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SPECTRAL THEOREM for SYMMETRIC OPERATORS with COMPACT RESOLVENT 1. Compact Operators Let X, Y Be Banach Spaces. a Linear Operato

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SPECTRAL THEOREM for SYMMETRIC OPERATORS with COMPACT RESOLVENT 1. Compact Operators Let X, Y Be Banach Spaces. a Linear Operato SPECTRAL THEOREM FOR SYMMETRIC OPERATORS WITH COMPACT RESOLVENT Abstract. These are the letcure notes prepared for the workshop on \Functional Analysis and Operator Algebras" to be held at NIT-Karnataka, during 2-7 June 2014. 1. Compact Operators Let X; Y be Banach spaces. A linear operator T from X into Y is said to be compact if for every bounded sequence fxng in X, fT xng has a convergent subsequence. Lemma 1.1. Compact operators form a subspace of B(X; Y ): Proof. Clearly, a scalar multiple of a compact operator is compact. Suppose S; T are com- pact operators and let fxng be a bounded sequence. Since S is compact, there exists a convergent subsequence fSxnk g of fSxng: Since fxnk g is bounded, by the compactness of T; we get a convergent subsequence fT xn g of fT xng: Thus we obtain a convergent kl subsequence f(S + T )xn g of f(S + T )xng; and hence S + T is compact. kl Remark 1.2 : If S is compact and T is bounded then ST; T S are compact. Thus compact operators actually form an ideal of B(X; Y ): Exercise 1.3 : Show that every finite-rank bounded linear map T : X ! Y is compact. Pk Hint. Note that there exist vectors y1; ··· ; yk 2 Y such that T x = i=1 λi(x)yi; where λ1; ··· ; λk are linear functionals on X. If fxng is bounded then so is each fλi(xn)g: Now apply Heine-Borel Theorem. The following proposition provides a way to generate examples of compact operators which are not necessarily of finite rank. Proposition 1.4. Compact operators form a closed subspace of B(X; Y ): Proof. In view of Lemma 1.1, it suffices to check that the space of compact operators is sequentially closed. Given > 0; find an integer N ≥ 1 such that kT − TN k < . Now find a convergent sequence fTN xnk g using compactness of TN : Check that fT xnk g is a Cauchy sequence. Since B(X; Y ) is complete, fT xnk g is convergent as required. Exercise 1.5 : Let a := fang be a bounded sequence. Show that the diagonal operator Da 2 on l with diagonal entries a1; a2; ··· is compact iff fang 2 c0: Hint. If Db is a diagonal operator with diagonal entries b1; b2; ··· ; then kDbk = supn jbnj: The sufficient part follows from limn!1 kDa;n − Dak = 0, where Da;n is the diagonal oper- ator with diagonal entries a1; ··· ; an; 0; 0; ··· : To see the necessary part, WLOG, assume 2 2 2 that fang is bounded from below, and note that kDaen − Daemk = an + am > c > 0 for a scalar c: 2 Example 1.6 : Let a := fang 2 c0 and let en denote the sequence in l with nth entry 1 and all other entries equal to 0: Define Uaen = anen+1, and extend Ua linearly and continuously 2 to l : Then Ua is compact. This follows from Remark 1.2 and the preceding exercise in view 2 of the decomposition Ua = UDa; where U 2 B(l ) is governed by Uen = en+1: The integral operators provide the most interesting examples of compact operators, which are of great importance as far as applications are concerned. 1 2 SPECTRAL THEOREM FOR SYMMETRIC OPERATORS R x Exercise 1.7 : Consider (T f)(x) = 0 f(y)dy (x 2 [0; 1]) and n−1 X Z k=n (Tnf)(x) = χ[k=n;(k+1)=n)(x) f(y)dy (x 2 [0; 1]): k=0 0 2 2 −1=2 Verify that T 2 B(L (0; 1);L (0; 1)) satisfies kT −Tnk ≤ n : Conclude that T is compact. n+m Exercise 1.8 : For t 2 C0(R ) (continuous with compact support), define a linear 2 m 2 n operator Tt : L (R ) ! L (R ) by Z 2 m n (Ttu)(x) = t(x; y)u(y)dy; u 2 L (R ); x 2 R : Rm Verify the following: (1) Tt is a bounded, linear operator. (2) Let fukg be a bounded sequence with bound M: Then m=2 j(Ttuk)(x)j ≤ ktk1(2c) M; where c is chosen so that t(x; y) = 0 if kxk2 > c or kyk2 > c: (3) Given > 0; there exists δ 2 (0; 1) such that 0 m=2 j(Ttuk)(x) − (Ttuk)(x )j ≤ (2c) M 0 whenever kx − x k2 < δ: (4) Tt is a compact operator. Hint. Recall the Arzela-Ascoli Theorem. Exercise 1.9 : The conclusion of the last exercise holds true for any measurable function 2 n+m t in L (R ): n+m 2 n+m n+m Hint. Since C0(R ) is dense in L (R ); there exists tk 2 C0(R ) such that ktk − tkL2(Rn+m) ! 