When products of projections diverge

TL;DR

This paper explores the relationship between the convergence and divergence of cyclic and random products of projections in Hilbert spaces, providing a characterization of when random products converge based on geometric and combinatorial structures.

Contribution

It establishes a parallel to known cyclic projection results by characterizing families of subspaces for which all random products converge.

Findings

Cyclic projections always converge in norm.

Random projections can diverge even when cyclic ones converge.

The paper provides a geometric and combinatorial characterization of convergence for random products.

Abstract

Slow convergence of cyclic projections implies divergence of random projections and vice versa. Let $L_1,L_2,\dots,L_K$ be a family of $K$ closed subspaces of a Hilbert space. It is well known that although the cyclic product of the orthogonal projections on these spaces always converges in norm, random products might diverge. Moreover, in the cyclic case there is a dichotomy: the convergence is fast if and only if $L_1^{\perp}+\dots+L_K^{\perp}$ is closed; otherwise the convergence is arbitrarily slow. We prove a parallel to this result concerning random products: we characterize those families $L_1,\dots,L_K$ for which all random products converge using their geometric and combinatorial structure.

Equations

$$z_n=P(L_{k_n})z_{n-1},$$

$$w_n=P(X_{k_n})w_{n-1}=P(L_{k_n})w_{n-1}.$$

$$I(m,\delta )=\{k\in [K]:B(z_m,\delta )\cap L_k\ne \varnothing \}.$$

$$0<A:=\underset{N\to \infty }{\lim }\sup \{|z_n-z_m|:N\le m\le n\}.$$

$$|z_i-z_{i+1}{|}^2=|z_i{|}^2-|z_{i+1}{|}^2\le {\delta }^2/4.$$

$$|z_{n-1}-z_m|\ge |z_n-z_m|-|z_{n-1}-z_n|>\delta -\delta /2.$$

$$|z_n-z_m{|}^2\le |z_n{|}^2-|z_m{|}^2\le {\delta }^2/4,$$

$$\underset{n\to \infty }{\lim }\max \{\mathrm{dist}(w_n,L_k):k\in I\}=0.$$

$$|z_i-z_{i+1}|\le \delta .$$

$$B(z_m,\delta )\cup B(z_{m+1},\delta )\subset B(z_m,2\delta ),$$

$$\{1,2,3\}=I(N,\delta )=I(N+1,\delta )=I(N+2,\delta )=\dots ,$$

$$\{1,2,3,4\}\subset I(m_n,1/n)$$

$$\underset{n\to \infty }{\lim }\max \{\mathrm{dist}(w_n,L_k):k\in I\}=0$$

$$\{k_n{\}}_{n=M_i}^{N_i}=[K].$$

$$\begin{array}{rl}{V}_i^m& =\{A\subset I(m,2/i):3\le |A|\},\\ V_i& =\bigcup _{m=M_i}^{N_i}{V}_i^m,\\ E_i& =\{\{A,B\}\in (\genfrac{}{}{0ex}{}{V_i}{2}):A\ne B\text{mbox}and3\le |A\cap B|\}.\end{array}$$

$$\bigcup V_i=[K].$$

$$\underset{n\to \infty }{\lim }\max \{\mathrm{dist}(z_{m_n},L_i):i\in A\}\le \underset{n\to \infty }{\lim }\max \{\mathrm{dist}(z_{m_n},{\tilde{L}}_i):i\in A\}=0$$

$$\Vert \psi (A_1,\dots ,A_n)E-\psi (B_1,\dots ,B_n)E\Vert \le \sum _{1\le i\le n}|{\psi }_i|\Vert A_{i}E-B_{i}E\Vert .$$

$$\bigl{\|}\psi(\tilde{Z},\tilde{X},\tilde{Y})u-v\bigr{\|}\leq 4\varepsilon.$$

$$\begin{split}\bigl{\|}&\psi(\tilde{Z},\tilde{X},\tilde{Y})P(E)-\psi(W,X,Y)P(E)\bigr{\|}\leq N\cdot\|P(W)-P(\tilde{Z})P(E)\|\\ &+|\psi|\cdot\|P(X)-P(\tilde{X})P(E)\|+|\psi|\cdot\|P(Y)-P(\tilde{Y})P(E)\|\\ &\leq 3\varepsilon;\end{split}$$

$$|\psi (\tilde{Z},\tilde{X},\tilde{Y})u-v|\le |\psi (\tilde{Z},\tilde{X},\tilde{Y})u-\psi (W,X,Y)u|+|\psi (W,X,Y)u-v|\le 4\epsilon .$$

$$g\in G_{R,t}=\underset{w_n\notin E_t}{\bigvee }R(w_n),$$

$$|{\psi }^j(\tilde{Z},\tilde{X},\tilde{Y})e_j-e_{j+1}|<30{\epsilon }_j.$$

$$\Vert P({\tilde{X}}_j)+P(\tilde{X}\cap G_{S,j})-P(\tilde{X})\Vert \le 4{\epsilon }_j/|{\psi }^j|.$$

$$\begin{array}{rl}\Vert P(\tilde{X})P(E_j)-P(X_j)\Vert \le & \Vert P({\tilde{X}}_j)+P(\tilde{X}\cap G_{S,j})-P(\tilde{X})\Vert \\ & +\Vert P({\tilde{X}}_j)-P(X_j)\Vert +\Vert P(\tilde{X}\cap {\tilde{G}}_{S,j})P(E_j)\Vert \\ & \le 6{\epsilon }_j/|{\psi }^j|.\end{array}$$

$$\Vert P({\tilde{Y}}_j)+P(\tilde{Y}\cap G_{T,j})-P(\tilde{Y})\Vert \le 4{\epsilon }_j/|{\psi }^j|,$$

$$\Vert P(\tilde{Y})P(E_j)-P(Y_j)\Vert \le 6{\epsilon }_j/|{\psi }^j|.$$

$$\begin{array}{rl}\Vert & P(\tilde{Z})P(E_j)-P(W_j)\Vert \le 13{\epsilon }_j/N_j,\end{array}$$

$$\bigl{\|}\psi_{j}(\tilde{Z},\tilde{X},\tilde{Y})e_{j}-e_{j+1}\|<30\varepsilon_{j}.$$

$$\Vert P({\tilde{Z}}_j)+P(\tilde{Z}\cap G_{U,j})-P(\tilde{Z})\Vert \le 4{\epsilon }_j/|{\psi }^j|.$$

Full text

Let $L_{1},L_{2},\dots,L_{K}$ be a family of $K$ closed subspaces of a Hilbert space. It is well known that although the cyclic product of the orthogonal projections on these spaces always converges in norm, random products might diverge. Moreover, in the cyclic case there is a dichotomy: the convergence is fast if and only if $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is closed; otherwise the convergence is arbitrarily slow.

We prove a parallel to this result concerning random products: we characterize those families $L_{1},\dots,L_{K}$ for which all random products converge using their geometric and combinatorial structure.

Research partially supported by NSF Grant No. 1440140 while the author was visiting MSRI, Berkeley, California in 2017

Introduction

Let $K$ be a fixed natural number and let $L_{1},L_{2},\dots,L_{K}$ be a family of $K$ closed subspaces of a Hilbert space $H$ such that, for simplicity, $\bigcap_{i=1}^{K}L_{i}=\{0\}$. Let $z_{0}\in H$ and $k_{1},k_{2},\dots\in[K]=\{1,2,\dots,K\}$ be an arbitrary sequence in which each $k\in[K]$ appears infinitely often. Consider the sequence of vectors $\{z_{n}\}_{n=1}^{\infty}$ defined by

[TABLE]

where $P(X)$ denotes the orthogonal projection of $H$ onto the subspace $X$. Then $\{z_{n}\}$ is a weakly-null sequence according to [AA]. If $H$ is infinite dimensional and $K\geq 3$, then the sequence $\{z_{n}\}$ does not, in general, converge in norm as exhibited in [KM, KP].

If the sequence $\{k_{n}\}$ is periodic, the product of the projections is called cyclic. In the cyclic case the sequence $\{z_{n}\}$ always does converge in norm [N, Ha]. The convergence is, however, either exponentially fast, or arbitrarily slow, depending on whether $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is closed or not. This dichotomy result was obtained in [BDH, DH] and independently in [BaGM1, BaGM2]; see also [BaS1, BaS2] for recent development. In this paper we show how slow convergence of cyclic iterates of projections corresponds to existence of non-converging iterates.

For the general, not necessarily cyclic case Bauschke showed in [B] that if $\sum_{j\in J}L_{j}^{\perp}$ is closed for each nonempty $J\subset[K]$, then $\{z_{n}\}$ converges in norm. In Example LABEL:slownono we show that a condition this strong is not necessary for the convergence.

Deutsch and Hundal ask in [DH] if just $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ being closed is enough to guarantee the norm convergence of $\{z_{n}\}$ in the general case. In Theorem LABEL:3-eq-positive we show that this is indeed the case for $K=3$ and $K=4$. For $K\geq 5$, we exhibit in Example LABEL:5 that this is not the case in general. For $K\geq 4$, we give in Theorem LABEL:hypothesis a condition which is equivalent to the norm convergence of $\{z_{n}\}$ and which involves connected subgraphs of the Johnson graph $J(K,4)$.

Here is an overview of the various conditions and their relations; for simplicity we assume that $\bigcap L_{k}=\{0\}$.

1. (i)

$L_{1}^{\perp}+L_{2}^{\perp}+\dots+L_{K}^{\perp}$ is closed in $H$; 2. (ii)

if $T=P(L_{K})\dots P(L_{2})P(L_{1})$, then $\|T^{n}\|\leq Cr^{n}$ for some $r\in[0,1)$; 3. (iii)

for all closed subspaces $\tilde{L}_{i}\subset L_{i}\subset H$, each starting point $z_{0}\in H$ and every sequence of indices $k_{1},k_{2},\dots\in\{1,\dots,K\}$, the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does converge in norm.

For $K=2$ condition (iii) is always satisfied [N]. If $K\geq 2$, then (i)$\Leftrightarrow$(ii) by [BDH, DH, BaGM1, BaGM2]. In Corollary LABEL:inspaces we prove that (iii)$\Rightarrow$(i) holds for $K\geq 3$ subspaces. In Theorem LABEL:3-eq-positive we show that (i)$\Leftrightarrow$(iii) for $K=3$ and $K=4$ subspaces.

Here is the complementary version of the above, enriched by the Johnson graphs condition (d) and examples (e) of subspaces that show that the condition (d) really takes place.

