NOTES ON UNBOUNDED OPERATORS MATH 581, SPRING 2017

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1 NOTES ON UNBOUNDED OPERATORS MATH 581, SPRING 17 Throughout, X will denote Bnch, possibly Hilbert spce. Some of this mteril drws from Chpter 1 of the book Spectrl Theory nd Differentil Opertors by E. Brin Dvies. 1. Introduction nd exmples Definition 1.1. A liner opertor on X is liner mpping A : D(A) X defined on some subspce D(A) X. A is densely defined if D(A) is dense subspce of X. An opertor A is sid to be closed if the grph of A Γ(A) = {(x, Ax) : x D(A)} X X, is closed subspce of X X. An opertor A is sid to be closble if Γ(A) defines the grph of liner opertor, tht is, Γ(A) = {(x, Āx) : x D(Ā)} for some liner opertor Ā which extends A. We cll Ā the closure of A. It is not hrd to check tht the grph Γ(A) nd its closure re indeed lwys subspces of X X. Thus closble mens tht Γ(A) is subspce of X X such tht whenever (x, y i ) Γ(A) for i = 1, then y 1 = y s this is the miniml requirement for Γ(A) to define the grph of function. But from here it cn be seen tht since Γ(A) is subspce, it is the grph of liner mp (Exercise: verify this). A few remrks re in order: (1) One cn clerly extend the definitions bove to define liner opertors between two Bnch spces X nd Y (or normed vector spces). () Up until now, when discussing liner opertors we hve lwys ssumed D(A) = X. Allowing D(A) to possibly be proper subspce llows us to consider lrger fmily of opertors of interest, primrily those which re unbounded. Perhps surprisingly, it is very difficult to define n unbounded liner opertor on Bnch spce with D(A) = X without using the xiom of choice. Hence nerly ll exmples of unbounded opertors re defined on proper subspce of X. (3) The utility of the closed opertor definition comes into ply when R(A) = X nd N(A) = {}, tht is, A is bijection from D(A) to X. In this cse, Γ(A 1 ) = {(y, A 1 y) : y X} = {(Ax, x) : x D(A)}, so tht Γ(A 1 ) is the imge of Γ(A) under the permuttion (x, y) (y, x). Thus if Γ(A) is closed, so is Γ(A 1 ) nd hence A 1 is bounded. In other words, the inverse of ny closed liner opertor, when it exists, is bounded. Proposition 1.. Suppose A is liner opertor on X. A is closble if nd only if whenever (, y) Γ(A), then y =. 1

2 NOTES ON UNBOUNDED OPERATORS Proof sketch. By definition, the opertor A is closble if nd only if Γ(A) is the grph of liner opertor. This mens tht if (x, y i ) Γ(A), i = 1,, then y 1 = y s this must occur for Ā to be well defined. Setting y = y 1 y, we cn tke the difference (x, y 1 ) (x, y ) = (, y) to see tht this is equivlent to the sttement tht (, y) Γ(A) implies y =. Exmple 1.3. In Assignment 4, Exercise #3 you considered the subspce of l 1 D(A) = x l1 : j x(j) < nd then considered the opertor A : D(A) l 1 defined by (Ax)(j) = j x(j) (though this ws clled T there). You showed tht A ws closed nd densely defined in l 1. Exercise 1.4. Generlize this exmple, considering l p for ny 1 p < nd tking ny sequence {α(j)} nd defining D(A) = x lp : α(j)x(j) p <, (Ax)(j) := α(j)x(j). Show tht A is lwys closed nd densely defined. Moreover, show tht D(A) = l p nd A is bounded if nd only if {α(j)} is bounded sequence. Which of these sttements re vlid when p =? Exmple 1.5. Consider l p but with 1 < p <. Fix nonzero vector y l p. Let D(A) = l 1 l p nd define 1 Ax = x(j) y x D(A). It is not hrd to verify tht A is liner opertor. However, it is not closble. To see this, define x n l 1 l p by { 1 x n (j) = n, 1 j n,, j > n. Then x n(j) = 1 so tht Ax n = y, but n x n l p = 1 n p 1/p Hence lim n (x n, Ax n ) = (, y ) Γ(A) but y. = ( n n p ) 1/p = n 1 p 1 s n. 1 We never proved it, but it is fct tht l q l p when 1 q p nd moreover x l p x l q. Hence l 1 l p = l 1.

