# 2.1 Dual spaces

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Definition of the dual space of a normed space containing all linear operators.

## Definition

The set of linear functionals $f$ on $X$ is itself a vector space:

• $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ (closed under addition),
• $\left(af\right)\left(x\right)=af\left(x\right)$ (closed under scalar multiplication),
• $z\left(x\right)=0$ is linear and $f+z=f$ (features an addition zero).

Definition 1 The dual space (or algebraic dual) ${X}^{*}$ of $X$ is the vector space of all linear function on $X$ .

Example 1 If $X={\mathbb{R}}^{N}$ then $f\left(x\right)={\sum }_{i=1}^{N}{a}_{i}{x}_{i}\in {X}^{*}$ and any linear functional on $X$ can be written in this form.

Definition 2 Let $X$ be a normed space. The normed dual ${X}^{*}$ of $X$ is the normed space of all bounded linear functionals with norm $\parallel f\parallel ={sup}_{\parallel x\parallel \le 1}|f\left(x\right)|$ .

Since in the sequel we refer almost exclusively to the normed dual, we will abbreviate to “dual” (and ignore the algebraic dual in the process).

Example 2 Let us find the dual of $X={\mathbb{R}}^{N}$ . In this space,

$x=\left[\begin{array}{c}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{n}\end{array}\right]\phantom{\rule{3.33333pt}{0ex}}\mathrm{and}\phantom{\rule{3.33333pt}{0ex}}\parallel x\parallel =\sqrt{{\sum }_{i=1}^{n}{|{x}_{i}|}^{2}},\phantom{\rule{3.33333pt}{0ex}}\mathrm{where}\phantom{\rule{3.33333pt}{0ex}}{x}_{i}\in \mathbb{R}.$

From Example [link] , a linear functional $f$ can be written as:

$f\left(x\right)=\sum _{i=1}^{n}{a}_{i}{x}_{i}=⟨x,a⟩,\phantom{\rule{3.33333pt}{0ex}}a\in {\mathbb{R}}^{n}$

There is a one-to-one mapping between the space $X$ and the dual of $X$ :

$f\in {\left({\mathbb{R}}^{n}\right)}^{*}⟷a\in {\mathbb{R}}^{n}$

For this reason, ${\mathbb{R}}^{n}$ is called self-dual (as there is a one-to-one mapping ${\left({\mathbb{R}}^{n}\right)}^{*}⟷{\mathbb{R}}^{n}$ ). Using the Cauchy-Schwartz Inequality, we can show that

$|f\left(x\right)|=|⟨x,a⟩|\le \parallel x\parallel \parallel a\parallel ,$

with equality if $a$ is a scalar multiple of $x$ . Therefore, we have that $\parallel f\parallel =\parallel a\parallel$ , i.e., the norms of $X$ and ${X}^{*}$ match through the mapping.

Theorem 1 All normed dual spaces are Banach.

Since we have shown that all dual spaces ${X}^{*}$ have a valid norm, it remains to be shown that all Cauchy sequences in in ${X}^{*}$ are convergent.

Let ${\left\{{f}_{n}\right\}}_{n=1}^{\infty }\subseteq {X}^{*}$ be a Cauchy sequence in ${X}^{*}$ . Thus, we have that for each $ϵ>0$ there exists ${n}_{0}\in {\mathbb{Z}}^{+}$ such that if $n,m>{n}_{0}$ then $\parallel {f}_{n}\left(x\right)-{f}_{m}\left(x\right)\parallel \to 0$ as $n,m\to \infty$ . We will first show that for an arbitrary input $x\in X$ , the sequence ${\left\{{f}_{n}\left(x\right)\right\}}_{n=1}^{\infty }\subseteq R$ is Cauchy. Pick an arbitrary $ϵ>0$ , and obtain the ${n}_{0}\in {\mathbb{Z}}^{+}$ from the definition of Cauchy sequence for ${\left\{{f}_{n}^{*}\right\}}_{n=1}^{\infty }$ . Then for all $n,m>{n}_{0}$ we have that $|{f}_{n}\left(x\right)-{f}_{m}\left(x\right)|=|\left({f}_{n}-{f}_{m}\right)\left(x\right)|\le \parallel {f}_{n}\left(x\right)-{f}_{m}\left(x\right)\parallel ·\parallel x\parallel \le ϵ\parallel x\parallel$ .