1. (a)

$L_{1}^{\perp}+L_{2}^{\perp}+\dots+L_{K}^{\perp}$ is not closed in $H$; 2. (b)

for every sequence $\{a_{n}\}\in c_{0}$ there exists a starting point $z\in H$ so that $|T^{n}z-P(\bigcap L_{k})z|\geq a_{n}$; 3. (c)

there exist closed subspaces $\tilde{L}_{i}\subset L_{i}\subset H$, a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in\{1,\dots,K\}$ so that the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does not converge in norm; 4. (d)

there exists $V\subset\binom{[K]}{4}$ such that $\bigcup V=[K]$, for $E=\{\{A,B\}\in\binom{V}{2}:\,|A\cap B|=3\}$ the graph $G=(V,E)$ is connected, and if $A\in V$, then $\bigcap_{i\in A}L_{i}$ is infinite dimensional, or $\sum_{i\in A}L_{i}^{\perp}$ is not closed; 5. (e)

for every $\emptyset\neq V\subset\binom{[K]}{4}$ so that $\bigcup V=[K]$, and for every $\alpha:V\to\{0,1\}$ there exist closed subspaces $L_{1},\dots,L_{K}\subset H$ with the following properties. If $A\in V$ and $\alpha(A)=0$, then $\bigcap_{i\in A}L_{i}$ is infinite dimensional. If $A\in V$ and $\alpha(A)=1$, then $\sum_{i\in A}L_{i}^{\perp}$ is not closed. If $B\in\binom{[K]}{4}\setminus V$, then $\bigcap_{i\in B}L_{i}$ is finite dimensional and $\sum_{i\in B}L_{i}^{\perp}$ is closed.

If $K\geq 2$, then (a)$\Leftrightarrow$(b) by [BDH, DH, BaGM1, BaGM2]. In Corollary LABEL:inspaces we prove that (a)$\Rightarrow$(c) holds for $K\geq 3$ spaces. In Theorem LABEL:3-eq-positive we show that (a)$\Leftrightarrow$(c) for $K=3$ and $K=4$ spaces if $\bigcap L_{k}=\{0\}$. In Theorem LABEL:hypothesis we show that (c)$\Leftrightarrow$(d) for $K\geq 4$. If $H$ is infinite dimensional and $K\geq 4$ then according to Proposition LABEL:johnbio, (e) is satisfied. That is, for each induced subgraph $G$ of the Johnson graph $J(K,4)$, there are $K$ closed subspaces of an infinite dimensional Hilbert space with the geometry of all their 4-tuples corresponding to $G$.

The paper is organized as follows. In Section LABEL:almortho we recall the closedness of $L_{1}^{\perp}+L_{2}^{\perp}+\dots+L_{K}^{\perp}$ as a geometric tool suitable for examining the convergence of products of projections. In Section LABEL:SJ we consider subspaces $L_{1},\dots,L_{K}$ for which there is a non-convergent sequence of products of projections. We use Johnson graphs to describe the family of sets of indices $A\subset[K]$ for which $\sum_{i\in A}L_{i}^{\perp}$ is not closed or $\bigcap_{i\in A}L_{i}$ is infinite dimensional, and answer a question of Deutsch and Hundal [DH]. Let $K\geq 3$ and let $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ be not closed. In Section LABEL:K34 we construct $\tilde{L}_{k}\subset L_{k}$, $k\in[K]$, and a product of projections on the spaces $\tilde{L}_{k}$ which does not converge in norm. For $K=3$ and $K=4$, we show in Theorem LABEL:3-eq that the existence of such $\tilde{L}_{k}$’s is equivalent to $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ not being closed. In Section LABEL:K5 we give for $K\geq 5$ a weaker condition involving Johnson graphs under which a non-converging product of projections exists. In Section LABEL:examples we give examples showing where the previously exhibited results cannot be strengthened or modified. In Section LABEL:appen we collect a couple of elementary tools used throughout the text.

Notation. In the entire paper $H$ is a Hilbert space; we will explicitly mention when we need $H$ to be infinite dimensional. For a closed subspace $X$ of $H$ we denote by $P(X)$ the orthogonal projection onto $X$. For $M,N\subset H$ we denote by $\bigvee M$ the closed linear span of $M$ and by $M\vee N$ the closed linear span of $M\cup N$. Similarly, we use $\vee x$ and $x\vee y$ for $x,y\in H$. By $|x|$ we denote the norm of $x$.

Let $\mathcal{B}(H)$ be the space of bounded linear operators from $H$ to $H$. For $m\in\mathbb{N}$, let $\mathcal{S}_{m}$ be the free semigroup with generators $a_{1},\dots,a_{m}$. If $\varphi=a_{i_{r}}\cdots a_{i_{1}}\in\mathcal{S}_{m}$ (for some $r\in\mathbb{N}$ and $i_{j}\in\{1,\dots,m\}$) and $A_{1},\dots,A_{m}\in\mathcal{B}(H)$, then we write $\varphi(A_{1},\dots,A_{m})=A_{i_{r}}\cdots A_{i_{1}}\in\mathcal{B}(H)$. If $X_{1},\dots,X_{m}$ are closed subspaces of $H$, then, by a slight abuse of notation, $\varphi(X_{1},\dots,X_{m})=P(X_{i_{r}})\cdots P(X_{i_{1}})\in\mathcal{B}(H)$ is the corresponding product of orthogonal projections.

Denote by $|\varphi|=r$ the “length” of the word $\varphi$ and by $|\varphi_{i}|$ or $|\varphi_{a_{i}}|$ the number of “occurrences” of $a_{i}$ in the word $\varphi$. Then $\sum_{i=1}^{m}|\varphi_{i}|=|\varphi|=r$.

We denote $[K]=\{1,\dots,K\}$. For a finite set $V$ and $k\in\mathbb{N}$ we denote by $\binom{V}{k}$ the set of all $k$-element subsets of $V$.

1. Common orthogonal structure

Suppose $L_{1},\dots,L_{K}$ are closed subspaces of a Hilbert space $H$. The existence of an orthonormal sequence which is almost contained in all of the spaces $L_{k}$ is the reason behind both the slow convergence of the cyclic products $(P(L_{K})\dots P(L_{1}))^{n}$ and the possible non-convergence of the random products of $P(L_{k})$’s.

Here is a brief heuristic. The larger the acute angle between two lines $L_{1}$ and $L_{2}$, the faster do the projections $(P(L_{2})P(L_{1}))^{n}z$ converge to the intersection of the lines. This simple geometric observation was developed in [BDH, DH, BaGM1, BaGM2] to the statement that if “the angles” between the subspaces $L_{1},\dots,L_{K}$ of a Hilbert space $H$ are positive, then $(P(L_{K})\dots P(L_{1}))^{n}$ converges “fast”. An economical way of stating that the “angles are positive” is to say that $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is closed. Bauschke and Borwein [BB] have given an equivalent geometric condition, $\neg$(ii) of the next lemma, that $L_{1},\dots,L_{K}$ are boundedly regular. In infinite dimensions even spaces which intersect only at the origin, can have “zero angle” between them. Geometrically this means that the spheres of these spaces almost touch (condition (ii) of Lemma LABEL:BB), or, more vividly, that the subspaces $L_{1},\dots,L_{K}$ almost share an orthonormal sequence (condition (iv) of Lemma LABEL:BB).

On the other hand, the iterative construction of [P, KM, KP] uses an orthonormal sequence $\{w_{n}\}$ to build three almost touching spaces $\tilde{L}_{1},\tilde{L}_{2},\tilde{L}_{3}$ for which certain random product of projections does not converge. If $\{w_{n}\}$ is very close to all of the spaces $L_{1},L_{2},L_{3}$, it is possible to build $\tilde{L}_{1}$ within $L_{1}$, and in general $\tilde{L}_{k}\subset L_{k}$.

Lemma 1.1**.**

Let $L_{1},\dots,L_{K}$, $K\in\mathbb{N}$ be closed subspaces of a Hilbert space $H$ and let $L=\bigcap_{k\in[K]}L_{k}$.

The following statements are equivalent:

1. (i)

$L_{1}^{\perp}+\dots+L_{K}^{\perp}$* is not closed in $H$;* 2. (ii)

there exists a bounded $A\subset H$ and $\varepsilon>0$ such that for all $\delta>0$, there exists $w\in A$ so that $\max\{{\rm dist\,}(w,L_{i}):\,i\in[K]\}<\delta$ and ${\rm dist\,}(w,L)>\varepsilon$; 3. (iii)

there exists a normalized weakly-null sequence $\{w_{n}\}$ in $L^{\perp}$ so that $\lim_{n\to\infty}\max\{{\rm dist\,}(w_{n},L_{i}):\,i\in[K]\}=0$; 4. (iv)

*there exists an orthonormal sequence $\{w_{n}\}$ in $L^{\perp}$ so *

that $\lim_{n\to\infty}\max\{{\rm dist\,}(w_{n},L_{i}):\,i\in[K]\}=0$. If $V\subset H$ is finite dimensional, then it is, moreover, possible to choose $\{w_{n}\}$ in $V^{\perp}$.

Proof.

The equivalence of (i) and (ii) appears in [BB]. If the negation of the property (ii) takes place, then the $K$-tuple of the subspaces $L_{1},\dots,L_{K}$ is in [BB] called boundedly regular.

Clearly, (iv) implies (ii) and (iii). Now assume (ii) is satisfied. To show (iii), for $\delta_{n}=1/n$, choose $w_{n}\in A$ so that $\max\{{\rm dist\,}(w_{n},L_{i}):\,i\in\{1,\dots,K\}\}<1/n$ and ${\rm dist\,}(w_{n},L)>\varepsilon$. Since $w_{n}-P(L)w_{n}$ also satisfies these inequalities, we can assume that $w_{n}\in L^{\perp}$. Since $A$ is bounded and the $L_{i}$’s are linear subspaces, by normalizing we can assume that $|w_{n}|=1$ for all $n$ and that $\{w_{n}\}$ converges weakly to some $w\in L^{\perp}$. Suppose $w\neq 0$. Then there is a weakly open halfspace $W\ni w$ so that, without loss of generality, $\bar{W}\cap L_{1}=\emptyset$. This means that $\lim_{n\to\infty}{\rm dist\,}(w_{n},L_{1})>0$, which is a contradiction.

Lemma LABEL:justforme0 implies that (iv) follows from (iii). ∎

We will also use the followig simple corollary.

Lemma 1.2**.**

Let $H$ be a Hilbert space and $L_{1},\dots,L_{K}$, $K\in\mathbb{N}$ its closed subspaces and let $L=\bigcap_{k\in[K]}L_{k}$. The following statements are equivalent:

1. (i)

$L$* is infinite dimensional, or $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is not closed in $H$;* 2. (ii)

there exists a normalized weakly-null sequence $\{w_{n}\}$ in $H$ so that $\lim_{n\to\infty}\max\{{\rm dist\,}(w_{n},L_{i}):\,i\in[K]\}=0$; 3. (iii)

*there exists an orthonormal sequence $\{w_{n}\}$ in $H$ so *

that $\lim_{n\to\infty}\max\{{\rm dist\,}(w_{n},L_{i}):\,i\in[K]\}=0$. If $V\subset H$ is finite dimensional, then it is, moreover, possible to choose $\{w_{n}\}$ in $V^{\perp}$.

Proof.

Assume (i) is satisfied. If $L$ is infinite dimensional, then it contains an orthonormal sequence, hence (ii) and (iii) are satisfied. If $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is not closed, then (ii) and (iii) are satisfied according to Lemma LABEL:BB. Clearly, (iii) implies (ii).

