Page 366 - DMTH401_REAL ANALYSIS

Real Analysis

Notes Definition: If f: [a, b]   is a function and P = {a = a  a  … a  a = b} is a partition of
0 1 n – 1 n
b
b
b
[a, b], let V (f, P) = å n i 1 |f(a ) – f(a i – 1 )|. Define the total variation of f as V (f) = sup{ V (f, P) : P
a
a
=
a
i
b
is a partition of [a, b]}. We say f is of bounded variation if V (f) < ¥. It is easy to see that if f is of
a
bounded variation, then f is bounded ( if x  [a, b], take P = {a  x  b} to see that |f(x) – f(a)| 
b
V (f)).
a
Examples:
b
(i) If f : [a, b]   is a monotone function, then V (f, P) = |f(b) – f(a)| for any partition P of
a
b
[a,b] and hence V (f) = |f(b) – f(a)| < ¥. So f is of bounded variation.
a
1
(ii) Suppose f : [a,b]   is Lipschitz continuous (this happens if f is C ) with Lipschitz constant
b
 > 0. Then, it may be seen that V (f)  (b – a) < ¥ and hence f is of bounded variation.
a
Example: A (uniformly) continuous function f : [a, b]   need not be of bounded
variation. Let f: [0, 1]   be the (uniformly) continuous function defined as f(0) = 0 and f(x) = x
sin (1/x) if x (0,1). Let a = 2/kp  [0,1] for k  . Observe that |f(a ) – f(a )| = |0 – a |
k 2k 2k – 1 2k – 1
1
= a . Let m   and P = {0  a  a  …  a  1}. Then V (f, P)  å m |f(a ) – f(a )| =
=
2k–1 2m 2m–1 1 0 k 1 2k 2k–1
1
–1
å m k 1 a 2k–1 = (2/p) å m k 1 (2K – 1)  ¥ as m  ¥. Hence V (f) = ¥, and thus f is not of bounded
=
=
0
variation. This example also shows that bounded  bounded variation.
Notes If f, g: [a, b]   are of bounded variation, r, s  , and h : [a, b]   is defined as
b
b
b
h = r f(x) + sg(x), then V (h)  |r| V (f) + |s| V (g) < ¥. Hence {f : [a,b]  : f is of
a a a
bounded variation } is a real vector space (in fact, it is a normed space with the norm ||f||
b
= |f(a)| + V (f)).
a
A function f : [a, b]   is of bounded variation iff there exist monotone functions g, h : [a, b]  
such that f(x) = g(x) – h(x) for every x  [a,b]. Consequently, for any function f : [a, b]   of
bounded variation, we have m * L,1 [{x  [a, b] : f is not differentiable at x}] = 0.
Proof: Suppose f = g – h, where g, h are monotone. Since g, h are of bounded variation, f is also of
bounded variation since the collection of functions of bounded variation on [a, b] is a vector
x
space. Conversely assume that f is of bounded variation and define g : [a, b]   as g(x) = V (f).
a
Then g is monotone increasing. Let h = g – f, and consider points x < y in [a, b]. We have g(y) – g(x)
y
= V (f)  |f(y) – f(x)|  f(y) – f(x), and therefore h(y)  h(x). Thus h is also monotone increasing.
x
Clearly, f = g – h.
Let [a, b] be a compact interval. Then for a function f : [a, b]  , we have the following
implications: f is Lipschitz continuous  f is absolutely continuous  f is of bounded variation.
Consequently, if f is either Lipschitz continuous or absolutely continuous, then m * L,1 [Y] = 0,
where Y = {x  [a, b] : f is not differentiable at x}.
Note However, there is a limit to these type of results; there are continuous functions f :
[a, b]   which are not differentiable at any point.
Now we mention a characterization of Riemann integrable functions in terms of small sets. For
simplicity, we restrict ourselves to dimension one, even though the corresponding result is true in
higher dimensions as well. If f : [a, b]   is a bounded function and if P = {a = a < a    a
0 1 n–1

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