Conditions for polynomials in right inverses with stationary and algebraic coefficients to be Yolterra operators

ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXX (1991)

Ng u y e n Va n Ma u

(Warszawa)

Conditions for polynomials in right inverses with stationary and algebraic coefficients

to be Yolterra operators

Abstract. The conditions for polynomials in right inverses with constant coefficients were obtained by Przeworska-Rolewicz and von Trotha (see Th. 2.4.1 in [6]). In the present note we generalize this result to the case of polynomials with algebraic and stationary coefficients.

Moreover, we also give a necessary and sufficient condition for

R1 + R2, R\Ri

0. Let X be a linear space over C. Denote by L(X) the set of all linear operators with domains and ranges in X. Let L0(X) = {A eL(X): domzl = X}.

Denote by R{X) the set of all right invertible operators belonging to L(X). For a DeR(X), we write

= { R e L

(X): DR = I},

= {.F

L 0(X): F

= F, F X = kerD, 3Re@ D FR = 0}.

If F e and FR = 0 for an R e &D then F is called an initial operator corresponding to R. In the sequel, we assume kerD Ф 0.

Following [6] we say that an operator A e L

(X) is a Volterra operator if for every scalar t the operator I — tR is invertible. We denote by V(X) the set of all Volterra operators from L

(X).

De f in it io n

1 [8]. An operator A e L

(X) is said to be stationary if DA = AD on domD, RA = AR.

For given D e R (X ) and R

@

we denote by SDtR the set of all stationary operators.

De f in it io n

2 [7]. We say that an operator A e L

(X) is algebraic if there exists a non-zero polynomial

P{t) =

p

, . . . , p NeC,

such that P(A) = 0.

Without loss of generality we can assume that pN = 1, i.e. P(t) is unitary.

We say that the algebraic operator A is of order N if no unitary polynomial (2(0 of degree m < N such that Q(A) = 0 on X exists. The minimal unitary polynomial annihilating A is called the characteristic polynomial of A and will be denoted by PA(t). Denote by sé the set of all algebraic operators.

De f in it io n

3 [1]. Let 38 c L0(X) be a commutative algebra. An operator S

L

(X) is said to be generalized algebraic over 38 if there exists a polynomial

Af(0 = A 0 + A l t + ... + Antn, An Ф 0,

with A0, ..., A„e38, AjS = SAj (j = 0, ..., n) such that M(S) = 0.

The following results were obtained by Przeworska-Rolewicz and von Trotha (see [6], Th. 2.4.1).

Th e o r e m

I. Write

(1) Q(t, s) = £ è (0 = Q(t, 1), P(t) = tMQ(t),

k = 0

where q0, ..., q^-x eC , qN = 1, M is a non-negative integer.

(i) (Przeworska-Rolewicz). I f there exists an Re3êDn V(X) (Le. R is a Volterra right inverse of D) then P(D)e R(X) and

(2) i^o = ^ M+iV[ 0 ( / , ^ ) ] _1 is a Volterra inverse of P(D).

(ii) (von Trotha). I f R 0 of the form (2) is a Volterra operator then R is a Volterra operator.

In the present note we generalize Theorem I to the case when Q(t, s) is a polynomial with algebraic and stationary coefficients (Theorems 1, 2 and 3).

Moreover, we give conditions for R l + R 2, R l R 2 to be Volterra operators provided that R lt R 2 are Volterra right inverses of a D

R(X) (Theorems 9,10).

The method used here is essentially based on the properties of generalized algebraic operators obtained in [l]-[5 ].

Th e o r e m

II [2]. Let S be an algebraic operator with characteristic polynomial of the form

p s(t) = f t ( t - t y

i = 1

Ch # tj for i r1+ ... +r„ = N 0, t j c C j = 1 ,..., и), and m

V:= V(S)= ^ AjSm- j

7 = 1

where A

E38 (see Definition 3), AjS = SAj, j = 1, ..., m. Then V is a generalized

algebraic operator.

