M∆tk 2 M∆tk osc(A, [tk 1, tk])∆tk+e 1 M∆tk osc(A, [tk 1, tk])∆tk+(M∆tk) e . ≤ − − − ≤ − Therefore we conclude that
M(b a) 2 M∆tk P (A, D) P (A, D∗) e − (osc(A, [tk 1, tk])∆tk + (M∆tk) e ). k − k ≤ −
Now, since the given partition D0 can be obtained from D by successively refining the subintervals [t0, t1],..., [tm 1, tm], we obtain − m M(b a) 2 M∆tk P (A, D) P (A, D0) e − (osc(A, [tk 1, tk])∆tk + (M∆tk) e ). k − k ≤ − =1 kX
1 n n Corollary 3.1.10. Consider a Riemann integrable function A :[a, b] R × . Then for every ε > 0 there exists δ > 0 such that →
P (A, D) P (A, D0) < ε k − k
whenever ν(D) < δ and D0 D. ≺ Proof. The statement follows from the previous lemma, Lemma 3.1.7 and the estimate
m m (M∆t )2eM∆tk ν(D)M 2eMν(D) ∆t = (b a)ν(D)M 2eMν(D). k ≤ k − =1 =1 kX kX
1 [LS1], p. 39–41
70 1 b Theorem 3.1.11. The product integral (I+A(x) dx) a exists for every Riemann n n integrable function A :[a, b] R × . → Q Proof. Take ε > 0. Corollary 3.1.10 guarantees the existence of a δ > 0 such that
P (A, D) P (A, D0) < ε/2 k − k whenever ν(D) < δ and D0 D. Consider a pair of tagged partitions D1, D2 of ≺ interval [a, b] satisfying ν(D1) < δ and ν(D2) < δ. These partitions have a common refinement, i.e. a partition D such that D D1, D D2 (the tags in D can be chosen arbitrarily). Then ≺ ≺
P (A, D1) P (A, D2) P (A, D1) P (A, D) + P (A, D) P (A, D2) < ε. k − k ≤ k − k k − k n n We have proved that every Riemann integrable function A :[a, b] R × satisfies a certain Cauchy condition and this is also the end of Schlesinger’s→ proof; the existence of product integral follows from the Cauchy condition in the same way as in the analoguous theorem for the ordinary Riemann integral (see e.g. [Sch2]). 2 n n Theorem 3.1.12. Consider a Riemann integrable function A :[a, b] R × . If c [a, b], then → ∈ b c b (I + A(x) dx) = (I + A(x) dx) (I + A(x) dx) . · a a c Y Y Y Proof. As Schlesinger remarks, the proof follows directly from the definition of product integral (see the proof in Chapter 2).
3.2 Matrix exponential function n n Let A :[a, b] R × be a constant function. If Dm is a partition of [a, b] to m subintervals of→ length (b a)/m, then − b a m P (A, D ) = I + − A . m m Since ν(D ) 0 as m , we have m → → ∞ b m b a (b a)A (I + A(x) dx) = lim I + − A = e − . m m a →∞ Y A m The last equality follows from the fact that e = limm (I + A/m) for every n n →∞ A R × ; recall that the matrix exponential was defined in Chapter 2 using the series∈ ∞ Am eA = . (3.2.1) m! m=0 X 1 [LS1], p. 41 2 [LS1], p. 41
71 n n n n Lemma 3.2.1. If A1,...,A R × and B1,...,B R × , then m ∈ m ∈ m m m i 1 m − A B = B (A B ) A . i − i j · i − i · j i=1 i=1 i=1 j=1 j=i+1 Y Y X Y Y Proof.
m m m Ai Bi = (B1 Bi 1Ai Am B1 BiAi+1 Am) = − ··· − ··· − ··· ··· i=1 i=1 i=1 Y Y X
m i 1 m − = B (A B ) A . j · i − i · j i=1 j=1 j=i+1 X Y Y
1 n n Theorem 3.2.2. Consider a Riemann integrable function A :[a, b] R × . Then →
m m b A(ξk)∆tk lim e = lim (I + A(ξk)∆tk) = (I + A(t) dt) . ν(D) 0 ν(D) 0 → =1 → =1 a kY kY Y
Proof. Since every Riemann integrable function is bounded, we have A(x) M for some M R and for every x [a, b]. The definition of matrixk exponentialk ≤ (3.2.1) implies∈ ∈
A(ξ )∆t 2 A(ξ ) ∆t 2 M∆t e k k (I + A(ξ )∆t ) ( A(ξ ) ∆t ) ek k k k (M∆t ) e k − k k ≤ k k k k ≤ k
for k = 1, . . . , m. According to Lemma 3.2.1,
m m A(ξk)∆tk e (I + A(ξk)∆tk) = − k=1 k=1 Y Y
m j 1 m − = (I + A(ξ )∆t ) (eA(ξj )∆tj I A(ξ )∆t ) eA(ξk)∆tk k k · − − j j · ≤ j=1 k=1 k=j+1 X Y Y m m M(b a) A(ξ )∆t M(b a) 2 2 M∆t e − e j j I A(ξ )∆t e − M (∆t ) e j ≤ − − j j ≤ j ≤ j=1 j=1 X X m M(b a) 2 Mν(D) M(b a) 2 Mν(D) e − M ν(D)e ∆t = (b a)e − M ν(D)e . ≤ j − j=1 X 1 [LS1], p. 42
72 By choosing a sufficiently fine partition D of [a, b], the last expression can be made arbitrarily small. Definition 3.2.3. The trace of a matrix A = a n is the number { ij}i,j=1 n
Tr A = aii. i=1 X
1 n n Theorem 3.2.4. If A :[a, b] R × is a Riemann integrable function, then →
b b det (I + A(x) dx) = exp Tr A(x) dx . a ! a ! Y Z
Proof.
b m m det (I + A(x) dx) = det lim eA(ξk)∆tk = lim det eA(ξk)∆tk = ν(D) 0 ν(D) 0 a ! → =1 ! → =1 Y kY kY
m m b Tr A(ξk)∆tk = lim e = lim exp Tr A(ξk)∆tk = exp Tr A(x) dx ν(D) 0 ν(D) 0 → =1 → =1 ! a ! kY kX Z (we have used a theorem from linear algebra: det exp A = exp Tr A).
