Chap5: Differentiation and integration
In this chapter, we want to deal with 2 questions
Z x
F (x) =
f (t)dt
a
when do we have
(i)F 0 (x) = f (x)?
Rb
(ii) a f 0 (x)dx = f (b) − f (a)?
1
Differentiation of Monotone functions
Let us define the following quantities:
D+ f (x) = lim+
h→0
f (x + h) − f (x)
h
and
f (x) − f (x + h)
.
h
h→0
If D+ f (x) = D− f (x) 6= ∞, we say that f is differentiable at x, and we
define f 0 (x) = D+ f (x) = D− f (x).
If D+ f (x) 6= ∞ then f is right differentiable at x.
If D− f (x) 6= ∞ then f is left differentiable at x.
D− f (x) = lim+
Proposition 1 If f is continuous on [a, b] and D+ (or D− ) is everywhere nonnegative on (a, b) then f is nondecreasing on [a, b], that is
f (x) ≤ f (y) for x ≤ y.
Theorem 2 Let f and increasing real-valued function on [a, b]. Then f is differentiable almost everywhere. The derivative f 0 is measurable and
Z b
f 0 (x)dx ≤ f (b) − f (a).
a
Proof. : See Royden
Example: Let
x sin(1/x) x ∈ (0, 1]
f (x) =
0
x 6= 0
Compute D+ f (0).
1
2
Functions of bounded variations
Let f be a real-valued function defined on the interval [a, b] and let a = x0 <
· · · < xk = b be any subdivision of [a, b]. Define
p :=
k
X
+
(f (xi ) − f (xi−1 ))
i=1
n :=
k
X
−
(f (xi ) − f (xi−1 ))
i=1
and
t := p + n =
k
X
|f (xi ) − f (xi−1 )|
i=1
where
r+ = max(r, 0), r− = max(−r, 0), r− = |r| − r+ .
We have f (b) − f (a) = p − n. Set
P :
=
sup p
N:
= sup n
xi
xi
T :
=
sup t
xi
the supremum is over all possible subdivisions of [a, b]. We have that
P ≤ T ≤ P + N.
P is called the positive part of f , N is called the negative part of f and T is
the total variation of f over [a, b] denoted by Tab (f ).
Definition 3 If Tab (f ) < ∞, we say that f is of bounded variation over [a, b].
We write f ∈ BV ([a, b]).
Lemma 4 If f ∈ BV ([a, b]) then
Tab (f ) = Pab + Nab and f (b) − f (a) = Pab − Nab .
Proof. For any subdivision of [a, b] we have that
f (b) − f (a) = p − n
=⇒ f (b) − f (a) = sup p − sup n = Pab − Nab .
xi
xi
On the other hand
p = f (b) − f (a) + n
2
=⇒ P = N + f (b) − f (a).
Moreover
t = p + n = p + p − (f (b) − f (a)) = 2p − (f (b) − f (a))
=⇒ T = 2P − (f (b) − f (a)) = 2P − (P − N ) = P + N.
Theorem 5 A function f ∈ BV ([a, b]) if and only if f is the difference of 2
monotone real-valued functions on [a, b].
Proof. Let f ∈ BV ([a, b]). Set g(x) = Pax , h(x) = Nax .
Pax
= sup
xk
k
X
+
(f (xi − f (xi−1 ) and
i=1
Nax
= sup
xk
k
X
(f (xi − f (xi−1 )−
i=1
Pax ≤ Pay for x ≤ y.
g and h are monotone increasing functions which are real valued and
0 ≤ Nax ≤ Tax ≤ Tab < ∞.
Moreover:
f (x) − f (a) = g(x) − h(x) =⇒ f (x) = g(x) − h(x) + f (a)(Lemma4).
The function h − f (a) is a monotone function then f is the difference of 2
monotone functions.
Viceversa, if f = g − h on [a, b] with g and h increasing then for any subdivision we have
k
X
|f (xi ) − f (xi−1 )|
≤
k
X
[g(xi ) − g(xi−1 )] +
i=1
i=1
=
k
X
[h(xi ) − h(xi−1 )]
i=1
(g(b) − g(a)) + (h(b) − h(a))
Taking the supremum over all the subdivisions we get
Tab (f ) ≤ (g(b) − g(a)) + (h(b) − h(a)).
Corollary 6 If f is of bounded variation on [a, b] then f 0 (x) exists for almost
all x ∈ [a, b].
Example 7 Show that
f (x) =
0
x2 cos(1/x2 )
is not of bounded variation on [0, 1].
3
x=0
0<x≤1
3
Differentiation of an integral
Lemma 8 If f ∈ L1 ([a, b]) then the function F defined by
Z x
F (x) :=
f (t)dt
a
is a continuous function of bounded variation on [a, b].
Proof. Let us show that F is of bounded variation. Let a = x0 < x1 . . . xk = b
be a subdivision of [a, b]. Then
k
X
Z
k
X
|F (xi ) − F (xi−1 )| =
|
i=1
i=1
=⇒
xi
f (t)dt| ≤
xi−1
Tab (f )
k Z
X
i=1
xi
Z
|f (t)|dt =
xi−1
b
|f (t)|dt
a
b
Z
≤
|f (t)|dt < ∞.
a
Lemma 9 If f ∈ L1 ([a, b]) and
Rx
a
f (t)dt = 0 ∀x ∈ [a, b] then f (t) = 0 a.e. on [a, b].
Proof. See Royden.
Lemma 10 If f is bounded and measurable on [a, b] and
Z x
F (x) :=
f (t)dt + F (a)
a
then F 0 (x) = f (x) for almost all x ∈ [a, b].
