ y = 0. Hence B = A ∗ . square Exercise 1.3 (Properties of the adjoint) . (a) If A ∈ B ( H,K ) then ( A ∗ ) ∗ = A . (b) If A , B ∈ B ( H,K ) and α , β ∈ F , then ( αA + βB ) ∗ = ¯ αA ∗ + ¯ βB ∗ . (c) If A ∈ B ( H 1 ,H 2 ) and B ∈ B ( H 2 ,H 3 ), then ( BA ) ∗ = A ∗ B ∗ . (d) If A ∈ B ( H ) is invertible in B ( H ) (meaning that there exists A − 1 ∈ B ( H ) such that AA − 1 = A − 1 A = I ), then A ∗ is invertible in B ( H ) and ( A − 1 ) ∗ = ( A ∗ ) − 1 . Note: By the Inverse Mapping Theorem, A is invertible in B ( H ) if and only if A is a bounded linear bijection. Proposition 1.4. If A ∈ B ( H,K ), then bardbl A bardbl = bardbl A ∗ bardbl = bardbl A ∗ A bardbl 1 / 2 = bardbl AA ∗ bardbl 1 / 2 ....
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This note was uploaded on 08/25/2011 for the course MATH 6338 taught by Professor Staff during the Summer '08 term at Georgia Tech.
- Summer '08
- Dot Product