Theorem

III

Let A and В be commuting algebraic operators with characteristic polynomials

П

PA(t) = П ui ^

f ° r i # j ( i j = !>•••, n),

i= 1

m

р в( 0 = П ^ vi for {i, j m),

j= i

respectively. Then A + B is an algebraic operator with characteristic roots belonging to the set

i =

n; j =

. . . m}.

1. Let D

R{X), R e M D. Let A0, A

^ n S DtR be mutually com­

muting and let A

= J.

Write

(4) Q(t,

= £ 0 (0 = Q(t, 1), p(t) =

7 = 0

where M is a non-negative integer.

Theorem

1. I f R

V(X) then the operator Q(I, R) is invertible and

(5) R

= R N+MIQ(I, V(X).

P ro o f. Write

^ 0 = Нп{Я*} (k = 0, 1,...).

Then J '

c= L0(X) is a commutative algebra. Hence, by Theorem II, Q(I, R) is a generalized algebraic operator with characteristic roots belonging to the set

N

№) {/+ I k = 1 , ,

k= 1

where { ^ _ м , • ••, ^-к,гк -к)к

are the characteristic roots of the operators

^N-k (k = 1 ,..., ЛГ), respectively.

Theorem I implies that every operator I + Y

!=

tN- ktjRk is invertible. It follows that Q(I, R) is invertible (cf. [2]).

Now we prove that R ± = R N[Q(I, Я)]-1 is a right inverse of Q(D). Indeed, Q(D)R

= Q(D)RN[Q(I, Я )]"1 = £ Я)]-1 = /.

k = 0

Consequently, the operator P(D) = DMQ(D) is also right invertible and has a right inverse of the form (5).

To finish the proof, we have to check that R

V(X). For tE C we

obtain I - t R

= [Q(I, P )]_1H(P), where H(R) = YJk=oAkRN~k- t R N+M. By

Theorem II, H(R) is a generalized algebraic operator with characteristic roots

belonging to ,the set

(7) {I + tR”+M + £ tN. k,jRk; j = 1 ,.... rN_t}.

k = 0

Theorem I shows that every operator I + tRN + M + YJk=otN-kjRk is invertible.

Hence, H(.R) is invertible (cf. [2]). We conclude that I — tR0, as a superposition of invertible operators, is again invertible, i.e.

Theorem 2.

Suppose that R e 0lD

V

for a D eR(X). Suppose, moreover, that Aestf n SD R. Then AR e V (X).

P ro o f. By Theorem II, I-{-AAR is a generalized algebraic operator over

= lin{Rk} (к = 0, 1, ...) with characteristic roots of the form {/ + 2uff?: i

= 1 ,..., n}. Since every operator / + Au}R is invertible for all А e C we conclude that I + AAR is invertible for all ДеС, i.e. AR is a Volterra operator.

The converse of Theorem 1 is given by the following

Theorem 3.

Let DeR(X), R

Md. Let A 0, ..., ANe j / n S DtR be mutually commuting and let AN = I. Suppose that Q{t, s), Q(t), P(t) are of the form (4). I f

Q(i, R) is invertible then

(8) R 0 = RM+Nm i , R ) T 1^^PiD).

Moreover, if R 0eV (X ) then ReV(X).

P ro o f. It is enough to check that R e V(X), provided that the operator R0 of the form (8) is a Volterra operator. Fix ДеС. Write A = ()(Д). Then A is an algebraic operator (cf. [6]) and AP(D) = P(D)A, AR0 = R 0A. By our assump­

tion, R0 is a Volterra operator. This implies that so is AR 0 (Theorem 2), and I — A R 0 is invertible.

On the other hand,

I - A R 0 = I - A R M+N = IQ(I, R ) - A R M+N][Q(I, R )YK From the last equality we obtain

/ = ( I - A R or l m , Д)]-1 [0 (/, R ) - A R M+*-]

= IQ(I, R ) - A R M+NW - A R or 1lQ(I, R ) T \ i.e. Q(I, R) — ARM+N is invertible.