Remark 3.2.5. This formula (sometimes called the Jacobi formula) appeared already in Volterra’s work. Schlesinger employs a different proof and his statement is also more general – it requires only the Riemann integrability of A, in contrast to Volterra’s assumption that A is continuous.
n n Corollary 3.2.6. If A :[a, b] R × is a Riemann integrable function, then the → b product integral (I + A(x) dx) a is a regular matrix. Recall that Volterra has also assignedQ meaning to product integrals whose lower limit is greater than the upper limit; his definition for the right integral was
a 1 (I + A(t) dt) = lim (I A(ξk)∆tk). ν(D) 0 − → = Yb kYm If A is Riemann integrable, we know that this is equivalent to
a 1 A(ξ )∆t (I + A(t) dt) = lim e− k k . ν(D) 0 → = Yb kYm 1 [LS1], p. 43–44
73 Thus m 1 A(ξ )∆t A(ξ )∆t I = lim e k k e− k k = ν(D) 0 · → =1 = ! kY kYm m 1 A(ξ )∆t A(ξ )∆t = lim e k k lim e− k k = ν(D) 0 · ν(D) 0 → =1 → = kY kYm b a (I + A(t) dt) (I + A(t) dt) , · a Y Yb a b which proves that (I +A(t) dt) b is the inverse matrix of (I +A(t) dt) a; compare with Volterra’s proof of Theorem 2.4.10. Q Q
3.3 The indefinite product integral Schlesinger now proceeds to study the properties of the indefinite product integral, x i.e. of the function Y (x) = (I + A(t) dt) a. 1 n n Theorem 3.3.1. If A :[a, b] R × is Riemann integrable, then the function x → Q Y (x) = (I + A(t) dt) a is continuous on [a, b]. Proof. We prove the right-continuity of Y at x0 [a, b); continuity from left is Q ∈ proved similarly. Let x0 x0 + h b. The function A is bounded: A(x) M ≤ ≤ k k ≤ for some M R. We now employ the inequality from Lemma 3.1.9. Let D0 be a ∈ partition of interval [x0, x0 + h]. Then
Mh 2 Mh I + A(x0)h P (A, D0) e (osc(A, [x0, x0 + h])h + (Mh) e ). k − k ≤
Passing to the limit ν(D0) 0 we obtain →
x0+h Mh 2 Mh I + A(x0)h (I + A(t) dt) e (osc(A, [x0, x0 + h])h + (Mh) e ), − ≤ x0 Y
which implies x0+h lim (I + A(t) dt) = I. h 0+ → x Y0 Therefore
x0+h lim (Y (x0 + h) Y (x0)) = Y (x0) (I + A(t) dt) I = 0. h 0+ − − → x ! Y0
1 [LS1], p. 44–46
74 1 n n Theorem 3.3.2. If A :[a, b] R × is Riemann integrable, then the function → x Y (x) = (I + A(t) dt) , a Y satisfies the integral equation
x Y (x) = I + Y (t)A(t) dt, x [a, b]. ∈ Za
Proof. It is sufficient to prove the statement for x = b. Let
D : a = t0 ξ1 t1 ξ2 tm 1 ξm tm = b ≤ ≤ ≤ ≤ · · · ≤ − ≤ ≤ be a tagged partition of interval [a, b]. We define
k k Y = (I + A(ξi)∆ti), k = 0, . . . , m. i=1 Y Then k k 1 k 1 Y Y − = Y − A(ξ )∆t , k = 1, . . . , m. (3.3.1) − k k Since Y 0 = I and Y m = P (A, D), adding the equalities (3.3.1) for k = 1, . . . , m yields m k 1 P (A, D) I = Y − A(ξ )∆t . − k k =1 kX The function A is bounded: A(x) M for some M R. We estimate k k ≤ ∈
b Y (b) I Y (t)A(t) dt Y (b) P (A, D) + − − a ≤ k − k Z
b + P (A, D) I Y (t)A(t) dt Y (b) P (A, D) + − − a ≤ k − k Z
m m k 1 + (Y − Y (tk 1))A(ξk)∆tk + (Y (tk 1) Y (ξk))A(ξk)∆tk + − − − − k=1 k=1 X X m b + Y (ξk)A(ξk)∆tk Y (t)A(t) dt . (3.3.2) − a k=1 Z X
1 [LS1], p. 46–47
75 Using the inequalities
m m k 1 k 1 (Y − Y (tk 1))A(ξk)∆tk M Y − Y (tk 1) ∆tk − − ≤ k − − k ≤ k=1 k=1 X X
m m M(b a) 2 M∆tj M e − ∆tk (osc(A, [tj 1, tj])∆tj + (M∆tj) e ) ≤ − ≤ =1 j=1 kX X m M(b a) 2 Mν(D) Me − (b a) osc(A, [tj 1, tj])∆tj + M ν(D)e ≤ − − j=1 X (we have used Lemma 3.1.9) and
m m
(Y (tk 1) Y (ξk))A(ξk)∆tk M osc(Y, [tk 1, tk])∆tk, − − ≤ − k=1 k=1 X X
we see that all terms on the right-hand side of (3.3.2) can be made arbitrarily small if the partition D is sufficiently fine. 1 n n Corollary 3.3.3. If A :[a, b] R × is continuous, then the function → x Y (x) = (I + A(t) dt) a Y provides a solution of the differential equation
Y 0(x) = Y (x)A(x), x [a, b] ∈ and satisfies the initial condition Y (a) = I. Remark 3.3.4. The function Y is therefore the fundamental matrix of the system
n
yi0(x) = aji(x)yj(x), i = 1, . . . , n. j=1 X
Schlesinger uses the notation DxY (x) = A(x), where
1 DxY = Y − Y 0,
i.e. Dx is exactly Volterra’s right derivative of a matrix function.
