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Cantor Set Construction through Ternary Expansions

Recall the Cantor (middle-third) set \Lambda  which is equal to

\displaystyle{\bigcap_{k=0}^{\infty}{E_k}} 

where E_0 = [0,1]  and E_k  for k \geq 1  is obtained by removing the middle-thirds of each of the disjoint closed intervals making up E_{k-1}  , i.e. each disjoint closed interval [a,b]  making up E_{k-1}  gets mapped to [a, a + 1/3^k] \cup [a+2/3^k, b]  .

For example, to obtain E_1  , [0,1]  gets mapped to [0,1/3] \cup [2/3,1]  , and this latter set is E_1  . To obtain E_2  , [0,1/3]  gets mapped to [0,1/9] \cup [2/9,1/3]  , and [2/3,1]  gets mapped to [2/3, 7/9] \cup [8/9, 1]  , and E_2  is the union of these disjoint intervals.

The purpose of this post is to show that in fact

\displaystyle{S := \left\{ \sum_{k=1}^{\infty}a_{k}3^{-k}:a_{k}\in\left\{ 0,2\right\}\right\} } = \Lambda 

This set S  is the collection of all ternary expansions with coefficients 0  and 2  .

First, it is important to note that the elements of S  all coverge to real numbers. This is true because

\displaystyle{\sum_{k=1}^{\infty}{a_k 3^{-k}} \leq 2 \sum_{k=1}^{\infty}{3^{-k}} <\infty} 

the latter series being geometric.

Next, it will be shown that \Lambda \subseteq S  . Let x \in \Lambda  . It is sufficient to construct a sequence a_k \subseteq \{0,2\}  corresponding to x  such that

\displaystyle{x = \sum_{k=1}^{\infty}{a_k 3^{-k}}} (*) 

By definition of \Lambda , x \in E_k  for all k  . First, x \in E_0 = [0,1]  . Second, x \in E_1  . Since E_1  is the result of removing the middle-third of [0,1]  , therefore necessarily x  is in the left or right remaining third. If it was the left, define a_1 = 0  , and otherwise define a_1 = 2  . Now let k \geq 1  . Since x \in E_{k-1}  , therefore it is in one of the disjoint closed intervals [a,b]  making up E_{k-1}  . Since x \in E_k  and [a,b]  has its left and right thirds mapped into E_k  , therefore necessarily x  is in the left third (a_k = 0  ) or the right third (a_k = 2  ).

Inductively this defines a sequence a_k \subseteq \{0,2\}  corresponding to x  . Define the partial sums

\displaystyle{S_k = \sum_{i=1}^k{a_i 3^{-i}}} 

To show (*) holds for our choice of a_k  , it is sufficient to show that \forall{k}  , a_k  and S_k  are in one of the disjoint closed intervals [a,b]  making up E_k  with a = S_k  . This is sufficient because a property of \Lambda  is that the length of the [a,b]  , excluding E_0  , is 3^{-k}  (e.g. by an inductive argument), so |S_k - x| \leq |a-b| = 3^{-k} \rightarrow 0 as k \rightarrow \infty  . This would imply

\displaystyle{\sum_{k=1}^\infty{a_k 3^{-k}} = \lim S_k = x} 

We proceed by induction. For k=1  , if a_1 = 0  , then by definition x \in [0,1/3]  and S_1 = 0  . Otherwise a_1 = 2  in which case x \in [2/3,1]  and S_1 = 2/3  . Hence the claim holds for k=1  .

Now suppose that x  and S_k  are both in one of the disjoint closed intervals [a,b]  making up E_k  and S_k = a  . Then [a,b]  is mapped to [a,a+1/3^{k+1}] \cup [a+2/3^{k+1},b]  for E_{k+1}  . If x  is in the left interval, then a_{k+1} = 0  , in which case S_{k+1} = S_k = a . Otherwise x  is in the right interval in which case a_{k+1} = 2  , so

S_{k+1} = S_k + 2/3^{k+1} = a + 2/3^{k+1} 

This completes showing (*) so that indeed \Lambda \subseteq S  .

