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 \(\{\|A x\|:\|x\|=1\}\) , for example, means the set of all real numbers of the form \(\|A x\|\) , considered for all \(x\) ’s for which \(\|x\|=1\) .

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

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

Similarly if \(x \neq 0\) and \(y \neq 0\) , 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, \(r \leq s\) , so that \(r=s\) .

To consolidate our position, we note that so far we have proved that \[p=q=\|A\| \quad \text { and } \quad r=s.\] Since \[\frac{|(A x, y)|}{\|x\| \cdot\|y\|} \leq \frac{\|A x\|\cdot\|y\|}{\|x\|\cdot\|y\|}=\frac{\|A x\|}{\|x\|},\] it follows that \(r \leq p\) ; we shall complete the proof by showing that \(p \leq r\) . For this purpose we consider any vector \(x\) for which \(A x \neq 0\) (so that \(x \neq 0\) ); for such an \(x\) we write \(y=A x\) and we have \[\|A x\| /\|x\|=|(A x, y)| /\|x\| \cdot\|y\|.\] 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 \(\|A\|\) 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 \(\|A\|=r\) , 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 \(\|A\| \leq \|A^{*}\|\) ; replacing \(A\) by \(A^{*}\) and \(A^{*}\) by \(A^{* *}=A\) , we obtain the reverse inequality.

EXERCISES