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 {xn} in X, {T xn} 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 compact operators and let {xn} be a bounded sequence. Since S is compact, there exists a convergent subsequence {Sxnk } of {Sxn}. Since {xnk } is bounded, by the compactness of T, we get a convergent subsequence {T xn } of {T xn}. Thus we obtain a convergent kl subsequence {(S + T )xn } of {(S + T )xn}, and hence S + T is compact. kl Remark 1.2 : If S is compact and T is bounded then ST,TS 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 ∈ Y such that T x = i=1 λi(x)yi, where λ1, ··· , λk are linear functionals on X. If {xn} is bounded then so is each {λi(xn)}. 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 {TN xnk } using compactness of TN . Check that {T xnk } is a Cauchy sequence. Since B(X,Y ) is complete, {T xnk } is convergent as required.

Exercise 1.5 : Let a := {an} be a bounded sequence. Show that the diagonal operator Da 2 on l with diagonal entries a1, a2, ··· is compact iﬀ {an} ∈ c0.

Hint. If Db is a diagonal operator with diagonal entries b1, b2, ··· , then kDbk = supn |bn|. The suﬃcient part follows from limn→∞ kDa,n − Dak = 0, where Da,n is the diagonal operator with diagonal entries a1, ··· , an, 0, 0, ··· . To see the necessary part, WLOG, assume 2 2 2 that {an} is bounded from below, and note that kDaen − Daemk = an + am > c > 0 for a scalar c.

2 Example 1.6 : Let a := {an} ∈ 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 ∈ 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 ∈ [0, 1]) and n−1 X Z k/n (Tnf)(x) = χ[k/n,(k+1)/n)(x) f(y)dy (x ∈ [0, 1]). k=0 0 2 2 −1/2 Verify that T ∈ B(L (0, 1),L (0, 1)) satisﬁes kT −Tnk ≤ n . Conclude that T is compact.

n+m Exercise 1.8 : For t ∈ 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 ∈ L (R ), x ∈ R . Rm Verify the following:

(1) Tt is a bounded, linear operator. (2) Let {uk} be a bounded sequence with bound M. Then m/2 |(Ttuk)(x)| ≤ ktk∞(2c) M,

where c is chosen so that t(x, y) = 0 if kxk2 > c or kyk2 > c. (3) Given > 0, there exists δ ∈ (0, 1) such that 0 m/2 |(Ttuk)(x) − (Ttuk)(x )| ≤ (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 ∈ C0(R ) such that ktk − tkL2(Rn+m) → 0 as k → ∞. 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, ∞) → L2(0, ∞) given by 1 Z x (Af)(x) = f(t)dt (x ∈ (0, ∞)). 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 ∈ L (0, ∞). 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 < ∞.

Recall that C[0, 1] is dense in Lp(0, 1) for 1 ≤ p < ∞. 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 ∈ ran(T ) be such that kynk ≤ 1. Consider the bounded linear surjection S : X → ran(T ) given by Sx = T x for x ∈ X. By the Open Mapping Theorem, there exists xn ∈ X such that yn = T xn for some bounded sequence {xn}. Since T is compact, {yn} 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 |M is compact. Clearly, T |ker(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.

∞ 2 ∗ ∗ The operator of multiplication by φ ∈ 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 ∈ 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. √ Hint. Suppose {xα} is orthonormal. Then {B(xα, 1/ 2)} is a collection√ of disjoint open balls in H. Now if {yn} 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 ∈ [0, 1]). 0 x Verify the following: (1) T ∈ 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. Note that kNk2 = kN ∗Nk = k(N ∗N)2k1/2 = kN 2k. Since any positive integral power of a normal operator is normal, one may check by induction that kN 2k k = kNk2k for n 1/n every positive integer k. Use now the formula: r(T ) = limn→∞ kT k . Exercise 2.9 : A normal operator N ∈ B(H) is invertible iﬀ there exists a positive scalar α such that kNxk ≥ αkxk for every x ∈ H.

