On the optimal error bound for the first step in the method of cyclic alternating projections

TL;DR

This paper investigates the optimal error bounds for the initial step in cyclic alternating projections in Hilbert spaces, providing explicit formulas and bounds for the associated functions that measure convergence rates.

Contribution

It introduces a novel approach linking the problem to Hermitian matrix optimization, deriving explicit formulas for three subspaces, and establishing bounds for general cases.

Findings

Explicit formula for f_3(c) derived.

Bounds for f_n(c) established for all n ≥ 4.

Connection made between error bounds and Hermitian matrix optimization.

Abstract

Let $H$ be a Hilbert space and $H_1,...,H_n$ be closed subspaces of $H$. Set $H_0:=H_1\cap H_2\cap...\cap H_n$ and let $P_k$ be the orthogonal projection onto $H_k$, $k=0,1,...,n$. The paper is devoted to the study of functions $f_n:[0,1]\to\mathbb{R}$ defined by $$ f_n(c)=\sup\{\|P_n...P_2 P_1-P_0\|\,|c_F(H_1,...,H_n)\leqslant c\},\,c\in[0,1], $$ where the supremum is taken over all systems of subspaces $H_1,...,H_n$ for which the Friedrichs number $c_F(H_1,...,H_n)$ is less than or equal to $c$. Using the functions $f_n$ one can easily get an upper bound for the rate of convergence in the method of cyclic alternating projections. We will show that the problem of finding $f_n(c)$ is equivalent to a certain optimization problem on a subset of the set of Hermitian complex $n\times n$ matrices. Using the equivalence we find $f_3$ and study properties of $f_n$, $n\geqslant 4$. Moreover, we show that $$ 1-a_n(1-c)-\widetilde{b}_n(1-c)^2\leqslant f_n(c)\leqslant 1-a_n(1-c)+b_n(1-c)^2 $$ for all $c\in[0,1]$, where $a_n=2(n-1)\sin^2(\pi/(2n))$, $b_n=6(n-1)^2\sin^4(\pi/(2n))$ and $\widetilde{b}_n$ is some positive number.

Equations

$$f_n(c)=\sup \{\Vert P_n...P_2{P}_1-P_0\Vert |c_F(H_1,...,H_n)⩽c\},c\in [0,1],$$

$$1-a_n(1-c)-{\tilde{b}}_n(1-c{)}^2⩽f_n(c)⩽1-a_n(1-c)+b_n(1-c{)}^2$$

$$c_F(H_1,H_2):=\sup \{|⟨x,y⟩||x\in H_1\ominus (H_1\cap H_2),\Vert x\Vert ⩽1,y\in H_2\ominus (H_1\cap H_2),\Vert y\Vert ⩽1\}$$

$$x_{2k}-P_{0}x=((P_2{P}_1{)}^k-P_0)x.$$

$$(P_2{P}_1{)}^k-P_0=0\oplus (P_2^{\prime }{P}_1^{\prime }{)}^k=(P_2{P}_1-P_0{)}^k$$

$$\Vert x_{2k}-P_{0}x\Vert =\Vert ((P_2{P}_1{)}^k-P_0)x\Vert =\Vert (P_2{P}_1-P_0{)}^{k}x\Vert ⩽\Vert P_2{P}_1-P_0{\Vert }^k\Vert x\Vert .$$

$$\Vert x_{2k}-P_{0}x\Vert ⩽(c_F(H_1,H_2){)}^k\Vert x\Vert .$$

$$\Vert (P_2{P}_1{)}^k-P_0\Vert ⩽(c_F(H_1,H_2){)}^{2k-1}.$$

$$\Vert x_{2k}-P_{0}x\Vert ⩽(c_F(H_1,H_2){)}^{2k-1}\Vert x\Vert .$$

$$\Vert (P_2{P}_1{)}^k-P_0\Vert =(c_F(H_1,H_2){)}^{2k-1},k⩾1.$$

$$x_0:=x,x_1:=P_1{x}_0,x_2:=P_2{x}_1,...,x_n:=P_n{x}_{n-1}$$

$$x_{n+1}:=P_1{x}_n,x_{n+2}:=P_2{x}_{n+1},...,x_{2n}:=P_n{x}_{2n-1},$$

$$c_F(H_1,H_2)$$

$$x_1\in H_1\ominus (H_1\cap H_2),x_2\in H_2\ominus (H_1\cap H_2),(x_1,x_2)\ne (0,0)\}=$$

$$=\sup \{\frac{⟨x_1,x_2⟩+⟨x_2,x_1⟩}{\Vert x_1{\Vert }^2+\Vert x_2{\Vert }^2}|$$

$$x_1\in H_1\ominus (H_1\cap H_2),x_2\in H_2\ominus (H_1\cap H_2),(x_1,x_2)\ne (0,0)\}$$

$$c_F(H_1,...,H_n)$$

$$x_i\in H_i\ominus (H_1\cap H_2\cap ...\cap H_n),i=1,2,...,n,(x_1,x_2,...,x_n)\ne (0,0,...,0)\}=$$

$$=\sup \{\frac{1}{n-1}\frac{{\sum }_{i\ne j}⟨x_i,x_j⟩}{\Vert x_1{\Vert }^2+\Vert x_2{\Vert }^2+...+\Vert x_n{\Vert }^2}|$$

$$x_i\in H_i\ominus (H_1\cap H_2\cap ...\cap H_n),i=1,2,...,n,(x_1,x_2,...,x_n)\ne (0,0,...,0)\}.$$

$$c_D(H_1,...,H_n)$$

$$x_i\in H_i,i=1,2,...,n,(x_1,x_2,...,x_n)\ne (0,0,...,0)\}=$$

$$=\sup \{\frac{1}{n-1}\frac{{\sum }_{i\ne j}⟨x_i,x_j⟩}{\Vert x_1{\Vert }^2+\Vert x_2{\Vert }^2+...+\Vert x_n{\Vert }^2}|$$

$$x_i\in H_i,i=1,2,...,n,(x_1,x_2,...,x_n)\ne (0,0,...,0)\}.$$

$$c_F(H_1,...,H_n)=c_D(H_1\ominus H_0,...,H_n\ominus H_0),$$

$$\Vert P_1+...+P_n\Vert =1+(n-1)c_D(H_1,...,H_n).$$

$$c_D(H_1,...,H_n)=\frac{1}{n-1}\Vert P_1+...+P_n\Vert -\frac{1}{n-1}$$

$$c_F(H_1,...,H_n)=c_D(H_1\ominus H_0,...,H_n\ominus H_0)=\frac{1}{n-1}\Vert P_1+...+P_n-n{P}_0\Vert -\frac{1}{n-1}.$$

$$\Vert (P_n...P_2{P}_1{)}^k-P_0\Vert ⩽q^k,k⩾1$$

$$\Vert (P_n...P_2{P}_1{)}^k-P_0\Vert =1,k⩾1.$$

$$f_n(c):=\sup \{\Vert P_n...P_2{P}_1-P_0\Vert |c_F(H_1,...,H_n)⩽c\},c\in [0,1].$$

$$f_n(c)=\sup \{\Vert P_n...P_2{P}_1\Vert |c_D(H_1,...,H_n)⩽c\},$$

$$f_n(c)=\sup \{\Vert P_n...P_2{P}_1\Vert |\Vert P_1+...+P_n\Vert ⩽1+(n-1)c\}.$$

$$\Vert (P_n...P_2{P}_1{)}^{k}x-P_{0}x\Vert =\Vert ((P_n...P_2{P}_1{)}^k-P_0)x\Vert ⩽\Vert (P_n...P_2{P}_1{)}^k-P_0\Vert \Vert x\Vert .$$

$$(P_n...P_2{P}_1{)}^k-P_0=0\oplus (P_n^{\prime }...P_2^{\prime }{P}_1^{\prime }{)}^k=(P_n...P_2{P}_1-P_0{)}^k$$

Full text

[TABLE]

where the supremum is taken over all systems of subspaces $H_{1},...,H_{n}$ for which the Friedrichs number $c_{F}(H_{1},...,H_{n})$ is less than or equal to $c$. Using the functions $f_{n}$ one can easily get an upper bound for the rate of convergence in the method of cyclic alternating projections. We will show that the problem of finding $f_{n}(c)$ is equivalent to a certain optimization problem on a subset of the set of Hermitian complex $n\times n$ matrices. Using the equivalence we find $f_{3}$ and study properties of $f_{n}$, $n\geqslant 4$. Moreover, we show that

[TABLE]

for all $c\in[0,1]$, where $a_{n}=2(n-1)\sin^{2}(\pi/(2n))$, $b_{n}=6(n-1)^{2}\sin^{4}(\pi/(2n))$ and $\widetilde{b}_{n}$ is some positive number.

