\documentclass[11pt]{article} \usepackage{amsmath, amssymb, amsthm, graphicx} \usepackage[authoryear, round]{natbib} \usepackage{color} \usepackage{float} \def\post{{p}} \def\bR{\mathbb{R}} % real line \def\defn{{\stackrel{\rm def}{=}}} \newcommand{\CN}{{\mathcal N}} \def\bE{\mathbb{E}} % expectation \newtheorem{lemma}{Lemma} \newtheorem{theorem}{Theorem} \newcommand{\eps}{\varepsilon} \def\vec#1{\mathchoice{\mbox{\boldmath$\displaystyle\bf#1$}} {\mbox{\boldmath$\textstyle\bf#1$}} {\mbox{\boldmath$\scriptstyle\bf#1$}} {\mbox{\boldmath$\scriptscriptstyle\bf#1$}}} \begin{document} \section{Introduction} {\color{red} Can I learn anything about optimization theory as it relates to (say) sequentially compact sequences? How about regarding functions of a Euclidean and functional variable?} \subsection{Kernel smoothing operators} Let $\Omega\in \bR^r$ be a compact set and let $L_1^+(\Omega)$ denote the set of positive functions that are Lebesgue integrable on $\Omega$. For some vector $\vec h\in \bR^r$ of positive values and $\vec x \in \Omega$, let \[ K_{\vec h}(\vec x) = \prod_{i=1}^r \frac1{h_i} K \left( \frac{x}{h_i} \right) \] be a product kernel density function, where $K(\cdot)$ is some fixed nonnegative kernel density function on $\bR$. Frequently, we take $\vec h$ to be $(h, \ldots, h)^\top$ and write $K_h(\vec x)$. For any $\phi\in L_1^+(\Omega)$, we define the operator ${\cal S}:L_1^+(\Omega)\to L_1^+(\Omega)$ by \[ {\cal S}\phi(\vec x) = \int_\Omega K_h(\vec x - \vec u) \phi(\vec u)\, d\vec u. \] By this definition, ${\cal S}$ is a linear operator on $L_1^+(\Omega)$ and it depends implicitly on $K(\cdot)$ and $h$ (or $\vec h$). Similarly, we define a nonlinear operator ${\cal N}:L_1^+(\Omega)\to L_1^+(\Omega)$ by \begin{eqnarray*} {\cal N}\phi(\vec x) &=& \exp\left\{ \int_\Omega K_h(\vec x - \vec u) \log \phi(\vec u) \, d\vec u \right\}. \end{eqnarray*} Each $\phi\in L_1^+(\Omega)$, since it is a positive function defined on a compact set, must be bounded away from zero, which means that $|\log\phi|$ is bounded and ${\cal N}\phi(\vec x)$ is finite. Here are two inequalities satisfied by these operators, as proved by Eggermont and Lariccia (1995, lemma 7.1). First, \begin{equation}\label{ELineq1} {\cal N}\phi(\vec x)\le \{S\sqrt{\phi} (\vec x)\}^{2} \le \{S\phi(\vec x)\} \end{equation} for all $\phi\in L_1^+(\Omega)$ and almost all $\vec x\in\Omega$. The outer inequality relating ${\cal N}\phi$ and ${\cal S}\phi$ follows from the arithmetic-geometric mean inequality, and it has $\int_\Omega {\cal N}\phi(\vec x)\,d\vec x \le \int_\Omega \phi(\vec x)\,d\vec x $ as a corollary. Secondly, we have \begin{equation}\label{ELineq2} {\cal S}\phi(\vec x) - 2\{S\sqrt{\phi} (\vec x)\}^{2} + {\cal N}\phi(\vec x) \ge 0 \end{equation} for all $\phi\in L_1^+(\Omega)$ and almost all $\vec x\in\Omega$. \subsection{A subset of functions} We will consider a subset ${\cal C} \subset L_1^+(\Omega)$ that consists of all $\phi\in L_1(\Omega)$ such that: \begin{enumerate} \item $\int_\Omega \phi(\vec x) \, d\vec x = 1$. \item $\epsilon\le \phi(\vec x)\le M$ for some positive constants $\epsilon$ and $M$. \end{enumerate} The rationale for this choice of ${\cal C}$ is that we wish to consider only density functions and we want ${\cal C}$ to be a compact subset of $L_1^+(\Omega)$. {\bf Remarks:\ } \begin{enumerate} \item Although action on ${\cal C}$ by the operator ${\cal S}$ does produce a function whose integral equals 1 (by Fubini's theorem), it is not necessarily the case that ${\cal C}$ is closed under action by ${\cal S}$ because of the condition $\epsilon\le \phi(\vec x)\le M$. \item The set ${\cal C}$ is not closed under action by ${\cal N}$, since ${\cal N}\phi(\vec x)$ does not necessarily integrate to one. \item The choice of $L_1(\Omega)$ is motivated by the fact that Cauchy sequences in ${\cal C}$ have limits that are in ${\cal C}$; on the other hand, it is possible to construct a sequence of density functions in, say, $L_2(\Omega)$ whose limit is not a density function even though the sequence is Cauchy (Eggermont \& Lariccia, 2001, p.