SOME PROPERTIES OF OPERATOR EXPONENT

We study operators given by series, in particular, operators of the form e = ∞ ∑ n=0 B/n!, where B 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.

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][4][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 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 interpreted as the convergence in the norm of this space.For a series In what follows, we will need the following auxiliary assertion.
Assertion 1.The following statements hold: (i) (An analog of Abel's test for numerical series).Suppose that a series ∞ n=0 a n converges in X and a sequence {α n } ∞ n=0 ⊂ R is monotonic and bounded.Then, the series a m,n converges absolutely.Then, every rearrangement of this series converges absolutely to the same sum.
(iii) Suppose that the terms of a series a m,n (a m,n ∈ X) converges absolutely, then the series a m,n also 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
is monotonic and bounded, and the series Then, the series ∞ n=0 α n A n x converges.If the operator A k is linear and closed, then the following equality holds: which is equivalent to the expression P r o o f.The following relations are valid: The latter series converges by the analog of Abel's test.Moreover, under the assumption that the operator A k is linear and closed, equality (2) holds.
Remark 1.The operator A k is linear and closed if the operator A is linear and its resolvent set ρ(A) = ∅ [2, VII.9.7].
Remark 2. The boundedness and monotonicity of the sequence α n+k α n starting from a certain index follow from the fact that the sequence α n+1 α n is monotonic and bounded.
This remark follows from the relation 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 α n+k α n is monotonic and bounded if the operator A k is linear and closed and both the series in (2) converge.
Corollary 1. Suppose that k ∈ N and the operator A k is linear and closed.Then, To prove this fact, it is sufficient to take α n = 1/n! in (3).
In Assertion 1, the sequence is required to be monotonic and bounded.Let us consider the conditions related to these properties.
α n is monotonic and α n ≤ Ca n for some C, a ∈ (0, +∞) and all n, then the sequence α n+1 α n is bounded.
Note that the requirement of monotonicity in the second part of the lemma is essential.
. For all m, we have Then, according to Lemma 1, there is a number C r ∈ (0, +∞) such that Moreover, Hence, according to Lemma 1, for some C r , a r ∈ (0, +∞) and arbitrary m.Therefore, for all n.The assertion is proved.
Assertion 4. Suppose that operators B 1 , . . ., B n act in X, B 1 , . . ., B n−1 are linear operators with nonempty resolvent sets, x ∈ X, the series converges absolutely, and the following condition holds: (v) for all k ∈ N and a set of natural indices i 1 , . . ., i k not exceeding n, the expression B i 1 . . .B i k x is valid and, if a set j 1 , . . ., j k is obtained from the set i 1 , . . ., i k by a rearrangement of its elements, then In this case, e B 1 . . .e Bn x = e B 1 +•••+Bn x (both parts of the relation are valid).

P r o o f. Let us first establish the equality
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 Taking into account the fact that the operators B m 1 1 (m 1 ∈ N) are closed, condition (v), and the induction hypothesis, we obtain i. e., equality ( 6) is proved.Using this equality, we obtain The assertion is proved.
Remark 4. Equality (5) holds if the operators B 1 , . . ., B n pairwise commute and the left-hand side of (5) is valid.
Corollary 2. Suppose that α 1 , . . ., α n ∈ C, x ∈ X, A is a linear operator acting in X, ρ(A) = ∅, and a series (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 (λ) by the equation 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) ⊂ X.The following estimate for the norm of the resolvent operator for some C 0 > 0 and γ ≤ 1 and all λ ∈ G.
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) ⊂ f (A), and the functions coincide if one of them is continuous.
We can take the function e −tλ (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).
Assertion 5. Let a curve L t (t > 0) be given by the equation β 2 = 2tpα ln α qt (α = Re λ, β = Im λ, and α ≥ qt), and let n(t) ∈ N satisfy the inequality Then e tA (e −tA ) P r o o f.Let us first consider the case t = 1.Note that L 1 = L .Denote by n 0 1 the value n(1).The following equalities hold for x ∈ D e A (e −A ) I : For every n ∈ N, consider n 1 ∈ N satisfying the inequality n − n 1 − γ < −1.Then, according to [4, Theorem 9], i. e., according to [5, Theorem 3], For the functions 10) and (11), we can take an arbitrary σ from (12).
Thus, the right-hand side of ( 16) is independent of n 1 ∈ N ∪ {0}.Therefore, for x ∈ D(A n 0 1 ), we have 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 and let λ = α + iβ ∈ L be arbitrary.Then (the limit and the convergence of the series are considered with respect to the operator norm).Let us show that the sequence { H n (λ) } is dominated by a Lebesgue integrable function in L : where is continuous in L , has a finite limit at infinity and, therefore, is bounded in L ).Using (8), we obtain that i. e., Hence, where C = C 0 C 1 q p .By (13) for t = 1, the integral (the limit is considered with respect to the operator norm).Therefore, where n 1 , 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 µ = tλ takes the curve L to the curve L t and the domain G = G 1 to the domain G t ∋ 0 such that ∂G t = L t .In addition, ρ(tA) = tρ(A) (ρ(A) and ρ(tA) are the regular sets of the operators A and tA, respectively) and the estimate for R tA (λ) in G t coincides with the estimate (9) for R A (λ) in G with certain constant C t instead of C 0 .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 e tA is closed, then it is invertible and (e tA ) −1 = (e −tA ) I .
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 = L p [1, +∞) and Ax(t) = tx(t) (x ∈ X).Let us show that D(e A ) = {x ∈ X : e t x ∈ X} and the equality e A x = e t x holds for x ∈ D(e A ).
Let x ∈ X ande t x ∈ X.Let us establish that x ∈ D(e A ) and e A x = e t x.To this end, we have to prove that the series x converges in X to e t x, i. e., that Since at every point t ≥ 1; i. e., y = e t x in X.Since e t x ∈ X, we have e A x = e t x.Note that equality (7) holds for the operator A if x, e αnt x, e (α n−1 +αn)t x, . . ., e (α 1 +•••+αn)t x ∈ 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 e A 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.

∞ n=0 A
n of operators A n acting in X, its sum is the operator A with the domain D(A) = x ∈ X : ∞ n=0 A n x converges and such that Ax = ∞ n=0 A n x for x ∈ D(A).The expression A ⊂ B (B ⊃ A) for operators A and B means that B is the extension of A [7, Ch. 7, Sect.6].Let us proceed to the results.
a m,n (a m,n ∈ X) are reindexed (with a single index) and the series ∞ k=0 b k is composed of them.If one of these two series or the repeated series ∞ m=0 ∞ n=0 a m,n converges absolutely, then the other two series converge absolutely to the same sum.(iv) If a series ∞ m,n=0 and (14) is proved.Let us show that (15) holds.For x ∈ D(e A ) ⊂ ∞ ∩ n=0 D(A n ), in view of continuity of the operator (e −A ) I , we have (e −A ) I e A x = (e −A ) I lim n→∞ n k=0 we obtain α n ≤ Cb n for all n.