I have been reading about shearlets & frames and there is a part that I'd like help with. I have only studied the basics of functional analysis and wavelets.
First few definitions:
Classical shearlets
Let $\psi \in L^2(\mathbb{R}^2)$ be defined the following way $$\hat{\psi}(\xi) = \hat{\psi}(\xi_1,\xi_2) = \hat{\psi}_1(\xi_1)\hat{\psi}_2\Big(\frac{\xi_2}{\xi_1}\Big)$$ where $\psi_1 \in L^2(\mathbb{R})$ is discrete wavelet in the sense that it satisfies the discrete Calderòn condition, given by \begin{equation}\label{10} \sum_{j\in\mathbb{Z}}|\hat{\psi}_1(2^{-j}\xi)|^2 = 1 \quad a.e \texttt{ } \xi\in\mathbb{R} \end{equation} with $\hat{\psi}_1\in C^\infty(\mathbb{R})$ and supp($\hat{\psi}_1$) $\subseteq [-\frac{1}{2}, -\frac{1}{16}] \cup [\frac{1}{16}, \frac{1}{2}]$, and $\psi_2 \in L^2(\mathbb{R})$ is a bump function in the sense that \begin{equation}\label{11} \sum_{k=-1}^1 |\hat{\psi}_2(\xi+k)|^2 = 1 \quad a.e \texttt{ } \xi \in [-1, 1] \end{equation} satisfying $\hat{\psi}_2 \in C^\infty(\mathbb{R})$ and supp($\hat{\psi}_2$) $\subseteq [-1, 1]$. Then $\psi$ is called a classical shearlet.
Shearlet system
$$\text{SH}(\psi) = \{\psi_{j,k,m} = 2^{-\frac{3}{4}j}\psi(S_kA_{2^j} \cdot -m) : j,k \in \mathbb{Z}, m\in\mathbb{Z}^2\}$$
With these we have the following proposition:
Let $\psi \in L^2(\mathbb{R}^2)$ be a classical shearlet. Then SH($\psi$) is a Parseval frame for $L^2(\mathbb{R}^2)$.
Proof. \begin{align*} \sum_{j\geq 0} \sum_{|k| \leq \lceil 2^{j/2} \rceil} |\hat{\psi}(S^\text{T}_{-k} A_{2^{-j}}\xi)|^2 &= \sum_{j\geq 0} \sum_{|k| \leq \lceil 2^{j/2} \rceil} |\hat{\psi}_1(2^{-j}\xi_1)|^2 |\hat{\psi}_2(2^{j/2}\frac{\xi_2}{\xi_1}-k)|^2\\ &= \sum_{j\geq 0} |\hat{\psi}_1(2^{-j}\xi_1)|^2 \sum_{|k| \leq \lceil 2^{j/2} \rceil} |\hat{\psi}_2(2^{j/2}\frac{\xi_2}{\xi_1}+k)|^2 = 1 \end{align*} The claim follows immediately from this observation and the fact that supp $\hat{\psi} \subset [-\frac{1}{2}, \frac{1}{2}]$. $\square$
The problem that I have is with the last sentence of the proof. Does the claim follow from direct computation and the definition of frames or is there something more?
Thank you in advance.
It does not follow immediately, but there are proofs in wavelettheory, which are quite similiar to this one.
