%gmChap14.tex

\graphicspath{{./figures/chapter14/}}

\chapter{Asymptotic Behavior of Affine-Invariant Filters}

We are going to analyze the asymptotic behavior of affine-invariant operators in much the same way we analyzed contrast-invariant isometric operators in Chapters 10 and 11. The analysis in this chapter will be in $\RR^2$. Recall that when we say an operator $T$ is affine invariant, we mean that $T$ commutes with all elements of the special linear group $SL(\RR^2)$. Thus, for $A\in SL(\RR^2)$, we have $ATu=TAu$ for all functions $u$ in the domain of $T$, where $Au$ is defined by $Au(\fxx)=u(A\fxx)$, $\fxx\in\RR^2$. At this point, there are two possible scenarios: Assume we are given an affine-invariant operator $T$ that is also contrast and translation invariant and then use Theorem 7.3 to conclude that $T$ can be represented as
\begin{equation}
Tu(\fxx)=\inf_{B\in\mathcal B}\sup_{\fxy\in\fxx+B}u(\fxy),
\end{equation}
where the set of structuring elements $\mathcal B$ may be taken to be $\{X\mid 0\in {\mathcal T}X\}$, ${\mathcal T}X=\levelset_1T\bmone_X$, and where (14.1) holds almost everywhere for $u$ in the domain of $T$. The other approach, which is the one we take, is to assume the set of structuring elements $\mathcal B$ is given and to define $T$ by (14.1). This places the focus on $\mathcal B$. With this approach, we know immediately that $T$ is 
contrast and translation invariant and that it is defined on all $u:\RR^2\to\RR$. We are, however, left with the task of proving that $T$ is affine invariant if and only if $\mathcal B$ is affine invariant. (This is the content of Exercise 14.1.). Again, it is understood that, in our context, ``affine invariant'' always means ``invariant with respect to $SL(\RR^2)$.'' 

We assume that $\mathcal B$ is a set of affine-invariant structuring elements, and we define for every $u:\RR^2\to\RR$,
\begin{equation}
\begin{aligned}\label{eq141a}
SI_hu(\fxx)&=\sup_{B\in \mathcal B}\inf_{\fxy\in\fxx+hB}u(\fxy);\\
IS_hu(\fxx)&=\inf_{B\in \mathcal B}\sup_{\fxy\in\fxx+hB}u(\fxy).
\end{aligned}
\end{equation}
$SI_hu$ is considered to be an \textit{affine erosion} of $u$, and $IS_hu$ is considered to be an \textit{affine dilation} of $u$. (Note that this nomenclature is consistent with the definitions of $\tilde{\mathcal E}_a$ and $\tilde{\mathcal D}_a$ in Chapter 12.) Since we have the relation $SI_hu=-IS_h(-u)$, it suffices to study just one of these operators, and we choose to investigate $IS_h$. Our main concern is the behavior of $IS_hu(\fxx)$ as $h\to0$ for $u\in C^3(\RR^2)$. We will prove that, if $\mathcal B$ is affine invariant and $1$-localizable, then 
\[
\lim_{h\to0}\frac{IS_hu(\fxx)-u(\fxx)}{h^{4/3}}=c_{\mathcal B}|Du(\fxx)|\Big(\frac1{2}{\rm curv}(u)(\fxx)^+\Big)^{1/3},
\]
where $c_{\mathcal B}$ is a suitable constant. (As before, $r^{1/3}$ means $(r/|r|)|r|^{1/3}$.)

\begin{exercise}Show that $T$ is affine invariant if and only if $\mathcal B$ is affine invariant. \finex
\end{exercise}

\section{The analysis of $IS_h$}

The analysis of affine-invariant operators $IS_h$ will follow the general plan outlined in section 10.1.1 and exemplified by Theorem 10.2, with the important difference that the structuring elements $\mathcal B$ are not bounded. They are, however, isometric, since the group of isometries is a subgroup of the $SL(\RR^2)$. This means that given a $C^3$ function $u$, we can expand it in the form
\[
u(\fxx+\fxy)=u(\fxx)+px+ax^2+by^2+cxy+R(\fxx,\fxy),
\] 
where $\fxy=(x,y)$ and the linear term is $px$. (We use the notation and conventions of section 4.5.) If we assume that this expansion holds for $\fxy\in D(0,r)$, then the analysis of the error term $R$ given in the proof of Theorem 10.2
implies that
\[
|u(\fxx+\fxy)-u(\fxx)-(px+ax^2+by^2+cxy)|\leq \sup_{|\fxy|\leq r}\|D^3(\fxx+\fxy)\||\fxy|^3.
\]
Define $v$ for $\fxy\in D(0,r)$ by $v(\fxy)=u(\fxx)+px+ax^2+by^2+cxy$. If $\mathcal B$ is $1$-localizable, and if $h\leq r^2$ and $h<1/R^2$, then we know from Lemma 13.6 that 
\[
|IS_hu(\fxx)-IS_hv(\fxx)|\leq (C+Kc)h^{3/2}.
\]
(Here and elsewhere we use the fact that, if $u$ is locally $C^3$, then it is locally Lipschitz. Also, refer to sections 13.1 and 13.2 for the meaning of the constants.) This implies that the analysis of $IS_h$ can be reduced to analyzing the action of $IS_h$ on polynomials of degree two. This analysis will be done in Theorem 14.4, but before we get there, we need to consider the action of $IS_h$ on two specific polynomials. Because the cases $a=c=0$ and $b=1$ or $b=-1$ play key roles, we introduce special notation:
\[
c_{\mathcal B}^+=\inf_{B\in\mathcal B}\sup_{\fxy\in B}(x+y^2)\quad\text{and}\quad c_{\mathcal B}^-=\inf_{B\in\mathcal B}\sup_{\fxy\in B}(x-y^2),
\]
where ``$IS_h(px+ax^2+by^2+cxy)$'' always means ``$IS_h(px+ax^2+by^2+cxy)(0)$.''
Since our main results use these constants, it is worth examining some examples.

