



%\graphicspath{{figures/}}



\chapter{From continuous to digital images, and back\label{blurassumption} }
Consider  a continuous and bounded image $u(\fxx)$  defined for every $\fxx=(x,y)\in \RR^2$.
 All continuous \textit{image} operators including the
sampling will be written in  capital letters $A$, $B$ and their composition as a mere juxtaposition $AB$.
  For any affine map $A$ of the plane consider the affine transform of a continuous image
$u$ defined by $A u(\fxx)=: u(A\fxx)$. For instance $H_\lambda u(\fxx)=:u(\lambda\fxx)$ denotes an expansion of $u$ by
a factor $\lambda^{-1}$. In the same way if $R$ is a rotation, $R u =: u\circ R$ is the image rotation by $R^{-1}$.

\subsection*{Sampling and interpolation}

Digital images, only defined for  $(n_1,n_2)\in\relative^2$, will be denoted by $\bu(n_1,n_2)$.
 The  $\delta$-sampled image
 $\bu=\SA_\delta u$ is  defined on $\relative^2$ by \begin{equation} \bu(n_1,n_2)=(\SA_\delta)u (n_1,n_2)=:u(n_1\delta,
n_2\delta); \label{defsampling}\end{equation} Conversely,  the  Shannon interpolate of a digital image $\bu$ is defined
as follows \cite{gasquet1999faa}. Let $\bu$ be a digital image, defined on $\relative^2$ and such that $\sum_{n\in
\relative^2}|\bu(n)|^2<\infty$ and $\sum_{n\in \relative^2}|\bu(n)|<\infty.$ (Of course, these conditions are
automatically satisfied if the digital has a finite number of non-zero samples, which is
 the case here.) We call Shannon
interpolate $I\bu$ of $\bu$ the only $L^2(\RR^2)$ function $u$ having $\bu$ as samples and with spectrum  support
contained in $(-\pi,\pi)^2$.   $u=I\bu$ is defined by the Shannon-Whittaker formula
$$I\bu(x,y)=:\sum_{(n_1,n_2)\in \relative^2} \bu(n_1,n_2)\hbox{sinc}
(x-n_1)\hbox{sinc} (y-n_2),$$ where $\hbox{sinc }x=:\frac{\sin \pi x}{\pi  x}.$
 The Shannon interpolation  has the fundamental property
$\SA_1I\bu=\bu$. Conversely, if $u$ is $L^2$ continuous image,   band-limited in $(-\pi,\pi)^2$, then
 \begin{equation}\label{interpolationrecovery}I\SA_1u = u.\end{equation}
 In that case we simply say that $u$ is {\it band-limited}. We shall also say that a digital image $\bu=\bS_1u$ is
  {\it well-sampled} if it was obtained from a band-limited image $u$.




\subsection{The practical Shannon interpolation: zero-padding} Of course, the Shannon interpolate is unpractical in that it
assumes the knowledge of infinitely many samples. In practice image samples and image interpolation will be performed
on rectangle. For a sake of simplicity we describe here what happens on a square. Let $a>0$ and consider a function $u$
from $[0,a]^2$ to $\real$ such that $u(x+a,y+a)=u(x,y)$. Fix an integer $N$, and consider the $N^2$ samples of $u$,
$\bu_{k,l}=(\SA u)\left(\frac{ka}{N},\frac{la}{N}\right)$ on $[0,a]^2$.

\begin{definition}\label{defDFT} The  discrete Fourier transform  (DFT) of  the $N^2$ samples
 $u=(u_{k,l})_{k,l=0,1,\dots, N-1}$ is  the double sequence of
discrete Fourier coefficients for $m,n \in \{-\frac N 2,...,\frac N 2 -1\}$ defined by
\begin{equation}
\label{TFD2} DFT(u)_{m,n}=
\tilde{u}_{m,n}=\frac{1}{N^2}\sum_{k=0}^{N\!-\!1}\sum_{l=0}^{N\!-\!1}u_{k,l}\omega_N^{-mk}\omega_N^{-nl },
\end{equation}
where $\omega_N=: e^{\frac{2i\pi}N}$ is the first $N$-root of 1.
\end{definition}

