MeromorphicFamiliesof CompactOperators STANLY STEINBERG

Communicated by M. M. SCHIFFER Let B be a Banach Space and O(B) be the bounded operators on B. Also, let 12 be a subset of the complex plane which is open and connected. We are going to consider functions T(z) mapping 12 into O (B). T(z) is said to be analytic in 12 if, for each Zoe 12, oo

T(z) = E T~(z - Zo)", 0

where T~eO(B) and where the series converges in the uniform operator topology in some neighborhood of Zo (the topology is not critical; see HILLE&PmLLn'S [4], page 92). T(z) is said to be meromorphic in f2 if it is defined and analytic in f2 except for a discrete set of points. If z o is one of the exceptional points, then T(z)= ~ Tn(z-zo) ~, -N

where TneO(B), N>O, and where the series converges in the uniform operator topology in some punctured neighborhood of z o . We will first state our main theorems, then give their proofs and after that give some corollaries and comments. Theorem 1. If T(z) is an analytic family of compact operators for ze12, then either (I-T(z)) is nowhere invertible in 12 or else (I-T(z))-1 is meromorphic in 12. The next theorem gives sufficient conditions for a pole of (1-T(z))- ~ to be of order 1. Theorem 2. Let T(z) be an analytic family of compact operators for z near z o.

Let oo

I - T ( z ) = • A n(z - Zo) n 0

and Po(z)=,, 1 . I ( 2 - T ( z ) ) - ' d2, Z~I

C

where C={2, 12-1[ =~/} for ~1 sufficiently small. Suppose also that (I-T(z)) is somewhere invertible but not invertible at Zo, and that i) Ag v=o=~Aov=O for all v e B and ii) Aov=O=:,P(zo)Alv4:O for all veB, v4:0.

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Then ( I - T(z))- ' has a pole of order 1 at Zo; that is, in a neighborhood of Zo, (I -

T(z)) -i = E B . ( z - Zo)n, --I

where B, ~ 0 (B) and B_, 4: O. The next theorem tells us how the poles of ( I - T(z)) behave under perturbation. Theorem 3. Suppose T(z, x) is a family of compact operators analytic in z and jointly continuous in (z, x)for each (z, x)ef2 x R (R denotes the set of real numbers). If ( I - T ( z , x)) is somewhere invertible for each x, then (1-T(z, x))-' is meromorphic in z for each x. If zo is not a pole of ( I - T ( z , Xo))-1, then ( I - T ( z , x)) -1 isjointly continuous in (z, x) at (z o , Xo). Moreover, the poles of ( I - T(z, x))- ' depend continuously on x and can appear and disappear only at the boundary of f2 (including ~ ). Proof of Theorem 1. Suppose Cep(T(z)), the resolvent set of T(z). If we define i) P ( z ) = 2 ~ c ( 2 - T ( z ) ) - I

d2,

ii) M(z) = Range P(z), M c(z) = Range ( I - P (z)), then a) P2(z)=P(z), P(z) T(z)= T(z) e(z) and b) T(z) M(z) ~_M(z), T(z) MC(z) ~_M~(z), H = M (z) + M ~(z), M (z) n M ~(z) = {0}. In addition, if iii) 7"(z) = r(z)[M(z); i.e., r(z) restricted to M(z), and TC(z)= T(z)[M~(z), then

c) a(T(z)) = a(T(z)) c~interior of C

(a = spectrum),

a(TC(z)) = a(T(z)) c~ exterior of C and d)

T(z) = Yr(z)P(z)+ fC(z)(r- P(z)), (2- T(z)) -I = (2- T(z))-I P(z) + (2- 7"C(z))- I( I - e(z))

if ( 2 - T(z))-I and (A- 7"C(z))-' exist. For a proof of these facts see HILLE& PHILLIPS[4], page 177. If we restrict C so that it encloses at most the point I in tr(T(z)), then HILLE & PHILLIPS([4], page 182) show that since T(z) is compact, M is finite dimensional and consists of all vectors veB such that (I-T(z))nv=0 for some n>0. If we let T* be the adjoint of T operating on B*, the adjoint space of B, then HILLE

374

S. STEINBERG:

& PHILLIPS also show that

P * ( z ) = ,, 1 - S ( 2 ZT~l c

T*(z))-I d2.