3 NOTES ON UNBOUNDED OPERATORS Differentil opertors. The exmples bove re resonbly simple exmples of unbounded opertors. However, severl of the unbounded opertors of interest re differentil opertors. The theory here intersects tht of so-clled Sobolev spces, nd importnt re of mthemtics, but perhps outside the scope of this course. We focus on exmples of function spces defined on compct nondegenerte subintervl [, b] R for simplicity. Exmple 1.6. The esiest exmple is to tke X = C([, b]). Consider the simple differentil opertor (Ax)(t) = dx(t) for t [, b]. The nturl choice for D(A) is simply the C 1 functions on [, b]. In this cse, checking tht A is closed is the sme s checking tht whenever x n x C([,b]) nd dxn y C([,b]), then x C 1 ([, b]) nd dx = y. But you did this work in Assignment 1, Exercise #5 when you proved tht C 1 ([, b]) ws complete! It is not hrd to see tht A is unbounded in this cse: simply tke trigonometric functions x n (t) = sin(α n t) for some choice of α n so tht x n C([,b]) = 1 but Ax n C([,b]). For pplictions in differentil equtions, one is often in defining differentil opertors on L p spces with 1 p < (or even p = ). For one, vrious inequlities nd conservtion lws tht rise in the nlysis of differentil equtions typiclly involve integrls of functions insted of the supremum of them. Moreover, C([, b]) lcks n inner product structure, nd our richest spectrl theory for these opertors is for Hilbert spces. In this clss, we hve defined L p ([, b]) s the completion of C([, b]) with respect to the norm ( b x L p ([,b]) := x(t) p In other words, functions in L p ([, b]) re those which re the limits of continuous functions with respect to this norm. For those who hve hd mesure theory, L p ([, b]) cn be equivlently defined s the spce of Lebesgue mesurble functions on [, b] such tht the integrl on the right here is finite (treting two functions s the sme if they only differ on set of mesure zero). Definition 1.7. A function y(t) is sid to be the wek derivtive of function x(t) if given ny ψ C 1 ([, b]) such tht ψ(t) vnishes in neighborhood of the endpoints, b of the intervl (i.e. there exists ɛ > such tht ψ(t) = for t / [ + ɛ, b ɛ]), we hve (1.1) b x(t) dψ(t) = b ) 1 p. y(t)ψ(t). Note tht integrtion by prts shows tht whenever x C 1 ([, b]), (1.1) is stisfied with y(t) = x (t). This is in some sense the rtionle for the definition. It is fct tht whenever the wek derivtive exists, then it is unique up to set of mesure zero, in other words if y 1, y re two wek derivtives of x, then the set {t [, b] : y 1 (t) y (t)} is of mesure zero. Since we re concerned with L p spces in this discussion, which trets two functions which differ only on set of mesure zero s equl, we consider the wek derivtive to be essentilly unique. Another wy to describe the ψ under considertion here is to define the support of function s supp(ψ) = {t : ψ(t) }, then set C 1 c (, b) := {ψ C 1 (, b) : supp(ψ) (, b)}.

4 4 NOTES ON UNBOUNDED OPERATORS This is the sme considering the C 1 functions on (, b) with compct support, in tht supp(ψ) is compct. The clss of functions in Definition 1.7 re ψ C 1 c (, b). Indeed, since supp(ψ) is disjoint from the closed set (, b) C, then such ψ must vnish in neighborhood of the endpoints. One cn define C k c (, b) nd C c (, b) similrly, simply replcing C 1 (, b) by C k (, b) in the bove definition. Exercise 1.8. Let [, b] = [ b, b] for ny b >. Show tht wek derivtive of x(t) = t is given by ny function y such tht y(t) = t/ t for t. Definition 1.7 gives us prospective definition of D(A) for A = d s (1.) {x L p ([, b]) : x hs wek derivtive y nd y L p ([, b])}. Alterntively, one cn ppel to the theory of bsolutely continuous functions to mke sense of the domin of A. Definition 1.9. A function x : [, b] C is sid to be bsolutely continuous if for every ɛ >, there exists δ > such tht n x(t i) x(t i ) < ɛ i=1 whenever (t 1, t 1),..., (t n, t n) [, b] is collection of disjoint intervls such tht n i=1 (t i t i) < δ. Theorem 1.1. Suppose x is bsolutely continuous on [, b]. Then x (t) exists in the clssicl sense for lmost every t [, b] (tht is, ll t outside mesure zero set) nd x (t) defines n integrble function such tht with y(t) = x (t), (1.3) x(t) = x() + t y(s)ds, t [, b]. Conversely, if (1.3) is stisfied for some y integrble on [, b], then x is bsolutely continuous. The theorem thus gives us fundmentl theorem of clculus for Lebesgue integrls. See Stein nd Shkrchi, Rel Anlysis, Theorem 3.11 for proof. We cn therefore lso define D(A) s { (1.4) x L p ([, b]) : x is bsolutely continuous nd dx } Lp ([, b]). We hve the following chrcteriztion of the domin of A = d, whose proof is beyond the scope of this course. Theorem Suppose 1 < p <. The two function spces (1.) nd (1.4) re identicl nd define dense subspces of L p ([, b]). Moreover, tking this subspce to be D(A), the liner opertor Ax = dx cn be defined equivlently s either the wek derivtive of x or the clssicl derivtive defined up to set of mesure zero. Moreover, A is closed. The common function spce (1.), (1.4) is typiclly clled the L p -Sobolev spce of order 1 on [, b], often denoted W 1,p (, b).