Now, since $R=\mathbb{R}$ and $R=\mathbb{C}$ are complete, we have that the sequence ${\left\{{f}_{n}\left(x\right)\right\}}_{n=1}^{\infty }$ converges to some scalar ${f}^{*}\left(x\right)$ . In this way, we can define a new function ${f}^{*}$ that collects the limits of all such sequences over all inputs $x\in X$ . We conjecture that the sequence ${f}_{n}\to {f}^{*}$ : we begin by picking some $ϵ>0$ . Since ${\left\{{f}_{n}\left(x\right)\right\}}_{n=1}^{\infty }$ is convergent to ${f}^{*}\left(x\right)$ for each $x\in X$ , we have that for $n>{n}_{0,x}$ , $|{f}_{n}\left(x\right)-{f}_{n}^{*}\left(x\right)|\le ϵ$ . Now, let ${n}_{0}^{*}={sup}_{x\in X,\parallel x\parallel =1}{n}_{0,x}$ . Then, for $n>{n}_{0}^{*}$ ,

$\parallel {f}_{n}-{f}^{*}\parallel =\underset{x\in X,\parallel x\parallel =1}{sup}|{f}_{n}\left(x\right)-{f}^{*}\left(x\right)|<ϵ.$

Therefore, we have found an ${n}_{0}^{*}$ for each $ϵ$ that fits the definition of convergence, and so ${f}_{n}\to {f}^{*}$ .

Next, we will show that ${f}^{*}$ is linear and bounded. To show that ${f}^{*}$ is linear, we check that

${f}^{*}\left(ax+by\right)=\underset{n\to \infty }{lim}{f}_{n}\left(ax+by\right)=a\underset{n\to \infty }{lim}{f}_{n}\left(x\right)+b\underset{n\to \infty }{lim}{f}_{n}\left(x\right)=a{f}^{*}\left(x\right)+b{f}^{*}\left(y\right).$

To show that ${f}^{*}$ is bounded, we have that for each $ϵ>0$ there exists ${n}_{0}^{*}\in {\mathbb{Z}}^{+}$ (shown above) such that if $n>{n}_{0}^{*}$ then

$|{f}^{*}\left(x\right)|\le |{f}_{n}\left(x\right)-{f}^{*}\left(x\right)|+|{f}_{n}\left(x\right)|\le ϵ||x||+||{f}_{n}||·||x||\le \left(ϵ+||{f}_{n}||\right)||x||.$

This shows that ${f}^{*}$ is linear and bounded and so ${f}^{*}\in {X}^{*}$ . Therefore, ${\left\{{f}_{n}^{*}\right\}}_{n=1}^{\infty }$ converges in ${X}^{*}$ and, since the Cauchy sequence was arbitrary, we have that ${X}^{*}$ is complete and therefore Banach.

## The dual of ${\ell }_{p}$

Recall the space ${\ell }_{p}=\left\{\left({x}_{1},{x}_{2},...\right):{\sum }_{i=1}^{\infty }|{x}_{i}{|}^{p}<\infty \right\}$ , with the ${\ell }_{p}$ -norm ${||x||}_{p}=\left({\sum }_{i=1}^{\infty }|{x}_{i}{{|}^{p}\right)}^{1/p}$ . The dual of ${\ell }_{p}$ is ${\left({\ell }_{p}\right)}^{*}={\ell }_{q}$ , with $q=\frac{p}{p-1}$ . That is, every linear bounded functional $f\in {\left({\ell }_{p}\right)}^{*}$ can be represented in terms of $f\left(x\right)={\sum }_{n=1}^{\infty }{a}_{n}{x}_{n}$ , where $a=\left({a}_{1},{a}_{2},...\right)\in {\ell }_{q}$ . Note that if $p=2$ then $q=2$ , and so ${\ell }_{2}$ is self-dual.

## Bases for self-dual spaces

Consider the example self-dual space ${\mathbb{R}}^{n}$ and pick a basis $\left\{{e}_{1},{e}_{2},...,{e}_{n}\right\}$ for it. Recall that for each $a\in {\mathbb{R}}^{n}$ there exists a bounded linear functional ${f}_{a}\in {\left({\mathbb{R}}^{n}\right)}^{*}$ given by ${f}_{a}\left(x\right)=⟨x,a⟩.$ Thus, one can build a basis for the dual space ${\left({\mathbb{R}}^{n}\right)}^{*}$ as $\left\{{f}_{e1},{f}_{e2},...,{f}_{en}\right\}$ .

## Riesz representation theorem

Since it is easier to conceive a dual space by linking it to elements of a known space (as seen above for self duals), we may ask if there are large classes of spaces who are self-dual.

Theorem 2 (Riesz Representation Theorem) If $f\in {H}^{*}$ is a bounded linear functional on a Hilbert space $H$ , there exists a unique vector $y\in H$ such that for all $x\in H$ we have $f\left(x\right)=⟨x,y⟩$ . Furthermore, we have $||f||=||y||$ , and every $y\in H$ defines a unique bounded linear functional ${f}_{y}\left(x\right)=⟨x,y⟩$ .