\begin{lemma}$(i)$ Let $\mathcal B$ be an affine-invariant family of open convex sets that have area one and that are symmetric with respect to the origin. Then $c_{\mathcal B}^+>0$ and $c_{\mathcal B}^-=0$. $(ii)$ If $\mathcal B={\mathcal B}_{\rm aff}$, then $c_{\mathcal B}^+>0$ and $c_{\mathcal B}^-=0$. $(iii)$ If a set of structuring elements $\mathcal B$ is affine invariant and contains one bounded element that is open and contains the origin, then $c_{\mathcal B}^-=0$. In particular, this is the case if $\mathcal B$ is affine invariant and 1-localizable.
\end{lemma}  
\begin{proof} We prove (i) first. Let $B$ be an element of $\mathcal B$, and let $l(\alpha)$ denote the line segment defined in polar coordinates by $(\alpha, \rho)$, $0\leq\rho<+\infty$. Both $B$ and $l(\alpha)$ are convex, so their intersection $B\cap l(\alpha)$ is convex. Since $B$ is open, this set has the form $[0,d(\alpha))$, where $d(\alpha)$ is the distance from the origin to the boundary of $B$ in the direction $\alpha$. The function $\alpha\mapsto d(\alpha)$ is continuous (Exercise 14.2). Since $B$ has area one, this function can be a constant only if $d(\alpha)=1/\sqrt{\pi}$. In all other cases, $d(\alpha)$ takes values greater than $1/\sqrt{\pi}$ and less than $1/\sqrt{\pi}$, and since it is continuous, it must take the value $1/\sqrt{\pi}$. In fact, since $B$ is symmetric, $d$ must assume the value $1/\sqrt{\pi}$ four times. In any case, there is a point $(x,y)\in\partial B$ such that $x>0$ and $x^2+y^2=1/\pi$.

Now consider the disk $D=D(0,1/\sqrt{\pi})$. We have just seen that there is a point $(x,y)\in\partial B\cap\partial D$ such that $x>0$. If $(x,y)\in D$ and $x>0$, then $x>x^2$, and we have the following inequalities:
\[
\sup_{(\fxx,\fxy)\in B}(x+y^2)\geq\sup_{(\fxx,\fxy)\in B\cap D, x>0}(x+y^2)\geq\sup_{(\fxx,\fxy)\in B\cap D, x>0}(x^2+y^2)=\frac1{\pi}.
\]
The right-hand term does not depend on $B$, so we have $c_{\mathcal B}^+\geq1/{\pi}$. The value $1/{\pi}$ is not significant for our purposes; we just wish to show that $c_{\mathcal B}^+>0$. 

To prove that $c_{\mathcal B}^-=0$, first note that since $B$ is open and contains the origin, there are points $(x,y)\in B$ with $x>0$ and $y=0$. Thus, $c_{\mathcal B}^-\geq0$. Fix $B\in\mathcal B$ and consider the sets obtained by ``squeezing'' $B$ onto the line $x=0$:
\[
B_{\varepsilon}=\{(x',y')\mid x'=\varepsilon x, y'=y/\varepsilon, (x,y)\in B\}.
\]
Then $B_{\varepsilon}$ is an affine transform of $B$, so $B_{\varepsilon}\in\mathcal B$. Therefore,
\[
c_{\mathcal B}^-\leq\sup_{(x,y)\in B_{\varepsilon}}(x-y^2)\leq\sup_{(x,y)\in B_{\varepsilon}}(x)\leq C\varepsilon.
\]
Thus, $c_{\mathcal B}^-=0$. 

We turn now to the proof of (ii). Assume that $B$ is in ${\mathcal B}_{\rm aff}$. Then by definition, $B$ is open, connected, and contains the origin, and the connected components of $B\cap\Delta^c$ that contain the origin in their boundaries have areas greater than or equal to $b=\delta(0,B^c)>1$. If we let $\Delta$ be the $y$-axis and $H$ be the open half-plane defined by $x>0$, then the definition implies that ${\rm area}(B\cap H)>1$. This implies that
\begin{equation}\label{eq152}
\Big(\sup_{(x,y)\in B\cap H}(x)\Big)\Big(\sup_{(x,y)\in B\cap H}(|y|)\Big)\geq\frac1{2}.
\end{equation}
We wish to find a lower bound for $\sup_{(x,y)\in B}(x+y^2)$, so we may  assume that $B$ is bounded, and define $\mu=\sup_{(x,y)\in B\cap H}(x)$ and $\nu=\sup_{(x,y)\in B\cap H}(y^2)$. Then
\[
\sup_{(x,y)\in B}(x+y^2)\geq\sup_{(x,y)\in B\cap H}(x+y^2)\geq\inf\{\mu,\nu\}.
\]
Thus, $c_{\mathcal B_{\rm aff}}^+\geq \inf\{\mu,\nu\}$. Using the constraint \eqref{eq152}, we conclude that 
\[
c_{\mathcal B_{\rm aff}}^+\geq \inf_{\mu\geq0}\Big\{\mu,\frac1{4\mu^2} \Big\}=2^{-2/3}>0.
\]
Finally, we must show that $c_{\mathcal B_{\rm aff}}^-=0$. If $B\in\mathcal B_{\rm aff}$ and if $B$ is bounded, then, as in the proof of (i),
\[
c_{\mathcal B_{\rm aff}}^-\leq\sup_{(x,y)\in B_{\varepsilon}}(x-y^2)\leq\sup_{(x,y)\in B_{\varepsilon}}(x)\leq C\varepsilon,
\]
and $c_{\mathcal B_{\rm aff}}^-=0$. The proof of $(iii)$ is exactly the same as showing that $c_{\mathcal B_{\rm aff}}^-=0$.
\end{proof}

\begin{exercise}The purpose of this exercise is to show that the function $\alpha\mapsto d(\alpha)$ in Lemma 14.1 is continuous. (Hint: The assumption that $d$ is not continuous leads to a contradiction of the fact that the line segment $l(\alpha)$ intersects $\partial B$ in one and only one point.) \finex
\end{exercise}

\begin{exercise}This exercise is to show that it is possible to have $c_{\mathcal B}^-<0$ for a simple set of structuring elements. Let $\mathcal B=\{AC\mid A\in SL(\RR^2)\}$, where $C$ is a square with one side missing defined as follows: $C=\{(x,y)\mid x=-2, -2\leq y\leq2; y=+2, -2\leq x\leq2; y=-2,-2\leq x\leq2 \}$. Show that $c_{\mathcal B}^+>0$ and $c_{\mathcal B}^-<0$. \finex
\end{exercise}

As one can imagine, the polynomial $px+by^2$, with $p>0$, is particularly important in the affine-invariant theory. Fortunately, an invariance argument allows us to compute explicitly the action of $IS_h$ on $px+by^2$.

