Frame operator and dual frames

Show that

We say that $\left\{ y_{n}\right\}$ has a lower frame bound of $B^{-1}$; compare Definition 8.2. Note that $\left\{ y_{n}\right\}$ need not be a Bessel sequence; see Exercise 7.1(h).

Show that for all $N \in \mathbb {N}$ and $c_{1}, \ldots , c_{N} \in \mathbf{F}$ we have

$\left\{ v_{1}, \ldots , v_{n}\right\}$ is a Bessel sequence in $\mathbf{F}^{d}$, its synthesis operator corresponds to multiplication by the matrix $R$ that has $v_{1}, \ldots , v_{n}$ as columns, and its analysis operator is $C = R^{*}$, the Hermitian of the matrix $R$.

The following are equivalent: (i) $\left\{ v_{1}, \ldots , v_{n}\right\}$ spans $\mathbf{F}^{d}$, (ii) $S = C^{*} C$ is positive definite, (iii) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is a frame for $\mathbf{F}^{d}$. Further, in case these hold, the optimal frame bounds for $\left\{ v_{1}, \ldots , v_{n}\right\}$ are $\lambda_{1}, \lambda_{d}$, where $\lambda_{1}$ is the smallest eigenvalue of $S$ and $\lambda_{d}$ is the largest eigenvalue.

The following are equivalent: (i) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is linearly independent, (ii) $G = C C^{*}$ is positive definite, (iii) $\left\{ v_{1}, \ldots , v_{n}\right\}$ is a Riesz sequence in $\mathbf{F}^{d}$.

Show that if $U, V \in \mathcal{B}(H)$ are positive and $U \leq V$ then $\left\| U\right\| \leq \left\| V\right\|$.

Let $S$ be the frame operator for a frame $\left\{ x_{n}\right\}$ that has frame bounds $A, B$. Prove the operator inequalities

and

and use either one of these to show that $S$ is a topological isomorphism.

If $\left\{ y_{n}\right\}$ is a Bessel sequence in $H$ that is an alternative dual for $\left\{ x_{n}\right\}$, then $\left\{ y_{n}\right\}$ is a frame and $\left\{ x_{n}\right\}$ is an alternative dual frame for $\left\{ y_{n}\right\}$.

A Bessel sequence $\left\{ y_{n}\right\}$ is the canonical dual frame for $\left\{ x_{n}\right\}$ if and only if it is an alternative dual and $\operatorname {range}(C) = \operatorname {range}(\widetilde{C})$, where $\widetilde{C}$ is the analysis operator of $\left\{ y_{n}\right\}$.

$\left\{ P x_{n}\right\}$ is a frame for $M$ with the same frame bounds as $\left\{ x_{n}\right\}$, and $\left\{ P \widetilde{x}_{n}\right\}$ is an alternative dual frame for $\left\{ P x_{n}\right\}$.

$\left\{ P \widetilde{x}_{n}\right\}$ is the canonical dual frame for $\left\{ P x_{n}\right\}$ if and only if $PS = SP$, where $S$ is the frame operator for $\left\{ x_{n}\right\}$.

Show that $\widetilde{C} = C S^{-1}$ and $\widetilde{R} = S^{-1} R$.

Show that the orthogonal projection $P$ of $\ell^{2}$ onto range $(C)$ is $P c = C \widetilde{R} c = C S^{-1} R c = \left\{ \left\langle \sum_{n} c_{n} \widetilde{x}_{n}, x_{k} \right\rangle \right\}_{k \in \mathbb {N}}, \quad c = \left(c_{n}\right) \in \ell^{2}$.

$\left\{ y_{n}\right\}$ is an alternative dual frame for $\left\{ x_{n}\right\}$.

There is a bounded left-inverse $V$ of $C$ such that $y_{n} = V \delta_{n}$ for all $n \in \mathbb {N}$ (that is, $V: \ell^{2} \rightarrow H$ is bounded, $V C = I$, and $y_{n} = V \delta_{n}$).