3.4 Product integral inequalities In this section we summarize various inequalities that will be useful later.
1 [LS1], p. 47–48
76 1 n n Lemma 3.4.1. If A :[a, b] R × is a Riemann integrable function, then →
b b (I + A(x) dx) exp A(x) dx . ≤ a k k ! a Z Y
Proof. Lemma 3.1.3 implies that
m m
(I + A(ξi)∆ti) exp A(ξi) ∆ti ≤ k k ! i=1 i=1 Y X
for every tagged partition D of interval [a, b]; the proof is completed by passing to the limit ν(D) 0. → 2 n n Lemma 3.4.2. Let m N, A ,B R × for every k = 1, . . . , m. Then ∈ k k ∈ m m m m (I + Bk) (I + Ak) exp Ak exp Bk Ak 1 . − ≤ k k! k − k! − ! k=1 k=1 k=1 k=1 Y Y X X
Proof. Define
k k k k Y = (I + Ai),Z = (I + Bi), k = 0, . . . , m i=1 i=1 Y Y (where the empty product for k = 0 equals the identity matrix). Then
k k 1 k 1 Y Y − = Y − A , − k k k 1 k 1 Z Z − = Z − B , − k for k = 1, . . . , m. This implies
k k k 1 k 1 Z Y = (Z − Y − )(I + B ) + E , (3.4.1) − − k k where k 1 E = Y − (B A ). k k − k Applying the equality (3.4.1) m times on the difference Zm Y m we obtain − m 1 m m − Z Y = E (I + B +1) (I + B ) + E . − k k ··· m m =1 kX 1 [LS1], p. 51 2 [LS1], p. 52–53
77 We also estimate k 1 − E exp A B A k kk ≤ k ik k k − kk i=1 ! X (the empty sum for k = 0 equals zero),
m 1 k 1 m − − Zm Y m exp A B A exp ( B A + A ) + k − k ≤ k ik k k − kk · k i − ik k ik =1 i=1 ! = +1 ! kX X i Xk m 1 − + exp A B A = k ik k m − mk i=1 ! X m 1 m − = exp A B A exp B A + k ik k k − kk · k i − ik k=1 i=k i=k+1 ! X X6 X m 1 − + exp A B A k ik k m − mk ≤ i=1 ! X m m m exp A B A exp B A . ≤ k ik k k − kk · k i − ik =1 i=1 ! = +1 ! kX X i Xk Since B A exp ( B A ) 1, k k − kk ≤ k k − kk − we conclude that
m m m m (I + Bk) (I + Ak) = Z Y − k − k ≤ k=1 k=1 Y Y m m m
exp A (exp ( B A ) 1) exp B A = ≤ k ik k k − kk − k i − ik i=1 ! =1 = +1 !! X kX i Xk m m m m = exp A exp B A exp B A = k ik k i − ik − k i − ik i=1 ! =1 = ! = +1 !! X kX Xi k i Xk m m = exp A exp B A 1 . k ik k i − ik − i=1 ! i=1 ! ! X X
1 n n Corollary 3.4.3. If A, B :[a, b] R × are Riemann integrable functions, then → b b (I + B(x) dx) (I + A(x) dx) − ≤ a a Y Y 1 [LS1], p. 53
78 b b exp A(x) dx exp B(x) A(x) dx 1 . ≤ k k k − k − Za ! Za ! !
Proof. The previous lemma ensures that for every tagged partition D of interval [a, b] we have
m m
P (B,D) P (A, D) = (I + B(ξk)∆tk) (I + A(ξk)∆tk) k − k − ≤ k=1 k=1 Y Y
m m exp A(ξ ) ∆t exp B(ξ ) A(ξ ) ∆t 1 . ≤ k k k k k k − k k k − =1 ! =1 ! ! kX kX The proof is completed by passing to the limit ν(D) 0. → Remark 3.4.4. Lemma 3.4.2 is not present in Schlesinger’s work, he proves directly the Corollary 3.4.3; our presentation is perhaps more readable.
3.5 Lebesgue product integral The most valuable contribution of Schlesinger’s paper is his generalized definition of product integral which is applicable to all matrix functions with bounded and measurable (i.e. bounded Lebesgue integrable) entries. From a historical point of view, such a generalization certainly wasn’t a straightfor- b ward one. Recall the original Lebesgue’s definition: To compute the integral a f of a bounded measurable function f :[a, b] [m, M], we choose a partition → R
D : m = m0 < m1 < < m = M, ··· p then form the sets
E0 = x [a, b]; f(x) = m , { ∈ } Ej = x [a, b]; mj 1 < f(x) mj , j = 1, . . . , p, { ∈ − ≤ } and compute the lower and upper sums
p p
s(f, D) = m0µ0 + mj 1µj,S(f, D) = m0µ0 + mjµj, (3.5.1) − j=1 j=1 X X where µj = µ(Ej) is the Lebesgue measure of the set Ej. Since
p
S(f, D) s(f, D) = (mj mj 1)µj ν(D)(b a), − − − ≤ − j=1 X 79 the sums in (3.5.1) approach a common limit as ν(D) 0 and we define → b f(x) dx = lim s(f, D) = lim S(f, D). ν(D) 0 ν(D) 0 Za → → Similar procedure cannot be used to define product integral of a matrix function n n n n A :[a, b] R × , because R × is not an ordered set. Schlesinger was instead inspired by→ an equivalent definition of Lebesgue integral which is due to Friedrich Riesz (see [FR, KZ]): A bounded function f :[a, b] R is integrable, if and only if there exists a uniformly bounded sequence of step (i.e.→ piecewise-constant) functions f ∞ such that f f almost everywhere on [a, b]; in this case, { n}n=1 n → b b f(x) dx = lim fn(x) dx. n Za →∞ Za To proceed to the definition of product integral we first recall that (see Theorem 3.2.2) b m (I + A(x) dx) = lim eA(ξk)∆tk ν(D) 0 a → =1 Y kY n n for every Riemann integrable function A :[a, b] R × . The product on the right side might be interpreted as →
m b A(ξk)∆tk e = (I + AD(t) dt) , =1 a kY Y where AD is a step function defined by
AD(t) = A(ξk), t (tk 1, tk) ∈ −
(the values A(t ), k = 0, . . . , m, might be chosen arbitrarily). If D ∞ is a k { k}k=1 sequence of tagged partitions of [a, b] such that limk ν(Dk) = 0, it is easily proved that →∞ lim AD (t) = A(t) (3.5.2) k k →∞ at every point t [a, b] at which A is continuous. Since Riemann integrable func- tions are continuous∈ almost everywhere, the Equation (3.5.2) holds a.e. on [a, b]. We are therefore led to the following generalized definition of product integral:
b b (I + A(x) dx) = lim (I + Ak(x) dx) , k a →∞ a Y Y where Ak k∞=1 is a suitably chosen sequence of matrix step functions that converge to A almost{ } everywhere. n n Definition 3.5.1. A function A :[a, b] R × is called a step function if there exist numbers → a = t0 < t1 < < t = b ··· m 80 such that A is a constant function on every interval (tk 1, tk), k = 1, . . . , m. − Clearly, a matrix function A = a n is a step function if and only if all the { ij}i,j=1 entries aij are step functions. n n Definition 3.5.2. A sequence of functions A :[a, b] R × , k N, is called k → ∈ uniformly bounded if there exists a number M R such that Ak(x) M for every k N and every x [a, b]. ∈ k k ≤ ∈ ∈ n n Definition 3.5.3. A function A :[a, b] R × is called measurable if all the → entries aij are measurable functions. n n Lemma 3.5.4. Let Ak :[a, b] R × , k N, be a uniformly bounded sequence of measurable functions such that→ ∈
