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.