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
Exercise 1. If B is invertible, then \|A B\| \geq\|A\| /\|B^{-1}\| for every A .
Exercise 2. Is it true for every linear transformation A that \|A^{*} A\|=\|A A^{*}\| ?
Exercise 3.
- If A is Hermitian and if \alpha \geq 0 , then a necessary and sufficient condition that \|A\| \leq \alpha is that -\alpha \leq A \leq \alpha .
- If A is Hermitian, if \alpha \leq A \leq \beta , and if p is a polynomial such that p(t) \geq 0 whenever \alpha \leq t \leq \beta , then p(A) \geq 0 .
- If A is Hermitian, if \alpha \leq A \leq \beta , and if p is a polynomial such that p(t) \neq 0 whenever \alpha \leq t \leq \beta , then p(A) is invertible.
