$X_i =P_i X_j P_i$ implies that $X_i$ converges in SOT?

Question

Let $X_i$ be a uniformly bounded net in $B(H)$, the set of all bounded linear operators on a hilbert space $H$. $P_i$ is a net of projections increasing to $1$. For any $j> i$, we have $X_i =P_i X_j P_i$. So, can I say that $X_i$ converges in the strong operator topology? I saw similar assertion in one paper without explanation.

Answer 1

I found a possible answer. Since $X_i$ is uniformly bounded in $B(H)$, it follows that there is a subnet of $\{X_i\}$ converging to $X\in B(H)$ in WOT. We assert that $ X_i =P_i X P_i$ for every $i$. Indeed, for every $i$, we can find a subnet $\{X_{i_k}\}$ of the subnet such that $i_k \ge i$ for every $k$. Since $\{X_{i_k}\}$ converging to $X$ in WOT, it follows that $\{P_iX_{i_k}P_i\}$ converging to $P_iXP_i$ in WOT. Note that $P_iX_{i_k}P_i = X_i$ for every $k$. We obtain that $X_i =P_i X P_i$.
Now, we only need to show that $X_i \rightarrow X$ in SOT. Since $X_i =P_i X P_i$, we have $$\|(X_i -X)\eta\|_H = \|(P_i X P_i -X)\eta\|_H \le \|(P_i X P_i - P_iX)\eta\|_H +\|(P_i X -X)\eta\|_H \le \|P_iX\|_\infty \|(P_i - 1)\eta\|_H + \|(P_i - 1)X\eta\|_H .$$ Note that $\|P_iX\|_\infty \le \|X\|_\infty <\infty.$ We obtain that $\|(X_i -X)\eta\|_H \rightarrow_i 0$.