\(T^ae_i=\mu _i^ae_i\) for every \(i\); in particular \(T^1=T\), \(T^aT^b=T^{a+b}\) for \(a,b{\gt}0\) and \((T^{1/2})^2=T\).

\(T^a\) is self-adjoint and positive semidefinite, \(\| T^a\| \le M^a\), and \(\langle T^ax,x\rangle =\sum _i\mu _i^a\langle e_i,x\rangle ^2\).

For \(a{\gt}0\), \(\operatorname {ran}T^a\subseteq (\ker T)^\perp =\overline{\operatorname {ran}T}\): if \(Tv=0\) then \(\mu _i\langle e_i,v\rangle =0\) for every \(i\), so each term of \(\langle v,T^ag\rangle =\sum _i\mu _i^a\langle v,e_i\rangle \langle e_i,g\rangle \) vanishes (\(0^a=0\)).

Let \(T\) be diagonal in the Hilbert basis \(e\) with eigenvalues \(0\le \mu _i\le M\), and \(a\ge 0\). If \(Tv=\lambda v\), then \(T^av=\lambda ^av\). (Coordinatewise: \(\langle e_i,T^av\rangle =\mu _i^a\langle e_i,v\rangle \) and \(\langle e_i,v\rangle =0\) unless \(\mu _i=\lambda \), since eigenvectors of the self-adjoint \(T\) for distinct eigenvalues are orthogonal.)

Let \(T\) be diagonal in two Hilbert bases \(e=(e_i)_{i\in I}\) and \(e'=(e'_j)_{j\in J}\) of \(H\), \(Te_i=\mu _ie_i\) with \(0\le \mu _i\le M\) and \(Te'_j=\mu '_je'_j\). Then for every \(a\ge 0\) the spectral powers of Definition 789 defined through \(e\) and through \(e'\) coincide: by Lemma 793 both map \(e'_j\) to \(\mu _j'{}^ae'_j\), and a bounded operator is determined by its values on a Hilbert basis.

For \(\lambda {\gt}0\), \(a{\gt}0\) and \(t\in [0,1]\), \(\frac{\lambda t^a}{t+\lambda }\le \lambda ^{\min (a,1)}\). For \(0{\lt}a\le 1\) the weighted arithmetic–geometric mean inequality \(\lambda ^{1-a}t^a\le (1-a)\lambda +at\le \lambda +t\) gives \(\frac{\lambda t^a}{t+\lambda }=\lambda ^a\frac{\lambda ^{1-a}t^a}{t+\lambda }\le \lambda ^a\); for \(a{\gt}1\), \(t^a\le t\) on \([0,1]\) gives \(\frac{\lambda t^a}{t+\lambda }\le \frac{\lambda t}{t+\lambda }\le \lambda \).

Let \(T\) be diagonal in \(e\) with eigenvalues \(0\le \mu _i\le 1\), \(\lambda {\gt}0\), \(a{\gt}0\) and \((T+\lambda )y=\lambda T^ag\), i.e. \(y=\lambda (T+\lambda )^{-1}T^ag\). Then \(\| y\| \le \lambda ^{\min (a,1)}\| g\| \) and \(\langle Ty,y\rangle \le \bigl(\lambda ^{\min (a+1/2,1)}\| g\| \bigr)^2\). (Coefficientwise \(\langle e_i,y\rangle =\frac{\lambda \mu _i^a}{\mu _i+\lambda }\langle e_i,g\rangle \) and \(\mu _i\langle e_i,y\rangle ^2=\bigl(\frac{\lambda \mu _i^{a+1/2}}{\mu _i+\lambda }\bigr)^2 \langle e_i,g\rangle ^2\); apply Lemma 795 and Parseval.)

Let \(T=B^*B\) be diagonal in \(e\) with eigenvalues \(\mu _i\in [0,M]\). Then \(\operatorname {ran}B^*\subseteq \operatorname {ran}T^{1/2}\): for \(g\) in the target space of \(B\), \(g_0:=\sum _{\mu _i\ne 0}\mu _i^{-1/2}\langle Be_i,g\rangle e_i\) satisfies \(T^{1/2}g_0=B^*g\) and \(\| g_0\| \le \| g\| \) (Bessel’s inequality for the orthonormal family \((\mu _i^{-1/2}Be_i)_{\mu _i\ne 0}\); \(Be_i=0\) when \(\mu _i=\| Be_i\| ^2=0\)).