I was thinking about this just now. If we are in a complex Hilbert space $H$ and we have a sequence of self-adjoint operators converging in norm, $A_n \to A$, is it also true that $A_n^m \to A^m$ for all natural numbers $m$? What about if $\|A_n\| \leq 1$? What if $A_n \geq 0$? Do we necessarily have convergence of natural powers? I thank all helpers.
2026-03-26 14:20:57.1774534857
Do natural powers of convergent operators also converge
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Yes, because multiplication of operators is a continuous function in the norm topology.
EDIT: To prove it, use $$\eqalign{\|A_1 B_1 - A_2 B_2\| &= \|A_1 (B_1 - B_2) + (A_1 - A_2) B_2\| \cr &\le \|A_1 (B_1 - B_2)\| + \|(A_1 - A_2) B_2\| \cr &\le \|A_1\| \|B_1 - B_2\| + \|A_1 - A_2\| \|B_2\|}$$