Procrustes Problems and Parseval Quasi-Dual Frames

Abstract

Parseval frames have particularly useful properties, and in some cases, they can be used to reconstruct signals which were analyzed by a non-Parseval frame. In this paper, we completely describe the degree to which such reconstruction is feasible. Indeed, notice that for fixed frames ${\mathcal{F}}$ and ${\mathcal{X}}$ with synthesis operators F and X, the operator norm of FX ^∗−I measures the (normalized) worst-case error in the reconstruction of vectors when analyzed with ${\mathcal{X}}$ and synthesized with ${\mathcal{F}}$. Hence, for any given frame ${\mathcal{F}}$, we compute explicitly the infimum of the operator norm of FX ^∗−I, where ${\mathcal{X}}$ is any Parseval frame. The ${\mathcal{X}}$’s that minimize this quantity are called Parseval quasi-dual frames of ${\mathcal{F}}$. Our treatment considers both finite and infinite Parseval quasi-dual frames.

Appendix

In this section we revise some facts related with unitary approximations of bounded operators in Hilbert spaces and we present the proof of Theorem 15. Throughout this section ${\mathcal{F}}$ denotes a frame with finite excess for an infinite dimensional Hilbert space $\mathcal{H}$. So that, if $F\in B(\ell _{2},\mathcal{H})$ denotes the frame operator of ${\mathcal{F}}$, dim(N(F))<∞.

Recall that given $T\in B(\mathcal{H})$, then m(T)=inf{∥Tx∥,∥x∥=1}=minσ(|T|), σ _e (T) denotes the essential spectrum of T, m _e (T)=minσ _e (|T|) and ∥T∥ _e =maxσ _e (|T|).

We begin by recalling one of the main results from [25].

Theorem 18

(Theorem 1.3 in [25])

Let $T\in B(\mathcal{H})$, then,

$$d_{\mathcal{U}(\mathcal{H})}(T):=\inf_{U\in\mathcal{U}(\mathcal {H})} \|T-U\|= \begin{cases} \max\{ \|T\|-1, 1-m(T)\} & \textit{if } \operatorname{ind}(T)=0\\ \max\{\|T\|-1, 1+m_e(T)\} &\textit{if } \operatorname{ind}(T)<0. \end{cases} $$

(the case $\operatorname{ind}(T>0)$ follows using T ^∗).

Let ${\mathcal{M}}$ be an infinite dimensional closed subspace of ℓ ₂ and consider a coisometry $Y\in B(\ell_{2},\mathcal{H})$ such that $Y^{*}Y=P_{\mathcal{M}}$. Recall that our interest in the unitary approximation problem is motivated by Eq. (13). Hence, in order to apply Theorem 18 to the operator $(Y^{*}F)|_{\mathcal{M}}\in B({\mathcal{M}})$, we need to relate the Fredholm index of this operator to that of F (which is finite, since we are assuming that F is surjective and that N(F) has finite dimension). In order to describe such a relation we introduce the following notation: given $T\in B(\mathcal{H})$ and ${\mathcal{M}}\subseteq\mathcal{H}$ a closed subspace then $T_{\mathcal{M}}=P_{\mathcal{M}}T|_{\mathcal{M}}\in B({\mathcal{M}})$ denotes the compression of T to ${\mathcal{M}}$.

Lemma 19

Let $T\in B(\mathcal{H})$.

• 1. Let ${\mathcal{M}}\subseteq\mathcal{H}$ be a closed subspace with $\dim{\mathcal{M}}^{\perp}<\infty$. If $\operatorname{ind}(T)\in\mathbb{Z}$ then $\operatorname{ind}(T_{\mathcal{M}})=\operatorname{ind}(T)$.

• 2. If T is a closed range operator with $\operatorname {ind}(T)=-\infty$, then $\operatorname{ind}(T_{R(T)})=-\infty$.

Proof

1. Let P be the orthogonal projection onto ${\mathcal{M}}$ and let $m=\dim N(P)=\dim{\mathcal{M}}^{\perp}<\infty$. Notice that $\dim N(T_{\mathcal{M}}) = \dim N(PTP)-m$ and $\dim N(T^{*}_{\mathcal{M}}) =\dim N(PT^{*}P) -m$ so that $\operatorname {ind}(PTP)=\operatorname{ind}(T_{\mathcal{M}})$. Moreover, T=PTP+PT(I−P)+(I−P)T=PTP+K, where K is a finite rank operator. Then, $\operatorname{ind}(T)=\operatorname{ind}(PTP)=\operatorname {ind}(T_{\mathcal{M}})$.

