I want to show:
Let $H$ be a complex Hilbert space and $(P_n)$ a sequence of orthogonal projections (bounded linear operators $P_n:H\to H$ s.t. $P_n=P_nP_n={P_n}^*$). Suppose $P$ is a bounded operator and $P_n\to P$ strongly, i.e. for all $\psi\in H$ we have $P_n\psi\to P\psi$ in H. Then $P$ is an orthogonal projection.
There are several questions discussing the analogous statement being wrong for weak operator convergence. However I haven't been able to find a proof of the above. Here is what I have done:
Proof. Since the adjoint operation is continuous even in the WOT [THIS IS NOT APPLICABLE HERE, AS POINTED OUT BELOW. Nonetheless, the self-adjointness of the limit is easy to prove], it suffices to show that $$\lim_{n\to\infty}\lim_{m\to\infty}P_n P_m \psi=\psi \qquad \text{for all }\psi\in H.$$
However, I'm stuck here. I know that the limits in the line above can be exchanged and that the left hand side exists. I would like to do a diagonal sequence argument but don't know how to show that those limits are equal. What am I missing?
Best Answer
Multiplication is continuous in SOT on bounded sets. In this case, we can show that $P_n^2\to P^2$ in SOT using the identity
$$P_n^2-P^2=P_n(P_n-P)+(P_n-P)P,$$ from which it follows that for all $x\in H$,
$$\|(P_n^2-P^2)x\|\leq \|(P_n-P)x\|+\|(P_n-P)Px\|,$$
and each term on the right goes to zero because $P_n\to P$ in SOT.
For self-adjointness, here's another question: SOT limit of Self-adjoint operators is self-adjoint?