Equivalence of strong operator and weak operator convergence for a sequence of orthoprojectors

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I want to prove the equivalence of strong operator convergence and weak operator convergence for a sequence of orthoprojectors {$P_n$}. These orthoprojectors act from one Hilbert space H into subspaces $H_n$ of the space H, which, as I understand it, do not have to coincide for different n.

Of course, if the dimension of the space H is finite, then a weak one will follow from strong convergence, (since then the dimensions of all $H_n$ are finite) but I do not quite understand how to prove that a strong convergence implies a weak one in the case when H is infinite-dimensional.

To prove that a weak convergence will be followed by a strong one, I think it's worth trying to work with scalar products. After all, we can consider the functional f(x) = <x,y> (Where <x,y> denotes a scalar product, and y is an arbitrary vector from H). Then f($P_n$(x)) converges to f(P(x)), that is, <$P_n$(x),y> converges to <P(x),y>, which implies that <$P_n$(x)-P(x),y> converges to 0 for any y of H. And from this we can conclude that $P_n$(x)-P(x) converges to zero, that is, $P_n$(x) converges to P(x).

Can you tell me how it can be proved that a strong convergence leads to a weak one in the infinite-dimensional case and to what extent the reasoning in the opposite direction is correct? After all, I did not use the properties of orthoprojectors in any way in it, and this is very strange. If my reasoning is wrong in the opposite direction, can you tell me how it can be proved?

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$\|P_nx-Px\|^{2}= \langle P_nx, P_nx \rangle +\langle Px, Px \rangle-2\Re \langle P_nx, Px \rangle$. By properties of projections we can write this as $\langle P_nx, Px \rangle +\langle Px, Px \rangle -2\Re \langle x, P_nPx \rangle$. By wek convergence this converges to $\langle Px, Px \rangle +\langle Px, Px \rangle -2\Re \langle x, Px \rangle=0$ since $ \langle x, Px \rangle =\langle Px, Px \rangle$.

I have used the facts an orthogonal projection $Q$ has the properties $Q^{2}=Q$ and $Q^{*}=Q$.