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Functional Analysis 10: Linear Functionals

Definition. Let $X$ be a vector space. A linear functional is a linear map $f:\mathcal{D}(f)\subset X\longrightarrow\mathbb{R}$ (or $f:\mathcal{D}(f)\subset X\longrightarrow\mathbb{C}$).

Definition. A linear functional $f:\mathcal{D}(f)\longrightarrow\mathbb{R}$ is said to be bounded if there exists a number $c$ such that $|f(x)|\leq c||x||$ for all $x\in\mathcal{D}(f)$. Just as in linear operators case $||f||$ is defined by
\begin{align*}
||f||&=\sup_{\begin{array}{c}x\in\mathcal{D}(f)\\x\ne O\end{array}}\frac{|f(x)|}{||x||}\\
&=\sup_{\begin{array}{c}x\in\mathcal{D}(f)\\||x||=1\end{array}}|f(x)|.
\end{align*}
Also we have the inequality holds
$$|f(x)|\leq ||f||||x||$$
for all $x\in\mathcal{D}(f)$.

Just as in linear operators case, we have the following theorem holds.

Theorem. A linear functional $f$ with domain $\mathcal{D}(f)$ in a normed space is continuous if and only if $f$ is bounded.

Example. Let $a=(\alpha_j)\in\mathbb{R}^3$. Define $f:\mathbb{R}^3\longrightarrow\mathbb{R}$ by
$$f(x)=x\cdot a=\xi_1\alpha_1+\xi_2\alpha_2+\xi_3\alpha_3$$ for each $x=(\xi_j)\in\mathbb{R}^3$. Then $f$ is a linear functional. By Cauchy-Schwarz inequality, we obtain
$$|f(x)|=|x\cdot a|\leq ||x||||a||$$
which implies $||f||\leq ||a||$. On the other hand, for $x=a$
$$||a||=\frac{||a||^2}{||a||}=\frac{|f(a)|}{||a||}\leq ||f||.$$
Hence, we have $||f||=||a||$.

Example. Define $f:\mathcal{C}[a,b]\longrightarrow\mathbb{R}$ by
$$f(x)=\int_a^b x(t)dt$$
for each $x(t)\in\mathcal{C}[a,b]$. Then $f$ is a linear functional.
\begin{align*}
|f(x)|&\leq\left|\int_a^b x(t)dt\right|\\
&\leq(b-a)\max|x(t)|\\
&=(b-a)||x||.
\end{align*}
So, $||f||\leq b-a$. Let $x=x_0=1$. Then
\begin{align*}
b-a&=\int_a^b dt\\
&=\frac{|f(x_0)|}{||x_0||}\\
&\leq ||f||.
\end{align*}
Hence, we have $||f||=b-a$.

Let $X^\ast$ be the set of all linear functionals. Then $X\ast$ can be made into a vector space. For any $f,g\in X^\ast$ and scalar $\alpha$, define addition $f+g$ and scalar multiplication $\alpha f$ as follows: For each $x\in X$,
\begin{align*}
(f+g)(x)&=f(x)+g(x),\\
(\alpha f)(x)&=\alpha f(x).
\end{align*}
$X^\ast$ is called the dual space of $X$. One may also consider $X^{\ast\ast}=(X^\ast)^\ast$, the dual space of $X^\ast$. Fix $x\in X$. Define a map $g_x: X^\ast\longrightarrow\mathbb{R}$ by
$$g_x(f)=f(x)$$
for each $f\in X^\ast$. For any $f_1,f_2\in X^\ast$,
\begin{align*}
f_1=f_2&\Longrightarrow f_1(x)=f_2(x)\\
&\Longrightarrow g_x(f_1)=g_x(f_2).
\end{align*}
so, $g_x$ is well-defined. Furthermore, $g_x$ is linear. To show this, for any $f_1,f_2\in X^\ast$ and scalars $\alpha,\beta$,
\begin{align*}
g_x(\alpha f_1+\beta f_2)&=(\alpha f_1+\beta f_2)(x)\\
&=\alpha f_1(x)+\beta f_2(x)\\
&=\alpha g_x(f_1)+\beta g_x(f_2).
\end{align*}
Define a map $C: X\longrightarrow X^{\ast\ast}$ by
$$Cx=g_x$$
for each $x\in X$. Then $C$ is a linear map. First let $x=y\in X$. Then for any $f\in X^\ast$, $g_x(f)=f(x)=f(y)=g_y(f)$, so $C(x)=g_x=g_y=C(y)$. Hence, $C$ is well-defined. To show that $C$ is linear, let $x,y\in X$ and $\alpha,\beta$ scalars. For any $f\in X^\ast$,
\begin{align*}
g_{\alpha x+\beta y}(f)&=f(\alpha x+\beta y)\\
&=\alpha f(x)+\beta f(y)\ (f\ \mbox{is linear})\\
&=\alpha g_x(f)+\beta g_y(f)\\
&=(\alpha g_x+\beta g_y)(f).
\end{align*}
Thus,
$$C(\alpha x+\beta y)=g_{\alpha x+\beta y}=\alpha g_x+\beta g_y=\alpha Cx+\beta Cy.$$
If $X$ is an inner product space or $X$ is a finite dimensional vector space, $C$ becomes oen-to-one. Let us assume that $X$ is equipped with an inner product $\langle\ ,\ \rangle$. Then for any fixed $a\in X$, the map $f_a: X\longrightarrow\mathbb{R}$ defined by
$$f_a(x)=\langle a,x\rangle\ \mbox{for each}\ x\in X$$
is a linear functional. Let $Cx=Cy$. Then $g_{x-y}=0$ and so $g_{x-y}(f_{x-y})=||x-y||^2=0$, hence $x=y$. Therefore, $C$ is one-to-one. We will discussed the case when $X$ is finite dimensional in the next lecture. If $C$ is one-to-one, $X$ is embedded into $X^{\ast\ast}$. We call $C:X\hookrightarrow X^{\ast\ast}$ the canonical embedding. (Here, the notation $\hookrightarrow$ means an embedding or a monomorphism.) If in addition $C$ is onto i.e. $X\stackrel{C}{\cong}X^{\ast\ast}$, then $X$ is said to be algebraically reflexive. If $X$ is finite dimensional, then $X$ is algebraically reflexive. This will be discussed in the next lecture as well.

Functional Analysis 9: Bounded and Continuous Linear Operators

Definition. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\longrightarrow Y$ be a linear operator where $\mathcal{D}(T)\subset X$. $T$ is said to be bounded if there exists $c\in\mathbb{R}$ such that for any $x\in\mathcal{D}(T)$,
$$||Tx||\leq c||x||.$$

Suppose that $x\ne O$. Then
$$\frac{||Tx||}{||x||}\leq c.$$
Let
$$||T||:=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\
x\ne O\end{array}}\frac{||Tx||}{||x||}.$$
Then $||T||$ is called the norm of the operator $T$. If $\mathcal{D}(T)=\{O\}$ then we define $||T||=0$.

Lemma. Let $T$ be a bounded linear operator. Then

  1. $||T||=\displaystyle\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||Tx||.$
  2. $||\cdot||$ defined on bounded linear operators satisfies (N1)-(N3).

Proof.

  1. \begin{align*}||T||&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\frac{||Tx||}{||x||}\frac{||Tx||}{||x||}\\&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\left|\left|\frac{||Tx||}{||x||}\right|\right|\\&=\sup_{\begin{array}{c}y\in\mathcal{D}(T)\\||y||=1\end{array}}||Ty||.\end{align*}
  2. \begin{align*}||T||=0&\Longleftrightarrow Tx=0,\ \forall x\in\mathcal{D}(T)\\&\Longleftrightarrow T=0.\end{align*} Since $$\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||(T_1+T_2)x||\leq \sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_1x||+\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\||x||=1\end{array}}||T_2x||,$$ $$||T_1+T_2||\leq ||T_1||+||T_2||.$$

Examples.

