CC BY
7URAL MATHEMATICAL JOURNAL, Vol. 4, No. 2, 2018, pp. 33-42
DOI: 10.15826/umj.2018.2.005
SOME PROPERTIES OF OPERATOR EXPONENT
Lyudmila F. Korkina^ and Mark A. Rekanttt
Ural Federal University, 51 Lenin aven., Ekaterinburg, Russia, 620000
tl.f.korkina@urfu.ru, ttm.a.rekant@urfu.ru
Abstract: We study operators given by series, in particular, operators of the form eB = Bn/n!, where B
n=0
is an operator acting in a Banach space X. A corresponding example is provided. In our future research, we will use these operators for introducing and studying functions of operators constructed (with the use of the Cauchy integral formula) on the basis of scalar functions and admitting a faster than power growth at infinity.
Keywords: Closed operator, Operator exponent, Multiplicative property.
The theory of functions of normal operators has been developed in Hilbert spaces [8, Ch. 12,13]. However, functions of an operator in Banach spaces are introduced under quite serious restrictions on the operator and the corresponding scalar functions (see e.g. [2, Ch. VII.3]). For a considerable class of operators, these scalar functions are assumed to be analytical with polynomial growth at infinity (see e.g. [1] and [6, Ch. 1, § 5]). The authors' papers [3-5] are in the same vein. In these papers, based on the Cauchy integral formula, functions of an operator were constructed in terms of natural powers of the operator. To introduce and study functions of an operator built constructed on the basis of scalar functions and admitting the growth at infinity faster than the power function but not faster than the exponential function have, we will need operators of the form
eB =
ro b n
n.
n=0
where B is an operator on a Banach space X. In this paper, we study the properties of such operators.
We will use series of elements of a Banach space X and operator series. The principal notions of numerical series (double series and repeated series) are naturally extended to series of elements of the space X [7, Ch. 2, § 2]. In this paper, the convergence of partial sums of series from X is
ro
interpreted as the convergence in the norm of this space. For a series ^ An of operators An acting
n=0 ro
in X, its sum is the operator A with the domain D(A) = \x € X: ^ Anx converges > and such
^ n=0 -1
ro
that Ax = Anx for x € D(A). The expression A c B (B d A) for operators A and B means
n=0
that B is the extension of A [7, Ch. 7, Sect. 6]. Let us proceed to the results.
In what follows, we will need the following auxiliary assertion.
Assertion 1. The following statements hold:
ro
(i) (An analog of Abel's test for numerical series). Suppose that a series ®n converges in X
n=0
ro
and a sequence {an}ro=0 C R is monotonic and bounded. Then, the series anan converges.
n=0 n=0
oo
ro
(ii) Suppose that {am,n}mn=o C X and the series am,n converges absolutely. Then, every
m,n=0
rearrangement of this series converges absolutely to the same sum.
ro
(iii) Suppose that the terms of a series am,n (am,n € X) are reindexed (with a single
m,n=0
ro
index) and the series bk is composed of them. If one of these two series or the repeated series k=o
ro ro
am,n converges absolutely, then the other two series converge absolutely to the same sum.
m=0n=0
ro ro ro
(iv) If a series am,n (am,n € X) converges absolutely, then the series am,n also
m,n=0 k=0 m+n=k
(m,n> 0)
converges absolutely to the same sum.
The proof of statement (i) is almost the same as the proof of Abel's test for numerical series. The proofs of statements (ii)—(iv) reduce to the use of the corresponding statements for numerical series after the application a continuous linear functional to the series under consideration. Here, we take into account the fact that if values of all such functionals coincide at two elements from X, then these elements are equal [2, Ch. II.3.15].
Assertion 2. Suppose that A is an operator acting in X, x € X, k € N, a sequence {an}ro=0 C R is such that the sequence j n+h | is monotonia and bounded, and the series ctnAn+kx converges.
^ an J n=0
ro
Then, the series anAnx converges. If the operator Ak is linear and closed, then the following
n=0
equality holds:
roro
Y^ anAn+kx = Ak ^ anAnx, (2)
n=0 n=0
which is equivalent to the expression
roro
an n
n=0 n=0
5>nAn+k C Ak^ a,nAn. (3)
Proof. The following relations are valid:
ro k—1 ro k—1 ro
]T anAnx = ]T anAnx + ^ an+kAn+kx = ]T anAnx + ^ ^(anAn+kx).
n=0 n=0 n=0 n=0 n=0 n
The latter series converges by the analog of Abel's test. Moreover, under the assumption that the operator Ak is linear and closed, equality (2) holds. □
Remark 1. The operator Ak is linear and closed if the operator A is linear and its resolvent set p(A) = 0 [2, VII.9.7].
