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Additional Notes on Adjoint and Unitary Operators

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Additional Notes on Adjoint and Unitary Operators AMATH 731: Applied Functional Analysis Fall 2017 Additional notes on adjoint and unitary operators (to supplement Section 4.5 of the Course Notes) Just to recall the basic facts: Let H be a Hilbert space and L : H H a bounded linear operator. → Then there exists a bounded linear operator L∗ : H H such that → Lx,y = x,L∗y . (1) h i h i L∗ is the adjoint of L. Here are a few additional properties not listed in Section 4.3 of the Course Notes. The proofs are left as an exercise. L∗ is unique: If Lx,y = x,L∗y = x,L∗y for all x,y H, then L∗ = L∗. h i h 1 i h 2 i ∈ 1 2 I∗ = I, 0∗ = 0, (S + T )∗ = S∗ + T ∗, (αT )∗ = αT ∗, (ST )∗ = T ∗S∗. Some additional examples of adjoint operators 4. Multiplication operator on L2( , ): Consider the operator −∞ ∞ F : x(t) f(t)x(t), (2) → where f(t) B < for all t. (Note that we do not require that f L2. After all, multiplication of | |≤ ∞ ∈ x(t) by a constant k R is a special case.) Then F is a bounded linear operator with ∈ F = sup f(t) = f ∞. k k t∈R | | k k Now consider the following, ∞ ∞ Fx,y = f(t)x(t)y(t) dt = x(t)f(t)y(t) dt h i Z−∞ Z−∞ = x, F ∗y . h i This implies that the associated adjoint operator is given by F ∗ : y(t) f(t)y(t). (3) → 1 The condition that f(t) be strictly bounded may be replaced by the condition f(t) B < for | |≤ ∞ almost all t R. Then ∈ F = ess. sup. R f(t) = f , k k t∈ | | k k∞ where denotes the L∞ norm, i.e., f(t) f almost everywhere. k · k∞ | | ≤ k k∞ 5. An interesting example involving Volterra integral operators: Let I = [0, T ] and consider the linear operator K : C(I) C(I) defined as follows: y = Kx implies that → t y(t) = (Kx)(t)= k(t,s)x(s) ds. (4) Z0 In the usual case of the Volterra integral operator, k(t,s) needs only to be defined in the region s t, ≤ t [0, T ]. However, if we wish to define an adjoint operator, essentially replacing k(t,s) with k(s,t), ∈ we shall have to extend the definition of k. If we set k(t,s)=0 for s>t, then Eq. (4) becomes y(t)= k(t,s)x(s) ds. (5) ZI Then, according to Example 3 in the Course Notes, p. 67, the adjoint is given by T (K∗y)(t)= k(s,t)y(s) ds = k(s,t)y(s) ds. (6) ZI Zt Thus, the adjoint of a Volterra integral operator is also a Volterra integral operator. But if K depends on the “past” (i.e., y(t) = (Kx)(t) is determined by x(s) for 0 s t), then K∗ depends on the ≤ ≤ “future” (i.e., (K∗y(t) is determined by y(s) for t s T ). ≤ ≤ The Volterra operator in Eq. (4) is an example of a causal operator and the adjoint Volterra operator in Eq. (6) is an example of an anticausal operator. 6. Adjoint operators of (unbounded) differential operators: We take a little detour here and consider second-order (unbounded) differential operators on the linear subspace M of (piecewise) C2 functions u : [a, b] C satisfying the boundary conditions u(a)= u(b) = 0. In what follows, let g(x) → be a real-valued function. For an u M, define Lu as follows, ∈ (Lu)(x)= u′′(x)+ g(x)u(x). (7) For u, v M, consider the inner product ∈ b Lu, v = v(x)[u′′(x)+ g(x)u(x)] dx. (8) h i Za Integrating the first term by parts yields b b ′′ ′ b ′ ′ v(x)u (x) dx = v(x)u (x) a v (x)u (x) dx. (9) Za | − Za The endpoint terms vanish because of the boundary conditions. Another integration by parts yields b b ′ ′ ′′ b ′′ v (x)u (x) dx = v (x)u(x) a v (x)u(x) dx. (10) Za | − Za 2 Once again, the endpoint terms vanish because of the boundary conditions. Substitution of this result into (8) yields b Lu, v = [v′′(x)+ g(x)v(x)]u(x) dx = u, L∗v = u, Lv . (11) h i Za h i h i In other words, L∗ = L, implying that the differential operator L is self-adjoint. Note that the boundary conditions obeyed by functions in this space play an important role in making possible the self-adjointness of L. We shall see