  1. The identity operator $I:X\longrightarrow X$ with $X\ne\{O\}$ is a bounded linear operator with $||I||=1$.
  2. Zero operator $O: X\longrightarrow Y$ is a bounded linear operator with $||O||=0$.
  3. Let $X$ be the normed space of all polynomials on $[0,1]$ with $||x||=\max_{t\in[0,1]}|x(t)|$. Differentiation$$T: X\longrightarrow X;\ Tx(t)=x’(t)$$ is not a bounded operator. To see this, let $x_n(t)=t^n$, $n\in\mathbb{N}$. Then $||x_n||=1$ for all $n\in\mathbb{N}$. $Tx_n(t)=nt^{n-1}$ and $||Tx_n||=n$, for all $n\in\mathbb{N}$. So, $\frac{||Tx_n||}{||x_n||}=n$ and hence $||T||$ is not bounded.
  4. Integral operator $$T:\mathcal{C}[0,1]\longrightarrow\mathcal{C}[0,1];\ Tx=\int_0^1\kappa(t,\tau)x(\tau)d\tau$$ is a bounded linear operator. The function $\kappa(t,\tau)$ is a continuous function on $[0,1]\times[0,1]$ called the kernel of $T$. \begin{align*}||Tx||&=\max_{t\in[0,1]}\left|\int_0^1\kappa(t,\tau)x(\tau)d\tau\right|\\&\leq\max_{t\in[0,1]}\int_0^1|\kappa(t,\tau)||x(\tau)|d\tau\\&\leq k_0||x||,\end{align*}where $k_0=\displaystyle\max_{(t,\tau)\in[0,1]\times[0,1]}\kappa(t,\tau)$.
  5. Let $A=(\alpha_{jk})$ be an $r\times n$ matrix of real entries. The linear map $T:\mathbb{R}^n\longrightarrow\mathbb{R}^r$ given by $Tx=Ax$ for each $x\in\mathbb{R}^n$ is bounded. To see this, Let $x\in\mathbb{R}^n$ and write $x=(\xi_j)$. Then $||x||=\sqrt{\displaystyle\sum_{m=1}^n\xi_m^2}$.\begin{align*}||Tx||^2&=\sum_{j=1}^r\left[\sum_{k=1}^n\alpha_{jk}\xi_k\right]^2\\&\leq\sum_{j=1}^r\left[\left(\sum_{k=1}^n\alpha_{jk}^2\right)^\frac{1}{2}\left(\sum_{m=1}^n\xi_m\right)^\frac{1}{2}\right]^2\\&=||x||^2\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2.\end{align*}By setting $c^2=\displaystyle\sum_{j=1}^r\sum_{k=1}^n\alpha_{jk}^2$, we obtain$$||Tx||^2\leq c^2||x||^2.$$

In general, if a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded. Before we discuss this, we first introduce the following lemma without proof.

Lemma. Let $\{x_1,\cdots,x_n\}$ be a linearly independent set of vectors in a normed space $X$. Then there exist a number $c>0$ such that for any scalars $\alpha_1,\cdots,\alpha_n$, we have the inequality
$$||\alpha_1x_1+\cdots+\alpha_nx_n||\geq c(|\alpha_1|+\cdots+|\alpha_n|).$$

Theorem. If a normed space $X$ is finite dimensional, then every linear operator on $X$ is bounded.

Proof. Let $\dim X=n$ and $\{e_1,\cdots,e_n\}$ be a basis for $X$. Let $x=\displaystyle\sum_{j=1}^n\xi_je_j\in X$. Then
\begin{align*}
||Tx||&=||\sum_{j=1}^n\xi_jTe_j||\\
&\leq\sum_{j=1}^n||\xi_j|||Te_j||\\
&\leq\max_{k=1,\cdots,n}||Te_k||\sum_{j=1}^n|\xi_j|.
\end{align*}
By Lemma, there exists a number $c>0$ such that
$$||x||=||\xi_1e_1+\cdots+\xi_ne_n||\geq c(|\xi_1|+\cdots+|\xi_n|)=c\sum_{j=1}^n|\xi_j|.$$
So, $\displaystyle\sum_{j=1}^n|\xi_j|\leq\frac{1}{c}||x||$ and hence
$$||Tx||\leq M||x||,$$ where
$M=\frac{1}{c}\max_{k=1,\cdots,n}||Te_k||$.

What is really nice about linear operators from a normed space into a normed space is that a linear operator being bounded is equivalent to it being continuous.

Theorem. Let $X,Y$ be normed spaces and $T:\mathcal{D}(T)\subset X\longrightarrow Y$ a linear operator. Then

  1. $T$ is continuous if and only if $T$ is bounded.
  2. If $T$ is continuous at a single point, it is continuous.

Proof.

  1. If $T=O$, then we are done. Suppose that $T\ne O$. Then $||T||\ne 0$. Assume that $T$ is bounded and $x_0\in\mathcal{D}(T)$. Let $\epsilon>0$ be given. Choose $\delta=\frac{\epsilon}{||T||}$. Then for any $x\in\mathcal{D}(T)$ such that $||x-x_0||<\delta$, $$||Tx-Tx_0||=||T(x-x_0)||\leq ||T||||x-x_0||<\epsilon.$$ Conversely, assume that $T$ is continuous at $x_0\in\mathcal{D}(T)$. Then given $\epsilon>0$ there exists $\delta>0$ such that $||Tx-Tx_0||<\epsilon$ whenever $||x-x_0||\leq\delta$. Take $y\ne 0\in\mathcal{D}(T)$ and set $$x=x_0+\frac{\delta}{||y||}y.$$ Then $x-x_0=\frac{\delta}{||y||}y$ and $||x-x_0||=\delta$. So,\begin{align*}||Tx-Tx_0||&=||T(x-x_0)||\\&=\left|\left|T\left(\frac{\delta}{||y||}y\right)\right|\right|\\&=\frac{\delta}{||y||}Ty\\&<\epsilon.\end{align*}Hence, for any $y\in\mathcal{D}(T)$, $||Ty||\leq\frac{\epsilon}{\delta}||y||$ i.e. $T$ is bounded.
  2. In the proof of part (a), we have shown that if $T$ is continuous at a point, it is bounded. If $T$ is bounded, then it is continuous by part (a).

Corollary. Let $T$ be a bounded linear operator. Then

  1. If $x_n\to x$ then $Tx_n\to Tx$.
  2. $\mathcal{N}(T)$ is closed.

Proof.

  1. If $T$ is bounded, it is continuous and so the statement is true.
  2. Let $x\in\overline{\mathcal{N}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{N}(T)$ such that $x_n\to x$. Since $Tx_n=0$ for each $n=1,2,\cdots$, $Tx=0$. Hence, $x\in\mathcal{N}(T)$.

Theorem. Let $X$ be a normed space and $Y$ a Banach space. Let $T:\mathcal{D}(T)\subset X\longrightarrow Y$ be a bounded linear operator. Then $T$ has an extension $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ where $\tilde T$ is a bounded linear operator of norm $||\tilde T||=||T||$.

Proof. Let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$. Since $T$ is bounded and linear,
\begin{align*}
||Tx_m-Tx_n||&=||T(x_m-x_n)||\\
&\leq||T||||x_m-x_n||,
\end{align*}
for all $m,n\in\mathbb{N}$. Since $(x_n)$ is convergent, it is Cauchy so given $\epsilon>0$ there exists a positive integer $N$ such that for all $m,n\geq N$, $||x_m-x_n||<\frac{\epsilon}{||T||}$. Hence, for all $m,n>N$,
\begin{align*}
||Tx_m-Tx_n||&\leq ||T||||x_m-x_n||\\
&<\epsilon.
\end{align*}
That is, $(Tx_n)$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, there exists $y\in Y$ such that $Tx_n\to y$. Define $\tilde T:\overline{\mathcal{D}(T)}\longrightarrow Y$ by $\tilde Tx=y$. In order for $\tilde T$ to be well- defined, its definition should not depend on the choice $(x_n)$. Suppose that there is a sequence $(z_n)\subset\mathcal{D}(T)$ such that $z_n\to x$. Then $x_n-z_n\to 0$. Since $T$ is bounded, it is continuous so $T(x_n-z_n)\to 0$. This means that $\displaystyle\lim_{n\to\infty}Tz_n=\lim_{n\to\infty}Tx_n=y$. $\tilde T$ is linear and $\tilde T|_{\mathcal{D}(T)}=T$. To show that $\tilde T$ is bounded, let $x\in\overline{\mathcal{D}(T)}$. Then there exists a sequence $(x_n)\subset\mathcal{D}(T)$ such that $x_n\to x$ as before. Since $T$ is bounded, for each $n=1,2,\cdots$,
$$||Tx_n||\leq ||T||||x_n||.$$ Since the norm $x\longmapsto||x||$ is continuous, as $n\to\infty$ we obtain
$$||\tilde Tx||\leq ||T||||x||.$$ Hence, $\tilde T$ is bounded and $||\tilde T||\leq ||T||$. On the other hand, since $\tilde T$ is an extension of $T$, $||T||\leq||\tilde T||$. Therefore, $||\tilde T||=||T||$.

Functional Analysis 8: Linear Operators

From here on, a map from a vector space into another vector space will be called an operator.

Definition. A linear operator $T$ is an operator such that

  1. $T(x+y)=Tx+Ty$ for any two vectors $x$ and $y$.
  2. $T(\alpha x)=\alpha Tx$ for any vector $x$ and a scalar $\alpha$.

Proposition. An operator $T$ is a linear operator if and only if
$$T(\alpha x+\beta y)=\alpha Tx+\beta Ty$$
for any vectors $x,y$ and scalars $\alpha,\beta$.

Denote by $\mathcal{D}(T)$, $\mathcal{R}(T)$ and $\mathcal{N}(T)$, the domain, the range and the null space, respectively, of a linear operator $T$. The null space $\mathcal{N}(T)$ is the kernel of $T$ i.e.
$$\mathcal{N}(T)=T^{-1}(0)=\{x\in \mathcal{D}(T): Tx=0\}.$$
Since the term kernel is reserved for something else in functional analysis, we call it the null space of $T$.