Assume that (ii) is satisfied and $L$ is finite dimensional. We write $w_{n}=x_{n}+y_{n}$, where $x_{n}\in L$ and $y_{n}\in L^{\perp}$. We may assume that $\{x_{n}\}$ converges in norm to $x\in L$, and $\{y_{n}\}$ converges weakly to $y\in L^{\perp}$. Then $\{w_{n}\}$ converges weakly to $0=x+y$; hence $x=y=0$ and $\lim_{n\to\infty}\max\{{\rm dist\,}(y_{n}/|y_{n}|,L_{i}):\,i\in[K]\}=0$. By the implication (iii)$\Rightarrow$(i) of Lemma LABEL:BB, $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is not closed. ∎

Here is an explanation, where normalized weakly-null sequences in $L^{\perp}$ naturally appear in connection to non-convergent products of projections.

Let $L_{k}$, $k\in[K]$, be as in the assumptions of Lemma LABEL:seqeq. We write each $L_{k}=L+X_{k}$, where $X_{k}$ is a closed subspace of $L^{\perp}$. Then $P(L_{k})=P(L)+P(X_{k})$ and $\bigcap_{k\in[K]}X_{k}=\{0\}$. Let $z_{0}=y+w_{0}$ for $y\in L$ and $w_{0}\in L^{\perp}$. Assume that $\{z_{n}\}$ is defined as in (LABEL:iter). Then $z_{n}=y+w_{n}$, where $w_{n}\in L^{\perp}$ is defined iteratively by

[TABLE]

If $\{z_{n}\}$ does not converge in norm, then neither does $\{w_{n}\}$. By [AA], $\{w_{n}\}$ is weakly null. Since the sequence $\{|w_{n}|\}$ is decreasing, there exists $\lim_{n\to\infty}|w_{n}|=c>0$. By considering $z_{0}/c$ instead of $z_{0}$, we can assume that $\lim_{n\to\infty}|w_{n}|=1$.

2. Divergent products and Johnson graphs

In this section we consider subspaces for which there is a divergent sequence of products of projections, and examine when the sum of the orthogonal complements of the subspaces is not closed.

Let $L_{1},\dots,L_{K}$, be closed subspaces of a Hilbert space $H$. Suppose there is a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm. By [N], $K\geq 3$. If $K=3$ or $K=4$, Corollary LABEL:n34 shows that $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is not closed, answering thus positively a question of Deutsch and Hundal [DH]. For $K\geq 5$, this is in general no longer true, but still $\sum_{i\in A}L_{i}^{\perp}$ is not closed for a rich family of sets $A\subset[K]$. To describe the structure of this family, Johnson graphs turn out to be the right tool.

Let $K\geq 4$ be a natural number. The vertices of the Johnson graph $J(K,4)$ are the 4-element subsets of [K]; two vertices are adjacent when the intersection of the two vertices (subsets) contains 3 elements.

In Theorem LABEL:johnson we show that there is a connected induced subgraph $G=(V,E)$ of $J(K,4)$ so, that $\bigcup V=[K]$ and if $A\in V$, then $\sum_{i\in A}L_{i}^{\perp}$ is not closed or $\bigcap_{i\in A}L_{i}$ is infinite dimensional. Conversely, in Proposition LABEL:johnbio we construct for each subgraph $G$ as above a corresponding configuration of subspaces of $H$.

Next we prove that a diverging sequence of iterates of projections needs at least three subspaces to “move along”.

Let $L_{1},\dots,L_{K}$ be closed subspaces of a Hilbert space $H$, $z_{0}\in H$ and $z_{n}=P(L_{k_{n}})z_{n-1}$ for a sequence of indices $k_{1},k_{2},\dots\in[K]$. We denote

[TABLE]

Lemma 2.1**.**

Let $L_{1},\dots,L_{K}$ be closed subspaces of a Hilbert space $H$ so that $\bigcap L_{k}=\{0\}$. Suppose there is a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm. Then for every $\delta>0$, there exists $M\in\mathbb{N}$ so that $|I(m,\delta)|\geq 3$ for every $m\geq M$.

Proof.

The orthogonal projection is linear, the sequence $\{|z_{n}|\}$ is decreasing, on the other hand the sequence $\{z_{n}\}$ does not converge in norm, so we can assume that $\lim|z_{n}|=1$. Since $\bigcap L_{k}=\{0\}$, according to [AA], the sequence $z_{n}$ converges weakly to zero.

Since $\{z_{n}\}$ does not converge in norm,

[TABLE]

We can assume $0<\delta<A/2$ and choose $M$ so that $|z_{i}|^{2}<1+\delta^{2}/4$ for all $M\leq i$. Then

[TABLE]

For a given $m>M$ choose the smallest $n>m$ so that $|z_{n}-z_{m}|>\delta/2$. Then $|z_{n}-z_{m}|<\delta$ by (LABEL:1step), as otherwise

[TABLE]

Hence $z_{m},z_{m+1},\dots,z_{n}\in B(z_{m},\delta)$. If all of the points $z_{m},z_{m+1},\dots,z_{n}$ consist of projections on only two of the subspaces $L_{1},\dots,L_{K}$, then by e.g. [KR], we have

[TABLE]

which is a contradiction. ∎

If $K\geq 4$ spaces are involved infinitely often in a diverging sequence of iterates of projections, then at least four of the spaces have to be arbitrarily close together. In Example LABEL:5 we show that for five spaces this is no longer true.

Proposition 2.2**.**

Let $L_{1},\dots,L_{K}$ be closed subspaces of a Hilbert space $H$, denote $L=\bigcap L_{k}$. Suppose there is a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm. Then there is $I\subset[K]$ with $|I|=\min\{4,K\}$ and a normalized weakly-null sequence $w_{n}$ in $L^{\perp}$ so that

[TABLE]

Proof.

As explained at the end of Section LABEL:almortho, we can assume that $\{z_{n}\}$ is a weakly-null sequence in $L^{\perp}$, and $\lim|z_{n}|=1$. Notice that according to von-Neumann’s theorem, only $K\geq 3$ is possible. If $K=3$, then by Lemma LABEL:3step it is enough to put $w_{n}=z_{n}/|z_{n}|$.

Suppose $4\leq K$. Let $0<\delta$ and $N\in\mathbb{N}$ be given. We will show that there is $m(\delta,N)>N$ so that $4\leq|I(m(\delta,N),2\delta)|$. Indeed, we can assume that for a given $\delta$ the given $N$ is at least as large as the $M$ of Lemma LABEL:3step and that for all $i\geq N$, we have

[TABLE]

For a contradiction assume that $|I(m,2\delta)|=3$ for all $m\geq N$. Since for all $m\geq N$,

[TABLE]

$3=|I(m,\delta)|=|I(m+1,\delta)|\leq|I(m,\delta)\cup I(m+1,\delta)|\leq|I(m,2\delta)|=3$. Then, without loss of generality,

[TABLE]

which means that $4$ appears in the given sequence $\{k_{i}\}$ only finitely many times. This contradicts our assumptions.

Consequently, we can choose an increasing sequence $m_{n}\geq n$ so that $4\leq|I(m_{n},1/n)|$. Since there are only finitely many 4-tuples of $K$ numbers, hence by dropping to a subsequence we can assume that for all $n\in\mathbb{N}$,

[TABLE]

and define $w_{n}=z_{m_{n}}/|z_{m_{n}}|$. ∎

Deutsch and Hundal ask in [DH] if $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ being closed is enough to guarantee the norm convergence of any product of projections onto the $L_{k}$’s. The next corollary shows that for $K=3$ and $K=4$ this is indeed the case. In Theorem LABEL:3-eq we modify the two conditions in the corollary so that they are actually equivalent.

Corollary 2.3**.**

Let $H$ be a Hilbert space, $K=3$ or $K=4$, and let $L_{1},\dots,L_{K}$ be closed subspaces of $H$. Suppose

1. (i)

there exist a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm.

Then

1. (ii)

$L_{1}^{\perp}+\dots+L_{K}^{\perp}$* is not closed in $H$.*

Proof.

Denote $L=\bigcap L_{i}$ and assume (i). By Proposition LABEL:34, there is a weakly-null sequence $w_{n}$ in $L^{\perp}$ so that

[TABLE]

and (ii) follows from Lemma LABEL:BB. ∎

The following theorem strengthens Proposition LABEL:34; it describes what exactly is happening when $K\geq 4$. Notice, that the graph $G$ in (ii) is the induced subgraph with vertices $V$ of the Johnson graph $J(K,4)$. In Proposition LABEL:johnbio we will construct for each subgraph $G$ as above a corresponding configuration of closed subspaces of $H$. In Theorem LABEL:hypothesis we will show that the statement (i) and (ii) are actually equivalent.

Theorem 2.4**.**

Let $H$ be a Hilbert space, $4\leq K$, and let $L_{1},L_{2},\dots,L_{K}$ be closed subspaces of $H$.

Suppose

1. (i)

*there exist a starting point $z_{0}\in H$, a sequence of indices *

$k_{1},k_{2},\dots\in[K]$* containing each of the numbers $1,\dots,K$ infinitely often, and for each $k\in[K]$, there exists a closed subspace $\tilde{L}_{k}\subset L_{k}\subset H$, so that the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does not converge in norm.*

Then

1. (ii)

there exists $V\subset\binom{[K]}{4}$ so that $\bigcup V=[K]$, for $E=\{\{A,B\}\in\binom{V}{2}:\,|A\cap B|=3\}$ the graph $G=(V,E)$ is connected, and if $A\in V$ then $\bigcap_{i\in A}L_{i}$ is infinite dimensional or $\sum_{i\in A}L_{i}^{\perp}$ is not closed.

Proof.

Denote this time $L=\bigcap_{k\in[K]}\tilde{L}_{k}$. As explained at the end of Section LABEL:almortho, we may assume that $\{z_{n}\}$ is a weakly-null sequence in $L^{\perp}$, and that $\lim|z_{n}|=1$. Define $N_{0}=1$. For $\delta=1/i$ choose $M_{i}>N_{i-1}$ as in Lemma LABEL:3step applied to $\tilde{L}_{k}$’s and so that $|z_{m}-z_{m+1}|<1/i$ for $M_{i}\leq m$. Let $N_{i}>M_{i}$ be the smallest number for which

[TABLE]

Let

[TABLE]

Then by (LABEL:sm),

[TABLE]

Consider the graph $G_{i}=(V_{i},E_{i})$. Clearly, each of its induced subgraphs $G^{m}_{i}=(V^{m}_{i},E_{i}\cap\binom{V_{i}^{m}}{2})$ is connected. Since $3\leq|I(m,1/i)|$ for all $m\geq M_{i}$ and $I(m+1,1/i)\subset I(m,2/i)$, the graph $G_{i}=(V_{i},E_{i})$ is connected as well. Indeed, if $A\in V_{i}^{m}$ and $B\in V_{i}^{m+1}$ are given, choose $D\in V_{i}^{m+1}$, $D\subset I(m+1,1/i)$. Since all $G^{m}_{i}$’s are connected, there is a path from $A$ to $D$ and from $B$ to $D$.

