Generalizing dual spaces- A study on functionals.

A functional is that which maps a vector space to a scalar field like $\Bbb{R}$ or $\Bbb{C}$ . If $X$ is the vector space under consideration, and $f_i:X\to \Bbb{R}$ (or $f_i:X\to\Bbb{C}$ ), then the vector space $\{f_i\}$ of functionals is referred to as the algebraic dual space $X^*$ . Similarly, the vector space of functionals $f'_i:X^*\to \Bbb{R}$ (or $f'_i:X^*\to\Bbb{C}$ ) is referred to as the second algebraic dual space. It is also referred to as $X^{**}$ .

How should one imagine $X^*$ ? Imagine a bunch of functionals being mapped to $\Bbb{R}$ . One way to do it is to make all of them map only one particular $x\in X$ . Hence, $g_x:X^*\to \Bbb{R}$ such that $g_x(f)=g(f(x))$ . Another such mapping is $g_y$ . The vector space $X^{**}$ is isomorphic to $X$ .

My book only talks about $X, X^*$ and $X^{**}$ . I shall talk about $X^{***}, X^{****}$ , and $X^{**\dots *}$ . Generalization does indeed help the mind figure out the complete picture.

Say we have $X^{n*}$ ( $n$ asterisks). Imagine a mapping $X^{n*}\to \Bbb{R}$ . Under what conditions is this mapping well-defined? When we have only one image for each element of $X^{n*}$ . Notice that each mapping $f:X^{n*}\to \Bbb{R}$ is an element of the vector space $X^{(n+1)*}$ . To make $f$ a well-defined mapping, we select any one element $a\in X^{(n-1)*}$ , and determine the value of each element of $X^{n*}$ at $a$ . One must note here that $a$ is a mapping ( $a: X^{(n-2)*}\to\Bbb{R}$ ). What element in $X^{(n-2)*}$ that $a$ must map to $\Bbb{R}$ should be mentioned in advance. Similarly, every element in $X^{(n-2)*}$ is also a mapping, and what element it should map from $X^{(n-3)*}$ should also be pre-stated.

Hence, for every element in $X^{n*}$ , one element each from $X^{(n-2)*}, X^{(n-3)*},X^{(n-4)*},\dots ,X$ should be pre-stated. For every such element in $X^{n*}$ , this $(n-2)$ -tuple can be different. To define a well-defined mapping $f:X^{n*}\to \Bbb{R}$ , we choose one particular element $b\in X^{(n-1)*}$ , and call the mapping $f_b$ . Hence,