If $A_n$ is a sequence of positive bounded linear operators converging in norm to $A$ on a Hilbert Space, show $\sqrt{A_n}\to\sqrt{A}$ in norm. I can show that $A$ would be positive and thus have a square root, but then I'm mostly stuck.
If $A_n=B_n^2$ and $A=B^2$, I have also shown that since $B_n$ is positive it is by definition self adjoint and so $\|B_n x\| = \sqrt{\langle B_n x, B_n x \rangle} = \sqrt{\langle A_n x, x\rangle} \to \sqrt{\langle Ax,x\rangle} = \|Bx\|2$ for all $x$ and so therefore $\|B_n\|\to \|B\|$. However, I am completely stuck on the desired result.
If I knew $B_n$ and $B$ would commute, then I'd use $\|A_n^2 - A\| = \|(B_n - B)(B_n + B) \|$ and play with the inner product, but I don't know this a priori.
Thanks for your help.
EDIT: For those wondering this question comes from Mathematical Physics I: Functional Analysis by Reed and Simon in chapter 7 question 14.