lim Ak(x) = A(x) k →∞ a.e. on [a, b]. Then A A in the norm of the space L1, i.e. k → b lim Ak(x) A(x) dx = 0. k k − k →∞ Za
Proof. Choose ε > 0. As Ak(x) M for every k N and every x [a, b], we can estimate k k ≤ ∈ ∈
b A (x) A(x) dx ε(b a) + 2Mµ( x; A (x) A(x) ε ). k k − k ≤ − { k k − k ≥ } Za 1 The convergence Ak A a.e. implies convergence in measure , i.e. for every ε > 0 we have → lim µ( x; A(x) Ak(x) ε ) = 0. k { k − k ≥ } →∞ Therefore b lim Ak(x) A(x) dx ε(b a) k k − k ≤ − →∞ Za for every ε > 0. 2 n n Theorem 3.5.5. Let Ak :[a, b] R × , k N, be a sequence of step functions such that → ∈ b lim Ak(x) A(x) dx = 0. k k − k →∞ Za Then the limit b
lim (I + Ak(x) dx) k →∞ a Y 1 [IR], Proposition 8.3.3, p. 256 2 [LS1], p. 55–56
81 exists and is independent on the choice of the sequence A ∞ . { k}k=1 b Proof. We verify that (I + Ak(x) dx) a is a Cauchy sequence. According to Corollary 3.4.3 we have Q b b
(I + Al(x) dx) (I + Am(x) dx) − ≤ a a Y Y
b b exp A (x) dx exp A (x) A (x) dx 1 . ≤ k m k k l − m k − Za ! Za ! ! b The assumption of our theorem implies that the sequence of numbers a Am(x) dx is bounded and that k k R b lim Al(x) Am(x) dx = 0, l,m k − k →∞ Za which proves the existence of the limit. To verify the uniqueness consider two sequences of step functions A , B that satisfy the assumption of the theorem. { k} { k} We construct a sequence Ck , where C2k 1 = Ak and C2k = Bk. Then Ck A a.e. and { } − → b lim Ck(x) A(x) dx = 0, k k − k →∞ Za b which means that limk (I + Ck(x) dx) a exists. Every subsequence of Ck must have the same limit,→∞ therefore { } Q b b
lim (I + Ak(x) dx) = lim (I + Bk(x) dx) . k k →∞ a →∞ a Y Y
n n Definition 3.5.6. Consider function A :[a, b] R × . Assume there exists a → n n uniformly bounded sequence of step functions A :[a, b] R × such that k →
lim Ak(x) = A(x) k →∞ a.e. on [a, b]. Then the function A is called product integrable and we define
b b
(I + A(x) dx) = lim (I + Ak(x) dx) . k a →∞ a Y Y n n We use the symbol L∗([a, b], R × ) to denote the set of all product integrable functions. Remark 3.5.7. The correctness of the previous definition is guaranteed by Lemma n n 3.5.4 and Theorem 3.5.5. Every function A L∗([a, b], R × ) is clearly bounded ∈ 82 and measurable (step functions are measurable and the limit of measurable func- n n tions is again measurable). Assume on the contrary that A :[a, b] R × is a measurable function on [a, b] such that →
a (x) M, x [a, b], i, j = 1, . . . , n. k ij k ≤ ∈ 1 There exists a sequence of step functions Ak k∞=1 which converge to A in the 1 2 { } L norm. This sequence contains a subsequence Bk k∞=1 of matrix functions B = bk n such that B A a.e. on [a, b]. Without{ } loss of generality we k { ij}i,j=1 k → can assume that the sequence Bk k∞=1 is uniformly bounded (otherwise consider k{ } the functions min(max( M, bij),M)). We have thus found a uniformly bounded sequence of step functions− which converge to A a.e. on [a, b]. This means that
n n n n L∗([a, b], R × ) = A :[a, b] R × ; A is measurable and bounded . → Schlesinger remarks that it is possible to further extend the definition of product integral to encompass all matrix functions with Lebesgue integrable (not necessarily bounded) entries, but he doesn’t give any details. We return to this question at the end of the chapter.
3.6 Properties of Lebesgue product integral After having defined the Lebesgue product integral in [LS1], Schlesinger carefully studies its properties. Interesting results may be found also in [LS2].
Lemma 3.6.1. Assume that Ak k∞=1 is a uniformly bounded sequence of functions n n { } from L∗([a, b], R × ), and that A A a. e. on [a, b]. Then k → b b A(x) dx = lim Ak(x) dx. k k k k k Za →∞ Za Proof. According to the Lebesgue’s dominated convergence theorem,
b b b lim Ak(x) dx = lim Ak(x) dx = A(x) dx k k k k k k k k →∞ Za Za →∞ Za (we have used continuity of the norm). Corollary 3.6.2. Inequalities 3.4.1 and 3.4.3 are satisfied for all step functions. As a consequence of the previous lemma we see they are valid even for functions n n from L∗([a, b], R × ). The next statement represents a dominated convergence theorem for the Lebesgue product integral.
1 [RG], Corollary 3.29, p. 47 2 [IR], Theorem 8.4.14, p. 267, and Theorem 8.3.6, p. 257
83 1 Theorem 3.6.3. Assume that Ak k∞=1 is a uniformly bounded sequence of func- n n { } tions from L∗([a, b], R × ) such that A A a. e. on [a, b]. Then k → b b
(I + A(x) dx) = lim (I + Ak(x) dx) . k a →∞ a Y Y
n n Proof. The function A is measurable and bounded, therefore A L∗([a, b], R × ). To complete the proof we use Corollary 3.4.3 in the form ∈
b b
(I + A(x) dx) (I + Ak(x) dx) − ≤ a a Y Y
b b exp A(x) dx exp A (x) A(x) dx 1 k k k k − k − Za ! Za ! ! and Lemma 3.5.4. Remark 3.6.4. The previous theorem holds also for Riemann product integral in case we add an extra assumption that the limit function A is Riemann product integrable. n n Definition 3.6.5. If M is a measurable subset of [a, b] and A L∗([a, b], R × ), we define ∈ b
(I + A(x) dx) = (I + χM (x)A(x) dx) a YM Y (where χM is the characteristic function of the set M).
