Page 24 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS

Unit 2: Functions of Bounded Variation

n 1 Notes
f x f x f x f x f x f x
r r 1 n 1 n
r 1 2
x
V f + f c f x
a 2
c x
V f V f , [by (i)]
a a
c x
0 V f V f ,
a a
x s.t. c x c,we have v c v x

lt v x = v c .
x c 0
v x is continuous on the left at x = c.

Similarly considering the partition of [c,b], one can show that v(x) is right continuous also at
x = c and hence v(x) is also continuous at x = c.

Converse of the above Theorem
If v(x) is continuous at x c [a,b] so is f also at x – c.

Proof: Since v(x) is continuous at x = c, for arbitrary small 0, a 0 such that

v x v c ,x c ,c ...(i)

Now let c x c . Then by Note (ii) of Theorem 4, we get

x c x
V f V f V f
a a c
x
v x v c V f
c
x
v x v c V f f x f c
c
f x f c v x v c [by (i)] ...(ii)

Similarly, we can show that f c f x ,if c x c. ...(iii)

(ii) and (iii) show that f(x) is also continuous at x = c.
Theorem 6: Let f and g be functions of bounded variation on [a,b] ; then prove that f+g, f-g, fg and
f/g g x 0, x and cf are functions of bounded variation, c being constant.

Proof:

(i) Set f + g = h, then

h x r 1 h x r f x r 1 g x r 1 f x r g x r
where a = x < x < x <...<x = b
0 1 2 n

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