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

**N****g u y e n**** V****a n**** M****a 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*

**to be Volterra**

**operators, provided R 2, R 2 are Volterra right inverses of a right invertible operator D. The method****used here is essentially based on the properties of generalized algebraic operators.**

*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*

*0*

*(X): DR = I},*

*= {.F*

*e*

*L 0(X): F*

*2*

* = 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*

*0*

*(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*

*0*

*(X).*

**D****e f in it io n**

*1 [8]. An operator A e L*

*0*

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

*For given D e R (X ) and R*

*e*

*@*

*d*

* we denote by SDtR the set of all stationary * operators.

**D****e f in it io n**

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

*0*

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

*P{t) = *

**P 0 + P i t + . . . + P Nt N,***p*

*0*

*, . . . , 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.*

**D****e f in it io n**

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

*e*

*L*

*q*

*(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).

**T****h 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*

*e*

*R(X) (Theorems 9,10). *

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

**T****h 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*

*j*

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

*algebraic operator.*

**T****heorem**

### III

**[3].**

*Let A and В be commuting algebraic operators with * *characteristic polynomials*

*П*

*PA(t) * = П *ui* ^

**u j***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 *

**{ щ + Vf.***i = *

1**,**

*n; j = *

1**,**

### . . . m}.

*1. Let D*

*e*

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

*n ej*

*^ n S DtR be mutually com*

*muting and let A*

*n*

### = J.

### Write

### (4) *Q(t, *

^{S) }*= £ * 0 (0 = Q(t, 1), p(t) =

7 = 0

### where M is a non-negative integer.

**T****heorem**

*1. I f R*

*e*

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

### (5) *R*

*0*

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

### P ro o f. Write

### ^ 0 = Нп{Я*} *(k = 0, 1,...).*

### Then J '

q* 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

=1### are the characteristic roots of the operators

### ^N-k (k = 1 ,..., ЛГ), respectively.

*Theorem I implies that every operator I + Y*

*j*

*!=*

*i*

* 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*

*1*

* = 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*

*0e*

*V(X). For tE C we *

*obtain I - t R*

*0*

* = [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.

**,R0 e F (X ).**

**T****heorem**** 2. **

*Suppose that R e 0lD *

**n**

*V*

**(X)**

*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

**T****heorem**** 3. **

*Let DeR(X), R*

*e*

*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*

*(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*

*a*

*& *

*s*

*)*

*= 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).*

**C****o r o l l a r y**** 1. **

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

**n**

*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 *

*1*7=1

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

*x — * *(/ — tjR)~*

*1*

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

*where*

*P ,= * П *] = 1 ,.... n.*

*k= l,k¥>j*

**C****o 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*

*e*

*R { X ) ,R*

*e*

*M*

*d*

* 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.

**D****efinition**** 4. **

*If A e jrf n SD<R *

**and**

*О ф xA e *

**ker**

*(D—A) *

**then**

*x A *

**is said to be**

**an**

*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.*

**T****heorem**** 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 *

g* sé n SDtR such that*

### This implies (

11### )

### fc + i

### E

*dj x Aj>*

*dj<=jrf r \ S DtR.*

*j = i*

*k+ 1*

### 0 = D (E ^

^{j x A j}### )» *d j e s / r\SDtR,*

*j = i*

### and

**(**12

**)**

### fe + i

*° = Z dJÂ l XAj- * 7=1

### The equalities (11) and (12) together imply

*0 = Z dM j - A i)**

*a*

*, = 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.*

**T****heorem**** 5. **

*I f RDn V*

**(X)**

*for a D e R(X), A e s é *

**n**

*SDtR and *

**dim ker**

*A — *

**0,**

*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.*

**D****efinition**** 5. **

*If D*

*e*

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

**(X),**

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

*tial operator.*

**T****heorem***6*

*. I f R e f%D *

**n**

*V*

**(X),**

*A e SDtR *

**n**

*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 ) ~*

*1*

*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.*

**C****orollary**** 4. **

*I f D eR(X), R e MD *

**п**

*V*

**(X),**

*then for every operator * 0

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

**C****orollary** **5. **

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

**F(X )**

*and * *A s s # n SD'R. Then*

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

**zek erZ ),**

*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).*

**D****efinition**

*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.*

**T****heorem**** 7. **

*The algebraic cosine and sine operators have the following * *properties:*

*c a*

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

*sa*

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

*casb*

* + *

*cbsa*

* = 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).

