SMA 5878 Functional Analysis II

SMA 5878 Functional Analysis II

Spectral Analysis of Linear Operators SMA 5878 Functional Analysis II Alexandre Nolasco de Carvalho Departamento de Matem´atica Instituto de Ciˆencias Matem´aticas and de Computa¸c˜ao Universidade de S˜ao Paulo March 27, 2019 Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Example 2 2 Let X = L (0,π) and D(A0)= C0 (0,π) the set of functions which are twice continuously differentiable functions and have compact support in (0,π). Define A0 : D(A0) ⊂ X → X by ′′ (A0φ)(x)= −φ (x), x ∈ (0,π). Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range 2 2 It is easy to see that A0 is symmetric and that hA0φ, φi≥ π2 kφkX for all φ ∈ D(A0). From Friedrichs Theorem, A0 has a self-adjoint extension A that 2 2 satisfies hAφ, φi≥ π2 kφkX for all φ ∈ D(A). 1 Note that, the space X 2 from Friedrichs theorem is, in this example the closure of D(A) in the norm H1(0,π) and therefore 1 1 X 2 = H0 (0,π). Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range On the other hand D(A∗) is characterised by ∗ ∗ ′′ ∗ D(A0)= {φ ∈ X : ∃ φ ∈ X such that h−u , φi = hu, φ i, ∀u ∈ D(A0)} ∗ ′′ ∗ and A0φ=−φ for all φ∈D(A0). 2 1 ′′ Hence, D(A)=H (0,π)∩H0 (0,π) and Au =−u for all u ∈D(A). Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range 1 Also from Friedrichs Theorem we know that (−∞, π ) ⊂ ρ(A). In particular 0 ∈ ρ(A) and if φ ∈ D(A), we have that 1 ′ 1 1 |φ(x) − φ(y)| ≤ |x − y| 2 kφ kL2(0,π) = |x − y| 2 hAφ, φi 2 . Hence, if B is a bounded subset of D(A) with the graph norm, ′ then supφ∈B kφ kL2(0,π) < ∞ and the family B of functions is equicontinuous and bounded in C([0,π], R) with the uniform convergence topology. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range It follows from the Arzel´a-Ascoli Theorem that B is relatively compact in C([0,π], R) and consequently B is relatively compact in L2(0,π). From a previous exercise it follows that A−1 is a compact operator. 2 It follows that σ(A)= {λ1, λ2, λ3, ···} where λn = n ∈ σp(A) 1 2 2 N with eigenfunctions φn(x)= π sen(nx), n ∈ . Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Min-Max Characterisation of Eigenvalues In this section we introduce min-max characterisations of eigenvalues of compact and self-adjoint operators. To present these characterisations we will employ the following result Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Lemma Let H be a Hilbert space over K and A ∈L(H) be a self-adjoint operator, then kAkL(H) = sup |hAu, vi| = sup |hAu, ui|. kuk=1 kuk=1 kvk=1 Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Proof: It is enough to prove that kAkL(H) = sup |hAu, vi| ≤ sup |hAu, ui| := a. kuk=1 kuk=1 kvk=1 If u, v ′ ∈ H, kuk = kv ′k = 1, |hAu, v ′i| eiα = hAu, v ′i and v = e−iαv ′, we have that 1 |hAu, v ′i| = hAu, vi = [hA(u + v), u + vi − hA(u − v), u − vi] 4 a ≤ [ku + vk2 + ku − vk2] ≤ a. 4 This completes the proof. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Exercise Show that, if 0 6= A ∈L(H) is self-adjoint, then A is not quasinilpotent. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Theorem Let H be a Banach space over K and A ∈ K(H) be a self-adjoint operator such that hAu, ui≥ 0 for all u ∈ H. Then, 1. λ1 :=sup{hAu, ui :kuk=1} is an eigenvalue and exists v1 ∈H, kv1k=1 such that λ1 =hAv1, v1i. Besides that Av1 =λ1v1. 