Example. [Differentiation] Let $X$ be the space of all polynomials on $[a,b]$. Define an operator $T: X\longrightarrow X$ by
$$Tx(t)=x’(t)$$
for each $x(t)\in X$. Then $T$ is linear and onto.

Example. [Integration] Recall that $\mathcal{C}[a,b]$ denotes the space of all continuous functions on the closed interval $[a,b]$. Define an operator $T:\mathcal{C}[a,b]\longrightarrow\mathcal{C}[a,b]$ by
$$Tx(t)=\int_a^tx(\tau)d\tau$$
for each $x(t)\in\mathcal{C}[a,b]$. Then $T$ is linear.

Example. Let $A=(a_{jk})$ be an $r\times n$ matrix of real entries. Define an operator $T: \mathbb{R}^n\longrightarrow\mathbb{R}^r$ by
$$Tx=Ax=(a_{jk})(\xi_l)=\left(\sum_{k=1}^na_{jk}\xi_k\right)$$
for each $n\times 1$ column vector $x=(\xi_l)\in\mathbb{R}^n$. Then $T$ is linear as seen in linear algebra.

Theorem. Let $T$ be a linear operators. Then

  1. The range $\mathcal{R}(T)$ is a vector space.
  2. If $\dim\mathcal{D}(T)=n<\infty$, then $\dim\mathcal{R}\leq n$.
  3. The null space $\mathcal{N}(T)$ is a vector space.

Proof. Parts 1 and 3 are straightforward. We prove part 2. Choose $y_1,\cdots,y_{n+1}\in\mathcal{R}(T)$. Then $y_1=Tx_1,\cdots,y_{n+1}=Tx_{n+1}$ for some $x_1,\cdots,\\x_{n+1}\in\mathcal{D}(T)$. Since $\dim\mathcal{D}(T)=n$, $x_1,\cdots,x_{n+1}$ are linearly dependent. So, there exist scalars $\alpha_1,\cdots,\alpha_{n+1}$ not all equal to 0 such that $\alpha_1x_1+\cdots+\alpha_{n+1}x_{n+1}=0$. Since $T(\alpha_1x_1+\cdots+\alpha_{n+1}x_{n+1})=\alpha_1y_1+\cdots+\alpha_{n+1}y_{n+1}=0$, $\mathcal{R}$ has no linearly independent subset of $n+1$ or more elements.

Theorem. $T$ is one-to-one if and only if $\mathcal{N}=\{O\}$.

Proof. Suppose that $T$ is one-to-one. Let $a\in\mathcal{N}$. Then $Ta=O=TO$. Since $T$ is one-to-one, $a=O$ and hence $\mathcal{N}=\{O\}$. Suppose that $\mathcal{N}=\{O\}$. Let $Ta=Tb$. Then by linearity $T(a-b)=O$ and so $a-b\in\mathcal{N}=\{O\}\Longrightarrow a=b$. Thus, $T$ is one-to-one.
Theorem.

  1. $T^{-1}: \mathcal{R}(T)\longrightarrow\mathcal{D}(T)$ exists if and only if $\mathcal{N}=\{O\}$ if and only if $T$ is one-to-one.
  2. If $T^{-1}$ exists, it is linear.
  3. If $\dim\mathcal{D}(T)=n<\infty$ and $T^{-1}$ exists, then $\dim\mathcal{R}(T)=\dim\mathcal{D}(T)$.

Proof. Part 1 is trivial. Part 3 follows from part 2 of the previous theorem. Let us prove part 2. Let $y_1,y_2\in\mathcal{R}(T)$. Then there exist $x_1,x_2\in\mathcal{D}(T)$ such that $y_1=Tx_1$, $y_2=Tx_2$. Now,
\begin{align*}
\alpha y_1+\beta y_2&=\alpha Tx_1+\beta Tx_2\\
&=T(\alpha x_1+\beta x_2).
\end{align*}
So,
\begin{align*}
T^{-1}(\alpha y_1+\beta y_2)&=T^{-1}(T(\alpha x_1+\beta x_2))\\
&=\alpha x_1+\beta x_2\\
&=\alpha T^{-1}y_1+\beta T^{-1}y_2.
\end{align*}

Functonal Analysis 7: Further Properties of Normed Spaces

Normed spaces are not necessarily finite dimensional. So it is important to understand the notion of a basis for an infinite dimensional normed space. Suppose that there is a basis of a normed space $X$ as an infinite sequence $(e_n)$ in $X$. Then any $x\in X$ can be represented as the infinite superposition of the $e_n$’s
\begin{equation}
\label{eq:superposition}
x=\sum_{j=1}^\infty\alpha_je_j,
\end{equation}
where $\alpha_1,\alpha_2,\cdots$ are scalars.
In order for this to make sense, we need to make sure that the infinite sum in \eqref{eq:superposition} converges. Thus we have the following definition of a basis for an infinite dimensional normed space.

Definition. Suppose that a normed space $X$ contains a sequence $(e_n)$ with property that $\forall x\in X$, there exists uniquely a sequence of scalars $(\alpha_n)$ such that
$$||x-\sum_{j=1}^n\alpha_je_j||\rightarrow 0\ \mbox{as}\ n\to\infty.$$
Then $(e_n)$ is called a Schauder basis for $X$. The infinite sum $\displaystyle\sum_{j=1}^\infty\alpha_je_j$ is called the expansion of $x$.

Example. $\ell^p$ has a Schauder basis $(e_n)$, where $e_n=(\delta_{nj})$.

Theorem. If a normed space has a Schauder basis, it is separable i.e. it has a countable dense subset.

Proof. Recall that $D$ is a dense subset of $X$ if $\bar D=X$. This equivalent to saying that $\forall \epsilon>0$, $\forall x\in X$, $B(x,\epsilon)\cap D\ne\emptyset$.

Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s. Then $D$ is countable. Let $x\in X$. Then $\exists$ a sequence of scalars $(\alpha_n)$ such that $x=\displaystyle\sum_{j=1}^\infty\alpha_je_j$. Given $\epsilon>0$, $\exists$ a positive integer $N$ such that
$$||x-(\alpha_1e_1+\cdots+\alpha_ne_n)||<\epsilon$$
for all $n\geq N$. This implies that $\alpha_1e_1+\cdots+\alpha_ne_n\in B(x,\epsilon)\cap D$ for all $n\geq N$.

One question mindful readers may have is does every separable Banach space have a Schauder basis? The answer is negative and a counterexample can be found in

Enflo, P. (1973), A counterexample to the approximation property. Acta Math. 130, 309–317.

We finish this lecture with the following theorem.

Theorem. [Completion] Let $X$ be a normed space. Then there exists a Banach space $\hat X$ and an isometry from $X$ onto $W\subset\hat X$ which is dense in $\hat X$. The space $\hat X$ is unique up to isometries.

Factor Groups (Quotient Groups)

Let $G$ be a group and $N\leq G$. Recall the equivalence relation $\sim$ on $G$ defined by
$$a\sim b\ \mbox{if}\ ab^{-1}\in N$$ for any $a,b\in G$. Each equivalence class $[a]$ is identified with right coset $Na$. The quotient set $G/\sim$ is the set of all equivalence classes or equivalently all right cosets of the subgroup $N$ in $G$. A question we can ask is can we give a group structure to the quotient set $G/\sim$? The answer is affirmative if $N$ is a normal subgroup of $G$. This is what we are going to discuss in this lecture.

Let $N\vartriangleleft G$. Denote by $G/N$ the set of all right cosets of $N$ in $G$ i.e.
$$G/N=\{Na:a\in G\}.$$
For any $Na, Nb\in G/N$,
\begin{align*}
(Na)(Nb)&=N(aN)b\\
&=N(Na)b\\
&=N(Nab)\\
&=Nab.
\end{align*}
So, it appears that if $N\vartriangleleft G$, we may define an operation $\cdot$ on $G/N$ naturally from the binary operation on $G$ by the equation \begin{equation}
\label{eq:cosetoper}
Na\cdot Nb=Nab
\end{equation} for any $Na,Nb\in G/N$. But is this operation well-defined? To see that let $Na=Nc$ and $Nb=Nd$. Then $Nac^{-1}=N$ and $Nbd^{-1}=N$. So $ac^{-1}\in N$ and $bd^{-1}\in N$. Now,
\begin{align*}
N(ac)(bd)^{-1}&=N(ac)(d^{-1}b^{-1})\\
&=Na(cd^{-1})b^{-1}\\
&=a(Ncd^{-1})b^{-1}\\
&=aNb^{-1}\\
&=Nab^{-1}\\
&=N.
\end{align*}
That is, $Na\cdot Nb=Nab=Ncd=Nc\cdot Nd$. Hence, right coset operation given by the equation \eqref{eq:cosetoper} is well-defined. Conversely, if $N\leq G$ and right coset operation given by the equation \eqref{eq:cosetoper} is well-defined, then $N$ must be a normal subgroup of $G$. The verification of this is left to readers as an exercise. Furthermore, $Ne=N$ is an identity element in $G/N$ and for each $Na\in G/N$, $Na^{-1}\in G$ and $NaNa^{-1}=N$, hence $(Na)^{-1}=Na^{-1}$. Therefore, $(G/N,\cdot)$ is a group. This group is called a factor group or a quotient group of $G$ modulo $N$.