\begin{lemma}Let $\mathcal B$ be an affine-invariant set of structuring elements and assume that at least one $B\in\mathcal B$ is bounded. Let $IS_h$ be the associate inf-sup operator and assume that $p>0$. Then
\begin{align*}
IS_h(px+by^2)&=c_{\mathcal B}^+\Big(\frac{b}{p}\Big)^{1/3}ph^{4/3}\kern6pt\quad\text{if}\quad b>0;\\
IS_h(px+by^2)&=c_{\mathcal B}^-\Big(\frac{-b}{p}\Big)^{1/3}ph^{4/3}\quad\text{if}\quad b\leq0.
\end{align*}
\end{lemma} 

\begin{proof}The existence of a bounded structuring element ensures that $\sup_{(x,y)\in B}(px+by^2)$ is not always infinite. If $b\not=0$, then
\[
B\in\mathcal B\Longleftrightarrow h
\begin{pmatrix}
h^{1/3}|b|^{1/3}&0\\
0&h^{-1/3}|b|^{-1/3}
\end{pmatrix}B\in h\mathcal B.
\]
Thus, 
\begin{align*}
\inf_{B\in h\mathcal B}\sup_{(x,y)\in B}(x+by^2)&=\inf_{B\in\mathcal B}\sup_{(x,y)\in B}(|b|^{1/3}h^{4/3}x+b(|b|^{-2/3}h^{4/3}y^2))\\
&=|b|^{1/3}h^{4/3}\inf_{B\in\mathcal B}\sup_{(x,y)\in B}(x+(b/|b|)y^2).
\end{align*}
Then we have
\begin{equation*}
IS_h(x+by^2)=
\begin{cases}
c_{\mathcal B}^+ \,b^{1/3}h^{4/3}\kern14pt\quad\text{if}\quad b>0;\\
c_{\mathcal B}^- (-b)^{1/3}h^{4/3}\quad\text{if}\quad b<0.
\end{cases}
\end{equation*}
Since $p>0$, $IS_h(px+by^2)=pIS_h(x+(b/p)y^2)$, and we deduce that 
\begin{equation*}
IS_h(px+by^2)=
\begin{cases}
c_{\mathcal B}^+ ({b}/{p})^{1/3}ph^{4/3}\kern8pt\quad\text{if}\quad b>0;\\
c_{\mathcal B}^- ({-b}/{p})^{1/3}ph^{4/3}\quad\text{if}\quad b<0.
\end{cases}
\end{equation*}

Finally, we must deal with the case $b=0$. Let $B$ be a bounded element of $\mathcal B$ and assume it is contained in the square $[-R,R]\times[-R,R]$. Then the element 
$h\Big(
\begin{smallmatrix}
h\varepsilon&0\\
0&h^{-1}\varepsilon^{-1}
\end{smallmatrix}
\Big)B$
belongs to $h\mathcal B$ and is contained in the rectangle $\mathcal R=[-R\varepsilon h^2,R\varepsilon h^2]\times[-R/\varepsilon,R/\varepsilon]$. Hence,
\[
0\leq IS_h(px)\leq\sup_{(x,y)\in\mathcal R}(px)\leq pRh^2\varepsilon.
\]
Since we can take $\varepsilon>0$ arbitrarily small, $IS_h(px)=0$.
\end{proof}

When we studied the asymptotic behavior of an operator $T$ applied to a smooth function $u$ in Chapters 10 and 11, we usually assumed that $Du(\fxx)\not=0$; $Du(\fxx)=0$ was a special case. This is also true for affine-invariant inf-sup operators, and the next lemma deals with this case.

\begin{lemma}Let $\mathcal B$ be an affine-invariant set of structuring elements, one of which is bounded and all of which contain the origin, and let $K$ be a compact subset of $\RR^2$. Then for $\fxx\in K$ the following inequality holds for every $C^3$ function $u\!:$
\[
|IS_hu(\fxx)-u(\fxx)|\leq C(\|D^2u(\fxx)\|+|Du(\fxx)|)h^{4/3}+C_Kh^2,
\]
where $0<h\leq1$, $C>0$ is a constant that depends only on $\mathcal B$, and the constant $C_K$ depends only on $\mathcal B$, $u$, and $K$. If $Du(\fxx)=0$, then
\[
|IS_hu(\fxx)-u(\fxx)|\leq C'\|D^2u(\fxx)\|h^2+C'_Kh^3,
\]
where $C'$ depends only on $\mathcal B$ and $C'_K$ depends only on $\mathcal B$, $u$, and $K$.
\end{lemma} 

\begin{proof}We use the notation of sections 4.5 and 10.1. Let $B$ be an arbitrary element of $\mathcal B$. Since $B$ contains the origin, $\sup_{\fxy\in\fxx+hB}u(\fxy)\geq u(\fxx)$, which implies that $IS_hu(\fxx)\geq u(\fxx)$. 

