# Strong limit of $A^n$ and $A^{1/n}$ for a positive operator

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!

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 $$\ker(I-A).$$