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 \(\mathcal{V}\) and if \(x\) is a vector in \(\mathcal{V}\) , 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 \(\mathcal{V}\) we make correspond the dual space \(\mathcal{V}^{\prime}\) consisting of all linear functionals on \(\mathcal{V}\) ; to every pair, \(x\) and \(y\) , where \(x\) is a vector in \(\mathcal{V}\) 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 \(\mathcal{V}\) and \(y\) in \(\mathcal{V}^{\prime}\) .

EXERCISES

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

  1. \(y(x)=\xi_{1}\) 
  2. \(y(x)=\xi_{2}\) ,
  3. \(y(x)=\xi_{1}^{2}\) ,
  4. \(y(x)=\xi_{1}-i \xi_{2}\) ,
  5. \(y(x)=\sqrt{\xi_{1}^{2}+\xi_{2}^{2}}\) . (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 \(x=(\xi_{1}, \xi_{2}, \xi_{3})\) in \(\mathbb{C}^{3}\) the function \(y\) is defined by

  1. \(y(x)=\xi_{1}+\xi_{2}\) ,
  2. \(y(x)=\xi_{1}-\xi_{3}^{2}\) ,
  3. \(y(x)=\xi_{1}+1\) ,
  4. \(y(x)=\xi_{1}-2 \xi_{2}+3 \xi_{2}\) .

In which of these cases is \(y\) a linear functional?

Exercise 3. Suppose that for each \(x\) in \(\mathcal{P}\) the function \(y\) is defined by

  1. \(\displaystyle y(x)=\int_{-1}^{+2} x(t) \,d t\) ,
  2. \(\displaystyle y(x)=\int_{0}^{2}(x(t))^{2} \,d t\) ,
  3. \(\displaystyle y(x)=\int_{0}^{1} t^{2} x(t) \,d t\) ,
  4. \(\displaystyle y(x)=\int_{0}^{1} x(t^{2}) \,d t\) ,
  5. \(\displaystyle y(x)=\frac{d x}{d t}\) ,
  6. \(\displaystyle y(x)=\frac{d^{2} x}{d t^{2}}\bigg|_{t=1}\) .

In which of these cases is \(y\) a linear functional?

Exercise 4. If \((\alpha_{0}, \alpha_{1}, \alpha_{2}, \ldots)\) is an arbitrary sequence of complex numbers, and if \(x\) is an element of \(\mathcal{P}\) , \(x(t)=\sum_{i=0}^{n} \xi_{i} t^{i}\) , write \(y(x)=\sum_{i=0}^{n} \xi_{i} \alpha_{i}\) . 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 \(\alpha\) ’s.

Exercise 5. If \(y\) is a non-zero linear functional on a vector space \(\mathcal{V}\) , and if \(\alpha\) is an arbitrary scalar, does there necessarily exist a vector \(x\) in \(\mathcal{V}\) such that \([x, y]=\alpha\) ?

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 \(\alpha\) such that \(y=\alpha z\) . (Hint: if \([x_{0}, z] \neq 0\) , write \(\alpha=[x_{0}, y] /[x_{0}, z]\) .)