Now expand $u$ in the familiar local coordinate system in a neighborhood of $\fxx$:
\[
u(\fxx+h\fxy)=u(\fxx)+phx+ah^2x^2+bh^2y^2+ch^2xy+R(\fxx,h\fxy),
\]
where $\fxy=(x,y)$. Then for any $B\in\mathcal B$,
\[
\sup_{\fxy\in B}u(\fxx+h\fxy)\leq u(\fxx)+h\sup_{\fxy\in B}(px+ahx^2+bhy^2+chxy)+\sup_{\fxy\in B}|R(\fxx,h\fxy)|.
\]
Now assume that $B^*$ is bounded. Then all of the suprema are finite, and $\sup_{\fxy\in B^*}u(\fxx+h\fxy)$ is a finite upper bound for $\inf_{B\in\mathcal B}\sup_{\fxy\in B}u(\fxx+h\fxy)=IS_hu(\fxx)$. Thus we have
\begin{equation}\label{eq14a}
0\leq IS_hu(\fxx)-u(\fxx)\leq h\sup_{\fxy\in B^*}(px+ahx^2+bhy^2+chxy)+\sup_{\fxy\in B^*}|R(\fxx,h\fxy)|,
\end{equation}
which is true for any bounded set $B^*\in\mathcal B$. We are now going to use the affine invariance of $\mathcal B$ to manipulate $B^*$ and thereby obtain a good estimate for the terms on the right-hand side of \eqref{eq14a}. Since $B^*$ is bounded, it is contained in the square $[-S,S]\times[-S,S]$, where $S=\sup_{\fxy\in B^*}|\fxy|$. By the affine invariance of $\mathcal B$, the set 
$B'=\Big(
\begin{smallmatrix}
h^{1/3}&0\\
0&h^{-1/3}
\end{smallmatrix}
\Big)B^*$
belongs to $\mathcal B$ and is contained in the rectangle
\[
\mathcal R=[-S h^{1/3},Sh^{1/3}]\times[-Sh^{-1/3},Sh^{-1/3}].
\]
We replace $B^*$ with $B'$ in \eqref{eq14a} and proceed to estimate the terms on the right-hand side.  
\begin{align*}
&\sup_{\fxy\in B'}(px+ahx^2+bhy^2+chxy)\leq\sup_{\fxy\in\mathcal R}(px+ahx^2+bhy^2+chxy)\\
&\leq pSh^{1/3}+|a|S^2h^{5/3}+|b|S^2h^{1/3}+|c|S^2h
\leq pSh^{1/3}+(|a|+|b|+|c|)S^2h^{1/3}\\
&=|Du(\fxx)|Sh^{1/3}+(1/2)\|D^2u(\fxx)\|S^2h^{1/3}
\leq C(\|D^2u(\fxx)\|+|Du(\fxx)|)h^{1/3},
\end{align*}
where $C=\max\{S,S^2/2\}$. Note that $C$ depends only on $\mathcal B$; in particular, it does not depend on $\fxx$. Note also that this holds for all $h$, $0<h\leq1$. We now turn to the other term:
\begin{align*}
\sup_{\fxy\in B'}|R(\fxx,h\fxy)|&\leq\sup_{\fxy\in\mathcal R}|R(\fxx,h\fxy)|\leq\sup_{\fxy\in\mathcal R}\|D^3u(\fxx+h\fxy)\|h^3|\fxy|^3\\
&\leq \sup_{\fxy\in\mathcal R}\|D^3u(\fxx+h\fxy)\|h^3(S^2h^{2/3}+S^2h^{-2/3})^{3/2}\\&\leq \sup_{\fxy\in\mathcal R}\|D^3u(\fxx+h\fxy)\|2^{3/2}S^3h^2.
\end{align*}
If $K$ is an arbitrary compact set, then $\sup_{\fxx\in K}\sup_{\fxy\in\mathcal R}\|D^3u(\fxx+h\fxy)\|2^{3/2}S^3\leq C_K$ for some constant that depends only on $K$, $u$, and $\mathcal B$. This proves that
\[
|IS_hu(\fxx)-u(\fxx)|\leq C(|Du(\fxx)|+\|D^2u(\fxx)\|)h^{4/3}+C_Kh^2.
\]

If $Du(\fxx)=0$, we do the same computation, but we treat both axes the same:   $\mathcal R$ is replaced with the square $[-Sh,Sh]\times[-Sh,Sh]$, and
\[
\sup_{\fxy\in\mathcal R}(ax^2+by^2+cxy)+\sup_{\fxy\in\mathcal R}|R(\fxx,h\fxy)|\leq C'\|D^2u(\fxx)\|h^2+C'_Kh^3. \qedhere
\]
\end{proof}

Since $SI_hu=-IS_h(-u)$, it is clear that the lemma is true for $SI_h$. Everything is now in place to state and prove the first major result of this chapter.  

\begin{theorem}Let $\mathcal B$ be a $1$-localizable affine-invariant family of structuring elements. Assume that $u:\RR^2\to\RR$ is a Lipschitz function that is $C^3$ in a neighborhood of $\fxx$. Then
\[
IS_hu(\fxx)-u(\fxx)=h^{4/3}c_{\mathcal B}|Du(\fxx)|\Big(\frac1{2}{\rm curv}(u)(\fxx)^+\Big)^{1/3}+o(\fxx,h^{4/3}),
\]
where $c_{\mathcal B}=c^+_{\mathcal B}$. If $u$ is $C^3$ in a neighborhood of a compact set $K$ and $Du(\fxx)\not=0$ on $K$, then the result holds for all $\fxx\in K$ and $o(\fxx,h^{4/3})/h^{4/3}\to0$ as $h\to0$ uniformly for $\fxx\in K$.
\end{theorem}

\begin{proof}The proof is rather involved, and it is difficult to motivate the various steps. The overall object is to ``eliminate'' the $x^2$-term and the $xy$-term and then to apply Lemma 14.2. This is done using the fact that 	$IS_h$ is contrast invariant. 

Since $IS_h$ is invariant under translation, we assume that $\fxx=0$, and since $IS_h$ commutes with the addition of constants, we assume that $u(0)=0$. Since $\mathcal B$ is isometric, we can represent $u$ in the disk $D(0,1)$ as
\begin{equation}\label{eq151}
u(\fxy)=px+ax^2+by^2+cxy+R(\fxy),
\end{equation}
where $|R(\fxy)|\leq C_Rr^3$ for $\fxy=(x,y)\in D(0,r)$, $0\leq r\leq1$. Then
\[
px+ax^2+by^2+cxy-C_Rr^3\leq u(\fxy)\leq px+ax^2+by^2+cxy+C_R(r^3).
\]
The operators $IS_h^r$ satisfy the local maximum principle in the sense that $v(\fxy)\leq w(\fxy)$ for $\fxy\in D(0,r)$ implies that $IS_h^rv(0)\leq IS_h^rw(0)$ for $0<h\leq1$. This implies that for all $h$, $0<h\leq1$,
\begin{equation}\label{eq153}
IS^r_hu(0)=IS^r_h[px+ax^2+by^2+cxy](0)+O(r^3),
\end{equation}
where $|O(r^3)|\leq C_Rr^3$. 
The proof now branches into two cases, the first being simple and the second being more complicated.

\medskip 

\noindent {\bf Case (1):} $p=|Du(0)|=0$.
\medskip 

From Lemma 14.3, $IS_hu(0)=O(h^2)$. Hence, $IS_hu(0)/h^{4/3}=O(h^{2/3})$, which proves the assertion.

\medskip 

\noindent {\bf Case (2):} $p=|Du(0)|\not=0$.