Theorem. If $N\vartriangleleft G$ then there exists an onto homomorphism (an epimorphism) $\psi:G\longrightarrow G/N$ such that $\ker\psi=N$, Such a homomorphism is called a natural homomorphism or a canonical homomorphism.

Proof. Define $\psi: G\longrightarrow G/N$ by
$$\psi(a)=Na$$
for any $a\in G$. If $a=b$ then $\psi(a)=Na=Nb=\psi(b)$ so $\psi$ is well-defined. Let $Na\in G/N$. Then $a\in G$ and $\psi(a)=Na$, so $\psi$ is onto. $\psi(ab)=Nab=NaNb=\psi(a)\psi(b)$, hence $\psi$ is a homomorphism. To show that $\ker\psi=N$,
\begin{align*}
a\in\ker\psi&\Longleftrightarrow \psi(a)=N\\
&\Longleftrightarrow Na=N\\
&\Longleftrightarrow a\in N.
\end{align*}

$G/N$ is the set of all right coset of $N$ in $G$, so $|G/N|=|G:N|=\frac{|G|}{|N|}$.

Example. Since $\mathbb{Z}$ is an abelian group, for any $n\in\mathbb{N}$, $n\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$. So, for any $n\in\mathbb{N}$, $\mathbb{Z}/n\mathbb{Z}$ is a factor group. Each $\mathbb{Z}/n\mathbb{Z}$ is indeed isomorphic to $\mathbb{Z}_n$.

The notion of factor groups can be used to prove an important theorem in group theory called Cauchy Theorem. First we study the following theorem as we will also need it (actually its corollary) to prove Cauchy Theorem.

Theorem. Let $G$ be a finite group and $a\in G$ with $|a|=n$. Then for any $k\in\mathbb{Z}$,

  1.  $|a^k|=\frac{n}{d}$ where $d=(k,n)$.
  2.  $|a^k|=n$ if and only if $(k,n)=1$.

Proof. Since 2 follows from 1, we prove 1 only. Since $d=(k,n)$, $n=n_1d$ and $k=k_1d$ for some $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$.
$$(a^k)^{n_1}=a^{k_1n}=(a^n)^{k_1}=e.$$ So, if we let $|a^k|=s$, then $s|n_1$. On the other hand, $e=|a^k|^s=a^{ks}$, so $n|ks$ which implies that $n_1|k_1s$. Since $(n-1,k_1)=1$, $n_1|s$. Therefore, $s=n_1=\frac{n}{d}$.

Corollary. If a finite group $G$ has no nontrivial subgroup, then $G$ must be a cyclic group of a prime order.

Proof. Let $a\ne e\in G$. Then $\langle a\rangle\leq G$. Since $G$ has no nontrivial subgroup and $a\ne e$, $\langle a\rangle=G$ i.e. $G$ is a cyclic group. Let $|a|=n$. If $k\ne n$ and $k|n$. Then by the above theorem $|a^k|=\frac{n}{k}$ since $(n,k)=k$. Since $a^k\ne e$ and $G$ has no nontrivial subgroup, $\langle a^k\rangle=G$. This implies that $|a^k|=n$ and so $k=1$. Therefore, $n$ is a prime.

Theorem. [Cauchy] If $G$ a finite abelian group and and $p||G|$ where $p$ is a prime, then $G$ has an element of order $p$.

Proof. We prove the theorem by induction on $|G|$. If $|G|=1$, then there is no prime that divides $|G|$, so the theorem is vacuously true. Suppose that the statement is true for all abelian groups whose order is less than $|G|$. Let $\{e\}\not\leq N\not\leq G$. If $p||N|$, then by induction hypothesis there exists $a\in N$ such that $|a|=p$. Since $N\subset G$, we are done in this case. Now suppose that $p\not|N|$. Since $G$ is abelian, $N\vartriangleleft G$. Since $p|G$ and $p\not| N$, $p||G/N|<|G|$. So by induction hypothesis, $G/N$ has an element $Na$ of order $p$.
$$(Na)^p=Na^p=N\Longrightarrow a^p\in N$$
but $a\not\in N$ since $Na\ne N$. Let $|N|=m$. Then $e=(a^p)^m=(a^m)^p$. Let $b=a^m$. If $b\ne e$, then $b$ is an element of order $p$. What if $b=a^m=e$? If so, $(Na)^m=N$. Since $|Na|=p$, $p|m=|N|$. but by assumption $p\not||N|$. A contradiction! So, we are done if $G$ has a nontrivial subgroup. What if $G$ does not have a nontrivial subgroup? If so, by corollary above $G$ must be a cyclic group of a prime order. Since $p||G|$, $|G|=p$. In this case, any $a\ne e\in G$ is an element of order $p$.

Remark. It should be noted that Cauchy Theorem still holds even if the group $G$ is not abelian. We will not however prove the generalized version of Cauchy Theorem here.

Group Theory 9: Normal Subgroups

In here, it was shown that the kernel $K=\ker\varphi$ of a homomorphism $\varphi: G\longrightarrow G’$ is a subgroup of $G$ and that it satisfies the property $aK=Ka$ for all $a\in G$. The kernel of a homomorphism is an example of a particular kind of subgroups of a group called normal subgroups.

Definition. The subgroup $N$ of $G$ is called a normal subgroup of $G$ if $a^{-1}Na\subset N$, $\forall a\in G$. If $N$ is a normal subgroup of $G$, we write $N\vartriangleleft G$.

Note that $\forall a\in G$, $a^{-1}Na\subset N$ implies that $\forall a\in G$, $a^{-1}Na=N$. Hence, we have

Theorem. $N\vartriangleleft G$ if and only if $\forall a\in G$, $aN=Na$.

If $G$ is abelian, every subgroup is a normal subgroup. But the converse need not be true. There are nonabelian groups in which every subgroup is normal. Such nonabelian groups are called Hamiltonian.

Example. Let $\mathbb{Q}_8=\{\pm 1,\pm, i, \pm j, \pm k\}$ where $1,i,j,k$ satisfy the multiplication laws
\begin{align*}
1^2&=1,\ i^2=j^2=k^2=-1,\\
ij&=-ji=k,\ jk=-kj=i,\ ki=-ik=j.
\end{align*}
Then $\mathbb{Q}_8$ forms a nonabelian group. There are four subgroups of $\mathbb{Q}_8$:
\begin{align*}
\langle -1\rangle&=\{1,-1\},\\
\langle i\rangle&=\{1,-1,i,-i\},\\
\langle j\rangle&=\{1,-1,j,-j\},\\
\langle k\rangle&=\{1,-1,k,-k\}.
\end{align*}
All these subgroups are normal subgroups of $\mathbb{Q}_8$, so $\mathbb{Q}_8$ is Hamiltonian. Note that $1,i,j,k$ form a 4-dimensional algebra of quaternions over $\mathbb{R}$ $$\mathbb{H}=\{a1+bi+cj+dk:a,b,c,d\in\mathbb{R}\}.$$ The real algebra of quaternions $\mathbb{H}$ is identified with the 4-dimensional Euclidean space $\mathbb{R}^4$.

Recall that the index $|G:H|$ of a subgroup $H$ in $G$ is the number of right cosets of $H$ in $G$ or equivalently the number of left cosets of $H$ in $G$ as discussed here.

Theorem. Let $G$ be a group and $N\leq G$. If $|G:N|=2$ then $N\vartriangleleft G$.

Proof. Let $a\in G$. If $a\in N$, $aN=Na$. Now suppose that $a\in G\setminus N$. Then
$$N\cap Na=\emptyset=N\cap aN.$$
Furthermore, since $G=N\cup Na=N\cup aN$, $Na=Na$.

Example. Recall that $|S_n|=n!$ and $|A_n|=\frac{n!}{2}$. Since $|S_n:A_n|=2$, $A_n\vartriangleleft S_n$.