Show that $\operatorname {range}(C) = \operatorname {range}(\widetilde{C})$.

Show that the pseudoinverse $R^{\dagger }$ of the synthesis operator $R$ is $R^{\dagger } = \widetilde{C}$, so

Show that the optimal frame bounds for $\left\{ x_{n}\right\}$ are

Show that $\left\{ x_{n}\right\}$ and $\left\{ y_{n}\right\}$ are equivalent if and only if range $\left(C_{X}\right) =$ range $\left(C_{Y}\right)$, where $C_{X}, C_{Y}$ are the analysis operators for $\left\{ x_{n}\right\}$ and $\left\{ y_{n}\right\}$.

Show that part (a) is equivalent to:

Exhibit a frame $\left\{ x_{n}\right\}$ and a permutation $\sigma$ of $\mathbb {N}$ that fixes all but finitely many elements of $\mathbb {N}$ such that $\left\{ x_{n}\right\}$ is not equivalent to $\left\{ x_{\sigma (n)}\right\}$. Show that this cannot happen if $\left\{ x_{n}\right\}$ is an exact frame.

Exhibit a frame $\left\{ x_{n}\right\}$ and a sequence of signs $\varepsilon_{n} = \pm 1$ such that $\left\{ x_{n}\right\}$ is not equivalent to $\left\{ \varepsilon_{n} x_{n}\right\}$. Show that this cannot happen if $\left\{ x_{n}\right\}$ is an exact frame.

$U\left(\operatorname {ker}(C)\right) \subseteq \operatorname {ker}(C)$.

$V^{*}\left(\operatorname {ker}(R)\right) \subseteq \operatorname {ker}(R)$.

$V\left(\operatorname {range}(C)\right) \subseteq \operatorname {range}(C)$.

$U^{*}\left(\operatorname {range}(R)\right) \subseteq \operatorname {range}(R)$.

Show that $S\left(T_{k} f\right) = T_{k}(S f)$ for each $f \in V_{0}(g)$. Use this to show that the canonical dual frame in $V_{0}(g)$ is $\mathcal{T}(h)$ where $h = S^{-1} g$.

Show that

Show that the frame operator $S$ commutes with $M_{b n}$ and $T_{a k}$ for all $k, n \in \mathbb {Z}$, and use this to show that $S^{-1}$ also commutes with $M_{b n}$ and $T_{a k}$.

Show that the canonical dual frame of $\mathcal{G}(g, a, b)$ is the Gabor frame $\mathcal{G}(\widetilde{g}, a, b)$ where $\widetilde{g} = S^{-1} g$.

Suppose that $\mathcal{G}(g, a, b)$ is a frame for $L^{2}(\mathbb {R})$. Show that $\mathcal{G}(g, a, b)$ is a Riesz basis if and only if $\left\langle g, \widetilde{g} \right\rangle = 1$.

Show that the canonical Parseval frame of a lattice Gabor frame is another lattice Gabor frame. Specifically, if $\mathcal{G}(g, a, b)$ is a frame for $L^{2}(\mathbb {R})$ and we set $g^{\sharp } = S^{-1 / 2} g$, where $S$ is the frame operator, then $\mathcal{G}\left(g^{\sharp }, a, b\right)$ is a Parseval frame for $L^{2}(\mathbb {R})$.

The frame operator is pointwise multiplication by $b^{-1} G_{0}$, i.e., $S f = b^{-1} G_{0} f$ for $f \in L^{2}(\mathbb {R})$.

The canonical dual frame is $\mathcal{G}(\widetilde{g}, a, b)$ where $\widetilde{g} = b g / G_{0}$.

If $a b = 1$ then $\mathcal{G}(g, a, b)$ is a Riesz basis for $L^{2}(\mathbb {R})$.

If $0 < a b < 1$ then $\mathcal{G}(g, a, b)$ is a redundant frame for $L^{2}(\mathbb {R})$.