Since $K$ is fixed, there are only finitely many such graphs $G_{i}$. By passing to a subsequence we can assume they are all the same $G=G_{1}=G_{2}=\dots$. Since $K\geq 4$, the graph $G$ has at least 2 vertices. Suppose $\{A,B\}$ is an edge of $G$ so that $|A|\leq|B|$. If $|A|=3$, then $A\subset B$ and $|B|\geq 4$. Hence we can drop all vertices of size 3 from the graph and all edges containing them and the graph still fulfills (LABEL:union) and stays connected. Since with each set $A$ of size at least 5, $V$ contains all of its 4-element subsets, by keeping in $V$ only the sets of size 4 we preserve (LABEL:union). We claim that such a graph is still connected. Suppose $P$ is a path from $C$ to $D$ in $G$ and $A$ is a vertex of $P$ with $|A|\geq 5$. Then $A$ can be replaced by at most three of its $4$-point subsets to create a walk from $C$ to $D$. Hence we can drop from $V$ all the vertices of size at least 5 and still preserve both connectivity and (LABEL:union), as claimed.

If $A\in V$, then

[TABLE]

for a subsequence $\{z_{m_{n}}\}$ of the weakly-null sequence $\{z_{m}\}$. Hence by Lemma LABEL:seqeq, $\bigcap_{i\in A}L_{i}$ is infinite dimensional or $\sum_{i\in A}L_{i}^{\perp}$ is not closed. ∎

We close with ideas that will enable us in Section LABEL:K5 to prove that (i) and (ii) in the above theorem are, in fact, equivalent.

Lemma 2.5**.**

Let $G=(V,E)$ be the Johnson graph $J(K,4)$ and let $A_{1},A_{2},\dots\in V$, $\mathcal{E}_{i}=\{A_{i},A_{i+1}\}\in E$ be a walk in $G$. For a given $k\in[K]$ define $a_{i}=1$ if $k\in A_{i}\cap A_{i+1}$ and $a_{i}=0$ otherwise. Then the sequence $\{a_{i}\}$ is either constant, or it consists of blocks of ones separated by blocs of zeros of length at least two.

Proof.

We can assume that $k=1$ and that $a_{i}=1$ and $a_{i+1}=0$. Then $1\in A_{i+1}$. Hence $1\notin A_{i+2}$, and $a_{i+2}=0$. ∎

Let $K\in\mathbb{N}$ be given. Let $G=(V,E)$ be the Johnson graph $J(K,4)$, and let $A_{1},A_{2},\dots\in V$, $\mathcal{E}_{i}=\{A_{i},A_{i+1}\}\in E$ be an infinite walk in $G$. With a slight abuse of notation we also denote by $\mathcal{E}_{i}$ the intersection $A_{i}\cap A_{i+1}$. We will iteratively define $K$ sequences $\{s^{k}_{i}\}_{i\in\mathbb{N}}$, $k\in[K]$, consisting of the symbols $X,Y,Z,0$, so that the $i$-th entry of exactly one of the $K$ sequences is $X$, of exactly one is $Y$, of exactly one is $Z$, and the $i$-th entries of all the other sequences are [math]. We proceed according to the following rules.

To start with, we set $s^{k}_{i}=0$ for all $k\in[K]$ and $i\in\mathbb{N}_{0}$. Assume $\mathcal{E}_{1}=\{a,b,c\}$. In the first step we re-define $s^{a}_{1}=X$, $s^{b}_{1}=Y$, $s^{c}_{1}=Z$. Here the order does not matter; we simply choose one, assuming, for example, that $a<b<c$.

In the $i$-th step we change the $i$-th coordinate from zero to $X$, $Y$, or $Z$, respectively, in three of the sequences corresponding to $\mathcal{E}_{i}$. We proceed as follows. Since $3\leq|\mathcal{E}_{i-1}\cup\mathcal{E}_{i}|\leq|A_{i}|=4$, we can assume $\mathcal{E}_{i-1}=\{a,b,c\}$ and $\mathcal{E}_{i}=\{a,b,d\}$, with possibly $c=d$. We define $s^{a}_{i}=s^{a}_{i-1}$, $s^{b}_{i}=s^{b}_{i-1}$, $s^{d}_{i}=s^{c}_{i-1}$. Each of the $K$ resulting sequences starts with [math] followed by, according to Lemma LABEL:walk, blocks of $X$’s, blocks of $Y$’s and blocks of $Z$’s (some or all of which can be missing) separated by blocks of zeros of length at least two. For example: $0,X,X,X,0,0,Y,0,0,Y,Y,0,0,0,Z,Z,Z,\dots$.

Moreover, for each $i\in\mathbb{N}$, there is exactly one $k_{i,X}\in[K]$ so that $s_{i}^{k_{i,X}}=X$. Similarly, there is exactly one $k_{i,Y}\in[K]$ and exactly one $k_{i,Z}\in[K]$ so that $s_{i}^{k_{i,Y}}=Y$ and $s_{i}^{k_{i,Z}}=Z$. Clearly, $\varepsilon_{i}=\{k_{i,X},k_{i,Y},k_{i,Z}\}$.

3. $K=3$ or $K=4$ spaces $L_{1},\dots,L_{K}$

Suppose $K\geq 3$ and let $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ be not closed. The key technical result of this section is Proposition LABEL:squeorthoblocs. With its help we will construct $\tilde{L}_{k}\subset L_{k}$, $k\in[K]$, and a product of projections on the spaces $\tilde{L}_{k}$ which does not converge in norm. For $K=3$ and $K=4$, we will show in Theorem LABEL:3-eq that the existence of such $\tilde{L}_{k}$’s is equivalent to $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ not being closed.

The following result on the continuous dependence of words on letters will enable us to build the spaces $X$ and $Y$ of Lemma LABEL:orthobuildingbloc using any normalized weakly-null sequence instead of an orthonormal basis.

Lemma 3.1**.**

[KP]** Let $\psi\in\mathcal{S}_{n}$ for some $n\in\mathbb{N}$. Assume $A_{i},B_{i},E\in\mathcal{B}(H)$, $i\in\{1,2,\dots,n\}$, are contractions so that each $A_{i}$ commutes with $E$. Then

[TABLE]

The following lemma expresses the essence of the construction of [P].

Lemma 3.2**.**

For every $\varepsilon>0$, there exists $N=N(\varepsilon)\in\mathbb{N}$, so that for every $\eta>0$, there exists $\psi\in\mathcal{S}_{3}$ such that $|\psi_{1}|\leq N\leq|\psi|$ with the following property.

Given subspaces $X\subset E$ of a Hilbert space $H$ so that $X$ is separable, $\dim X\geq N$, and $\dim(X^{\perp}\cap E)\geq\dim X$ and given $u,v\in X$ so that $\|u\|=\|v\|=1$ and $u\perp v$, there exists a subspace $Y\subset E$ such that

1. (i)

$X\cap Y=\{0\}$; 2. (ii)

$\|P(X)-P(Y)\|<\eta$; 3. (iii)

$|\psi(W,X,Y)u-v|<\varepsilon$*, where $W=u\vee v$. *

If, moreover, $\tilde{X},\tilde{Y},\tilde{Z}$ are subspaces of $H$ such that

1. (iv)

$\|P(W)-P(\tilde{Z})P(E)\|\leq\varepsilon/N$; 2. (v)

$\|P(X)-P(\tilde{X})P(E)\|\leq\varepsilon/|\psi|$; 3. (vi)

$\|P(Y)-P(\tilde{Y})P(E)\|\leq\varepsilon/|\psi|$,

then

[TABLE]

Proof.

The statements (i)-(iii) are basically Lemma 2.3 of [KP]. To see that one can work with finite dimensional spaces and the dependence of the dimension $N$ on $\varepsilon$, one just needs to read carefully the proofs of Lemma 2.1, Lemma 2.2, and Lemma 2.3 of [KP], or, alternatively, Lemma 2.1, Lemma 2.4, and Lemma 2.5 of [KM]. To show we have a slight freedom of choice concerning the 3 spaces onto which we project, we use Lemma LABEL:wordcont:

[TABLE]

hence

[TABLE]

∎

A careful reading of the proof of Lemma 2.5 of [KP] or of Theorem 2.6 of [KM] implies the following result.

Lemma 3.3**.**

For every sequence of positive numbers $\varepsilon_{i}>0$, $i\in\mathbb{N}$, there exist $N_{i}\in\mathbb{N}$ and $\Psi^{i}\in\mathcal{S}_{3}$ with the following property.

Let $\{e_{i}\}_{i=1}^{\infty}$ be an orthonormal sequence in a Hilbert space $H$ and $W_{i}=e_{i}\vee e_{i+1}$. Let $\{E_{i}\}_{i=1}^{n}$ be a sequence of subspaces of $H$ so that

1. (i)

$\dim E_{i}\geq 2N_{i}$; 2. (ii)

$e_{i},e_{i+1}\in E_{i}$; 3. (iii)

$\vee e_{i+1}=P(E_{i})E_{i+1}=P(E_{i+1})E_{i}$; 4. (iv)

$E_{i}\perp E_{j}$* if $|i-j|\geq 2$.*

Then there exist sequences $X_{i},Y_{i}$ of subspaces of $H$ so that

1. (v)

$X_{i},Y_{i}\subset E_{i}$; 2. (vi)

$e_{i},e_{i+1}\in X_{i}$; 3. (vii)

$|\Psi^{i}(Z,X,Y)e_{i}-e_{i+1}|<\varepsilon_{i}$*; *

where $Y_{0}=\vee e_{1}$ and $X=\bigvee_{i=1}^{\infty}X_{i},\ Y=\bigvee_{k=0}^{\infty}Y_{2k+1},\ Z=\bigvee_{k=0}^{\infty}Y_{2k}$.

We allow a slight skewing in the above rigid orthonormal structure. This will enable us later to fit the spaces $X,Y,Z$ inside given subspaces of $H$.

Proposition 3.4**.**

For every sequence of positive numbers $\varepsilon_{i}>0$, $i\in\mathbb{N}$, there exist $N_{i}\in\mathbb{N}$ and $\psi^{i}\in\mathcal{S}_{3}$ so that $N_{1}\leq N_{2}\leq\dots$ and $|\psi^{1}|\leq|\psi^{2}|\leq\dots$ with the following property.

Let $\{w_{n}\}$ be an orthonormal sequence in a Hilbert space $H$ and $e_{1}=w_{1},e_{2},e_{3},\dots$ be a subsequence of $\{w_{n}\}$. Let $\{E_{i}\}_{i=1}^{n}$ be a sequence of spans of consecutive, with one vector $e_{i}$ overlapping, blocks of $\{w_{n}\}$ so that

1. (i)

$\dim E_{i}\geq 2N_{i}$; 2. (ii)

$e_{i},e_{i+1}\in E_{i}$; 3. (iii)

$\vee e_{i+1}=P(E_{i})E_{i+1}=P(E_{i+1})E_{i}$; 4. (iv)

$E_{i}\perp E_{j}$* if $|i-j|\geq 2$.*

Then there exist subspaces $X_{i},Y_{i}$ of $H$ so that

1. (v)

$X_{i},Y_{i}\subset E_{i}$; 2. (vi)

$e_{i},e_{i+1}\in\chi_{i}\subset\{w_{n}\}$, where $\chi_{i}$ is an orthonormal basis of $X_{i}$; 3. (vii)

$|\psi^{i}(W_{i},X_{i},Y_{i})e_{i}-e_{i+1}|<\varepsilon_{i}$, where $W_{i}=e_{i}\vee e_{i+1}$.