Hint. Suppose kNxk ≥ αkxk for every H and for some α > 0. Then ker(N) = {0}. If y ∈ ran(N) then for some sequence {xn}, T xn → y. Check that {xn} is Cauchy, and hence 4 SPECTRAL THEOREM FOR SYMMETRIC OPERATORS converges to some x. This shows that y = T x, so that ran(T ) is closed. On the other hand, since ker(N) = ker(N ∗), ran(N) is dense in H.

Lemma 2.10. Let λ be a non-zero complex number and let T be a compact normal operator on a separable Hilbert space. If T − λ is not invertible then T − λ is not one-to-one. In particular, σ(T ) \{0} = σp(T ) \{0}.

Proof. Since T −λ is not invertible, by the preceding exercise, there exists xn ∈ H such that kxnk = 1 and k(T −λ)xnk < 1/n. Since T is a compact operator, there exists a subsequence

{xnl } such that T xnl → y for some y ∈ H. It follows that λxnl → y, and hence T y = λy with kyk = |λ| > 0. It follows that σ(T ) \{0} ⊆ σp(T ) \{0}. The reverse inclusion holds for any bounded linear operator. Remark 2.11 : The conclusion of the lemma holds for any compact operator.

Now we state and proof the main result of this section:

Theorem 2.12. Let T be a non-zero normal operator on a complex Hilbert space H. If T ⊥ is compact then there exists an orthonormal basis {en} of ker(T ) and a sequence {λn} of non-zero complex numbers (possibly repeated) such that T en = λnen. Moreover, the following statements hold:

(1) For each n ≥ 1, ker(T − λn) is finite-dimensional. (2) inf |λn| = 0. n (3) {λn} has no accumulation point except 0. (4) For any > 0, the annulus A(0, , r(T )) := {z ∈ C : ≤ |z| ≤ r(T )} intersects σ(T ) in finitely many points. In particular, ∆ := {n ∈ N : |λn| ≥ } contains λn for finitely many values of n. X (5) If x ∈ H, then T x = λnhx, enien in the following sense: For any > 0,

n∈σp(T )\{0}

X T x − λnhx, enien < . n∈∆ Proof. We will prove the result for a separable Hilbert space H. In view of Exercise 2.6, we may enumerate all the eigenvalues of T as {λn}. We may choose an orthonormal set {en} of corresponding eigenvectors: T en = λnen. Note that (1) is already obtained in Corollary

2.2. Let us see (2). Suppose {λn} has a subsequence {λnk } which is bounded from below in modulus by c > 0. Then

2 2 2 2 kT enk − T enl k = |λnk | + |λnl | ≥ 2c .

Thus we have the bounded sequence {enk } for which {T enk } has no convergent subsequence. Since T is compact, this is not possible unless infn |λn| = 0. This also shows that {λn} has no non-zero accumulation point. Further, (4) follows from (3), Lemma 2.10, and Bolzano- Weierstrass Theorem. Indeed, since any infinite subset of the compact set A(0, , r(T )) has a limit point, A(0, , r(T )) contains λn for finitely many values of n. To see (5), let > 0. By (4), ∆ is a finite set. Consider the sequence {Tk} of non-zero compact normal operators Tk defined by X Tkx := T x − λnhx, enien (x ∈ H).

n∈∆

We check that kTkk < . By Exercise 2.8, there exists λ ∈ σ(Tk) such that kTkk = |λ|. By the preceding lemma, λ must belong to σp(Tk). Thus, for some y 6= 0,Tky = λy and ∗ Tk y = λy. We now complete the proof of (5). Note that X (2.1) λy = T y − λnhy, enien.

n∈∆ SPECTRAL THEOREM FOR SYMMETRIC OPERATORS 5

Moreover, for any m ∈ ∆, one has X λhy, emi = hT y − λnhy, enien, emi

n∈∆ ∗ X = hy, T emi − hλnhy, enien, emi,

n∈∆

= hy, λmemi − λmhy, emi = 0.