\medskip 

{\bf Eliminating the ${\bm x\bm y}$-term.} For any $\varepsilon>0$, 
\[
-|c|\varepsilon y^2-\frac{|c|}{\varepsilon}x^2\leq cxy\leq |c|\varepsilon y^2+\frac{|c|}{\varepsilon}x^2.
\]

Then using \eqref{eq153} and the fact that the $IS_h^r$ are localized, we deduce the following inequalities:
\begin{align}\label{eq154}
IS_h^ru(0)&\leq IS_h^r\big[px+(a+|c|\varepsilon^{-1})x^2+(b+|c|\varepsilon)y^2\big](0) +C_Rr^3;\\
IS_h^ru(0)&\geq IS_h^r\big[px+(a-|c|\varepsilon^{-1})x^2+(b-|c|\varepsilon)y^2\big](0) +C_Rr^3.\label{eq155}
\end{align}
These inequalities hold for all $\varepsilon>0$, $0<h\leq1$, and $0<r\leq1$. This leads us to study the expressions
\[
IS_h^r\big[px+(a\pm|c|\varepsilon^{-1})x^2+(b\pm|c|\varepsilon)y^2\big](0),
\]
which do not contain $xy$-terms. To keep the notation simple, we will write $A=a\pm|c|\varepsilon^{-1}$ and $B=b\pm|c|\varepsilon$. The important point for making the estimates is that $A=O(\varepsilon^{-1})$ and $B=O(1)$. For typographical reasons, we will write $IS_h[P(\fxy)]$ for $IS_h[P(\fxy)](0)$ for polynomials $P$. We will also assume that $b\not=0$; the case $b=0$ leads to similar computations and the same conclusion.

\medskip 

{\bf Eliminating the ${\bm x^2}$-term.} We are going to use the fact that the operators are contrast invariant, and the contrast change that we choose is defined by $g_{\varepsilon}(s)=s-(A/p^2)s^2$. For $|s|$ sufficiently small, $g_\varepsilon$ is increasing, and we wish to say that
\[
IS_h^r[px+Ax^2+By^2]=g_\varepsilon^{-1}\bigg(IS_h^r\Big[px+By^2-\frac{A}{p^2}\big((Ax^2+By^2)^2+2px(Ax^2+By^2)\big)\Big]\bigg),
\]
but to do so, we must be sure that the argument of $g_\varepsilon^{-1}$ is in a neighborhood of zero where $g_\varepsilon$ is increasing. If we let $t$ denote the argument of $g_\varepsilon^{-1}$, this means that $|t|$ must be less than $C{\varepsilon}$ for some constant $C>0$. (Note that there are a couple of cases involving the values of $a$ and $|c|$ that must be investigated, but the controlling condition is that $|t|$ is less than $C{\varepsilon}$ for some constant $C>0$.)

The next step is then to estimate the term $({A}/{p^2})\big((Ax^2+By^2)^2+2px(Ax^2+By^2)\big)$. Using the fact that $A=O(\varepsilon^{-1})$ and $B=O(1)$, a straightforward computation shows that
\[
\frac{A}{p^2}\big((Ax^2+By^2)^2+2px(Ax^2+By^2)\big)=O(\varepsilon^{-2}r^3)
\]
for $(x,y)\in D(0,r)$. At this point, it is necessary to relate the three parameters $\varepsilon$, $r$, and $h$. Our point of view is that $\varepsilon$ and $r$ are functions of $h$, and we let $r=h^{1/2}$ and $\varepsilon=h^\theta$, where $\theta>0$ is a small number that must meet several conditions. The first of these is that $|IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]|< C{\varepsilon}=Ch^{\theta}$.

We know from Lemma 14.2 that $IS_h[px+By^2]=O(h^{4/3})+O(\varepsilon^{1/3}h^{4/3})=O(h^{4/3})$, so we know from Lemma 13.4$(i)$ that 
\[
IS^r_h[px+By^2]=IS_h[px+By^2]+O(h^{3/2})=O(h^{4/3}).
\]
This implies that $IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]=O(h^{4/3})+O(h^{(3/2)-2\theta})$, which means that we must choose $\theta$ such that $O(h^{4/3})+O(h^{(3/2)-2\theta})<Ch^{\theta}$ to be certain that
\[
IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]
\]
is in the domain of $g_\varepsilon^{-1}$. Clearly, any value of $\theta$ less than $1/2$ will work.

Now that we have $IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]$ in the domain of $g_\varepsilon^{-1}$, we use that fact that 
 $g_\varepsilon^{-1}(t)=t+O(\varepsilon^{-1}t^2)$ to deduce that 
\begin{align*}
IS_h^r[px+Ax^2+By^2]&=g_\varepsilon^{-1}\big(IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]\big)\\
&=IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]+O(\varepsilon^{-1}h^{8/3})\\
&=IS_h^r[px+By^2]+O(\varepsilon^{-2}h^{3/2})+O(\varepsilon^{-1}h^{8/3})\\
&=IS_h^r[px+By^2]+O(\varepsilon^{-2}h^{3/2}),
\end{align*}
where for these estimates we must have $\theta\leq 1/12$, which is the second condition on $\theta$.  

\medskip

{\bf Final estimates.} Use Lemma 13.4$(i)$ to replace $IS_h^r[px+By^2]$ with $IS_h[px+By^2]$:
\[
IS_h^r[px+Ax^2+By^2]=IS_h[px+By^2]+O(\varepsilon^{-2}h^{3/2}).
\]
Using inequalities \eqref{eq154} and \eqref{eq155} and another call on Lemma 13.4$(i)$ shows that
\[
IS_h[px+(b-|c|\varepsilon)y^2]-O(\varepsilon^{-2}h^{3/2})\leq IS_hu(0)\leq IS_h[px+(b+|c|\varepsilon)y^2]+O(\varepsilon^{-2}h^{3/2}).
\]

We now apply Lemma 14.2 to the terms $IS_h\big(px+(b\pm|c|\varepsilon)y^2\big)$ in this last expression and use the inequalities $(\alpha+\beta)^{1/3}\leq \alpha^{1/3}+ \beta^{1/3}$ and $\alpha^{1/3}-\beta^{1/3}\leq (\alpha-\beta)^{1/3}$, for $\alpha,\beta\geq0$, to deduce that
\[
IS_hu(0)=c_{\mathcal B}^+\Big(\frac{b}{p}\Big)^{1/3}ph^{4/3}+O(h^{(3/2)-2\theta})+O(h^{(4+\theta)/3})\quad\text{\,\, if}\quad b>0.
\] 

This essentially concludes the pointwise part of the proof. We simply replace $b$ with $(1/2)|Du(0)|{\rm curv}(u)(0)$ and $p$ with $|Du(0)|$ in these last equations and, once again, choose a suitable $\theta$. If we choose $\theta$ so that $(3/2)-2\theta=(4+\theta)/3$, that is $\theta=1/14$, then it satisfies the two conditions on already encountered and gives an error term $O(h^{19/14})$. Recall that when we apply Lemma 13.4$(i)$, there is the restriction that $h<r/R$. Since $r=h^{1/2}$, this means that we must have $h<1/R^2$. To avoid repeating this restriction, we write the error term as $o(h^{4/3})$ rather than as $O(h^{19/14})$ with the caveat ``for small h.'' This is completely consistent with the fact that we are only interested in the behavior of $(IS_hu(\fxx)-u(\fxx))/h^{4/3}$ as $h\to0$. 