Suppose, moreover, that $S,T,U:{\rm span\,}\{w_{n}\}\to H$ are linear so that

1. (viii)

$\|P(R(E))-P(E)\|\leq(\varepsilon_{i+1}/|\psi^{i+1}|)^{2}$* for all subspaces $E\subset E_{i}$ and all linear mappings $R$ so that $R(w_{n})\in\{S(w_{n}),T(w_{n}),U(w_{n})\}$.* 2. (ix)

Let $\tilde{X}_{i}\subset S(E_{i})$, $\tilde{Y}_{i}\subset T(E_{i})$, $\tilde{Z}_{i}\subset U(E_{i})$ and let $\tilde{X}=\bigvee_{i=1}^{\infty}\tilde{X}_{i}$, $\tilde{Y}=\bigvee_{i=1}^{\infty}\tilde{Y}_{i}$, and $\tilde{Z}=\bigvee_{i=1}^{\infty}\tilde{Z}_{i}$.

*For $i\leq 0$ define $e_{i}=0$ and $X_{i}=Y_{i}=\{0\}$.

Assume that for a fixed $j\in\mathbb{N}$,*

1. (x)

$\tilde{X}_{j}=S(X_{j})$; 2. (xi)

$\tilde{Y}_{j}=T(Y_{j})$, $\tilde{Y}_{j-1}\subset\vee T(e_{j-1})$, and $\tilde{Y}_{j+1}\subset\vee T(e_{j+2})$; 3. (xii)

There are four options for $\tilde{Z}_{j}$:

$\tilde{Z}_{j}=\vee\{U(e_{j}),U(e_{j+1})\}$, $\tilde{Z}_{j-1}=\tilde{Z}_{j+1}=\{0\}$; 2. 2)

$\tilde{Z}_{j}=\vee U(e_{j})$, $\tilde{Z}_{j-1}=\{0\}$, $\tilde{Z}_{j+1}=U(Y_{j+1})$ and $\tilde{Z}_{j+2}\subset\vee U(e_{j+3})$; 3. 3)

$\tilde{Z}_{j}=\vee U(e_{j+1})$, $\tilde{Z}_{j+1}=\{0\}$, $\tilde{Z}_{j-1}=U(Y_{j-1})$ and $\tilde{Z}_{j-2}\subset\vee U(e_{j-2})$; 4. 4)

$\tilde{Z}_{j}=\{0\}$, $\tilde{Z}_{j-1}=U(Y_{j-1})$, $\tilde{Z}_{j-2}\subset\vee U(e_{j-2})$, and, similarly $\tilde{Z}_{j+1}=U(Y_{j+1})$ and $\tilde{Z}_{j+2}\subset\vee U(e_{j+3})$. 4. (xiii)

If $t\in\{j-1,j,j+1\}$, then $|\langle f,g\rangle|\leq(\varepsilon_{j+1}/|\psi^{j+1}|)^{2}$ for all $|f|=|g|=1$ so that $f\in E_{t}\cup S(E_{t})\cup T(E_{t})\cup U(E_{t})$ and

[TABLE]

where $R(w_{n})\in\{w_{n},S(w_{n}),T(w_{n}),U(w_{n})\}$. Then

[TABLE]

Proof.

Assume $1>\varepsilon_{1}>\varepsilon_{2}>\dots$. We define $N_{0}=1$ and for each $i\in\mathbb{N}$ we choose $N_{i}=N(\varepsilon_{i})\geq N_{i-1}$ according to Lemma LABEL:orthobuildingbloc. Then we put $\eta_{i}=\varepsilon_{i+1}/N_{i+1}$ for each $i\in\mathbb{N}$. Again, we use Lemma LABEL:orthobuildingbloc to choose $\psi^{i}\in\mathcal{S}_{3}$. We define $X_{i}$ as an $N_{i}$-dimensional subspace of $E_{i}$ so that $e_{i},e_{i+1}\in\chi_{i}\subset\{w_{n}\}$ for an orthonormal basis $\chi_{i}$ of $X_{i}$. We choose $Y_{i}\subset E_{i}$ according to Lemma LABEL:orthobuildingbloc.

Since $\tilde{X}=\tilde{X}_{j}\vee(\tilde{X}\cap G_{S,j})$ by (vi), Lemma LABEL:justforme1 and (xiii) imply that

[TABLE]

Hence by (viii) and (xiii),

[TABLE]

Similarly, by (xi), $\tilde{Y}=\tilde{Y}_{j}\vee(\tilde{Y}\cap G_{T,j})$; hence

[TABLE]

and

[TABLE]

It remains to prove that

[TABLE]

as then (LABEL:X), and (LABEL:Y) and Lemma LABEL:orthobuildingbloc imply that

[TABLE]

If 1) of (xii) is satisfied, then $\tilde{Z}=\tilde{Z}_{j}\vee(\tilde{Z}\cap G_{U,j})$ hence by (xiii) and Lemma LABEL:justforme1,

[TABLE]

Hence

[TABLE]

by (viii) and (xiii).

In the other extreme case, when 4) of (xii) is satisfied, we have $\tilde{Z}=\tilde{Z}_{j-1}\vee\tilde{Z}_{j+1}\vee(\tilde{Z}\cap G)$, where

[TABLE]

Hence by (xiii) applied to $G_{U,t}$ for $t\in\{j-1,j,j+1\}$ and Lemma LABEL:justforme1,

[TABLE]

and

[TABLE]

Since $X_{j-1}$ and $Y_{j-i}$ were chosen so that $\|P(X_{j-1})-P(Y_{j-1})\|\leq\eta_{j-1}$, by (viii),

[TABLE]

Similarly,

[TABLE]

By the above four inequalities,

[TABLE]

In 2) and 3) of (xii) one proceeds similarly. ∎

The following corollary is interesting in particular in the case where $L=\bigcap_{k\in[K]}L_{k}$ is finite dimensional. If it is infinite dimensional, the existence of the spaces $\tilde{L}_{k}\subset L$ was established in [KM].

Corollary 3.5**.**

Let $H$ be a Hilbert space and $\{w_{n}\}_{n\in\mathbb{N}}$ a normalized weakly-null sequence in $H$. Let $K\geq 3$ and let $L_{1},\dots,L_{K}$ be closed subspaces of $H$ so that

[TABLE]

Then there exist closed subspaces $\tilde{L}_{k}\subset L_{k}$, a starting point $z_{0}\in H$, and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each $i\in{K}$ infinitely often so that the sequence of iterates defined by $z_{n}=\tilde{L}_{k_{n}}z_{n-1}$ does not converge in norm.

Proof.

According to Lemma LABEL:seqeq, we may assume that $\{w_{n}\}$ is an orthonormal sequence. For $\varepsilon_{i}=200^{-i}$ we choose $N_{i}\in\mathbb{N}$, $\psi^{i}\in\mathcal{S}_{3}$ according to Proposition LABEL:squeorthoblocs and define $\beta_{n}=(\varepsilon_{n+1}/|\psi^{n+1}|)^{2}/8$. We choose a decreasing sequence $0<\alpha_{n}<1/(K\cdot 200^{n})$ of numbers small enough to accommodate the property described below and so that $2\sum_{n}^{\infty}\alpha_{i}<\beta_{n}$.

Assume $x_{n}\in H$ satisfy $|w_{n}-x_{n}|\leq\alpha_{n}$ for all $n\in\mathbb{N}$ and define a linear mapping $S$ by $S(w_{n})=x_{n}$. Then

[TABLE]

for all subspaces $E\subset H_{n}:=\bigvee_{j=n}^{n+2N_{n}}w_{j}$ and for all $n\in\mathbb{N}$.

By taking a subsequence, we call it $\{w_{n}\}$ again, we can assume that for a suitably choosen $w^{k}_{n}\in L_{k}$

[TABLE]

Then, clearly, (LABEL:8) is also satisfied if $n\leq M$ and $E\subset\bigvee_{j=M}^{M+2N_{n}}w_{j}$.

Assume, moreover, we have chosen the subsequence so that $|\langle w_{n},g\rangle|\leq\alpha_{n}$ for all

[TABLE]

with $|g|\leq 1$ and all $n\in\mathbb{N}$. This is possible, because $G_{n}$ is finite dimensional and $\{w_{n}\}$ converges weakly to zero. Suppose $x_{j}\in\{w_{j},w_{j}^{k}:\,k\in[K]\}$, $j\in\mathbb{N}$. Then according to Lemma LABEL:justforme3 and the choice of $\{\alpha_{n}\}$ we have that $|\langle f,g\rangle|\leq 2\beta_{n}$ for all $f\in\bigvee_{n\leq j}x_{j}$ and $g\in G_{n}$ with $|f|=|g|=1$. If $n<N$, then by Lemma LABEL:justforme1,

[TABLE]

In order to apply Proposition LABEL:squeorthoblocs, we define $e_{1}=w_{1}$ and we group $\{w_{n}\}$ into successive blocks of length $2N_{1},2N_{2},\dots$. The neighboring blocks overlap by exactly one vector which we call $e_{2},e_{3},\dots$. The spans of the blocks we call $E_{1},E_{2},\dots$. By (LABEL:app) we have $|P(L_{k})e_{n}-e_{n}|\leq\alpha_{n}$ for all $k\in[K]$ and $n\in\mathbb{N}$.

First assume $K=3$. We define linear mappings $S,T,U:{\rm span\,}\{w_{n}\}\to H$ so that

[TABLE]

Then by (LABEL:8) and (LABEL:13), assumption (viii) and the assumption (xiii) of Proposition LABEL:squeorthoblocs are satisfied for any $j\in\mathbb{N}$. Choose $X_{i},Y_{i}\subset E_{i}$ according to Proposition LABEL:squeorthoblocs and for $i\in\mathbb{N}$, define $\tilde{X}_{i}=S(X_{i})$ and

[TABLE]

except for $Z_{1}$ which we define as $\vee U(e_{1})$. Assumptions (x), (xi), and (xii) of Proposition LABEL:squeorthoblocs are satisfied for all odd $j\in\mathbb{N}$. If we switch between $\tilde{Y}_{j}$ and $\tilde{Z}_{j}$, they are also satisfied for all even $j\in\mathbb{N}$.