2. Suppose now that T is a semi-Fredholm operator with $\operatorname {ind}(T)=-\infty$. Let ${\mathcal{M}}=R(T)$, $\mathcal{N}=N(T)$ and notice that by hypothesis $\dim{\mathcal{M}}^{\perp}=\dim N(T^{*})=\infty$, $\dim \mathcal{N}<\infty$. This last fact shows that $\dim{\mathcal{M}}=\infty$.

In this case we have that $N(T_{\mathcal{M}})=\mathcal{N}\cap{\mathcal{M}}$, and $\dim N(T_{\mathcal{M}} ^{*})=\dim\mathcal{N}^{\perp}\cap{\mathcal{M}}^{\perp}$; indeed, the first identity easily follows from the definition of $T_{\mathcal{M}}$. For the second identity, notice that $T_{\mathcal{M}} ^{*}=P_{\mathcal{M}} T^{*}|_{\mathcal{M}}$ and $N(T_{\mathcal{M}}^{*})\subseteq{\mathcal{M}}=N(T^{*})^{\perp}$; hence $T^{*}|_{N(T_{\mathcal{M}} ^{*})}:N(T_{\mathcal{M}} ^{*})\rightarrow R(T^{*})\cap{\mathcal{M}}^{\perp}= \mathcal{N}^{\perp}\cap{\mathcal{M}}^{\perp}$ is a linear isomorphism i.e. a linear transformation with bounded inverse.

Let $X, Y\in B(\mathcal{H})$ be coisometries with initial space ${\mathcal{M}}$ and $\mathcal{N}^{\perp}$ respectively. Hence, $\operatorname{ind}(X)=\dim{\mathcal{M}}^{\perp}=\infty$ and $\operatorname{ind}(Y)=\dim\mathcal{N}$. Then, by the additivity property of the index for (left) semi-Fredholm operators, $\operatorname{ind}(YX^{*})=\operatorname{ind}(Y)+\operatorname {ind}(X^{*})=-\infty$. On the other hand, arguing as before it is easy to see that $\dim N(XY^{*}) =\dim{\mathcal{M}}\cap\mathcal{N}$ and $\dim N(YX^{*}) =\dim {\mathcal{M}}^{\perp}\cap\mathcal{N}^{\perp}$. Hence, $\operatorname {ind}(T_{\mathcal{M}})=\dim{\mathcal{M}}\cap\mathcal{N}- \dim{\mathcal{M}}^{\perp}\cap\mathcal{N}^{\perp}=-\infty$. □

Given $T\in B(\mathcal{H})$, a useful way to compute m _e (T) and ∥T∥ _e is by using the maps

$$U_k(T)=\sup_{E\in\mathcal{P}(\mathcal{H}), \operatorname{tr}E\leq k} \operatorname{tr}(|T|E) \quad \text{and}\quad L_k(T)=\inf_{ E\in\mathcal{P}(\mathcal{H}), \operatorname{tr}E\leq k} \operatorname{tr}(|T|E) , $$

where $\mathcal{P}(\mathcal{H})$ denotes the set of orthogonal projections in $B(\mathcal{H})$ and $\operatorname{tr}(\cdot)$ denotes the usual (semifinite) trace in $B(\mathcal{H})$. Indeed, by [2, Proposition 3.5] we have that

$$ m_e(T)=\lim_{k\rightarrow\infty} \frac{L_k(T)}{k} \quad\text{and} \quad\|T\|_e=\lim_{k\rightarrow\infty} \frac{U_k(T)}{k}. $$

(14)

On the other hand, for $n\in\mathbb{N}$ we let

$$\begin{aligned} &u_n(T)=\sup\bigl\{ \min\sigma(|T|_{\mathcal{M}}) : \dim( {\mathcal{M}})=n\bigr\} \quad\text{and} \\ & l_n(T)=\inf\bigl\{ \max \sigma(|T|_{\mathcal{M}}) : \dim({\mathcal{M}})=n\bigr\} . \end{aligned}$$

In this way we obtain the non-increasing sequence $(u_{n}(T))_{n\in \mathbb{N}}$ and the non-decreasing sequence $(l_{n}(T))_{n\in\mathbb{N}}$. Denote by E(⋅) the spectral measure of |T|. Then, it is easy to see that u _n =∥T∥ _e if the range of the projection E((∥T∥ _e ,∥T∥]) is a subspace of dimension k<n. Otherwise u _n =λ _n if $\lambda=(\lambda_{i})_{i=1}^{m}$ are the eigenvalues (counting multiplicity) arranged in a decreasing order of the finite rank operator E(I)|T|E(I), where I⊂(∥T∥ _e ,∥T∥] is any interval such that $n<\operatorname{rk}(E(I))=m<\infty$. There is an obvious analogue for l _n using the eigenvalues of |T| strictly smaller than m _e (T).

