Csdrhrt

Suppose we have a Hilbert space $mathcal H$, with projections $P,Q$, and let $R$ be the projection onto $Pmathcal Hcap Qmathcal H$, then I would like to show that $(PQ)^nto R$ strongly.

We can clearly reduce to the case where $Pmathcal Hcap Qmathcal H=0$ and $Pmathcal H+Qmathcal H=mathcal H$, in which case we must show that $(PQ)^nto0$ strongly. Now, if the limit $y=lim_{ntoinfty}(PQ)^n x$ exists, then if it is nonzero, then
$$|y| = |PQ y|<|Q y|le| y|$$ which is a contradiction, so once we have convergence, we're done. However, I don't see quite how to show convergence in a general infinite dimensional case.

I work in your reduced case and consider instead the operator $A = QPQ$ and note that
$$|(PQ)^nx| = |PA^{n-1}x| leq |A^{n-1}x|$$
so it will suffice to show that $A^n to 0$ strongly as $n to infty$.

The reason it is more convenient to consider $A$ is that $A$ is a self-adjoint operator and so we can apply the spectral theorem in the following form.

Theorem: Let $A$ be a bounded, self-adjoint operator on a Hilbert space $mathcal{H}$. Then there is a finite measure space $(Omega,Sigma,mu)$ and a real valued, essentially bounded function $f$ on $Omega$ and a unitary operator $U:mathcal{H} to L_{mu}^2(Omega)$ such that $$U^*MU = A$$ where $M$ is the multiplication operator defined by $$[Mphi](x)=f(x)phi (x)$$ We also have $|M| = |f|_{infty}$.

It is not so hard to check that if $A$ is additionally a positive operator, as in our case, then we must have $f geq 0$ a.e. and so since for $A = QPQ$ we have that $$1 geq |A| = |M| = |f|_infty,$$ we have here that $f(x) in [0,1]$ a.e.

Now it is clear that $P mathcal{H} cap Q mathcal{H} subseteq ker(A-I)$. In fact, this is an equality since $QPQx = x$ implies that
$$|x| = |QPQx| leq |PQ x| leq |Qx| leq |x|$$
and so $|Qx| = |x|$ and hence by, by pythagoras, we must have $Qx = x$. Then by a similar argument with
$$|x| = |QPQx| = |QPx| leq |Px| leq |x|$$
we see that $Px = x$ also so that $x in ker(A-I)$ implies that $x in P mathcal{H} cap Q mathcal{H}$.

In particular, $P mathcal{H} cap Q mathcal{H} = emptyset$ implies that $ker(M-I) = emptyset$ where $M$ is the multiplication operator associated to $A$. This means that $mu({s: f(s) = 1}) = 0$ so now $f(x) in [0,1)$ a.e.

We are finally ready to complete the proof. By conjugating with our unitary operators it is enough to show that $|M^k phi| to 0$ for every $phi in L_mu^2(Omega)$. This is easy.

$$|M^k phi|^2 = int_Omega |f^k phi|^2 to 0$$
by the D.C.T. since $|f^k phi| to 0$ and $|f^k phi| leq |phi| in L_mu^2(Omega)$.