Docstoc

The spectral theory for compact self adjoint operators

Document Sample
The spectral theory for compact self adjoint operators Powered By Docstoc
					   • The spectral theory for compact, self-adjoint operators

        – Spectrum
        – Compact operators
        – The spectral theorem
        – Hilbert-Schmidt operators
        – Functions of operators

   Definitions:
In finite dimensions:

   • eigenvalue, eigenvector, diagonalizable, spectrum, normal matrix (n ×
     n matrix is normal iff Cn has an orthonormal basis of eigenvectors),
     multiplicity
Bounded linear operators on infinite-dimensional Hilbert spaces:

   • Resolvant Set: A ∈ B(H), ρ(A) = {λ ∈ C|(A−λI) : H → H is one-to-one and onto.}

   • Spectrum: A ∈ B(H), σ(A) = C\ρ(A).

   • A ∈ B(H),

        – The point spectrum of A is all λ ∈ σ(A) such that A − λI is not
          one-to-one, and λ is an eigenvalue of A.
        – The continuous spectrum of A is all λ ∈ σ(A) such that A − λI is
          one-to-one but not onto and ran(A − λI) is dense in H.
        – The residual spectrum is λ ∈ σ(A) such that A − λI is one-to-one
          but not onto and ran(A − λI) is not dense in H.

   • If λ ∈ ρ(A) then A−λI has an everywhere defined and bounded inverse,
     Rλ = (λI−A)−1 is called the resolvant of A, which is an operator-valued
     function defined on ρ(A) ∈ C.

   • The spectral radius of A, r(A), is the radius of the smallest disk centered
     at zero which contains σ(A),

                            r(A) = sup{|λ| | λ ∈ σ(A)}.
• A linear subspace M of H is called an invariant subspace of a linear
  operator A on H if Ax ∈ M for all x ∈ M.

• A bounded linear operator A on a separable Hilbert space H is Hilbert-
  Schmidt if there is an ONB {en | n ∈ N} such that
                                ∞
                                            2
                                      Aen       < ∞.
                                n=1

  Furthermore, if A is a Hilbert-Schmidt operator then

                                            ∞
                            A         =           Aen    2
                                HS
                                            n=1


  is called the Hilbert-Schmidt norm of A.

Theorems:

• If A ∈ B(H), then ρ(A) is an open subset of C that contains the
  exterior disc {λ ∈ C | |λ| > A }. The resolvant Rλ is an operator
  valued analytic function of λ defined on ρ(A).
  For any λ0 ∈ ρ(A) write

              λI − A = (λ0 I − A)[I − (λ0 − λ)( 0I − A)−1 ].

  Then for small enough |λ0 − λ| invert the operator on RHS above by
  the Neumann series and the resolvant is given by an operator-norm
  convergent Taylor series in the disc so it is analytic where defined. If
  | > A then the same Neumann series shows that Rλ = λ(I − A/λ)
  is invertible so λ ∈ ρ(A).

• If A ∈ B(H) then
                            r(A) = lim An          1/n
                                      n→∞

  and if A is self-adjoint then r(A) = A .

• The spectrum of a bounded linear operator on a Hilbert space is nonempty
  (may just be zero, if r(A) = 0 then A is called nilpotent.)
  Proof uses Liouville’s theorem from complex analysis which states that
  a bounded entire function is constant. If A ∈ B(H) then Rλ the re-
  solvant is an analytic function on the resolvant set. Define f (λ) =
   x, Rλ y , for all x, y ∈ H. then f is analytic in ρ(A) and limλ→∞ f (λ) =
  0. If σ(A) is empty f is a constant function but by it’s limiting value
  it must be zero. This would imply Rλ = 0 for all λ ∈ C which is a
  contradiction, so σ(A) is not empty.

• The eigenvalues of a bounded, self-adjoint operator are real, and eigen-
  vectors associated with different eigenvalues are orthogonal.

