Spectral Theory
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Spectral Theory Kim Klinger-Logan November 25, 2016 Abstract: The following is a collection of notes (one of many) that compiled in prepa- ration for my Oral Exam which related to spectral theory for automorphic forms. None of the information is new and it is rephrased from Rudin's Functional Analysis, Evans' Theory of PDE, Paul Garrett's online notes and various other sources. Background on Spectra To begin, here is some basic terminology related to spectral theory. For a continuous linear operator T 2 EndX, the λ-eigenspace of T is Xλ := fx 2 X j T x = λxg: −1 If Xλ 6= 0 then λ is an eigenvalue. The resolvent of an operator T is (T − λ) . The resolvent set ρ(T ) := fλ 2 C j λ is a regular value of T g: The value λ is said to be a regular value if (T − λ)−1 exists, is a bounded linear operator and is defined on a dense subset of the range of T . The spectrum of is the collection of λ such that there is no continuous linear resolvent (inverse). In oter works σ(T ) = C − ρ(T ). spectrum σ(T ) := fλ j T − λ does not have a continuous linear inverseg discrete spectrum σdisc(T ) := fλ j T − λ is not injective i.e. not invertibleg continuous spectrum σcont(T ) := fλ j T −λ is not surjective but does have a dense imageg residual spectrum σres(T ) := fλ 2 σ(T ) − (σdisc(T ) [ σcont(T ))g Bounded Operators Usually we want to talk about bounded operators on Hilbert spaces. Recall that a Hilbert space is a vector space with an inner product space that is complete with respect to the distance function induced by the metric. We can sometimes deduce nice properties for operators on Banach spaces (a complete normed metric space) when the operator satisfies more specific conditions but most of the time we will stick with Hilbert spaces. 1 A linear map T : X ! Y is bounded if for all > 0 there is a δ > 0 such that if jxjX < δ then jT xjY < . For a linear map of Hilbert spaces, the following are equivalent: (i) continuous (ii) continuous at 0 (iii) bounded. ∗ ∗ The adjoint of a map T : X ! Y is the map T : Y ! X so that hT x; yiY = hx; T yiX : Any continuous linear map T have a unique adjoint T ∗ and T ∗∗ = T . For a self-adjoint maps, all eigenvalues are real. If T 2 EndX and T ∗T = TT ∗ then T is normal. For a normal operator there is no residual spectrum{it is all discrete or continuous. Any normal operator has σres(T ) = ;. A normal operator is unitary if T ∗T = TT ∗ = I. Compact Operators All compact operators are bounded operators. Thus everything that has previously been said still applies. There are many ways to define a compact operator T : X ! Y : • maps the unit ball in X to a precompact (has compact closure) set in Y • maps bounded subsequences in X to sequences in Y with convergent subsequences. Let fTng be a sequence of compact operators on a normed linear space X. Suppose that Tn ! T (in the space of bounded operators). Then T is also compact. For compact operators we do not need to restrict the domain to Hilbert spaces in order to have some spectral theorem. We have a spectral theorem for compact operators on Banach spaces. One method of proof involves the following two Lemmas: Reisz's Lemma: Let X be a Banach space and Y ⊂ X a proper closed subspace. For all > 0 there is an x 2 X with jjxjj = 1 and 1 ≥ d(x; Y ) ≥ 1 − (We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaces where there is no notion of orthogonality.) proof: Let x1 2= Y and set R := infy2Y jy − x1j. Note that R > 0. For > 0 let y1 2 Y so that jx1 − y1j < R + . x − y Set x := 1 1 and note that jjxjj = 1 and y1 − x1 x1 − y1 y1 x1 infy2Y jy − x1j R inf jy − xj = inf y − = inf y + − = = y2Y y2Y y1 − x1 y2Y y1 − x1 y1 − x1 jy1 − x1j R + We can make this quotient arbitrarily close to 1 by taking smaller . 2 We can think of this as the analogue to orthogonality in Hilbert spaces but for Banach spaces where there is no notion of orthogonality. Lemma 2: If T is compact then Im(T − I) is closed. We can also use the Fredholm Alternative to prove the first part of the Spectral Theorem for Compact Operators. Fredholm Alternative: Let T : X ! X be a compact operator on Banach spaces and λ 6= 0, either (A) T − λ is a bijection or (B) Im(T − λ) is closed and dim (X=Im(T − λ)) = dim (ker(T − λ)) : We can also use Reisz' Lemma to prove the Fredholm Alternative. Spectral Theorem for Compact Operators on Banach Spaces: Let T be a compact on an infinite-dimensional Banach space then (1) the nonzero spectrum is discrete (for both T and T ∗) (2) λi ! 