Lecture Notes in Functional Analysis

by William L. Paschke

edition 0.9

Contents

Frontmatter

I. Hilbert Space

II. Bounded Operators

III. Compact Operators

IV. The Spectral Theorem

Index and References

## II. Bounded Operators

Definition II.1

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 $$$x\in X$$$.

Proposition II.2

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.

Proof.

1. ((b) ⇒ (c)) There exists a $$$δ\gt 0$$$ such that $$$\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\lt 1$$$ for all $$$\mathopen{}\left\lVert{}x\right\rVert\mathclose{}\lt δ$$$. So for any $$$y$$$ with $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\leq 1$$$ we have $$$\mathopen{}\left\lVert{}\frac{δ}{2}y\right\rVert\mathclose{}\leq \frac{δ}{2}\lt δ$$$, and thus $$$\mathopen{}\left\lVert{}T\mathopen{}\left( \frac{δ}{2}y\right)\mathclose{}\right\rVert\mathclose{}\lt 1$$$. This makes $$$\mathopen{}\left\lVert{}T\mathopen{}\left( y\right)\mathclose{}\right\rVert\mathclose{}\lt \frac{2}{δ}$$$ whenever $$$\mathopen{}\left\lVert{}y\right\rVert\mathclose{}\leq 1$$$.
2. ((d) ⇒ (e)) For $$$x\neq 0$$$, $$$\mathopen{}\left\lVert{}T\mathopen{}\left( \frac{x}{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}}\right)\mathclose{}\right\rVert\mathclose{}\leq r$$$ (where $$$r$$$ is the supremum on the unit sphere) implies $$$\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq r\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$.
3. ((e)⇒(a)) $$$\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}-T\mathopen{}\left( {x}_{n}\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}T\mathopen{}\left( x-{x}_{n}\right)\mathclose{}\right\rVert\mathclose{}\leq r\mathopen{}\left\lVert{}x-{x}_{n}\right\rVert\mathclose{}$$$. So $$${x}_{n}$$$ converging to $$$x$$$ makes $$$T\mathopen{}\left( {x}_{n}\right)\mathclose{}$$$ converge to $$$T\mathopen{}\left( x\right)\mathclose{}$$$.

Example II.3

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.

Example II.4

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.

Example II.5

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 $$$T\mathopen{}\left( {f}_{r}\right)\mathclose{}= r{f}_{r}$$$.

Example II.6

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. $$$\mathopen{}\left\lVert{}T\mathopen{}\left( f\right)\mathclose{}\right\rVert\mathclose{}\leq \sqrt{ 2N }\mathopen{}\left\lVert{}f\right\rVert\mathclose{}$$$.

Definition II.7

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

Remark II.8

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 $$\mathopen{}\left\lVert{}T\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{} \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} = \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1 }{} \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} = \sup_{ x\in X\setminus \mathopen{}\left\{\, 0\,\right\}\mathclose{} }{} \frac{\mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}{\mathopen{}\left\lVert{}x\right\rVert\mathclose{}} \text{.}$$

Example II.9

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 $$${\mathopen{}\left\lVert{}T\right\rVert\mathclose{}}^{2}\geq {r}^{2}$$$.

Proposition II.10

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.

Proof. Let $$$\mathopen{}\left( {T}_{n} \right)\mathclose{}$$$ be a Cauchy sequence in $$$\mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$. For $$$x\in X$$$, the sequence $$$\mathopen{}\left( {T}_{n}\mathopen{}\left( x\right)\mathclose{} \right)\mathclose{}$$$ is Cauchy. Hence $$$\mathopen{}\left\lVert{}{T}_{n}\mathopen{}\left( x\right)\mathclose{}-{T}_{m}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}{T}_{n}-{T}_{m}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$$ and so $$$\mathopen{}\left( {T}_{n}\mathopen{}\left( x\right)\mathclose{} \right)\mathclose{}$$$ is convergent in $$$Y$$$. Define $$$T : X \to Y$$$ by $$$T\mathopen{}\left( x\right)\mathclose{}= \lim_{n\to\infty}{} {T}_{n}\mathopen{}\left( x\right)\mathclose{}$$$. Note that $$$T$$$ is plainly linear. We have $$\mathopen{}\left\lVert{}\mathopen{}\left(T-{T}_{n}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}\mathopen{}\left(T-{T}_{m}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}\mathopen{}\left({T}_{m}-{T}_{n}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}-{T}_{m}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}+\mathopen{}\left\lVert{}{T}_{m}-{T}_{n}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}$$ for all $$$m$$$ and $$$n$$$. Take the limit supremum: $$\mathopen{}\left\lVert{}\mathopen{}\left(T-{T}_{n}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}\leq \limsup_{m}{} \mathopen{}\left\lVert{}T\mathopen{}\left( x\right)\mathclose{}-{T}_{m}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} +\mathopen{}\left(\limsup_{m}{} \mathopen{}\left\lVert{}{T}_{m}-{T}_{n}\right\rVert\mathclose{} \right)\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{}= \limsup_{m}{} \mathopen{}\left\lVert{}{T}_{m}-{T}_{n}\right\rVert\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{} \leq \mathopen{}\left(\sup_{ m\geq n }{} \mathopen{}\left\lVert{}{T}_{m}-{T}_{n}\right\rVert\mathclose{} \right)\mathclose{}\mathopen{}\left\lVert{}x\right\rVert\mathclose{} \text{.}$$ Therefore $$\mathopen{}\left\lVert{}T-{T}_{n}\right\rVert\mathclose{}\leq \sup_{ m\geq n }{} \mathopen{}\left\lVert{}{T}_{m}-{T}_{n}\right\rVert\mathclose{}$$ and the latter approaches zero as $$$n \to \infty$$$ by the Cauchy property. Thus $$$T-{T}_{n}\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$ (so $$$T\in \mathcal{L}\mathopen{}\left( X, Y\right)\mathclose{}$$$) and $$$\mathopen{}\left\lVert{}T-{T}_{n}\right\rVert\mathclose{} \to 0$$$.

Remark II.11

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 $$$\mathopen{}\left\lVert{}T\mathopen{}\left( S\right)\mathclose{}\right\rVert\mathclose{}\leq \mathopen{}\left\lVert{}T\right\rVert\mathclose{}\mathopen{}\left\lVert{}S\right\rVert\mathclose{}$$$.

Definition II.12

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.

Definition II.13

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 $$$\mathopen{}\left\lVert{}φ\right\rVert\mathclose{}= \sup_{ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq 1 }{} \mathopen{}\left\lvert{}φ\right\rvert\mathclose{}$$$.