Proof. We know that F is a continuous function of bounded variation on [a, b]
and so F 0 (x) exists a.e. on [a, b]. Let |f | ≤ K and set
fn (x) =
F (x + h) − F (x)
1
, h=
h
n
then
fn (x) =
thus |fn | ≤ K. Moreover
1
h
Z
x+h
f (t)dt
x
fn (x) −→ F 0 (x) a.e.
Using the Bounded Convergence Theorem and for every c ∈ [a, b]
4
c
Z
c
Z
F 0 (x)dx =
lim
fn (x)dx
a
a
=
1
h
lim
=
=
=
c
Z
(F (x + h) − F (x))dx
a
1
lim
h
Z
1
lim
h
Z
1
lim
h
Z
c+h
Z
F (x)dx −
!
c
F (x)dx
a
a+h
c+h
Z
a+h
F (x)dx −
c
Z
F (x)dx −
c
c+h
Z
F (x)dx −
c
!
c
F (x)dx
a
!
a+h
F (x)dx
a
Using the fact that F is continuous on [a, b], the integral is differentiable
Z c
Z c
=⇒
F 0 (x)dx = F (c) − F (a) =
f (t)dt.
a
a
Z
=⇒
c
(F 0 (x) − f (x))dx = 0 ∀c ∈ [a, b]
a
=⇒ F 0 (x) = f (x) a.e. on [a, b].
Theorem 11 Let f be an integrable function on [a, b] and
Z x
F (x) :=
f (t)dt + F (a)
a
0
then F (x) = f (x) a.e. on [a, b].
Proof. See Royden
4
Absolute continuity
Definition 12 f : [a, b] 7−→ R is absolutely continuous on [a, b] if for every
n
> 0, there exists a δ > 0 such that for every finite collection {(xi , x0i )}i=1 of
disjoint open subintervals of [a, b]
n
X
i=1
|x0i − xi | < δ =⇒
n
X
|f (x0i ) − f (xi )| < .
i=1
Remark 13 (i) An absolutely continuous function is continuous.
(ii) Every indefinite integral is absolutely continuous.
(iii) The sum and the difference of absolutely continuous functions is absolutely
continuous.
5
Lemma 14 If f is an absolutely continuous function on [a, b] then it is of
bounded variation on [a, b].
Proof. One can prove that for c ∈ [a, b], Tab = Tac + Tcb .
Assume that for = 1, I can find δ > 0 such that the absolute continuity holds.
Now let n a natural number such that
b−a
<δ
n
and let a = x0 < x1 · · · < xn = b be the partition of [a, b] with xi −xi−1 =
Tab (f ) =
n
X
b−a
.
n
i
Txxi−1
(f ).
i=1
Now, using the absolute continuity of f
|xi − xi−1 | =
b−a
<δ
n
=⇒ |f (xi ) − f (xi−1 )| < 1
i
(f ) < 1
=⇒ Txxi−1
=⇒ Tab (f ) ≤ n < ∞.
Corollary 15 If f is absolutely continuous, then f has a derivative almost
everywhere.
Lemma 16 If f ∈ AC([a, b]) and f 0 (x) = 0 a.e. then f is constant.
Theorem 17 A function F is an indefinite integral if and only if F ∈ AC([a, b]).
Proof. If F ia an indefinite integral then F is absolutely continuous. Viceversa,
if F ∈ AC([a, b]) then F ∈ BV ([a, b]) and F can be written as the difference of
2 monotone increasing functions;
F (x) = F1 (x) − F2 (x),
and moreover F 0 (x) exists a.e. on [a, b] and
|F 0 (x)| ≤ F10 (x) + F20 (x)
using Theorem 2
Z b
Z
0
=⇒
|F (x)|dx ≤
a
b
F10 (x)dx
a
Z
+
b
F20 (x)dx
a
≤ (F1 (b) − F1 (a)) + (F2 (b) − F2 (a)) < ∞
6
=⇒ F 0 ∈ L1 ([a, b]).
Let us define
Z
x
G(x) =
F 0 (t)dt,
a
then
G ∈ AC([a, b]) =⇒ f (x) = F (x) − G(x) ∈ AC([a, b]).
Using Theorem 11
=⇒ G0 (x) = F 0 (x) a.e. on [a, b]
hence
f 0 (x) = G0 (x) − F 0 (x) = 0 a.e.
Using Lemma 16 we get that f is a constant function.
Z x
=⇒ F (x) =
F 0 (t)dt + F (a).
a
Z
Corollary 18 f ∈ AC([a, b]) =⇒ f (x) =
x
f 0 (t)dt.
a
5
Convex functions
Definition 19 Let φ defined on (a, b). φ is said to be convex on (a, b) if ∀x, y ∈
(a, b), 0 ≤ λ ≤ 1
φ(λx + (1 − λ)y) ≤ λφ(x) + (1 − λ)φ(y).
Geometrically, this means that each point on the chord between (x, φ(x)) and
(y, φ(y)) is above the graph of phi.
Proposition 20 If φ is convex on (a, b) then phi is absolutely continuous on
each closed subinterval of (a, b).
Proposition 21 (Jensen Inequality) Let phi be a convex function on (−∞, ∞)
and f an integrable function on [0, 1] then
Z
Z
φ
f (t)dt ≤ φ (f (t)) dt.
Proof. See Royden.
Corollary 22 Let f ∈ L1 (0, 1) then
0Z 1
1
@ f (t)dtA Z 1
• e 0
≤
ef (t) dt
0
Z
•
0
1
2 Z
f (t)dt ≤
1
f 2 (t)dt.
0
7