Choose closed subspaces $\tilde{X},\tilde{Y},\tilde{Z}\subset H$ according to (ix) of Proposition LABEL:squeorthoblocs and call them $\tilde{L}_{1},\tilde{L}_{2},\tilde{L}_{3}$. Then $\tilde{L}_{k}\subset L_{k}$. Write for short $R_{j}=\psi^{j}(\tilde{Z},\tilde{X},\tilde{Y})$ if $j$ is odd and $R_{j}=\psi^{j}(\tilde{Y},\tilde{X},\tilde{Z})$ if $j$ is even. Since $\|R_{j}\|\leq 1$, we have by induction

[TABLE]

for all $n\in\mathbb{N}$. Since $\{e_{i}\}$ is an orthonormal sequence, the norm-limit $\lim_{n\to\infty}R_{n}R_{n-1}\dots R_{1}e_{1}$ does not exist.

If $K>3$ define $\tilde{L}_{k}=L_{k}$ for $k\in\{4,\dots,K\}$ and if $j$ is odd $R_{j}=\psi_{j}(\tilde{Z},\tilde{X},\tilde{Y})P(L_{K})\dots P(L_{4})$; for even $j$ we define $R_{j}$ similarly. Then

[TABLE]

The statement then again follows from (LABEL:11). ∎

We show next that for $K=3$ and $K=4$ the existence of a diverging product of projections is equivalent to $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ not being closed.

Notice that for (i) of the next theorem to be true, $H$ has to be infinite dimensional. If $L=\bigcap_{k\in[K]}L_{k}$ is finite dimensional, in particular if $L=\{0\}$, then it is enough to require $\tilde{L}_{k}\subset L_{k}$ in (iii). If $L$ is infinite dimensional this condition cannot be dropped, as there always exist $\tilde{L}_{k}\subset L$ as required in (iii) [KM, KP].

Theorem 3.6**.**

Let $H$ be a Hilbert space, $K=3$ or $K=4$, and let $L_{1},\dots,L_{K}$ be closed subspaces of $H$; denote $L=\bigcap_{k\in[K]}L_{k}$. The following statements are equivalent:

1. (i)

$L_{1}^{\perp}+\dots+L_{K}^{\perp}$* is not closed in $H$;* 2. (ii)

for every sequence $\{a_{n}\}\in c_{0}$, there exists a starting point $z\in H$ so that $|T^{n}z-P(\bigcap L_{k})z|\geq a_{n}$; 3. (iii)

there exist closed subspaces $\tilde{L}_{k}\subset L_{k}\cap L^{\perp}$, a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each $k\in[K]$ infinitely often, so that the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does not converge in norm.

Proof.

The equivalence of (i) and (ii) is due to [BDH, DH, BaGM1, BaGM2]. That (i) implies (iii) follows from Lemma LABEL:BB and Corollary LABEL:inspaces applied to $L_{k}\cap L^{\perp}$.

Assume (iii) is satisfied. Corollary LABEL:n34 implies that $\tilde{L}_{1}^{\perp}+\dots+\tilde{L}_{K}^{\perp}$ is not closed. By Lemma LABEL:BB (iii), there exists a normalized weakly-null sequence $\{w_{n}\}$ so that

[TABLE]

Since $\tilde{L}_{k}\subset L^{\perp}$ for all $k\in[K]$, we can assume that $\{w_{n}\}$ is in $L^{\perp}$ as well. By Lemma LABEL:BB, $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ is not closed. ∎

Here is a positive version of the above equivalence.

Theorem 3.7**.**

Let $H$ be a Hilbert space, $K=3$ or $K=4$, and let $L_{1},\dots,L_{K}$ be closed subspaces of $H$, $\bigcap_{k\in[K]}L_{k}=\{0\}$. The following statements are equivalent:

1. (i)

$L_{1}^{\perp}+\dots+L_{K}^{\perp}$* is closed in $H$;* 2. (ii)

if $T=P_{L_{K}}\dots P_{L_{1}}$, then $\|T^{n}\|\leq Cr^{n}$ for some $r\in[0,1)$; 3. (iii)

for all closed subspaces $\tilde{L}_{i}\subset L_{i}\subset H$, each starting point $z_{0}\in H$ and every sequence of indices $k_{1},k_{2},\dots\in[K]$, the sequence of iterates defined by $z_{n}=P({\tilde{L}_{k_{n}}})z_{n-1}$ does converge in norm.

Proof.

The equivalence of (i) and (ii) is due to [BDH, DH, BaGM1, BaGM2]; explicit values for $C$ and $r$ in (ii) can be found in [BaGM2] and [PRZ]. The rest is contained in Theorem LABEL:3-eq. ∎

4. $K\geq 4$ spaces $L_{1},\dots,L_{K}$

In this section we give a version of Theorem LABEL:3-eq which is suitable for $K\geq 4$ spaces. We give a condition involving Johnson graphs which is equivalent to the existence of a divergent product of projections.

The assumption of Corollary LABEL:inspaces implies by Lemma LABEL:seqeq that $\bigcap_{i\in A}L_{i}$ is infinite dimensional, or that $\sum_{i\in A}L_{i}^{\perp}$ is not closed for each $A\in\binom{[K]}{4}$. This in turn implies that (ii) of the following theorem is satisfied for $V=\binom{[K]}{4}$. Hence the implication (ii)$\Rightarrow$(i) of the following theorem generalizes Corollary LABEL:inspaces.

Theorem 4.1**.**

Let $H$ be a Hilbert space, $4\leq K$, and let $L_{1},L_{2},\dots,L_{K}$ be closed subspaces of $H$. The following statements are equivalent:

1. (i)

there exist a starting point $z_{0}\in H$, a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often, and for each $k\in[K]$, there exists a closed subspace $\tilde{L}_{k}\subset L_{k}\subset H$, so that the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does not converge in norm; 2. (ii)

there exists $V\subset\binom{[K]}{4}$ so that $\bigcup V=[K]$, for $E=\{\{A,B\}\in\binom{V}{2}:\,|A\cap B|=3\}$ the graph $G=(V,E)$ is connected, and if $A\in V$, then $\bigcap_{i\in A}L_{i}$ is infinite dimensional, or $\sum_{i\in A}L_{i}^{\perp}$ is not closed.

Proof.

The implication (i)$\Rightarrow$(ii) is the statement of Theorem LABEL:johnson.

The proof of (ii)$\Rightarrow$(i) is a refined version of the proof of Corollary LABEL:inspaces. Inductively, we will build a structure as in the assumptions of Proposition LABEL:squeorthoblocs.

For $\varepsilon_{i}=200^{-i}$ we choose $N_{i}\in\mathbb{N}$, $\psi^{i}\in\mathcal{S}_{3}$ according to Proposition LABEL:squeorthoblocs and define $\beta_{n}=(\varepsilon_{n+1}/|\psi^{n+1}|)^{2}/8$.

We choose a decreasing sequence $\{\alpha_{n}\}$ of numbers satisfying $0<\alpha_{n}<1/(K\cdot 200^{n})$ and $2\sum_{n}^{\infty}\alpha_{i}<\beta_{n}$, which are also small enough to accommodate the following property: Assume $\{w_{n}\}$ is an orthonormal sequence, $x_{n}\in H$ satisfy $|w_{n}-x_{n}|\leq\alpha_{n}$ for all $n\in\mathbb{N}$ and a linear mapping $R$ is defined by $R(w_{n})=x_{n}$. Then

[TABLE]

for all subspaces $E\subset H_{n}:=\bigvee_{j=n}^{n+2N_{n}}w_{j}$ and for all $n\in\mathbb{N}$.

We choose an infinite walk in $G$, namely a sequence of vertices $A_{i}\in V$ so that $\{A_{i},A_{i+1}\}\in E$ and $\bigcup_{i=n}^{\infty}A_{i}=[K]$ for all $n\in\mathbb{N}$.

Next we repeatedly use (i)$\Rightarrow$(iii) of Lemma LABEL:seqeq to inductively construct an orthonormal sequence $\{w_{n}\}$. We block the sequence $\{w_{n}\}$ into blocks of length $2N_{i}$ so that all vectors in the $i$-th block are very near to all the spaces $L_{k}$ with $k\in A_{i}$.

To start the construction we choose an orthonormal family $\{w_{n}\}$ so that

[TABLE]

for all $n\in[2N_{1}]$. For each $n\in[2N_{1}]$ and $k\in A_{1}$, we choose $w^{k}_{n}\in L_{k}$ so that $|w_{n}-w^{k}_{n}|\leq\alpha_{n}$. In the second step, we enlarge the orthonormal family $\{w_{n}\}$ so that

[TABLE]

for all $n\in\{2N_{1}+1,\dots,2N_{1}+2N_{2}\}$. For each such $n$ and $k\in A_{2}$, we choose $w_{n}^{k}\in L_{k}$ so that $|w_{n}-w^{k}_{n}|\leq\alpha_{n}$.

We define $E_{1}=\vee\{w_{n}\}_{n=1}^{2N_{1}+1}$ and $e_{1}=w_{1}$, $e_{2}=w_{2N_{1}+1}$.

As with the second step, in the third one we enlarge the orthogonal family $\{w_{n}\}$ to be of size $2(N_{1}+N_{2}+N_{3})$, and afterwards set

[TABLE]

and $e_{3}=w_{2(N_{1}+N_{2})+1}$. In the next step we define

[TABLE]

$e_{4}=w_{2(N_{1}+N_{2}+N_{3})+1}$, and so on. Moreover, during the above process, when choosing every single $w_{n}$ we make sure that

[TABLE]

with $|g|=1$. In the above set $G_{n}$ we include only those $w_{m}^{k}$ which were defined, that is, where $k$ is contained in the appropriate set $A_{i}$.

We choose $X_{i},Y_{i}\subset E_{i}$ as in Proposition LABEL:squeorthoblocs.

For each $k\in[K]$, we define a linear mapping $S^{k}:{\rm span\,}\{w_{n}\}\to H$ as follows. First we set $S^{k}(w_{n})=w_{n}$ for all $n\in\mathbb{N}$. If $w_{n}^{k}$ has already been chosen for some $n$, then we re-define $S^{k}(w_{n})=w_{n}^{k}$. According to (LABEL:8g), $S^{k}$ in place of $S$, $T$, or $U$ satisfies (viii) of Proposition LABEL:squeorthoblocs.

Let $\{A_{i}\}$ be the infinite walk in $G$ chosen above. For $k\in[K]$, let $\{s_{i}^{k}\}_{i=0}^{\infty}$ be the sequence of $X,Y,Z,0$ as described at the end of Section LABEL:SJ. If $k\in A_{i}\cap A_{i+1}$, that is, when $s_{i}^{k}\neq 0$, then $S^{k}(E_{i})\subset L_{k}$.

Next we define a sequence of subspaces $L_{k,i}\subset L_{k}$ as described below and set

[TABLE]

To start with, we set $L_{k,i}=\{0\}$ for all $i\in\mathbb{N}$. Some of the subspaces we re-define as explained below.

