Proving square roots of convergent positive operators are convergent

Question

In my functional analysis class, we are studying positive self-adjoint operators and their square roots. I have the following exercise

1. Let $A_n \to A$ be a sequence of positive self-adjoint operators converging in norm to $A$. We are asked to show $\sqrt{A_n} \to \sqrt{A}$ in norm.

2. Let $A_n \to A$ be a sequence of positive self-adjoint operators converging strongly to $A$, meaning that for all vectors $x$ we have $\lVert (A_n - A)x \rVert \to 0$. We are asked to show that $\sqrt{A_n} \to \sqrt{A}$ strongly, meaning that $\lVert (\sqrt{A_n} - \sqrt{A})x \rVert \to 0$.

Here is what I have from class. The power series for $f(z)=\sqrt{1-z} = \sum_{i=0}^{\infty} c_i z^i$ converges absolutely for $|z| \leq 1$ so if we WLG scale our operators such that $ \lVert A_n \rVert \leq 1$ and thus $\lVert A \rVert \leq 1$ one has $\lVert I-A_n \rVert \leq 1$ and $\lVert I-A \rVert \leq 1$ (not difficult to show) and we may construct the square root of the operators as $$ \sqrt{A_n} = \sqrt{I-(I-A_n)} = I - \sum_{i=1}^{\infty} c_i (I-A_n)^i$$ and $$ \sqrt{A} = \sqrt{I-(I-A)} = I - \sum_{i=1}^{\infty} c_i (I-A)^i$$ That is how we constructed the square root of a positive operator in the lecture, but I do not know how to use this or any other method to show $\lVert \sqrt{A_n} - \sqrt{A} \rVert \to 0$ and $\lVert (\sqrt{A_n} - \sqrt{A})x \rVert \to 0$. I am stuck and would appreciate a solution for this. I thank all helpers.

******* Progress: for 1 I can look at the following difference in norm $$ \lVert \sqrt{A_n}-\sqrt{A} \rVert = \left\lVert \sum_{i=1}^{\infty} c_i \left[ (I-A_n)^i - (I-A)^i \right] \right\rVert $$ and using the triangle inequality $$ \lVert \sqrt{A_n}-\sqrt{A} \rVert \leq \sum_{i=1}^{\infty} |c_i| \left\lVert (I-A_n)^i - (I-A)^i\right\rVert $$ and I can show that $\left\lVert (I-A_n)^i - (I-A)^i\right\rVert \leq 1$ and thus I can use Tannery's theorem to take the limit inside the sum so all I would need is that for all $i$ one has $\left\lVert (I-A_n)^i - (I-A)^i\right\rVert \to 0$ as $n \to \infty$. Is it possible to show this?

Answer 1

For $s > 0$ and $0 < \alpha < 1$, $$ \Gamma(\alpha)=\int_{0}^{\infty}e^{-u}u^{-1+\alpha}du \\ = \int_0^{\infty}e^{-su}(su)^{-1+\alpha}sdu \\ = s^{\alpha}\int_0^{\infty}e^{-su}u^{-1+\alpha}du \\ = s^{\alpha}\int_0^{\infty}\frac{e^{-su}-1}{s}u^{-2+\alpha}(-1+\alpha)du \\ =(1-\alpha) s^{\alpha}\int_0^{\infty}\frac{1-e^{-su}}{s}u^{-2+\alpha}du $$ Therefore, $$ \frac{1-\alpha}{\Gamma(\alpha)}\int_0^{\infty}(1-e^{-us})u^{-2+\alpha}du=s^{1-\alpha}. $$ From this, the fractional powers $0 < \alpha < 1$ of a positive operator $A$ are obtained from the positive semigroup $S_A(u)=e^{-uA}$: $$ A^{1-\alpha}=\frac{1-\alpha}{\Gamma(\alpha)}\int_0^{\infty}(1-e^{-uA})u^{-2+\alpha}du $$ $0 \le (I-e^{-uA}) \le I$ because $0 \le e^{-uA} \le I$, which gives $0 \le A^{1-\alpha} \le I$. The question of norm convergence of $A_n^{r}$ to $A^{r}$ is reduced to looking at the operator norm convergence of $e^{-uA_n}$ to $e^{-uA}$.