Remark. The converse of the above theorem need not be true, namely the index $|G:N|$ of a normal subgroup $N$ in $G$ is not necessarily 2. The 4th dihedral group (we introduced it here)
$$D_4=\{1,\sigma,\sigma^2,\sigma^3,\tau,\sigma\tau,\sigma^2\tau,\sigma^3\tau\}$$
has the following nontrivial subgroups:
\begin{align*}
&\{1,\sigma^2,\sigma\tau,\sigma^2\tau\},\ \{1,\sigma,\sigma^2,\sigma^3\},\ \{1,\sigma^2,\tau,\sigma^2\tau\},\\
&\{1,\sigma^3\tau\},\ \{1,\sigma\tau\},\ \{1,\sigma^2\},\ \{1,\tau\},\ \{1,\sigma^2\tau\}.
\end{align*}
The subgroups of order 4 are normal subgroups of $D_4$ because their indices are 2. The center of $D_4$ is $\{1,\sigma^2\}$, so it is also a normal subgroup of $D_4$. However, $|D_4:Z(D_4)|=4$. The rest of nontrivial subgroups are not normal subgroups of $D_4$.

Remark. $H\vartriangleleft N, N\vartriangleleft G\not\Longrightarrow H\vartriangleleft G$. In the previous remark, $H=\{1,\tau\}$ is a normal subgroup of $\{1,\sigma^2,\tau,\sigma^2\tau\}$ but $H\not\vartriangleleft D_4$.

Example. The center $Z(G)$ of a group $G$
$$Z(G)=\{x\in G: \forall g\in G, xg=gx\}$$ is a normal subgroup of $G$.

Example. If $N\leq Z(G)$ then $N\vartriangleleft G$.

Example. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\vartriangleleft G$ as seen here.

Example. Let $\mathbb{R}^\ast=\mathbb{R}\setminus\{0\}$. Define $f: \mathrm{GL}(2,\mathbb{R})\longrightarrow\mathbb{R}^\ast$ by
$$f(A)=\det A,\ \mbox{for any}\ A\in\mathrm{GL}(2,\mathbb{R}).$$ Then $f$ is a homomorphism because for any $A,B\in\mathrm{GL}(2,\mathbb{R})$, $f(AB)=\det(AB)=(\det A)(\det B)=f(A)f(B)$. $f$ is also onto: for any $r\in\mathbb{R}^\ast$ let $A=\begin{pmatrix}
r & 0\\
0 & 1 \end{pmatrix}$. Then $A\in\mathrm{GL}(2,\mathbb{R})$ and $f(A)=r$. For any $A\in\mathrm{GL}(2,\mathbb{R})$, $A\in\ker f$ if and only if $\det A=1$ and so $\ker f=\mathrm{SL}(2,\mathbb{R})$. Hence, $\mathrm{SL}(2,\mathbb{R})\vartriangleleft\mathrm{GL}(2,\mathbb{R})$.

Functional Analysis 6: Normed Spaces and Banach Spaces

From here on, I assume a background of undergraduate level Linear Algebra. Readers should be familiar with notions such as vector spaces, subspaces, linear combination, linear dependence and linear independence, basis, and dimension.

In this lecture, we begin with the notion of a normed space. A normed space is a vector space with a norm defined. So what is a norm? A norm $||\cdot||$ is a real-valued function $||\cdot||: V\longrightarrow\mathbb{R}^+\cup\{0\}$ such that for $x,y\in V$ and for $\alpha$ a scalar,

(N1) $||x||=0\Longleftrightarrow x=O$.

(N2) $||\alpha x||=|\alpha|||x||$.

(N3) $||x+y||\leq ||x||+||y||$. (Triangle Inequality.)

A norm on $X$ defines a metric $d$ on $X$
\begin{equation}
\label{eq:metric}
d(x,y)=||x-y||.
\end{equation}
\eqref{eq:metric} is called a metric induced by the norm $||\cdot||$. So, a normed space is a metric space but the converse need not be true.

A complete normed space is called a Banach space.

Example. $\mathbb{R}^n$ and $\mathbb{C}^n$ with the Euclidean norm
$$||x||=\left(\sum_{j=1}^n|\xi_j|^2\right)^{\frac{1}{2}}$$ are Banach spaces.

Example. $\ell^p$ with the norm
$$||x||=\left(\sum_{j=1}^n|\xi_j|^p\right)^{\frac{1}{p}}$$ is a Banach space.

Example. $\ell^\infty$ with the norm
$$||x||=\sup_{j\in\mathbb{N}}|\xi_j|$$ is a Banach space.

Example. $\mathcal{C}[a,b]$ with the norm
$$||x||=\max_{t\in[a,b]}|x(t)|$$ is a Banach space.

What follows next is an example of a normed space which is not complete. $\mathcal{C}[a,b]$, the vector space of all continuous real-valued functions on $[a,b]$ forms a normed space with the norm defined by
\begin{equation}
\label{eq:intnorm}
||x||=\left(\int_a^b x(t)^2dt\right)^{\frac{1}{2}}.
\end{equation}
Let $[a,b]=[0,1]$. for each $n=1,2,\cdots$, let $a_n=\frac{1}{2}+\frac{1}{n}$. Define a sequence $(x_n)$ in $\mathcal{C}[0,1]$ by
$$x_n(t)=\left\{\begin{array}{ccc}
0 & \mbox{if} & t\in\left[0,\frac{1}{2}\right],\\
nt-\frac{n}{2} & \mbox{if} & t\in\left[\frac{1}{2},a_n\right],\\
1 & \mbox{if} & t\in[a_n,1].
\end{array}\right.$$

nobanachnobanach2Let us assume that $n>m$. Then
\begin{align*}
||x_m-x_n||^2&=\int_0^1(x_m(t)-x_n(t))^2dt\\
&=\int_0^{\frac{1}{2}}(x_m(t)-x_n(t))^2dt+\int_{\frac{1}{2}}^{a_n}(x_m(t)-x_n(t))^2dt\\&+\int_{a_n}^{a_m}(x_m(t)-x_n(t))^2dt+\int_{a_m}^1(x_m(t)-x_n(t))^2dt\\
&=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left(nt-\frac{n}{2}-mt+\frac{m}{2}\right)^2dt+\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}\left(1-mt+\frac{m}{2}\right)^2dt\\
&=\frac{(n-m)^2}{3mn^2}\\
&<\frac{1}{3m}-\frac{1}{3n}.
\end{align*}
Given $\epsilon>0$, choose a positive integer $N$ so that $N>\frac{1}{3\epsilon^2}$. Then for all $m,n\geq N$,
\begin{align*}
||x_n-x_m||^2&<\frac{1}{3m}-\frac{1}{3n}\\
&<\frac{1}{3m}\\
&\leq\frac{1}{3N}\\
&<\epsilon^2.
\end{align*}
Therefore, $(x_n)$ is a Cauchy sequence in $\mathcal{C}[0,1]$. For any $x(t)\in\mathcal{C}[0,1]$,
\begin{align*}
||x_n-x||^2&=\int_0^1(x_n(t)-x(t))^2dt\\
&=\int_0^{\frac{1}{2}}x(t)^2dt+\int_{\frac{1}{2}}^{a_m}(x_n(t)-x(t))^2dt+\int_{a_n}^1(1-x(t))^2dt.
\end{align*}
Suppose that $x_n\rightarrow x$ as $n\to\infty$. The by the continuity of $x_n(t)$ and $x(t)$, $x(t)$ must satisfy that
$$x(t)=\left\{\begin{array}{ccc}
0 & \mbox{if} & t\in\left[0,\frac{1}{2}\right),\\
1 & \mbox{if} & t\in\left(\frac{1}{2},1\right].
\end{array}\right.$$
But this is impossible since $x(t)$ is continuous. Hence, $(x_n)$ is not convergent in $\mathcal{C}[0,1]$.

In here, we studied completion of metric spaces. Since a normed space is a metric space, we also have completion of normed spaces. The completion of $\mathcal{C}[a,b]$ with the norm \eqref{eq:intnorm} is denoted by $L^2[a,b]$. More generally, for any $p\geq 1$, the Banach space $L^p[a,b]$ is the completion of the normed space $\mathcal{C}[a,b]$ with the norm
\begin{equation}
||x||_p=\left(\int_a^b|x(t)|^pdt\right)^{\frac{1}{p}}.
\end{equation}

Lemma. A metric $d$ induced by a norm on a normed space $X$ satisfies

  1. $d(x+a,y+a)=d(x,y)$, $a,x,y\in X$. This means that $d$ is translation invariant.
  2. $d(\alpha x,\alpha y)=|\alpha|d(x,y)$, $\alpha$, a scalar.

Proof.
\begin{align*}
d(x+a,y+a)&=||x+a-(y+a)||=||x-y||=d(x,y),\\
d(\alpha x,\alpha y)&=||\alpha x-\alpha y||=|\alpha|||x-y||=|\alpha|d(x,y).
\end{align*}

We know that a norm on a vector space $V$ defines a metric on $V$. Can every metric on a vector space $V$ be obtained from a norm? The answer is negative. Let $V$ be the set of all bounded or unbounded sequences of complex numbers. Then $V$ is a vector space. The metric $d$ on $V$ defined by
$$d(x,y)=\sum_{j=1}^\infty\frac{1}{2^j}\frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}$$ is not translation invariant, so it cannot be obtained from a norm.