Thus, since there is a one-to-one mapping between the dual of the Hilbert space and the Hilbert space ( ${H}^{*}\approx H$ ), we say that all Hilbert spaces are self-dual. Pick a linear bounded functional $f\in {H}^{*}$ and let $\mathcal{N}\subseteq H$ be the set of all vectors $x\in H$ for which $f\left(x\right)=0$ . Note that $\mathcal{N}$ is a closed subspace of $H$ , for if $\left\{{x}_{n}\right\}\subseteq \mathcal{N}$ is sequence that converges to $x\in H$ then, due to the continuity of $f$ , $f\left({x}_{n}\right)\to f\left(x\right)=0$ and so $x\in \mathcal{N}$ . We consider two possibilities for the subspace $\mathcal{N}$ :

• If $N=H$ then $y=0$ and $f\left(x\right)=⟨x,0⟩=0$ .
• If $N\ne H$ then we can partition $H=\mathcal{N}+{\mathcal{N}}^{\perp }$ and so there must exist some ${z}_{0}\in {\mathcal{N}}^{\perp }$ , ${z}_{0}\ne 0$ . Then, define $z=\frac{{z}_{o}}{|f\left({z}_{o}\right)|}$ to get $z\in {\mathcal{N}}^{\perp }$ such that $f\left(z\right)=1$ . Now, pick an arbitrary $x\in H$ and see that
$f\left(x-f\left(x\right)z\right)=f\left(x\right)-f\left(x\right)f\left(z\right)=f\left(x\right)-f\left(x\right)=0.$
Therefore, we have that $x-f\left(x\right)z\in \mathcal{N}$ . Since $z\perp \mathcal{N}$ , we have that
$\begin{array}{cc}\hfill ⟨x-f\left(x\right)z,z⟩& =0,\hfill \\ \hfill ⟨x,z⟩-f\left(x\right)⟨z,z⟩& =0,\hfill \end{array}$
and so
$f\left(x\right)=\frac{⟨x,z⟩}{⟨z,z⟩}=\frac{⟨x,z⟩}{{||z||}^{2}}=〈x,,,\frac{z}{{||z||}^{2}}〉.$
Thus, we have found a vector $y=\frac{z}{{\parallel z\parallel }^{2}}$ such that $f\left(x\right)=⟨x,y⟩$ for all $x\in H$ . Now, to compute the norm of the functional, consider
$|f\left(x\right)|=|⟨x,y⟩|\le ||x||·||y||,$
with equality if $x=\alpha y$ for $\alpha \in R$ . Thus, we have that $\parallel f\parallel =\parallel y\parallel$ .

We now show uniqueness: let us assume that there exists a second $\stackrel{^}{y}\ne y$ , $\stackrel{^}{y}\in H$ , for which $f\left(x\right)=⟨x,\stackrel{^}{y}⟩$ for all $x\in H$ . Then we have that for all $x\in H$ ,

$\begin{array}{cc}\hfill ⟨x,y⟩& =⟨x,\stackrel{^}{y}⟩,\hfill \\ \hfill ⟨x,y⟩-⟨x,\stackrel{^}{y}⟩& =0,\hfill \\ \hfill ⟨x,y-\stackrel{^}{y}⟩& =0,\hfill \\ \hfill y-\stackrel{^}{y}& =0,\hfill \\ \hfill y& =\stackrel{^}{y}.\hfill \end{array}$

This contradicts the original assumption that $y\ne \stackrel{^}{y}$ . Therefore we have that $y$ is unique, completing the proof.

## Linear functional extensions

When we consider spaces nested inside one another, we can define pairs of functions that match each other at the overlap.

Definition 3 Let $f$ be a linear functional on a subspace $M\subseteq X$ of a vector space $X$ . A linear functional $F$ is said to be an extension of $f$ to $N$ (where $N$ is another subspace of $X$ that satisfies $M\subseteq N\subseteq X$ ) if $F$ is defined on $N$ and $F$ is identical to $f$ on $M$ .

Here is another reason why we are so interested in bounded linear functionals.

Theorem 3 (Hahn-Banach Theorem) A bounded linear functional $f$ on a subspace $M\subseteq X$ can be extended to a bounded linear functional $F$ on the entire space $X$ with

$\parallel F\parallel =\underset{x\in X}{sup}\frac{|F\left(x\right)|}{\parallel x\parallel }=\parallel f\parallel =\underset{x\in M}{sup}\frac{|f\left(x\right)|}{\parallel x\parallel }=\underset{x\in M}{sup}\frac{|F\left(x\right)|}{\parallel x\parallel }.$

The proof is in page 111 of Luenberger.

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