Spectral Theory in Hilbert Space

122 B. STIELTJES INTEGRALS
(4) C
(5)
b
a
b
a
f dg =
f dg =
d
a
b
a
f d(Cg),
f dg +
b
d
f dg for a < d < b.
where f, f1, f2, g, g1 and g2 are functions, C a constant and the
formulas should be interpreted to mean that if the integrals to the left
of the equality sign exist, then so do the integrals to the right, and
equality holds.
Proposition B.2 (Change of variables). Suppose that h is continuous
and increasing and f is integrable with respect to g over [h(a), h(b)].
Then the composite function f ◦h is integrable with respect to g ◦h over
[a, b] and
h(b)
h(a)
f dg =
b
a
f ◦ h d(g ◦ h).
We leave the proof also of this proposition to the reader (Exercise
B.3). The formula for integration by parts takes the following
nicely symmetric form in the context of the Stieltjes integral.
Theorem B.3 (Integration by parts). If f is integrable with respect
to g, then g is also integrable with respect to f and
b
a
g df = f(b)g(b) − f(a)g(a) −
b
a
f dg.
Proof. Let a = x0 < x1 < · · · < xn = b be a partition ∆ of [a, b]
and suppose xk−1 ≤ ξk ≤ xk, k = 1, . . . , n. Set ξ0 = a, ξn+1 = b. Then
a = ξ0 ≤ ξ1 ≤ · · · ≤ ξn+1 = b gives a partition ∆ ′ (one discards any
ξk+1 which is equal to ξk) of [a, b] for which |∆ ′ | ≤ 2|∆| (check this!).
We have ξk ≤ xk ≤ ξk+1 and
s =
n
g(ξk)(f(xk) − f(xk−1)) =
k=1
n
n−1
g(ξk)f(xk) − g(ξk+1)f(xk)
k=1
= f(b)g(b) − f(a)g(a) −
k=0
n
f(xk)(g(ξk+1) − g(ξk)).
If |∆| → 0 we have |∆ ′ | → 0, so the last sum converges to b
f dg
a
(note that if ξk+1 = ξk then the corresponding term in the sum is 0).
It follows that s converges to f(b)g(b) − f(a)g(a) − b
f dg and the
a
theorem follows.
k=0