Recall the Cantor (middle-third) set which is equal to
where and for is obtained by removing the middle-thirds of each of the disjoint closed intervals making up , i.e. each disjoint closed interval making up gets mapped to .
For example, to obtain , gets mapped to , and this latter set is . To obtain , gets mapped to , and gets mapped to , and is the union of these disjoint intervals.
The purpose of this post is to show that in fact
This set is the collection of all ternary expansions with coefficients and .
First, it is important to note that the elements of all coverge to real numbers. This is true because
the latter series being geometric.
Next, it will be shown that . Let . It is sufficient to construct a sequence corresponding to such that
By definition of , for all . First, . Second, . Since is the result of removing the middle-third of , therefore necessarily is in the left or right remaining third. If it was the left, define , and otherwise define . Now let . Since , therefore it is in one of the disjoint closed intervals making up . Since and has its left and right thirds mapped into , therefore necessarily is in the left third () or the right third ().
Inductively this defines a sequence corresponding to . Define the partial sums
To show (*) holds for our choice of , it is sufficient to show that , and are in one of the disjoint closed intervals making up with . This is sufficient because a property of is that the length of the , excluding , is (e.g. by an inductive argument), so as . This would imply
We proceed by induction. For , if , then by definition and . Otherwise in which case and . Hence the claim holds for .
Now suppose that and are both in one of the disjoint closed intervals making up and . Then is mapped to for . If is in the left interval, then , in which case . Otherwise is in the right interval in which case , so
This completes showing (*) so that indeed .
To show that as well, let . Then
and . The latter fact and an earlier argument show that the partial sums of are still a left endpoint of one of the disjoint closed intervals making up for all . Consequently . Since is (topologically) closed, it then follows that
is in . This completes the proof.
In elementary analysis one might have come across the problem of showing
for . It turns out (Real and Complex Analysis, Rudin) that there is a more advanced proof using Lebesgue integration with respect to a counting measure!
Specifically, we can consider the counting measure where is the power set of , so that is the cardinality of if is finite, and otherwise it is (e.g. while ). Indeed the domain is so that in fact all functions are measurable.
It turns out that integration of such a function with respect to the counting measure is
That is, integrals with respect to the counting measure are just sums (if you are interested, see this document for a proof of this). A consequence (Real and Complex Analysis, Rudin) of Lebesgue’s Monotone Convergence Theorem is: If (are measurable) and
From what was noted earlier the integrals can be replaced with sums:
Consequently simply choosing , which is measurable because as noted earlier all functions are measurable with respect to the counting measure, does the trick! This completes the proof.
Consider a function . Recall is periodic (with period ) iff , . For example, is periodic:
recalling that for . Assume is periodic and continuous. Define, for ,
The are called the Fourier coefficients of , while the Fourier series is
Let’s try an example. Take . Of course, this isn’t periodic. However implicity it is meant that , and is the periodic extension of this mapping, i.e. , , etc. For ,
Evaluating these (real) integrals individually, it is found that
Also . Hence the Fourier series of is
implicitly having the -th term to be , not . However, immediately this a bit difficult to deal with, say if pointwise convergence were in question!
To deal with this, we derive an equivalent representation of the Fourier series. Notice that in general,
Continuing, for , . Similarly,
The same work shows
It is then immediate that . Hence
recalling that , . Consequently the equivalent expression
for the partials sums is obtained, thereby obtaining the equivalent expression
for the Fourier series of . This is much easier to work with for calculations. For example the recently calculated Fourier series can now be written as
Since and are decreasing, for pointwise convergence it is sufficient to show that
are bounded. But this is true, and a common proof is due to Dirichlet (e.g. as explained in this post–easier to show is bounded and substitute ). But more than pointwise convergence is true: Uniform convergence of the Fourier series of is guaranteed, and in fact to . This by a theorem since is continuous, periodic, and piecewise continuously differentiable.