Monketoo's picture
Add files using upload-large-folder tool
f6876fa verified

Lemma 2.11 The operator $G^\dagger : \ell^2(\mathbb{N}) \to \ell^2(\mathbb{N})$ is the same as $\tilde{G}$. It is therefore self-adjoint and has the following properties:

  1. $GG^\dagger = Q = G^\dagger G$.

  2. $G^\dagger (I - Q) = 0$.

  3. $G^\dagger Q = QG^\dagger = G^\dagger$.

The following corollary is the same as [8], Proposition 5.3.6.

Corollary 2.12 Let ${f_k}{k=1}^\infty$ be a frame sequence in H. For any $c \in \ell^2(\mathbb{N})$, $Qc = \sum{k=1}^\infty \langle c, G^\dagger T^* f_k \rangle \epsilon_k$.

We may also write $Qc = \sum_{k,j=1}^\infty \langle c, G^\dagger \epsilon_j \rangle \langle f_j, f_k \rangle \epsilon_k$.

Lemma 2.13 We have the following properties:

  1. $(T^*)^\dagger = TG^\dagger$.

  2. $T^\dagger (T^\dagger)^* = G^\dagger$.

  3. $T^\dagger = G^\dagger T^*$.

  4. $|T|^2 = |G|$.

  5. $|T^\dagger|^2 = |G^\dagger|$.

Proof. (i): Since $Q = T^TG^\dagger = T^(T^*)^\dagger$, it follows from the uniqueness of the Moor-Penrose pseudo-inverse that $(T^*)^\dagger = TG^\dagger$.

(ii) follows immediately from (1.) by multiplying on the left by $T^\dagger$ and using Property 3 of Lemma 1.1.

(iii) follows from (1.) by taking adjoints.

(iv): We have for every $c \in \ell^2(\mathbb{N})$, $|Tc|^2 = \langle Gc, c \rangle \leq |G||c|^2$. Thus $|T|^2 \leq |G|$. On the other hand, $|G| = |T^*T| \leq |T|^2$, which yields $|G| = |T|^2$.

(v): As $T^\dagger = \tilde{T}$ and $G^\dagger = \tilde{G}$, (5) is the same as (4) for the dual frame. $\blacksquare$

Corollary 2.14 $|G| = |S|$ and $|G^\dagger| = |S^\dagger|$.

Theorem 2.6 (or rather, Corollary 2.7) can also be reformulated as

Theorem 2.15 The following are equivalent:

  1. ${f_k}_{k=1}^\infty$ is a frame sequence in H.