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.

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||}\\&=\sup_{\begin{array}{c}x\in\mathcal{D}(T)\\x\ne O\end{array}}\left|\left|T\frac{x}{||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, $||\tilde T||=||T||$.

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 operator. Then

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

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. Let $X$ be a normed space with a Schauder basis $(e_n)$. Let $D$ be the set of all possible finite linear combinations (superpositions) of the $e_n$’s with rational scalar coefficients, i.e. $$D=\{\sum_{j=1}^n q_je_j: n\in\mathbb{N},\ q_j\in\mathbb{Q}\}$$Then $D$ is countable. (Why?)

Now, we show that $D$ is dense in $X$. 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 $x\in X$. Since $(e_n)$ is a Schauder basis,  $\exists$ 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 $||x-\sum_{j=1}^n\alpha_je_j||<\frac{\epsilon}{2}$ for all $n\geq N$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for each $j=1,\cdots,N$, $\exists q_j\in\mathbb{Q}$ such that $|q_j-\alpha_j|<\frac{\epsilon}{2N(||e_j||+1)}$. \begin{align*}||x-\sum_{j=1}^N q_je_j||&<||x-\sum_{j=1}^N\alpha_je_j||+\sum_{j=1}^N|\alpha_j-q_j|( ||e_j||+1)\\&<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\end{align*} This means that $\sum_{j=1}^Nq_je_j\in B(x,\epsilon)\cap D\ne\emptyset$ and hence the proof is complete.

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.

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 $V$ defines a metric $d$ on $V$
\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
$$x_n(t)=\left\{\begin{array}{ccc}
0 & \mbox{if} & t\in\left[0,\frac{1}{2}\right],\\
nt-\frac{n}{2} & \mbox{if} & t\in\left[\frac{1}{2},a_n\right],\\
1 & \mbox{if} & t\in[a_n,1].
\end{array}\right.$$

Let 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_n}(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.