Proposition 20

Let T be a positive operator with dim(N(T))=n and let ${\mathcal{M}}\subseteq\mathcal{H}$ be a closed subspace with $\dim{\mathcal{M}}=\infty$. Let u _n+1=u _n+1(T) and l _n+1=l _n+1(T). Then,

• 1. $m_{e}(T)\leq m_{e}(T_{\mathcal{M}})$.

• 2. If we assume further that $\dim{\mathcal{M}}^{\perp}=n$ then $m(T_{\mathcal{M}})\leq l_{n+1}\leq u_{n+1}\leq\|T_{\mathcal{M}}\|$.

Moreover, for every ε>0 there exist infinite dimensional closed subspaces $\mathcal{N}_{\varepsilon}$ and ${\mathcal{M}}$ such that $\dim({\mathcal{M}}^{\perp})=n$ and

• 3. $m(T_{{\mathcal{M}}})= l_{n+1}$ and $\|T_{{\mathcal{M}}}\|= u_{n+1}$.

• 4. $m_{e}(T_{\mathcal{N}_{\varepsilon}})=m_{e}(T)$ and $\|T_{\mathcal{N}_{\varepsilon}}\|\leq m_{e}(T)+\varepsilon$.

Proof

Let P be the orthogonal projection onto ${\mathcal{M}}$. Then, by Eq. (14) we get that

$$\begin{aligned} m_e(T_{\mathcal{M}})&=\lim_{k\rightarrow\infty} \frac{1}{k} \inf\bigl\{ \operatorname{tr}(T_{\mathcal{M}}E) : E\in \mathcal{P}({\mathcal{M}}), \operatorname{tr}(E)\leq k\bigr\} \\ &=\lim_{k\rightarrow\infty} \frac{1}{k} \inf\bigl\{ \operatorname {tr}(TE) : E\in\mathcal{P}(\mathcal{H}), \operatorname {tr}(E)\leq k, R(E) \subset{\mathcal{M}}\bigr\} \\ &\geq\lim_{k\rightarrow\infty} \frac{1}{k} \inf\bigl\{ \operatorname {tr}(TE) : E\in\mathcal{P}(\mathcal{H}) \operatorname {tr}(E)\leq k\bigr\} =m_e(T). \end{aligned}$$

Assume further that $\dim{\mathcal{M}}^{\perp}=n$. Given ε>0, let ${\mathcal{S}}_{\varepsilon}$ be a n+1-dimensional subspace of $\mathcal{H}$ such that $\min\sigma(T_{{\mathcal{S}}_{\varepsilon}})>u_{n+1}-\varepsilon$. We claim that ${\mathcal{S}}_{\varepsilon}\cap {\mathcal{M}}\neq\{0\}$: indeed, if ${\mathcal{S}}_{\varepsilon}\cap{\mathcal{M}}= \{0\}$ then $P_{{\mathcal{M}}^{\perp}}|_{{\mathcal{S}}_{\varepsilon}}:{\mathcal{S}}_{\varepsilon}\rightarrow{\mathcal{M}}^{\perp}$ is an injection, which contradicts the fact that $\dim{\mathcal{S}}_{\varepsilon}>\dim{\mathcal{M}}^{\perp}$. Therefore, if $x\in{\mathcal{S}}_{\varepsilon}\cap{\mathcal{M}}$ with ∥x∥=1 then

$$\langle T_{\mathcal{M}} x , x \rangle= \langle T x , x \rangle= \langle T_{{\mathcal{S}}_\varepsilon} x , x \rangle \geq u_{n+1}-\varepsilon. $$

Since ε was arbitrary, we see that $\|T_{\mathcal{M}}\|\geq u_{n+1}$. The proof for the lower bound is similar.

In order to finish the proof, we exhibit the subspaces ${\mathcal{M}}_{\varepsilon}$ and $\mathcal{N}_{\varepsilon}$ as above.