5 NOTES ON UNBOUNDED OPERATORS Domins formed by tking closures. One finl pproch to finding suitble domin for A = d on Lp ([, b]) is to consider subspce of C 1 ([, b]) nd show tht the resulting opertor is closble. One cse of interest is to begin with D(A) = {x C 1 ([, b]) : x() = x(b) = }, tht is, the subspce of C 1 functions which vnish t the endpoints. To see tht A is closble, suppose tht (, y) Γ(A), so there exists sequence of functions x n D(A) such tht lim x n Lp ([,b]) =, lim dx n n n y =. Lp ([,b]) We now recll Hölder s inequlity, which sttes tht for two functions w(s), z(s), then for 1 < p, q < stisfying 1 p + 1 q = 1, we hve b ( b ) 1 ( b p ) 1 b q w(s)z(s)ds w(s)z(s) ds w(s) p ds z(s) q ds. In prticulr, if we tke z(s) 1, then we obtin b ( b b w(s)ds w(s) ds (b ) 1 1 p w(s) p ds We now define x(t) := t y(s)ds nd observe tht t ( x n (t) x(t) = dx n t (s) y(s)ds ds (t 1 )1 p dx n (s) y(s) ds (b ) 1 1 p dx n ds y. This shows tht s n, sup x n (t) x(t) (b ) 1 1 p t b ) 1 p. Lp ([,b]) dx n ds y L p ([,b]) hence x n x L p ([,b]) s n since ( ) 1 b p x n (t) x(t) p (b ) 1 p sup x n (t) x(t). t b p ) 1 p ds But the limit of ny sequence in L p is unique nd hence we must hve x = in L p ([, b]). But this mens tht t y(s)ds = for ny t [, b] nd hence in fct t y(s)ds = for ny t 1, t [, b]. This llows us to deduce tht y(t) = on t 1 [, b] (t lest up to set of mesure zero). Hence (, y) Γ(A) implies tht y =, showing A is closed. The domin of the closure of A in this cse is often denoted s W 1,p ([, b]), the subscript denoting the vnishing of these functions t the endpoints. Indeed, there re meningful restriction mps R (x) := x(), R b (x) := x(b) initilly defined for C 1 ([, b]) functions which extend to W 1,p ([, b]) functions. It cn be seen tht membership in W 1,p ([, b]) is equivlent to R (x) =, R b (x) =. For problems involving higher derivtives nd for problems in higher dimensionl spce, the wek derivtive pproch is esier to generlize nd hence more common.