If {vi} is a basis for M, then there is a basis {v*} for M * = R a n g e ( P * ) such that v*(vj) = 0 if i~j and v* (vi)= 1. Consequently

P(z) v= E v*(v) v,. For the following argument we need only assume that z is in some topological space and that T(z) is continuous in z. We consider some fixed Zo and some fixed curve C enclosing at most the point 1 in a(T(zo)). Such a curve can be found, since T(zo) is compact and thus either 1 ~a(T(zo)) or 1 is an isolated point of a(T(zo)). In fact, we can take C={z, [z-11 =t/} for t/sufficiently small. We now observe that 2 - T(z) is jointly continuous in (2, z). Thus, if 20 - T(zo) is invertible, then for (2, z) sufficiently close to (20, Zo) we can make 11(2- T(z))-(20 - Z(zo))ll < 11(2o- Z(zo))-111-1. Consequently,

(2- T(z)) -1 =(20 - T(zo))-~ E

[{20 -

T(zo)-(2-

T(z)} (20 -

T(zo))- 1]n,

where the Neumann series is absolutely uniformly convergent. It is now clear that (2-T(z))-1 is jointly continuous at (2, z) for 2 e C and z near z o . Also, since C is a compact set, P(z) is continuous in z for z near Zo. If we now introduce iv)

U(z)= P(zo) P(z) + (1- P(zo)) ( I -

P(z)),

then for z near z o we have

e) U(z) is

1 - 1 and onto; i.e.,

U-leO(B)

and

f) U(z) M(z)=M(zo), U(z) Me(z)=MC(zo), re(z) e(z) = P(zo) V(z), V(z) ( i - e(z)) = q - P(Zo)) V(z). The operator U(z) has been introduced by WOLF [7]. Since P(z) is continuous, there is an M and a neighborhood of zo where IIP(z)ll < M and Ile(z)-P(zo)P (Zo)II < (2 M ) - I Consequently III- u (z)ll = IIP(zo)

(P(z)- P(zo)) + e((z)- e(zo)) e

< lle(zo)ll Thus

IIP(z)-e(zo)ll + IIe(z)-P(zo)ll

U(z)=I-(I- U(z)) is invertible and

(z)lI IIP(z)ll < 1.

U - l ( z ) is given by the Neumann series

or)

U -~ ( z ) = ~ ( I - U ( z ) ) " . This proves e). The proof of f) is clear. 0

We now assume v) B ( z ) =

U(z)(I- T(z)) U-1(z), B(z)=B(z)lM(zo), Br

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Then in a sufficiently small neighborhood of Zo, g) / ~ ( g ) - - U ( z ) ( I - Z ( g ) )

u-l(z), BC(z)=U(z)(I- TC(z)) U-I(z),

h) /~(z) M(zo) ~_M(zo), BC(z) MC(zo) ~_MC(zo), k) U (z) (2- ( I - T(z))) -' U - '(z) = (2- B(z)) -' = (2 - / ~ (z))-I p (Zo) + (2 - / ~ (z))-i (I - P (Zo)), provided the inverses exist, and l) a ( I - T(z))=a(B(z)), a ( I - :F(z))=a([~(z)), a ( I - T~(z))=a([~(z)). We shall say that the operators B(z) are equivalent to the operators (I-T(z)). Since the operators B(z) are all reduced by the same subspaces M(zo) and M*(zo), where B = M ( z o ) + MC(zo)and M(zo)c~ MC(zo)={0}, we shall say that the operators B(z) are simultaneously in block form. The proofs of g) through I) follow easily from what we have done. Moreover, it is clear that if T(z) is analytic in z, then so are all the other operators introduced. We now show that if (I-T(z)) has an accumulation point of "zeros", then (I-T(z)) is nowhere invertible. Suppose z,,--',zo and lea(T(z,,)) for n > 0 . By h), B(z) is an operator on a fixed finite dimensional space. Hence A (z), the determinant (B(z)), is a well defined analytic function of z. Since 1 ea(T(z,,)) we see by c) and 1) that 0ea(/'~(zn) ), and thus A (zn)=0. Because A(z) is analytic, and zn--*Zo, we infer that A(z)=O in a neighborhood of Zo. Again by c) and 1), this implies 1 ~a(T(z)) for z near Zo. We see, as in the theory of analytic continuation, that the component of the set S={z, zef2, 1ea(T(z))} containing z o is both open and closed in f2. Also f2 is connected. Therefore S must be all of f2, so that ( I - T(z)) is nowhere invertible. Hence, if we assume that (I-T(z)) is somewhere invertible, then the points where l ea(T(z)) are isolated. We now show that such points are poles of ( I - T(z))- 1. Let z o be one of these points and define P, U, B,/~, and/~c as above. We apply c) and 1) several times. We conclude t h a t / ~ ( z ) is invertible and or)

B (z) = E C:(z - Zo) 0

Also, B(z) is invertible for z:l=zo and M is finite dimensional. From Cramer's rule it is clear that oo

Z C.(z- zo)

-N

where C_N~eO and N > 0 . Thus (I -- T(z))- ' = ~ U - ' (z) ( C~ Po + C~(I - Po)) U (z) (z - Zo) ~,

-N

376

S. STEINBERG:

where CLn=O for n > O and Po =P(zo). It follows that oo

(I -- T(z)) -1 = ~ Bn(z - Zo)~,

(.)