1. Introduction

1.1. The Friedrichs number of a pair of subspaces and

the method of alternating projections for two subspaces

Let $H$ be a complex Hilbert space and $H_{1},H_{2}$ be two closed subspaces of $H$. The number $c_{F}(H_{1},H_{2})$ defined by

[TABLE]

is called the Friedrichs number (more precisely, the cosine of the Friedrichs angle) of subspaces $H_{1},H_{2}$. Why is $c_{F}$ important? A few properties of a pair $H_{1},H_{2}$ can be formulated in terms of the Friedrichs number, for example

1. (1)

the orthogonal projections onto $H_{1}$ and $H_{2}$ commute if and only if $c_{F}(H_{1},H_{2})=0$; 2. (2)

the sum $H_{1}+H_{2}$ is closed if and only if $c_{F}(H_{1},H_{2})<1$,

see, e.g., [6]. Also, the Friedrichs number is closely related to the rate of convergence in the method of alternating projections. This is a well-known method of finding the orthogonal projection of a given element $x\in H$ onto the intersection $H_{1}\cap H_{2}$ when the orthogonal projections $P_{1}$ and $P_{2}$ onto $H_{1}$ and $H_{2}$ are assumed to be known. Define the sequence $x_{0}:=x$, $x_{1}:=P_{1}x_{0}$, $x_{2}:=P_{2}x_{1}$, $x_{3}:=P_{1}x_{2}$, $x_{4}:=P_{2}x_{3}$ and so on. Back in 1933 von Neumann [10] proved that $x_{k}\to P_{0}x$ as $k\to\infty$, where $P_{0}$ is the orthogonal projection onto $H_{1}\cap H_{2}$. What can be said about the rate of convergence? Since $x_{2k}=(P_{2}P_{1})^{k}x$, we see that

[TABLE]

With respect to the orthogonal decomposition $H=(H_{1}\cap H_{2})\oplus(H\ominus(H_{1}\cap H_{2}))$ we have $P_{1}=I\oplus P_{1}^{\prime}$, $P_{2}=I\oplus P_{2}^{\prime}$ and $P_{0}=I\oplus 0$, where $I$ is the identity operator and $P_{1}^{\prime},P_{2}^{\prime}$ are orthogonal projections. Hence

[TABLE]

and

[TABLE]

But $\|P_{2}P_{1}-P_{0}\|=c_{F}(H_{1},H_{2})$ (see, e.g., [6]) and therefore we get estimate

[TABLE]

This estimate is not sharp. Aronszajn [1] proved that

[TABLE]

Therefore we get

[TABLE]

It is worth mentioning that this estimate is sharp because Kayalar and Weinert [8] proved that

[TABLE]

1.2. The method of cyclic alternating projections for $n$ subspaces

Let $H$ be a complex Hilbert space and $H_{1},...,H_{n}$ be closed subspaces of $H$. The method of cyclic alternating projections is a well-known method of finding the orthogonal projection of a given element $x\in H$ onto the intersection $H_{1}\cap H_{2}\cap...\cap H_{n}$ when the orthogonal projections $P_{i}$ onto $H_{i}$, $i=1,2,...,n$ are assumed to be known. The method plays an important role in many areas of mathematics, see, e.g., [5].

Define the sequence

[TABLE]

and after this

[TABLE]

and so on. Back in 1962 Halperin [7] proved that $x_{k}\to P_{0}x$ as $k\to\infty$, where $P_{0}$ is the orthogonal projection onto the intersection $H_{1}\cap H_{2}\cap...\cap H_{n}$. A simple and elegant proof of the result can be found in [9]. In particular, the subsequence $x_{nk}=(P_{n}...P_{2}P_{1})^{k}x\to P_{0}x$ as $k\to\infty$. What can be said about the rate of convergence of $\{x_{nk}|k\geqslant 1\}$ to $P_{0}x$? To answer this question Badea, Grivaux and Müller in [2], [3] introduced the Friedrichs number of $n$ subspaces, $c_{F}(H_{1},...,H_{n})$.

1.3. The Friedrichs number of $n$ subspaces

Badea, Grivaux and Müller noticed that for two subspaces $H_{1},H_{2}$

[TABLE]

and defined

[TABLE]

Since this definition seems to be rather difficult, we will present a more simple formula for $c_{F}$. But first we define the Dixmier number of $n$ subspaces, $c_{D}(H_{1},...,H_{n})$. Following [3], set

[TABLE]

It is clear that

[TABLE]

where $H_{0}=H_{1}\cap H_{2}\cap...\cap H_{n}$.

The Dixmier number of $n$ subspaces is closely related to the sum of the corresponding orthogonal projections.

Proposition 1.1**.**

The following equality holds:

[TABLE]

As a corollary, we see that

[TABLE]

and consequently

[TABLE]

This equality is not new, see [3, Proposition 3.7].

1.4. The rate of convergence in the method of cyclic alternating projections

Let us return to the question on the rate of convergence in the method of cyclic alternating projections. In [3] Badea, Grivaux and Müller showed that

1. (1)

if $c_{F}(H_{1},...,H_{n})<1$, i.e., if the angle between $H_{1},...,H_{n}$ is positive, then

[TABLE]

for some $q=q(c_{F}(H_{1},...,H_{n}))\in[0,1)$. The inequality means that the sequence of operators $(P_{n}...P_{2}P_{1})^{k}$ converges “quickly” to $P_{0}$ as $k\to\infty$. 2. (2)

if $c_{F}(H_{1},...,H_{n})=1$, i.e., if the angle between $H_{1},...,H_{n}$ equals zero, then

[TABLE]

Moreover, the sequence of operators $(P_{n}...P_{2}P_{1})^{k}$ converges strongly to $P_{0}$ as $k\to\infty$ and we have “arbitrarily slow” convergence of $(P_{n}...P_{2}P_{1})^{k}$ to $P_{0}$ (see [3]).

For more complete picture of the quick uniform convergence/arbitrarily slow convergence dichotomy see [3] and [4].

1.5. What this paper is about.

Let $H$ be a complex Hilbert space and $H_{1},...,H_{n}$ be closed subspaces of $H$. Denote by $P_{i}$ the orthogonal projection onto $H_{i}$, $i=1,...,n$. Set $H_{0}:=H_{1}\cap H_{2}\cap...\cap H_{n}$. Denote by $P_{0}$ the orthogonal projection onto $H_{0}$. This paper is devoted to the study of functions $f_{n}:[0,1]\to\mathbb{R}$, $n\geqslant 2$, defined by

[TABLE]

The supremum is taken over all systems of subspaces $H_{1},...,H_{n}$ with $c_{F}(H_{1},...,H_{n})\leqslant c$, where $c\in[0,1]$ is a given number.