Remark 2. The boundedness and monotonicit.y of the sequence J CXn+k I starting from a certain
I an J
index follow from the fact that, the sequence / CXn+l 1 monotonic and bounded.
an
This remark follows from the relation
Q^+fc _ cxn+\ an-1-2 «ra+fc
— X X * * * X
an an an+1 an+k—1
and the fact that terms of a monotonic sequence of reals are of the same sign starting from a certain index.
Remark 3. Equality (2) holds without the assumption that the sequence J CXn+k I ¿s monotonic
I an J
and bounded if the operator Ak is linear and closed and both the series in (2) converge.
Corollary 1. Suppose that k € N and the operator Ak is linear and closed. Then,
eA Ak c Ak eA.
To prove this fact, it is sufficient to take an = 1/n! in (3).
In Assertion 1, the sequence J CXn+k I required to be monotonic and bounded. Let us
I an J n=0
consider the conditions related to these properties.
an, i
Lemma 1. If a £ R and a sequence {a„} C (0, +oo) is such that —— < a for all n, then
an
ara < Can for all n, where C = a\/a. Conversely, if the sequence { \ is monotonic and
I a„, J
an >
ara < Can for some C,a € (0, +oo) and alln, then the sequence { CXn+l I %s bounded.
L a„ j
(y I 1
Proof. Suppose that, a, € M is such that. —— < a, for all n. Then,
an
an a2 a3 a^ n-1
— = — x — x • • • x-< a ;
ai ai a2 an-i
i.e., an < Can for C = a1/a.
Conversely, suppose that. an < Can for some C, a G (0, +oo), and all n and the sequence { CXn+l
I an
is monotonic. Denote by d the limit, of this sequence, d € [0, +00]. Assume that, d = +00. Then lim v/a^ = +00. This contradicts the inequality v/a^ < \[Ca. Therefore, d € R; i. e., the sequence
{ CXn+l I is bounded. The lemma is proved. □
I an J
Note that the requirement of monotonicity in the second part of the lemma is essential.
Example 1. Suppose that a sequence } C N is such that nm+1 > + 1, 2nm > m!, and
a« = {ni — 1)! for nm-1 < 11 < nm (no = 0) for all m € N. In this case, the sequence J CXn+l I
L an J
unbounded, although an < 2n. (Indeed, if nm-1 < n < then an = (m — 1)! < 2nm-1 < 2n).
Assertion 3. Suppose that k € N and {«„}~=0 C (0, +00). If the sequence j n+k | is bounded,
{ an
then an < Cbn for some b,C € (0, +00) and alln. Conversely, if the sequence { n+k } is monotonic,
I an J
C,b € (0, +00), and an < Cbn for all n, then the sequence { n+k } is bounded.
I an J
a
Proof. Suppose that, a € R is such that. ——- < a for all n. Let. us consider the subsequences
an
of {an} with = amk+r (r = 0,1,..., k — 1). For all m, we have
Pm+l a{m+l)k+r a(mk+r)+k ^
Then, according to Lemma 1, there is a number Cr € (0, such that
Cr
ß{„[> < Cram = -^r( ^)mfc+r
I m — -r/k v '
for all m. Setting C = max
Cr
0<r<k-ia
-i ar/k
and b = yfa, we obtain an < Cbn for all n.
Conversely, suppose that, the sequence / CXn+k 1 js monotonie, C, b € (0,+oo), and an < Cbn
I an j
for all n. If lim CXn+k is finite, then there is nothing to prove. Assume that, lim CXn+k = +oo.
Again, introducing = amk+r (r = 0,1,... ,k — 1), we conclude that the sequence }ro=0 is monotonic because
3(r)
[m+1 _ a(mk+r)+k
ß(r)
ßm
amk+r
Moreover,
ß(m) = Mmk+r < Cbmk+r = Cibm (Ci = Cbr, bi = bk). Hence, according to Lemma 1,
ß(r)
a(mk+r)+k _ Pm+1 Oimk+r ߣ>
for some Cr, ar € (0, +rc>) and arbitrary m. Therefore,
ar
«ra+fc n
< max{a0, ai,..., ak-i} = a
for all n. The assertion is proved.