The previous definition is correct, because the product χM A is obviously a mea- surable bounded function. Remark 3.6.6. The following theorem is proved in the theory of Lebesgue inte- gral2: For every f L1([a, b]) and every ε > 0 there exists δ > 0 such that ∈ f(x) dx < ε ZM
whenever M is a measurable subset of [a, b] and µ(M) < δ. Schlesinger proceeds to prove an analoguous theorem for the product integral (he speaks about “total continuity”). 3 n n Theorem 3.6.7. For every A L∗([a, b], R × ) and every ε > 0 there exists δ > 0 such that ∈ (I + A(x) dx) I < ε − M Y 1 [LS1], p. 57–58 2 [RG], theorem 3.26, p. 46 3 [LS1], p. 59
84 whenever M is a measurable subset of [a, b] and µ(M) < δ. Proof. Substituting B = 0 to Corollary 3.4.3 we obtain
(I + A(x) dx) I exp A(x) dx exp A(x) dx 1 , − ≤ M k k M k k − M Z Z Y
which completes the proof (see Remark 3.6.6). Schlesinger now turns his attention to the indefinite product integral. Recall that if f L1([a, b]), then the indefinite integral ∈ x F (x) = f(t) dt, x [a, b] ∈ Za
is an absolutely continuous function and F 0(x) = f(x) a. e. on [a, b]. Before looking at a product analogy of this theorem we state the following lemma. n n Lemma 3.6.8. If A L∗([a, b], R × ), then ∈ 1 x+h lim A(t) A(x) dt = 0 h 0 h k − k → Zx for almost all x (a, b). ∈ Proof. If f L1([a, b]), then1 ∈ 1 h lim f(x + t) f(x) dt = 0 h 0 h | − | → Z0 for almost all x (a, b) (every such x is called the Lebesgue point of f). Applying this equality to the∈ entries of A we obtain
1 x+h 1 h lim A(t) A(x) dt = lim A(x + t) A(x) dt = 0 h 0 h k − k h 0 h k − k → Zx → Z0 for almost all x (a, b). ∈ 2 n n Theorem 3.6.9. If A L∗([a, b], R × ), then the indefinite integral ∈ x Y (x) = (I + A(t) dt) . a Y 1 satisfies Y − (x)Y 0(x) = A(x) for almost all x [a, b]. ∈ Proof. According to the definition of derivative,
1 1 Y − (x)Y (x + h) I Y − (x)Y 0(x) = lim − . h 0 h → 1 [IR], Theorem 6.3.2, p. 194 2 [LS1], p. 60–61
85 We now prove that
1 x+h Y − (x)Y (x + h) I 1 lim − = lim (I + A(t) dt) I = A(x) (3.6.1) h 0+ h h 0+ h − → → x ! Y for almost all x [a, b]; the procedure is similar for the limit from left. We estimate ∈ 1 x+h 1 x+h (I + A(t) dt) I A(x) (I + A(t) dt) eA(x)h + h − ! − ≤ h − ! x x Y Y k k x+h 1 ∞ A (x)h 1 A(x)h 2 A (x) h + (I + A(t) dt) e + A(x) h ek k h k! ≤ h − ! k k | | k=2 x X Y (3.6.2)
Since A(x) M for some M R, the Corollary 3.4.3 yields k k ≤ ∈ 1 x+h 1 x+h x+h (I + A(t) dt) eA(x)h = (I + A(t) dt) (I + A(x) dt) h − ! h − ! ≤ x x x Y Y Y
1 x+h x+h exp A(x) dt exp A(t) A(x) dt 1 = ≤ h k k k − k − | | Zx ! Zx ! ! k x+h 1 ∞ 1 = exp( A(x) h) A(t) A(x) dt k k h k! k − k ≤ =1 x ! | | kX Z 1 x+h exp(Mh) A(t) A(x) dt + (2M)2h exp(2Mh) . (3.6.3) ≤ h k − k | | Zx ! Equations (3.6.2), (3.6.3), and Lemma 3.6.8 imply Equation (3.6.1). Remark 3.6.10. In the previous theorem we have tacitly assumed that the matrix
x Y (x) = (I + A(t) dt) a Y n n is regular for every x [a, b]. Schlesinger proved it only for A R([a, b], R × ) ∈ ∈ n n (see Corollary 3.2.6), but the proof is easily adjusted to A L∗([a, b], R × ): ∈ If Ak k∞=1 is a uniformly bounded sequence of step functions such that Ak A a.{ e. on} [a, b], then (using 3.2.4 and Lebesgue’s dominated convergence theorem)→
b b b
det(I + A(t) dt) = det lim (I + Ak(t) dt) = lim det(I + Ak(t) dt) = k k a →∞ a →∞ a Y Y Y b b = lim exp Tr Ak(t) dt = exp Tr A(t) dt > 0. k →∞ Za ! Za ! 86 1 n n Theorem 3.6.11. If A L∗([a, b], R × ), then ∈ b b x x ∞ k 2 (I + A(x) dx) = I + A(x1) A(x ) dx1 dx ··· ··· k ··· k a =1 a a a Y kX Z Z Z (where the integrals on the right side are taken in the sense of Lebesgue).
Proof. Let Ak k∞=1 be a uniformly bounded sequence of step functions such that A A a. e.{ on} [a, b]. Every function A is associated with a partition k → k D : a = tk < tk < < tk = b k 0 1 ··· m(k) such that k k k Ak(x) = Aj , x (tj 1, tj ). ∈ − According to the definition of Lebesgue product integral,
b b m(k) k k (I + A(x) dx) = lim (I + Ak(x) dx) = lim exp(Aj ∆tj ). k k a →∞ a →∞ j=1 Y Y Y Schlesinger proves2 first that the product integral might be also calculated as
b m(k) k k (I + A(x) dx) = lim (I + Ai ∆ti ), (3.6.4) k a →∞ i=1 Y Y provided that lim ν(Dk) = 0 (3.6.5) k →∞ (which can be assumed without loss of generality); note that if (3.6.5) is not satis- k k fied, (3.6.4) need not hold (consider A = Ak = I and the partitions a = t0 < t1 = b for every k N). Schlesinger’s proof of (3.6.4) seems too complicated and even faulty; we instead∈ argue similarly as in the proof of Theorem 3.2.2: Take a positive number M such A (x) M for every k N and x [a, b]. Then k k k ≤ ∈ ∈ k k k k k 2 M∆tk exp(A ∆t ) I A ∆t (M∆t ) e j j j − − j j ≤ j for every k N and j = 1, . . . , m(k). According to Lemma 3.2.1, ∈ m(k) m(k) exp(Ak∆tk) (I + Ak∆tk) = j j − j j j=1 j=1 Y Y
m(k) j 1 m(k) − = (I + Ak∆tk) (exp(Ak∆tk) I Ak∆tk) exp(Ak∆tk) l l · j j − − j j · l l ≤ j=1 l=1 l=j+1 X Y Y 1 [LS2], p. 487 2 [LS2], p. 485–486