Remark 1.1**.**

The reader may wonder why we do not write $c_{F}(H_{1},...,H_{n})=c$. Answer: we believe that the assumption $c_{F}(H_{1},...,H_{n})\leqslant c$ is more convenient for applications. Indeed, finding the exact value of $c_{F}(H_{1},...,H_{n})$ is usually much more difficult than obtaining the inequality $c_{F}(H_{1},...,H_{n})\leqslant c$.

1.6. An equivalent problem

Let us present a problem which is equivalent to the problem of finding $f_{n}(c)$. The fact that these problems are equivalent will be used in the sequel.

Proposition 1.2**.**

For every $c\in[0,1]$

[TABLE]

where the supremum is taken over all systems of subspaces $H_{1},...,H_{n}$ with $c_{D}(H_{1},...,H_{n})\leqslant c$.

Now from Propositions 1.2 and 1.1 it follows that

[TABLE]

1.7. An application of $f_{n}$

Using the functions $f_{n}$ one can easily estimate the rate of convergence in the method of cyclic alternating projections. Indeed, we have

[TABLE]

With respect to the orthogonal decomposition $H=H_{0}\oplus(H\ominus H_{0})$ we have $P_{i}=I\oplus P_{i}^{\prime}$, $i=1,2,...,n$ and $P_{0}=I\oplus 0$. Hence

[TABLE]

and

[TABLE]

where $c_{F}(H_{1},...,H_{n})\leqslant c$.

1.8. Notation

Throughout this paper $H$ is a complex Hilbert space. The inner product in $H$ is denoted by $\langle\cdot,\cdot\rangle$ and $\|\cdot\|$ stands for the corresponding norm, $\|x\|=\sqrt{\langle x,x\rangle}$. The identity operator on $H$ is denoted by $I$ (throughout the paper it is clear which Hilbert space is being considered). All vectors are vector-columns; the letter ”t” means transpose.

2. Results and Questions

Our Main Problem is the following: find $f_{n}(c),c\in[0,1]$ for $n\geqslant 2$. It is trivial that $f_{2}(c)=c,c\in[0,1]$ (this follows from the equality $\|P_{1}P_{2}-P_{0}\|=c_{F}(H_{1},H_{2})$). But what about $f_{n}$, $n\geqslant 3$? Or, at least, what about $f_{3}$?

2.1. The functions $f_{n}$ and an optimization problem

We will show that our Main Problem is equivalent to a certain optimization problem on a subset of the set of Hermitian complex $n\times n$ matrices. For two Hermitian $n\times n$ matrices $A,B$ we will write $A\leqslant B$ if $\langle Ax,x\rangle\leqslant\langle Bx,x\rangle$ for every $x\in\mathbb{C}^{n}$, where $\langle\cdot,\cdot\rangle$ is the standard inner product in the space $\mathbb{C}^{n}$. Equivalently, $A\leqslant B$ if the matrix $B-A$ is positive semidefinite.

Theorem 2.1**.**

The following equality holds:

[TABLE]

where the maximum is taken over all Hermitian complex matrices $A=(a_{ij}|i,j=1,...,n)$ such that $a_{ii}=1$, $i=1,...,n$ and $0\leqslant A\leqslant(1+(n-1)c)I$.

Now it’s time for some notation. For an $n\times n$ matrix $A$ set

[TABLE]

For a real number $t\geqslant 1$ denote by $\mathcal{H}_{n}(t)$ the set of all Hermitian matrices $A=(a_{ij}|i,j=1,...,n)$ such that $a_{ii}=1$ for $i=1,2,...,n$ and $0\leqslant A\leqslant tI$. Then Theorem 2.1 says that

[TABLE]

The following natural problem arises.

Problem 1: find an optimal matrix $A$ for the optimization problem above, i.e., a matrix $A\in\mathcal{H}_{n}(1+(n-1)c)$ such that $\Pi(A)=f_{n}(c)$. Or, at least, find 1-diagonal $(a_{12},a_{23},...,a_{n-1,n})$ of an optimal matrix.

It is natural to try to reduce the set of matrices on which the function $\Pi$ is considered. To this end we will use the following lemma.

Lemma 2.1**.**

Let $t\geqslant 1$. For arbitrary matrix $A\in\mathcal{H}_{n}(t)$ there exists a matrix $B\in\mathcal{H}_{n}(t)$ such that

1. (1)

$b_{ij}\in\mathbb{R}$* for all $i,j=1,2,...,n$ and $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$;* 2. (2)

$b_{i,j}=b_{n+1-i,n+1-j}$* for all $i,j=1,2,...,n$;* 3. (3)

$\Pi(B)\geqslant\Pi(A)$.

Denote by $\mathcal{H}_{n}^{\prime}(t)$ the set of all matrices $A\in\mathcal{H}_{n}(t)$ such that

1. (1)

$a_{ij}\in\mathbb{R}$ for all $i,j=1,...,n$ and $a_{i,i+1}\geqslant 0$ for all $i=1,2,...,n-1$; 2. (2)

$a_{ij}=a_{n+1-i,n+1-j}$ for all $i,j=1,2,...,n$.

For example, matrices from $\mathcal{H}_{3}^{\prime}(t)$ have the form

[TABLE]

where $x\geqslant 0$ and $y\in\mathbb{R}$; matrices from $\mathcal{H}_{4}^{\prime}(t)$ have the form

[TABLE]

where $x\geqslant 0$, $y\geqslant 0$, $w,z\in\mathbb{R}$.

Using Lemma 2.1, we can write

[TABLE]

Now it is natural to specify Problem 1.

Problem 1′**: find an optimal matrix $A\in\mathcal{H}_{n}^{\prime}(1+(n-1)c)$ for the optimization problem above. Or, at least, find a 1-diagonal $(a_{12},a_{23},...,a_{n-1,n})$ of an optimal matrix $A\in\mathcal{H}_{n}^{\prime}(1+(n-1)c)$.

Remark 2.1**.**

It is worth mentioning that there exists a unique optimal 1-diagonal

$(a_{12},a_{23},...,a_{n-1,n})$ (i.e., 1-diagonal of an optimal matrix) for which $a_{i,i+1}\geqslant 0$, $i=1,2,...,n-1$. Indeed, assume that $A,B\in\mathcal{H}_{n}(1+(n-1)c)$ are optimal and $a_{i,i+1}\geqslant 0$, $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$. We claim that $a_{i,i+1}=b_{i,i+1}$, $i=1,2,...,n-1$.

If $c=0$, then $\mathcal{H}_{n}(1)=\{I\}$ and our assertion is clear. Assume that $c>0$. Then $a_{i,i+1}>0$ and $b_{i,i+1}>0$ for $i=1,2,...,n-1$. Consider the matrix $C:=1/2(A+B)$. It is clear that $C\in\mathcal{H}_{n}(1+(n-1)c)$. Since $c_{i,i+1}=1/2(a_{i,i+1}+b_{i,i+1})\geqslant\sqrt{a_{i,i+1}b_{i,i+1}}$, we conclude that $\Pi(C)\geqslant\sqrt{\Pi(A)\Pi(B)}=f_{n}(c)$. If follows that $\Pi(C)=f_{n}(c)$ and consequently $a_{i,i+1}=b_{i,i+1}$ for all $i=1,2,...,n-1$.

2.2. The function $f_{3}$

Now we are ready to find $f_{3}$.