\medskip
 
{\bf Uniform convergence.} It remains to argue that everything we have done is ``uniform on compact sets on which $Du(\fxx)\not=0$.'' Based on previous work, particularly in the proof of Theorem 10.2, this should be familiar territory: Given a compact set $K$ on which $Du(\fxx)\not=0$, the functions $a$, $b$ and $c$ are bounded on $K$, $p$ is bounded away from zero, and the various estimates can be made to hold uniformly on $K$. The new element in the current proof is the family of functions $g_\varepsilon$: It must be shown that  $IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]$ is in the domain of $g_\varepsilon$ uniformly for $(x,y)\in K$. The problem is not to have the estimate 
$IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]\leq Ch^\theta$ uniform, but rather the issue is to establish a uniform domain for the functions $g_\varepsilon^{-1}$. Thus, it is necessary to show that the domain of $g_\varepsilon^{-1}$ is $\{t\mid |t|<Ch^\theta\}$, where the constant $C>0$ does not depend on the point $(x,y)\in K$. Once stated, this is easy to demonstrate. 

Write $g_\varepsilon(s)=s(1-ks)$, where $k=(a\pm|c|\varepsilon^{-1})/p^2$. Since $a$ and $c$ are continuous on $K$ and since $p$ is bounded away from zero on $K$, there is a constant $C>0$ such that 
\[
|k|\leq C\Big(1+\frac1{\varepsilon}\Big)
\]
uniformly on $K$. The domain of $g_\varepsilon^{-1}$ is the set $\{t\mid |t|<1/4|k|\}$, so the set $\{t\mid |t|< (8C)^{-1}{\varepsilon}\}$ forms a ``uniform'' domain for $g_\varepsilon^{-1}$, that is, $IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]\leq (8C)^{-1}h^\theta$ implies that $IS_h^r[px+By^2-O(\varepsilon^{-2}r^3)]$ is in the domain of $g_\varepsilon^{-1}$.
\end{proof}

\begin{exercise}Lemma 13.4$(i)$ was used twice in the proof. Check to be sure the functions involved were locally Lipschitz. \finex
\end{exercise}

\section{Alternate schemes}

One of the main difficulties in mathematical morphology is that erosions and dilations do not commute, and this means that openings and closings do not commute. Here is the reason this is a problem: If we perform an opening of a shape in $\RR^2$ using a disk of radius $h$, then the external peaks are removed and their radii of curvature become greater than $h$. A closing with a disk of radius $h$ removes the peaks pointing inward making their radii of curvature greater than $h$. Thus, the aim of opening and closing is to smooth the peaks in exactly the same way, and it would be quite desirable to do both operations simultaneously. In other words, it would be nice if openings and closings commuted so we could speak of a ``curvature thresholding'' operator. This kind of commutation is roughly attained asymptotically by alternating openings and closings \cite{Serra:1982:IAM}. The idea is to choose a small scale $h$ and to alternate the openings $O$ and the closings $F$ as follows:
\[
O_hF_hO_{h/2}F_{h/2}\cdots O_{h/2^n}F_{h/2^n}
\]
Since $O_h$ and $F_h$ are idempotent, the growth in scale is necessary to perform a progressive smoothing up to scale $h$. In the same way, affine erosions and dilations do not commute and should be used alternately. Since these operators are not idempotent, the situation is simpler. We need only choose a scale small enough and alternate $\tilde{\mathcal E}_h$ and $\tilde{\mathcal D}_h$, that is, compute $(\tilde{\mathcal E}_h\tilde{\mathcal D}_h)^n$. In Chapter 16, we will prove that the iterated filter $(\tilde{\mathcal E}_h\tilde{\mathcal D}_h)^n$ converges to the affine shortening equations when $h\to0$ and $n\to\infty$ appropriately. 

We are now going to extend the results of section 14.1, particularly Theorem 14.4, to alternate schemes, that is, to products like $IS_hSI_h$. The alternate schemes are easier to implement and numerically more efficient if we want to have the property $T(-u)=-Tu$. It was precisely because this property is not satisfied by erosions and dilations that it has been proposed, with some experimental success, to build alternating operators like $T=IS_hSI_h$. In this case, it is not true that $T(-u)=-Tu$. We will show, however, that $(IS_hSI_h)^n$ tends to an operator that does satisfy this condition if $n\to\infty$ and $h\to0$ appropriately.

It is not possible to obtain an asymptotic result for $IS_hSI_h$ by applying Theorem 14.4 twice. Indeed, there is no guarantee (and it is generally false) that $IS_hu$ is $C^3$ if $u$ is $C^3$. The next lemma shows, however, how it is possible to extend to alternate operators convergence results like the ones given in Theorem 14.4.

Here and in the next chapter, we will be dealing with real-valued functions of several variables $F:(A,p,\fxx)\mapsto F(A,p,\fxx)$, where $A$ is a symmetric matrix, $p$ is a vector, and $\fxx\in\RR^N$. We will say more about these functions in the next chapter, but, in general, they will be continuous with respect to all arguments, except where $p=0$.

