Decomposition to compact operator and commutative operator in Hilbert space Let $H$ Hilbert space, and $H_n$ are increasing(i.e. $H_n\subset H_{n+1}$) finite dimensional subspaces such that $\bigcup _{n=1}^{\infty} H_n$ is dense in $H$. Suppose $P_n$ be projection $H$ to $H_n$ and bounded linear operator $T:H\to H $ s.t., $||TP_n-P_nT||\leq \frac{1}{2^{n+1}}$
Then there exist  linear operator $S$ which is commutative with all $P_n$, and compact operator $K$ s.t., $||K||<1$ , $T=S+K$
I know that $P_nP_m=P_mP_n$ for all $m,n$, so I guess $S$ is linear combination of $P_n$. But I can't pick up $S$ and $K$.
 A: Note that, by definition, the projections $P_n$ converge strongly to $I$. 
Let $r\in\mathbb N$ (to be determined later), and define
$$
Q_n=P_{n+r}-P_n. 
$$
The projections $Q_n$ are finite-rank, and pairwise orthogonal. 
Let 
$$
S=\sum_n Q_nTQ_n,\ \ \ \ K=T-S.
$$
Let us check first that $SP_n=P_nS$ for all $n$. It is obvious that $SQ_n=Q_nS$ for all $n$. We have 
$$
SP_n-P_nS=-(SQ_n-Q_nS)+SP_{n+r}-P_{n+r}S=SP_{n+r}-P_{n+r}S.
$$
Repeat the argument, to get 
$$
SP_n-P_nS=SP_{n+kr}-P_{n+kr}S,\ \ \ k\in\mathbb N. 
$$
As $P_n\nearrow I$, we get $SP_n-P_nS=0$. 
As for $K$, we have 
$$
K=T-\sum_n Q_nSQ_n=\sum_n TQ_n-Q_nTQ_n=\sum_n (TQ_n-Q_nT)Q_n.
$$
Also,
$$
\|TQ_n-Q_nT\|\leq\|TP_n-P_nT\|+\|TP_{n+r}-P_{n+r}T\|\leq\frac1{2^{n+1}}+\frac1{2^{n+r+1}}
$$
$$\tag1
\left\|\sum_{n=m+1}^\infty (TQ_n-Q_nT)Q_n\right\|
\leq\sum_{n=m+1}^\infty \|TQ_n-Q_nT\|\leq\sum_{m+1}^\infty \frac1{2^{n+1}}+\frac1{2^{n+r+1}}\xrightarrow[m\to\infty]{}0
$$
The estimate $(1)$ shows that $K$ is of the form $\sum_{n=1}^m(TQ_n-Q_nT)Q_n$, which is finite-rank, plus an arbitrarily small operator; that is, $K$ is a limit of finite-rank operators, and thus compact. 
Finally, using $(1)$ with $m=1$, we get 
$$
\|K\|\leq\sum_{m=1}^\infty \frac1{2^{n+1}}+\frac1{2^{n+r+1}}
=\frac12+\frac1{2^{r+1}}.
$$
So any $r\geq3$ will give us $\|K\|<1$. 
As a final note, a very small tweak of the argument allows one to get $\|K\|<\varepsilon$ for any fixed $\varepsilon>0$. 
