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• 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.

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The spectral theory for compact self adjoint operators

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