Expressions for the norm

To facilitate working with the norm of a transformation, we consider the following four expressions: \begin{align} p & =\sup \big\{\|A x\| /\|x\|: x \neq 0\big\}, \\ q & =\sup \big\{\|A x\|:\|x\|=1\big\}, \\ r & =\sup \big\{|(A x, y)| /\|x\| \cdot\|y\|: x \neq 0, y \neq 0\big\}, \\ s & =\sup \big\{|(A x, y)|:\|x\|=\|y\|=1\big\}. \end{align}In accordance with our definition of the brace notation, the expression $\{\Vert Ax\Vert :\Vert x\Vert =1\}$ , for example, means the set of all real numbers of the form $\Vert Ax\Vert$ , considered for all $x$ ’s for which $\Vert x\Vert =1$ .

Since $\Vert Ax\Vert \le K\Vert x\Vert$ is trivially true with any $K$ if $x=0$ , the definition of supremum implies that $p=\Vert A\Vert$ ; we shall prove that, in fact, $p=q=r=s=\Vert A\Vert$ . Since the supremum in the expression for $q$ is extended over a subset of the corresponding set for $p$ (that is, if $\Vert x\Vert =1$ , then $\Vert Ax\Vert /\Vert x\Vert =\Vert Ax\Vert$ ), we see that $q\le p$ ; a similar argument shows that $s\le r$ .

For any $x\ne 0$ we consider $y=\frac{x}{\Vert x\Vert }$ (so that $\Vert y\Vert =1$ ); we have $\Vert Ax\Vert /\Vert x\Vert =\Vert Ay\Vert$ . In other words, every number of the set whose supremum is $p$ occurs also in the corresponding set for $q$ ; it follows that $p\le q$ , and consequently that $p=q=\Vert A\Vert$ .

Similarly if $x\ne 0$ and $y\ne 0$ , we consider x^{\prime} = x /\|x\| and y^{\prime} = y /\|y\| ; we have |(A x, y)| /\|x\|\cdot\|y\|=|(A x^{\prime}, y^{\prime})|, and hence, by the argument just used, $r\le s$ , so that $r=s$ .

To consolidate our position, we note that so far we have proved that $$p=q=\Vert A\Vert \text{ and }r=s.$$ Since $$\frac{|(Ax,y)|}{\Vert x\Vert \cdot \Vert y\Vert }\le \frac{\Vert Ax\Vert \cdot \Vert y\Vert }{\Vert x\Vert \cdot \Vert y\Vert }=\frac{\Vert Ax\Vert }{\Vert x\Vert },$$ it follows that $r\le p$ ; we shall complete the proof by showing that $p\le r$ . For this purpose we consider any vector $x$ for which $Ax\ne 0$ (so that $x\ne 0$ ); for such an $x$ we write $y=Ax$ and we have $$\Vert Ax\Vert /\Vert x\Vert =|(Ax,y)|/\Vert x\Vert \cdot \Vert y\Vert .$$ In other words, we proved that every number that occurs in the set defining $p$ , and is different from zero, occurs also in the set of which $r$ is the supremum; this clearly implies the desired result.

The numerical function of a transformation $A$ given by $\Vert A\Vert$ satisfies the following four conditions: \begin{align} \|A+B\| & \leq\|A\|+\|B\|, \tag{1}\\ \|A B\| & \leq\|A\| \cdot\|B\|, \tag{2}\\ \|\alpha A\| & =|\alpha| \cdot\|A\| , \tag{3}\\ \|A^{*}\| & =\|A\|. \tag{4} \end{align}The proof of the first three of these is immediate from the definition of the norm of a transformation; for the proof of (4) we use the equation $\Vert A\Vert =r$ , as follows. Since \begin{align} |(A x, y)| &= |(x, A^{*} y)|\\ &\leq\ |x\| \cdot\|A^{*} y\| \\ &\leq \|A^{*}\| \cdot\|x\| \cdot\|y\|, \end{align}we see that $\Vert A\Vert \le \Vert A^{\ast }\Vert$ ; replacing $A$ by $A^{\ast }$ and $A^{\ast }$ by $A^{\ast \ast }=A$ , we obtain the reverse inequality.

EXERCISES