-N

where B~eO(H) and B_N=C_NPo~O. If z o is not one of the above mentioned points, then ( I - T ( z ) ) - ' exists and is analytic in a neighborhood of Zo. Hence (I-T(z))-1 is meromorphic in 12. Remark. In view of (.), we define the order of the pole at Zo to be N. We define the weak order of the pole at z o to be the order of the zero of A(z)=det(B(z)) at Zo. Corollary 1. If T(z) is an analytic 0 (H)-valued function for zeO ~_if2, if T"(z) is compact for some n > 1, and if (I- T" (z)) is somewhere invertible, then ( I - T(z))-I is a meromorphic O(H)-valued function. Here we assume that f2 is open and connected. Proof. Use the identity

( I - T ( z ) ) - I = (I + T(z) + . . . + T "-1 (z)) ( I - T"(z))- 1. Proof of Theorem 2. We know from Theorem 1 that ( I - T ( z ) ) - x is meromorphic, which implies co

( I - T(z)) -I = E B.(z - Zo)~ -N

where B-N 4: 0. Since (1-T(zo) ) is not invertible, N > 1. We now assume that N > 2 and obtain a contradiction. F r o m the proof of T h e o r e m 2 , PoB_N=B_NPo where eo = P (Zo). Multiplying out the series for ( I - T(z))- i ( I - T(z)) = I and collecting powers of z, we find B-sAI+B-N+IAo=O, which is true only if N > I . Recalling that v e M implies (I-T(zo))mv=Ar~v=O for some m > 0 , then by i), Aov =0. Thus 0 =PoB-NAlV =B-N(PoAx) v. But by ii), PoA1 is 1 - 1 on the finite dimensional space M. Hence B_ N = 0 on M. Also B-N = 0 on M c and M + M ~=H. Therefore B_ N =0. Corollary 2. Under the conditions of Theorem 2,

(I -- T(z))- I = B_ , Po + T, (z), Z--Z 0

where T1 (z) is analytic at Zo, T1 (Zo) = ( I - T(zo))- i (I - Po), and B_ 1 is invertible on M. Proof. The value of T1 (Zo) is clear from the proof of Theorem 1 and does not depend on the fact that the pole is of order 1. If we multiply out the series for ( I - T ( z ) ) - l ( I - T ( z ) ) = I and collect powers of z, we obtain B _ I A I + B o A o = I . Multiplying this relation by Po on the left and right, and recalling that Po B_ 1 = B _ I P o and A o P o = 0 , we obtain B_IPoAtPo=Po . Since both Po and PoA1Po are 1 - 1 operators on a finite dimensional space M, B_ 1 must be 1 - 1 ; in fact, BZ~ =(PoA1Po) -1 as an operator on M.

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Proof of Theorem 3. To be more precise about the statement of the theorem, we will prove the existence of a set K of continuous functions z(x). Each z(x) is defined on an interval (as, b~)_ R which depends on z(x). Each z(x) has its range in f2, and z(x) must be a pole of (I-T(z,x)) -1. Also, if Zo is a pole of (I-T(z, x0))-1, then there are as many functions z(x)eK with Z(Xo)=Zo as the weak order of the pole of (I-T(z, Xo))-1 at z o. In addition, if z(x)eK and is defined on (a, b) and a (or b) is finite, then for any compact subset of f2 and for x near enough to a (or b) we require that z(x) lie outside of the compact subset. Let A (z, x)=I-T(z, x). As before, we define for any point (z o, Xo) 1

P(z,

r - l ( z , x)) -1

U (z, x)= P(zo, Xo) P(z, x) + ( I - P(z o , Xo)) ( I - P(z, x)), B (z, x) = U (z, x) A (z, x) U- 1 (z, x),

M=R(P(zo, Xo)), M~=R(I-P(zo, Xo)) ( R = R a n g e ) , and

(z, x) = B (z, x) [M,

BC(z,x) = B (z, x) IU ~,

where C encloses at most the point 1 in a(T(zo, Xo)). From Theorem 1 and its proof it follows that A - 1(z, x) is meromorphic in z and jointly continuous in (z, x) at (Zo, Xo) if z0 is not a pole of A - 1 (z, Xo). Statements a) through 1) in the proof of Theorem 1 are still true if we replace (z) by (z, x). We first show that the poles of ( I - T(z, x))- ~ cannot " a p p e a r " o r " disappear" except possibly at the boundary of f2. We suppose there are sequences x, ~ Xo and z,~zoef2 such that A-l(z, x,) has a pole at z, for n > 0 . If z o is not a pole of A-l(z, Xo), then P(zo, Xo)=0. Since P(z, x) is continuous in (z, x), [[P(z, x ) P(zo, Xo)[] = []P(z, x)[] _-<1 for (z, x) near (z o, Xo). Since P(z, x) is an idempotent, liP(z, x)][ =
378