6 6 NOTES ON UNBOUNDED OPERATORS In prticulr, on n intervl [, b] R, one cn define the k-th wek derivtive of function x s n integrble function y stisfying b x(t) dk ψ(t) k = ( 1) k b y(t)ψ(t) for ll ψ C k ([, b]) vnishing in neighborhood of the endpoints s in Definition 1.7. One cn then define D(A k ) s the set of ll function whose wek derivtives of order up to k exist nd re functions in L p ([, b]), nd then tke A k x = dk x s this wek k-th derivtive. Similrly, one cn mke sense of wek prtil derivtives on n open domin in R n. 1.. Adjoints. Definition 1.1. Given densely defined liner opertor A on X, we sy liner functionl f X is in D(A ) if there exists g X such tht f(ax) = g(x) for every x D(A) nd define the djoint of A for f D(A ) by A f = g. Note tht if g 1, g X ech stisfy g i (x) = f(ax), i = 1, for every x D(A) then since D(A) is dense the two functionl coincide on dense set, so by continuity we must hve g 1 = g in X. It is then verified tht D(A ) is subspce of X nd tht A defines liner opertor on X. Note tht if one hs n inequlity of the form f(ax) C x for some uniform C nd ll x D(A), then the mpping x f(ax) extends to bounded liner functionl by the Hhn-Bnch theorem, which must be unique since D(A) is dense. Exmple Return to the sitution of Exmple 1.4 with 1 < p <, but now express the dependence on the order of the l p spce under considertion, so tht D(A p ) denotes the domin of A p s liner opertor on l p given by pointwise multipliction by the {α(j)} sequence. Suppose q is the Hölder conjugte of p, stisfying q = p p 1. We know tht ny liner functionls f, g (lp ) cn be identified with sequences y, z l q stisfying for every x X, f(x) = x(j)y(j), nd g(x) = x(j)z(j). So in order for there to exist liner functionl g (l p ) such tht f(a p x) = g(x) for every x D(A p ), we must hve α(j)x(j)y(j) = x(j)z(j). Now revisit the sequences e k defined by e k (j) = δ jk (Kronecker delt), which re esily seen to be in D(A p ). Tking x = e k for every k N, we see tht in order for f (l p ) to be in D(A p), we must hve α(k)y(k) = z(k) for ech k while z l q, showing tht D(A p) cn be identified with the set of sequences y lq : α(j)y(j) q <. In other words, the domin of A p is identified with D(A q ), nd A p is lso identified with pointwise multipliction by the sequence.

7 NOTES ON UNBOUNDED OPERATORS Hilbert spce djoints. Let X be Hilbert spce. Recll from Assignment #3, Exercise #, tht if T : X X is bounded liner mp nd V : X X is the isometric conjugte liner bijection furnished by the Riesz representtion such tht V y(x) = x, y then the Hilbert spce djoint of T stisfies T = V 1 T V, where T is the djoint. It is chrcterized by T x, y = x, T y for ll x, y X. Alredy this instructs us how to define the Hilbert spce djoint of densely defined liner opertor A s defined in Definition 1.1: we set D(A ) s the imge of D(A ) under V 1, or equivlently, D(A ) is the unique subspce of X such tht V (D(A )) = D(A ). Hence D(A ) is the set of ll y X such tht g(x) = (V y)(ax) = Ax, y extends to liner functionl on D(A). The liner opertor A : D(A ) X is then V 1 A V. Equivlently, A cn be defined without reference to X by defining D(A ) s the set of ll y X such tht there exists z X stisfying (1.5) Ax, y = x, z for ll x D(A), then defining A y = z, so tht A is chrcterized by Ax, y = x, A y for ll x D(A) nd y D(A ). Note tht since D(A) is dense, if there exists z X stisfying (1.5), then z is the unique vector stisfying the identity, for if there exists nother vector z stisfying (1.5), we would hve x, z = x, z for every x D(A) so tht z z D(A) = {}. Exercise Verify tht the two chrcteriztions of A given here re indeed equivlent. Before proceeding, we recll the Hilbert spce structure on X X. (x, y), ( x, ỹ) X X, we tke (1.6) (x, y), ( x, ỹ) = x, x + y, ỹ so tht the induced norm on X X is (x, y) = ( x + y ) 1/. Given Proposition Suppose A is closed nd densely defined on Hilbert spce X, then so is A. Proof. Let M = {(Ax, x) : x D(A)}, which is seen to define subspce of X X. We clim tht (y, z) Γ(A ) = {(w, A w) : w D(A )} if nd only if (y, z) M (using the inner product in (1.6)). Since this is equivlent to sying Γ(A ) = M, nd the orthogonl complement of ny subspce is lwys closed (regrdless of whether or not the originl subspce is), this will show tht Γ(A ) closed nd hence A is closed. To see the clim, note tht (y, z) Γ(A ) if nd only if Ax, y = x, z for every x D(A), nd this in turn is equivlent to sying tht in the product spce X X (Ax, x), (y, z) = for every x D(A), which is precisely the clim we wnted to estblish.