0 as i ! 1 (and this is the only accumulation point) (3) the spectrum is countable (4) the number of eigenvalues outside a disk jλj ≤ r is finite for r ≥ 0 proof: (1) This follows nicely from the Fredholm Alternative: Suppose that λ 6= 0 is in σ(T ) but is not an eigenvalue for T . Then T − λ is injective. Thus dim (ker(T − λ)) = 0 and by the Fredholm Alternative, dim (X=Im(T − λ)) = 0. Thus T − λ is surjective and so λ must be in the resolvent. We can also prove this result using Reisz' Lemma: The idea of the proof is by contradiction. Also WLOG, assume λ = 1. Assume λ 2 σ(T ) is not an eigenvalue. Thus T − λ = T − I is injective but not surjective. Since T is compact, by Lemma 2, Im(T − I) is a closed proper subspace of X. Call this Y1 := Im(T − I). Since T − I is injective, Y2 := (T − I)Y1 is a closed proper subspace of Y1. n Create a strictly decreasing sequence of Yn := Im(T − I) so Y1 ⊃ Y2 ⊃ · · · ⊃ Yn ⊃ · · · ⊃ Ym ⊃ ::: Now apply Reisz's Lemma for Y := Yn+1 and = 1=2: Choose yn 2 Yn so that jjynjj = 1 and d(yn;Yn+1) > 1=2: Also note that by compactness of T , fT yng must contain a norm convergent subsequence. 3 But for n < m, jjT yn − T ymjj = jj(T − I)yn + yn − (T − I)ym − ymjj and (T −I)yn −(T −I)ym −ym 2 Yn+1 and thus jjT yn −T ymjj > 1=2 which is our contradiction. Thus λ must be an eigenvalue. (2) This will also proceed by contradiction using Reisz' Lemma. Assume for the sake of contradiction that there are infinitely many distinct eigenvalues λn outside a ball centered at the origin of radius and each λn has associated eigenvector xn. Define Yn := spanfx1; x2; : : : ; xng. This gives a strictly increasing sequence Y1 ⊂ Y2 ⊂ · · · ⊂ Yn ⊂ · · · ⊂ Ym ⊂ ::: Now apply Reisz' Lemma for Y = Yn−1 and = 1=2: Choose yn 2 Yn so that jjynjj = 1 and d(yn;Yn−1) > 1=2. Again by compactness of T , fT yng must contain a norm convergent subsequence. But for n < m, jjT yn − T ymjj = jj(T − λn)yn + λnyn − (T − λm)ym + λmymjj Pm Pm−1 where (T − λn)yn + λnyn − (T − λm)ym 2 Ym−1 since ym = i=1 cixi where i=1 cixi 2 Ym−1 and (T − λm)xm = 0. Thus jjT yn − T ymjj > /2. This is a contradiction. Thus there are only finitely many eigenvalues outside a ball centered at 0. This gives us that zero is the only accumulation point. (3) From (2), [ fλng = fjλj > 1=ng n where each of these sets is finite. Thus σ(T ) is countable. Let T : X ! Y be a compact operator where X is a Banach space and Y is a Hilbert space, Then T is the limit (in operator norm) of a sequence of finite-dimensional oper- ators. More can be said about the operators on Hilbert spaces{especially when they are self-adjoint. For self-adjoint operators (or even symmetric) on Hilbert spaces all of the eigenvalues are real: Let T be a self-adjoint operator on a Hilbert space with eigenvalue λ so that T x = λx. Then λhx; xi = hλx, xi = hT x; xi = hx; T xi hT x; xi = hλx, xi = λhx; xi Thus λ = λ and so λ 2 R. 4 Spectral Theorem for Self-Adjoint Compact Operators on a Hilbert Space: Let T be a compact self-adjoint operator on a Hilbert space H, then L • H has an orthonormal basis vi of eigenvectors of T and H = the completion of λ Hλ. • λi ! 0 as i ! 1 (and this is the only accumulation point) • every eigenspace Hλ is finite-dimensional • (Ralleigh-Ritz) either ±|T jop is an eigenvalue. Recall that jT jop := inffc ≥ 0 j jT xj ≤ cjxj for all x 2 Xg. Example: Finite-dimensional Let T : X ! Y be a continuous linear mapping between normed spaces. If Ran(T ) has finite dimension then T is called a finite-dimensional operator. Finite dimensional operators are compact. For a bounded set B ⊆ X, T (B) is closed and bounded in the finite-dimensional subspaces Ran(T ) ⊆ Y . Heine-Borel says that T (B) is compact in Ran(T ). Example: Hilbert-Schmidt Hilbert Schmidt Theorem: Let X be a locally compact space endowed with a positive Borel measure and assume that L2(X) is a separable Hilbert space.