**C****orollary**

*6. DcA = —A*

*s*

*a, *

*D*

*sa*

*= A*

*c*

*a.*

**T****heorem**** 8. **

*Let F be an initial operator for a D*

*e*

*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*

*sa*

*(*

*z*

*) = 0 and cA(z) ф 0.*

### P ro o f. By (14),

*F*

*sa*

*(*

*z*

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

*z*

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

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

*A 2R 2)*

*ca*

*(*

*z*

*)*

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

**3. Operations on Volterra right inverses. Let D**e

*R{X), R r, R 2*

*e*

*M*

*d*

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

*i*

*_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*

*i*

* + R 2, ^*

1^2### to be Volterra operators, we obtain the following conditions:

**T****heorem**** 9. **

*Let D*

*e*

*R{X), R x, R 2*

*e*

*$ 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*

*e*

*V(X). Write*

*C*

*rx*

*R*

*z*

* = У ^*

1### ^

2### ) ’ *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*

*2*

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

*2*

*R i is a Volterra operator.*

**T****h e o r e m**

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

*2*

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

*2*

* to be a Volterra * *operator is that*

### (17) *F2(I — tR2)~1z *

*Ф*

### 0, 0 ^ zekerD , *where F*

*2*

* is an initial operator for D corresponding to R 2.*

*P ro o f. Note that R*

*1*

*R*

*2e& D2*

* and R*

*2*

### is a Volterra right inverse of D2.

*Hence, if R i R*

*2*

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

*1*

### z, zekerD 2, *z ф 0.*

*If v *

*Ф*

*t, veC then*

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

*1*

*{tRl - v R*

*2*

*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*

*2*

### has no eigenvectors.

*If v = teC , then we have и = (/ — tR*

*1*

*R 2)(I — tRj)~*

*1*

*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

j### M # 0 for 0

*Ф*

### zekerD . Thus

*и Ф*

*0, i.e. R t R*

*2*

### has no eigenvector in kerD.

*(ii) Let z = R*

*1*

*z*

*1*

*+ z2, where z1? z2ekerD , z l *

*Ф*

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

*2*

*D2)(I — tR2)~l z *

*= z + tR*

*1*

*F*

*2*

*R 1(I — t R l y*

*1*

* z.*

*If z2 Ф 0 then F^u = F*

*1*

*R l z*

*1*

*+ F*

*1*

*z*

*2*

* — z*

*2*

* ф 0, which implies и ф 0. If *

*z2*

* — 0, i.e. z = R l zi , then*

*u = R*

*1*

*z*

*1*

* + tR*

*1*

*F*

*2*

*R*

*21*

*( I - t R l y*

*1*

*z l = R*

*1*

*z*

*1*

* + R l F*

*2*

*[I—(I—tRlJ] (I — t R j y*

*1*

* Zl*

*= R*

*1*

*zl + R*

*1*

*F2(I — tR*

*2*

*)~*

*1*

*zl — R*

*1*

*F*

*2*

*z l = R*

*1*

*F2(I — tR*

*2*

*)~*

*1*

*z*

*1Ф*

### 0.

*Hence, и *

*Ф 0*

* 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

**T****h e o r e m**

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

*e*

*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.*

**E****x 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 =*

**J ,**

*x x Ф x 2i x x,*

**x**

**2**

**g**

**[0 , 1].**

XI *X 2*

### It is easy to check that

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

*c e*

*F, j = l , 2. Hence*

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

*If t0 = — ^(*

*x*

*2 — *

*x x*

*)~2*

*k*

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

**T****h e o r e m** **11. **

*Suppose that D *

*e*

* R(X) and R x, R 2*

*e*

*n V*

**(X ).**

*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*

*e*

*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 *

*e*

*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 *

*e*

### 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).

**E****x a m p l e**

*2. Let X , D, R lt R*

*2*

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

### (/ — *= cexp[t(x — x})~\ * *f o r c e d , *

**j***= 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*

*1*

*+ 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*

*2*

### 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