# Weak convergence

Let $H$ be a Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and let $V,W$ be two closed subspaces. For $x_0\in H$ we may define the sequence of projections $$x_{2n+1}=P_W(x_{2n}), \qquad x_{2n+2}=P_V(x_{2n+1}).$$ I intend to prove that if $V\cap W=\{0\}$, then $x_n\rightarrow 0$.

In the first place, one can easily show that $||x_{n+1}||^2=\langle x_{n+1},x_n\rangle$, from which one easily deduces that the sequence of norms is decreasing, so that $\lVert x_n\rVert\to\ell\geq 0$.

Then one can also show that if $V\cap W=\{0\}$ and a subsequemce $\{x_{2n_k}\}_k$ converges weakly to some $x$, then the sequence $\{x_{2n_k+1}\}_k$ also converges weakly to $x$, whence $x=0$.

So now it suffices to prove that a subsequence $\{x_{2n_k+1}\}$ converges weakly to something. For this, I first note that $$\lVert x_n\rVert^2=\langle x_{n+1},x_{n-2}\rangle=\cdots=\langle x_{2n-1},x_0\rangle$$ And now I would like to use that the sequence of norms converges and Riesz's representation theorem to conclude weak convergence but I am unsure as to how to proceed, since while it is true that any functional in $H$ is of the for $\langle \cdot,x_0\rangle$, varying $x_0$ also varies the sequence $\{x_{2n-1}\}$.

Thanks in advance for any insight.

• Alaoglu's theorem will guarantee a weakly convergent subsequence. – Nate Eldredge May 1 '12 at 23:24
• @NateEldredge: why a subsequence as opposed to a subnet? – user16299 May 2 '12 at 1:03
• @Yemon: because we're only working in the span of the $x_{n+1}$, so we have metrizability of the unit ball in the weak$^\ast$ topology by separability. – t.b. May 2 '12 at 1:05
• @t.b. of course - thanks! – user16299 May 2 '12 at 3:58

Your result is a particular case of something more general: if $P,Q$ are projections onto closed subspaces $M$ and $N$ of $H$ respectively, then $$\lim_{n\to\infty}(QPQ)^n=P\wedge Q,$$ where $P\wedge Q$ is the projection onto $M\cap N$ and the limit is taken in the strong operator topology (i.e. with the pointwise convergence).
To see this, fix any $z\in H$; then $$\langle (QPQ)^{2n+1}z,z\rangle=\langle (QPQ)^n(QPQ)(QPQ)^{n}z,z\rangle=\langle (QPQ)(QPQ)^{n}z,(QPQ)^nz\rangle=\langle PQ(QPQ)^{n}z,PQ(QPQ)^nz\rangle= \|PQ(QPQ)^nz\|^2\leq\|Q(QPQ)^nz\|^2=\|(QPQ)^nz\|^2 =\langle(QPQ)^nz,(QPQ)^nz\rangle=\langle(QPQ)^{2n}z,z\rangle.$$ This shows that $(QPQ)^{2n+1}\leq(QPQ)^{2n}$ as operators. A very similar argument shows that $(QPQ)^{2n+2}\leq(QPQ)^{2n+1}$. So the sequence of selfadjoint operators $\{(QPQ)^n\}$ is monotone, and so it converges in the strong operator topology to a selfadjoint operator $R$.
Edit: here's a short justification of the last sentence. That $\{(QPQ)^n\}_n$ is monotone decreasing (and positive) implies that for any $z\in H$ the sequence of numbers $\{\langle(QPQ)^nz,z\rangle\}_n$ is positive and decreasing, so convergent to a number $R_{x,x}$; using polarization, one gets that $\langle(QPQ)^nz,w\rangle\to R_{z,w}$ in such a way that the map $(z,w)\mapsto R_{z,w}$ is a sesquilinear form; it is then standard that any sesquilinear form defines an operator, which we call $R$ in this case.
We also have, since the sequence is bounded, that the product of the limit is the limit of the product, so $$R^2=\lim_{n\to\infty}(QPQ)^n\lim_{n\to\infty}(QPQ)^n =\lim_{n\to\infty}(QPQ)^n(QPQ)^n=\lim_{n\to\infty}(QPQ)^{2n}=R.$$ This shows that $R$ is a projection. Also, since it is the infimum of the sequence, $$R\leq QPQ\leq Q^2=Q,$$ and similarly $R\leq P$. So $R\leq P\wedge Q$. On the other hand, $P\wedge Q\leq P$ and $P\wedge Q\leq Q$, so $$PQP\geq P(P\wedge Q)P=P\wedge Q,$$ and similarly $(PQP)^n\geq P\wedge Q$ for all $n$. So $P\wedge Q\leq R$. Thus $R=P\wedge Q$.