A sufficient condition for comapct operators? I am reading a paper about the classification of essentially normal operators right now, this paper is available in https://link.springer.com/chapter/10.1007/BFb0080022.
In this paper, when he tries to prove Prop 4.1 in page 38, He claims that $T=R\oplus S+K$ where K is a compact operator. I am kind of confused here.
Here is my attempt.
I guess he wants to use the lemma before to get the fact that $K$ is a compact operator, however, I think it is not sufficient. To be more specific, if we want to get this claim we may need to use a proposition like this: If $v_n$ is an orthnomal basis for $H$ and $||Tv_n||\rightarrow 0$, then $T$ is compact.
However, this proposition is FALSE in general. I have found a counterexaple by consider $Te_n=\sqrt{\frac{1}{2^k}}e_{k+1}$ when $2^k< n\leq 2^{k+1}$ and $Te_1=e_1$. Then $||Te_n||\rightarrow 0$ but we can show that $T$ is not compact!
So I wonder how do we get the claim here?
Any help will be truly grateful!
 A: The Lemma gives us an orthonormal sequence $\{f_k'\}$ with $\|(T-\mu_k)f_k\|\to0$, $\|(T-\mu_k)^*f_k'\|\to0$. Choose a subsequence $\{f_{k_n}'\}$ such that $$\|(T-\mu_n)f_{k_n}'\|^2<2^{-n}.$$
We can put $\mu_n$ instead of $\mu_{k_n}$ because the $\mu$ are repeated infinitely many times, and so it is enough, when we choose $f_{n_k}$, to move further along the sequence so that $\mu_{k_n}=\mu_n$.
Now choose a subsequence $\{f_{k_{n_j}}'\}$ such that
$$ % essentially normal
\|(T-\mu_j)^*f_{k_{n_j}}'\|^2<2^{-j}
$$
and $\mu_{k_{n_j}}=\mu_j$. Let $f_j=f_{k_{n_j}}'$. We get automatically that
$$
\|(T-\mu_j)^*f_{j}\|^2<2^{-j}.
$$
And, since $\mu_j=\mu_{k_{n_j}}$,
$$
\|(T-\mu_j)f_{j}\|^2=\|(T-\mu_{k_{n_j}})f_{k_{n_j}}'\|^2<2^{-n_j}<2^{-j}.
$$
In summary, we have an orthonormal sequence $\{f_k\}$ with
$$
\|(T-\mu_k)f_k\|<2^{-k},\qquad \|(T-\mu_k)^*f_k\|<2^{-k},\qquad k\in\mathbb N. 
$$
Let $H_1=\{f_k\}^{\perp\perp}$, and $R\in B(H_1)$ with $Rf_k=\mu_kf_k$. Let $K_1=T|_{H_1}-R$. This is an operator $H_1\to H$. With $P_n$ the orthogonal projection onto $\operatorname{span}\{f_1,\ldots,f_n\}$,
\begin{align}
\Big\|(K_1-K_1P_n)\sum_k\alpha_kf_k\Big\|^2
&=\Big\|\sum_{k>n}\alpha_k(T-\mu_k)f_k\Big\|^2\\[0.3cm]
&\leq \Big(\sum_{k>n}|\alpha_k|\,\|(T-\mu_k)f_k\|\Big)^2\\[0.3cm]
&\leq \Big(\sum_{k>n}|\alpha_k|^2\Big)\sum_{k>n}\,\|(T-\mu_k)f_k\|^2\\[0.3cm]
&\leq \Big\|\sum_{k}\alpha_kf_k\Big\|^2\sum_{k>n}\,\|(T-\mu_k)f_k\|^2\\[0.3cm]
&\leq \Big\|\sum_{k}\alpha_kf_k\Big\|^2\sum_{k>n}\,2^{-k}\\[0.3cm]
&=2^{-n}\, \Big\|\sum_{k}\alpha_kf_k\Big\|^2.
\end{align}
This shows that $\|K_1-K_1P_n\|\to0$; hence $K_1$ is compact, being a limit of finite-rank operators.
Since the roles of $T$ and $T^*$ are symmetric, we also get that $K_2=T^*|_{H_1}-R$ is compact. With $P$ the orthogonal projection onto $H_1$, we also have that $K_3=PT|_{H_1}-R$ is compact. Let us think of $T$ in block from with respect to $H_1$, say,
$$
T=\begin{bmatrix} A&B\\ C&S\end{bmatrix}. 
$$
We have
$$
K_1=T|_{H_1}-R=\begin{bmatrix} A-R&0\\ C&0\end{bmatrix},
\qquad
K_2=T^*|_{H_1}-R^*=\begin{bmatrix} A^*-R^*&0\\ B^*&0\end{bmatrix},
$$
and
$$
K_3=PT|_{H_1}-R=\begin{bmatrix} A-R&0\\ 0&0\end{bmatrix}.
$$
We have shown that all three are compat. Then
\begin{align}
T&=\begin{bmatrix}R&0\\0&S\end{bmatrix}
+\begin{bmatrix} A-R&0\\ C&0\end{bmatrix}
+\begin{bmatrix} A^*-R^*&0\\ B^*&0\end{bmatrix}^*
-\begin{bmatrix} A-R&0\\ 0&0\end{bmatrix}\\[0.3cm]
&=(R\oplus S)+(K_1+K_2^*-K_3).
\end{align}