Group Theory 8: Homomorphisms

In this lecture, we study maps from a group into another group that carry over the same group structure i.e preserve group operations. Such maps are called homomorphisms. It turns out that it suffices to require a homomorphism to preserve a binary operation only. Then the unary operation (inverses) and the nullary operation (the identity element) are automatically preserved.

Definition. Let $(G,\ast)$ and $(G’,\sharp)$ be two groups. A map $\varphi: G\longrightarrow G’$ is called a homomorphism if $\varphi(a\ast b)=\varphi(a)\sharp\varphi(b)$.

Example. Define $\varphi: (\mathbb{R}^+,\cdot)\longrightarrow(\mathbb{R},+)$ by
$$\varphi(x)=\log_{10}x,\ \forall x\in\mathbb{R}^+.$$
Then $\varphi$ is a homomorphism. In addition, $\varphi$ is one-to-one and onto. A homomorphism which is one-to-one and onto is called an isomorphism. If there is an isomorphism from a group onto another group, they are the same group meaning we do not distinguish them.

Example. The map $\varphi: \mathbb{Z}\longrightarrow\mathbb{Z}_n$ defined by
$$\varphi(x)=[x],\ \forall\ x\in\mathbb{Z}$$
is a homomorphism called the canonical or the natural homomorphism. The canonical homomorphism is onto.

Example. Let $G$ be a group and $A(G)$ the set of all bijective (i.e. one-to-one and onto) maps of $G$ onto itself. Recall that $A(G)$ forms a group with composition. Let $a\in G$. Define $T_a: G\longrightarrow G$ by
$$T_ax=ax\ \forall\ x\in G.$$ Hence $\forall a\in G$, $T_a\in A(G)$. Furthermore, $\forall a,b\in G$, $T_aT_b=T_{ab}$: $\forall x\in G$,
$$T_aT_b(x)=T_a(T_bx)=T_a(bx)=a(bx)=(ab)x=T_{ab}x.$$
Define $\varphi: G\longrightarrow A(G)$ by
$$\varphi(a)=T_a,\ \forall\ a\in G.$$ Then $\varphi$ is a homomorphism. In addition, it is one-to-one. If $|G|=n$, then $|A(G)|=n!$, so $\varphi$ cannot be onto.

Definition. A homomorphism which is one-to-one (injective) is called a monomorphism. A homomorphism which is onto (surjective) is called an epimorphism. A homomorphism which is one-to-one and onto (bijective) is called an isomorphism as mentioned above. $G$ is said to be isomorphic to $G’$ if there exists an isomorphism from $G$ onto $G’$. If $G$ is isomorphic to $G’$, we write $G\cong G’$. An isomorphism from a group $G$ onto itself is called an automorphism.

Theorem. [Cayley's Theorem] Every group $G$ is isomorphic to some subgroup of $A(S)$, for an appropriate set $S$.

We in fact proved this theorem in the previous example by taking $S$ to be the group $G$ itself. But there may be other choices for $S$. If $G$ is finite, $S$ may be taken to be a finite set in which case $A(S)$ is $S_n$, the group of permutations of $\{1,2,\cdots,n\}$. In this case, Cayley’s Theorem is stated as: A finite group can be represented as a group of permutations.

Example. Let $G$ be a group and $a\in G$ be fixed. The map $\varphi: G\longrightarrow G$ defined by
$$\varphi(x)=a^{-1}xa,\ \forall x\in G$$
is an automorphism called the inner automorphism of $G$ induced by $a$.

Example. Define $\varphi: (\mathbb{R},+)\longrightarrow (\mathbb{C}\setminus\{0\},\cdot)$ by
$$\varphi(x)=e^{ix},\ \forall x\in\mathbb{R}.$$
While $\varphi$ is a homomorphism, it is neither one-to-one nor onto.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then

(a) $\varphi(e)=e’$.

(b) $\varphi(a^{-1})=\varphi(a)^{-1}$.

Proof. (a) Let $a\in G$. Then $ae=a$. Since $\varphi$ is a homomorphism, $$\varphi(ae)=\varphi(a)\varphi(e)=\varphi(a)=\varphi(a)e’.$$ So by the cancellation law, we obatin $\varphi(e)=e’$.

(b) Let $a\in G$. Then
\begin{align*}
\varphi(a)\varphi(a^{-1})&=\varphi(aa^{-1})\ (\mbox{because $\varphi$ is a homomorphism})\\
&=\varphi(e)\\
&=e’\ (\mbox{by part (a)})
\end{align*}
Hence, $\varphi(a)^{-1}=\varphi(a^{-1})$.

Lemma. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\varphi(G)\leq G’$.

In here, the kernel of a function was introduced. Let $\varphi: G\longrightarrow G’$ be a homomorphism and $K=\varphi^{-1}(e’)=\{a\in G: \varphi(a)=e’\}$. Then $\ker\varphi$ and $K=\varphi^{-1}(e’)$, the preimage of $e’$ under $\varphi$ are closely related as:
\begin{align*}
(a,b)\in\ker\varphi&\Longleftrightarrow\varphi(a)=\varphi(b)\\
&\Longleftrightarrow\varphi(ab^{-1})=e’\\
&\Longleftrightarrow ab^{-1}\in K.
\end{align*}
So, in group theory we define $K=\varphi^{-1}(e’)$ to be $\ker\varphi$, the kernel of $\varphi$.

Theorem. $\varphi: G\longrightarrow G’$ is one-to-one if and only if $\ker\varphi=\{e\}$.

Proof. ($\Rightarrow$) Suppose that $\varphi$ is one-to-one and let $a\in\ker\varphi$. Then $\varphi(a)=\varphi(e)=e’$. Since $\varphi$ is one-to-one, $a=e$.

($\Leftarrow$) Suppose that $\ker\varphi=\{e\}$ and let $\varphi(a)=\varphi(b)$. Then
\begin{align*}
\varphi(ab^{-1})=e’&\Longrightarrow ab^{-1}\in\ker\varphi\\
&\Longrightarrow ab^{-1}=e\\
&\Longrightarrow a=b.
\end{align*}
So, $\varphi$ is one-to-one.

Theorem. Let $\varphi: G\longrightarrow G’$ be a homomorphism. Then $\ker\varphi\leq G$.

Theorem. Let $K=\ker\varphi$. Then $\forall a\in G$, $aK=Ka$.

Proof. It suffices to show (why?) that $\forall a\in G$, $a^{-1}Ka\subset K$. Let $a\in G$. Then $\forall k\in K$,
$$\varphi(a^{-1}ka)=\varphi(a)^{-1}\varphi(k)\varphi(a)=e’\Longrightarrow a^{-1}ka\in K.$$

Group Theory 7: Congruence Modulo $n$

In this lecture, we study the congruence modulo $n$, $\equiv\mod n$ on $\mathbb{Z}$. As we shall see, $\equiv\mod n$ is an equivalence relation on $\mathbb{Z}$. It is indeed more than just an equivalence relation. $\equiv\mod n$ also preserves operations $+$ and $\cdot$ on $\mathbb{Z}$. Because of this, we can define an operation $+$ on the quotient set $\mathbb{Z}/\equiv\mod n$ so that it becomes a quotient group. In general, if an equivalence relation preserves operations, it is called a congruence relation and a congruence relation allows us to define a quotient algebra e.g. a quotient group. Let $G$ be a group and $H\leq G$. We studied the equivalence relation $\sim$ on $G$: $\forall a,b\in G$,
$$a\sim b \Longleftrightarrow ab^{-1}\in H.$$ It turns out that $\sim$ is a congruence relation i.e. it preserves the group operation $\cdot$ if $H$ is a normal subgroup of $G$. I am not going to talk about normal subgroups here but we will study details about this next time.