                   λ x, x = x, Ax = Ax, x = λ x, x

  and

                    λ x, y = Ax, y = x, Ay = µx, y

  So if λ = µ then x, y = 0

• If A is a bounded, self-adjoint operator on H and M is an invariant
  subspace of A then M⊥ is also invariant.
  If x ∈ M⊥ and y ∈ M then

                            y, Ax = Ay, x = 0

  so Ax ∈ M⊥ .

• If λ is in the residual spectrum of a bdd operator A on a Hilbert space
  then λ is an eigenvalue of A∗ .
  This follows from a general theorem about bounded linear operators,
  namely
                  ranA = (kerA∗ )⊥ , kerA = (ranA∗ )⊥ .

• If A is bdd, self-adjoint on a Hilbert space then the spectrum is real
  and contained in the interval [− A , A ].

• The residual spectrum of a bounded, self-adjoint operator is empty.
• A non-zero eigenvalue of a compact, self-adjoint operator has finite
  multiplicity. A countably infinite set of nonzero eigenvalues has zero as
  an accumulation point and no other accumulation points.
  If λ = 0 has infinite multiplicity then there is a sequence (en ) of or-
  thonormal eigenvectors, which is bounded but (Aen ) has no convergent
  subsequence since Aen = λen and this contradicts that A is compact.
  Furthermore if A has a countably infinite set {λn } of non-zero eigen-
  values then since they are bdd by A there is a convergent subseq.
  (λnk ). If λnk → λ = 0 then the orthogonal sequence of eigenvectors
  ((fnk ) where fnk = λ−1 enk with enk = 1) would be bdd. But (Afnk )
                       nk
  has no convgt subsequence since Afnk = enk .

• Spectral theorem for compact, self-adjoint operators: Let A : H → H
  be a compact, self-adjoint operator on a Hilbert space H. There is an
  orthonormal basis of H consisting of eigenvectors of A. The nonzero
  eigenvalues of A form a finite or countably infinite set {λk } or real
  numbers and
                              A=       λk Pk
                                     k

  where Pk is the orthogonal projection onto the finite-dimensional eigenspace
  of eigenvectors with eigenvalue λk . If the number of nonzero eigenval-
  ues is countably infinte then the series above converges to A in the
  operator norm .
  Idea: First prove that if A is compact and self-adjoint then λ = A
  or − A (or both) is an eigenvalue of A. Do this by writing

                            A = sup | x, Ax |,
                                   x =1


  and look at sequence leading to the sup. By self-adjoint xn , Axn is
  real so taking a further subsequence have

                             lim xn , Axn = λ
                            n→∞

  where λ = ± A . The sequence is bounded and so by compactness of A
  there is a subsequence such that (Axn ) converges. Let y = limn→∞ Axn .
  Now show that y is an eigenvector of A with eigenvalue λ.
  Compute:
                    2                                 2
     (A − λI)y          = lim (A − λI)Axn
                          n→∞
                             2                                 2
                        ≤ A      lim (A − λI)xn
                                 n→∞
                                                 2
                        = A lim        Axn           + λ 2 xn      2
                                                                       − 2λ xn , Axn
                              n→∞
                              2              2            2
                        ≤ A      lim    A        xn           + λ 2 xn   2
                                                                             − 2λ xn , Axn   .
                                n→∞
                              2
                        = A       2
                                 λ + λ2 − 2λ2 .

  Finish the proof by using an orthogonal decomposition of H into in-
  variant subspaces and apply the above result to smaller and smaller
  subspaces. Using previous results the eigenvalues have finite multiplic-
  ities and limn→∞ λn = 0 so can construct the series
                                        n
                                 A=          λk Pk + An+1 ,
                                       k=1

  where An+1 is zero on the subspace spanned by the previously con-
  structed {e1 , ...., en }. Furthermore An+1 = |λn+1 | → 0.
  Generalizations for compact, normal operators hold, eigenvectors are
  orthogonal but may be complex. Also can generalize for bounded,
  self-adjoint or normal operators. There the sum must be replaced by
  an integral with respect to an appropriate measure to account for the
  possibility of a continuous spectrum.
• To show an operator is compact, one can verify directly that if E is
  bounded in H then A(E) is precompact (using Arzela-Ascoli or Rellich’s
  theorem or... )
  other criteria:
• Let E be a subset of an infinite dimensional separable Hilbert space.
  If E is precompact, then for every orthonormal set {en | n ∈ N]} and
  every ε > 0 there is an N such that

                        sum∞ +1 | en , x |2 < ε for allx ∈ E.
                           n=N

  Also if E is bounded and there is an ONB such with the abover property
  then E is precompact.
  first show if E is bounded and the condition doesn’t hold then E is not
  precompact. Construct an explicit subsequence using Parseval. For the
  second part use a diagonalization argument to contruct a convergent
  subsequence showing precompactness.

