Archive | Analysis

# 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.

# 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.