Brackets

Before studying linear functionals and dual spaces in more detail, we wish to introduce a notation that may appear weird at first sight but that will clarify many situations later on. Usually we denote a linear functional by a single letter such as $y$ . Sometimes, however, it is necessary to use the function notation fully and to indicate somehow that if $y$ is a linear functional on $𝒱$ and if $x$ is a vector in $𝒱$ , then $y (x)$ is a particular scalar. According to the notation we propose to adopt here, we shall not write $y$ followed by $x$ in parentheses, but, instead, we shall write $x$ and $y$ enclosed between square brackets and separated by a comma. Because of the unusual nature of this notation, we shall expend on it some further verbiage.

As we have just pointed out $[x, y]$ is a substitute for the ordinary function symbol $y (x)$ ; both these symbols denote the scalar we obtain if we take the value of the linear function $y$ at the vector $x$ . Let us take an analogous situation (concerned with functions that are, however, not linear). Let $y$ be the real function of a real variable defined for each real number $x$ by $y (x) = x^{2}$ . The notation $[x, y]$ is a symbolic way of writing down the recipe for actual operations performed; it corresponds to the sentence [take a number, and square it].

Using this notation, we may sum up: to every vector space $𝒱$ we make correspond the dual space \mathcal{V}^{\prime} consisting of all linear functionals on $𝒱$ ; to every pair, $x$ and $y$ , where $x$ is a vector in $𝒱$ and $y$ is a linear functional in \mathcal{V}^{\prime} , we make correspond the scalar $[x, y]$ defined to be the value of $y$ at $x$ . In terms of the symbol $[x, y]$ the defining property of a linear functional is

=\alpha_{1}[x_{1}, y]+\alpha_{2}[x_{2}, y] \tag{1}

and the definition of the linear operations for linear functionals is

=\alpha_{1}[x, y_{1}]+\alpha_{2}[x, y_{2}]. \tag{2}

The two relations together are expressed by saying that $[x, y]$ is a bilinear functional of the vectors $x$ in $𝒱$ and $y$ in \mathcal{V}^{\prime} .

EXERCISES

Exercise 1. Consider the set $ℂ$ of complex numbers as a real vector space (as in Section: Examples , (9)). Suppose that for each $x = ξ_{1} + i ξ_{2}$ in $ℂ$ (where $ξ_{1}$ and $ξ_{2}$ are real numbers and $i = \sqrt{- 1}$ ) the function $y$ is defined by

$y (x) = ξ_{1}$
$y (x) = ξ_{2}$ ,
$y (x) = ξ_{1}^{2}$ ,
$y (x) = ξ_{1} - i ξ_{2}$ ,
$y (x) = \sqrt{ξ_{1}^{2} + ξ_{2}^{2}}$ . (The square root sign attached to a positive number always denotes the positive square root of that number.)

In which of these cases is $y$ a linear functional?

Exercise 2. Suppose that for each $x = (ξ_{1}, ξ_{2}, ξ_{3})$ in $ℂ^{3}$ the function $y$ is defined by

$y (x) = ξ_{1} + ξ_{2}$ ,
$y (x) = ξ_{1} - ξ_{3}^{2}$ ,
$y (x) = ξ_{1} + 1$ ,
$y (x) = ξ_{1} - 2 ξ_{2} + 3 ξ_{2}$ .

In which of these cases is $y$ a linear functional?

Exercise 3. Suppose that for each $x$ in $𝒫$ the function $y$ is defined by

$y (x) = \int_{- 1}^{+ 2} x (t) d t$ ,
$y (x) = \int_{0}^{2} (x (t))^{2} d t$ ,
$y (x) = \int_{0}^{1} t^{2} x (t) d t$ ,
$y (x) = \int_{0}^{1} x (t^{2}) d t$ ,
$y (x) = \frac{d x}{d t}$ ,
$y (x) = \frac{d^{2} x}{d t^{2}} |_{t = 1}$ .

In which of these cases is $y$ a linear functional?

Exercise 4. If $(α_{0}, α_{1}, α_{2}, \dots)$ is an arbitrary sequence of complex numbers, and if $x$ is an element of $𝒫$ , $x (t) = \sum_{i = 0}^{n} ξ_{i} t^{i}$ , write $y (x) = \sum_{i = 0}^{n} ξ_{i} α_{i}$ . Prove that $y$ is an element of \mathcal{P}^{\prime} and that every element of \mathcal{P}^{\prime} can be obtained in this manner by a suitable choice of the $α$ ’s.

Exercise 5. If $y$ is a non-zero linear functional on a vector space $𝒱$ , and if $α$ is an arbitrary scalar, does there necessarily exist a vector $x$ in $𝒱$ such that $[x, y] = α$ ?

Exercise 6. Prove that if $y$ and $z$ are linear functionals (on the same vector space) such that $[x, y] = 0$ whenever $[x, z] = 0$ , then there exists a scalar $α$ such that $y = α z$ . (Hint: if $[x_{0}, z] \neq 0$ , write $α = [x_{0}, y] / [x_{0}, z]$ .)

EXERCISES

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