later that this self-adjointness property also applies to Sturm- Liouville differential operators. If the differential operator L on M contains a first order derivative, i.e., (Lu)(x)= a(x)u′′(x)+ b(x)u′(x)+ c(x)v(x), (12) where a, b, c are real-valued functions, then (Exercise) (L∗v)(x) = (a(x)v(x))′′ (b(x)v(x))′ + c(x)v(x). (13) − Invariant subspaces Definition 1 Let X be a vector space. If T : X X is a linear transformation and M is a linear → subspace of X such that T (M) M, then M is said to be invariant under T . ⊂ Example: Let X = L2( , ) and let I be an interval in R. Let −∞ ∞ M = x L2( , ) : x(t)=0 for t / I . (14) { ∈ −∞ ∞ ∈ } Then given the multiplication operator F defined in (4) above, F (M) M. ⊂ This idea of invariant subspaces is used to prove the following result: Theorem 1 If a bounded linear operator L is causal, then its adjoint L∗ is anticausal. Proof: See Naylor and Sell, p. 356. We present below some results that will be useful in future discussions in this course. Theorem 2 Let L be a bounded (therefore continuous) linear transformation of a Hilbert space H into itself. A closed linear subspace M of H is invariant under L if and only if M ⊥ is invariant under L∗. Proof: If M is invariant under L, then L(M) M. This means that x,Ly = 0 for all y M and ⊂ h i ∈ x M ⊥. But this means that L∗x,y = 0 for all y M and x M ⊥. Thus, L∗x M ⊥ for all ∈ h i ∈ ∈ ∈ x M ⊥, implying that L∗(M ⊥) M ⊥. If M ⊥ is invariant under L∗, the same type of argument ∈ ⊂ shows that M is invariant under L. Let H be a Hilbert space and L : H H a bounded linear operator. Recall the following definitions: → 3 1. Range of L: R(L)= Lx : x H , { ∈ } 2. Null space of L: N(L)= x H : Lx = 0 . { ∈ } The following theorem is presented as a problem in Problem Set 6 of the Course Notes. Theorem 3 Let H be a Hilbert space and L : H H a bounded linear operator with adjoint L∗. Let → N(L) and N(L∗) denote the null spaces of, respectively, L and L∗. Also let R(L) and R(L∗) denote the closures of the ranges of, respectively, L and L∗. Then R(L) = N(L∗)⊥ R(L∗) = N(L)⊥. (15) Unitary operators Definition 2 Let U : H H be a bounded linear operator on a Hilbert space H. Then U is said to → be isometric if Ux,Uy = x,y , for all x,y H. (16) h i h i ∈ If R(U)= H, then U is called a unitary operator. Theorem 4 Let U : H H be a bounded linear operator on a Hilbert space H. Then U is unitary → if and only if UU ∗ = U ∗U = I, i.e., U ∗ = U −1. Proof: If U is unitary, then U ∗Ux,y = Ux,Uy = x,y , h i h i h i for all x,y H. Therefore U ∗U = I. Similarly, UU ∗ = I. ∈ Now assume that U ∗U = I. Then x,y = U ∗Ux,y = Ux,Uy , h i h i h i for all x,y H, implying that U is unitary. ∈ Note: Setting x = y in Eq. (16) gives Ux 2 = x 2, implying that Ux = x , i.e., the operator k k k k k k k k norm of a unitary operator is 1. Examples: Delay or shift operator: Let H = L2( , ) and consider the delay or shift operator S : H H, −∞ ∞ τ → τ R, defined as ∈ (S x)(t)= x(t τ), for t ( , ). τ − ∈ −∞ ∞ −1 Note that Sτ has an inverse: Sτ = S−τ . Sτ is also a unitary operator, since ∞ ∞ Sτ x,Sτ y = y(t τ)x(t τ) dt = y(t)x(t) dt, (17) h i Z−∞ − − Z−∞ 4 for all x,y H. ∈ Note also that ∞ ∞ Sτ x,y = y(t)x(t τ) dt = y(t + τ)x(t) dt, (18) h i Z−∞ − Z−∞ for all x,y H. This implies that the adjoint of S is defined by ∈ τ (S∗y)(t)= y(t + τ), t ( , ). (19) τ ∈ −∞ ∞ Therefore, as expected (since Sτ was found to be unitary), ∗ −1 Sτ = S−τ = Sτ . (20) ∗ If we interpret Sτ , for τ > 0, as a causal operator, then Sτ is anticausal. Fourier transform on R: There are various definitions of the Fourier transform (depending upon the coefficients in front of the integrals and the sign of the exponents). Here we define the Fourier transform F (ω) of a function f L1(R) L2(R) as follows, ∈ ∩ 1 ∞ F (ω) = ( f)(ω)= e−iωxf(x) dx. (21) F √2π Z−∞ is a unitary mapping of L1(R) L2(R) onto itself.
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