Write

HA(t, s) = Q(t, s ) - A s M+N, HA(t) = HA(t, 1).

Then H

(A) = HA(Д, 1) = Q(A, 1) — A = Q(A) — A = 0 and HA(I, R) is invertible for HA(I, R) = Q(I, R ) ~ A R m+n.

Hence

HA(t) = (t-AI)Q A(t), where QA(t) = £ Bjtj

j = о

(9)

and B jeS D>R n jrf(j = 0, . . N — 1) are mutually commuting. From (9) we get HA(t, s) = (t-Às)QA(t, s),

where

Q

&

= V B j t W - 1 4 QA(t) = 0 Ж 1).

7=i

Thus HA(I, R) = R), i.e.

I = ( / - А К Ш / , Л)[ЯА(/, Я )]"1 = [Я А(/, я ) Г ^ ( J , Я)(1-ЛЯ).

This shows that I — XR is invertible for all XeC, i.e. ReV(X).

Co r o l l a r y 1.

Suppose that DeR(X), R e $ Dn V ( X ) and A e S DyR

sf.

Then every solution of the equation

(10) (D — A)x = y, y e X

is of the form

x = (I — AR)~1(Ry + z), where zekerD is arbitrary.

In particular, if A has the characteristic polynomial of the form П

PA(t) = П (f - ^ ) (*« ф h f ° r

t/геп the solutions of (10) are of the form П

x — (/ — tjR)~

Pj(Ry + z), zekerD , 7=i

where

P ,= П ] = 1 ,.... n.

k= l,k¥>j

Co r o l l a r y

2. dim ker 0(D) = iV dim kerf).

Indeed,

Q(D) = DN ^ AjRN~j = DNQ(I, R).

7 = 0

By Theorem 1, the operator Q(I, R) is invertible. Hence dim ker Q(D)

= dim ker DN = N dim ker D.

As a corollary, we obtain the formula

dim DMQ{D) = (M + AO dim ker D.

R em ark 1. In general, the converse to Theorem 2 is not true. Further­

more, for every D

R { X ) ,R

M

there exists an Aestf n S DyR such that

12 - Comment. Math. 30.2

AR e V (X), i.e. AR e V (X) does not imply R e V (X). Indeed, if A e stf n SDtR and A 2 — 0 then I — XAR is invertible for all Xe C and {I — XAR)"1 = I+ XAR. This implies A R e V ( X ) for every Re@tD. However, we have the following

C^orollary 3.

Suppose that DeR(X), Re£%D. Suppose, moreover, that A e s / n S DiR and A is invertible. I f AR is a Volterra operator then so is R.

Indeed, by Theorem 2, A ~ 1(AR) = R is Volterra, provided that AReV(X).

2. Algebraic exponentials.

Definition 4.

If A e jrf n SD<R

О ф xA e

(D—A)

x A

algebraic exponential element corresponding to A.

It is easy to check that if О Ф x^eker (I — AR) for an A e S D>Rn sé then xAeker(D — A), i.e. x A is an algebraic exponential and FxA = 0, where F is an initial operator for D corresponding to the right inverse R. By Theorems 1 and 3, in that case, R is not a Volterra operator.

Theorem 4.

Suppose that {An} c SDfR n sé is a sequence of algebraic operators such that Aj, А { — А-} are invertible for i Ф j . Then for an arbitrary positive integer n the algebraic exponentials xAl, ..., x An are linearly independent over the set srf r\S D>R.