\begin{proposition}\label{inverseDFT}
Consider the trigonometric polynomial
\begin{equation}\label{definitionofIonasquare}I\bu(x,y)=\sum_{m,n=-\frac
N2}^{\frac N2-1}\tilde u_{m,n}\exp(\frac{2i\pi mx}{a})\exp(\frac{2i\pi ny}{a}).
\end{equation}
 Then its coefficients
$\tilde u_{m,n}$ are the only complex numbers such that for every  $k, l \in \{0, ..., N\!-\!1\}$,
$I\bu\left(\frac{ka}{N},\frac{la}{N}\right)=\bu\left(\frac{ka}{N},\frac{la}{N}\right)$. In consequence, the discrete
inverse transform of the DFT $\bu\to \tilde u$ is nothing but the calculation of the value of the polynomial at the
samples $(\frac {ka}N, \frac {la}N)$, $0\leq k,l\leq N\!-\!1$. In other terms, setting $\tilde{u}=DFT(\bu)$, the
inverse transform $DFT^{-1}$ is given by
$$\bu(k,l)=\sum_{-\frac N2}^{\frac N2-1}\sum_{-\frac N2}^{\frac
N2-1}\tilde \bu_{m,n}\omega_N^{km+ln},$$ for every $k,l =0,1,\dots N-1$.
\end{proposition}


\begin{exercise}  Recall that $\omega_N=\exp\left(\frac{2i\pi}{N}\right)$,  $N$-th root of 1. Show that
$\sum_{k=0}^{N\!-\!1}\omega_N^k=0$, and that $\sum_{k=0}^{N\!-\!1}\omega_N^{kl}=0$ for $l\neq 0$ modulo $N$ and finally
that for every $k_0$, $\sum_{k=k_0}^{k_0+N\!-\!1}\omega_N^{kl}=0$ for all $l\neq 0$ modulo $N$. Using these relation
show the above proposition, namely that $DFT(DFT^{-1})=Id$.\finex
\end{exercise}

In conclusion, the interpolation and sampling operators we shall consider both in theory and practice are the usual
sampling $\SA$, implicitly restricted to a square. The inverse interpolation operator $I$  is  defined by
\eqref{definitionofIonasquare}, and Proposition \ref{inverseDFT} tells us that $\SA I\bu=\bu$. The next statement gives
the converse statement.

\begin{proposition}\label{mainsamplinginterpolationframework} If $u(x,y)$ is a $a$-periodic  band-limited function, then it is a trigonometric polynomial.  If
 its highest degree is
$\frac N2-1$, with $N$ even, then its coefficients  $\tilde u_{m,n}$ are obtained by DFT from the samples
$\bu(k,l)=u(\frac{ka}N, \frac{la}N)$.
 In consequence for such functions we have $I\SA u=u$.
\end{proposition}
\begin{exercise} Give a detailed proof of Proposition \ref{mainsamplinginterpolationframework}. It is a direct
consequence of Proposition \ref{inverseDFT}. \finex \end{exercise}
In the rest of this chapter and of the book, we
shall always take the functional setting of Proposition \ref{mainsamplinginterpolationframework}.

\subsubsection*{Zoom in by zero-padding} Let $(u_{k,l})$ be a digital image and define its zoomed version
$\left(v_{i,j}\right)_{i,j=0,\;...,\; 2N\!-\!1}$ as the inverse discrete Fourier transform of   $\tilde v_{i,j}$
defined for $i,j=-N,\;...,\;N\!-\!1$ by

\begin{equation}
\label{TFDzoom2}
 \tilde{v}_{m,n} = \tilde u_{m,n} \hbox{  if  } -\frac N2\leq m, n\leq \frac N2-1, \;\; \tilde
v_{m,n}=0
 \hbox {  otherwise}.
\end{equation}


\begin{proposition} \index{zoom discret par Fourier} \label{zeropadding}
The image $v$ whose Discrete Fourier Transform  is given by (\ref{TFDzoom2}) satisfies $v_{2k,2l}=u_{k,l}$, for $k,l=0,
..., N\!-\!1$.
\end{proposition}

\begin{proof} Here is the proof in dimension 1:  $$v_{2k}=\sum_{-N}^{N\!-\!1}\tilde v_n\omega_{2N}^{2nk}=\sum_{-\frac
N2}^{\frac N2-1}\tilde u_n\omega_N^{nk} =u_k.$$ Indeed, $\omega_{2N}^{2nk}=\omega_N^{nk}.$
\end{proof}

\begin{exercise}
Prove Proposition \eqref{zeropadding} in two dimensions. \finex
\end{exercise}

