Why is the WOT-limit of alternating projections a projection in uniformly convex Banach spaces?

Question

In this question it was shown that in uniformly convex spaces the intersection of two $1$-complemented subspaces is still $1$-complemented.

Let $X$ be a uniformly convex Banach space, let $Y$ and $W$ be $1$-complemented subspaces in $X$, that is, there exist continuous projections $P: X \to Y$ and $Q: X \to W$ such that $\|P\| \leq 1$ and $\|Q\| \leq 1$.

Define a sequence of operators $(P_n)$ on $X$ by the method of alternating projections:

$$ P_1 = P, \quad P_2 = QP, \quad P_3 = PP_2 = PQP, \quad P_4 = QP_3 = QPQP, \quad \text{and so on}. $$

That is,

$$ P_{2n} = QP_{2n-1}, \quad P_{2n+1} = PP_{2n}. $$

Let $x \in X$, and consider the sequence $\|P_n x\|$. Because $\|P\| = \|Q\| = 1$, this sequence is non-increasing and bounded below, hence it converges. Denote the limit by $a(x) = \lim \|P_n x\|$.

One can show that:

$$ \|P_{n+1}x - P_n x\| \to 0. $$

Indeed, $\frac{1}{2} \|P_{n+1}x + P_nx\|$ also converges to $a$, so the claim follows from the uniform convexity of $X$. Let $V$ be a limit in the weak operator topology of some subnet of $P_{2n}$. By the claim, the corresponding subnet of $P_{2n+1}$ also converges to $V$ in the weak operator topology. From this, it is evident that $V$ is a norm one projection onto $PX \cap QX$.

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Question:

It is a standard fact that the WOT-limit of a sequence of projections is not necessarily a projection, see this question. For example, one can construct sequences of orthogonal projections in Hilbert spaces whose WOT-limits are not idempotent. So this leads me to ask:

Why is the WOT-limit $V$ of this particular sequence $(P_n)$ a projection onto $PX \cap QX$?

How can I prove that it is a projection onto $PX \cap QX$?

I would like to understand the details of why this works in this setting

Thanks in advance for any insight!

Answer 1

It's not just that we have a wot limit of projections. Without other requirements, for operators on an infinite-dimensional Hilbert spaces, the norm-one idempotents are wot-dense in the unit ball of positive operators. But the sequence in your question is more specialized.

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First a bit of background. Since $X$ is uniformly convex, it is reflexive. This implies that the unit ball of $B(X)$ is compact in the weak operator topology (wot). The wot is the topology in $B(X)$ where $T_j\to T$ means that $\varphi(T_jx)\to\varphi(Tx)$ for all $x\in X$ and $\varphi\in X^*$. The reason the unit ball of $B(X)$ is wot-compact is that it can be seen that $B(X)$ is a dual and that the weak$^*$-topology is the wot; hence Banach-Alaoglu applies. The predual is the space $$ D=\overline{\operatorname{span}}^{\|\cdot\|}\{g_{x,\varphi}:B(X)\to\mathbb C\}, $$ where $g_{x,\varphi}(T)=\varphi(Tx)$. With a bit of work one can show that $D^*$ can be naturally identified with $B(X)$ via $\hat T(g_{x,\varphi})=g_{x,\varphi}(T)$ and that the weak$^*$ topology is precisely the wot.

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Now the actual answer. We know that $\|P_xx\|\to a(x)$ since the sequence is positive and non-increasing. If $a(x)=0$, this means that $P_nx\to0$ and so the sequence converges. So we assume $a(x)>0$.

We have $P_{2k}=(QP)^k$, $P_{2k+1}=PP_{2k}$. Then \begin{align} \|P_{2k+1}x\|&=\frac12\|P_{2k+1}x+PP_{2k}x\|=\frac12\,\|P(P_{2k+1}x+P_{2k}x)\|\leq\frac12\,\|P_{2k+1}x+P_{2k}x\|\\[0.2cm] &\leq\frac12\Big(\|P_{2k+1}x\|+\|P_{2k}x\|\Big). \end{align} This shows that $\frac12\|P_{2k+1}x+P_{2k}x\|\to a(x)$. Working similarly with the case $P_{2k+2}x+P_{2k+1}x$, we get that $$ \frac12\,\|P_{n+1}x+P_nx\|\to a(x). $$ Normalizing by $a(x)$, we get two sequences $\{x_n\}$ and $\{y_n\}$ such that $\|x_n\|\to1$, $\|y_n\|\to1$ and $\frac12\|x_n+y_n\|\to1$. The uniform convexity then gives us $\|x_n-y_n\|\to0$. That is, $$\tag1 \|P_{n+1}x-P_nx\|\to0. $$

Let $R$ be a cluster point of the sequence $\{P_{2n}\}$ in the weak operator topology. That is, there exists a subnet $\{P_{2n_k}\}$ such that $P_{2n_k}x\to R$ weakly for all $x\in X$. Since $P_{2n_k+1}x-P_{2n_k}x\to0$, we also get that $P_{2n_k+1}x\to Rx$ weakly for all $x$. From these two limits we get that $PR=QR=R$. Knowing that $PR=QR=R$, it follows that $P_{n_k}R=R$ for all $k$. Taking the wot limit we get that $R^2=R$. Also, $$ \varphi(Rx)=\lim_{n_k}\varphi(P_{n_k}x)\leq\|\varphi\|\,\|x\|, $$ so $\|R\|\leq1$ and, being a projection, $\|R\|=1$.

Finally, if $x\in PX\cap QX$, then $P_{n_k}x=x$ for all $k$, and therefore $Rx=x$; that is, $PX\cap QX\subset RX$. But $PR=RQ=R$, which implies that $RX\subset PX\cap QX$. Therefore $R$ is a norm-one projection onto $PX\cap QX$.

Answer 2

There is a general theorem about limits of alternating projections that is due to Bruck and Reich [1]:

Theorem. Suppose that $X$ is uniformly convex. Let $P_1, \ldots, P_k$ be linear projections with norm one on the closed subspaces $M_1, \ldots, M_k$ respectively. Then, the strong limit $$ \lim_{n \to \infty} (P_k \cdots P_1)^n x = Px $$ of alternating projections exists for every $x \in X$ and defines a linear projection $P$ with norm one on the intersection of the subspaces $\bigcap_{j=1}^k M_j$.

[1]: Bruck, Ronald E., and Simeon Reich. "Nonexpansive projections and resolvents of accretive operators in Banach spaces." Houston J. Math 3.4 (1977): 459-470.