P ro o f. By Definition 4, x Al Ф 0. Suppose that for a fixed к (к ^ 1) the algebraic exponentials x Al, ..., x Ak are linearly independent over sé n S DtR. If x Ak+1 is linearly dependent on {x^, •••,xAk} then there exists an operator 0 ф dk + 1

sé n SDtR such that

This implies (

)

fc + i

E

j = i

k+ 1

0 = D (E ^

)» d j e s / r\SDtR,

j = i

and

fe + i

° = Z dJÂ l XAj- 7=1

The equalities (11) and (12) together imply

0 = Z dM j - A i)*

, = Z d№ j ~ A i)x Aj-

j = 1 J = 2

By our assumption, dj(Aj — A 1) = 0 (j = 2, . . . , k + \ ) and the operators

Aj — A iij = 2, ..., k+1) are invertible. Hence dj = 0 {j = 2, ..., k+ 1), which

contradicts our assumption. We therefore conclude that the set {xAl, ..., xAk+1} is linearly independent over sé r\S DtR for an arbitrary k e N.

Theorem 5.

I f RDn V

for a D e R(X), A e s é

SDtR and

A —

then the operator eA = (I — AR)~i has no eigenvectors.

P ro o f. Suppose that there exists а и ф 0 such that for a t ф 0 we have eAu = tu. This implies

(13) [(\ — t)I + tRA~\u = 0.

By Theorem 2, AR e V(X). Hence, by (13), и — 0 for t Ф 1. This contradicts the assumption. If t — 1 then from (13) we get ARu = 0. Our assumption, that dim ker A = 0 implies и = 0, again yields a contradiction.

Definition 5.

If D

R(X) and there is an R e 01D n V

then every operator eA = (I — AR)- 1 , where A e SDtR n stf, is called an algebraic exponen­

tial operator.

Theorem6

. I f R e f%D

V

A e SDtR

s f and F is an initial operator for D corresponding to R, then the algebraic exponentials eA(z) are uniquely determined by their initial values:

eA(z) = ( I - A R ) ~

F(eA{z)), i.e. F(eA(z)) = z for zeker D.

P ro o f. Suppose that zekerD . Then (I — AR)eA{z) = z, i.e. eA(z)

= z + AReA{z) and DeA(z) = AeA(z). This means that eA(z) is an algebraic eigenvector of D corresponding to A. Thus

FeA(z) = (I-R D )e A(z) = eA(z)-R D eA(z)

= eA( z )-R A e A{z) = (I - A R ) e A(z), i.e. F(eA{z)) = z.

Corollary 4.

I f D eR(X), R e MD

V

then for every operator 0

é A e S DR n sé there exist non-trivial exponentials.

Corollary 5.

Suppose that DeR(X), {Rv}veIo <= n

and A s s # n SD'R. Then

(Fv- F J e A(z) = AFvR/ieA(z) for

D, v, p e l 0.

Indeed,

AFvR„eAz) = р Л цЛеА(2) = FvR»DeA(Z)

= F f l - F J e A z ) = (Fv — F VF f)eA(z) = {Fy- F f}eA{z).

Definition

6. If D eR(X ) and there is an R e ^ Dn V(X), then the operators

£ a ^ ( e i A F e —î a) , s a 2 j ( @ i A F e —i A ) ,

where Aestf n S D R, are called the algebraic cosine and sine operator, respec­

tively. The elements cA(z) and sA(z), where zekerZ), are called the algebraic cosine and sine elements, respectively.

Theorem 7.

The algebraic cosine and sine operators have the following properties:

c a

= {I + A 2R 2)~1,

= A(I + A 2R 2)~1R, c2 A + s2 A = eiAe - iA, (14)

+

= i(eiAeiB- e - iAe - iB),

CA CB ~ S A S B = 2.(e i A e iB + e - i A e ~ i B ) -

P r o o f is the same as for “scalar” cosine and sine operators (cf. [6], p. 99).

Corollary

6. DcA = —A

a,

= A

a.

Theorem 8.

Let F be an initial operator for a D

R(X) corresponding to R e & Dr\V(X). Then for every A e s / n S D<R and O ^ z e k e r D we have F

(

) = 0 and cA(z) ф 0.

P ro o f. By (14),

F

(

) = FA(I + A 2R2)~1R

= FRA{I + A 2R 2)~1z = 0.