[TABLE]

If $i$ is even and $s_{i-1}^{k}=s_{i}^{k}=s_{i+1}^{k}=Y$, then $L_{k,i}$ remains $\{0\}$. If the neighbor $s_{i-1}^{k}$ or $s_{i+1}^{k}$ of $s_{i}^{k}$ is [math], we redefine $L_{k,i}$ as follows. If $i$ is odd and $s_{i}^{k}=Z$, then we proceed similarly:

[TABLE]

Let $j\in\mathbb{N}$ be given. Recall from the very end of Section LABEL:SJ that there is exactly one $k_{j,X}\in[K]$ so that $s_{j}^{k_{j,X}}=X$. Similarly, there is exactly one $k_{j,Y}\in[K]$ and exactly one $k_{j,Z}\in[K]$ so that $s_{j}^{k_{j,Y}}=Y$ and $s_{j}^{k_{j,Z}}=Z$. We apply Proposition LABEL:squeorthoblocs with $S$, $T$, and $U$ replaced by $S^{k}$ with $k\in A_{j}\cap A_{j+1}=\{k_{j,X},k_{j,Y},k_{j,Z}\}$ and $\tilde{X}$, $\tilde{Y}$, $\tilde{Z}$ replaced by $\tilde{L}_{k_{j,X}},\tilde{L}_{k_{j,Y}},\tilde{L}_{k_{j,Z}}$, respectively. Then (ix) of Proposition LABEL:squeorthoblocs is clearly satisfied for $\tilde{X}_{i}=L_{k_{j,X},i}$, $\tilde{Y}_{i}=L_{k_{j,Y},i}$, and $\tilde{Z}_{i}=L_{k_{j,Z},i}$.

From the inductive construction of $E_{i}$’s and from (LABEL:8g), (LABEL:13g), Lemmma LABEL:justforme3 and Lemma LABEL:justforme1 it follows that (xiii) of Proposition LABEL:squeorthoblocs is satisfied as well. From the definition of the spaces $L_{k,i}$ it follows that $\tilde{X}_{j}=L_{k_{j,X},j}$ satisfies (x) of Proposition LABEL:squeorthoblocs. If $j$ is odd, then $\tilde{Y}_{j}=L_{k_{j,Y},j}$ satisfies (xi) of Proposition LABEL:squeorthoblocs and $\tilde{Z}_{j}=L_{k_{j,Z},j}$ satisfies (xii) of Proposition LABEL:squeorthoblocs. Hence by Proposition LABEL:squeorthoblocs, we have

[TABLE]

for odd $j$. If $j$ is even, then we switch $Y$ and $Z$.

We define $z_{0}=e_{1}$ and write for short $R_{i}=\psi^{i}(\tilde{L}_{k_{i,Z}},\tilde{L}_{k_{i,X}},\tilde{L}_{k_{i,Y}})$; again, we switch $Y$ and $Z$ if $j$ is even. Since $\|R_{i}\|\leq 1$, we have by induction

[TABLE]

for all $n\in\mathbb{N}$. Since $\{e_{i}\}$ is an orthonormal sequence, the norm-limit $\lim_{n\to\infty}R_{n}R_{n-1}\dots R_{1}e_{1}$ does not exist.

To make sure that all the $K$ spaces $\tilde{L}_{1},\dots,\tilde{L}_{K}$ appear in the product infinitely often, we re-define $R_{i}=\psi_{i}(\tilde{L}_{k_{i,Z}},\tilde{L}_{k_{i,X}},\tilde{L}_{k_{i,Y}})\Pi_{k\in A_{i}}P(\tilde{L}_{k})$. Then

[TABLE]

Now the statement is again seen to follow from (LABEL:glue). ∎

In Example LABEL:not3 we exhibit that reducing the 4-tuple condition to a 3-tuple does not work.

5. Examples

We dedicate this section to examples which show where the previously presented results cannot be strengthened or modified.

Let $L_{1},\dots,L_{K}$ be closed subspaces of a Hilbert space $H$. Suppose there is a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in\{1,\dots,K\}$ containing each of the numbers $1,\dots,K$ infinitely often so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm.

In Theorem LABEL:johnson we have shown that there is a connected induced subgraph $G=(V,E)$ of $J(K,4)$ so that $\bigcup V=[K]$ and if $A\in V$, then $\bigcap_{i\in A}L_{i}$ is infinite dimensional or $\sum_{i\in A}L_{i}^{\perp}$ is not closed. Here we construct for each subgraph $G$ as above a corresponding configurations of subspaces of $H$.

Proposition 5.1**.**

Let $H$ be an infinite dimensional Hilbert space. Let $K\geq 4$ and $V\subset\binom{[K]}{4}$ be so that $\bigcup V=[K]$, and let $\alpha:V\to\{0,1\}$. Then there exist closed subspaces $L_{1},\dots,L_{K}\subset H$ with the following properties. If $A\in V$ and $\alpha(A)=0$, then $\bigcap_{i\in A}L_{i}$ is infinite dimensional. If $A\in V$ and $\alpha(A)=1$, then $\sum_{i\in A}L_{i}^{\perp}$ is not closed. If $B\in\binom{[K]}{4}\setminus V$, then $\bigcap_{i\in B}L_{i}$ is finite dimensional and $\sum_{i\in B}L_{i}^{\perp}$ is closed.

Proof.

Choose $|V|$ pairwise orthogonal closed infinite dimensional subspaces $H_{A}$ of $H$ so that

[TABLE]

In each $H_{A}$ choose 5 pairwise orthogonal orthonormal sequences $\{e^{A}_{n}\}_{n\in\mathbb{N}}$ and $\{e^{A}_{i,n}\}_{n\in\mathbb{N}}$, $i\in A$. For $k\in[K]$, we define

[TABLE]

Note that if $k\notin A$, then $L_{k}\perp H_{A}$.

Assume $A\in V$. If $\alpha(A)=0$, then $\vee\{e_{n}^{A}\}_{n\in\mathbb{N}}\subset\bigcap_{i\in A}L_{i}$. If $\alpha(A)=1$, then $\bigcap_{i\in A}L_{i}=\{0\}$ and $\max\{{\rm dist\,}(e_{n}^{A},L_{i}:\,i\in A\}\leq 1/n$. Hence the subspace $\sum_{i\in A}L_{i}^{\perp}$ is not closed by Lemma LABEL:BB.

Let $B\in\binom{[K]}{4}\setminus V$ be given. Assume $0\neq w\in\bigcap_{i\in B}L_{i}$. Then there is $A\in V$ so that $P(A)w\neq 0$. Choose $j\in B\setminus A$. Then $L_{j}\perp H_{A}$, hence $w\notin L_{j}$, which is a contradiction. Thus $\bigcap_{i\in B}L_{i}=\{0\}$.

Assume $\sum_{i\in B}L_{i}^{\perp}$ is not closed for some $B\in\binom{[K]}{4}\setminus V$. By Lemma LABEL:BB, there is an orthornormal sequence $\{w_{n}\}$ so that

[TABLE]

Write each $w_{n}=\sum_{A\in V}P_{A}(w_{n})$, where $P_{A}$ is the orthoprojection onto $H_{A}$. For each $n\in\mathbb{N}$, there is an $A\in V$ for which $|P_{A}(w_{n})|\geq 1/|V|$. Since $V$ is finite, by replacing $\{w_{n}\}$ with a suitable subsequence, we can assume this $A$ is the same for all $n\in\mathbb{N}$. Choose $j\in B\setminus A$. Then $L_{j}\perp H_{A}$, hence ${\rm dist\,}(w_{n},L_{j})\geq|P_{A}(w_{n})|\geq 1/|V|$ for all $n\in\mathbb{N}$, which is a contradiction. ∎

Bauschke showed in [B] that if $\sum_{j\in J}L_{j}^{\perp}$ is closed for each nonempty $J\subset[K]$, then $\{z_{n}\}$ converges in norm. Reich and Zaslavski show in [RZ], that $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ being closed is highly not hereditary with respect to subsets $J\subset[K]$. A simple example for $K=3$ spaces follows.

Example 5.2**.**

Let $H$ be an infinite dimensional Hilbert space. Then there exist three closed subspaces $L_{1},L_{2},L_{3}$ of $H$ so that $L_{1}^{\perp}+L_{2}^{\perp}+L_{3}^{\perp}$ is closed, but $L_{1}^{\perp}+L_{2}^{\perp}$ is not closed in $H$.

Proof.

It is enough to construct the example in a separable Hilbert space, so assume $H=\ell_{2}$ and write

[TABLE]

where the $j$-th copy of $\mathbb{R}^{3}$ has an orthonormal basis $\{e_{3j},e_{3j+1},e_{3j+2}\}$. Define

[TABLE]

Clearly, $L_{1}\cap L_{2}=\{0\}$, and $L_{1}^{\perp}+L_{2}^{\perp}+L_{3}^{\perp}=H$. Since

[TABLE]

it follows from Lemma LABEL:BB that the subspace $L_{1}^{\perp}+L_{2}^{\perp}$ is not closed. ∎

On the one hand, the condition that $L_{1}^{\perp}+L_{2}^{\perp}+L_{3}^{\perp}$ is closed is, according to Theorem LABEL:3-eq-positive, enough to ensure norm convergence even when projecting onto subspaces of the given spaces, but on the other hand, it is not really necessary to ensure norm convergence when projecting just on the $L_{k}$’s themselves, as the next example shows.

Example 5.3**.**

Let $H$ be an infinite dimensional Hilbert space. Then there exist three closed subspaces $L_{1},L_{2},L_{3}$ of $H$ so that

1. (i)

$L_{1}^{\perp}+L_{2}^{\perp}+L_{3}^{\perp}$* is not closed in $H$;* 2. (ii)

for every $0\neq z_{0}\in H$ and $k_{1},k_{2},\dots\in\{1,2,3\}$, the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ converges in norm (possibly very slowly).

Proof.

It is enough to construct the example in a separable Hilbert space, so assume $H=\ell_{2}$ and write

[TABLE]

where the $j$-th copy of $\mathbb{R}^{2}$ has an orthonormal basis $\{e_{2j-1},e_{2j}\}$. Define

[TABLE]

It is easy to see that $L_{1}\cap L_{2}\cap L_{3}=\{0\}$. Since

[TABLE]

it follows from Lemma LABEL:BB that the subspace $L_{1}^{\perp}+L_{2}^{\perp}+L_{3}^{\perp}$ is not closed.

Let $z_{0}\in H$ and $k_{n}\in\{1,2,3\}$ be given. Write $z_{0}=\sum_{j=1}^{\infty}z_{0}^{j}$, where $z_{0}^{j}$ is the projection of $z_{0}$ onto the $j$-th copy of $\mathbb{R}^{2}$. For every $j\in\mathbb{N}$, define the sequence $z^{j}_{n}=P(L_{k_{n}})z^{j}_{n-1}\in\bigvee\{e_{2j-1},e_{2j}\}$. This is a sequence of iterates of projections on the three lines $\vee e_{2j-1}$, $\vee(e_{2j-1}+e_{2j}/j)$ and $\vee(e_{2j-1}+2e_{2j})/j$ in the $j$-th copy of $\mathbb{R}^{2}$, which intersect at the origin. Hence

[TABLE]

By, e.g., [KKM], $z_{n}=\sum_{j=1}^{\infty}z_{n}^{j}$. Hence for every $\varepsilon>0$, there exists $N\in\mathbb{N}$ so that for all $n$ large enough,

[TABLE]

∎

Deutsch and Hundal ask in [DH] if just $L_{1}^{\perp}+\dots+L_{K}^{\perp}$ being closed is enough to guarantee the norm convergence of $\{z_{n}\}$. In Theorem LABEL:3-eq-positive we show this is indeed the case for $K=3$ and $K=4$. For $K\geq 5$ the next example shows that this is not the case in general. In other words, Proposition LABEL:34 is no longer true for $|I|\geq 5$.

