I want to prove the following fact: Let $A$ be a positive operator (thus self-adjoint) over a Hilbert space $H$. (Notation: $s-\lim$: strong limit of operators; $P_K$: orthogonal projection onto a closed subspace $K\subset H$; $ker(A)$, $R(A)$: kernel and range)

(1) $s-\lim_{n\rightarrow \infty} A^{1/n}=P_{\overline{R(A)}}$;

(2) If $A\leq I$, then $s-\lim_{n\rightarrow \infty} A^{n}=P_{ker(I-A)}$.

By the spectral theorem for bounded self-adjoint operators, $A=\int_\mathbb{R} \lambda dE_\lambda$. Meanwhile, $A$ is positive and $dE_\lambda$ is supported on a compact set, say an interval $[0,a]$. Then I think from the functional calculus, I can obtain $$A^{1/n}=\int_{[0,a]} \lambda^{1/n} dE_\lambda,~A^{n}=\int_{[0,1]} \lambda^{n} dE_\lambda~(A\leq I),$$ and the convergence of the spectral integral is in the sense of the strong convergence. However, I don't know how to explain the validity of taking $s-\lim_n$ now: here you are taking the limit with respect to some Stieljes integral in the space of operators, and to prove the strong convergence of these two sequences, I have to take an arbitrary element $x\in H$ and to use some trick. Can anyone give me some hint/suggestions? Thanks a lot in advance!


1 Answer 1


We can avoid applying the spectral theorem. Instead we will use a well known fact that a monotonic and bounded sequence of positive operators is strongly convergent.

Concerning $(a)$ we may assume that $0\le A\le I.$ Then the sequence $A^{1/n}$ is increasing and bounded above by $I.$ Hence it is strongly convergent. Let $P$ denote the strong limit. The sequence $A^{1+1/n}$ is norm convergent to $A.$ Indeed, by applying the continuous functional calculus we get $$\|A-A^{1+1/n}\|=\max_{x\in \sigma(A)}|x-x^{1+1/n}|\le \max_{0\le x\le 1}x(1-x^{1/n})\\ =\max_{0\le t\le 1}t^n(1-t)={n\over n+1}\left [1- \left ({n\over n+1}\right )^{1/n}\right ]\underset{n\to\infty}{\longrightarrow} 0$$ We get $AP=PA=A.$ Therefore $Px=x$ for $x\in R(A).$ Moreover if $x\in R(A)^\perp =\ker A$ then $Ax=0$ and $A^{1/n}x=0,$ i.e. $Px=0.$ Therefore $P$ is the orthogonal projection onto $\overline{R(A)}.$

Concerning $(b),$ the sequence $A^n$ is decreasing and positive, therefore strongly convergent. Let $Q$ denote the strong limit. Then $AQ=QA=Q$ and $Q^2=Q.$ Thus $Q$ vanishes on $R(I-A).$ If $x\in R(I-A)^\perp =\ker (I-A)$ then $Ax=x.$ Thus $A^nx=x$ and $Qx=x.$ Hence $Q$ is the orthogonal projection onto $\ker(I-A).$


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