Theorem 2.2**.**

We have

[TABLE]

For $c\in[0,1/4]$ the matrix

[TABLE]

is optimal, for $c\in[1/4,1]$ the matrix

[TABLE]

is optimal.

2.3. On the functions $f_{n}$ with $n\geqslant 4$

For $n=4$, to find $f_{4}(c)$ one have to consider matrices of the form

[TABLE]

where $x\geqslant 0$, $y\geqslant 0$, $w,z\in\mathbb{R}$, and have to maximize $x^{2}y$. We could not find $f_{4}$ (and $f_{n}$ for $n\geqslant 4$). Nevertheless we have the following theorem.

Theorem 2.3**.**

Let $n\geqslant 2$ and $c\in[0,1/(n-1)^{2}]$. Then $f_{n}(c)=(n-1)^{n-1}c^{n-1}$ and the matrix $A\in\mathcal{H}_{n}^{\prime}(1+(n-1)c)$ defined by

[TABLE]

is optimal.

Although we could not find $f_{n}$ for $n\geqslant 4$, we know some properties of the function. Firstly, note that $f_{n}$ is non-decreasing on $[0,1]$ (it follows directly from the definition of $f_{n}$).

Theorem 2.4**.**

The function $f_{n}^{1/(n-1)}$ is concave on $[0,1]$.

Corollary 2.1**.**

The function $f_{n}$ is continuous on $[0,1]$.

Theorem 2.5**.**

The function $f_{n}$ satisfies the following functional equation:

[TABLE]

for $c\in[1/(n-1)^{2},1]$.

Regarding Problem 1*′*, we have the following criterion for a matrix to be optimal.

Proposition 2.1**.**

Let $c>0$ and a matrix $A\in\mathcal{H}_{n}(1+(n-1)c)$ be such that $a_{i,i+1}>0$ for $i=1,2,...,n$. Then $A$ is optimal, i.e., $\Pi(A)=f_{n}(c)$ if and only if

[TABLE]

for arbitrary matrix $B\in\mathcal{H}_{n}(1+(n-1)c)$ with $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$.

2.4. Bounds for $f_{n}(c)$: known upper bounds

In this subsection we present known upper bounds for $f_{n}(c)$. The upper bounds are of great interest because using them one can easily estimate the rate of convergence in the method of cyclic alternating projections (see subsection 1.7).

Let $H$ be a Hilbert space and $H_{1},...,H_{n}$ be closed subspaces of $H$. Denote by $P_{i}$ the orthogonal projection onto $H_{i}$, $i=1,...,n$. Set $H_{0}:=H_{1}\cap H_{2}\cap...\cap H_{n}$. Denote by $P_{0}$ the orthogonal projection onto $H_{0}$. Set $c_{F}:=c_{F}(H_{1},...,H_{n})$. Let $c\in[0,1]$. In what follows we assume that $c_{F}\leqslant c$.

Badea, Grivaux and Müller [3] showed that

[TABLE]

It follows that

[TABLE]

Badea and Seifert [4] showed that

[TABLE]

It follows that

[TABLE]

2.5. Our upper bound for $f_{n}(c)$

Theorem 2.6**.**

The following inequality holds:

[TABLE]

for every $c\in[0,1]$.

By using Theorem 2.6 one can get a more simple estimate for $f_{n}(c)$ (a more simple than the estimate given by Theorem 2.6). One can easily check that $a/b\leqslant(a+x)/(b+x)$, where $0\leqslant a\leqslant b$ and $x\geqslant 0$. Setting $a=n-4(n-1)(\sin^{2}(\pi/(2n)))(1-c)$, $b=n+4(n-1)^{2}(\sin^{2}(\pi/(2n)))(1-c)$ and $x=4(n-1)(\sin^{2}(\pi/(2n)))(1-c)$, we get

[TABLE]

Using Taylor’s theorem with the Lagrange form of the remainder one can easily check that

[TABLE]

for $u\geqslant 0$. Thus

[TABLE]

where $a_{n}=2(n-1)\sin^{2}(\pi/(2n))$ and $b_{n}=6(n-1)^{2}\sin^{4}(\pi/(2n))$. Note that $a_{2}=1$ and $a_{3}=1$.

Question 1. Is it true that $f_{n}(c)\leqslant 1-a_{n}(1-c)$ for all $c\in[0,1]$? Or, at least, for all $c$ which are sufficiently close to $1$?

2.6. Bounds for $f_{n}$: lower bounds

Theorem 2.7**.**

For every $n\geqslant 2$ there exists a positive constant $\widetilde{b}_{n}$ such that

[TABLE]

for all $c\in[0,1]$.

Consequently, we have

[TABLE]

for all $c\in[0,1]$. These inequalities mean that the estimate for $f_{n}(c)$ given by Theorem 2.6 is optimal for $c\approx 1$, up to $O((1-c)^{2})$, $c\to 1-$.

3. Proofs

3.1. Proof of Proposition 1.1

Set $c=c_{D}(H_{1},...,H_{n})$. Since

[TABLE]

we conclude that

[TABLE]

Consider an operator $S:H_{1}\oplus H_{2}\oplus...\oplus H_{n}\to H$ defined by

[TABLE]

Then $1+(n-1)c=\|S\|^{2}$. It is easy to check that $S^{*}:H\to H_{1}\oplus...\oplus H_{n}$ acts as follows: $S^{*}x=(P_{1}x,...,P_{n}x)^{t}$, $x\in H$. Thus $SS^{*}=P_{1}+...+P_{n}$ and

[TABLE]

The proof is complete.

3.2. Proof of Proposition 1.2

Define a function $g_{n}:[0,1]\to\mathbb{R}$ by

[TABLE]

We have to prove that $f_{n}(c)=g_{n}(c)$ for every $c\in[0,1]$.

First, we will show that $f_{n}(c)\leqslant g_{n}(c)$, $c\in[0,1]$. Consider arbitrary system of subspaces $H_{1},...,H_{n}$ of a Hilbert space $H$ such that $c_{F}(H_{1},...,H_{n})\leqslant c$. Set $H_{0}:=H_{1}\cap...\cap H_{n}$ and denote by $P_{0}$ the orthogonal projection onto $H_{0}$. Let us prove that $\|P_{n}...P_{2}P_{1}-P_{0}\|\leqslant g_{n}(c)$. To this end consider the orthogonal decomposition $H=H_{0}\oplus(H\ominus H_{0})=:H_{0}\oplus H^{\prime}$. With respect to this orthogonal decomposition $H_{i}=H_{0}\oplus(H_{i}\ominus H_{0})=:H_{0}\oplus H_{i}^{\prime}$, $i=1,2,...,n$. Thus

[TABLE]

Therefore $c_{D}(H_{1}^{\prime},...,H_{n}^{\prime})\leqslant c$. Further, with respect to the orthogonal decomposition $H=H_{0}\oplus H^{\prime}$ we have $P_{i}=I\oplus P_{i}^{\prime}$, where $P_{i}^{\prime}$ is the orthogonal projection onto $H_{i}^{\prime}$ in $H^{\prime}$, $i=1,2,...,n$, and $P_{0}=I\oplus 0$. Thus $P_{n}...P_{2}P_{1}-P_{0}=0\oplus P_{n}^{\prime}...P_{2}^{\prime}P_{1}^{\prime}$ whence

[TABLE]

because $c_{D}(H_{1}^{\prime},...,H_{n}^{\prime})\leqslant c$. It follows that $f_{n}(c)\leqslant g_{n}(c)$.

