A linear map $\( T : X \to Y \)$ between normed linear spaces is called bounded if there exists an $\( r\gt 0 \)$ such that $\( \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq r\mathopen{}\left\lVert{}T\right\rVert\mathclose{} \)$ for all $\( x\in X \)$.

For a linear map $\( T : X \to Y \)$, the following are equivalent.

(a)

$\( T \)$ is continuous.

(b)

$\( T \)$ is continuous at $\( 0\in X \)$.

(c)

$\( \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1}{}\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\lt \infty \)$.

(d)

$\( \sup_{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1}{}\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\lt \infty \)$.

(e)

$\( T \)$ is bounded.

Let $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω, M, μ\right)\mathclose{} \)$, and let $\( φ : Ω \to \mathbb{C} \)$ be a bounded measurable function on $\( Ω \)$. Define $\( T : H \to H \)$ by $\( T\mathopen{}\left( φ\right)\mathclose{}= φf \)$, i.e., $\( \mathopen{}\left(T\mathopen{}\left( φ\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= φ\mathopen{}\left( t\right)\mathclose{}f\mathopen{}\left( t\right)\mathclose{} \)$. We have $$\[ {\mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \int _{Ω}{} {\mathopen{}\left\lvert{}φ\right\rvert\mathclose{}}^{2}{\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2} \leq {\mathopen{}\left(\sup{}\mathopen{}\left\lvert{}φ\right\rvert\mathclose{}\right)\mathclose{}}^{2}\int _{Ω}{} {\mathopen{}\left\lvert{}f\right\rvert\mathclose{}}^{2} \leq {\mathopen{}\left(\sup{}\mathopen{}\left\lvert{}φ\right\rvert\mathclose{}\right)\mathclose{}}^{2}\int _{Ω}{} {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2} \text{.} \]$$ The operator $\( T \)$ so defined is called a multiplication operator.

Let $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ and define $\( T : H \to H \)$ by $$\[ \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{} f\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t= \int _{ \mathopen{}\left[0, x\right]\mathclose{} }{}f \text{.} \]$$ Note that $\( T\mathopen{}\left( f\right)\mathclose{} \)$ is a well-defined function of $\( x \)$ as $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\subseteq \mathrm{L}^{\mathrm{1}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$. Furthermore, $\( T \)$ maps $\( \mathrm{L}^{\mathrm{2}} \)$ into $\( \mathrm{L}^{\mathrm{2}} \)$ in a bounded fashion because $\( \mathopen{}\left\lvert{}\mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lVert{}\mathopen{}\left\langle{}f, \chi_{ \mathopen{}\left[0, x\right]\mathclose{} }\right\rangle\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}f\right\rVert\mathclose{}\mathopen{}\left\lVert{}\chi_{ \mathopen{}\left[0, x\right]\mathclose{} }\right\rVert\mathclose{}= \mathopen{}\left\lVert{}f\right\rVert\mathclose{}{x}^{\frac{1}{2}} \)$, and hence $$\[ {\mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}}^{2}= \int _{0}^{1}{} {\mathopen{}\left\lvert{}T\mathopen{}\left( f\mathopen{}\left( x\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}}^{2} \,\mathrm{d}x\leq {\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2}\int _{0}^{1}{}x\,\mathrm{d}x= \frac{1}{2}{\mathopen{}\left\lVert{}f\right\rVert\mathclose{}}^{2} \text{,} \]$$ so $\( \mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}\leq \frac{1}{\sqrt{2}}\mathopen{}\left\lVert{}f\right\rVert\mathclose{} \)$. This is known as the Volterra Operator.

We next present an example of an unbounded operator. Consider $\( D= \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$, the space of continuously differentiable complex functions on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$ with the integral inner product. Define $\( T : D \to \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ by $\( T\mathopen{}\left( f\right)\mathclose{}= f' \)$. Note $\( f' \)$ is continuous and is therefore in $\( \mathrm{L}^{\mathrm{2}} \)$. To see that $\( T \)$ is not bounded, notice that $\( {f}_{r} \)$ defined by $\( {f}_{r}= {\mathrm{e}}^{rt} \)$ gives $\( T\mathopen{}\left( {f}_{r}\right)\mathclose{}= r{f}_{r} \)$.

