Page 280 - DMTH401_REAL ANALYSIS

Real Analysis

Notes To go farther it will be useful to have some more notation. If  is a partition of [a, b] we will
write

n 
V = V (f, [a, b]) := å|f(x ) – f(x )|,
  i i – 1
1
so that V = sup V (here, f and [a, b] are “assumed”).

 |[a,b]
We can call V the “-variation” of f over [a, b]. Since f has a finite value for each |[a, b], V is
 
always finite. However, V can be infinite. This is so, for example, if f is the Dirichlet function.
Each V pays attention only to the absolute value of the difference between the values at the

opposite ends of an interval of the partition . We will need to take the signs of those differences
into account, and they will lead to two new “variations.”
For a real number x we define its positive part to be x := max{0, x} and we define its negative part
+
–
+
–
to be x := max{0, –x}. Both “parts” are non-negative, and we have x + x = |x| and x – x = x.
–
+
+
–
–
–
+
+
Example: Prove that for all real numbers x and y, (x + y)  x + y and (x + y)  x + y .
+
–
These are “triangle inequalities!” What can be said about (xy) and (xy) ?
We now define the “positive” and “negative” “ -variations” of f over [a, b]:
n  n 
+
–
P = P (f, [a, b]) := å(f(x ) – f(x )) and N = N (f, [a, b]) := å(f(x ) – f(x )) .
  i i – 1   i i – 1
1 1
Definition: The positive variation, P = P(f, [a, b]) and the negative variation N = N(f, [a, b]) of f
over [a, b] are given by P = sup P and N = sup N respectively.


 |[a,b]  |[a,b]
For example, if f increases on [a, b], P = V = f(b) – f(a) and N = 0. If we look at f(x) := |x| on
  
[–1, 1] we will always have 0  P  1 and 0  N  1, and 0  V  2.
  
+
Because of how x and x were defined, we always have (for any function)
–
P + N = V and P – N = f(b) – f(a)
    
If  is a refinement of , we always have O  O , where O stands for any of the letters N, P or V.
 
This follows from several applications of the triangle inequality.
22.6 Some Properties of Functions of Bounded Variation
If f Î BV[a, b] then f is bounded on [a, b].
Proof: Suppose a  x  b. Then, if we let := {a, x, b},
|f(x)| = |f(x) – f(a) + f(a)|  |f(a)| + |f(x) – f(a)| +|f(b) – f(x)| = |f(a)| + V  |f(a)| + V.

The space BV[a, b] is a vector space. For all c Î  and all f Î BV[a, b], V(cf, [a, b]) = |c|V(f, [a, b]).
For all f Î BV[a, b] and g Î BV[a, b], V(f, [a, b])  V(f, [a, b]) + V(g, [a, b]); V(f, [a, b]) = 0 if and only
if f is constant.
Proof: The second assertion follows from these facts: for all | [a, b], V (cf, [a, b]) = |c|V (f,
 
[a, b]); sup{|c|x: x Î E} = |c| sup{x: x Î E} = |c| sup E. The first assertion and the first part of the
third one follow from the second one and the triangle inequality. Finally, suppose that V(f, [a,
b]) = 0 and that a  x  b. Then, with := {a, x, b}, |f(x) – f(a)| |f(x) – f(a)| + |f(b) – f(x)| = V = 0.

Therefore f(x)  f(a).
If f Î BV[a, b] and a < c <b then f Î BV[a, c] and f Î BV[c, b], and conversely. Moreover, V =
V(f, [a, b]) = V(f, [a, c]) + V(f, [c, b]).

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