Convergence of square root operators

Question

Let $Q_n$ and $Q$ be compact positive and symmetric operators. Let $A_n = {Q_n}^{\frac12}$ and $A=Q^{\frac12}$. Given $Q_n$ converges to $Q$ w.r.t. operator norm. Does $A_n$ converges to $A$? Thanks.

Answer 1

Sketch: The square root is operator monotone. Now, if $Q_n\rightarrow Q$, then, given $\epsilon>0$, there is an $N$ such that for all $n\geq N$, $$ (Q-\epsilon I)_+\leq Q_n\leq Q+\epsilon I. $$ It follows that $$ (Q-\epsilon I)_+^{1/2}\leq Q_n^{1/2}\leq (Q+\epsilon I)^{1/2}. $$ It therefore suffices to show that $$ (Q\pm\epsilon I)_+^{1/2}\rightarrow Q^{1/2}, $$ in norm, as $\epsilon\rightarrow 0$. But this convergence we can show by working in an orthonormal eigenbasis for $Q$!

Here is a sketch of a more elementary proof, with wider applicability: Let $f\in C(\mathbb R)$ be any continuous function (for the square root, put $f(x)=0$ for $x\leq 0$). Since $\|Q_n\|\leq \|Q-Q_n\|+\|Q\|$, it suffices to consider $f$ restricted to $[0,2\|Q\|]$ for $n$ large enough. Then we can pick a polynomial $p$ which approximates $f$ uniformly, say $\|p-f\|_{[0,2\|Q\|]}<\epsilon$, and thus \begin{align*} \|f(Q)-f(Q_n)\|&\leq\|f(Q)-p(Q)\|+\|p(Q)-p(Q_n)\|+\|p(Q_n)-f(Q_n)\|\\ &\leq 2\epsilon+ \|p(Q)-p(Q_n)\|, \end{align*} whenever $n$ is large enough, say $n\geq N$. But it is not hard to show that $p(Q_n)\rightarrow p(Q)$ in norm!