If cA(z) = 0 then z = (/ + A 2R 2)(I + A 2R 2)~1z = (I+

= 0. which contradicts our assumption that z ф 0.

3. Operations on Volterra right inverses. Let De

R{X), R r, R 2

M

n V(X). It is well known that R XR 2 Ф P 2^ i (cf- [6]). The following question arises: Are R

_R2 and R 1 + R 2 Volterra operators, provided that R1? R 2 are Volterra operators? In general, the answer in negative. In order for R

+ R 2, ^

to be Volterra operators, we obtain the following conditions:

Theorem 9.

Let D

R{X), R x, R 2

$ D. Then R2 is a Volterra operator if and only if so is R 2R 1.

P ro o f. Suppose that R 1R 2

V(X). Write

C

R

= У ^

^

) ’ E = I t R 2eRlR2R l , tE C.

Then E is well defined on X and

(15) (I tR 1R 2)E = (I tR2R l)(I + tR2eBiB2R 1)

— I — tR2R l + t(I — tR2R 1)R2eRiR2R 1

= / — tR2R 1 + tR2{I — tR^R2)eRlRiR^ — I — tR2R^-\-tR2Ri = /.

Similarly, we find

(16)

(I — tR2R})E = /.

Thus I~ tR

R , is invertible for all te C , i.e. R

R i is a Volterra operator.

Th e o r e m

10. Suppose that D eR(X) and R t , R

are Volterra right inverses of D. Then a necessary and sufficient condition for R t R

to be a Volterra operator is that

(17) F2(I — tR2)~1z

0, 0 ^ zekerD , where F

is an initial operator for D corresponding to R 2.

P ro o f. Note that R

R

and R

is a Volterra right inverse of D2.

Hence, if R i R

has an eigenvector then it must be of the form (18) q = (I — t R l y

z, zekerD 2, z ф 0.

If v

t, veC then

и — (I — vR t R 2)(I — tR2) ~1 z = U - t R 2 l + R

{tRl - v R

f t ( I - t R 2l)~l z

= z + R^tRj^ — vR2)(I — tRl)~l z.

Hence D2u = (t — v)(I — tR2)~1z ф 0 for all 0 ^ zek erD 2, i.e. R t R

has no eigenvectors.

If v = teC , then we have и = (/ — tR

R 2)(I — tRj)~

z. Consider two cases:

(i) zekerD and (ii) zek e rD 2\kerD.

(i) In this case, we get

Fxu = F l ( I - t R l ) ~ 1z = l I - t R l + R ^ t R ^ D m i - t R l y ' z

= z + R y t R ^ - D y i - t R l y ' z .

Since domD = kerD + /?1X (cf. [6]), we conclude that F

M # 0 for 0

zekerD . Thus

0, i.e. R t R

has no eigenvector in kerD.

(ii) Let z = R

z

+ z2, where z1? z2ekerD , z l

0. It is easy to check that и = ( I - t R ^ J q = (I — R l R

D2)(I — tR2)~l z

= z + tR

F

R 1(I — t R l y

z.

If z2 Ф 0 then F^u = F

R l z

+ F

z

— z

ф 0, which implies и ф 0. If

— 0, i.e. z = R l zi , then

u = R

z

+ tR

F

R

( I - t R l y

z l = R

z

+ R l F

[I—(I—tRlJ] (I — t R j y

Zl

= R

zl + R

F2(I — tR

)~

zl — R

F

z l = R

F2(I — tR

)~

z

0.

Hence, и

for all v eC and zek erD 2 if and only if the condition (17) is

satisfied, which was to be proved. »

Similarly, we obtain the following

Th e o r e m

10'. I f R 1 and R 2 are Volterra right inverses of D

R(X) then a necessary and sufficient condition for R X R 2 to be a Volterra operator is that (19) FX(I — tR2)~l z ф 0 for all te C , zekerT),

where Fx is an initial operator for D corresponding to R x.