To show that S \subseteq \Lambda  as well, let x \in S  . Then

\displaystyle{\sum_{k=1}^\infty{a_k 3^{-k}} = x} 

and a_k \in \{0,2\}  . The latter fact and an earlier argument show that the partial sums S_k  of x are still a left endpoint of one of the disjoint closed intervals making up E_k  for all k  . Consequently S_k \in \Lambda  . Since \Lambda  is (topologically) closed, it then follows that

\lim S_k = x 

is in \Lambda  . This completes the proof.

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Applications of Lebesgue integration to elementary analysis problems

In elementary analysis one might have come across the problem of showing

{\displaystyle \sum_{n=1}^{\infty}\sum_{k=1}^{\infty}a_{nk}=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}a_{nk}}

for a_{nk} \geq 0 . 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 \mu : P(\mathbb{Z}^+) \rightarrow [0,\infty] where P(\mathbb{Z}^+) is the power set of \mathbb{Z}^+ , so that \mu(E) is the cardinality of E if E is finite, and otherwise it is \infty (e.g. \mu(\{1,5,8\})=3 while \mu(\{1,3,5,\ldots\})=\infty ). Indeed the domain is P(\mathbb{Z}^+) so that in fact all functions f : \mathbb{Z}^+ \rightarrow [0,\infty] are measurable.

It turns out that integration of such a function f with respect to the counting measure is

\displaystyle \int_{\mathbb{Z}^+} f d\mu = \sum_{k=1}^{\infty} f(k)

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 f_n : \mathbb{Z}^+ \rightarrow [0,\infty] (are measurable) and

\displaystyle f(x) = \sum_{n=1}^\infty f_n(x)

then

\displaystyle \int_{\mathbb{Z}^+} f d\mu = \sum_{n=1}^\infty \int_{\mathbb{Z}^+} f_n d\mu

From what was noted earlier the integrals can be replaced with sums:

\displaystyle \sum_{k=1}^\infty \sum_{n=1}^\infty f_n(k) = \sum_{k=1}^\infty f(k) = \sum_{n=1}^\infty \sum_{k=1}^\infty f_n(k) 

Consequently simply choosing f_n(k) = a_{nk} , which is measurable because as noted earlier all functions are measurable with respect to the counting measure, does the trick! This completes the proof.

Continuous derivative of the Fourier transform of a function

Define L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right) to be the set of functions f:\mathbb{R}\rightarrow\mathbb{C} which are continuous, bounded, and satisfy
{\displaystyle \left\Vert f\right\Vert _{1}=\int_{-\infty}^{\infty}\left|f\right|<\infty}
Then given f\in L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right) , define its Fourier transform \hat{f} to be
{\displaystyle \hat{f}\left(y\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi ixy}dx}
In my readings of the Fourier transform I encountered details about its derivative. In particular, if g\left(x\right)=-2\pi ixf\left(x\right) \in L_{\mathrm{bc}}^{1}\left(\mathbb{R}\right) , then \hat{f} is continuously differentiable and \hat{f}\,^{\prime}=\hat{g} . The proof I read of it left much to be desired, or at least much was left to the reader. I ended up putting considerable effort into a complete proof, using various resources from internet searches, and I thought I would give back with this proof itself. Link to proof: Continuous derivative of the Fourier transform of a function

Complex Fourier series calculations

Consider a function f:\mathbb{R}\rightarrow\mathbb{C} . Recall  f is periodic (with period  1 ) iff  \forall x f(x+1)=f(x) . For example,  e_{k}=e_{k}(x)=e^{2\pi ikx} is periodic:

e_{k}\left(x+1\right)=e^{2\pi ikx}e^{2\pi ik}=e^{2\pi ikx}\cdot1=e_{k}\left(x\right)

recalling that e^{iy}=\cos y+i\sin y for y\in\mathbb{R} . Assume  f is periodic and continuous. Define, for  k\in\mathbb{Z} ,

{\displaystyle c_{k}=c_{k}(f)=\int_{0}^{1}f(x)\overline{e_{k}}dx=\int_{0}^{1}f(x)e^{-2\pi ikx}dx}

The c_{k} are called the Fourier coefficients of  f , while the Fourier series is