\subsection{Shannon or zero-padding?}
In what follows, most of the formal reasoning will be done with the Shannon interpolation of the digital image on $\real^2$.
Thus, the operators $I$ and $S_1$ refer to the Shannon framework developed above. However, the practical interpolation is
made by the discrete Fourier transform. Thus the practical DFT interpolation from a digital image $\bu$ is NOT 
the Shannon interpolation, simply because the extension of the samples outside the square is not the same. In the Shannon-Whittaker interpolation the extension is made with zero samples while in the DFT interpolation, the sampled are
periodized and the image becomes a trigonometric polynomial, not even belonging to $L^2(\real^2).$  
We shall therefore denote this second interpolation periodic of a digital image by $\tilde I \bu$.   Of course the values of both
interpolations only make sense  inside the image domain. Outside, $Iu$ has zero samples and oscillates around the zero value, and $\tilde Iu$ is simply a periodic repetition of the restriction of $\tilde Iu$ in the image domain.
Thus, it is crucial for the forthcoming reasoning to prove by numerical and theoretical arguments that $I\bu$ and $\tilde \bu$
are in fact extremely close to each other, so that their difference is actually negligible with respect to other 
image perturbations such as noise and quantization.

To that effect, an obvious experimental device is to compare $I\bu$ and $\tilde\bu$ on generic digital images, and to compute
their quadratic distance per pixel. Even if $I\bu$ cannot be computed exactly, it can be approximated by  extending $\bu$
by zero samples on a white band around the image domain before applying $\tilde I$ to the resulting image.
Another more theoretical way is to give a formula for the quadratic distance of $I\bu$ and $\tilde I\bu$
in the case where $u$ is a white noise. The white noise is here the most difficult case because it has a flat spectrum,
far less regular than any typical digital image.
\medskip

(...)
\medskip The above table and formula lead to the conclusion that the difference between $I\bu(\bx)$ and $\tilde I\bu(\bx)$ is 
negligible in general, and still more negligible when $\bx$ stands at a reasonable and fixed distance from the image boundary.
Thus, in the sequel, we shall reason with operators $I$ and $\tilde I$ depending on what kind of property for image operators
we have in mind. If for example geometric invariance properties are at stake, the operator $I$ will be preferred.

\section{The Gaussian semigroup}
  For a sake of simple notation,  $G_\sigma$
denotes the  convolution operator on $\RR^2$  with  the gauss kernel $G_\sigma(x_1,x_2)=\frac
1{2\pi(\sigma)^2}e^{-\frac{x_1^2+x_2^2}{2(\sigma)^2}}$, namely $G_\sigma \bu(x,y)=: (G_\sigma \ast \bu)(x,y).$ Notice
that the paramaterization of the gaussian is {\it not} the same as the parameterization used for the heat equation. To
make a difference in notation, we use Greek letters for the new parameter. It is easily checked that $G_\sigma$
satisfies the semigroup property
\begin{equation}
\label{continuoussemigroupproperty}G_{\sigma} G_{\beta}= G_{\sqrt{\sigma^2+ \beta^2}}.\end{equation}

\begin{exercise} Prove \eqref{continuoussemigroupproperty}.
\finex
\end{exercise}
 The proof of the
next formula is a mere change of variables in the integral defining the convolution.
\begin{equation}\label{commutationconvolutionzoomcontinus}G_\sigma  H_\gamma u=
H_\gamma G_{\sigma\gamma}  u.\end{equation}

\begin{exercise} Prove \eqref{commutationconvolutionzoomcontinus}.
\finex
\end{exercise}

\subsection*{Discrete Gaussians \label{condenseddescription}}
Many algorithms in computer vision and image processing assume that all blurs are gaussian. Thus,  it will
be crucial to prove that gaussian blur gives {\it in practice} well-sampled images. Thus, in all that follows, we are
dealing with  initial digital
 images obtained by sampling a continuous image with gaussian blur, $u=\bS_1 G_{\bc}u_0$;

Another question we need to deal with is how to perform a gaussian convolution on a discrete (digital) image.   This is
valid if and only if a discrete convolution can give an account of the underlying continuous one.