Example 5.4**.**

Let $H$ be an infinite dimensional Hilbert space. For every $K\geq 5$, there exist closed subspaces $L_{1},\dots,L_{K}\subset H$, so that $\bigcap_{k\in[K]}L_{k}=\{0\}$ with the following properties:

1. (i)

there is a starting point $z_{0}\in H$ and a sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often so that the sequence of iterates defined by $z_{n}=P(L_{k_{n}})z_{n-1}$ does not converge in norm; 2. (ii)

the subspaces $L_{4},L_{5},\dots,L_{K}$ are pairwise orthogonal;

in particular, $L_{1}^{\perp}+\dots+L_{K}^{\perp}=H$.

Proof.

For $\varepsilon_{i}=4^{-i}$ consider the spaces $X,Y,Z$ as defined in Lemma LABEL:orthoblocs, and define $z_{0}=e_{1}$, $L_{1}=X$, $L_{2}=Y$, $L_{3}=Z$. For $k\in\{4,\dots,K\}$, define

[TABLE]

Then $L_{4},L_{5},\dots,L_{K}$ are pairwise orthogonal; hence

[TABLE]

Also $P(L_{k})e_{i}=e_{i}$ if $i=k\mod(K-3)$, and

[TABLE]

Since $\|A_{i}\|\leq 1$, we have by induction

[TABLE]

for all $n\in\mathbb{N}$. Since $\{e_{i}\}$ is an orthonormal sequence, the norm-limit $\lim_{n\to\infty}A_{n}A_{n-1}\dots A_{1}e_{1}$ does not exist. ∎

To conclude, we give in the above spirit an example showing that for $K\geq 4$, the Johnson graph $J(K,3)$ can not replace $J(K,4)$ in Theorem LABEL:hypothesis.

Example 5.5**.**

Let $H$ be an infinite dimensional Hilbert space and let $4\leq K$. Then there exist $L_{1},L_{2},\dots,L_{K}$, closed subspaces of $H$, $\bigcap_{k\in[K]}L_{k}=\{0\}$, so that

1. (i)

for all closed subspaces $\tilde{L}_{i}\subset L_{i}\subset H$, any starting point $z_{0}\in H$ and any sequence of indices $k_{1},k_{2},\dots\in[K]$ containing each of the numbers $1,\dots,K$ infinitely often, the sequence of iterates defined by $z_{n}=P(\tilde{L}_{k_{n}})z_{n-1}$ does converge in norm; 2. (ii)

there exists $V\subset\binom{[K]}{3}$, so that $\bigcup V=[K]$, for $E=\{\{A,B\}\in\binom{V}{2}:\,A\neq B,\,|A\cap B|=2\}$ the graph $G=(V,E)$ is connected, and if $A\in V$, then $\sum_{i\in A}L_{i}^{\perp}$ is not closed.

Proof.

It is enough to construct the example in a separable Hilbert space, so assume $H=\ell_{2}$ and write again

[TABLE]

where the $j$-th copy of $\mathbb{R}^{2}$ has an orthonormal basis $\{e_{2j-1},e_{2j}\}$. For $i\in\{3,\dots,K\}$, define

[TABLE]

Then $L_{1}\cap L_{2}=\{0\}$. The spaces $L_{3},\dots,L_{K}$ are pairwise orthogonal, hence $\bigcap_{k\in A}L_{k}=\{0\}$, and $\sum_{i\in A}L_{i}^{\perp}=H$ for each $A\in\binom{[K]}{4}$. By Proposition LABEL:34 and Lemma LABEL:BB, for all closed subspaces $\tilde{L}_{i}\subset L_{i}\subset H$, each starting point $z_{0}\in H$ and every sequence of indices $k_{1},k_{2},\dots\in[K]$ containing all of the numbers in $[K]$ infinitely often, the sequence of iterates defined by $z_{n}=P({\tilde{L}_{k_{n}}})z_{n-1}$ does converge in norm.

Define $V=\{\{1,2,i\}:\,i\in\{3,\dots,K\}\}$. Then $\bigcup V=[K]$ and $|A\cap B|=2$ for all different $A,B\in V$. If $A\in V$, then $\sum_{i\in A}L_{i}^{\perp}$ is not closed. ∎

6. Appendix

For an easy reference, this section contains a few elementary tools we use in the above text.

Lemma 6.1**.**

Let $\{f_{n}\}$ be an orthonormal sequence in a Hilbert space $H$. Let $\{\alpha_{n}\}$ be a sequence of positive numbers so that $\sum_{n=1}^{\infty}\alpha_{n}<\beta<1/4$. Let $w_{n}\in H$ be so that $|w_{n}-f_{n}|\leq\alpha_{n}$ for all $n\in\mathbb{N}$. Let $T:\vee\{f_{n}\}\to\vee\{w_{n}\}$ be a linear operator defined by $Tf_{n}=w_{n}$. Then

1. (i)

$T$* is a surjective isomorphism;* 2. (ii)

$\|T\|\leq 1+\beta$; 3. (iii)

$\|T^{-1}\|\leq 1/(1-\beta)$; 4. (iv)

$|Tx-x|\leq\beta|x|$* for $x\in\vee\{f_{n}\}$.*

Proof.

Let $x=\sum_{n=1}^{N}a_{n}f_{n}$, $|x|\leq 1$, be given. Then

[TABLE]

Hence $(1-\beta)|x|\leq|Tx|\leq(1+\beta)|x|$ and $T:{\rm span\,}\{f_{n}\}\to{{\rm span\,}}\{w_{n}\}$ satisfies (ii)-(iv). Thus $T$ is an isomorphism of $\vee\{f_{n}\}$ onto $\vee\{w_{n}\}$. ∎

Next we observe that within a normalized weakly-null sequence there exist almost pairwise orthogonal blocks of almost orthogonal vectors.

Lemma 6.2**.**

Let $\{w_{n}\}$ be a normalized weakly-null sequence in a Hilbert space $H$ and let $0<a_{n}\leq 1/2$ be so that $\lim a_{n}=0$. Let $V\subset H$ be a finite dimensional subspace. Then there exists an orthonormal sequence $\{e_{i}\}_{i=1}^{\infty}$ in $V^{\perp}$ and a subsequence $\{w_{n_{i}}\}$ of $\{w_{n}\}$ so that for all $i\in\mathbb{N}$,

1. (i)

$|w_{n_{i}}-e_{i}|\leq a_{i}$; 2. (ii)

$|\langle f,g\rangle|\leq a_{i}$* for all $f\in F_{i}=V\vee\{e_{j},w_{n_{j}}:\,1\leq j\leq i\}$ and $g\in G_{i}=\vee\{w_{n_{j}}:\,i<j\}\cup\vee\{e_{j}:\,i<j\}$ with $|f|=|g|=1$.*

Proof.

We can assume that $a_{n+1}\leq a_{n}/2$ and choose a decreasing sequence $\alpha_{n}>0$ so that $\sum_{i=n}^{\infty}\alpha_{i}<a_{n}$ for all $n\in\mathbb{N}$. We define $e_{0}:=w_{n_{0}}:=w_{1}$ and continue by induction. In the $(i+1)$-st step we choose an orthonormal basis $\{f_{1},\dots,f_{m}\}\supset\{e_{0},\dots,e_{i}\}$ of $Z=V\vee\{e_{j},w_{n_{j}}:\,0\leq j\leq i\}$. Next, we choose $M\geq m$ large enough for our considerations below and $n_{i+1}>n_{i}$ so that

[TABLE]

The projection $e$ of $w_{n_{i+1}}$ onto $Z^{\perp}$ has norm at least $\sqrt{1-\alpha_{i+1}^{2}/M}$. Define $e_{i+1}=e/|e|$. Then $|e_{i+1}-w_{n_{i+1}}|<\alpha_{i+1}<a_{i+1}$ if $M>0$ is large enough.

To verify (ii), let $i\in\mathbb{N}$ and $f\in F_{i}$ with $|f|=1$ be given. If $g\in\vee\{e_{j}:\,i<j\}$, then $\langle f,g\rangle=0$. Let $T:\vee\{e_{j}:\,i<j\}\to\vee\{w_{n_{j}}:\,i<j\}$ be a linear operator defined by $Te_{j}=w_{n_{j}}$. Let $g\in\vee\{w_{n_{j}}:\,i<j\}$ with $|g|\leq 1$ be given. Then by Lemma LABEL:justforme3,

[TABLE]

∎

If the angle between two closed subspaces of a Hilbert space is large, then the sum of the spaces is closed. Moreover, the projection on the sum is approximately equal the sum of the projections on the two spaces.

Lemma 6.3**.**

Let $H$ be a Hilbert space, $F$ and $G$ be closed subspaces of $H$ and let $0<\alpha<1/3$. Assume $|\langle f,g\rangle|\leq\alpha$ for all $f\in F$ and $g\in G$ with $|f|\leq 1$ and $|g|\leq 1$. Then

1. (i)

$\frac{1}{2}B_{F\vee G}\subset B_{F}+B_{G}$; 2. (ii)

$F\vee G=F+G$, hence $F+G$ is closed; 3. (iii)

$\|P(F)+P(G)-P(F\vee G)\|\leq 4\sqrt{\alpha}$; 4. (iv)

$\|P(F)P(G)\|\leq\sqrt{\alpha}$.

Proof.

From the definition of the closed linear span of two linear spaces it follows that $F\vee G=\overline{F+G}$. Notice also that $B_{F}+B_{G}$ is closed as a sum of two weakly compact sets. Let $x\in F\vee G$ be so that $|x|<1/2$. By the above discussion, arbitrarily close to $x$ there exists a point $y=f+g$, where $f\in F$ and $g\in G$. Assume $|f|\leq|g|$. Then

[TABLE]

hence $|f|\leq|g|\leq 1$.

Statement (ii) follows immediately from (i).

To show (iii), let $h\in H$ with $|h|\leq 1$ be given. First we show that $h-(P(F)h+P(G)h)$ is almost orthogonal to $F\vee G$, meaning that $P(F)h+P(G)h$ is a good candidate for an approximation of $P(F\vee G)h$.

Let $x\in F\vee G$, $|x|\leq 1$, be given. According to (i), $x=f+g$ for some $f\in F$ and $g\in G$ with $|f|\leq 2$ and $|g|\leq 2$. Then

[TABLE]

Hence

[TABLE]

To show (iv), let $h\in H$ with $|h|=1$ be given. Then

[TABLE]

∎