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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?

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    $\begingroup$ The adjoint operation is not continuous in SOT. If $V$ is the unlateral shift, then $(V^*)^n\to 0$ in SOT, but $V^n$ does not converge. $\endgroup$ – Jonas Meyer Mar 7 '17 at 3:46
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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?

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  • $\begingroup$ What is meant by "Multiplication is continuous on bounded sets"? I have heared similar statements with compact sets but don't understand what it means. What sets are bounded here? Also how is that relevant to the proof you presented? $\endgroup$ – Adomas Baliuka Mar 7 '17 at 4:08
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    $\begingroup$ Multiplication of operators is a binary operation on $B(H)$, hence a map $B(H)\times B(H)\to B(H)$. It is not a continuous map in SOT, although it is continuous in the norm topology. Reference: math.stackexchange.com/questions/687067/…. However, if you restrict to bounded subsets of $B(H)\times B(H)$, multiplication is continuous in SOT. (In this case perhaps this remark is overkill because for sequences there is no problem by Uniform Boundedness Principle, as mentioned in the link.) $\endgroup$ – Jonas Meyer Mar 7 '17 at 4:13
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    $\begingroup$ "how is that relevant" It isn't used in my proof, it is just a remark intended to provide context for how greatly it generalizes. If you have $A_n\to A$ and $B_n\to B$ in SOT, then can you expect $A_n B_n\to AB$? Yes, if $(A_n)$ and $(B_n)$ are bounded. In this case, $A_n=B_n = P_n$, and $(P_n)$ is certainly bounded. The generalization would use a similar trick, $A_nB_n-AB=A_n(B_n-B) + (A_n-A)B$ then apply to $x$, then bound. $\endgroup$ – Jonas Meyer Mar 7 '17 at 4:15

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