Now we will show that $g_{n}(c)\leqslant f_{n}(c)$. Let us prove this inequality for $c\in[0,1)$. Consider arbitrary system of subspaces $H_{1},...,H_{n}$ of a Hilbert space $H$ such that $c_{D}(H_{1},...,H_{n})\leqslant c$. Let us prove that $\|P_{n}...P_{2}P_{1}\|\leqslant f_{n}(c)$. Since $c_{D}(H_{1},...,H_{n})<1$, we conclude that $H_{1}\cap...\cap H_{n}=\{0\}$. Indeed, assume that $H_{1}\cap...\cap H_{n}\neq\{0\}$. Take a vector $u\in H_{1}\cap...\cap H_{n}$, $u\neq 0$ and set $x_{i}=u$, $i=1,2,...,n$. Then

[TABLE]

whence $c_{D}(H_{1},...,H_{n})=1$, contradiction. Therefore $H_{1}\cap...\cap H_{n}=\{0\}$. Thus $c_{F}(H_{1},...,H_{n})=c_{D}(H_{1},...,H_{n})\leqslant c$ and $P_{0}=0$. Hence

[TABLE]

It follows that $g_{n}(c)\leqslant f_{n}(c)$.

Let us show that $g_{n}(1)\leqslant f_{n}(1)$. It is clear that

[TABLE]

(just take $H_{i}=H$, $i=1,2,...,n$, then $\|P_{n}...P_{2}P_{1}\|=\|I\|=1$). So we have to show that $f_{n}(1)\geqslant 1$. To this end we will show that the number $\|P_{n}...P_{2}P_{1}-P_{0}\|$ can be arbitrarily close to $1$. Let $H=\mathbb{C}^{2}$ be the two-dimensional Hilbert space. For an angle $\varphi\in(0,\pi/2]$ define two subspaces

[TABLE]

and

[TABLE]

Then $M\cap N=\{0\}$ and for the orthogonal projections $P_{M}$ and $P_{N}$ onto the subspaces $M$ and $N$, respectively, we have $\|P_{N}P_{M}\|=\|P_{N}(\cos\varphi,\sin\varphi)^{t}\|=\cos\varphi$. Thus for a system of $n$ subspaces $H_{1}=M$, $H_{i}=N$, $i=2,3,...,n$ we have

[TABLE]

can be arbitrarily close to $1$. Therefore $f_{n}(1)\geqslant 1$.

So, we proved that $f_{n}(c)\leqslant g_{n}(c)$ and $g_{n}(c)\leqslant f_{n}(c)$. It follows that $f_{n}(c)=g_{n}(c)$, $c\in[0,1]$.

3.3. Proof of Theorem 2.1

First, note the maximum $\max\{|a_{12}a_{23}...a_{n-1,n}|\,|A\in\mathcal{H}_{n}(1+(n-1)c)\}$ exists, i.e., is attained. This is a direct consequence of the following two facts: the function $A\mapsto|a_{12}a_{23}...a_{n-1,n}|$ is continuous and the set $\mathcal{H}_{n}(1+(n-1)c)$ is compact.

Let us show that

[TABLE]

To this end we consider arbitrary matrix $A\in\mathcal{H}_{n}(1+(n-1)c)$. We have to show that $f_{n}(c)\geqslant|a_{12}a_{23}...a_{n-1,n}|$. Since $A$ is Hermitian and positive semidefinite, we conclude that $A=B^{*}B$ for some $n\times n$ matrix $B$. Let $v_{1},...,v_{n}$ be the columns of $B$, i.e., $B=(v_{1}v_{2}...v_{n})$. We have $a_{ij}=\sum_{k=1}^{n}\overline{b_{ki}}b_{kj}=\langle v_{j},v_{i}\rangle$. This means that $A$ is the Gram matrix of the vectors $v_{1},...,v_{n}$. Since $a_{ii}=1$, we see that $\|v_{i}\|=1$, $i=1,2,...,n$. Consider the system of one dimensional subspaces $H_{i}=\{av_{i}|a\in\mathbb{C}\}$, $i=1,2,...,n$. We claim that $c_{D}(H_{1},...,H_{n})\leqslant c$ and $\|P_{n}...P_{1}\|=|a_{12}a_{23}...a_{n-1,n}|$. It will follow that $f_{n}(c)\geqslant|a_{12}a_{23}...a_{n-1,n}|$. First consider

[TABLE]

Since $P_{i}x=\langle x,v_{i}\rangle v_{i}$, $x\in\mathbb{C}^{n}$, one can easily check that

[TABLE]

It follows that

[TABLE]

Thus $\|P_{n}...P_{1}\|=|a_{12}a_{23}...a_{n-1,n}|$.

Let us show that $c_{D}(H_{1},...,H_{n})\leqslant c$. For arbitrary vectors $x_{1}=a_{1}v_{1},...,x_{n}=a_{n}v_{n}$ we have

[TABLE]

It follows that $\sum_{i\neq j}\langle x_{j},x_{i}\rangle\leqslant(n-1)c\sum_{i=1}^{n}\|x_{i}\|^{2}$. Therefore

[TABLE]

Let us show that

[TABLE]

Define $K:=\max\{|a_{12}a_{23}...a_{n-1,n}||A\in\mathcal{H}_{n}(1+(n-1)c)\}$ and consider arbitrary system of subspaces $H_{1},...,H_{n}$ of a Hilbert space $H$ such that $c_{D}(H_{1},...,H_{n})\leqslant c$. We have to prove that $\|P_{n}...P_{2}P_{1}\|\leqslant K$. Let $v_{1}\in H_{1},...,v_{n}\in H_{n}$ be arbitrary elements with $\|v_{i}\|=1$, $i=1,...,n$. Denote by $G$ the Gram matrix of these elements, i.e., $G=(g_{ij}=\langle v_{j},v_{i}\rangle|i,j=1,...,n)$. We claim that $G\in\mathcal{H}_{n}(1+(n-1)c)$. Indeed, it is clear that $G^{*}=G\geqslant 0$ and $g_{ii}=\|v_{i}\|^{2}=1$, $i=1,...,n$. It remains to show that $G\leqslant(1+(n-1)c)I$. For arbitrary scalars $a_{1},...,a_{n}$ we have

[TABLE]

It follows that $G\leqslant(1+(n-1)c)I$. (It is worth mentioning that this follows also from [3, Proposition 3.4] formulated for the nonreduced configuration constant and [3, Proposition 3.6(f)].) Since $G\in\mathcal{H}_{n}(1+(n-1)c)$, we conclude that $|g_{12}g_{23}...g_{n-1,n}|\leqslant K$, i.e., $|\langle v_{1},v_{2}\rangle\langle v_{2},v_{3}\rangle...\langle v_{n-1},v_{n}\rangle|\leqslant K$. It follows that for arbitrary elements $u_{1}\in H_{1},...,u_{n}\in H_{n}$ we have

[TABLE]

Now consider arbitrary $x\in H$ and set $u_{i}:=P_{i}P_{i-1}...P_{1}x$, $i=1,...,n$. Then

[TABLE]

Thus by (3.1) we get

[TABLE]

that is, $\|u_{n}\|\leqslant K\|u_{1}\|$. Since $u_{1}=P_{1}x$ and $u_{n}=P_{n}P_{n-1}...P_{1}x$, we see that

[TABLE]

Therefore $\|P_{n}...P_{2}P_{1}\|\leqslant K$. This completes the proof.

