# Question about SOT and compact operators

I need some help with functional analysis / Hilbert space theory. If you have a favorite text to recommend, please let me know~

Here is my question:

Given $v_t$ be the "squeeze operator" on $H=L^2[0, 1]$, where $v_t: L^2[0,1] \to L^2[0, \frac{2-t}{2}]$ acts on $f \in L^2[0,1]$ by squeezing the domain of the function. We have that $\{v_t\}$ for $t \in [0, 1]$ is a family of SOT continuous operators. I am wondering why given any $p \in \mathbb{K}(H)$, we have $\{ v_tpv_t^* \}$ is continuous in norm.

I found the following facts on a reference suggested by Wikipedia (Hilbert Space Operators in Quantum Physics), but I am not sure how to prove them or how I may use them...

1. $T_n \to^{SOT} T$ implies that for any $p \in \mathbb{K}(H)$, we have $T_np \to Tp$ in norm.

Hints or suggestions would be greatly appreciated!

This is essentially the definition of the strong operator topology; the SOT is the topology generated by the evaluation maps $\mathcal{L}(X,Y) \ni T \mapsto Tx \in Y$ (for a fixed $x$). We want all of these maps to be continuous (w.r.t. the topology induced by the norm on $Y$), or equivalently that $T_n \to T$ in this topology on $\mathcal{L}(X,Y)$ if and only if $T_nx \to Tx$ in the topology on $Y$.

Other good references off the top of my head: Folland Real Analysis, ch. 5; Rudin Functional Analysis, maybe Reed and Simon Functional Analysis.

EDIT: ok, I misunderstood the question a bit.

If $T_n \to T$ strongly, and $A$ is compact, we need to show that $$\sup_{x \in B_X}\left||(T_nA - TA)x \right|| \to 0$$ where $B_X = \{x\in X: |x| = 1\}$. Equivalently, $$(*):\quad\quad\sup_{y \in A(B_X)}\left||(T_n - T)y \right|| \to 0.$$ Now, we know that, for each $y$, $||(T_n - T)y|| \to 0$, but things could go wrong if this doesn't happen uniformly in $Y$. But fortunately, $A(B_X)$ is compact, so we have better control of the convergence.

In particular, given a fixed $\epsilon$ for each $y \in A(B_X)$, there is a neighborhood of $y$ and $N_y$ so that $||(T_n - T)\tilde y|| < \epsilon$ when $\tilde y$ is in this neighborhood, and when $n > N_y$. Then, because the set is compact, a finite number of these neighborhoods cover $A(B_X)$, and taking $N = \max\{N_y\}$ among the finite number of $N_y$'s associated to these neighborhoods demonstrates $(*)$.

• Hi BaronVT, thanks for your reply. I am sorry but I am a bit confused - the $p$ in my question above is not a vector in $H$, but a compact operator. I am not sure I see how the $T_nx \to Tx$ on the space level translates to the algebra level. Let me know if that doesn't make sense. Thanks! Oct 16, 2013 at 18:16
• Ok, I only responded to the 'I am not sure how to prove them or how I may use them' part of your post on why strong operator convergence gives the convergence in norm. Let me think about the compact operator part for a minute. Oct 16, 2013 at 18:39
• @FanFeiChong ok, hopefully your question is addressed now. sorry for misreading it the first time around. Oct 16, 2013 at 19:12
• Thanks for your very helpful response! I understand my minor result now - but would you mind giving a bit more hint regarding how I may apply that to show that $\{v_t p v_t^* \}$ is continuous in norm. Thanks again! Oct 16, 2013 at 20:16