140 CHAPTER 5. FUNCTIONS ON NORMED LINEAR SPACES
If f ≥ 0, then the mean value theorem implies that for some
t ∈ (a,b) ,F (b)−F (a) =∫ b
af dx = f (t)(b−a)≥ 0.
Thus ∫ b
a(| f |− f )dx≥ 0,
∫ b
a(| f |+ f )dx≥ 0
and so∫ b
a | f |dx≥∫ b
a f dx and∫ b
a | f |dx≥−∫ b
a f dx so this proves∣∣∣∫ b
a f dx∣∣∣≤ ∫ b
a | f |dx. This,
along with part 2 implies the other claim that∣∣∣∫ b
a f dx∣∣∣≤ ∣∣∣∫ b
a | f |dx∣∣∣.
The last claim is obvious because an antiderivative of 1 is F (x) = x. ■Note also that the usual change of variables theorem is available because if F ′ = f , then
f (g(x))g′ (x) = ddx F (g(x)) so that, from the above proposition,
F (g(b))−F (g(a)) =∫ g(b)
g(a)f (y)dy =
∫ b
af (g(x))g′ (x)dx.
We usually let y = g(x) and dy = g′ (x)dx and then change the limits as indicated above,equivalently we massage the expression to look like the above. Integration by parts alsofollows from differentiation rules.
Consider the iterated integral∫ b1
a1· · ·∫ bp
apαxα1
1 · · ·xα pp dxp · · ·dx1. It means just what it
meant in calculus. You do the integral with respect to xp first, keeping the other variablesconstant, obtaining a polynomial function of the other variables. Then you do this one withrespect to xp−1 and so forth. Thus, doing the computation, it reduces to
α
p
∏k=1
(∫ bk
ak
xαkk dxk
)= α
p
∏k=1
(bαk+1
αk +1− aαk+1
αk +1
)and the same thing would be obtained for any other order of the iterated integrals. Sinceeach of these integrals is linear, it follows that if (i1, · · · , ip) is any permutation of (1, · · · , p) ,then for any polynomial q,∫ b1
a1
· · ·∫ bp
ap
q(x1, ...,xp)dxp · · ·dx1 =∫ bi1
aip
· · ·∫ bip
aip
q(x1, ...,xp)dxip · · ·dxi1
Now let f : ∏pk=1 [ak,bk]→ R be continuous. Then each iterated integral results in a con-
tinuous function of the remaining variables and so the iterated integral makes sense. Forexample, by Proposition 5.9.5,
∣∣∣∫ dc f (x,y)dy−
∫ dc f (x̂,y)dy
∣∣∣=∣∣∣∣∫ d
c( f (x,y)− f (x̂,y))dy
∣∣∣∣≤ maxy∈[c,d]
| f (x,y)− f (x̂,y)|< ε
if |x− x̂| is sufficiently small, thanks to uniform continuity of f on the compact set [a,b]×[c,d]. Thus it makes perfect sense to consider the iterated integral
∫ ba∫ d
c f (x,y)dydx. Thenusing Proposition 5.9.5 on the iterated integrals along with Theorem 5.7.1, there exists asequence of polynomials which converges to f uniformly {pn} . Then applying Proposition5.9.5 repeatedly,∣∣∣∣∣
∫ bi1
aip
· · ·∫ bip
aip
f (x)dxp · · ·dx1−∫ bi1
aip
· · ·∫ bip
aip
pn (x)dxp · · ·dx1
∣∣∣∣∣