Let $H$ be a Hilbert space with projections $P,Q$, then $\lim_{n\to\infty}(PQ)^nx= Rx$ for all $x\in H$, where $R$ is the projection to $PH\cap QH$

Question

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

We can clearly reduce to the case where $P\mathcal H\cap Q\mathcal H=0$ and $P\mathcal H+Q\mathcal H=\mathcal H$, in which case we must show that $(PQ)^n\to0$ strongly. Now, if the limit $y=\lim_{n\to\infty}(PQ)^n x$ 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.

Answer 1

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 $$[M\phi](x)=f(x)\phi (x)$$ 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, $P \mathcal{H} \cap Q \mathcal{H} = \{0\}$ implies that $\ker(M-I) = \{0\}$ 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 $|f^k \phi| \leq |\phi| \in L_\mu^2(\Omega)$.