Consider a directed graph with edge set $\( E \)$ and vertex set $\( V \)$. Define maps $\( i\text{,}t : E \to V \)$ by letting $\( i\mathopen{}\left( e\right)\mathclose{} \)$ be the initial point of edge $\( e \)$ and $\( t\mathopen{}\left( e\right)\mathclose{} \)$ the terminal point of edge $\( e \)$. Assume the graph is of bounded degree, i.e., $\( {\#} {i}^{-1}\mathopen{}\left( v\right)\mathclose{} +{\#} {t}^{-1}\mathopen{}\left( v\right)\mathclose{} \leq N \)$ for all $\( v\in V \)$ for some positive integer $\( N \)$. Define $\( T : \mathrm{l}^{0}\mathopen{}\left( V\right)\mathclose{} \to \mathrm{l}^{0}\mathopen{}\left( E\right)\mathclose{} \)$ by $\( \mathopen{}\left(T\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{}= f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}-f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{} \)$. We show this goes into $\( \mathrm{l}^{0}\mathopen{}\left( E\right)\mathclose{} \)$ in a bounded fashion: $$\[ {\mathopen{}\left\lvert{}T\mathopen{}\left( f\right)\mathclose{}\mathopen{}\left( e\right)\mathclose{}\right\rvert\mathclose{}}^{2}\leq {\mathopen{}\left(\mathopen{}\left\lvert{}f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}+\mathopen{}\left\lvert{}f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}\right)\mathclose{}}^{2}\leq 2\mathopen{}\left({\mathopen{}\left\lvert{}f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}}^{2}+{\mathopen{}\left\lvert{}f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}}^{2}\right)\mathclose{} \text{.} \]$$ Therefore, $$\[ {\mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}}^{2}\leq 2\sum_{e\in E}{} \mathopen{}\left({\mathopen{}\left\lvert{}f\mathopen{}\left( i\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}}^{2}+{\mathopen{}\left\lvert{}f\mathopen{}\left( t\mathopen{}\left( e\right)\mathclose{}\right)\mathclose{}\right\rvert\mathclose{}}^{2}\right)\mathclose{} \leq 2N\sum_{v\in V}{} {\mathopen{}\left\lvert{}f\mathopen{}\left( v\right)\mathclose{}\right\rvert\mathclose{}}^{2} \text{.} \]$$ i.e. $\( \mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}\leq \sqrt{ 2N }\mathopen{}\left\lVert{}f\right\rVert\mathclose{} \)$.

For a bounded linear map $\( T : X \to Y \)$ between normed linear spaces define $$\[ \mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \inf{} \mathopen{}\left\{\, r\gt 0\,\middle\vert\, , \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq r\mathopen{}\left\lVert{}x\right\rVert\mathclose{}, , \forall{}\mathopen{}\left( x\in X\right)\mathclose{}, \,\right\}\mathclose{} \text{.} \]$$

We have $\( \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{} \)$ for all $\( x\in X \)$. If we denote by $\( \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$ the space of bounded linear maps from $\( X \)$ to $\( Y \)$, $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \)$ so defined is a norm on $\( \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$. Notice that $$\[ \mathopen{}\left\lVert{}\mathopen{}\left({T}_{1}+{T}_{2}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{T}_{1}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{2}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{T}_{1}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{2}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{} \]$$ for all $\( x \)$. So, $\( \mathopen{}\left\lVert{}{T}_{1}+{T}_{2}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{T}_{1}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{2}\right\rVert\mathclose{} \)$. Observe that $$\[ \mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{} \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} = \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1 }{} \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} = \sup_{ x\in X\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{} }{} \frac{\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}} \text{.} \]$$

