As usual we can say a little more about the special case of self-adjoint transformations than in the general case. We consider, for any self-adjoint transformation \(A\) , the sets of real numbers \[\Phi=\{(A x, x) /\|x\|^{2} : x \neq 0\}\] and \[\Psi=\{(A x, x):\|x\|=1\}.\] It is clear that \(\Psi \subset \Phi\) . If, for every \(x \neq 0\) , we write \(y=x /\|x\|\) , then \(\|y\|=1\) and \((A x, x) /\|x\|^{2}=(A y, y)\) , so that every number in \(\Phi\) occurs also in \(\Psi\) and consequently \(\Phi=\Psi\) . We write \begin{align} & \alpha=\inf \Phi=\inf \Psi, \\ & \beta=\sup \Phi=\sup \Psi, \end{align} and we say that \(\alpha\) is the lower bound and \(\beta\) is the upper bound of the self-adjoint transformation \(A\) . If we recall the definition of a positive transformation, we see that \(\alpha\) is the greatest real number for which \(A-\alpha \geq 0\) and \(\beta\) is the least real number for which \(\beta-A \geq 0\) . Concerning these numbers we assert that \[\gamma=\max \{|\alpha|,|\beta|\}=\|A\|.\] 

Half the proof is easy. Since \[|(A x, x)| \leq\|A x\| \cdot\|x\| \leq\|A\| \cdot\|x\|^{2},\] it is clear that both \(|\alpha|\) and \(|\beta|\) are dominated by \(\|A\|\) . To prove the reverse inequality, we observe that the positive character of the two linear transformations \(\gamma-A\) and \(\gamma+A\) implies that both \[(\gamma+A)^*(\gamma-A)(\gamma+A)=(\gamma+A)(\gamma-A)(\gamma+A)\] and \[(\gamma-A)^{*}(\gamma+A)(\gamma-A)=(\gamma-A)(\gamma+A)(\gamma-A)\] are positive, and, therefore, so also is their sum \(2 \gamma(\gamma^{2}-A^{2})\) . Since \(\gamma=0\) implies \(\|A\|=0\) , the assertion is trivial in this case; in any other case we may divide by \(2\) and obtain the result that \(\gamma^{2}-A^{2} \geq 0\) . In other words, \[\gamma^{2}\|x\|^{2}=\gamma^{2}(x, x) \geq(A^{2} x, x)=\|A x\|^{2},\] whence \(\gamma \geq\|A\|\) , and the proof is complete.

We call the reader’s attention to the fact that the computation in the main body of this proof could have been avoided entirely. Since both \(\gamma-A\) and \(\gamma+A\) are positive, and since they commute, we may conclude immediately ( Section: Functions of transformations ) that their product \(\gamma^{2}-A^{2}\) is positive. We presented the roundabout method in accordance with the principle that, with an eye to the generalizations of the theory, one should avoid using the spectral theorem whenever possible. Our proof of the fact that the positiveness and commutativity of \(A\) and \(B\) imply the positiveness of \(A B\) was based on the existence of square roots for positive transformations. This fact, to be sure, can be obtained by so-called "elementary" methods, that is, methods not using the spectral theorem, but even the simplest elementary proof involves complications that are purely technical and, for our purposes, not particularly useful.