{\displaystyle \sum_{k=-\infty}^{\infty}c_{k}e_{k}}

Let’s try an example. Take f\left(x\right)=x . Of course, this isn’t periodic. However implicity it is meant that \left.f\right|_{\left[0,1\right]}\left(x\right)=x , and f is the periodic extension of this mapping, i.e. \left.f\right|_{[1,2]}=x-1 \left.f\right|_{[-1,0]}=x+1 , etc. For k\neq0 ,

\begin{array}{ccc} c_{k} & = & {\displaystyle \int_{0}^{1}xe^{-2\pi ikx}}\\ \\ & = & {\displaystyle \int_{0}^{1}x\left(\cos\left(2\pi kx\right)-i\sin\left(2\pi kx\right)\right)}\\ \\ & = & {\displaystyle \int_{0}^{1}x\cos\left(2\pi kx\right)-i\int_{0}^{1}x\sin\left(2\pi kx\right)}\end{array}

Evaluating these (real) integrals individually, it is found that

{\displaystyle c_{k}=0-i\left(-\frac{1}{2\pi k}\right)=\frac{i}{2\pi k}}

Also c_{0}=\int_{0}^{1}x=1/2 . Hence the Fourier series of f is

{\displaystyle \sum_{k=-\infty}^{\infty}\frac{i}{2\pi k}e^{2\pi ikx}}

implicitly having the c_{0} -th term to be 1/2 , not i/\left(2\pi\cdot0\right) . 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,

{\displaystyle \sum_{k=-n}^{n}c_{k}e_{k}=c_{0}+\sum_{k=1}^{n}(c_{-k}e_{-k}+c_{k}e_{k})}

Continuing, for k\ne0 , \overline{e_{k}}=e^{-2\pi kx}=e_{-k} . Similarly,

\begin{array}{ccc}c_{-k} & = & {\displaystyle \int_{0}^{1}f(x)e^{-2\pi i(-k)x}}\\ \\ & = & {\displaystyle \int_{0}^{1}f(x)e_{k}}\\ \\ & = & {\displaystyle \int_{0}^{1}f(x)\cos\left(2\pi kx\right)+i\int_{0}^{1}f(x)\sin\left(2\pi kx\right)}\end{array}

The same work shows

{\displaystyle c_{k}=\int_{0}^{1}f(x)\cos\left(2\pi kx\right)-i\int_{0}^{1}f(x)\sin\left(2\pi kx\right)}

It is then immediate that  c_{-k}=\overline{c_{k}} . Hence

c_{-k}e_{-k}+c_{k}e_{k}=\overline{c_{k}e_{k}}+c_{k}e_{k}=2\mathrm{Re}(c_{k}e_{k})

recalling that  \forall z\in\mathbb{C} , \overline{z}+z=2\mathrm{Re}(z) . Consequently the equivalent expression

{\displaystyle c_{0}+\sum_{k=-n}^{n}2\mathrm{Re}(c_{k}e_{k})}

for the partials sums is obtained, thereby obtaining the equivalent expression

{\displaystyle c_{0}+\sum_{k=1}^{\infty}2\mathrm{Re}(c_{k}e_{k})}

for the Fourier series of f . This is much easier to work with for calculations. For example the recently calculated Fourier series can now be written as

{\displaystyle \sum_{k=1}^{\infty}2\mathrm{Re}\left(\frac{1}{2\pi k}\left(i\cos\left(2\pi kx\right)+i^{2}\sin\left(2\pi kx\right)\right)\right)=\sum_{k=1}^{\infty}-\frac{1}{\pi k}\sin\left(2\pi kx\right)}

Since 1/k\rightarrow0 and are decreasing, for pointwise convergence it is sufficient to show that

{\displaystyle \sum_{k=1}^{n}\sin\left(2\pi kx\right)}

are bounded. But this is true, and a common proof is due to Dirichlet (e.g. as explained in this post–easier to show \sum_{k=1}^{n}\sin\left(ky\right) is bounded and substitute y=2\pi x ). But more than pointwise convergence is true: Uniform convergence of the Fourier series of f\left(x\right)=x is guaranteed, and in fact to f . This by a theorem since f is continuous, periodic, and piecewise continuously differentiable.