I'm trying to prove the next:
a) If $A_{n}\geq 0,$ $A_{n}\rightarrow A$ in norm, then $\sqrt{A_{n}}\rightarrow\sqrt{A}$ in norm,
b) Suppose that $A_{n}\rightarrow A$ strongly for a sequence $\{A_{n}\}.$ Then $\sqrt{A_{n}}\rightarrow\sqrt{A}$ strongly.
I'm stuck prove this. I've seen the proof of a) using spectral theorem, but I don't familiar with this. I was thinking in a proof more elementary.
I was thinking in something of the form $$||\sqrt{A_{n}}-\sqrt{A}||=||\frac{A_{n}-A}{\sqrt{A_{n}}+\sqrt{A}}||,$$ and then bounding denominator an use the hypotesis of convegence in norm, but I guess this is not correct and useless.
Any kind of help is thanked in advanced.
Start with the binomial series $$ (1-z)^{\alpha}=1-\sum_{n=1}^{\infty}c_n z^n,\;\;\;\; |z| < 1, \;\; 0 < \alpha < 1. $$ where $c_n \ge 0$ for all $n \ge 1$. It follows that $\sum_{n=1}^{\infty}c_n=1$, by considering the limit as $z\rightarrow 1$ on the real line. This provides one way to construct the positive square root of an operator $0 \le A \le I$: $$ A^{1/2} = (I-(I-A))^{1/2}=1-\sum_{n=1}^{\infty}c_n(I-A)^{n} $$ By scaling, you can assume your sequence $\{ A_n \}$ satisfies $0 \le A_n \le I$, with $0 \le A \le I$. With that you can show that $A_n^{1/2}\rightarrow A^{1/2}$ in operator norm.