Thus y is orthogonal to en ∈ ker(T − λn) provided n ∈ ∆. By (2.1), λy = T y, and hence λ = λm and y ∈ ker(T − λm) for some m ≥ 1. Since y 6= 0, m∈ / ∆. That is, kTkk = |λ| = |λm| < . ⊥ ⊥ Finally, we check that {en} of is an orthonormal basis of ker(T ) . Let x ∈ ker(T ) be such that hx, eni = 0 for all integers n ≥ 1. But then by (5), for any > 0, kT xk < , that is, T x = 0. It follows that x ∈ ker(T ), and hence it must be 0. ⊥ Remark 2.13 : Let z = x + y, where x ∈ ker(T ) and y ∈ ker(T ). Since {en} of is an ⊥ P∞ orthonormal basis of ker(T ) , we can write x = n=1hx, enien. By continuity of T, ∞ ∞ X X T z = T x = λnhx, enien = λnhz, enien. n=1 n=1 3. Unbounded Operators with Compact Resolvent Let X be a complex normed linear space. By a linear operator S in X with domain D(S), we mean a linear transformation S defined on some linear subspace D(S) of X. If S is a linear operator in X with domain D(S), then the resolvent set of S, denoted by ρ(S), is defined to be the set of all scalars µ ∈ C for which there exists a bounded linear operator R(µ) on X such that (R1) for every y ∈ X we have that R(µ)y ∈ D(S) and (S − µ)R(µ)y = y, (R2) R(µ)(S − µ)x = x for all x ∈ D(S). Let R(µ) := (S − µ)−1. Thus −1 ρ(S) := µ ∈ C :(S − µ) exists as a bounded linear operator on X . Define the spectrum σ(S) := ρ(S)c, and the point spectrum

σp(S) := {µ ∈ C : Sx = µx for some non-zero x ∈ D(S)} . Remark 3.1 : We note the following:

(1) σp(T ) ⊆ σ(T ). (2) If (R1) holds and µ∈ / σp(T ) then (R2) is satisﬁed.

Deﬁnition 3.2 : A linear operator S in X is closed if for every sequence {xn} ⊂ D(S) such that xn → x, Sxn → y for some y ∈ X, one has x ∈ D(S) and Sx = y.

Exercise 3.3 : Let T be a closed linear operator in a Banach space X. Then µ ∈ ρ(T ) iﬀ T − µ is bijective.

Hint. Use the Closed Graph Theorem. Exercise 3.4 : If T is one-one and closed then T −1 is also closed.

−1 −1 Hint. Let {xn} be such that xn → x and T xn → y. Then yn := T xn ∈ D(T ), so that yn → y and T yn = xn → x. Exercise 3.5 : If ρ(T ) 6= ∅ then show that T is closed.

Hint. If µ ∈ ρ(T ) then (T − µ)−1 is closed; so is T − µ and hence T. Exercise 3.6 : If A and B are linear operators such that A = B on D(A) ⊆ D(B). If ρ(A) ∩ ρ(B) 6= ∅ then D(A) = D(B), and hence A = B. 6 SPECTRAL THEOREM FOR SYMMETRIC OPERATORS

Hint. If µ ∈ ρ(A) ∩ ρ(B) and x ∈ D(B) then (B − µ)x ∈ H = ran(A − µ), and hence for some y ∈ D(A), (B − µ)x = (A − µ)y. Exercise 3.7 : Let T be the linear operator of diﬀerentiation with domain D(T ) in Lp(0, 1) for p ∈ [1, ∞]. Verify: 1 (1) If D(T ) = {f ∈ C [0, 1] : f(0) = 0} then σ(T ) = C. (2) If D(T ) = {f ∈ AC[0, 1] : f 0 ∈ Lp(0, 1), f(0) = 0} then σ(T ) = ∅. (3) If D(T ) = {f ∈ AC[0, 1] : f 0 ∈ Lp(0, 1), f(0) = f(1)} then

σ(T ) = σp(T ) = {2πin : n ∈ Z}. (4) For the operator T in part (3), (T − 1)−1 is compact. Show that all the operators T are densely deﬁned.