On $\mathbb{Z}$, define $a\equiv b\mod n$, we say $a$ is congruent to $b$ modulo $n$, if $n|b-a$. The class of $a$
$$[a]=\{a+nk: k\in\mathbb{Z}\}$$ is called the congruence class of $a$. By Euclid’s algorithm, $\forall b\in\mathbb{Z}$, $b=qn+r$ where $0\leq r<n$. This implies that $b\equiv r\mod n$and so $[b]=[r]$. Hence, the $n$ classes $[0],[1],\cdots,[n-1]$ give us all the congruence classes and they are all distinct. Let
$$\mathbb{Z}_n=\{[0],[1],\cdots,[n-1]\}.$$ That is, $\mathbb{Z}_n$ is the quotient set modulo the equivalence relation $\equiv\mod n$, $\mathbb{Z}/\equiv\mod n$. Define $+$ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]+[b]:=[a+b].$$ Then $+$ is well-defined on $\mathbb{Z}_n$. What this means is that $+$ is well-defined as a function $+:\mathbb{Z}_n\times\mathbb{Z}_n\longrightarrow\mathbb{Z}_n$. In other words, if $[a]=[a']$ and $[b]=[b']$ then $[a+b]=[a'+b']$. This can be shown straightforward and is left for the readers. One can also readily see that $(\mathbb{Z}_n,+)$ is an abelian group. Define $\cdot $ on $\mathbb{Z}_n$: $\forall [a],[b]\in\mathbb{Z}_n$,
$$[a]\cdot[b]:=[ab].$$ Then $\cdot$ is also well-defined on $\mathbb{Z}_n$. However, $(\mathbb{Z}_n,\cdot)$ is not a group. $\forall [a]\in\mathbb{Z}_n$, $[0][a]=[0]$ and $[1]$ is the identity under multiplication, so $[0]$ does not have a multiplicative inverse. Would nonzero elements of $\mathbb{Z}_n$ form a group then? The answer is no. For instance, in $\mathbb{Z}_6$, $[2],[3]\ne 0$ but $[2][3]=[6]=[0]$. So, is there a subset of $\mathbb{Z}_n$ that forms a group under $\cdot$? The answer is affirmative. Let $\mathbb{Z}_n^\ast=\{[a]\in\mathbb{Z}_n: (a,n)=1\}$. First, note that for $[a]=[b]$, $(a,n)=1$ if and only if $(b,n)=1$. Recall the property that if $(a,n)=1$ and $(b,n)=1$, then $(ab,n)=1$. This implies that $[a][b]=[ab]\in\mathbb{Z}_n^\ast$. Now suppose that $[a][b]=[a][c]$. Then
\begin{align*}
[ab]=[ac]&\Longrightarrow ab-ac=nk\\
&\Longrightarrow n|a(b-c)\\
&\Longrightarrow n|b-c\ (\mbox{since}\ (a,n)=1)\\
&\Longrightarrow [b]=[c].
\end{align*}
That is, $\mathbb{Z}_n^\ast$ has the cancellation property, so $(\mathbb{Z}_n^\ast,\cdot)$ is a group. In some textbooks, $\mathbb{Z}_n^\ast$ is also denoted by $U_n$. The order of $\mathbb{Z}_n^\ast$ is the number of integers $1\leq m<n$ such that $(m,n)=1$.

Definition. The Euler $\varphi$-function, $\varphi(n)$, is defined by $\varphi(1)=1$ and for $n>1$, $\varphi(n)=$ the number of positive integers $m$ with $1\leq m<n$ such that $(m,n)=1$. Hence, $|\mathbb{Z}_n^\ast|=\varphi(n)$. If $n=p$, a prime, then $\varphi(p)=p-1$.

Note that the Euler $\varphi$-function is multiplicative, namely if $(m,n)=1$, then $\varphi(mn)=\varphi(m)\varphi(n)$.

Example. $\varphi(8)=4$, $\varphi(15)=8$.

Example.
\begin{align*}
\mathbb{Z}_8^\ast&=\{[1],[3],[5],[7]\},\\
\mathbb{Z}_{15}^\ast&=\{[1],[2],[4],[7],[8],[11],[13],[14]\}.
\end{align*}

Example. $\mathbb{Z}_9^\ast=\{[1],[2],[4],[5],[7],[8]\}$.
\begin{align*}
[1]^1&=[2],\ [2]^2=[4],\ [2]^3=[8],\ [2]^4=[16]=[7],\\
[2]^5&=[32]=[5],\ [2]^6=[2][2]^5=[2][5]=[10]=[1].
\end{align*}
Hence, $\mathbb{Z}_9^\ast$ is a cyclic group $\mathbb{Z}_9^\ast=\langle[2]\rangle$.

As a summary,

Theorem. $\mathbb{Z}_n^\ast$ forms an abelian group under the product $[a][b]=[ab]$, of order $\varphi(n)$, where $\varphi(n)$ is the Euler $\varphi$-function.

Theorem. [Euler] If $(a,n)=1$ then $a^{\varphi(n)}\equiv 1\mod n$.

Proof. Since $\mathbb{Z}_n^\ast$ is a group of order $\varphi(n)$, $a^{\varphi(n)}=1$, $\forall a\in\mathbb{Z}_n^\ast$. So,
$$[a^{\varphi(n)}]=[a]^{\varphi(n)}=[1]\Longrightarrow a^{\varphi(n)}\equiv 1\mod n.$$

Corollary. [Fermat] If $p$ is a prime and $p\not | a$, then
$$a^{p-1}\equiv 1\mod p.$$
For any integer $b$, $b^p\equiv b\mod p$. This corollary is usually called Fermat’s Little Theorem.

Proof. $\varphi(p)=p-1$. If $(a,p)=1$, then by the previous theorem,
\begin{align*}
a^{p-1}\equiv 1\mod p &\Longrightarrow a^{p-1}-1=pk\\
&\Longrightarrow a^p-a=p(ak)\\
&\Longrightarrow a^p\equiv a\mod p.
\end{align*}
If $p\not| b$, then $b^p\equiv b\mod p$. If $p|b$, then $b\equiv 0\mod p\Longrightarrow b^p\equiv 0\mod p$. So for any $b\in\mathbb{Z}$, $b^p\equiv b\mod p$.

Example. Compute the remainder of $8^{103}$ when divided by 13.

Solution. Since $13\not|8$, by Fermat’s Little Theorem we have $8^{13-1}=8^{12}\equiv 1\mod 13$. Dividing 103 by 13, we obtain $103=12\cdot 8+7$. So,
\begin{align*}
8^{103}&\equiv (8^{12})^88^7\equiv 8^7\equiv (-5)^7\ (8\equiv -5\mod 13)\\
&\equiv (25)^3(-5)\equiv (-1)^3(-5)\equiv 5\mod 13\ (25\equiv -1\mod 13).
\end{align*}

Example. Show that $2^{11,213}-1$ is not divisible by 11.

Solution. Since $11\not| 2$, by Fermat’s Little Theorem we have $2^{10}\equiv 1\mod 11$. $11,213=10\cdot 1,121+3$. So,
\begin{align*}
2^{11,213}-1&\equiv (2^{10})^{1,121}\cdot 2^3-1\equiv 2^3-1\\
&\equiv 7\mod 11.
\end{align*}
Hence, when $2^{11,213}-1$ is divided by 11, the remainder is 7.

The number $11,213$ is a prime and $2^{11,213}-1$ is also a prime. In number theory, primes of the form $2^p-1$ where $p$ is a prime are called Mersenne primes.

Functional Analysis 5: Completion of Metric Spaces

This lecture will conclude our discussion on metric spaces with completion of metric spaces.

Definition. Let $X=(X,d)$ and $\tilde X=(\tilde X,d)$ be two metric space. A mapping $T: X\longrightarrow\tilde X$ is said to be an isometry of $T$ preserves distances, i.e.
$$\forall x,y\in X,\ \tilde d(Tx,Ty)=d(x,y).$$ The space $X$ is said to be isometric with $\tilde X$ if there exists a bijective isometry of $X$ onto $\tilde X$.

Theorem [Completion] For a metric space $(X,d)$ there exists a complete metric space $\hat X=(\hat X,\hat d)$ which has s subspace $W$ that is isometric with $X$ and is dense in $\hat X$. $\hat X$ is called the completion of $X$ and it is unique up to isometries.

Proof. This will be a lengthy proof and I have divided it into steps.

Step 1. Construction of $\hat X=(\hat X,\hat d)$.

Let $(x_n)$ and $(y_n)$ be Cauchy sequences in $X$. We say $(x_n)$ is equivalent to $(x_n’)$, and write $(x_n)\sim (x_n’)$, if
$$\lim_{n\to\infty}d(x_n,x_n’)=0.$$ $\sim$ is actually an equivalence relation on the set of all Cauchy sequences of $X$. Clearly $\sim$ is reflexive and symmetric. Let us show that $\sim$ is also transitive. Let $(x_n)\sim (x_n’)$ and $(x_n’)\sim (x_n^{\prime\prime})$. Then
$$\lim_{n\to\infty}d(x_nx_n’)=0\ \mbox{and}\ \lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})=0.$$
\begin{align*}
\lim_{n\to\infty}d(x_n,x_n^{\prime\prime})&\leq\lim_{n\to\infty}d(x_n,x_n’)+\lim_{n\to\infty}d(x_n’,x_n^{\prime\prime})\\
&=0
\end{align*}
Let $\hat X$ be the set of all equivalence classes $\hat x,\hat y,\cdots$ of Cauchy sequences. Define
$$\hat d(\hat x,\hat y)=\lim_{n\to\infty}d(x_n,y_n),$$
where $x_n\in\hat x$ and $y_n\in\hat y$. We claim that $\hat d$ is a metric on $\hat X$. It suffices to show that $\hat d$ is well-defined. The conditions (M1)-(M3) hold due to the fact that $d$ is a metric. First we show that $\hat d(\hat x,\hat y)$ exists. It follows from (M3) that
$$d(x_n,y_n)\leq d(x_n,x_m)+d(x_m,y_m)+d(y_m,y_n)$$ and so we have
$$d(x_n,y_n)-d(x_m,y_m)\leq d(x_n,x_m)+d(y_m,y_n).$$ Similarly, we obtain
$$d(x_m,y_m)-d(x_n,y_n)\leq d(x_n,x_m)+d(y_m,y_n).$$ Hence,
$$|d(x_n,y_n)-d(x_m,y_m)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n,m\rightarrow\infty$, i.e.
$$\lim_{n,m\to\infty}|d(x_n,y_n)-d(x_m,y_m)|=0.$$
Since $\mathbb{R}$ is complete, $\displaystyle\lim_{n\to\infty}d(x_n,y_n)$ exists. Now we show that the limit is independent of the choice of representatives $(x_n)$ and $(y_n)$. If $(x_n)\sim (x_n’)$ and $(y_n)\sim(y_n’)$, then
$$|d(x_n,y_n)-d(x_n’,y_n’)|\leq d(x_n,x_m)+d(y_m,y_n)\rightarrow 0$$ as $n\to\infty$.

