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The set of linear functionals $f$ on $X$ is itself a vector space:
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 $\parallel f\parallel ={sup}_{\parallel x\parallel \le 1}\left|f\left(x\right)\right|$ .
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,
From Example [link] , a linear functional $f$ can be written as:
There is a one-to-one mapping between the space $X$ and the dual of $X$ :
For this reason, ${\mathbb{R}}^{n}$ is called self-dual (as there is a one-to-one mapping ${\left({\mathbb{R}}^{n}\right)}^{*}\u27f7{\mathbb{R}}^{n}$ ). Using the Cauchy-Schwartz Inequality, we can show that
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 $\u03f5>0$ 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 $\u03f5>0$ , 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)|=|({f}_{n}-{f}_{m})\left(x\right)|\le \parallel {f}_{n}\left(x\right)-{f}_{m}\left(x\right)\parallel \xb7\parallel x\parallel \le \u03f5\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 $\u03f5>0$ . 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 $n>{n}_{0,x}$ , $|{f}_{n}\left(x\right)-{f}_{n}^{*}\left(x\right)|\le \u03f5$ . Now, let ${n}_{0}^{*}={sup}_{x\in X,\parallel x\parallel =1}{n}_{0,x}$ . Then, for $n>{n}_{0}^{*}$ ,
Therefore, we have found an ${n}_{0}^{*}$ for each $\u03f5$ 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
To show that ${f}^{*}$ is bounded, we have that for each $\u03f5>0$ there exists ${n}_{0}^{*}\in {\mathbb{Z}}^{+}$ (shown above) such that if $n>{n}_{0}^{*}$ then
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.
Recall the space ${\ell}_{p}=\{({x}_{1},{x}_{2},...):{\sum}_{i=1}^{\infty}|{x}_{i}{|}^{p}<\infty \}$ , with the ${\ell}_{p}$ -norm ${\left|\right|x\left|\right|}_{p}=({\sum}_{i=1}^{\infty}|{x}_{i}{{|}^{p})}^{1/p}$ . 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 $a=({a}_{1},{a}_{2},...)\in {\ell}_{q}$ . Note that if $p=2$ then $q=2$ , and so ${\ell}_{2}$ is self-dual.
Consider the example self-dual space ${\mathbb{R}}^{n}$ and pick a basis $\{{e}_{1},{e}_{2},...,{e}_{n}\}$ 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 ${f}_{a}\left(x\right)=\u27e8x,a\u27e9.$ Thus, one can build a basis for the dual space ${\left({\mathbb{R}}^{n}\right)}^{*}$ as $\{{f}_{e1},{f}_{e2},...,{f}_{en}\}$ .
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 $f\left(x\right)=\u27e8x,y\u27e9$ . Furthermore, we have $\left|\right|f\left|\right|=\left|\right|y\left|\right|$ , and every $y\in H$ defines a unique bounded linear functional ${f}_{y}\left(x\right)=\u27e8x,y\u27e9$ .
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 $\mathcal{N}$ :
We now show uniqueness: let us assume that there exists a second $\widehat{y}\ne y$ , $\widehat{y}\in H$ , for which $f\left(x\right)=\u27e8x,\widehat{y}\u27e9$ for all $x\in H$ . Then we have that for all $x\in H$ ,
This contradicts the original assumption that $y\ne \widehat{y}$ . Therefore we have that $y$ is unique, completing the proof.
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
The proof is in page 111 of Luenberger.
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