R λ(x−s) Hint. In (1), T −λ is never surjective. For (2), consider (R(λ)g)(x) = (0,x) e g(s)ds p 0 for g ∈ L (0, 1) and λ ∈ C. To see (3), note first that for f ∈ D(T ), T f = λf iff f = λf a.e. −λt 0 iff (e f) = 0 a.e. To see that σp(T ) = {2πin : n ∈ Z}, use now the fundamental theorem of calculus [2, Theorem 7.18], and the boundary conditions f(0) = f(1). If λ∈ / σp(T ) then R(λ) is given by Z λ(x−y) e p (R(λ)g)(x) = λ t(x, y)g(y)dy (g ∈ L (0, 1)), (0,1) 1 − e where t(x, y) = eλ if x ≤ y; and t(x, y) = 1 if x > y. The last part may be concluded from the Arzela-Ascoli Theorem. Definition 3.8 : A linear operator T in a Hilbert space is said to have compact resolvent −1 if there exists λ0 in the resolvent ρ(T ) of T such that R(λ0) := (T − λ0) is compact.

Exercise 3.9 : If T has compact resolvent then show that R(λ) is compact for every λ ∈ ρ(T ).

Hint. Derive the resolvent identity: R(λ) − R(µ) = (λ − µ)R(λ)R(µ) for λ, µ ∈ ρ(T ). If R(λ0) is compact then so is R(λ) = R(λ0) + (λ − λ0)R(λ)R(λ0) for any λ ∈ ρ(T ). Exercise 3.10 : Suppose that T has compact resolvent in X. For every λ 6= 0, show that ker(T − λ) is a closed subspace of X.

−1 Hint. Suppose (T − λ0) is compact for some λ0 ∈ C. Observe that ker(T − λ) = ker((λ − λ0)R(λ0) − 1), and apply Corollary 2.2.

4. Spectral Theorem for Symmetric Operators with Compact Resolvent A densely deﬁned linear operator S in a Hilbert space H is said to be symmetric if hSx, yi = hx, Syi for all x, y ∈ D(S).

−1 Exercise 4.1 : Let λ ∈ ρ(T ) ∩ R. Show that T is symmetric iﬀ R(λ) := (T − λ) is self-adjoint.

Hint. If T is symmetric then hR(λ)x, yi = hR(λ)x, (T − λ)R(λ)yi = h(T − λ)R(λ)x, R(λ)yi = hx, R(λ)yi for any x, y ∈ H. Similarly, one can check that if R(λ) is symmetric then h(T − λ)x, yi = hx, (T − λ)yi for any x, y ∈ D(T ). Exercise 4.2 : Show that T given by T f = if 0 (f ∈ D(T )) is a symmetric operator with compact resolvent, where D(T ) = {f ∈ AC[0, 1] : f 0 ∈ L2(0, 1), f(0) = f(1)}. SPECTRAL THEOREM FOR SYMMETRIC OPERATORS 7

Hint. Let f, g ∈ D(T ). Integration by parts gives Z Z hT f, gi = i f 0(t)g(t)dt = −i f(t)g0(t)dt = hf, T gi. (0,1) (0,1) Also, T has compact resolvent, note that T − I is invertible with inverse R(1) given by Z ex−y (R(1)g)(x) = −i t(x, y)g(y)dy (g ∈ Lp(0, 1)), (0,1) 1 − e where t(x, y) = e if x ≤ y; and t(x, y) = 1 if x > y. The second part may be concluded from Exercise 1.9.

Exercise 4.3 : Show that the point-spectrum σp(T ) of a symmetric operator T is a subset of the real line R.

−1 Exercise 4.4 : Let T be a symmetric operator in a Hilbert space H such that (T − λ0) is compact. Show that −1 −1 σ(T ) = σp(T ) = {λ0 + µ : µ ∈ σp((T − λ0) )}.

Hint. Use Lemma 2.10.

