a $K_1\times K_2$ filter. Then for an orthogonal matrix $\bY\in\C^{M_2\times M_2}$, we have $$\begin{aligned}
\bmH\left(\La\bsv\right)=\sum_{j=1}^{M_2}\left(\bmH\left(\La\bsv\right)\bY^{(:,j)}\right)\left(\bY^{(:,j)}\right)^*=\sum_{j=1}^{M_2}\left(\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot]\right)\left(\bY^{(:,j)}\right)^*.\end{aligned}$$ In addition, it is not hard to see that $$\begin{aligned}
\bmH^*\left(\w\bq^*\right)=\w\ast\overline{\bq}:=\left[\begin{array}{c}
\w_1\ast\overline{\bq}\vspace{0.25em}\\
\w_2\ast\overline{\bq}
\end{array}\right]~~~~~~\text{for}~~~\w=\left[\begin{array}{c}
\w_1\vspace{0.25em}\\
\w_2
\end{array}\right]\in\C^{2M_1}~~\text{and}~~\bq\in\C^{M_2},\end{aligned}$$ and the Moore-Penrose pseudoinverse of $\bmH$ is proportional to $\bmH^*$. Hence, for an orthogonal matrix $\bY\in\C^{M_2\times M_2}$, we have $$\begin{aligned}
\label{HankelTightFrame}
\La\bsv\propto\sum_{j=1}^{M_2}\left(\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot]\right)\ast\overline{\bY}^{(:,j)},\end{aligned}$$ where $\propto$ means the pointwise proportionality.

We further introduce $$\begin{aligned}
\bsW&=\left[\bsS_{\bq_1[-\cdot]}^T,\bsS_{\bq_2[-\cdot]}^T,\cdots,\bsS_{\bq_{M_2}[-\cdot]}^T\right]^T,\label{OurAnalysis}\\
\bsW^*&=\left[\bsS_{\overline{\bq}_1},\bsS_{\overline{\bq}_2},\cdots,\bsS_{\overline{\bq}_{M_2}}\right],\label{OurSynthesis}\end{aligned}$$ where $\bq_j=M_2^{-1/2}\bY^{(:,j)}$, and $\bsS_{\bq}$ is defined as in \[DiscreteConv\]. Then up to an appropriate scaling, we can rewrite \[HankelTightFrame\] as $$\begin{aligned}
\label{OrthogonalFilterBanks}
\La\bsv=\sum_{j=1}^{M_2}\bsS_{\overline{\bq}_j}\left(\bsS_{\bq_j[-\cdot]}\left(\La\bsv\right)\right)=\bsW^*\bsW\left(\La\bsv\right),\end{aligned}$$ which implies that the orthogonal matrix $\bY\in\C^{M_2\times M_2}$ indeed generates a set of filter banks of a tight frame system.

To reconstruct $\bsv=\msF(u)\big|_{L^{-1}\OO}$ corresponding to a piecewise constant function $u$ in \[uModel\], we assume that $\rank\left(\bmH\left(\La\bsv\right)\right)=r\ll2M_1\wedge M_2$, following \[SRLACSMRI\]. Consider its SVD $$\begin{aligned}
\label{HankelSVD}
\bmH\left(\La\bsv\right)=\left[\begin{array}{c}
\bmH\left(\La_1\bsv\right)\\
\bmH\left(\La_2\bsv\right)
\end{array}\right]=\bX\Sig\bY^*=\sum_{j=1}^r\Sig^{(j,j)}\bX^{(:,j)}\left(\bY^{(:,j)}\right)^*,\end{aligned}$$ with $\Sig^{(1,1)}\geq\Sig^{(2,2)}\geq\cdots\geq\Sig^{(r,r)}>0$ and $\Sig^{(j,j)}=0$ for $j>r$. Since we have $$\begin{aligned}
\label{HankelSVDConv}
\Sig^{(j,j)}\bX^{(:,j)}=\bmH\left(\La\bsv\right)\bY^{(:,j)}=\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot],\end{aligned}$$ it follows that $$\begin{aligned}
\label{HankelSVDSparse}
\begin{split}
\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot]&\neq\0,~~~~~~~j=1,\cdots,r,\vspace{0.25em}\\
\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot]&=\0,~~~~~~~j=r+1,\cdots,M_2,
\end{split}\end{aligned}$$ which leads to $$\begin{aligned}
\label{LowRankHankelSVD}
\La\bsv\propto\sum_{j=1}^r\left(\left(\La\bsv\right)\ast\bY^{(:,j)}[-\cdot]\right)\ast\overline{\bY}^{(:,j)}.\end{aligned}$$ Hence, if we define $\bsW$ and $\bsW^*$ as in \[OurAnalysis,OurSynthesis\] using $\bY\in\C^{M_2\times M_2}$ in \[HankelSVD\], we have $$\begin{aligned}
\label{TightFrameSparse}
\begin{split}
\bsS_{\bq_j[-\cdot]}\left(\La\bsv\right)&\neq\0,~~~~~~~j=1,\cdots,r,\vspace{0.25em}\\
\bsS_{\bq_j[-\cdot]}\left(\La\bsv\right)&=\0,~~~~~~~j=r+1,\cdots,M_2,
\end{split}\end{aligned}$$ which finally leads to $$\begin{aligned}
\label{LowRankHankelTightFrame}
\La\bsv=\sum_{j=1}^r\bsS_{\overline{\bq}_j}\left(\bsS_{\bq_j[-\cdot]}\left(\La\bsv\right)\right)=\bsW^*\bsW\left(\La\bsv\right).\end{aligned}$$ This means that, if $\rank\left(\bmH\left(\La\bsv\right)\right)=r$, then 