2. Inductively, λn :=sup{hAu, ui:kuk=1 and u ⊥vj , 1≤j ≤n−1}∈σp(A) (1) and exists vn ∈ H, kvnk = 1, vn ⊥ vj , 1 ≤ j ≤ n − 1, such that λn = hAvn, vni. Besides that Avn = λnvn. 3. If Vn = {F ⊂ H : F is a vec. subspace of dimension n of H}, λn = inf sup{hAu, ui : kuk = 1, u ⊥ F }, n ≥ 1 and (2) F ∈Vn−1 λn = sup inf{hAu, ui : kuk = 1, u ∈ F }, n ≥ 1. (3) F ∈Vn Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Proof: We consider only the case K = C and λ1 > 0 leaving the remaining cases as exercises for the reader. n→∞ 1.Let {un} be a sequence in H with kunk=1 and hAun, uni −→ λ1. Taking subsequences if necessary, {un} converges weakly to v1 ∈ H and {Aun} converges strongly to Av1. Hence hAv1, v1i = λ1. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Let us show that the sequence {un} converges strongly. From the previous lemma we know that 0 < λ1 = kAkL(H) and from the fact that {un} converges weakly to v1 we have that 0 < kv1k≤ 1. Hence, 2 2 2 lim kAun − λ1unk = lim kAunk − 2λ1 lim hAun, uni + λ n→∞ n→∞ n→∞ 1 2 2 = kAv1k − λ1 ≤ 0. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Since {Aun} converges strongly to Av1, {Aun − λ1un} converges strongly to zero and λ1 > 0, it follows that {un} converges strongly to v1, kv1k = 1 and Av1 = λ1v1. This concludes the proof of 1. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range 2. The proof of this item follows from 1. simply noting that the orthogonal of Hn−1 = span{v1, · · · , vn−1} is invariant by A and ⊥ repeating the procedure for the restriction of A to Hn−1, n ≥ 2. This concludes the proof of 2. Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range 3. We first prove expression (2). If G = span{v1, · · · , vn−1} we have, from (1), that λn = sup{hAu, ui : kuk = 1, u ⊥ G} ≥ inf sup{hAu, ui : kuk = 1, u ⊥ F }. F ∈Vn−1 Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range On the other hand, let F ∈Vn−1 and w1, · · · , wn−1 an n orthornormal set of F . Choose u = Pi=1 αi vi such that kuk = 1 n 2 and u ⊥ wj , 1 ≤ j ≤ n − 1. Hence Pi=1 |αi | = 1 and n 2 hAu, ui = X |αi | λi ≥ λn. i=1 Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range This implies sup{hAu, ui : kuk = 1, u ⊥ F }≥ λn, for all F ∈Vn−1 and completes the proof of (2). Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range We now prove (3). If G = span{v1, · · · , vn} and u ∈ G, kuk = 1, n n 2 we have that u = Pi=1 αi vi with Pi=1 |αi | = 1 e n 2 hAu, ui = X |αi | λi ≥ λn. i=1 This implies that sup inf{hAu, ui : kuk = 1, u ∈ F }≥ λn. F ∈Vn Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Conversely, given F ∈Vn choose u ∈ F , kuk = 1, such that u ⊥ vj , 1 ≤ j ≤ n − 1. It follows, from 2., that hAu, ui≤ λn and consequently inf{hAu, ui : kuk = 1, u ∈ F }≤ λn, for all F ∈Vn. This completes the proof of (3). Alexandre Nolasco de Carvalho ICMC - USP SMA 5878 Functional Analysis II Min-Max Characterisation of Eigenvalues Spectral Analysis of Linear Operators Dissipative operators and numerical range Exercise If A : D(A) ⊂ H → H is positive, self-adjoint and (hAu, ui > 0 for all u ∈ D(A)) and has compact resolvent, find the min-max characterisation for the eigenvalues of A.

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