Polynomial estimates for the method of cyclic projections in Hilbert spaces

Abstract

We study the method of cyclic projections when applied to closed and linear subspaces ${M_{i}}, i=1,\ldots, m$, of a real Hilbert space $\mathcal {H}$. We show that the average distance to individual sets enjoys a polynomial behavior $o(k^{-1/2})$ along the trajectory of the generated iterates. Surprisingly, when the starting points are chosen from the subspace $\sum _{i=1}^{m}M_i^{\perp }$, our result yields a polynomial rate of convergence $\mathcal {O}(k^{-1/2})$ for the method of cyclic projections itself. Moreover, if $\sum _{i=1}^{m} M_i^{\perp }$ is not closed, then both of the aforementioned rates are best possible in the sense that the corresponding polynomial $k^{1/2}$ cannot be replaced by $k^{1/2+\varepsilon }$ for any ${\varepsilon } >0$.

Appendix

In this section we sketch how to derive Theorems 1.4 and 1.5 in a real Hilbert space, having in mind that the corresponding results of [4] were established in a complex Hilbert space. We also present an alternative proof of Theorem 1.4 by using [16, Lemma 5.2]. For this purpose, we use a complexification argument. For more details concerning the complexification, we refer the reader to [26].

To this end, let $\textbf{H}_{\mathbb C}:= \mathcal {H} + i\mathcal {H}$ be the (external) complexification of $\mathcal H$ with scalar multiplication given by

$$\begin{aligned} (\alpha + i\beta )(x+iy) := \alpha x - \beta y +i (\alpha y + \beta x) \end{aligned}$$

(4.74)

and inner product $\langle \cdot , \cdot \rangle _{\mathbb C}$ defined by

$$\begin{aligned} \langle x+iy, x'+iy'\rangle _{\mathbb C} := \langle x,x'\rangle + \langle y,y'\rangle + i(\langle x',y\rangle - \langle x,y'\rangle ), \end{aligned}$$

(4.75)

where $\alpha ,\beta \in \mathbb R$ and $x,y,x',y'\in \mathcal {H}$. Thus, the induced norm on $\textbf{H}_{\mathbb C}$, denoted by $\Vert \cdot \Vert _{\mathbb C}$, satisfies

$$\begin{aligned} \Vert x+iy\Vert _{\mathbb C}^2 = \Vert x\Vert ^2 + \Vert y\Vert ^2 \end{aligned}$$

(4.76)

for all $x+iy\in \textbf{H}_{\mathbb C}$. It is not difficult to see that $( \textbf{H}_{\mathbb C}, \langle \cdot , \cdot \rangle _{\mathbb C})$ is indeed a complex Hilbert space.

Proof of Theorem 1.4

For each $j = 1, \ldots , m$, let $\textbf{M}_j:= M_j + i M_j$. Observe that $\textbf{M}_j$ is a closed linear subspace of $\textbf{H}_{\mathbb C}$. Denote by $P_{\textbf{M}_j }$ the orthogonal projection onto $\textbf{M}_j$. Then, for each $\textbf{z}= x+iy \in \textbf{H}_{\mathbb C}$, we have $P_{\textbf{M}_j}(\textbf{z}) = P_{M_j}(x) + i P_{M_j}(y)$. This implies that the product $\textbf{T}:= P_{\textbf{M}_m} \ldots P_{\textbf{M}_1}$ satisfies $\textbf{T}( \textbf{z}) = T(x) +iT(y)$, where T is defined as in (1.2). Using induction, we get

$$\begin{aligned} \textbf{T}^k(\textbf{z}) = T^k(x) + i T^k(y). \end{aligned}$$

(4.77)

By [4, Remark 4.2(b) and Theorem 2.1], we have

$$\begin{aligned} k \Vert \textbf{T}^k(\textbf{z}) - \textbf{T}^{k-1}(\textbf{z}) \Vert _{\mathbb C} \rightarrow 0 \quad \text {as } k \rightarrow \infty \end{aligned}$$

(4.78)

for all $\textbf{z}\in \textbf{H}_{\mathbb C}$. In particular, by taking $\textbf{z}:= y_0 +i0$, we obtain

$$\begin{aligned} \Vert y_k - y_{k-1}\Vert = \Vert \textbf{T}^k(\textbf{z}) - \textbf{T}^{k-1}(\textbf{z}) \Vert _{\mathbb C}. \end{aligned}$$

(4.79)