8 8 NOTES ON UNBOUNDED OPERATORS To see tht A is densely defined, suppose tht x X nd x D(A ), tht is x, y = for every y D(A ). Then with respect to the inner product on X X, (x, ), (y, A y) = for every y D(A ), which shows tht (x, ) Γ(A ) = (M ) = M (note tht this is essentilly Lemm 3.5 nd 3.6 in Schechter, fter mking the identifiction of X with X). Reclling the definition of M, there must exist sequence {z n } such tht (Az n, z n ) (x, ) in X X, which is equivlent to sying tht (z n, Az n ) (, x). But since (z n, Az n ) is in the closed subspce Γ(A), we must hve (, x) Γ(A), nd hence x = A() =. This shows tht D(A ) = {}, nd hence D(A ) = (D(A ) ) = {} = X, which shows tht A is indeed densely defined.. The spectrum of closed opertor Definition.1. Suppose A is liner opertor on X. We sy λ C is in the resolvent set of A, denoting this s λ ρ(a), if N(λ A) = {}, R(λ A) = X nd its inverse (λ A) 1 is bounded liner trnsformtion defined on ll of X, tht is, D((λ A) 1 ) = X. The opertor (λ A) 1 defined on ρ(a) is then clled the resolvent of A. The spectrum of A is defined s the complement of ρ(a) in C nd denoted s σ(a). Note tht we consider the domin of λ A to be identicl to tht of A. It is not hrd to check tht if A is closed then so is λ A. So for closed opertors, s soon s it is seen tht λ A : D(A) X is bijective, the inverse is bounded (see the remrks on p.1). There is slight disgreement between this definition nd the one given on p. 171 of Schechter, which is resolved by the following theorem. Theorem.. Suppose A is closed liner opertor on X. Then λ ρ(a) if nd only if R(λ A) is dense nd there exists bounded liner trnsformtion T : X X such tht T (λ A) = I on D(A) nd (λ A)T = I on R(λ A). Proof. Given our definition of the resolvent set of A, the forwrd impliction is cler. For the converse, first note tht N(λ A) = {} s T is left inverse for (λ A): indeed, if x N(λ A), then x = T (λ A)x = T () =. We next show R(λ A) = X which mens λ A is bijection from D(A) = D(λ A) to X. Suppose x X is rbitrry. Since R(λ A) is dense in X, there exists sequence {x n } in R(λ A) such tht lim n x n = x. Moreover, since T is bounded, lim n T x n = T x. But this mens (T x n, x n ) = (T x n, (λ A)T x n ) Γ(λ A), nd since (T x, x) = lim n (T x n, x n ), we hve (T x, x) Γ(λ A) = Γ(λ A). But this implies tht T x D(A) nd (λ A)T x = x, which shows tht x R(λ A) nd hence R(λ A) = X s climed. Since T is now shown to be left nd right inverse for (λ A), we re done. Theorem.3. Suppose A is closed. If λ ρ(a), then (.1) { ν C : λ ν < (λ A) 1 1} ρ(a),

9 NOTES ON UNBOUNDED OPERATORS 9 nd hence σ(a) C is closed. Moreover for λ, ν ρ(a), (.) (.3) (.4) (λ A) 1 (ν A) 1 = (ν A) 1 (λ A) 1 (λ A) 1 (ν A) 1 = (λ ν)(λ A) 1 (ν A) 1 d dλ (λ A) 1 = (λ A) Proof. Given λ ρ(a), let B = (λ A) 1 nd for ν in the bll on the left hnd side of (.1) write λ = ν + µ, so tht ν < B 1 = (λ A) 1 1. Thus (.5) C := ( µ) n B n+1 n= is bounded opertor defined on ll of X which commutes with B. Observe tht µbc = ( µ) n+1 B n+ = ( µ) n B n+1 = ( µ) n B n+1 B = C B, n= where the second equlity is result of chnging the index of summtion. Hence (.6) C = B µbc = B(I µc) nd B = C + µcb = C(I + µb). The first of these two identities show tht N(C) N(B) nd R(C) R(B). Indeed, if x N(C), then = Cx = Bx µbcx = Bx, nd if Cz R(C), then Cz = B(z µcz) R(B). In prticulr, since N(B) = {}, we hve tht N(C) = {}. The second identity in (.6) then shows tht