• Hilbert-Schmidt operators are compact.
                            2             2
  Diagonal operators A :        (N) →         (N) defined by

                       A(x1 , x2 , ...) = (λ1 x1 , λ2 x2 , ...)

  with λn ∈ C are compact iff λn → 0 as n → ∞. Any compact, normal
  operator on a separable Hilbert space is unitarily equivalent to such a
  diagonal operator. A diagonal operator is H-S iff

                                sum∞ |λn |2 < ∞.
                                   n=1

  An operator A is called trace class if
                                    ∞
                                         |λn | < ∞.
                                   n=1

  A trace class operator is Hilbert-Schmidt and a Hilbert-Schmidt oper-
  ator is compact.

• A BLO on a H-space is compact iff it maps weakly convergent sequences
  into strongly convergent sequences.

• Spectral Mapping: If A is a compact, self-adkoint operator on a Hilbert
  space and f : σ(A) → C is continuous then

                                σ(f (A)) = f (σ(A)).
Problems:

1. Suppose that P : H → H is a bounded, not necessarily self-adjoint
   projection (meaning that P 2 = P ) and P = 0, I. For λ ∈ C, compute
   (λI − P )−1 explicitly when it exists, and show that the spectrum of P
   is the set {0, 1}. HINT. Consider the series expansion of (λI − P )−1 .

2. Let U = 2 (Z) → L2 (T) be the unitary map from a complex sequence
   c = (cn )n∈Z to the function f : T → C whose sequence of Fourier
   coefficients is c :
                                     1
                         (U c)(x) = √        cn einx .
                                      2π n∈Z
                                  2           2
  Define the right-shift map S :       (Z) →       (Z) by

             Sc = b, where c = (cn ), b = (bn ), and bn = cn−1 .

  Define the multiplication operator M : L2 (T) → L2 (T) by

                            (M f )(x) = eix f (x).

   (a) Show that S = U −1 M U.
   (b) Determine the spectrum of M, and deduce the spectrum of S.
       Classify the spectrum.

3. (W’08) Suppose that A : H → H is a bounded linear operator on a
   (complex) Hilbert space H with spectrum σ(A) ⊂ C and resolvant set
   ρ(A) = C\σ(A). For µ ∈ ρ(A), let

                           R(µ, A) = (µI − A)−1

  denote the resolvant operator of A.

   (a) If µ ∈ ρ(A) and
                                            1
                              |ν − µ| <           ,
                                          R(µ, A)
        prove that ν ∈ ρ(A) and

                    R(ν, A) = [I − (µ − ν)R(µ, A)]−1 R(µ, A).
   (b) If µ ∈ ρ(A), prove that
                                                   1
                             R(µ, A) ≥
                                              d(µ, σ(A))

       where
                           d(µ, σ(A)) = inf |µ − λ|
                                             λ∈σ(A)

       is the distance of µ from the spectrum of A.

4. (W ’07) Let U be a unitary operator on a Hilbert space. Prove that
   the spectrum of U lies on the unit circle.

5. (W’05) Let 2 (Z) denote the complex Hilbert space of sequences xn ∈ C,
   n ∈ Z, such that
                                 ∞
                                      |xn |2 < ∞.
                             n=−∞
                                 2           2
  Define the shift operator S :       (Z) →       (Z) by

                             S((xn )) = (xn+1 ).

  Show that S has no eigenvalues.

				
DOCUMENT INFO
Shared By:
Categories:
Stats:
views:16
posted:7/2/2010
language:English
pages:8
Description: The spectral theory for compact self adjoint operators