How to prove the norm equality $\|T\|=$sup$\|P_nTP_n\|$ for Hilbert space operator $T$?

Question

Let $H$ be a separable Hilbert space and $P_1\leq P_2\leq ...$ be finite-rank projections in $B(H)$ such that $P_n$ has rank $n$ and $||P_n(h)-h||\rightarrow 0$ for all $h\in H$. How to prove $\|T\|=$sup$\|P_nTP_n\|$ for a bounded linear operator $T$ on $H$?

It is easy to see $\|T\|\geq$sup$\|P_nTP_n\|$. But how about the "$\leq$"?

Answer 1

Sketch: Assume $\|T\|<\infty$. Observe for each $x \in H$ and every $\varepsilon>0$ there exists $N=N(\varepsilon, x)$ such that for all $n\ge N$ we have that \begin{align} \| x-P_n x\|_H <\frac{\varepsilon}{\|T\|},\\ \| \underbrace{T x}_y-\underbrace{P_n Tx}_{P_ny}\|_H<\varepsilon. \end{align}

Next, observe \begin{align} Tx =&\ P_nTx+(I-P_n)Tx\\ =&\ P_n TP_nx+P_nT(x-P_nx)+(I-P_n)Tx \end{align} which means \begin{align} \|Tx\|_H \le&\ \|P_nTP_n x\|_H+\|P_nT(x-P_nx)\|_H+\varepsilon \\ \le&\ \|P_nTP_n x\|_H+\|T(x-P_nx)\|_H+\varepsilon\\ \le&\ \|P_nTP_n x\|_H+2\varepsilon. \end{align}