\begin{definition}\label{definitiondiscretegaussianconvolution}
The {\bf discrete gaussian convolution} applied to a digital image $\bu$ is defined as a digital operator by
\begin{equation}\label{defGdelta} \bG_\delta \bu= : \SA_1 G_\delta I\bu.\end{equation}
\end{definition}

 \begin{proposition} This definition maintains the gaussian semi-group property,
\begin{equation}\label{discretesemigroupproperty}\bG_{\delta} \bG_{\beta}= \bG_{\sqrt{\delta^2+ \beta^2}}.\end{equation}
\end{proposition}
\paragraph{Proof}
Indeed, using  twice \eqref{defGdelta} and once \eqref{continuoussemigroupproperty} and \eqref{interpolationrecovery},
$$\bG_{\delta} \bG_{\beta} \bu=\SA_1G_{\delta} I\SA_1G_{\beta} Iu=\SA_1G_{\delta}  G_{\beta}
  I\bu=\SA_1G_{\sqrt{\delta^2+ \beta^2}}  I\bu=\bG_{\sqrt{\delta^2+ \beta^2}}\bu.$$
\endproof
\begin{remark} The above formula is valid with $I$ replaced by $\tilde I$ and the proof is the same.
\end{remark}
The  SIFT method that we will study in detail uses repeatedly the semi-group formula and a 2-sub-sampling of images
with a gaussian blur larger than 1.6. These SIFT sampling manoeuvres are valid if and  only if the {\it empirical
proposition} below is true.
\begin{proposition}
 For every $\sigma$  larger than 0.8 and every continuous and bounded image $u_0$, the gaussian blurred image $G_\sigma
 u_0$ is well sampled, namely $I\bS_1G_\sigma u_0=G_\sigma u_0$.
\end{proposition}
This proposition is not a mathematical statement, but it will be checked experimentally in the next section, where we
shall see that  a $0.6$ blur is enough to ensure good sampling in practice.




  \section{\label{blurassumption}The right gaussian blur  for well-sampling}

Images need to be blurred before they are sampled. In principle gaussian blur cannot lead to a good sampling because it
is not {\it stricto sensu} band limited. Therefore the Shannon-Whittaker formula does not apply.  However, in practice
it does. The aim in this section is to define a procedure that checks that a gaussian blur works and to fix the minimal
variance of the blur ensuring well-sampling (up to a minor mean square and visual error).

\begin{figure}[]
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=4cm]{./figures/Lena.eps}
&
\includegraphics[width=4cm]{./figures/gaussian_MSE_Lena_t4.eps}
\\
\includegraphics[width=4cm]{./figures/Lena_c0p3_t4_blur_face.eps}
&
\includegraphics[width=4cm]{./figures/Lena_c0p3_t4_face.eps}
\\
\includegraphics[width=4cm]{./figures/Lena_c0p6_t4_blur_face.eps}
&
\includegraphics[width=4cm]{./figures/Lena_c0p6_t4_face.eps}
\end{tabular}
\end{center}
\caption{Top left: $\bu$. Top right: MSE($\bu_1,\bu_2$) vs $\ft2_r(4)$. Middle (from left to right): $\bu_1$ and
$\bu_2$ with $\ft2_r(4)=1.2$. MSE($\bu_1,\bu_2$)=17.5. Bottom (from left to right): $\bu_1$ and $\bu_2$ with $
\ft2_r(4)=2.4$. MSE($\bu_1,\bu_2$)=0.33. Digital images are always displayed by coloring  each square pixel with its
central sample value.} \label{fig:guassian:MSELena}
\end{figure}

One must distinguish two types of  blur: The \textit{absolute} blur with standard deviation $\ft2_a$ is the one that
must be applied to an ideal infinite resolution (blur free) image to create an approximately band-limited image before
1-sampling.   The \textit{relative} blur
 $\sigma=\ft2_r(t)$ is the one that must be applied to a well-sampled image before a sub-sampling by a $t$ factor.
In the case of gaussian blur, because of the semi-group formula \eqref{continuoussemigroupproperty}, the relation
between the absolute and relative blur is
$$t^2 \ft2_a^2 = \ft2_r^2(t) + \ft2_a^2,$$ which yields
\begin{equation}
\ft2_r(t) = \ft2_a \sqrt{t^2-1}. \label{eqn:cr:ca}
\end{equation}
 In consequence, if $t \gg 1$, then $\ft2_r(t)
\approx \ft2_a t$.