□
Assertion 4. Suppose that operators B\,..., Bn act in X, B\,..., Bn-\ are linear operators with nonempty resolvent sets, x € X, the series
^ B;i ... Bm-
mi,...,mn=0
mi!... mn!
(4)
converges absolutely, and the following condition holds :
(v) for all k € N and a set of natural indices il,... not exceeding n, the expression Bil ... Bik x is valid and, if a set ji,..., jk is obtained from the set il,... ,ik by a rearrangement of its elements, then
(5)
Bil ■ ■ ■ Bikx = Bji ■ ■ ■ Bjk
In this case,
pBi eBn x = eBi+—+Bn x
( both parts of the relation are valid).
Proof. Let us first establish the equality
> > • • • > -^—x = > 1 , 2 ,-V® (6)
mi! m2! m,! ^ „ mi!m2!...m, !
mi =0 m2=0 mn=0 mi,m2,...,mn=0
by induction on n. For n = 1, (6) holds. Assume that, under the conditions of the assertion, equality (6) holds for n = k — 1 (k > 2). Now, let n = k. Note that the absolute convergence of series (4) implies the absolute convergence of the series
ro R m2 Rm„
y^ £2
^ m2\...mn\
m2,...,m„=0
Taking into account the fact that the operators B^7"1 (m1 € N) are closed, condition (v), and the induction hypothesis, we obtain
> -^-r- > -*-r • • • > -^-rx = > -L-r > -¡¡-¡-x :
^ m 1! m2! ^„ffln! m 1! ^ №2! ...ffln!
mi =0 m2 =0 mn=0 mi=0 m2,...,mn=0
Z.^ mjlm^!... m„! ' ^ milm^l... mn!
mi=0 m2,...,mn=0 mi,m2,...,mn=0
i.e., equality (6) is proved. Using this equality, we obtain
ro £
_ R mi R mn ^TO ^^ R mi Rmn
pSi pi?IiT_ v 1 ' ' " T - V V 1 " " " " T -
C- ♦ ♦ ♦ C- «X/ --/ J, «X/ - 7 7 || «X/ -
m1!... mn! ^ ^^ m1!... mn!
mi,...,mn=0 s=0 mi+-----|-mn=s
S = 0 mi + ---+mn =s 1 " s=0
(mi,...,m„>0)
The assertion is proved. □
Remark 4. Equality (5) holds if the operators B1,..., Bn pairwise commute and the left-hand side of (5) is valid.
Corollary 2. Suppose that a1,..., an € C, x € X, A is a linear operator acting in X, p(A) = 0, and a series
a™1 ... cff
m1! . . . mn!
Ectl • • • y^mi H-----\-mn
m 1 ! m „ !
mi.....mn =0
converges absolutely. Then,
e«i a eu" Ax = e(ai+-^n)A x (7)
(both sides of the equality are valid).
To formulate the next assertion, let us introduce some definitions and make a number of assumptions.
Suppose that L = L(p, q) (p > 0, q > 0) is a curve given in the complex plane (A) by the equation
f32 = 2'pcx In — (a = Re A, /3 = Im A, a> q); (8)
q
Let G = G(p, q) be a domain containing the origin with the boundary L; let the direction of L be chosen so that the domain G is on the right; and let A be an injective linear operator with domain D(A) dense in X and range Im(A) c X. The following estimate for the norm of the resolvent operator R(A) = RaW = (A — AE)~l of the operator A in G is known:
for some Co > 0 and 7 < 1 and all A € G.
For f € (0, n), we denote by A(f) the domain in C that contains the negative real semiaxis and its boundary is L(f) = L1(f) U L2(f), where
L1(f) = {X € C : A = tet > 0}, L2(f) = {X € C : A = te-it > 0}.
Suppose that Q(a, f) = A(f) U B(0, a) (a > 0, f € (0, n), and B(0, a) is the open disk of radius a centered at the origin), and the direction of r(a,, <p) = dQ,(a,<p) is chosen so that. Q(a,<p) is on the right.. Given p and q, we chose a and <p so that. Q(a, <p) C G(p, q).