\begin{lemma}Let $S_h$ and $T_h$ be scaled versions of two inf-sup operators $S$ and $T$. Let $F_i:(A,p,\fxx)\mapsto F_i(A,p,\fxx)$, $i=1,2$, be two functions that are continuous on every compact set on which $p\not=0$. Assume that for some exponent $\alpha$ the operators $S_h$ and $T_h$ are such that 
\begin{equation}\label{eq148}
\begin{aligned}
S_hu(\fxx)-u(\fxx)&=h^\alpha F_1(D^2u,Du,\fxx)+o_1(\fxx,h^\alpha),\\
T_hu(\fxx)-u(\fxx)&=h^\alpha F_2(D^2u,Du,\fxx)+o_2(\fxx,h^\alpha),
\end{aligned}
\end{equation}
for any $u$ that is $C^3$ in a neighborhood of $\fxx$.
Assume that $S_h$ is localizable in the sense that there is an exponent $\gamma$ such that 
\begin{equation}\label{eq149}
|S_hu(\fxx)-S_h^{h^\gamma}u(\fxx)|\leq o_3(\fxx,h^\alpha)
\end{equation}
if $u$ is a Lipschitz function. These relations are assumed to hold uniformly on any compact set $K$ on which $Du(\fxx)\not=0$, that is, $o_j(\fxx,h^\alpha)/h^\alpha\to 0$, $j\in\{1,2,3\}$, uniformly for $\fxx\in K$.
Then the alternate operator $S_hT_h$ satisfies
\[
S_hT_hu(\fxx)-u(\fxx)=h^\alpha \big(F_1(D^2u,Du,\fxx)+F_2(D^2u,Du,\fxx)\big)+o(\fxx,h^\alpha).
\]
for any $u:\RR^2\to\RR$ that is $C^3$ in a neighborhood of $\fxx$ and is Lipschitz on $\RR^2$. If, in addition, $u$ is $C^3$ in a neighborhood of a compact set $K$ on which $Du(\fxx)\not=0$, then this relation is true uniformly for $\fxx\in K$, that is, $o(\fxx,h^\alpha)/h^\alpha\to 0$ uniformly for $\fxx\in K$.
\end{lemma}
\begin{proof}
Assume that $u$ is $C^3$ in a neighborhood of a compact set $K$ on which $Du(\fxx)\not=0$. ($K$ may be a single point.) Let $K_r=\overline{D}_r(K)$ be the closure of the $r$-dilation of $K$, where $r$ is chosen so that $|Du(\fxx)|$ is bounded away from zero for $\fxx\in K_r$. For $\fxx\in K$, define
\begin{align*}
f^+(\fxx,r)&=\sup_{\fxy\in\overline{D}(\fxx,r)}F_2(D^2u,Du,\fxy)\\
f^-(\fxx,r)&=\inf_{\fxy\in\overline{D}(\fxx,r)}F_2(D^2u,Du,\fxy),
\end{align*}
and define $o_2(\fxx,r,h^\alpha)=\sup_{\fxy\in\overline{D}(\fxx,r)}|o_2(\fxy,h^\alpha)|$. Then we have
\[ 
u(\fxy)+h^\alpha f^-(\fxx,r)-o_2(\fxx,r,h^\alpha)\leq T_hu(\fxy)\leq u(\fxy)+h^\alpha f^+(\fxx,r)+o_2(\fxx,r,h^\alpha)
\]
for $\fxy\in\overline{D}(\fxx,r)$. Applying $S_h^{h^\gamma}$ to this relation with $h^\gamma=r$ and using the fact that  $S_h^{h^\gamma}$ is monotone and commutes with constants, and more significantly, that $S_h^{h^\gamma}$ satisfies the local maximum principle with zero error, we have 
\[
S_h^{h^\gamma}u(\fxx)+h^\alpha f^-(\fxx,r)-o_2(\fxx,r,h^\alpha)\leq S_h^{h^\gamma}T_hu(\fxx)\leq S_h^{h^\gamma}u(\fxx)+h^\alpha f^+(\fxx,r)+o_2(\fxx,r,h^\alpha).
\]
By Lemma 6.5, the function $T_hu$ is Lipschitz. Thus, we can use \eqref{eq149} to replace $S_h^{h^\gamma}T_hu(\fxx)$ with $S_hT_hu(\fxx)$ and $S_h^{h^\gamma}u(\fxx)$ with $S_hu(\fxx)$ to obtain
\begin{align*}
S_hu(\fxx)-o'_3(\fxx,h^\alpha)&+h^\alpha f^-(\fxx,r)-o_2(\fxx,r,h^\alpha)\leq S_hT_hu(\fxx)\\
&\leq S_hu(\fxx)+o'_3(\fxx,h^\alpha)+h^\alpha f^+(\fxx,r)+o_2(\fxx,r,h^\alpha),
\end{align*}
where $o'_3(\fxx,h^\alpha)$ is a nonnegative error term that incorporates all of the errors accumulated in making these replacements. Written slightly differently, we have
\begin{align*}
S_hu(\fxx)-S_hT_hu(\fxx)\leq o'_3(\fxx,h^\alpha)-h^\alpha f^-(\fxx,r)+o_2(\fxx,r,h^\alpha),\\
S_hT_hu(\fxx)-S_hu(\fxx)\leq o'_3(\fxx,h^\alpha)+h^\alpha f^+(\fxx,r)+o_2(\fxx,r,h^\alpha).
\end{align*}
Now use \eqref{eq148} to replace $S_hu(\fxx)$ with $u(\fxx)+h^\alpha F_1(D^2u,Du,\fxx)+o_1(\fxx,h^\alpha)$. This shows that
\begin{align*}
u(\fxx)-S_hT_hu(\fxx)&\leq -h^\alpha F_1(D^2u,Du,\fxx)-h^\alpha f^-(\fxx,r)+o^-(\fxx,r,h^\alpha),\\
S_hT_hu(\fxx)-u(\fxx)&\leq h^\alpha F_1(D^2u,Du,\fxx) +h^\alpha f^+(\fxx,r)+o^+(\fxx,r,h^\alpha),
\end{align*}
where $o^\pm(\fxx,r,h^\alpha)=\pm\, o_1(\fxx,h^\alpha)+o_2(\fxx,r,h^\alpha)+o'_3(\fxx,h^\alpha)$. As a last step, we replace $f^-(\fxx,r)$ with $F_2(D^2u,Du,\fxy)+o^-_4(\fxx,r)$ and $f^+(\fxx,r)$ with $F_2(D^2u,Du,\fxy)+o^+_4(\fxx,r)$, where $o^\pm_4(\fxx,r)\to0$ as $r\to0$ because $F_2$ is continuous on $K_r$. Thus,
\[
S_hT_hu(\fxx)-u(\fxx)=h^\alpha \big(F_1(D^2u,Du,\fxx)+F_2(D^2u,Du,\fxx)\big)+E(\fxx,r,h,\alpha,\gamma),
\]
and it remains to show that the error term is $o(\fxx,h^\alpha)$ and that this error is uniform on $K$. Having followed all of the individual error terms, we have the estimate
\[
|E(\fxx,r,h,\alpha,\gamma)|\leq|o_1(\fxx,h^\alpha)|+o_2(\fxx,r,h^\alpha)+o'_3(\fxx,h^\alpha)+h^\alpha |o^\pm_4(\fxx,r)|
\]
The first term, $|o_1(\fxx,h^\alpha)|$, satisfies the conditions by hypothesis. By definition, 
\[
o_2(\fxx,r,h^\alpha)=\sup_{\fxy\in\overline{D}(\fxx,r)}|o_2(\fxy,h^\alpha)|\leq \sup_{\fxy\in K_r}|o_2(\fxy,h^\alpha)|,
\]
and, again,   $\sup_{\fxy\in K_r}|o_2(\fxy,h^\alpha)|/h^\alpha\to0$ as $h\to0$ by hypothesis. Similarly, the third term is uniform. The last term $h^\alpha |o^\pm_4(\fxx,r)|=h^\alpha |o^\pm_4(\fxx,h^\gamma)|$ requires a slightly different argument: $|o^\pm_4(\fxx,h^\gamma)|\to0$ uniformly on $K$, for $\gamma>0$, because $F_1$ is uniformly continuous on $K_r$. Thus,  
$E(\fxx,r,h,\alpha,\gamma)=o(\fxx,h^\alpha)$ and $o(\fxx,h^\alpha)/h^\alpha\to0$ uniformly on $K$.
\end{proof}