3.4. Proof of Lemma 2.1

First, note that the set $\mathcal{H}_{n}(t)$ has the following properties:

1. (1)

if $A\in\mathcal{H}_{n}(t)$ and $U$ is a diagonal unitary matrix, i.e., $U=diag(u_{1},...,u_{n})$, where $u_{1},...,u_{n}$ are scalars with $|u_{i}|=1$, $i=1,2,...,n$, then $U^{*}AU\in\mathcal{H}_{n}(t)$; 2. (2)

if $A\in\mathcal{H}_{n}(t)$, then $A^{\top}\in\mathcal{H}_{n}(t)$. Here $(A^{\top})_{ij}=a_{ji}$, $i,j=1,2,...,n$; 3. (3)

if $A\in\mathcal{H}_{n}(t)$, then $\overleftarrow{A}\in\mathcal{H}_{n}(t)$. Here $(\overleftarrow{A})_{ij}=a_{n+1-i,n+1-j}$, $i,j=1,2,...,n$. 4. (4)

the set $\mathcal{H}_{n}(t)$ is convex.

Now we are ready to prove the needed assertion. Let $A\in\mathcal{H}_{n}(t)$. For a diagonal unitary matrix $U=diag(u_{1},u_{2},...,u_{n})$ define $B:=U^{*}AU$. Then $B\in\mathcal{H}_{n}(t)$. Moreover, since $b_{i,i+1}=a_{i,i+1}\overline{u_{i}}u_{i+1}$ one can choose scalars $u_{1},...,u_{n}$ so that $b_{i,i+1}=|a_{i,i+1}|$ for $i=1,2,...,n-1$. Then $\Pi(B)=\Pi(A)$.

Further, consider the matrix $B^{\top}$ and set $C:=1/2(B+B^{\top})$. Then $C\in\mathcal{H}_{n}(t)$. We have $c_{ij}=1/2(b_{ij}+b_{ji})=Re(b_{ij})\in\mathbb{R}$ and $c_{i,i+1}=b_{i,i+1}\geqslant 0$. Therefore $\Pi(C)=\Pi(B)$.

Finally, consider the matrix $\overleftarrow{C}$ and set $D:=1/2(C+\overleftarrow{C})$. Then $D\in\mathcal{H}_{n}(t)$. The matrix $D$ has the following properties:

1. (1)

$d_{ij}=1/2(c_{ij}+c_{n+1-i,n+1-j})\in\mathbb{R}$ for all $i,j$ and $d_{i,i+1}=1/2(c_{i,i+1}+c_{n+1-i,n-i})=1/2(c_{i,i+1}+c_{n-i,n-i+1})\geqslant 0$ for $i=1,2,...,n-1$; 2. (2)

$d_{n+1-i,n+1-j}=d_{ij}$ for all $i,j=1,2,...,n$; 3. (3)

since $d_{i,i+1}=1/2(c_{i,i+1}+c_{n-i,n-i+1})\geqslant\sqrt{c_{i,i+1}c_{n-i,n-i+1}}$ for $i=1,2,...,n-1$, we conclude that

[TABLE]

Thus $D$ is a needed matrix.

3.5. Proof of Theorem 2.2

To find $f_{3}(c)$ one can consider matrices of the form

[TABLE]

where $x\geqslant 0$ and $y\in\mathbb{R}$. We have to maximize $x^{2}$ under the condition $0\leqslant A\leqslant(1+2c)I$.

Consider the condition $A\geqslant 0$. It is well-known that a Hermitian matrix is positive semidefinite if and only if every principal minor of the matrix (including its determinant) is nonnegative. (Recall that a principal minor is the determinant of a principal submatrix; a principal submatrix is a square submatrix obtained by removing certain rows and columns with the same index sets.) Using this criterion one can easily check that $A\geqslant 0$ if and only if

[TABLE]

Consider the condition $A\leqslant(1+2c)I$ $\Leftrightarrow$ $(1+2c)I-A\geqslant 0$. Now one can easily check that $A\leqslant(1+2c)I$ if and only if

[TABLE]

Hence, $0\leqslant A\leqslant(1+2c)I$ if and only if

[TABLE]

We have to maximize $x^{2}$ under these conditions.

Define two linear functions $\varphi(y)=(1+y)/2$ and $\psi(y)=c(2c-y)$. It is clear that $\varphi$ is increasing and $\psi$ is nonincreasing. Consider the equation $\varphi(y)=\psi(y)$. The unique solution is $y=2c-1$. Therefore

[TABLE]

This minimum attains its maximum value $c$ at the point $y=2c-1$. Thus $x^{2}\leqslant c$ and $x\leqslant\sqrt{c}$. Let us check for which $c\in[0,1]$ the values $x=\sqrt{c}$ and $y=2c-1$ are permissible. First consider the inequality $|y|\leqslant\min\{1,2c\}$. It is clear that $-1\leqslant 2c-1\leqslant 1$ and $2c-1\leqslant 2c$. However, the inequality $2c-1\geqslant-2c$ holds only for $c\geqslant 1/4$. For such $c$ we have $\sqrt{c}\leqslant 1$ and $\sqrt{c}\leqslant 2c$. Conclusion: for $c\in[1/4,1]$ the optimal values $x=\sqrt{c}$, $y=2c-1$, the optimal matrix is equal to

[TABLE]

and $f_{3}(c)=c$.

Consider the case $c\in[0,1/4)$. Then $2c-1<-2c$ and hence the conditions for $x$ and $y$ can be rewritten as

[TABLE]

Now it is easy to see that the optimal values of $x$ and $y$ are $x=2c$ and $y=-2c$. Therefore the optimal matrix is equal to

[TABLE]

and $f_{3}(c)=4c^{2}$.

3.6. Proof of Theorem 2.3

Consider an arbitrary matrix $A\in\mathcal{H}_{n}(1+(n-1)c)$. Since $A\leqslant(1+(n-1)c)I$, we conclude that the matrix $(1+(n-1)c)I-A$ is positive semidefinite. It follows that the determinant of every $2\times 2$ submatrix

[TABLE]

is nonnegative, i.e., $(n-1)^{2}c^{2}-|a_{ij}|^{2}\geqslant 0$, $|a_{ij}|\leqslant(n-1)c$. Therefore $|a_{12}a_{23}...a_{n-1,n}|\leqslant(n-1)^{n-1}c^{n-1}$.

On the other hand, consider the matrix $J$ where each entry is equal to $1$, i.e.,

[TABLE]

It is easily seen that $J$ is positive semidefinite and the largest eigenvalue of $J$ equals $n$. Thus $0\leqslant J\leqslant nI$, $-nI\leqslant-J\leqslant 0$, $-(n-1)I\leqslant I-J\leqslant I$, $-(n-1)^{2}cI\leqslant(n-1)c(I-J)\leqslant(n-1)cI$ and

[TABLE]

Set $M:=I+(n-1)c(I-J)$. Since $c\in[0,1/(n-1)^{2}]$ and

[TABLE]

we see that $M\in\mathcal{H}_{n}(1+(n-1)c)$ and $\Pi(M)=(n-1)^{n-1}c^{n-1}$. Therefore $f_{n}(c)=(n-1)^{n-1}c^{n-1}$.

Finally, define $U:=diag(-1,1,-1,1,...)$ and consider the matrix $A:=U^{*}MU$. Since

[TABLE]

we conclude that $A\in\mathcal{H}_{n}^{\prime}(1+(n-1)c)$ and $\Pi(A)=(n-1)^{n-1}c^{n-1}$. Thus $A$ is optimal.