If ∥T∥=∥T∥ _e we just take ${\mathcal{M}}_{\varepsilon}=N(T)^{\perp}$ and we are done. In case that ∥T∥>∥T∥ _e , define r=dimR(E((∥T∥ _e ,∥T∥])) and let k=min{n,r}. Notice that in this case, u ₁≥⋯≥u _k are eigenvalues of T. Denote by ${\mathcal{S}}=N(T)\oplus\mathcal{E}$, where $\mathcal{E}$ is the k-dimensional subspace generated by eigenvectors associated to u ₁,…,u _k . Notice that $\|T_{{\mathcal{S}}^{\perp}}\|=u_{k+1}$ and that the n+k eigenvalues of $T_{\mathcal{S}}$ (counting multiplicities and arranged in non-increasing order) are u ₁,…,u _k ,0,…,0. Therefore, by Theorem 6 there exists a k-dimensional subspace of ${\mathcal{S}}$, denoted by ${\mathcal{T}}$ such that $P_{\mathcal{T}}TP_{\mathcal{T}}=u_{k+1}P_{\mathcal{T}}$. Thus, if we define ${\mathcal{M}}={\mathcal{S}}^{\perp}\oplus{\mathcal{T}}$ we obtain a subspace with $\dim{\mathcal{M}}^{\perp}=n$ and such that $\|T_{\mathcal{M}}\| =u_{k+1}$. Therefore, if n<r (and hence k=n) we see that $\| T_{\mathcal{M}}\|=u_{k+1}=u_{n+1}$; otherwise n≥r (so that k=r) and hence $\|T_{\mathcal{M}}\|=u_{r+1}=\|T\|_{e}=u_{n+1}$.

Finally, if Q=E((m _e (T)−ε,m _e (T)+ε)) then Q is a orthogonal projection with infinite dimensional range $\mathcal{N}_{\varepsilon}=R(Q)$ and $m_{e}(T_{\mathcal{N}_{\varepsilon}})=m_{e}(T)$, $\| T_{\mathcal{N}_{\varepsilon}}\|\leq m_{e}(T)+\varepsilon$. □

Next we present the proof of the main result of Sect. 4.2. Recall that the Fredholm index of T is defined as $\operatorname{ind}(T)=\dim N(T)-\dim N(T^{*})$ if at least one of these numbers is finite.

Proof of Theorem 15

Let $Y\in B(\ell_{2},\mathcal{H})$ be a fixed coisometry with initial space ${\mathcal{M}}\subset\ell_{2}$ i.e. $YY^{*}=I_{\mathcal{H}}$ and $Y^{*}Y=P_{\mathcal{M}}$, where $P_{\mathcal{M}}$ denotes the orthogonal projection onto ${\mathcal{M}}$. Then, as explained at the beginning of Sect. 4.2 (see Eq. (13))

$$\inf\bigl\{ \|FX^*-I\| : X^*X=P_{{\mathcal{M}}}, XX^*=I_\mathcal{H}\bigr\} = d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}}\bigr). $$

Therefore, we have that

$$ \alpha(\mathcal{F})=\inf\bigl\{ d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F \bigr)_{\mathcal{M}}\bigr): YY^*=I_\mathcal{H} , \ Y^*Y=P_{\mathcal{M}}\bigr\} . $$

(15)

Notice that $N((Y^{*}F)_{\mathcal{M}})=N(F)\cap{\mathcal{M}}$ and that $\dim N((Y^{*}F)_{\mathcal{M}}^{*})=\dim R(F^{*})\cap{\mathcal{M}}^{\perp}$. In particular, $\dim N((Y^{*}F)_{\mathcal{M}})\leq n$ and therefore $\operatorname{ind}((Y^{*}F)_{\mathcal{M}})\leq n$ and it is well defined.

Now, we claim that $\operatorname{ind}((Y^{*}F)_{\mathcal{M}})=0$ if and only if $\dim({\mathcal{M}}^{\perp})=n$. Indeed, if we assume that $\dim({\mathcal{M}}^{\perp})=\infty$, then $R(Y^{*}F)={\mathcal{M}}$ and $\operatorname{ind}(Y^{*}F)=-\infty$; hence, by Lemma 19 we see that $\operatorname {ind}((Y^{*}F)_{\mathcal{M}})=-\infty$ in this case. On the other hand, if $\dim{\mathcal{M}}^{\perp}=m$ then, by Lemma 19 and the additivity of the Fredholm index for (left) semi-Fredholm operators, $\operatorname {ind}((Y^{*}F)_{\mathcal{M}})=\operatorname{ind}(Y^{*}F)=\operatorname {ind}(Y^{*})+\operatorname{ind}(F)=n-m$.