R(B) R(C) by similr rgument, nd we conclude tht R(C) = R(B) = D(A). Hence ν = λ + µ ρ(a) nd C = (ν A) 1. The identities (.), (.3), nd (.4), then follow by the sme proof s in the cse of bounded A: (.) is consequence of the fct tht λ A nd ν A commute, the identity (.3) follows from the sme computtion in the bounded cse, nd (.4) follows from (.3) nd the fct tht the series for C = (ν A) 1 in (.5) shows tht lim ν λ (ν A) 1 = B = (λ A) 1. n= 3. Self-Adjoint opertors In this section, we specilize to the cse where X is Hilbert spce, so tht in prticulr we hve n inner product on X. Definition 3.1. Let A be densely defined liner opertor on X. The opertor A is sid to be symmetric on X if Ax, y = x, Ay for ny x, y D(A). Exmple 3.. We discuss n exmple tht we be revisited in this section. Consider L ([, π]), nd consider the liner opertor d. We will consider two domins for this opertor (3.1) (3.) D(A D ) := {x(t) C ([, π]) : x() = x(π) = }, D(A N ) := {x(t) C ([, π]) : x () = x (π) = }, We denote A D = d when the domin is D(A D ) nd A N = d when the domin is D(A N ) (even though these two functions re determined by the sme rule, it is worthwhile to seprte them since their domins re different). In either cse, the liner opertor is well defined on ech domin nd we hve s consequence of

10 1 NOTES ON UNBOUNDED OPERATORS Green s identity (which in turn is strightforwrd consequence of integrtion by prts): π x (t)y(t) x(t)y (t) = x (t)y(t) y (t)x(t). t=π t= For ny choice of x, y D(A D ) or x, y D(A N ), the right hnd side vnishes here due to the boundry conditions imposed. Thus A D nd A N re both symmetric: A D x, y = x, A D y nd A N x, y = x, A N y. Proposition 3.3. Every densely defined symmetric opertor A is closble nd its closure is lso symmetric. Proof. To see tht A is closble, we ppel to Proposition 1., showing tht if (, y) Γ(A), then y =. To this end, let {x n } be sequence in D(A) such tht x n nd Ax n y. By continuity of the inner product, we hve for ny z D(A), y, z = since y, z = lim n Ax n, z = lim n x n, Az =, Az =. But since D(A) is dense in X, we my tke sequence of vectors {z n } in D(A) such tht z n y nd hence = lim n y, z n = y, y showing tht y =. We hve now shown A is closble, nd s usul we denote its closure s Ā. Recll this extends A to domin D(Ā) stisfying Γ(Ā) = {(x, Āx) : x D(Ā)} = Γ(A). The closure is seen to be symmetric by tking limits. Tke sequences {x n }, {z n } in D(A) such tht (x n, Ax n ) (x, Āx) nd (z n, Az n ) (z, Āz) in X X, leding to Āx, z = lim Ax n, z n = lim x n, Az n = x, Āz. n n Definition 3.4. Let A be densely defined liner opertor on X. We sy A is selfdjoint if it is symmetric, closed, nd D(A) = D(A ). We sy A is essentilly self-djoint if it is symmetric nd its closure Ā is self-djoint. Theorem 3.5. Let A by symmetric, densely defined liner opertor on X nd suppose tht there exists n countble orthonorml bsis {u n } such tht for ech n N, u n D(A), nd there exists λ n such tht Au n = λ n u n. Then A is essentilly self-djoint nd σ(a) = S where S := {λ n : n N}. Proof. Note tht since A is symmetric, then λ n R for ech N. Indeed, If x = x, u n u n D(A), then λ n = Au n, u n = u n, Au n = λ n. Ax, u n = x, Au n = λ n x, u n = λ n x, u n, nd hence Ax = Ax, u n u n = λ n x, u n u n. This shows (3.3) (1 + λ n) x, u n ( = x, un + Ax, u n ) = x + Ax <. We now define n opertor à by tking } (3.4) {x D(Ã) = X : (1 + λ n) x, u n <, Ãx := λ n x, u n u n.