S. STEINBERG :

z(x) for x =b. The case x = a will follow in the same manner. If lim z(x) exists and x"*b

belongs to f2, we define z(b)=lim z(x). The first part of this proof implies that x"~b

z(b) is a pole of A-1 (z, b). Then using the preceeding part of this proof, where we replace (Zo, Xo) by (z(b), b), we can prolong z(x) for x>b. In general, this limit may not exist. If for each sequence xn ~ b, z (x,) eventually lies outside of every compact subset of f2, then z(x) approaches the boundary of f2 as x ~ b and we cannot prolong z(x) any further. On the other hand, if there is some sequence x , ~ b with z(x,) lying inside some compact subset of f2, then there is a subsequence of xn, which we call xn, such that x, ~ b and z(x,)~zoef2. We want to show that lim z(x) exists and equals Zo. If lim z(x) does not exist, x"*b

x-~b

then there must be another sequence x', with x" ~ b and [z ( x ' , ) - Z o [ > 6 > 0. By the first part of the proof, z o must be a pole of A-1 (z, Xo). Such poles are isolated for fixed x =Xo, so we can find a 6 ' < 6 so that A - x (z, Xo) is invertible for 0 < Iz - z o [ < 6'. Now, since the curve Iz - z0 [ = 6' is a compact set and A (z, x) is jointly continuous in (z, x), there is a ~ so that A (z, x) is invertible for Ix - x o l <~ and Iz-zol =~'. Next, we choose an n so large that Ix.-xo I<~ and Ix'-xol <~. Recall that z(x) is a continuous curve joining z(x,) to z(x'n) and hence must pass through the set [Z-Zo] = 6 ' , contradicting the invertability of A(z, x) at suet[ points. Lemma. Let f(z, x) be a function mapping C x R ~ ~ which is analytic in z and jointly continuous in (z, x). Let Zo be a zero off(z, Xo) of multiplicity m. Also, let zi(x), i = 1 . . . . . m'
I - T ( z , x) depend continuously on x. Also, if T(z, x ) = z - i T(x) where T(x) is a continuous family of compact operators, then ( I - T ( z , x))-1 = z ( z _ T ( x ) ) - i and thus the poles of ( I - T ( z , x)) correspond to the eigenvalues of T(x). We can thus assert. Theorem 4. If T(x) is a family of compact operators depending continuously on x, then the nonzero eigenvalues of T(x) depend continuously on x. Moreover, the eigenspace of T(x) corresponding to ,~(x) depends continuously on x if x is not a point where 2(x) coalesces with another distinct eigenvalue. We note that ATKINSON [1] has given a proof of essentially the same fact. Theorem 5. Suppose T(z, x) is a family of compact operators which are analytic in z for fixed x and analytic in x for fixed z. In addition, if T(z, x) is somewhere

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invertible for each x, then the poles of ( I - T ( z , x)) -1 are analytic functions of x 1/p for some integer p, and the poles can appear and disappear only at the boundary of f2. Proof. We only need to replace the use of the lemma in the proof of Theorem 3 with the discussion on page 237 of GOLrRSAT[3]. Remarks. TAMARKIN[6] has proved a special case of Theorem 1 using the theory of Fredholm determinents. Also, a theorem somewhat weaker than Theorem 1 appears in DLrNFORD& SCHWARTZ,[2], page 592. The results in this paper in the author's Ph.D. Thesis [5] written under the direction of Professor R. PHILLIPS.

References 1. ATKINSON,K. E., The numerical solution of the eigenvalue problem for compact integral operators. Trans. Amer. Math. Soc. 129, 458--465 (1967). 2. DUNFORD,N., & J. SCHWARTZ,Linear Operators, Vol. 1. New York: Interscience 1958. 3. GOURSAT,E., Functions of a Complex Variable. Boston: Ginn 1916. 4. HILLE,E., & R. S. PnILL:PS, Functional Analysis and Semi-Groups. Am. Math. Soc. Coll. Publ. XXXI, Providence, (1957). 5. STEINBERG,S. L., On scattering theory for the Schroedinger equation. Ph.D. Thesis, Stanford University 1968. 6. TAMARrdN,J. D., On Fredholm's integral equations, whose kernels are analytic in a parameter. Ann. of Math. 28, 127-- 152 (1927). 7. WOLF,F., Analytic perturbations of operators in a Banach space. Math. Ann. 124, 317--333 (1952). Purdue University Lafayette, Indiana

(Received October 1, 1968)

27 Arch.Rational Mech.Anal., Vol.31