87 m(k) m(k) M(b a) k k k k M(b a) 2 k 2 M∆tk e − exp(A ∆t ) I A ∆t e − M (∆t ) e j ≤ j j − − j j ≤ j ≤ j=1 j=1 X X m(k) M(b a) 2 Mν(D ) k M(b a) 2 Mν(D ) e − M ν(D )e k ∆t = e − M ν(D )e k (b a). ≤ k j k − j=1 X This completes the proof of (3.6.4). Schlesinger now states that
m(k) m(k) (I + Ak∆tk) = I + Ak Ak ∆tk ∆tk i i i1 ··· is i1 ··· is i=1 s=1 1 i1<
k k k k lim Ai Ai ∆ti ∆ti = k 1 ··· s 1 ··· s →∞ 1 i1<
b xs x2 = A(x1) A(x ) dx1 dx ··· ··· s ··· s Za Za Za The last step perhaps deserves a better explanation: Denote
s s X = (x1, . . . , x ) R ; a x1 < x2 < < x b , { s ∈ ≤ ··· s ≤ } and s Xk = [ti1 1, ti1 ] [ti2 1, ti2 ] [tis 1, tis ], − × − × · · · × − 1 i1<
m(k) k k k k lim I + Ai Ai ∆ti ∆ti = k 1 ··· s 1 ··· s →∞ s=1 1 i1<
1 d d d 1 (CD− ) = D C D D− . dx dx − dx Consider two continuous matrix functions A, B defined on [a, b]. Using the previous formula and also the convention that 1 x y − (I + A(t) dt) = (I + A(t) dt) y x ! Y Y for y > x, we infer the equality
x b d (I + B(t) dt) (I + A(t) dt) = dx x ! Yb Y x b = (I + A(t) dt) (B(x) A(x))(I + A(t) dt) − x Yb Y for every x [a, b]. Denoting S(x) = (I + A(t) dt) x we obtain ∈ b x b Q d 1 (I + B(t) dt) (I + A(t) dt) = S(x)(B(x) A(x))S− (x), dx − x ! Yb Y and consequently (since the left hand side is equal to I for x = b)
x b x 1 (I + B(t) dt) (I + A(t) dt) = (I + S(t)(B(t) A(t))S− (t) dt) . − x Yb Y Yb Substituting x = a and inverting both sides of the equation yields
a b b 1 (I + A(t) dt) (I + B(t) dt) = (I + S(t)(B(t) A(t))S− (t) dt) . (3.6.6) − a a Yb Y Y A similar theorem (concerning the left product integral) was already present in Volterra’s work1. Schlesinger proves2 that the statement remains true even if A, n n B L∗([a, b], R × ). The proof is rather technical and we don’t reproduce it here. ∈ 1 [VH], p. 85–86 2 [LS2], p. 488–489
89 1 n n Theorem 3.6.14. Let A :[a, b] [c, d] R × be such that the integral × → b P (t) = (I + A(x, t) dx) a Y exists for every t [c, d] and that ∈ ∂A (x, t) M, x [a, b], t [c, d], ∂t ≤ ∈ ∈
for some M R. Then ∈ b d 1 ∂A 1 P = P − (t)P 0(t) = S(x, t) (x, t)S− (x, t) dx, dt ∂t Za x where S(x, t) = (I + A(u, t) du) b . Proof. The definition of derivative gives Q a b 1 1 P − (t)P 0(t) = lim (I + A(x, t) dx) (I + A(x, t + h) dx) I . h 0 h − → a ! Yb Y Using Equation (3.6.6) we convert the above limit to
b 1 1 lim (I + S(x, t)(A(x, t + h) A(x, t))S− (x, t) dx) I . h 0 h − − → a ! Y Expanding the product integral to Peano series (see Theorem 3.6.11) we obtain
b x x 1 ∞ k 2 lim ∆(x1, t, h) ∆(xk, t, h) dx1 dxk, (3.6.7) h 0 h ··· ··· ··· → =1 a a a kX Z Z Z where 1 ∆(x, t, h) = S(x, t)(A(x, t + h) A(x, t))S− (x, t). − As the Peano series converges uniformly (the Weierstrass M-test, see Theorem 2.4.5), we can interchange the order of limit and summation. According to the mean value theorem there is a ξ(h) [t, t + h] such that ∈ A(x, t + h) A(x, t) ∂A − = (x, ξ(h)) M. h ∂t ≤
The dominated convergence theorem therefore implies b b 1 ∆(x1, t, h) lim ∆(x1, t, h) dx1 = lim dx1 = h 0 h h 0 h → Za Za → 1 [LS2], p. 490–491
90 b ∂A 1 = S(x1, t) (x1, t)S− (x1, t) dx1, ∂t Za and for k 2 ≥ 1 b xk x2 lim ∆(x1, t, h) ∆(xk, t, h) dx1 dxk = h 0 h ··· ··· ··· → Za Za Za b xk x2 k 1 ∆(x1, t, h) ∆(xk, t, h) = lim h − dx1 dxk = 0, ··· h 0 h ··· h ··· Za Za Za → which completes the proof. The following statement generalizes Theorem 2.5.12; Schlesinger replaces Volterra’s n n n n assumption A ([a, b], R × ) by a weaker condition A L∗([a, b], R × ). ∈ C ∈ 1 n n n n Theorem 3.6.15. If A L∗([a, b], R × ) and C R × is a regular matrix, then ∈ ∈ b b 1 1 (I + C− A(x)C dx) = C− (I + A(x) dx) C. a a Y Y 1 k 1 k Proof. Since (C− AC) = C− A C for every k N, we have ∈ 1 1 exp(C− AC) = C− exp(A)C. If A is a step function, then
b m 1 1 C− A(ξ )C∆t (I + C− A(x)C dx) = e i i = a i=1 Y Y m b 1 A(ξ )∆t 1 = C− e i i C = C− (I + A(x) dx) C. i=1 a Y Y n n In the general case when A L∗([a, b], R × ), we rewrite the above equation with simple functions A in place∈ of A, and then pass to the limit k . k → ∞
3.7 Double and contour product integrals A considerable part of the paper [LS2] is devoted to double and contour product integrals (in R2 as well as in C). Probably the most remarkable achievement is Schlesinger’s proof of the “Green’s theorem” for product integral, which is repro- duced in the following text. 2 n n Definition 3.7.1. Let G be the rectangle [a, b] [c, d] in R and A : G R × a matrix function on G. The double product integral× of A over G is defined→ as
b d (I + A(x, y) dx dy) = I + A(x, y) dx dy , a ! ! c YG Z Y 1 [LS2], p. 489
91 provided both integrals on the right hand side exist (in the sense of Lebesgue). 2 n n Definition 3.7.2. Let G be the rectangle [a, b] [c, d] in R and P,Q : G R × continuous functions on G. We denote × →
x y U(x, y) = (I + P (t, c) dt) (I + Q(x, t) dt) , a c Y Y y x T (x, y) = (I + Q(a, t) dt) (I + P (t, y) dt) c a Y Y for every x [a, b], y [c, d]. The contour product integral over the boundary of rectangle G∈is defined∈ as the matrix
1 (I + P (x, y) dx + Q(x, y) dy) = U(b, d)T (b, d)− . (3.7.1) Y∂G