This implies Theorem 1.4.$\square$

Alternative proof of Theorem 1.4

By [16, Lemma 5.2] applied to the operator $\textbf{T}$, we have

$$\begin{aligned} \sum _{k=1}^{\infty } k \Vert \textbf{T}^k(\textbf{z}) - \textbf{T}^{k-1}(\textbf{z}) \Vert _{\mathbb C}^2 < \infty \end{aligned}$$

(4.80)

for all $\textbf{z}\in \textbf{H}_{\mathbb C}$. In particular, by taking $\textbf{z}:= y_0 +i0$, and knowing that the sequence $\{\Vert y_k - y_{k-1}\Vert \}_{k=1}^\infty$ is decreasing, we have

$$\begin{aligned} \nonumber k^2 \Vert y_k - y_{k-1}\Vert ^2&\le 2 k \lceil k/2 \rceil \Vert y_k - y_{k-1}\Vert ^2 \le 4 \sum ^{k}_{n= \lfloor k/2\rfloor + 1} \frac{k}{2} \Vert y_n - y_{n-1}\Vert ^2 \\&\le 4 \sum ^{k}_{n= \lfloor k/2\rfloor + 1} n \Vert y_n - y_{n-1}\Vert ^2 = 4 \sum ^{{k}}_{n= \lfloor k/2\rfloor + 1} n \Vert \textbf{T}^n(\textbf{z}) - \textbf{T}^{n-1}(\textbf{z}) \Vert ^2_{\mathbb C} \rightarrow 0 \end{aligned}$$

(4.81)

as $k \rightarrow \infty$. Thus we have shown that $\Vert y_k-y_{k-1}\Vert = o(k^{-1})$ for all $y_0\in \mathcal H$.$\square$

Proof of Theorem 1.5

It is not difficult to see that $\textbf{M}_j^\perp = M_j^\perp + i M_j^\perp$ for all $j = 1,\ldots ,m$; compare with (2.5). Consequently,

$$\begin{aligned} \textstyle \sum _{j=1}^{m}\textbf{M}_\textbf{j}^\perp = \sum _{j=1}^{m}M_j^\perp + i \sum _{j=1}^{m}M_j^\perp \end{aligned}$$

(4.82)

and thus

$$\begin{aligned} \textstyle \sum _{j=1}^{m} M_j^\perp \text { is (not) closed} \Longleftrightarrow \sum _{j=1}^{m}\textbf{M}_j^\perp \text { is (not) closed}. \end{aligned}$$

(4.83)

Consider now the subspaces of $\textbf{H}_{\mathbb C}$ given by $\textbf{X}_p:= \textbf{M}\oplus (\textbf{I}- \textbf{T})^p(\textbf{H}_{\mathbb C})$ and $\textbf{X}:= \textstyle \bigcap _{p=1}^\infty \textbf{X}_p$, where $\textbf{M}:= M+iM$ and $\textbf{I}$ is the identity operator on $\textbf{H}_{\mathbb C}$, $p=1,2,\ldots$. Obviously, $\textbf{X}_p$ and $\textbf{X}$ are the analogues of $X_p$ and X considered in Theorem 1.5. In fact, we have

$$\begin{aligned} \textbf{X}_p = X_p + i X_p \quad \text {and} \quad \textbf{X}= X + i X. \end{aligned}$$

(4.84)

Consequently, we obtain

$$\begin{aligned} X_p \ (X) \text { is dense in } \mathcal H \Longleftrightarrow \textbf{X}_p \ ( \textbf{X}) \text { is dense in } \textbf{H}_{\mathbb C}. \end{aligned}$$

(4.85)

By [4, Theorem 4.3], for each $\textbf{z}\in \textbf{X}_p$, we get

$$\begin{aligned} \Vert \textbf{T}^k(\textbf{z}) - P_{\textbf{M}}(\textbf{z})\Vert = o(k^{-p}) \end{aligned}$$

(4.86)

where $p = 1,2,\ldots$. Thus, for each $y_0 \in X_p$ it suffices to take $\textbf{z}:= y_0 +i0 \in \textbf{X}_p$, to see that

$$\begin{aligned} \Vert T^k(y_0) - P_M(y_0)\Vert = \Vert \textbf{T}^k(\textbf{z}) - P_{\textbf{M}}(\textbf{z})\Vert = o(k^{-p}), \end{aligned}$$

(4.87)

which shows (1.8). Similarly, for each $y_0 \in X$, it suffices to take $\textbf{z}:= y_0 +i0 \in \textbf{X}$ to see that (4.87) holds for all $p > 0$. Moreover, by [4, Theorem 4.3], we know that $\textbf{X}_p$ and $\textbf{X}$ are dense in $\textbf{H}_{\mathbb C}$.$\square$