11 NOTES ON UNBOUNDED OPERATORS 11 Note tht by (3.3), D(A) D(Ã) nd by the preceding observtions Ax = Ãx when x D(A). Since S R, we hve tht C\S. Thus there exists µ / S. Since S is closed set there exists ɛ > so tht B ɛ (µ) S =. In prticulr, this implies µ λ n ɛ for every n N. Observe tht by (3.4), (µ Ã)x = (µ λ n) x, u n u n. For z X, it is not hrd to see tht T z := (µ λ n) 1 z, u n u n defines bounded liner mp defined on ll of X with norm bounded bove by ɛ 1. It is verified tht it furnishes left nd right inverse for (µ Ã) : D(Ã) X s soon s it is shown tht R(T ) D(Ã). To see this, note tht 1 + λ n (µ λ n ) = λ n + 1 (µλ 1 n 1) mx(4(µ + 1), (1 + 4µ )ɛ ). Indeed, if λ n µ, then the second expression for the rtio shows tht it is bounded by the first expression inside the mximum. Otherwise if λ n µ, the the first expression for the rtio nd the bound µ λ n ɛ shows the rtio is bounded by the second expression in the mximum. The comprison test for series now shows 1 + λ n (µ λ n ) z, u n <. Since T = (µ Ã) 1 is bounded nd defined on ll of X, its grph is closed. Therefore, Γ(µ Ã) is lso closed since it is the imge of Γ((µ Ã) 1 ) under the permuttion (x, y) (y, x) (cf. the remrk following Definition 1.1). Thus if (x n, Ãx n) Γ(Ã) nd (x n, Ãx n) (x, y) in X X, then (x n, (µ Ã)x n) (x, µx y). But Γ(µ Ã) is closed, so x Γ(µ Ã) nd (µ Ã)x = µx y which implies tht Ãx = y. This shows tht Γ(Ã) is closed. We now show tht Γ(Ā) = Γ(A) = Γ(Ã). Indeed, we hve shown there is closed opertor à which extends A nd hence since Γ(Ã) is closed subpce contining Γ(A), we hve Γ(A) Γ(Ã). However, we still need to see the opposite continment. To tht end, suppose x D(Ã) nd write x = x, u n u n in terms of the orthonorml bsis. The series converges in norm, in other words, if x m := m x, u n u n then x m x. But the u n re in D(A), which is subspce so x m D(A) for every m, (x m, Ax m ) Γ(A) nd since D(A) D(Ã), lim Ax m = lim Ãx m = lim m m m m λ n x, u n u n = Ãx. Since Γ(Ã) is closed, (x m, Ax m ) (x, Ãx) in X X, implying tht (x, Ãx) Γ(A) proving the clim tht Γ(Ã) Γ(A). We cn now conclude tht à = Ā is self djoint. The continment D(Ã) D(à ) follows since the closure of ny symmetric opertor is symmetric. To see the opposite continment, suppose y D(à ), tht is, there exists z X such tht Ãx, y = x, z for ny x D(Ã). Tking x = u n in this identity, we see tht λ n u n, y = u n, z which llows us to conclude y D(Ã) since (1 + λ n) y, u n = y, u n + z, u n <

12 1 NOTES ON UNBOUNDED OPERATORS Exmple 3.6. We now return to the setting of Exmple 3., reclling the two domins for d given there. Consider the following sequences of functions { ( ) 1 } { ( ) 1 } sin(nt) nd cos(nt), π π which we denote s {u n,d (t)} nd {u n,n (t)}. In the ltter cse only, we include -th term in the sequence given by the constnt function u,n (t) 1 π. It is esy to check tht d u n,d(t) = n, d u n,n (t) = n, nd in the ltter cse, the identity holds even for n =. It is stndrd result from the theory of Fourier series tht {u n,d (t)}, {u n,n (t)} n= re orthonorml with respect to the inner product in L ([, π]). Moreover, they ctully form n orthonorml bsis for L ([, π]). To see this, recll tht the union of the functions u n,d, u n,n form bsis for L ([ π, π]) (see e.g. Chpter 8 of Rudin s Principles of Mthemticl Anlysis). One cn then tke odd (or even) extensions of ny function in L ([, π]) nd the coefficients in the bsis with respect to u n,n (resp. u n,d ) will vnish, leving convergent sum in terms of the u n,d (resp. u n,n ). Hence Theorem 3.5 implies tht A D, A N extend to closed self-djoint opertors Ā D, Ā N on L ([, π]). However, the extensions re indeed different! For one, the theorem implies σ(ād) = { n : n N} while σ(ān ) = { n : n N} {} nd the inclusion of in the ltter set cn mke substntive difference in the nlysis of these extensions. For exmple, Ā D is invertible, while ĀN is not. Moreover, u,n D(ĀN), while u,n / D(ĀD). Indeed, π u n,d (t)u,n (t) = π π sin(nt) = { πn n odd, n even, but since 8(1+n 4 ) n=k+1 πn =, u,n / D(ĀD) (cf. (3.4)). This is perhps not so surprising since the initl domin D(A D ) consists of functions which vnish t the endpoints nd bsed on our experience in 1.1.1, we expect tht functions in the domin of the closure D(ĀD) should in some sense vnish t the endpoints s well. Theorem 3.7. Suppose A is densely defined, self-djoint, closed opertor on X. Then σ(a) R nd σ(a). Moreover, (( z A) 1 ) = (z A) 1 for z / R nd (z A) 1 Im z 1, z / R. Proof. We first dispose of the identity (( z A) 1 ) = (z A) 1 ssuming tht z ρ(a) for z / R. To this end, suppose x 1, x D(A) nd y 1 = (z A)x 1, y = ( z A)x. The crucil observtion now is tht (z A)x 1, x = x 1, ( z A)x = y 1, ( z A) 1 y = (z A) 1 y 1, y, nd the clim now follows since x 1, x D(A) re rbitrry nd we re ssuming (z A) 1, ( z A) 1 re bijections.

