The Cantor function is a function that is continuous, differentiable,
increasing, non-constant, and the derivative is zero everywhere
except at a set with length zero. It is the most difficult function
in our repertoire and can be found, for example, in Kolmogorov
and Fomin.
Recall the definition of the Cantor set: Let
= [ 1/3, 2/3]
be the middle third of the interval [0, 1]. Let
= [ 1/9, 2/9],
= [ 7/9, 8/9 ]
be the middle thirds of the intervals remaining after deleting
from [0, 1]. Let
= [2/27, 3/27],
= [7/27, 8/27],
= [19/27, 20/27],
= [ 25/27, 26,27]
be the middle thirds of the intervals remaining after deleting
,
,
and
from [0, 1].
Continue in this fashion so that at the n-th stage we have the
intervals
Then the complement of the union of all these intervals
is the Cantor set without
the endpoints. Now define the following function
if t is in the interval
Then, for example, we have that
Then F(t) is defined everywhere in [0, 1] except at the Cantor
set minus the end points 0, 1, 1/3, 2/3, 1/9, 2/9, 7/9, 8/9, ...
If t is a number where F is not defined, then there exists an
increasing sequence {
} of
these endpoints converging to t, and a decreasing sequence
{
' }
of these endpoints converging to t. Since F is defined at those
endpoints
and
',
we define
Now we have defined completely the Cantor function. It has the
following properties:
- F is defined everywhere in the interval [0, 1]
- F is not constant
- F is increasing
- F is continuous in the interval [0, 1]
- F is differentiable in the interval [0, 1]
- F' is zero at every interior point of the intervals
In particular, F' is zero at points of total length 1 in the interval
[0, 1], yet F it is not constant.
Proof
Some of these properties are obvious, and some require more thought.
In particular, why does the above limit of endpoints exist ? That
is a crucial point, because we used this limit to extend the function
to the whole interval [0, 1]. For details and hints about the
Cantor Function, please consult Kolmogorov and Fomin, p 334 ff.
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