Ex a m p l e

1. Let X = të([0, 1], where F = C or F — R, and let

d x x

D

, R t = $, R 2 =

x x Ф x 2i x x,

XI X 2

It is easy to check that

(7 — tR2 ) ~1 с = c cos y / —t(x — Xj) for

F, j = l , 2. Hence

F 2(I — tR2)~1 c = ccoSyf--t(x2 — x 1).

If t0 = — ^(

2 —

)~2

2 then F2(I — toR 2)~1c = 0. Theorem 10 shows that R XR2 is not a Volterra operator.

Th e o r e m 11.

Suppose that D

R(X) and R x, R 2

n V

Then a neces­

sary and sufficient condition for R x + R 2 to be a Volterra operator is that (20) (7 — tjR1)_1z + (7 — tR2)~l z Ф 0 for all zeker7)\{0}.

P ro o f. Let R = %(RX+ R 2). Then R

Md. Hence every non-trivial eigen­

vector of R (if exists) must be of the form

q = (I — tRf)~x z, 0 # z e k e r D . If v

C and v ф t then

и = (I — vR)q = I - ^ v ( R x + R 2) ( I - t R x)~xz

= I — tR x + (t — %v)(Rx — jv R 2)(I — tR x)~l z

= z + ((t — v/2)R1—(v/2)R2)(I — tR 1)~1z.

This implies that

Du = (t — v){I — tRx)~xz # 0 for 0 # zekerD . If v = t

C, then

и = (I — tR)q = z + (t/2)(Rx—R 2)(I — tR x)~xz = ^z + ^(I — tR2)(I — tR x)~xz.

Hence

2 и = z + (7 — tR2)(I — zR x)~x z, i.e.

2(1 — tR2)~xu = (I — tR x)~x z + (I — tR2)~x z .

From this equality we conclude that и Ф 0 if and only if the right hand side of this last equality is different from zero, i.e. we get the condition (20).

Ex a m p l e

2. Let X , D, R lt R

be the same as in Example 1. It is easy to check that

(/ — = cexp[t(x — x})~\ f o r c e d ,

= 1,2.

Hence

u(x) = (I — tR 1)~1c + (I — tR2)~1c = cetx[e~tXl +etX2].

This equality implies that u(x) Ф 0 for all t e R. By Theorem 11, ^(R

+ R 2) is a Volterra operator in the space X = ^([0, 1], R).

Note that in the space X = ^ [ [ 0 , 1], C ], u(x) = 0 for t = (x2 — x ^ 1^ . Hence, by Theorem 11, R t + R

is not a Volterra operator.

References

[1] N g u y e n V an M au, On algebraic properties of differential and singular integral operators with shift, Differentsial’nye Uravneniya 10 (1986), 1799-1805 (in Russian).

[2] —, Characterization o f polynomials in algebraic elements with commutative coefficients, Acta Univ. Lodz. Folia Math., to appear.

[3] —, Characterization o f polynomials in algebraic elements with constant coefficients, Demon- stratio Math. 16 (1983), 375-405.

[4] —, Arithmetical operations on algebraic operators, ibid. 22 (1989), 1109-1119.

[5] —, On solvability in closed form o f a class o f singular integral equations, Differentsial’nye Uravneniya 25 (1989), 307-311.

[6] D. P r z e w o r s k a - R o le w ic z , Algebraic Analysis, PWN-Polish Scientific Publishers and D. Reidel, Warszawa-Dordrecht 1987.

[7] —, Equations avec opérations algébriques, Studia Math. 22 (1963), 337-368.

[8] M. T a sc h e , Abstrakte lineare Differentialgleichungen mit stationàren Operator en, Math. Nachr.

78 (1977), 21-36.

INSTITUTE OF MATHEMATICS, TECHNICAL UNIVERSITY OF WARSAW PL. POLITECHNIKI 1, 00-661 WARSZAWA, POLAND