Back to $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( Ω, M, μ\right)\mathclose{} \)$, and $\( φ : Ω \to \mathbb{C} \)$ a bounded measurable function on $\( Ω \)$. Define $$\[ \mathopen{}\left\lVert{}φ\right\rVert_\infty\mathclose{}= \sup{} \mathopen{}\left\{\, r\geq 0\,\middle\vert\, , μ\mathopen{}\left\{\, t\,\middle\vert\, , \mathopen{}\left\lvert{}φ\mathopen{}\left( t\right)\mathclose{}\right\rvert\mathclose{}\geq r, \,\right\}\mathclose{}\gt 0, \,\right\}\mathclose{} \text{.} \]$$ This is the essential supremum of $\( \mathopen{}\left\lvert{}φ\right\rvert\mathclose{} \)$. Note that $\( \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}φ\right\rVert_\infty\mathclose{} \)$ ( $\( T\mathopen{}\left( f\right)\mathclose{}= φ\mathopen{}\left( f\right)\mathclose{} \)$, $\( R\gt \mathopen{}\left\lVert{}φ\right\rVert_\infty\mathclose{} \)$ implies $\( \mathopen{}\left\lvert{}φ\right\rvert\mathclose{}\lt R \)$ almost everywhere, which gives $\( \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\leq R \)$). For the reverse inequality, we must assume that for every $\( B\in M \)$ with $\( μ\mathopen{}\left( B\right)\mathclose{}= \infty \)$, there exists $\( A\in M \)$ with $\( A\subseteq B \)$ and $\( 0\lt μ\mathopen{}\left( A\right)\mathclose{}\lt \infty \)$. Thus for any $\( 0\lt r\lt \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}_{\infty} \)$, we get $\( A\in M \)$ where $\( 0\lt μ\mathopen{}\left( A\right)\mathclose{}\lt \infty \)$ and $\( \mathopen{}\left\lvert{}φ\right\rvert\mathclose{}\geq r \)$ on $\( A \)$. So, $$\[ {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}{\mathopen{}\left\lVert{}{χ}_{A}\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}T\mathopen{}\left( {χ}_{A}\right)\mathclose{}\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}φ{χ}_{A}\right\rVert\mathclose{}}^{2}\geq {r}^{2}{\mathopen{}\left\lVert{}{χ}_{A}\right\rVert\mathclose{}}^{2} \text{,} \]$$ which gives $\( {\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}\geq {r}^{2} \)$.

If $\( X \)$ is a normed linear space and $\( Y \)$ is a Banach space, then $\( \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$ is a Banach space.

If $\( Y \)$ is complete and $\( \overline{X} \)$ is the completion of $\( X \)$, then $\( \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}\simeq \mathcal{L}\mathopen{}\left( \overline{X}, Y\right)\mathclose{} \)$ (if $\( T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$ and $\( x\in \overline{X} \)$, we get $\( \mathopen{}\left( {x}_{n} \right)\mathclose{} \)$ in $\( X \)$ with $\( {x}_{n} \to x \)$ and $\( \mathopen{}\left( T\mathopen{}\left( {x}_{n}\right)\mathclose{} \right)\mathclose{} \)$ is Cauchy). Hence $\( \lim_{n}{} T\mathopen{}\left( {x}_{n}\right)\mathclose{} \in Y \)$. We write $\( \overline{T}\mathopen{}\left( x\right)\mathclose{}= \lim_{n}{} T\mathopen{}\left( {x}_{n}\right)\mathclose{} \)$. If $\( W \)$, $\( X \)$, and $\( Y \)$ are normed linear spaces and $\( S\in \mathcal{L}\mathopen{}\left( W, X\right)\mathclose{} \)$, $\( T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{} \)$, then $\( TS= T\circ S\in \mathcal{L}\mathopen{}\left( W, Y\right)\mathclose{} \)$ with $\( \mathopen{}\left\lVert{}T\mathopen{}\left( S\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}S\right\rVert\mathclose{} \)$.

Write $\( \mathcal{L}\mathopen{}\left( X\right)\mathclose{}= \mathcal{L}\mathopen{}\left( X, X\right)\mathclose{} \)$, which is a normed algebra, i.e, an algebra over $\( \mathbb{C} \)$ equipped with a linear space norm that is furthermore submultiplicative.

Write $\( {X}^{*}= \mathcal{L}\mathopen{}\left( X, \mathbb{C}\right)\mathclose{} \)$ for the dual space (or the conjugate space) of $\( X \)$. These are the bounded linear functionals on $\( X \)$. $\( {X}^{*} \)$ is always a Banach space, with norm $\( \mathopen{}\left\lVert{}φ\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{} \mathopen{}\left\lvert{}φ\right\rvert\mathclose{} \)$.