Under these assumptions, the authors studied [5] the operator functions
f(A) = f MR{X)d\, (10)
2ni Ma,,?) Xn
№ = ~Tli An (11)
constructed on the basis of corresponding scalar functions /(A) continuous in C\Q(a, <p) and analytic in C \ Q(a,, (p); in addition, for every such function / there exist. C € (0, +00) and a € M such that.
If (A)|< C|Ar (12)
for all A € C \ Q(a, <p). The number n € N U {0} in (10) and (11) is chosen so that a — n — y < -1.
It was proved that the right-hand sides of these representations are independent of such n, the operator functions f (A) and f(A) are densely defined, f(A) c f (A), and the functions coincide if one of them is continuous.
We can take the function e-tx (t > 0) as the function f and consider two operator functions, one of which is given by series according to formula (1) and the other is given by relations (10) and (11) for n = 0 (these relations yield the same result because their right-hand sides are continuous). Denoted by (e-tA)i the function given by formulas (10) and (11).
Lemma 2. Let a € R, a — 7 < —1, and let a function f be continuous in C\Q(a,ip), analytic in C\Q(a, f) and such that (12) holds for some C € (0, +rc>) and all X € C\Q(a, f). Then
I f (X)R(X)dX = ! f (X)R(X)dX.
JT(a,tp) JL(p,q)
The proof of this lemma is similar to the proof of [3, Lemma 1].
cx
Assertion 5. Let a ™ « (i > 0) fc s«,en by the mat,on ? = 2iiMl„- «:, = Re A, [ = Im X, and a > qt), and let n(t) € N satisfy the inequality
tp — n(t) — y< —1. (13)
Then
etA(e tA )i\D(An(t)) = E\D(An(t)), (14)
(e-tA)j etA = E\D{etAy (15)
Proof. Let us first consider the case t = 1. Note that L1 = L. Denote by n0 the value n(1). The following equalities hold for x € D {eA (e-A)/):
1 ^ An r 1 n Ak r
eA{e-A)lX =--V — / e~xR(X)d,X x =--lim V — / e~xR(X)dX x.
v ; 2m ^ n\ .h' 2m oo k\ J<» K '
n=0 J L k=0 L
For every n € N, consider n1 € N satisfying the inequality n — n1 — y < —1. Then, according to [4, Theorem 9],
^k Ani f v^ Xk-ni
n
^ k\ 2m Jjf k\ v ; '
k=0 ^ k=0 i.e., according to [5, Theorem 3],
n Ak f f n Xk-n1
^ — / e~xR{X)dX = Ani / ^ ^pe~xR{X)dX. (16)
k=0 ' J L k=0 '
n Xk
For the functions fn(X) = E ¿Te_A ^^^ an<^ we can an arbitrary a from (12).
k=0 '
Thus, the right-hand side of (16) is independent of n1 € N U {0}. Therefore, for x € D(An1), we have
n Xk-ni 0 n Xk-ni Xk-ni 0
Ani / ^ ^pe~xR{X)dX x = Ani / ^ ^-e~xR{X)dX x= ^-e~xR{X)dX ■'L k=0 ' L k=0 ' L '
1 n \k-n1
eA(e-A)Ix = —— lim / V e~xR{X)dX An°x
v ; 2m oo J«, ^ k\ y '
u L ]„_n
i. e.,
k=0
whenever this limit exists. Let us establish the existence of this limit and find its value using the Lebesgue (dominated convergence) theorem on passing to the limit under the integral sign. Let us check the conditions of this theorem. Let
n Xk-ni k=0 '
and let A = a + ift € L be arbitrary. Then
lim Hn(X) = V e"AE(A) = X~n"R(X)
n^x ¿—' n!
n=0
( An-n1
(the limit and the convergence of the series are considered with respect to the operator norm). Let us show that the sequence [\\Hn(A)||} is dominated by a Lebesgue integrable function in L:
n I A|fc-n1 c I A|-n°e|A|-a
\\Hn(X)W < g^L_e-tt||ü(A)|| < °°^|+ei)7 < CoCilXr^eM-*,
(I I \ Y / I I \ Y
——— I (the function ( ——— ) is continuous in jSf, has a finite limit at
M + 1) VH + V
infinity and, therefore, is bounded in L). Using (8), we obtain that
02 2palnf , a
— a = \J a2 + B2 — a = —, y - < —-- = p In —;
■\J a2 + (32 + a~ 2 a ^ q'
i. e.,
Hence,
where C theorem,
,|A|-a
<
a
<
\Hn(X)\\ <c\x\
p—nj—Y
C0C1
By (13) for t = 1, the integral / |A|p—ni—Y|dA| converges. By the Lebesgue
JL
1
2ni
- lim [ Hn(X)d\ = —— [ X~n°R{X)dX = A~
i n^ro JL 2ni JL
(the limit is considered with respect to the operator norm). Therefore,
eA(e—A)/x = A-n0 An° x = x,
and (14) is proved.