Remark 3.7.3. Schlesinger refers to the matrices U(b, d) and T (b, d) as to the “integral over the lower step” and “integral over the upper step” of the rectangle G. They are clearly a special case of the contour product integral as defined by Volterra (see definition 2.6.8); the matrix (3.7.1) corresponds to the value of contour product integral along the (anticlockwise oriented) boundary of G. 1 2 n n Theorem 3.7.4. Let G be the rectangle [a, b] [c, d] in R and P,Q : G R × continuous matrix functions on G. Assume that× the derivatives → ∂P ∂Q , ∂y ∂x
exist and are continuous on G. Then
1 (I + P (x, y) dx + Q(x, y) dy) =(I + T ∆∗(P,Q) T − dx dy) , · · Y∂G YG where ∂Q ∂P ∆∗(P,Q) = + PQ QP, ∂x − ∂y − y x T (x, y) = (I + Q(a, t) dt) (I + P (t, y) dt) . c a Y Y Proof. A simple calculation reveals that (compare to Lemma 2.6.4)
1 ∂ d 1 T ∆∗(P,Q) T − = T Q T T − . · · ∂x − dy 1 [LS2], p. 496–497
92 Taking the product integral over G we obtain
1 (I + T ∆∗(P,Q) T − dx dy) = · · G Y b d d 1 = I + T (x, y) Q(x, y) T (x, y) T (x, y)− dy . (3.7.2) − dy a ! c Y According to the rules for differentiating a product of functions (see Theorem 2.3.2),
d a x x d T = (I + P (t, y) dt) Q(a, y)(I + P (t, y) dt) + (I + P (t, y) dt) . dy dy x a a ! Y Y Y Theorem 3.6.14 on differentiating the product integral with respect to a parameter yields x d lim (I + P (t, y) dt) = 0, x a dy → a ! Y and consequently d lim T = Q(a, y). (3.7.3) x a d → y The equalities (3.7.2) and (3.7.3) imply
1 (I + T ∆∗(P,Q) T − dx dy) = · · YG d d 1 = I + lim T (x, y) Q(x, y) T (x, y) T (x, y)− dy = x b − dy → c Y d d 1 = lim I + T (x, y) Q(x, y) T (x, y) T (x, y)− dy (3.7.4) x b − dy → c Y (we have used Theorem 3.6.3 on interchanging the order of limit and integral). For every x [a, b] we have ∈
d 1 d T (x, y) = T (x, d)− T (x, y) dy dy and also
y y 1 d 1 d T (x, d)− T (x, y) = I + T (x, u) du = I + T (x, d)− T (x, u) du . du du d d Y Y Using Theorem 3.6.15 and Equation (3.6.6) we arrive at
d d 1 I + T (x, y) Q(x, y) T (x, y) T (x, y)− dy = T (x, d) − dy · c Y 93 d 1 1 d 1 I + T (x, d)− T (x, y) Q(x, y) T (x, d)− T (x, y) T (x, y)− T (x, d) dy · − dy · c Y c 1 1 d T (x, d)− = T (x, d) I + T (x, d)− T (x, y) dy · dy · d Y d d 1 1 (I + Q(x, y) dy) T (x, d)− = T (x, d) T (x, d)− T (x, c)(I + Q(x, y) dy) · · c c Y Y d 1 1 T (x, d)− = T (x, c)(I + Q(x, y) dy) T (x, d)− . · c Y Finally, Equation (3.7.4) gives
d 1 1 (I + T ∆∗(P,Q) T − dx dy) = lim T (x, c)(I + Q(x, y) dy) T (x, d)− = · · x b → c ! YG Y d 1 = T (b, c)(I + Q(b, y) dy) T (b, d)− = (I + P (x, y) dx + Q(x, y) dy) . c Y Y∂G
Remark 3.7.5. The previous theorem represents an analogy of Green’s theorem for the product integral; we have already encountered a similar statement when discussing Volterra’s work. Volterra’s analogy of the curl operator was
∂Q ∂P ∆(P,Q) = + QP P Q, ∂x − ∂y − while Schlesinger’s curl has the form
∂Q ∂P ∆∗(P,Q) = + PQ QP. ∂x − ∂y −
The reason is that Volterra stated his theorem for the left product integral, while Schlesinger was concerned with the right product integral (see Theorem 2.6.15 and Remark 2.6.7). Whereas Volterra worked with a simply connected domain G (see definition 2.6.12), Schlesinger considers only rectangles. Consider functions P , Q that satisfy assumptions of Theorem 3.7.4 and such that
∆∗(P,Q) = 0 (3.7.5) everywhere in G. Then
(I + P (x, y) dx + Q(x, y) dy) = I, Y∂G 94 which in consequence means that the values of contour product integral over the lower step and over the upper step are the same. Schlesinger then denotes the common value of the matrices U(x, y) and T (x, y) (see definition 3.7.2) by the symbol (b,d) (I + P (x, y) dx + Q(x, y) dy) . (Ya,c) Clearly
(x,y) d d (I + P (u, v) du + Q(u, v) dv) = T (x, y) = P (x, y), dx dx (Ya,c) (x,y) d d (I + P (u, v) du + Q(u, v) dv) = U(x, y) = Q(x, y). dy dy (Ya,c) Schlesinger now proceeds to define product integral along a contour and shows that (in a simply connected domain) the condition (3.7.5) implies that the value of product integral depends only on the endpoints of the contour. His method is almost the same as Volterra’s and we don’t repeat it here. At the end of paper [LS2] Schlesinger treats matrix functions of a complex variable. He defines the contour product integral in complex domain and recapitulates the results proved earlier by Volterra (theorems 2.7.4, 2.7.7, and 2.7.6).
3.8 Generalization of Schlesinger’s definition Thanks to the definition proposed by Ludwig Schlesinger it is possible to extend the class of product integrable functions and to work with bounded measurable functions instead of Riemann integrable functions. At this place we remind the notation
n n n n L∗([a, b], R × ) = A :[a, b] R × ; A is measurable and bounded . → Schlesinger was aware that his definition might be extended to all matrix functions with Lebesgue integrable (not necessarily bounded) entries, i. e. to the class
b n n n n L([a, b], R × ) = A :[a, b] R × ;(L) A(t) dt < , → k k ∞ ( Za ) where the symbol (L) emphasizes that we are dealing with the Lebesgue integral.