Step 2. Construction of an isometry $T: X\longrightarrow W\subset \hat X$.

For each $b\in X$, let $\hat b$ be the equivalence class of the Cauchy sequence $(b,b,b,\cdots)$. Then $T(b):=\hat b\in\hat X$. Now,
$$\hat d(Tb,Tc)=\hat d(\hat b,\hat c)=d(b,c).$$ So, $T$ is an isometry. An isometry is automatically injective. $T$ is onto since since $T(X)=W$. Let us show that $W$ is dense in $\hat X$. Let $\hat x\in \hat X$ and let $(x_n)\in\hat x$. Since $(x_n)$ is Cauchy, given $\epsilon>0$ $\exists N$ such that $d(x_n,x_N)<\frac{\epsilon}{2}$ $\forall n\geq N$. Let $(x_N,x_N,\cdots)\in\hat x_N$. Then $\hat x_N\in W$.
\begin{align*}
\hat d(\hat x,\hat x_N)=\lim_{n\to\infty}d(x_n,x_N)\leq\frac{\epsilon}{2}<\epsilon&\Longrightarrow \hat x_N\in B(\hat x,\epsilon)\\
&\Longrightarrow B(\hat x,\epsilon)\cap W\ne\emptyset.
\end{align*}
Hence, $\bar W=\hat X$ i.e. $W$ is dense in $\hat X$.

Step 3. Completeness of $\hat X$.

Let $(\hat x_n)$ be any Cauchy sequence in $\hat X$. Since $W$ is dense in $\hat X$, $\forall \hat x_n$, $\exists\hat z_n\in W$ such that $\hat d(\hat x_n,\hat z_n)<\frac{1}{n}$.
\begin{align*}
\hat d(\hat z_m,\hat z_n)&\leq \hat d(\hat z_n,\hat x_m)+\hat d(\hat x_m,\hat x_n)+\hat d(\hat x_n,\hat z_n)\\
&<\frac{1}{m}+\hat d(\hat x_m,\hat x_n)+\frac{1}{n}.
\end{align*}
Given $\epsilon>0$ by Archimedean property $\exists$ a positive integer $N_1$ such that $N>\frac{\epsilon}{3}$. Since $(\hat x_n)$ is a Cauchy sequence, $\exists$ a positive integer $N_2$ such that $\hat d(\hat x_m,\hat x_n)<\frac{\epsilon}{3}$ $\forall m,n\geq N$. Let $N=\max\{N_1,N_2\}$. Then $\forall m,n\geq N$, $\hat d(\hat z_m,\hat z_n)<\epsilon$ i.e. $(\hat z_m)$ is Cauchy. Since $T: X\longrightarrow W$ is an isometry and $\hat z_m\in W$, the sequence $(z_m)$, where $z_m=T^{-1}\hat z_m$, is Cauchy in $X$. Let $\hat x\in\hat X$ be the class to which $(z_m)$ belongs. Show that $\hat x$ is the limit of $(\hat x_n)$. For each $n=1,2,\cdots$,
\begin{align*}
\hat d(\hat x_n,\hat x)&\leq \hat d(\hat x_n,\hat z_n)+\hat d(\hat z_n,\hat x)\\
&<\frac{1}{n}+\hat d(\hat z_n,\hat x)\\
&=\frac{1}{n}+\lim_{m\to\infty}d(z_n,z_m)
\end{align*} since $(z_m)\in\hat x$ and $(z_n,z_n,\cdots)\in\hat z_n\in W$. This implies that $\displaystyle\lim_{n\to\infty}\hat d(\hat x_n,\hat x)=0$ i.e. the Cauchy sequence $(\hat x_n)$ in $\hat X$ has the limit $\hat x\in\hat X$. Therefore, $\hat X$ is complete.

Step 4. Uniqueness of $\hat X$ up to isometries.

Suppose that $(\tilde X,\tilde d)$ is another completion of $X$ i.e. it is a complete metric space with a subspace $\tilde W$ dense in $\tilde X$ and isometric with $X$. We show that $\hat X$ is isometric with $\tilde X$. Let $X$ is isometric with $W$ and $\tilde W$ via isometries $T$ and $\tilde T$, respectively. Then $W$ is isometric with $\tilde W$ via the isometry $\rho=\tilde T\circ T^{-1}$.

$$\begin{array}{ccc}
& & W\\
& \nearrow &\downarrow\\
X & \longrightarrow & \tilde{W}
\end{array}$$

Let $\hat x\in\hat X$. Then $\exists$ a sequence in $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. $(\hat x_n)$ is a Cauchy sequence and $\rho$ is an isometry, so $(\tilde x_n)$, where $\tilde x_n:=\rho\hat x_n$, is a Cauchy sequence in $\tilde W\subset \tilde X$. Since $\tilde X$ is complete, $\exists\tilde x\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. Define a mapping $\psi:\hat X\longrightarrow\tilde X$ by $\psi\hat x=\tilde x$. Then we claim that $\hat X$ is isometric with $\tilde X$ via $\psi$.

Step A. $\psi$ is well-defined.

It suffices to show that $T\hat x$ does not depend on the choice of $(\hat x_n)\in W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Let $(\hat x_n’)$ be another sequence in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n’=\hat x$. Then $(\tilde x_n’)$, where $\tilde x_n’=\rho\hat x_n’$, is a Cauchy sequence in $\tilde W$ and so $\exists\tilde x’\in\tilde X$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n’=\tilde x’$. Now,
\begin{align*}
\tilde d(\tilde x,\tilde x’)&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde x_n’)\\
&=\lim_{n\to\infty}\hat d(\hat x_n,\hat x_n’)\ (\rho\ \mbox{is an isometry})\\
&=\hat d(\hat x,\hat x)\\
&=0.
\end{align*}
Hence, $\tilde x=\tilde x’$.

Step B. $\psi$ is onto.

Let $\tilde x\in\tilde X$. Then $\exists$ a sequence $(\tilde x_n)$ in $\tilde W$ such that $\displaystyle\lim_{n\to\infty}\tilde x_n=\tilde x$. $(\tilde x_n)$ is Cauchy (since it is a convergent sequence) and $\rho^{-1}$ is an isometry, so the sequence $(\hat x_n)\subset \hat X$, where $\hat x_n=\rho^{-1}\tilde x_n$, is Cauchy. Since $\hat X$ is complete, $\exists\hat x\in\hat X$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$. Clearly $\psi\hat x=\tilde x$ and hence $\psi$ is onto.

Step C. $\psi$ is an isometry.

Let $\hat x,\hat y\in\hat X$. Then $\exists$ sequences $(\hat x_n)$, $(\hat y_n)$ in $W$ such that $\displaystyle\lim_{n\to\infty}\hat x_n=\hat x$ and $\displaystyle\lim_{n\to\infty}\hat y_n=\hat y$, respectively.
\begin{align*}
\hat d(\hat x,\hat y)&=\lim_{n\to\infty}\hat d(\hat x_n,\hat y_n)\\
&=\lim_{n\to\infty}\tilde d(\tilde x_n,\tilde y_n)\ (\tilde x_n:=\rho\hat x_n,\ \tilde y_n:=\rho y_n)\\
&=\tilde d(\tilde x,\tilde y)\ (\lim_{n\to\infty}\tilde x_n=\tilde x,\ \lim_{n\to\infty}\tilde y_n=\tilde y)\\
&=\tilde d(\psi\hat x,\psi\hat y).
\end{align*}
Thus, $\psi$ is an isometry.

Remember that an isometry from a metric space into another metric space is automatically one-to-one. Therefore, $\hat X$ is isometric with $\tilde X$ via $\psi$.

Intuitively speaking, the completion of a metric space $X$ can be achieved by adding to $X$ all its limit points. Recall that if $x$ is a limit point of $X$, then there exists a sequence $(x_n)$ in $X$ such that $\displaystyle\lim_{n\to\infty}x_n=x$. This is a reminiscence of extending from rational numbers $\mathbb{Q}$ to real numbers $\mathbb{R}$ (which is complete) by adding to $\mathbb{Q}$ all its limit points (irrational numbers).