Linear functionals

Many systems in engineering can be characterized as linear systems, taking inputs in one signal space into outputs in another. The most common type of system maps into a space of scalars, defined as maps $f:X\to R$ . We will study these maps (known in the mathematics literature as functionals ) during the next few lectures.

Definition 1 A functional $f$ on $X$ is linear if for all $x,y\in X,a,b\in R$ , we have that $f(ax+by)=af\left(x\right)+bf\left(y\right)$ .

Example 1 In the case $X={\mathbb{R}}^n$ , all linear functionals can be written using the form $f\left(x\right)={\sum }_{i=1}^n{a}_i{x}_i$ for some set of scalars $\{a_1,...,a_n\}\subseteq \mathbb{R}$ .

Example 2 For a Hilbert space $X$ , $f\left(x\right)=〈x,,,y〉$ for any $y\in X$ is a linear functional.

Example 3 For the space $X=C\left(\right[0,1\left]\right)$ , the sampling functionals $f\left(x\right)=x\left(t_0\right)=〈x,,,\delta ,(,t,-,t_0,)〉$ are linear.

Example 4 $f\left(x\right)=\parallel x\parallel$ is not a linear functional (since the triangle inequality is not a strict equality).

Linear functionals are particularly amenable to analysis.

Theorem 1 If a linear functional on a normed space $X$ is continuous at a point $x_0\in X$ , then it is continuous on the entire space $X$ .

Assume $f\left(x\right)$ is contiuous at $x_0$ . Let $\left\{x_n\right\}$ be a sequence of points converging to $x$ . Then,

It is easy to show that $\{x_n-x+x_0\}$ is a sequence that converges to $x_0$ . Therefore, $f(x_n-x+x_0)$ has to converge to $f\left(x_0\right)$ as $f$ is continuous at $x_0$ :

This implies that the sequence $\left\{f\left(x_n\right)\right\}$ converges to $f\left(x\right)$ and so $f$ is continuous at $x$ . Since $x$ was arbitrary, then $f$ is continuous on $X$ .

Fact 1 For $f\left(x\right)$ a linear functional, what is the value of $f\left(0\right)$ ? For any such $f$ ,

Definition 2 A linear functional $f$ on a normed space $X$ is bounded if there exists a constant $M<\infty$ such that $\left|f\right(x\left)\right|\le M\parallel x\parallel$ for all $x\in X$ . The smallest such element is denoted as the norm of the function $\parallel f\parallel$ :

There are several ways in which we can write this norm:

Before continuing, we should verify that the defined $\parallel f\parallel$ is a valid norm.

• $\parallel f\parallel =0\iff f=z\left(x\right)=0$ for all $x\in X$ .
• $\parallel f\parallel \ge 0$ : $M$ has to be positive since $\left|f\right(x\left)\right|$ and $\parallel x\parallel$ are always positive.
• $\parallel \alpha f\parallel =\left|\alpha \right|\parallel f\parallel$ : can be trivially shown.
• $\parallel f_1+f_2\parallel \le \parallel f_1\parallel +\parallel f_2\parallel$ : by definition,
  $$\begin{array}{cc}\hfill \parallel f_1+f_2\parallel & =\underset{\parallel x\parallel \le 1}{sup}|(f_1+f_2)\left(x\right)|=\underset{\parallel x\parallel \le 1}{sup}|f_1\left(x\right)+f_2\left(x\right)|\le \underset{\parallel x\parallel \le 1}{sup}\left(\right|f_1\left(x\right)|+|f_2\left(x\right)\left|\right),\hfill \\ & \le \underset{\parallel x\parallel \le 1}{sup}|f_1\left(x\right)|+\underset{\parallel x\parallel \le 1}{sup}|f_2\left(x\right)|=\parallel f_1\parallel +\parallel f_2\parallel .\hfill \end{array}$$

The following theorem can save us much work in checking for continuity of linear functionals.

Theorem 2 A linear functional on a normed space is bounded if and only if it is continuous.

( $\Rightarrow$ ) Assume $f$ is bounded, and let $M$ be such that $\left|f\right(x\left)\right|\le M\parallel x\parallel$ for all $x\in X$ . Then for a sequence $\left\{x_n\right\}\to 0$ we have $|f\left(x_n\right)|\le M\parallel x_n\parallel$ .Therefore, $\left|f\right(x\left)\right|\to 0=f\left(0\right)$ , which implies that $f$ is continuous at $x=0$ . Using Theorem  [link] , $f$ is continuous on $X$ .

( $\Leftarrow$ ) Assume $f$ is continuous. If we set $ϵ=1$ there exist $\delta >0$ such that if $\parallel x-0\parallel <\delta$ then $\left|f\right(x)-f(0\left)\right|<1$ , i.e.

Since $f$ is linear, we can write this as

So, $\parallel f\parallel \le \frac{1}{\delta }<\infty$ and $f$ is bounded.

Example 5 Let $X$ be a space of finitely nonzero sequences with ${\parallel x\parallel }_{\infty }={max}_i\left|x_i\right|$ . Define a functional $f$ as

$f$ is linear because

If we want to show that $f$ is unbounded, we must show that for every $M>0$ there exists $x\in X$ such that $\left|f\right(x\left)\right|>M\parallel x\parallel$ . Let $x=({\underbrace{1,1,...,1}}_{\lceil M\rceil \text{times}},0,0,...)\in X$ , where $\lceil M\rceil$ is the smallest integer greater or equal to $M$ . Then,

Therefore, $f$ is unbounded.