Clearly L∗ L. If Ak k∞=1 is a uniformly bounded sequence of functions which converge to⊂A almost{ everywhere,} then according to lemma 3.5.4
b lim Ak A 1 = lim Ak(x) A(x) dx = 0, k k − k k k − k →∞ →∞ Za 95 n n i.e. Ak converge to A also in the norm of space L([a, b], R × ). Taking account of Theorem 3.5.5 it is natural to state the following definition. n n Definition 3.8.1. A function A :[a, b] R × is called product integrable if → there exists a sequence of step functions A ∞ such that { k}k=1
lim Ak A 1 = 0. k k − k →∞ We define b b
(I + A(t) dt) = lim (I + Ak(t) dt) . k a →∞ a Y Y
Remark 3.8.2. The correctness of the previous definition is ensured by theorem n n 3.5.5. Since step functions belong to the space L([a, b], R × ), which is complete, every product integrable function also belongs to this space. Moreover, step func- 1 b tions form a dense subset in this space , and therefore (I + A(t) dt) a exists iff n n b A L([a, b], R × ), i. e. iff the integral (L) A(t) dt exists. ∈ a Q Interested readers are referred to the book [DF]R for more details about the theory of product integral based on definition 3.8.1. As an interesting example we present the proof of theorem on differentiating the product integral with respect to the upper bound of integration. We start with a preliminary lemma (which follows also from Theorem 3.3.2, but we don’t want to use it as we are seeking another way to prove it). n n Lemma 3.8.3. If A :[a, b] R × is a step function, then → x Y (x) = I + Y (t)A(t) dt, x [a, b]. ∈ Za
Proof. There exist a partition a = t0 < t1 < < tm = b and matrices n n ··· A1,...,A R × such that m ∈
A(x) = Ak, x (tk 1, tk). ∈ − Then
x A1(t2 t1) Ak 1(tk 1 tk 2) Ak(x tk 1) Y (x) = (I + A(t) dt) = e − e − − − − e − − ··· a Y
for every x [tk 1, tk]. The function Y is continuous on [a, b] and differentiable except a finite∈ number− of points; we have
Y 0(x) = Y (x)A(x) (3.8.1)
1 [RG], Corollary 3.29, p. 47
96 for x [a, b] t0, t1, . . . , t . This implies ∈ \{ m} x Y (x) = I + Y (t)A(t) dt, x [a, b]. ∈ Za
1 n n Theorem 3.8.4. Consider function A L([a, b], R × ). For every x [a, b] the integral ∈ ∈ x Y (x) = (I + A(t) dt) (3.8.2) a Y exists and the function Y satisfies the equation x Y (x) = I + Y (t)A(t) dt, x [a, b]. (3.8.3) ∈ Za
n n Proof. Let A :[a, b] R × , k N be a sequence of step functions such that k → ∈ b lim Ak A 1 = lim Ak(t) A(t) dt = 0. (3.8.4) k k − k k k − k →∞ →∞ Za Then clearly x lim Ak(t) A(t) dt = 0, x [a, b], k k − k ∈ →∞ Za i. e. the definition (3.8.2) is correct. Denote
x
Yk(x) = (I + Ak(t) dt) . a Y Because Ak are step functions, Lemma 3.8.3 implies x Y (x) = I + Y (t)A (t) dt, x [a, b]. (3.8.5) k k k ∈ Za According to Corollary 3.4.3,
b b Y (x) Y (x) exp A (t) dt exp A (t) A (t) dt 1 = k l − m k ≤ k l k k l − m k − Za ! Za ! !
= exp A 1 (exp A (t) A (t) 1 1) . k lk k l − m k − From Equation (3.8.4) we see that A ∞ is a bounded and Cauchy sequence { k}k=1 with respect to the norm 1. The previous inequality therefore implies that Yk converge uniformly to Y , i.k· e. k
Yk Y = sup Yk(x) Y (x) 0 pro k . k − k∞ x [a,b] k − k → → ∞ ∈ 1 [DF], p. 54–55
97 We now estimate x x Y (t)A (t) dt Y (t)A(t) dt Y A YA 1 k k − ≤ k k k − k ≤ Za Za
(Yk Y )Ak 1 + Y (Ak A) 1 Ak 1 Y k Y + (Ak A) 1 Y , ≤ k − k k − k ≤ k k k − k∞ k − k k k∞ and consequently x x lim Yk(t)Ak(t) dt = Y (t)A(t) dt. k →∞ Za Za The equality (3.8.3) is obtained by passing to the limit in equation (3.8.5). n n Corollary 3.8.5. If A L([a, b], R × ), then the function ∈ x Y (x) = (I + A(t) dt) , x [a, b], ∈ a Y is absolutely continuous on [a, b] and 1 Y (x)− Y 0(x) = A(x) · almost everywhere on [a, b]. Remark 3.8.6. In our proof of the previous theorem we have employed Schlesin- ger’s estimate from Corollary 3.4.3, whose proof is somewhat laborious. The authors of [DF] instead make use of a different inequality, which is easier to demonstrate. n n Let A, B :[a, b] R × be two step functions. Denoting → x x Y (x) = (I + A(t) dt) ,Z(x) = (I + B(t) dt) , a a Y Y 1 we see that the function YZ− is continuous on [a, b] and differentiable except a finite number of points. Using Equation (3.8.1) we calculate 1 1 1 1 1 1 1 1 (YZ− )0 = Y 0Z− + Y (Z− )0 = YY − Y 0Z− YZ− Z0Z− = Y (A B)Z− , − − and consequently x x 1 1 1 Y (x)Z− (x) = I + (YZ− )0(t) dt = I + Y (t)(A(t) B(t))Z− (t) dt. − Za Za Multiplying this equation by Z from right and substituting x = b we obtain
b b (I + A(t) dt) (I + B(t) dt) = Y (b) Z(b) − k − k ≤ a a Y Y b 1 2 B 1 A 1 Y ( t) A(t) B(t) Z− (t) dt Z (b) e k k ek k A B 1 ≤ k k · k − k · k k · k k ≤ k − k Za 1 (we have used Lemma 3.4.1 to estimate Y (t) , Z− (t) and Z(b) ). The meaning of the last inequality is similar to thek meaningk k of inequalityk k fromk Corollary 3.4.3: “If two step functions A, B are close with respect to the norm 1, then the values of their product integrals are also close to each other.” k·k
98