I wrote a complete proof in my Bachelor thesis. What you need is the following result about the Fourier transform of a classical shearlet:
Let $\psi\in L^2(\mathbb{R}^2)$ be a classical shearlet and SH($\psi$) be the associated discrete shearlet system. Then the Fourier transfom of an element $\psi_{j,k,m}$ of SH($\psi$) ($j,k\in\mathbb{Z}, m\in\mathbb{Z}^2$) has the following form: \begin{align} \nonumber \hat{\psi}_{j,k,m}(\xi)=e^{-2\pi i\langle m,{(M_{j,k}^{-1})}^{T}\xi\rangle}\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\quad\text{for }\xi\in\mathbb{R}^2, \end{align} where \begin{align} M_{j,k}=S_kA_{2^j}=\begin{pmatrix}2^j & 2^{j/2}k\\0 & 2^{j/2}\end{pmatrix}. \end{align}
Armed with this Lemma, we can proof the theorem:
Let $f\in L^2(\mathbb{R}^2)$ by the Plancherel theorem it holds \begin{align} \nonumber \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,\vert\langle f,\psi_{j,k,m}\rangle\vert^2=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,|\langle \hat{f},\hat{\psi}_{j,k,m}\rangle\vert^2. \end{align} And by the previous lemma, the Transformationssatz and the fact that supp$\hat{\psi}\subset[-\frac{1}{2},\frac{1}{2}]^2$ we can conclude that \begin{align} \nonumber \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,|\langle \hat{f},\hat{\psi}_{j,k,m}\rangle\vert^2 &=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,\Bigl|\int_{\mathbb{R}^2}\hat{f}(\xi)\hat{\psi}({(M_{j,k}^{-1})}^T\xi)e^{-2\pi i\langle m,{(M_{j,k}^{-1})}^T\xi\rangle}\mathrm{d}\xi\Bigl|^2&&\\\nonumber &=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,\Bigl|\int_{M_{j,k}^T([-\frac{1}{2},\frac{1}{2}]^2)}\,\hat{f}(\xi)\hat{\psi}({(M_{j,k}^{-1})}^T\xi)e^{-2\pi i\langle m,{(M_{j,k}^{-1})}^T\xi\rangle}\mathrm{d}\xi\Bigl|^2&&\\\nonumber &=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\text{det}({(M_{j,k}^{-1})}^T)\sum_{m\in\mathbb{Z}^2}\,\Bigl|\int_{[-\frac{1}{2},\frac{1}{2}]^2}\hat{f}(M_{j,k}^T\xi)\hat{\psi}(\xi)e^{-2\pi i\langle m,\xi\rangle}\mathrm{d}\xi\Bigl|^2. \end{align} Since $\{e^{-2\pi i\langle m,(\cdot)\rangle}\,\vert\,m\in\mathbb{Z}^2\}$ in an orthonormal basis for $L^2([-\frac{1}{2},\frac{1}{2}]^2)$, we can apply the Parseval equality. Hence \begin{align} \nonumber \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,|\langle \hat{f},\hat{\psi}_{j,k,m}\rangle\vert^2&=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\text{det}({(M_{j,k}^{-1})}^T)\int_{[-\frac{1}{2},\frac{1}{2}]^2}\vert\hat{f}(M_{j,k}^T\xi)\vert^2\vert\hat{\psi}(\xi)\vert^2\mathrm{d}\xi&&\\\nonumber &=\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\int_{M_{j,k}^T([-\frac{1}{2},\frac{1}{2}]^2)}\vert\hat{f}(\xi)\vert^2\vert\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\vert^2\mathrm{d}\xi&&\\\nonumber &=\int_{M_{j,k}^T([-\frac{1}{2},\frac{1}{2}]^2)}\vert\hat{f}(\xi)\vert^2\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\vert\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\vert^2\mathrm{d}\xi&&\\\nonumber &=\int_{\mathbb{R}^2}\vert\hat{f}(\xi)\vert^2\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\vert\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\vert^2\mathrm{d}\xi. \end{align} We now show, that $\sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\vert\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\vert^2=1$, which almost completes the proof. By the definition of $M$ and the classical shearlet we have \begin{align} \nonumber \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\vert\hat{\psi}({(M_{j,k}^{-1})}^T\xi)\vert^2&= \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\vert\hat{\psi_1}(2^{-j}\xi_1)\vert^2\vert\hat{\psi_2}(2^{\frac{1}{2}j}\frac{\xi_2}{\xi_1}-k)\vert^2&&\\\nonumber &=\sum_{j\in\mathbb{Z}}\vert\hat{\psi_1}(2^{-j}\xi_1)\vert^2\sum_{k\in\mathbb{Z}}\vert\hat{\psi_2}(2^{\frac{1}{2}j}\frac{\xi_2}{\xi_1}+k)\vert^2=1. \end{align} Finally we conclude again by Plancherel \begin{align} \nonumber \sum_{j\in\mathbb{Z}}\sum_{k\in\mathbb{Z}}\sum_{m\in\mathbb{Z}^2}\,\vert\langle f,\psi_{j,k,m}\rangle\vert^2=\int_{\mathbb{R}^2}\vert\hat{f}(\xi)\vert^2\mathrm{d}\xi=\vert\vert\hat{f}\vert\vert^2=\vert\vert f\vert\vert^2. \end{align}