Two experiments have been designed to calculate the anti-aliasing absolute gaussian blur $\ft2_a$ ensuring that an
image is approximately well-sampled. The first experiment compares for several values of $\ft2_r(t)$ the digital images
$$\bu_1 =:\bG_{\ft2_r(t)} u=\bS_1 G_{\ft2_r(t)} Iu\;\; \hbox{ and }\;\;
\bu_2 =: (\bS_{1/t}I) \bS_t \bG_{\ft2_r(t)} \bu=(\bS_{1/t}I) \bS_t  G_{\ft2_r(t)} Iu,$$ where $\bu$ is an initial
digital image that is (intuitively) well-sampled, $\bS_{t}$ is a $t$ sub-sampling operator, $\bS_{\frac 1t}$ a $t$
over-sampling operator, and
 $I$  a Shannon-Whitakker  interpolation operator. The discrete convolution by a gaussian is defined in \eqref{defGdelta}.
  Since $t$ is an integer, the  $t$ sub-sampling is
  trivial\footnote{The Shannon over-sampling
 $\bS_{1/t}I$ with an integer zoom factor $t$ is
 obtained by the classic zero-padding method. Zero-padding is exactly a Shannon interpolation
  if the interpolated digital image was obtained from a band-limited and periodic continuous image \cite{gasquet1999faa}.
  Since, given a digital image, these properties on the underlying continuous image are far from granted, it can be
  taken as an assumption. In other terms, given a digital image $\bu_0$ with $(M\times N)$ samples we {\it assume}
  that its interpolated continuous image $u_0$  is
  the unique trigonometric polynomial with $(M\times N)$ degree interpolating $\bu_0$.  Once this convention has been
  adopted, it is indeed true that $\bu_0$ represents the (periodic) samples of $u_0$, and we can obtain the Fourier coefficients of
   $u_0$ from
  $\bu_0$ by Fast Fourier Transform.}.


  If the anti-aliasing filter size $\ft2_r(t)$ is too
small, $\bu_1$ and $\bu_2$ can be very different. The right value of  $\ft2_r(t)$ should be the
 smallest value permitting $\bu_1 \approx \bu_2$.
Fig.~\ref{fig:guassian:MSELena} shows $\bu_1$ and $\bu_2$ with $t=4$ and plots their mean square error
MSE($\bu_1,\bu_2$). An anti-aliasing filter with $\ft2_r(4)=1.2$ is clearly not broad enough:  $\bu_2$ presents strong
ringing artifacts. The ringing artifact is instead hardly noticeable with $\ft2_r(4)=2.4$. The value $\ft2_r(4)\simeq
2.4$ is a good visual candidate, and this choice is confirmed by the curve showing that MSE($\bu_1,\bu_2$) decays
rapidly until $\ft2_r(4)$ gets close to $2.4$, and is stable and small  thereafter. By \eqref{eqn:cr:ca}, this value of
$\ft2_r$ yields $\ft2_a = 0.62$.  This value has been confirmed by experiments on ten digital images. A doubt can be
cast on this experiment, however: Its result slightly depends on the assumption that the  initial blur on $\bu$ is
equal to $\ft2_a$.

In a second experiment, $\ft2_a$ has been evaluated directly by using a binary image $\bu_0$ that does not contain any
blur. As illustrated in Fig.~\ref{fig:guassian:MSELena:binary}, $\bu_0$ is obtained by binarizing  the digital test
image Lena (Fig.~\ref{fig:guassian:MSELena}), the threshold being the median value of Lena. Since $\bu_0$ is now
blur-free, we can compare for several values of $\ft2_a$ and for $t=4$, which is large enough, the digital images
$$\bu_1 =:\bG_{t\ft2_a} \bu_0=\bS_1 G_{t\ft2_a} Iu_0\;\; \hbox{ and }\;\;
\bu_2 =: (\bS_{1/t}I) \bS_t \bG_{t\ft2_a} \bu_0=(\bS_{1/t}I) \bS_t  G_{t\ft2_a} Iu_0,$$ As shown in
Fig.~\ref{fig:guassian:MSELena:binary}, $\ft2_a=0.6$ is the smallest value ensuring no visual ringing in $\bu_2$. Under
this value, for example for $\ft2_a=0.3$, clear ringing artifacts are present in $\bu_2$. That $\ft2_a=0.6$ is the
correct value is confirmed by the MSE($\bu_1,\bu_2$) curve showing that the mean square error decays rapidly until
$\ft2_a$ goes down to $0.6$, and is stable  and small thereafter. The result, confirmed in ten experiments with
different initial images, is consistent with the value obtained in the first experimental setting.


