Classical Shearlet system forms a Parseval frame. 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.
 A: 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}
