Let $A_n$ a norm convergent sequence of bounded operators in a Hilbert space with limit $A$. Let $|B|$ denote the absolute value of $B$ i.e. $\sqrt{B^* B}$. How to prove that $|A_n|$ norm converges to $|A|$?
Convergence Absolute Value Operators
functional-analysishilbert-spacesoperator-theory
Related Solutions
You can use the property of the compact operators: that they map bounded sequences in X to sequences in Y with convergent subsequences. So you can choose your sequence to be an orthonormal basis in your Hilbert space and then you can advance with your proof.
I think what you are saying is true. Never thought about it since i've always pre-assumed that the weakly-operator limit $A$ of the $A_n's$ was always in $A\in \mathfrak L(X,Y)$. Am writing the argument just to convience ourselfs. Indeed, we only need to assume that $Y$ has a norm, not necessarily a complete one.
So, lets suppose that $A_n\overset{\text{wo}}{\to}A$ in the weak operator topology where $A:X\to Y$ is a linear operator, not necessarily bounded. Convergence in the weak operator topology is described by $h(A_n x)\to h(A x)$ for every $x\in X$ and $h\in Y^*$. This implies that the set $\{A_n x: n\in \mathbb{N}\}$ is weakly bounded in $Y$, hence it is also bounded in $Y$. By the Banach-Steinhaus it follows that $\sup_{n}||A_n||=M<\infty$. Now, for $x\in X$ with $||x||=1$ we have $$||Ax||=\max_{h\in Y^*,\, ||h||=1}|h(Ax)|$$ So, there is some $||h||=1$ in $Y^*$ such that $||Ax||=|h(Ax)|$. Using the weak convergence for $A_nx$ we end up with \begin{align} ||Ax||&=|h(Ax)|\\ &=\lim_{n\to \infty}|h(A_nx)|\\ &\leq \underbrace{||h||}_{=1}\liminf_{n\to \infty}||A_n||\cdot \underbrace{||x||}_{=1} \end{align} Hence, $||Ax||\leq M$ for every $||x||=1$ and therefore, $||A||\leq M<\infty$.
Edit: (Responding to the comment)
The existence of such $A$ is trickier. To ensure such existence we need another assumption for $Y$, since there is a counter example in here where $X=Y=c_0$. The only natural that i could think while i was trying to prove it is that $Y$ has to be reflexive (from not being a Banach space we went straight out to reflexivity :P). In the case where $X=Y=H$ is a Hilbert space things were slightly more easier since we can identify $H^*$ with $H$ and dont need to mess with the second duals.
The argument in the case where $Y$ is reflexive is the following:
Suppose that $\lim_{n}\langle A_n x, h \rangle$ exists for every $x\in X$ and $h\in Y^*$. For fixed $x\in X$ let $f_x:Y^*\to \mathbb{R}$ defined by $$\langle h, f_x\rangle =\lim_{n\to \infty}\langle A_n x, h\rangle$$ Its easy to check that $f_x$ is a linear functional and by the previous discussion it is also bounded. Meaning, $f_x \in Y^{**}$. By reflexivity, there is some $y_x\in Y$ such that $\langle h, f_x\rangle =\langle y_x, h\rangle$ for all $h\in Y^*$. Now, let $x\overset{A}{\longmapsto} y_x$. Now, its easy to check that $A:X\to Y$ is a linear operator. By the previous discussion it is also bounded.
Best Answer
If $A_n\to A$, then certainly $A^*_nA_n\to A^*A$, and $p(A^*_nA_n)\to p(A^*A)$ for any polynomial $p$. If now $\sigma(A^*A)\subset[0,b)$, there is some $N_1\in\mathbb N$ such that $\sigma(A^*_nA_n)\subset[0,b)$ for all $n\geq N_1$ (for a proof, see Rudin's Functional Analysis, 2nd edition, Theorem 10.20). Fix $\varepsilon>0$, and choose a polynomial $p$ such that $$\sup_{x\in[0,b]}|x^{1/2}-p(x)|<\varepsilon.$$ There exists $N_2\geq N_1$ such that $$\|p(A^*_nA_n)-p(A^*A)\|<\varepsilon$$ for $n\geq N_2$. Then for $n\geq N_2$, we have $$\||A_n|-|A|\|<3\varepsilon.$$