3.7. Proof of Theorem 2.4

Let $c_{1},c_{2}\in[0,1]$ and $\lambda\in(0,1)$. We have to show that

[TABLE]

Let $A\in\mathcal{H}_{n}(1+(n-1)c_{1})$ be such that $a_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$ and $\Pi(A)=f_{n}(c_{1})$. Let $B\in\mathcal{H}_{n}(1+(n-1)c_{2})$ be such that $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$ and $\Pi(B)=f_{n}(c_{2})$. Consider the matrix $\lambda A+(1-\lambda)B$. It is clear that $\lambda A+(1-\lambda)B\in\mathcal{H}_{n}(1+(n-1)(\lambda c_{1}+(1-\lambda)c_{2}))$. Thus

[TABLE]

whence

[TABLE]

Now we will use the inequality

[TABLE]

where $m$ is a natural number and numbers $s_{1},...,s_{m},t_{1},...,t_{m}$ are nonnegative. We have

[TABLE]

The proof is completed.

3.8. Proof of Corollary 2.1

Define the function $g_{n}:=f_{n}^{1/(n-1)}$. Let us prove that $g_{n}$ is continuous on $[0,1]$. It will follow that $f_{n}=g_{n}^{n-1}$ is also continuous on $[0,1]$.

We will use the following well-known fact: if a function $\varphi:(a,b)\to\mathbb{R}$ is convex on $(a,b)$, then $\varphi$ is continuous on $(a,b)$. Since $g_{n}$ is concave on $[0,1]$ (by Theorem 2.4), we conclude that $g_{n}$ is continuous on $(0,1)$. Theorem 2.3 implies that $g_{n}(c)=(n-1)c$ for $c\in[0,1/(n-1)^{2}]$. Thus $g_{n}$ is continuous at the point [math]. Let us show that $g_{n}$ is continuous at the point $1$. We have $g_{n}(1)=(f_{n}(1))^{1/(n-1)}=1$ (Proposition 1.2 implies that $f_{n}(1)=1$) and $g_{n}(0)=0$. Since $g_{n}$ is concave on $[0,1]$, we conclude that $g_{n}(c)\geqslant c$ for all $c\in[0,1]$. Since $g_{n}$ is non-decreasing on $[0,1]$, we conclude that $g_{n}(c)\leqslant 1$ for all $c\in[0,1]$. Thus $c\leqslant g_{n}(c)\leqslant 1$ for $c\in[0,1]$. It follows that $\lim_{c\to 1-}g_{n}(c)=1=g_{n}(1)$. Therefore $g_{n}$ is continuous at the point $1$.

We proved that the function $g_{n}$ is continuous at every point of the segment $[0,1]$. Thus $g_{n}$ is continuous on $[0,1]$.

3.9. Proof of Theorem 2.5

Fix $c\in[1/(n-1)^{2},1]$. Consider arbitrary matrix $A\in\mathcal{H}_{n}(1+(n-1)c)$. Then $0\leqslant A\leqslant(1+(n-1)c)I$, $0\leqslant(1+(n-1)c)I-A\leqslant(1+(n-1)c)I$ and

[TABLE]

Define

[TABLE]

then $b_{ii}=1$, $i=1,2,...,n$ and $b_{ij}=-a_{ij}/((n-1)c)$ for $i\neq j$. It follows that $B\in\mathcal{H}_{n}(1+(n-1)/((n-1)^{2}c))$ and $\Pi(B)=\Pi(A)/((n-1)^{n-1}c^{n-1})$. Since the mapping $A\mapsto B$ from $\mathcal{H}_{n}(1+(n-1)c)$ to $\mathcal{H}_{n}(1+(n-1)/((n-1)^{2}c))$ is one-to-one and onto, we conclude that

[TABLE]

3.10. Proof of Proposition 2.1

First assume that

[TABLE]

for arbitrary matrix $B\in\mathcal{H}_{n}(1+(n-1)c)$ with $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$. Then

[TABLE]

It follows that

[TABLE]

and therefore $f_{n}(c)=\Pi(A)$.

Now assume that a matrix $A$ is optimal, i.e., $\Pi(A)=f_{n}(c)$. Consider arbitrary matrix $B\in\mathcal{H}_{n}(1+(n-1)c)$ with $b_{i,i+1}\geqslant 0$ for $i=1,2,...,n-1$. For arbitrary number $\alpha\in[0,1]$ the matrix $(1-\alpha)A+\alpha B$ belongs to $\mathcal{H}_{n}(1+(n-1)c)$. Define the function

[TABLE]

Since $A$ is optimal, we conclude that $\varphi(\alpha)\leqslant\Pi(A)=\varphi(0)$ for $\alpha\in[0,1]$. It follows that $\varphi^{\prime}(0)\leqslant 0$, i.e.,

[TABLE]

Thus

[TABLE]

3.11. Proof of Theorem 2.6

For $n\geqslant 2$ set $D_{n}:=n/(4\sin^{2}(\pi/(2n)))$.

Lemma 3.1**.**

For arbitrary real numbers $a_{1},...,a_{n}$ the following inequality holds:

[TABLE]

Proof.

Consider the inequality

[TABLE]

where $D>0$ and $a_{1},...,a_{n}\in\mathbb{R}$. We have to show that this inequality is valid for $D=D_{n}$ and arbitrary $a_{1},...,a_{n}\in\mathbb{R}$. Inequality (3.3) does not change after substitution $a_{i}\to a_{i}+b$, $i=1,2,...,n$, where $b\in\mathbb{R}$. Therefore without loss of generality we can and will assume that $a_{1}+...+a_{n}=0$. Then the left side of inequality (3.3) is equal to

[TABLE]

Thus inequality (3.3) is equivalent to the inequality

[TABLE]

which is equivalent to

[TABLE]

Define the matrix

[TABLE]

corresponding to the quadratic form $\sum_{i=1}^{n-1}(a_{i}-a_{i+1})^{2}$. The matrix $L$ is the Laplacian matrix of the graph $P_{n}$ with vertices $1,2,...,n$ and edges $\{1,2\},\{2,3\},...,\{n-1,n\}$ (the path of length $n-1$). Let $\lambda_{1}\leqslant\lambda_{2}\leqslant...\leqslant\lambda_{n}$ be the spectrum of $L$. It is clear that the eigenvalue $\lambda_{1}=0$ (with a corresponding eigenvector $(1,1,...,1)^{t}$) and the multiplicity of $\lambda_{1}$ is equal to $1$. Inequality (3.4) can be written as $\langle La,a\rangle\geqslant(n/D)\|a\|^{2}$, where a vector $a=(a_{1},...,a_{n})^{t}$ is orthogonal to the vector $(1,1,...,1)^{t}$. Therefore this inequality will be valid if $n/D=\lambda_{2}$, i.e., if $D=n/\lambda_{2}$. It is well-known that $\lambda_{2}=4\sin^{2}(\pi/(2n))$. Thus inequality (3.3) will be valid with $D=n/(4\sin^{2}(\pi/(2n)))=D_{n}$. ∎

Lemma 3.2**.**

For arbitrary vectors $v_{1},...,v_{n}\in H$ the following inequality holds:

[TABLE]

Proof.

