MTL 411: Functional Analysis Lecture D: Spectral theory April 18, 2020 Let T : V ! V be a linear operator on a vector space V over C. We say that a scalar λ 2 C is said to be an eigenvalue of T if there exists a non-zero vector x 2 V such that T x = λx. If V is finite dimensional space, then T has atleast one eigenvalue (over C). But in the infinite dimensional case, this may not be true. For example, consider the right shift operator 2 T (x1; x2;:::) = (0; x1; x2;:::) on ` . It is easily verified that if T x = λx, then x = 0: Hence T has no eigenvalues. So the notion of eigenvalue should be treated in more general way which is called spectrum. Exercise. Show that the Volterra operator T : C[0; 1] ! C[0; 1] is defined by x Z (T f)(x) = f(t)dt; x 2 [0; 1] 0 has no eigenvalues. Remark. A scalar λ is an eigenvalue of an operator T iff N (T ) 6= f0g. Recall, if λ is an eigenvalue of an operator T on a finite dimensional space V . Then λI −T is not invertible on V . Alternatively, if λ is not an eigenvalue of T , then λI − T is invertible on V since that 1-1 implies onto in the finite dimensional space. Moreover, (λI − T )−1 is bounded linear operator because of every linear operator is bounded in the finite dimensional space. Of course, this may not be true in the infinite dimensional space. Throughout this section we assume that X is a Banach space and T 2 B[X]: Definition 0.1 (Resolvent). The resolvent set ρ(T ) of T is defined by −1 ρ(T ) = fλ 2 C :(λI − T ) exists and bounded operator g: Here I denotes the identity operator on X. Definition 0.2 (Spectrum). The spectrum of T (denoted by σ(T )) is defined by σ(T ) = C n ρ(T ) = fλ 2 C : N (λI − T ) 6= f0g or R(λI − T ) 6= Xg: Lemma 0.3. If jjT jj < 1, then I − T is invertible, and 1 X (I − T )−1 = T n: (Neumann series) n=0 1 Pn k Proof. Consider the partial sum Sn = k=0 T . Then we have n X k jjSn − Smjj = jj T jj (n > m) k=m+1 n X ≤ jjT jjk: k=m+1 Pn k Since jjT jj < 1, we get the sequence of operators (Sn) is Cauchy in B[X]. Hence Sn = k=0 T P1 k converges in B[X], say S = k=0 T . It is easy to verify that (I − T )S = S(I − T ) = I. Theorem 0.4 (Properties). (a) If jλj > jjT jj, then λ 2 ρ(T ): 1 (b) If T is invertible and jjSjj < jjT −1jj , then T − S has the bounded inverse. (c) The resolvent set ρ(T ) is open, and the spectrum σ(T ) is compact. T T Proof. (a) If jλj > jjT jj, then jj λ jj < 1. Hence by above lemma, we get I − λ has the bounded inverse =) (λI − T )−1 2 B[X], i.e., λ 2 ρ(T ): (b) Since jjST −1jj ≤ jjSjj jjT −1jj < 1, we get I − ST −1 has the bounded inverse. Then T − S = (I − ST −1)T . So T − S has the bounded inverse. (c) Let λ 2 ρ(T ). We have to show λ is an interior point of ρ(T ), i.e., there exists a δ > 0 such that λ − µ 2 ρ(T ) whenever jµj < δ: 1 Choose δ = jj(λI−T )−1jj : Then by above result (b), we get (λ−µ)I −T has the bounded inverse whenever jµj < δ: Thus λ − µ 2 ρ(T ) whenever jµj < δ: Hence ρ(T ) is open in C and it's complement σ(T ) is closed and also bounded (why?). Therefore, σ(T ) is a compact subset of C: Remark. The spectrum σ(T ) is nonempty. The proof involves complex analysis techniques, we omit the proof here. Examples. 2 (1) Right shift operator: T (x1; x2;:::) = (0; x1; x2;:::), x 2 ` . Since jjT jj = 1, the spectrum σ(T ) ⊆ fλ 2 C : jλj ≤ 1g: For jλj ≤ 1; (λI − T )x = (λx1; λx2 − x1; λx3 − x2;:::): 1 1 1 2 Observe that if w = ( λ ; λ2 ; λ3 ;:::) then (λI − T )w = (1; 0; 0;:::) and w2 = ` . So λI − T is not onto for jλj ≤ 1. Hence σ(T ) = fλ 2 C : jλj ≤ 1g: x2 x3 2 (2) Diagonal operator: T (x1; x2;:::) = (x1; 2 ; 3 ;:::), x 2 ` . 1 1 1 Since ( n I −T )en = 0, we get N (( n I −T )) 6= f0g for every n 2 N. Hence n 2 σ(T ): For λ = 0, 1 T does not have bounded inverse (why?). On the other hand, for λ2 = f n : n 2 Ng [ f0g, we 1 can show that λ2 = σ(T ): Therefore, σ(T ) = f0g [ f n : n 2 Ng: Exercise. Consider the Multiplication operator: T : f 7! mf from the space (C[a; b]; jj · jj1) to itself, where m 2 C[a; b]: Show that the spectrum σ(T ) = m([a; b]): 2.