We were very careful in this proof to show in detail how the various assumptions were used to establish the properties of the final error term. In contrast, we only mention that the proof of the next result uses the same techniques. In fact, it is slightly simpler. 

\begin{lemma}Assume all of the hypotheses of Lemma 14.5, except for these changes: The functions $G_i$, $i=1,2$, satisfy the same conditions as the $F_i$. Equations \eqref{eq148} are replaced with the following inequalities:
\begin{equation}\label{}
\begin{aligned}
0\geq S_hu(\fxx)-u(\fxx)\geq h^\alpha G_1(D^2u,Du,\fxx)+o(\fxx,h^\alpha),\\
0\leq T_hu(\fxx)-u(\fxx)\leq h^\alpha G_2(D^2u,Du,\fxx)+o(\fxx,h^\alpha).
\end{aligned}
\end{equation}
Then the alternate operators $S_hT_h$ satisfy
\[
h^\alpha G_1(D^2u,Du,\fxx)+o(\fxx,h^\alpha)\leq S_hT_hu(\fxx)-u(\fxx)\leq h^\alpha G_2(D^2u,Du,\fxx)+o(\fxx,h^\alpha)
\]
for any Lipschitz function $u$ that is $C^3$ in a neighborhood of $\fxx$. If $u$ is $C^3$ in the neighborhood of a compact set $K$ on which $Du$ does not vanish, then $o(\fxx,h^\alpha)/h^\alpha\to0$ uniformly on $K$ as $h\to0$.
\end{lemma}

We are now in position to state and prove an asymptotic behavior theorem for the main affine-invariant alternate filter, namely, the filter obtained by alternating the affine erosions and dilations generated with $\mathcal B_{\rm aff}$.

\begin{theorem}Let $\mathcal B=\mathcal B_{\rm aff}$, and let $SI_h$ and $IS_h$ be the affine erosions and affine dilations defined by equations \eqref{eq141a}. Then for every Lipschitz function $u$ that is $C^3$ in a neighborhood of $\fxx$ and $Du(\fxx)\not=0$,
\begin{equation}\label{eq1510}
\lim_{h\to0}\frac{T_hu(\fxx)-u(\fxx)}{h^{4/3}}=c_{\mathcal B}|Du(\fxx)|g({\rm curv}(u)(\fxx)),
\end{equation}
where 
\begin{equation*}
g(r)=
\begin{cases}
(r^+/2)^{1/3} \text{ if } T_h=IS_h,\\
(r^-/2)^{1/3} \text{ if } T_h=SI_h,\\
(r/2)^{1/3} \kern5pt\text{ if } T_h=SI_hIS_h.
\end{cases}
\end{equation*}
If $u$ is $C^3$ in the neighborhood of a compact set $K$ on which $Du(\fxx)\not=0$, then the convergence is uniform on $K$. Also, there are continuous functions $G^\pm$ such that $G^\pm(0, 0,\fxx)=0$ and
\begin{equation}\label{eq1511}
h^{4/3}G^-(D^2u, Du,\fxx)+o(\fxx,h^{4/3})\leq T_hu(\fxx)-u(\fxx)\leq h^{4/3}G^+(D^2u, Du,\fxx)+o(\fxx,h^{4/3}).
\end{equation}
Furthermore, $o(\fxx,h^{4/3})/ h^{4/3}\to0$ uniformly on every compact set $K$.
\end{theorem}
\begin{proof} We know from Theorem 14.4 that \eqref{eq1510} is true for $T_h=IS_h$ and $T_h=SI_h$, since $\mathcal B_{\rm aff}$ satisfies the hypothesis, namely, $\mathcal B_{\rm aff}$ is 1-localizable. We know from Lemma 14.1 that $c^+_{\mathcal B_{\rm aff}}>0$ and $c^-_{\mathcal B_{\rm aff}}=0$, so $c_{\mathcal B}=c^+_{\mathcal B_{\rm aff}}$ in \eqref{eq1510}. We obtain \eqref{eq1510} for the alternate operator by applying Lemma 14.5 with $\gamma=1/2$ and $\alpha=4/3$.

The two inequalities \eqref{eq1511} are read directly from Lemma 14.3 for $T_h=IS_h$ and $T_h=SI_h$. This is possible since the elements of $\mathcal B_{\rm aff}$ are open, contain the origin, and there is a $B\in\mathcal B_{\rm aff}$ that is bounded. Once we have the inequalities for these two operators, we appeal to Lemma 14.6 to have the result for the alternate operators. This is possible since both $IS_h$ and $SI_h$ are localizable by Lemma 13.4 with $\gamma =1/2$.
\end{proof}

As a general rule, affine-invariant families of structuring whose members are open, connected, contain the origin, and some of which are bounded with areas bounded above, are localizable. In these cases, it is possible to define alternate schemes that have the properties that we have described in this chapter.

\begin{exercise}Check that the last proof and the results apply to the affine-invariant, localizable families considered in Exercises 13.3 and 13.4. \finex
\end{exercise}

\section{Comments and references}

Early versions of the results and proofs contained in this chapter were given in \cite{Guichard:1996:PDE, Guichard:1995:PDE, Guichard:1994:AAM}. The version presented here is shorter and simpler. Serra provides a good introduction to alternate sequential filters in \cite{Serra:1986:IMM}. A general axiomatic theory of self-dual morphological operators is given by Heijmans in \cite{Heijmans:1996:SDM}. Applications of alternate sequential filters are discussed in \cite{Pei:1997:ECA}.