\begin{figure}[]
\begin{center}
\begin{tabular}{cc}
\includegraphics[width=4cm]{./figures/LenaBinary.eps}
&
\includegraphics[width=4cm]{./figures/gaussian_MSE_LenaBianry_t2.eps}
\\
\includegraphics[width=4cm]{./figures/LenaBinary_c0p3_t2_blur_hat.eps}
&
\includegraphics[width=4cm]{./figures/LenaBinary_c0p3_t2_hat.eps}
\\
\includegraphics[width=4cm]{./figures/LenaBinary_c0p6_t2_blur_hat.eps}
&
\includegraphics[width=3cm]{./figures/LenaBinary_c0p6_t2_hat.eps}
\end{tabular}
\end{center}
\caption{Top left: $\bu$. Top right: MSE($\bu_1,\bu_2$) vs $\ft2_a$. Middle (from left to right): $\bu_1$ and $\bu_2$
with $\ft2_a=0.3$. MSE($\bu_1,\bu_2$)=7.46. Bottom (from left to right): $\bu_1$ and $\bu_2$ with $ \ft2_a=0.6$.
MSE($\bu_1,\bu_2$)=0.09.} \label{fig:guassian:MSELena:binary}
\end{figure}


\subsection{Discrete sampling\label{discretesetting}}

\begin{itemize}
\item  If $\bu$ is digital and therefore only defined on  $\relative^2$ and if $\delta$ is an integer, then one can
define any sub- or over-sampling operations on $\bu$. But this requires  interpolating $\bu$ first.   Thus we define a
{\it digital subsampling operator} by
\begin{equation}\label{defdiscretesampling}\SAD_\delta \bu=: \SA_\delta I\bu.\end{equation}   Notice  that
 $\SAD_\delta$  is a
discrete filter which can be interpreted as a {\rm re-sampling} operator.


\item Zooming in by oversampling. For every $\gamma\leq 1,$
\begin{equation}\label{commutationinterpolationsampling}
I\SAD_\gamma \bu=H_\gamma I\bu.\end{equation}
 Proof: $I\bu$ is well sampled, with spectrum  in
$[0,2\pi]^2$. Thus since $\gamma<1$, $I\SAD_\gamma \bu$ is over-sampled: it has spectrum in $[0, 2\pi\gamma]^2$. Thus
 $I\SAD_\gamma \bu$ is band limited, as $H_\gamma I\bu$. Since $$I\SAD_\gamma \bu(n_1,n_2)=\SAD_\gamma \bu(n_1,n_2)
 =I\bu(n_1\gamma, n_2\gamma) \hbox{ and }
H_\gamma I\bu(n_1,n_2)=I\bu(n_1\gamma, n_2\gamma),$$ both functions have the same $\relative^2$ samples and therefore
coincide.


\item Corollary: If $\gamma\leq 1$, then \begin{equation}\label{compositionofsampling}
\SAD_\beta\SAD_\gamma=\SAD_{\beta\gamma}.\end{equation}
\newline Proof: using once \eqref{commutationinterpolationsampling} and twice \eqref{defdiscretesampling},
$$\SAD_\beta\SAD_\gamma \bu=\SA_\beta I\SAD_\gamma \bu=\SA_\beta \bH_\gamma I\bu=\SA_{\beta\gamma}I\bu
=\SAD_{\beta\gamma}\bu.$$


\item a discrete commutation formula :  Assume $\bu$ is a digital image. Then
  for $\gamma<1$,
\begin{equation}\label{discretecommutation}\bG_\beta \SAD_\gamma \bu=\SAD_\gamma \bG_{\beta\gamma}\bu.\end{equation}
\item proof :
$$\bG_\beta \SAD_\gamma \bu\stackrel{\eqref{defGdelta}}{=}
\SA_1(G_{\beta}  I \SAD_\gamma  \bu)\stackrel{\eqref{commutationinterpolationsampling}}{=}
 \SA_1(G_\beta  H_\gamma I\bu)$$
$$ \stackrel{\eqref{commutationconvolutionzoomcontinus}}{=}
\SA_1(H_\gamma(G_{\beta\gamma}  I\bu)) \stackrel{(\ref{defsampling})}{=} \SA_\gamma(G_{\beta\gamma}
I\bu)\stackrel{\eqref{interpolationrecovery}}{=}\SA_\gamma I\SA_1(G_{\beta\gamma}  I\bu)
\stackrel{(\ref{defGdelta},\ref{defdiscretesampling})}{=}\SAD_\gamma \bG_{\beta\gamma}\bu.$$  Notice that we use
$I\SA_1 u=u$ with $u=G_{\beta\gamma}  I\bu$. Indeed, this last function is well sampled, because $Iu$ is.


\end{itemize}

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