0 as k ! 1: If Ttk is the integral operator associated with the kernel tk then kTtk − Ttk ≤ ktk − tkL2(Rn+m): Exercise 1.10 : Consider the linear operator A : L2(0; 1) ! L2(0; 1) given by 1 Z x (Af)(x) = f(t)dt (x 2 (0; 1)): x 0 Show that A is bounded linear but not compact. 2 Hint. Consider fn(t) = n if 0 < t ≤ 1=n ; and 0 otherwise. Note that kAfnk ≥ 1 for all ∗ 2 n: However, hAfn; gi = hfn;A gi ! 0 for all g 2 L (0; 1): One of the objectives of these notes is to give an elementary proof of spectral theorem for compact normal operators. This result is usually attributed to Hilbert and Schmidt. Although our treatment is along the lines of [1, Chapter 2], for simplicity, we will work only with complex infinite-dimensional separable Hilbert spaces. The separability assumption makes life simple in many instances (see, for example, Exercise 2.6). Exercise 1.11 : Show that Lp(0; 1) is separable for 1 ≤ p < 1: Recall that C[0; 1] is dense in Lp(0; 1) for 1 ≤ p < 1: By the Weierstrass Approximation Theorem, continuous functions can be approximated in Lp(0; 1) by polynomials, and hence by polynomials with rational coefficients. 2. Spectral Theorem for Compact Normal Operators We begin with some elementary properties of compact operators. Proposition 2.1. Let T be a compact operator with closed range ran(T ). Then the closed unit ball in ran(T ) is sequentially compact. In particular, ran(T ) is finite-dimensional. SPECTRAL THEOREM FOR SYMMETRIC OPERATORS 3 Proof. Let yn 2 ran(T ) be such that kynk ≤ 1: Consider the bounded linear surjection S : X ! ran(T ) given by Sx = T x for x 2 X: By the Open Mapping Theorem, there exists xn 2 X such that yn = T xn for some bounded sequence fxng: Since T is compact, fyng has a convergent subsequence, and the unit ball in ran(T ) is sequentially compact. The remaining part follows from the fact that a Banach space X is finite-dimensional iff the unit ball in X is sequentially compact. Corollary 2.2. Let µ be a non-zero complex number and T : X ! Y be compact. Then ker(T − µ) is finite-dimensional. Proof. If M is a closed subspace of X such that TM ⊆ M then T jM is compact. Clearly, T jker(T −µ) is a compact operator with closed range ker(T −µ): Now apply the last result. Exercise 2.3 : Let T : X ! X be a compact operator. Show that either X is finite dimensional or T is non-surjective. Definition 2.4 : A bounded linear operator on a Hilbert space H is normal if T ∗T = TT ∗: We say that T is self-adjoint if T ∗ = T: 1 2 ∗ ∗ The operator of multiplication by φ 2 L (0; 1) on L (0; 1) is normal. In fact, Mφ = Mφ¯; ¯ where φ(z) = φ(z): In particular, Mφ is self-adjoint iff φ is real-valued. Exercise 2.5 : If N is normal then show that so is N − λ for any scalar λ. Use this to deduce that the eigenspaces corresponding to distinct eigenvalues are orthogonal. Hint. First part follows from the definition. Check that M is normal iff kMhk = kM ∗hk for every h 2 H: Letting M := N − λ, we obtain Nx = λx iff N ∗x = λx.Finally, if Nx = λx and Ny = µy then λhx; yi = hNx; yi = hx; N ∗yi = µhx; yi: Exercise 2.6 : Let X be a separable inner-product space. Show that any orthonormal subset of X is countable. Conclude that the point-spectrum (that is, the set of eigenvalues) of any normal operator on X is countable. p Hint. Suppose fxαg is orthonormal. Then fB(xα; 1= 2)g is a collectionp of disjoint open balls in H: Now if fyng is countable dense subset of H then each B(xα; 1= 2) must contain at least one yn: Exercise 2.7 : Consider the linear operator T on L2[0; 1]: Z x Z 1 (T f)(x) = (1 − x) yf(y)dy + x (1 − y)f(y)dy (x 2 [0; 1]): 0 x Verify the following: (1) T 2 B(L2(0; 1);L2(0; 1)) is compact and self-adjoint. (2) If T f = λf then for some integer n ≥ 1; f(x) = c sin(nπx) for some scalar c and λ = 1=n2π2. Hint. For (1), use Exercise 1.7 and Remark 1.2. For (2), note that (T f)(0) = 0 = (T f)(1); and λf 00 = (T f)00 = −f a.e. Exercise 2.8 : Show that for any normal operator N on a Hilbert space, norm kNk of N is equal to the spectral radius r(N) of N: Hint.
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