Increasing Sequence of projection $P_n \rightarrow 1_H$ then for any compact operator $K, \|[P_n,K]\|=\|P_nK-KP_n\| \rightarrow 0$

Question

I want to prove that T is quasidiagonal iff $T=D+K$ where $D$ is a diagonal operator and $K$ is a compact operator. If $T=D+K$ then I want to show that $T$ is quasidiagonal. For this I want to show the following:

Let $\{P_n\}$ be an increasing sequence of projections in $\mathcal{B}(H)$ such that $P_n \rightarrow 1_H$ in strong operator topology. Then for any compact operator $K \in \mathcal{K}(H)$ $$\|P_n \mathcal{K} -\mathcal{K}P_n\| \rightarrow 0$$

Answer 1

Hints: From the fact that $\|P_nx -x\| \to 0$ for each $x$ it is easy to show that this convergence is uniform on compact subsets. [Cover the compact set by a finite union of $\epsilon$ balls and note that $\|P_n\| \leq 1$ for all $n$].

Since the closure of $\{Kx: \|x\| \leq 1\}$ is compact it follows that $\|P_nKx-Kx\|\to 0$ uniformly for $\|x\| \leq 1$ which means $\|P_nK-K\| \to 0$.

Now apply this to the compact operator $K^{*}$ to get $\|P_nK^{*}-K^{*}\| \to 0$. Conclude that $\|KP_n-K\| \to 0$.

Finally, $\|P_nK-KP_n\| \leq \|P_nK-K\|+\|KP_n-K\| \to 0$

Answer 2

Thanks for your answer @Kavi Rama Murty. I am putting the details here.

Since $\{P_n\}$ converges to $1_H$ in strong topology $P_n(x) \rightarrow x$ for all $x \in X$ that is $\|P_n(x)-x\| \rightarrow 0$ for all $x \in X$.

Claim 1: For any compact set $A \subset H$, $\|P_nx-x\| \rightarrow 0$ uniformly on $A$.

Let $\epsilon>0$. Let $x \in A$ be arbitrary. Since $A$ is compact $A$ is totally bounded and hence there exists $x_1,x_2, \dots,x_n$ such that $A \subseteq \cup_{i=1}^nB(x_i/3,\epsilon)$. So $x \in B(x_i,\epsilon/3)$ for some $i, 1 \leq 1 \leq n$. Since $P_n(x_i) \rightarrow x_i$, there exists $N \in \mathbb{N}$ such that for all $n \geq N$ $$\|P_n(x_i)-x_i\|<\epsilon/3$$ Thus for all $n \geq N$ \begin{align} \|P_n(x)-x\| &\leq \|P_n(x)-P_n(x_i)+P_n(x_i)-x_i+x_i-x\|\\ \\ ~~~~~~~~~~~~~~~~~~~~~&\leq \|P_n(x-x_i)\|+\|P_n(x_i)-x_i\|+\|x_i-x\| \\ \\ ~~~~~~~~~~~~~~~~~~~~~&\leq \|P_n\|\|x-x_i\|+\|P_n(x_i)-x_i\|+\|x_i-x\|\\ \\ ~~~~~~~~~~~~~~~~~~~~~&< 1\cdot \epsilon/3+\epsilon/3+\epsilon/3\\ \\ ~~~~~~~~~~~~~~~~~~~~~&< \epsilon \end{align}

Thus $P_n(x) \rightarrow x$ uniformly on $A$.

Claim 2: For any compact operator $K \in \mathcal{B}(H)$, $\|P_nK-K\| \rightarrow 0$.

Since $K$ is a compact operator the closure of the image of closed unit ball is compact. Thus $\overline{\{Kx:\|x\| \leq 1\}} \subseteq H$ is compact and hence $\|P_n(K(x)) -K(x)\|\rightarrow 0$ uniformly on $\overline{\{x:\|x\| \leq 1\}}$. Thus $\|P_nK-K\| \rightarrow 0$.

Claim 3: For any compact operator $K \in \mathcal{B}(H)$, $\|KP_n-K\| \rightarrow 0$.

Since $K$ is a compact operator $K^*$ is also a compact operator. Using Claim 2 $\|P_nK^*-K^*\| \rightarrow 0$ that is $\|(KP_n-K)^*\| \rightarrow 0$ and hence $\|KP_n-K\| \rightarrow 0$.

Now using all the claims

\begin{align} \|P_nK-KP_n\| &=\|P_nK-K+K-KP_n\|\\ \\ & \leq \|P_nK-K\| + \|K-KP_n\| \rightarrow 0 \end{align}

Thus $\|P_nK-KP_n\| \rightarrow 0$ as $n \rightarrow \infty$ .