Lemma 4.5. If T is a symmetric operator with compact resolvent then there exists λ1 ∈ −1 R ∩ ρ(T ) such that (T − λ1) is a compact self-adjoint operator. −1 Proof. Let λ0 ∈ C be such that R(λ0) := (T − λ0) is a compact operator. We claim that σ(T ) intersects Dr := {z ∈ C : |z| < r} in finitely many points. By Exercise 4.4, if −1 −1 λ ∈ σ(T ) ∩ Dr then λ = λ0 + µ for some µ ∈ σp(R(λ0)). Also, |µ | = |λ − λ0| < r + |λ0|, −1 so that |µ| > (r + |λ0|) . However, by Theorem 2.12(4), σp(R(λ0)) intersects C \ Dr in finitely many points. Thus there are finitely many choices for µ, and hence for λ. Hence the claim stands verified. −1 Choose λ1 ∈ (−r, r)∩ρ(T ). By Exercises 3.9 and 4.1, (T −λ1) is a compact self-adjoint operator as required. Theorem 4.6. Let T be a symmetric operator with compact resolvent in a Hilbert space H. For each λ ∈ σp(T ), define n(λ) = dim ker(T − λ), and choose an orthonormal basis {φλ,1, ··· , φλ,n(λ)} of ker(T − λ). Then {φλ,i : λ ∈ σp(T ), i = 1, ··· , n(λ)} forms a complete orthonormal set for H. Moreover, we have the following: (1) x ∈ D(T ) iff P Pn(λ) |λ|2|hx, φ i|2 < ∞. λ∈σp(T ) i=1 λ,i (2) For any x ∈ D(T ), T x = P Pn(λ) λhx, φ iφ . λ∈σp(T ) i=1 λ,i λ,i −1 Proof. By the last lemma, there exists λ0 ∈ R ∩ ρ(T ) such that R(λ) := (T − λ0) is a compact self-adjoint operator. By Theorem 2.12, n(µ) X X (4.2) R(λ0)x = µhx, φµ,iiφµ,i (x ∈ H). −1 µ∈σp((T −λ0) ) i=1 −1 −1 Also, by Exercise 4.4, σ(T ) = σp(T ) = {λ0 + µ : µ ∈ σp((T − λ0) )}. Thus (4.2) can be rewritten as m(λ) X X hx, ψλ,ii R(λ0)x = ψλ,i (x ∈ H), λ − λ0 λ∈σp(T ) i=1 where m(λ) := n(µ) and ψλ,i := φµ,i. Since ker(R(λ0)) = {0}, it follows from the completeness part of Theorem 2.12 that {φλ,i : λ ∈ σp(T ), i = 1, ··· , n(λ)} is a complete orthonormal set for H. Let x ∈ D(T ). Since T is symmetric, hT x, ψλ,ii = hx, T ψλ,ii = λhx, ψλ,ii. It is easy to see from the Parseval’s identity that P Pn(λ) |λ|2hx, φ i|2 = kT xk2 < ∞. Conversely, λ∈σp(T ) i=1 λ,i suppose that P Pn(λ) |λ|2|hx, φ i|2 < ∞, and set y = P Pn(λ) λhx, φ iφ . λ∈σp(T ) i=1 λ,i λ∈σp(T ) i=1 λ,i λ,i 8 SPECTRAL THEOREM FOR SYMMETRIC OPERATORS

It is easy to see that R(λ0)(y − λ0x) = x. In particular, x ∈ D(T ) and (T − λ0)x = y − λ0x. Finally, note that T x = y = P Pn(λ) λhx, φ iφ . λ∈σp(T ) i=1 λ,i λ,i Remark 4.7 : As a part of the deﬁnition of symmetric operators, we assumed that T is densely deﬁned. The conclusion of the theorem holds without this assumption.

Example 4.8 : Recall that the operator T as deﬁned in Exercise 4.2 is a symmetric operator with compact resolvent. It may be concluded from the last theorem that the eigenvectors 1, e2πix, e−2πix, e4πix, e−4πix, ··· of T form a complete orthonormal set in L2[0, 1] (for details, see Exercise 3.7). This provides (possibly the most elegant) proof of the completeness of the Fourier series in L2(−π, π) : For any f ∈ L2(−π, π), X fˆ(n)enπix converges to f in L2(−π, π), n∈Z ˆ 1 R nπix where f(n) = 2π (−π,π) f(x)e dx for n ∈ Z.

There are numerous applications of Theorem 4.6 to the theory of PDE’s including Sturm- Liouville problems (refer, for instance, to [1, Section 2.7]). Acknowledgements. I express my sincere thanks to the organizers Sam and Murugan for their warm hospitality during the stay 01-06 June, 2014 at NIT-Karnataka.

References [1] Milan Miklavˇciˇc, Applied Functional Analysis and Partial Diﬀerential Equations, World Scientiﬁc, Singapore, 1998. [2] W. Rudin, Real and Complex Analysis, McGraw-Hill Book Co. New York, 1987.