Appendix 1
Some examples in bounded variation and quadratic variation
Example 1. We discussed in class that the function
 x sin(1/ x) if x  0
f ( x)  
if x  0
0
does not have bounded variation on an interval containing zero. Notice that the problem
only occurs near zero. Below is a graph of this function.
Computations done with Maple for a partition of the interval [0,1] into 10,000
subintervals returned a value of approximately 3.465 for the sum
n 1
j 1
n
j
n
sin(
)  sin( )  3.465, for n  10000 . (1)

n
j 1 n
j
j 0
This may not be convincing enough to “see” that the total variation of the function goes
to infinity. We will get back to this example with a rigorous proof.
Example 2. A more convincing example is provided by the function
sin(1/ x) if x  0
f1 ( x)  
if x  0
0
Below is its graph. Notice the same “wild” oscillations of the graph as x approaches zero,
only that the sine wave is not “tamed” by the factor x as in the previous example.
Computations done with Maple for a partition of the interval [0,1] into 100 subintervals
returned a value of approximately 15.716for the sum
n 1
n
n
sin(
)  sin( )  15.716, for n  100 . (2)

j 1
j
j 0
Let us prove rigorously that f1 ( x) has unbounded variation on any interval containing
zero. The idea of the proof is to show that a sum like the one in (2) can be made as large
as possible, thus eliminating the possibility of the function having bounded variation. In
order to show that the sum can grow indefinitely it will be enough to show that we can
find certain partitions on which the sum will grow indefinitely. Now, looking back at the
definition of bounded variation, we need partitions for which   max( t j 1  t j )  0 .
For our partition we will choose points of the form
2
tj 
, for j  n, n  1,..., n  N
 (2 j  1)
where n and N are positive integers. Notice that the partition is a bit unusual. First, the
first value for j is not zero but n, then, as j increases t j decreases. The choice of n is to
guarantee that the norm of the partition can be made as small as possible. We have
4
4
1
t j 1  t j 

 2 
0
n 
2
 (2 j  1)(2 j  3)  4n  n
Now as one can observe the t j ’s don’t form a partition of the interval [0,1] but of the


2
2
,
interval [tn N , tn ]= 
 . But if we can prove that the total
  (2n  2 N  1)  (2n  1) 
variation corresponding to such partitions grows without bound then we prove that the
actual variation on the interval [0,1] must be bigger, so it cannot be bounded. Notice also
that the points in the partition are not equidistant, but we chose these points to capture the
maximum increments in the function sin(1/ x) . We have
n  N 1

sin(
j n
But since sin
n  N 1
1
 (2 j  3)
 (2 j  1)
)  sin( )   sin
 sin
(3)
t j 1
tj
2
2
j n
1
 (2 j  3)
2
 (1) j sin
the sum in (3) is
n  N 1

3
 (2 j  1)

 (1) j , and sin
 (1) j sin  (1) j ,
2
2
2
 (2 j  3)
 sin
 (2 j  1)
 2 N 
 ,
N 
2
2
thus proving that the total variation of the function can grow without bound.
sin
j n
Example 3. A much simpler example of functions that do not have bounded variation is
provided by functions which are unbounded themselves. For example:
0 if x  0

g ( x)   1
 x if x  0
We have
n 1
n n n
n
 j 1 
 j

 n
g
   2n  1 


 g   nn    
n 
2  2 3
 n 
n

 n 1 n 
j 0
Let us now get back to Example 1 and prove that the function has unbounded variation
on the interval [0,1] . We will use the same partition as in the proof in Example 2. Again,
it will be enough to show that the function has a total variation that grows without bound
on a subinterval of [0,1] , for a particular set of partitions. With the partition given by
2
tj 
, for j  n, n  1,..., n  N
 (2 j  1)
we have
n  N 1

j n

t j 1 sin(
n  N 1

j n

n  N 1
1
2
 (2 j  3)
2
 (2 j  1)
)  t j sin( )  
sin

sin
t j 1
tj
2
 (2 j  1)
2
j  n  (2 j  3)
1
2
2

4 n  N 1
   (2 j  3)   (2 j  1)    

j n
1


(2 j  3) N 