Set $a_{1}:=0$ and $a_{i}:=\|v_{1}-v_{2}\|+...+\|v_{i-1}-v_{i}\|$ for $i\geqslant 2$. For $i<j$ we have $a_{i}-a_{j}=-(\|v_{i}-v_{i+1}\|+...+\|v_{j-1}-v_{j}\|)$. Using Lemma 3.1 we get

[TABLE]

It follows that

[TABLE]

∎

Now we are ready to prove Theorem 2.6. The proof of Theorem 2.6 is based on Proposition 1.2. Let $H$ be a complex Hilbert space and $H_{1},...,H_{n}$ be closed subspaces of $H$. Denote by $P_{i}$ the orthogonal projection onto $H_{i}$, $i=1,...,n$. Assume that $c_{D}(H_{1},...,H_{n})\leqslant c$. We have to prove that

[TABLE]

By the definition of $c_{D}$ for arbitrary vectors $x_{1}\in H_{1},...,x_{n}\in H_{n}$ we have

[TABLE]

It follows that

[TABLE]

By Lemma 3.2 we get

[TABLE]

Now consider arbitrary $x\in H$ and set $x_{i}:=P_{i}...P_{2}P_{1}x$, $i=1,...,n$. Then $x_{i+1}=P_{i+1}x_{i}$, $i=1,...,n-1$. It follows that $\|x_{i+1}\|\leqslant\|x_{i}\|$ and

[TABLE]

Thus $\|x_{1}\|\geqslant\|x_{2}\|\geqslant...\geqslant\|x_{n}\|$ and $\sum_{i=1}^{n-1}\|x_{i}-x_{i+1}\|^{2}=\|x_{1}\|^{2}-\|x_{n}\|^{2}$. Using (3.5) we get

[TABLE]

We rewrite this inequality as follows:

[TABLE]

i.e.,

[TABLE]

that is,

[TABLE]

Since $x_{n}=P_{n}...P_{2}P_{1}x$ and $x_{1}=P_{1}x$, we conclude that

[TABLE]

It follows that $\|P_{n}...P_{2}P_{1}\|\leqslant\sqrt{\dfrac{D_{n}-\varepsilon}{D_{n}+(n-1)\varepsilon}}$. Finally, note that

[TABLE]

and the proof of Theorem 2.6 is complete.

3.12. Proof of Theorem 2.7

The proof of Theorem 2.7 is based on Proposition 1.2. Consider the two-dimensional Hilbert space $H=\mathbb{C}^{2}$. For a number $\alpha\in\mathbb{R}$ let $L(\alpha)=\{(\cos\alpha,\sin\alpha)^{t}z\,|z\in\mathbb{C}\}$ be the one-dimensional subspace spanned by the vector $(\cos\alpha,\sin\alpha)^{t}$. Let $\alpha_{1},...,\alpha_{n}$ be real numbers such that for some $i$ and $j$ $\alpha_{i}\neq\alpha_{j}$. For each $\tau\geqslant 0$ consider the system of one-dimensional subspaces $H_{k}:=L(\alpha_{k}\tau)$, $k=1,...,n$. Let us find

[TABLE]

By Proposition 1.1 we have $\|P_{1}+...+P_{n}\|=1+(n-1)c(\tau)$, where $P_{k}$ is the orthogonal projection onto $L(\alpha_{k}\tau)$, $k=1,2,...,n$. We have

[TABLE]

for $k=1,2,...,n$. Therefore

[TABLE]

Let us find $\|P_{1}+...+P_{n}\|=\|M(\tau)\|$. Since the matrix $M(\tau)$ is Hermitian and positive semidefinite, we conclude that $\|M(\tau)\|$ is equal to the largest eigenvalue of $M(\tau)$. The characteristic polynomial of $M(\tau)$ is equal to $\lambda^{2}-tr(M(\tau))\lambda+\det(M(\tau))$. It is clear that trace of $M(\tau)$ is equal to $n$. Consider

[TABLE]

Now we have the following equation for the eigenvalues of $M(\tau)$:

[TABLE]

The largest root is equal to $(n+\sqrt{n^{2}-4d(\tau)})/2$. Therefore

[TABLE]

Now we note a few properties of the functions $c(\tau)$ and $d(\tau)$:

(1) $d(0)=0$ and $c(0)=1$;

(2) the functions $d$ and $c$ are continuous on $[0,+\infty)$;

(3) there exists $\tau_{0}=\tau_{0}(\alpha_{1},...,\alpha_{n})>0$ such that $d$ is increasing on $[0,\tau_{0}]$. Consequently, $c$ is decreasing on $[0,\tau_{0}]$.

(4) Since $\sin^{2}(\alpha\tau)=\alpha^{2}\tau^{2}+O(\tau^{4})$ as $\tau\to 0+$, we conclude that

[TABLE]

where $s_{1}=s_{1}(\alpha_{1},...,\alpha_{n})=\sum_{i<j}(\alpha_{i}-\alpha_{j})^{2}$.

(5) Since $\sqrt{1+u}=1+u/2+O(u^{2})$ as $u\to 0$, we conclude that

[TABLE]

as $\tau\to 0+$. Thus for $c(\tau)$ we have

[TABLE]

i.e.,

[TABLE]

Now consider $\|P_{n}...P_{2}P_{1}\|$. We have

[TABLE]

Thus for small enough $\tau$ we have

[TABLE]

where $s_{2}=\sum_{i=1}^{n-1}(\alpha_{i}-\alpha_{i+1})^{2}$. So

[TABLE]

From (3.6) it follows that

[TABLE]

and

[TABLE]

Now using Proposition 1.2, (3.7), (3.9) and (3.8) we get

[TABLE]

for $\tau\in(0,\tau_{1}]$, where $\tau_{1}=\tau_{1}(\alpha_{1},...,\alpha_{n})>0$ and $K=K(\alpha_{1},...,\alpha_{n})$. Thus

[TABLE]

for all $c\in[c(\tau_{1}),1]$.

Now we want to choose $\alpha_{1},...,\alpha_{n}$ for which the value of $s_{2}/s_{1}$ is as small as possible. Consider $s_{2}/s_{1}$. Since the value of $s_{2}/s_{1}$ does not change under substitution $\alpha_{i}\to\alpha_{i}+a$, $i=1,2,...,n$, $a\in\mathbb{R}$, we can and will assume that $\alpha_{1}+...+\alpha_{n}=0$. This equality means that the vector $\overline{\alpha}=(\alpha_{1},...,\alpha_{n})^{t}$ is orthogonal to the vector $e=(1,...,1)^{t}$. For such $\overline{\alpha}$ we have

[TABLE]

Also $s_{2}=\langle L\overline{\alpha},\overline{\alpha}\rangle$, where

[TABLE]

and $\langle\cdot,\cdot\rangle$ is the standard inner product in $\mathbb{R}^{n}$. Note that the matrix $L$ is the Laplacian matrix of the graph $\mathcal{P}_{n}$ with vertices $1,2,...,n$ and edges $\{1,2\},\{2,3\},...,\{n-1,n\}$ (the path of length $n-1$). Let $\lambda_{1}\leqslant\lambda_{2}\leqslant...\leqslant\lambda_{n}$ be the spectrum of $L$. It is clear that $L$ is positive semidefinite and $\ker(L)$ is the one-dimensional subspace spanned by the vector $e$. Thus $\lambda_{1}=0$ and $\lambda_{2}>0$. Note that $\lambda_{2}$ is called the algebraic connectivity of the graph $\mathcal{P}_{n}$ and is denoted by $a(\mathcal{P}_{n})$. It is well-known that $\lambda_{2}=a(\mathcal{P}_{n})=4\sin^{2}(\pi/(2n))$.

Now we return to the problem of minimizing the value of $s_{2}/s_{1}$. We have

[TABLE]

The minimum value of $\langle L\overline{\alpha},\overline{\alpha}\rangle/\|\overline{\alpha}\|^{2}$ under conditions $\langle\overline{\alpha},e\rangle=0$, $\overline{\alpha}\neq 0$ is equal to $\lambda_{2}$ (and it is attained when $\overline{\alpha}$ is an eigenvector of $L$ corresponding to the eigenvalue $\lambda_{2}$). So, let $\overline{\alpha}$ be an eigenvector of $L$ corresponding to the eigenvalue $\lambda_{2}$, then from (3.10) it follows that

[TABLE]

for all $c\in[c_{n},1]$, where $c_{n}<1$ and $K=K_{n}$. By enlarging $K$, if necessary, we get the inequality

[TABLE]

for all $c\in[0,1]$.