Hence, if we take ${\mathcal{M}}\subseteq\ell_{2}$ such that $\dim ({\mathcal{M}}^{\perp})=n$ then the previous facts together with Theorem 18 imply that

$$d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}}\bigr)=\max\bigl\{ \| |F|_{\mathcal{M}}\| -1 , 1-m( |F|_{\mathcal{M}})\bigr\} $$

since $|(Y^{*}F)_{\mathcal{M}}|^{2}=(|F|^{2})_{\mathcal{M}}$ so that $\| |(Y^{*}F)_{\mathcal{M}}| \|= \| |F|_{\mathcal{M}}\|$ and $m(|(Y^{*}F)_{\mathcal{M}}|)=m(|F|_{\mathcal{M}})$. Moreover, using the fact that $l_{n+1}(|F|)=A_{\mathcal{F}}$ and $C_{\mathcal{F}}=u_{n+1}(|F|)$, then items 1 and 3 in Proposition 20 show that

$$\begin{aligned} & \inf\bigl\{ d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}} \bigr) : YY^*=I_\mathcal{H} , Y^*Y=P_{\mathcal{M}} , \operatorname{ind} \bigl(\bigl(Y^*F\bigr)_{\mathcal{M}}\bigr)=0 \bigr\} \\ &\quad=\max\bigl\{ 1-A_\mathcal{F}^{1/2}, C_\mathcal{F}-1\bigr\} . \end{aligned}$$

(16)

On the other hand, if $\dim({\mathcal{M}}^{\perp})\neq n$, then $\operatorname{ind}((Y^{*}F)_{\mathcal{M}})\neq0$. Thus, by Theorem 18 and Proposition 20 we conclude that

$$ \inf\bigl\{ d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}} \bigr) : YY^*=I_\mathcal{H} , Y^*Y={\mathcal{M}} , \operatorname{ind}\bigl( \bigl(Y^*F\bigr)_{\mathcal{M}}\bigr)<0 \bigr\} =1+m_e(F). $$

(17)

Finally, if we assume that $\operatorname{ind}((Y^{*}F)_{\mathcal{M}})>0$ then, as shown above, $\dim{\mathcal{M}}^{\perp}=m<n<\infty$. Thus $|(Y^{*}F)_{\mathcal{M}}^{*}|^{2}= Y^{*}(F P_{\mathcal{M}}F^{*})Y|_{\mathcal{M}}$ and hence, since Y is a coisometry with initial space ${\mathcal{M}}$, we see that $m_{e}((Y^{*}F)_{\mathcal{M}}^{*})=m_{e}(F P_{\mathcal{M}} F^{*})^{1/2}$. Now, notice that dimN(F)<∞ implies that m _e (FF ^∗)=m _e (F ^∗ F) (e.g. [2, Proposition 4.5] shows that (FF ^∗⊕0 _n ) and F ^∗ F are unitarily equivalent, where 0 _n is the zero operator acting on an n-dimensional Hilbert space). Then,

$$m_e\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}}^*\bigr)= m_e\bigl(F P_{\mathcal{M}} F^*\bigr)^{1/2}= m_e\bigl(FF^*\bigr)^{1/2}= m_e\bigl(F^*F \bigr)^{1/2}= m_e(F) , $$

since $FF^{*}=F P_{\mathcal{M}} F^{*}+F P_{{\mathcal{M}}^{\perp}} F^{*}$ and $F P_{{\mathcal{M}}^{\perp}} F^{*}$ is a finite rank operator. Thus, by Theorem 18 we get that

$$ \inf\bigl\{ d_{\mathcal{U}({\mathcal{M}})}\bigl(\bigl(Y^*F\bigr)_{\mathcal{M}} \bigr) : YY^*=I_\mathcal{H} , Y^*Y={\mathcal{M}} , \operatorname{ind}\bigl( \bigl(Y^*F\bigr)_{\mathcal{M}}\bigr)>0 \bigr\} \geq1+m_e(F). $$

(18)

The result follows by combining Eqs. (15), (16), (17) and (18). □