A ro
Let us show that (15) holds. For x € D(eA) C n^D(An), in view of continuity of the operator (e—A)/, we have
n=0
(e~A)ieAx = {e~A)i lim V ^
n^-ro —^
n^ro ^-' k!
fc=0
L
n A^—ni
j^-R(X)d,X Anix,
fc=0
where n1, as before, satisfies inequality (13) for t = 1. Arguing similarly to the proof of formula (14), we obtain (15). Thus, the assertion holds for t = 1.
Let us now consider an arbitrary t > 0. The mapping p = tA takes the curve L to the curve Lt and the domain G = G1 to the domain Gt 9 0 such that dGt = Lt. In addition, p(tA) = tp(A) (p(A) and p(tA) are the regular sets of the operators A and tA, respectively) and the estimate for ||RiA(A)|| in Gt coincides with the estimate (9) for ||Ra(A)|| in G with certain constant Ct instead of C0. The analysis of the proof for t = 1 shows that formulas (14) and (15) remain valid for t > 0 under condition (13). The assertion is proved. □
Corollary 3. If t > 0 and the operator eiA is closed, then it is invertible and (eiA)—1 = (e—iA)/.
The corollary follows from (14), (15), and the fact that if a closed operator coincides with a continuous operator on a dense set, then they coincide in the entire space.
Example 2. Let X = Lp[1, and Ax(t) = tx(t) (x € X). Let us show that D(eA) = {x € X: e*x € X} and the equality eAx = e*x holds for x € D(eA).
Let x € X ande*x € X. Let us establish that x € D(eA) and eAx = e*x. To this end, we have ro tn
to prove that the series 7 —x converges in X to etx, i.e., that.
n=0
n!
ex
£
fc=0
Ak x
k!
e*x —
£
fc=0
tkx ~k\
fc^ro
0.
/+ro
epi|x(t)|pdt < there exists a function a(t) defined on [1, such that a(t) > a0 for some a0 > 0, a(t) -> (in particular, we can take a continuous positive function a with
i^-ro
/+ro
|a(t)eix(t)|p dt < Then
ex
E
fc=0
tkx
Jd
1 — e—^n=0 tk/k! *
ae x
a
< sup i>1
a(t)
|aei x|| = Yn|aeix|
p
p
q
q
p
q
0
i
Let us show that
1 — e—^n=0 tk/k!
In = sup---->• o.
i>1 a(t) n^ro
Take an arbitrary e > 0. Since a(t) -> there is a number A > 1 such that 1/a(t) < e for
i^-ro
all t > A; i. e.,
a(t)
for all t > A and n € N. Since the power series ^ro=0tn/n! uniformly converges to ei on [1, A], the sequence of functions
a(t)
uniformly tends to zero on [1, A]. Hence, there exists a number N such that
i-e-'ELo tk/k\
a(t)
for all t € [1, A] and n > N. Thus, Yn < e for all n > N, i. e., Yn-> 0 and, consequently,
n^-ro
ro tn
E—:X = elx in X. n!
n=0
Conversely, suppose that x € D(eA) and eAx = y € X, i.e.,
n t^
Sn = ^ J1x-► V in x' t\ n^-ro
k! n^ro fc=0
Then, there exists a subsequence {Snfc} of {Sn} such that
But
Snfc(t) y(t).
fe^ro
Sn(t)x(t) -► eix(t)
n^ro
at every point t > 1; i. e., y = eix in X. Since eix € X, we have eAx = eix. Note that equality (7) holds for the operator A if
x e""tx e'Q"_i +an)ix e^ai +—)tx € X
Conclusion
We have considered some natural properties of exponential operator defined by power series (Corollary 1 and Assertion 4). The main result of the paper is the connection (under certain conditions) of the exponential operator eA in the form of power series with the exponential function e—A defined on the basis of the Cauchy integral formula (Assertion 5). These facts may give an impulse to obtaining further results on functional calculus of operators.
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