13 NOTES ON UNBOUNDED OPERATORS 13 We begin by climing tht A + i : D(A) X is invertible. First observe (A ± i)x = Ax ± ix = Ax + ix ± Re Ax, ix = Ax + x ± Im Ax, x = Ax + x where we hve used the identity Re ( iz) = Im z in the second to lst equlity nd tht Ax, x R, by self djointness in the lst equlity. Alredy this shows tht N(A ± i) = {}, for if x N(A ± i), we hve = (A ± i)x = Ax + x x, nd hence x =. The fct tht N(A i) = {} will be used momentrily. We now clim tht R(A + i) is closed nd tht R(A + i) = {} which will imply tht R(A + i) = {} = X nd hence the opertor is invertible. Suppose y R(A + i), so tht there exists sequence x n D(A) such tht (A + i)x n y. In prticulr, {(A + i)x n } is Cuchy sequence nd since (A + i)(x n x m ) = Ax n Ax m + x n x m we see tht {x n }, {Ax n } re both Cuchy sequences s well. Denote their limits s x, z respectively so tht (x n, Ax n ) (x, z) in X X. But since Γ(A) is closed, we must hve x D(A) nd z = Ax nd this in turn implies tht y = lim n (A + i)x n = (A + i)x R(A + i), showing tht R(A + i) is indeed closed. Now suppose z R(A + i), so tht for every x D(A), (A + i)x, z = which mens Ax, z x, iz =, tht is, Ax, z = x, iz for every x D(A). But this is mens tht z D(A ) = D(A) nd Az = A z = iz, tht is (A i)z =. Since N(A i) = {}, we hve z = which shows R(A + i) is indeed trivil. We hve shown tht i ρ(a) nd if x = (A + i) 1 y, then y = (A + i)x = Ax + x x = (A + i) 1 y, which implies (A+i) 1 1, showing the conclusion of the theorem in the specil cse z = i. For other z = α + βi with β nd α, β R, we note (( ) ) A + α z A = (α + βi A) = β + i. β Since A is self-djoint, so is 1 β ( A + α), so the rguments bove show tht the opertor in prentheses on the right is invertible nd (( ) 1 A + α + i), (z A) 1 = 1 β (z A) 1 = 1 β (( A + α β β ) 1 + i) 1 β = Im z 1. We conclude this section with somewht curious exmple, which shows tht if A is not symmetric but not self-djoint, then A is not necessrily symmetric.

14 14 NOTES ON UNBOUNDED OPERATORS Exmple 3.8. Let X be the complex vector spce X = L ([, )), let D(A) = C 1 c (, ) = {x C 1 (, ) : supp(x) is compct subset of (, )}, which is known to be dense in X. Note tht A is symmetric: for x, y D(A), Ax, y = ix (t)y(t) = ix(t)y (t) = x(t)iy (t) = x, Ay, where we hve used tht the compct support in (, ) mens tht the boundry terms in the integrtion by prts in the second equlity vnish. By Proposition 3.3, A is closble nd s usul, Ā denotes its symmetric closure. Thus if x D(Ā), there exists sequence {x n } in D(A) such tht x n x nd Ax n Āx. Fix z(t) = e t nd since e t = 1 <, z L ([, )). We clim tht z D(Ā ) nd Ā z = iz. If this is true, then Ā is not symmetric, even though Ā is! Indeed, it mens tht Ā z, z = i z i z = z, Ā z. To see tht indeed z D(Ā ) nd Ā z = iz, first suppose x D(A). We hve Ax, z = ix (t)z(t) = x(t)iz (t) = x, iz where the second identity follows from nerly the sme integrtion by prts s bove, using tht x Cc 1 (, ) implies tht x()z() = nd lim t x(t)z(t) =. The third identity is clculus computtion iz (t) = ie t = iz(t). This isn t quite enough to show tht z D(Ā ) nd A z = iz, since we hve ssumed x D(A), but we cn tke limits of elements in D(A). To this end, given x D(Ā) tke n pproximting sequence (x n, Ax n ) (x, Āx) with x n D(A) given bove. Hence Āx, z = lim Ax n, z = lim x n, iz = x, iz, n n showing tht Āx, z = x, iz for ll x D(Ā) which shows the clim.