~16). \end{enumerate} \subsection{Semiparametric mixtures} In the semiparametric multivariate mixture model setting, the parameters consist of a vector $\vec\lambda\in \bR^m$ of mixture weights, satisfying $\lambda_j\ge0$ and $\sum_j\lambda_j=1$, and corresponding functions $f_1, \ldots, f_m\in{\cal C}$. We shall write $\vec f=(f_1, \ldots, f_m)$. For a particular mixture weight vector $\vec\lambda$, we define the mixture operator ${\cal M}_{\vec\lambda}$, applied to a vector ${\cal N}\vec f = ({\cal N} f_1, \ldots, {\cal N} f_m)$ of nonlinearly smoothed functions, as \[ {\cal M}_{\vec\lambda} {\cal N} \vec f (\vec x) = \sum_{j=1}^m \lambda_j {\cal N} f_j (\vec x). \] Let us take $g(\vec x)\in{\cal C}$. %to be some density on $\Omega$. Given parameters $\vec\lambda$ and $\vec f$, we define \begin{equation}\label{KL} L_{\infty}(\vec\lambda, \vec f) \defn -\int_\Omega g(\vec x)\log ([{\cal M}_{\vec\lambda} {\cal N}(\vec f)](\vec x))\,d\vec x, %+\int_\Omega ([{\cal M}_{\vec\lambda} {\cal N}(\vec f)](\vec x))\,d\vec x , \end{equation} a function that may be viewed as a penalized Kullback-Leibler divergence %, up to an additive constant depending only on $g$, between $g$ and the mixed, nonlinearly smoothed $\vec f$. This is due to the fact that this divergence may be expressed as \[ \int_\Omega g(\vec x)\log \left( \frac{g(\vec x)} {[{\cal M}_{\vec\lambda} {\cal N}(\vec f)](\vec x)} \right) \,d\vec x +\int_\Omega \left( [{\cal M}_{\vec\lambda} {\cal N}(\vec f)](\vec x) - g(\vec x) \right)\,d\vec x, \] where the second integral is negative due to Inequality~\ref{ELineq1} and may be viewed as the penalization term. {\color{red}{\em from Dave: I wonder whether this interpretation of the $L_\infty$ function makes any sense. What types of functions $f$ will be more penalized, and what types will be less penalized?}} {\bf I believe this means that the difference between the target density $g(\vec x)$ and a sum of reweighted smoothed density components is being penalized. Depending on the choice of bandwidth $h,$ the second integral term in the above can be more or less smooth and, therefore, the difference between $g(x)$ and the reweighted ``mixture" will be smaller or larger, respectively. Since it is impossible to fit "spikes" for individual components in the "continuous" case, this argument seems to work well here} We may also view $-L_\infty$ as a smoothed log-likelihood function corresponding to an infinitely large sample from $g$. Our general aim will be to minimize $L_\infty$ as a function of $\vec \lambda$ and $\vec f$. Starting with parameters $\vec\lambda^0$ and $\vec f^0$, Levine et al (2011, section 3) show that the iterative algorithm \begin{eqnarray}\label{algorithm} f^{p+1}_j(\vec u) &\equiv& \left[\int_\Omega \frac{g(\vec x)}{{\cal M}_{\vec\lambda^p}{\cal N} \vec f^p(\vec x)}\lambda_j^p{\cal N}f^p_{j}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right] \end{eqnarray} for $j=1, \ldots, m$ and $p=0, 1, \ldots$ possesses the descent property \begin{equation*} L_\infty(\vec\lambda^p, \vec f^{p+1}) \le L_\infty(\vec\lambda^p, \vec f^p). \end{equation*} {\color{red}{\em from Dave: Equation~(\ref{algorithm}) probably needs to be modified because the $f_j^{p+1}$ must integrate to one. Also, even after normalizing, $\vec f^{p+1}$ need not be contained in $\cal{C}$. I'm not sure how to get around this problem.}} Furthermore, additional development in Levine et al (2011, section 3) shows that if we update the $\vec\lambda$ parameter as \begin{eqnarray} \lambda_j^{p+1} &=& \left[\int_\Omega \frac{g(\vec x)}{{\cal M}_{\vec\lambda^p}{\cal N}\vec f^{p+1}(\vec x)}\lambda_j^p{\cal N}f^{p+1}_{j}(\vec x)\,d\vec x\right], \label{algorithm2} \end{eqnarray} we obtain \begin{equation}\label{descent} L_\infty(\vec\lambda^{p+1}, \vec f^{p+1}) \le L_\infty(\vec\lambda^p, \vec f^{p+1}) \le L_\infty(\vec\lambda^p, \vec f^p). \end{equation} The fact that Kullback-Leibler divergence between two distributions is zero if and only if the distributions coincide implies that equality in equation~(\ref{descent}) may only be attained if $(\vec\lambda^p, \vec f^p) =(\vec\lambda^{p+1}, \vec f^{p+1})$. %{\bf What about $\vec \lambda_{p}=\vec \lambda_{p+1}$} So if we assume the existence of $(\vec\lambda^S, \vec f^S)$, a global minimizer of $L_\infty$, then %letting $(\vec\lambda^p, \vec f^p)=(\vec\lambda^S, \vec f^S)$ %implies equality in~(\ref{descent}), it must be the case that \begin{equation}\label{fixedpt} f^S_j(\vec u) \equiv \left[\int_\Omega \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^S(\vec x)}\lambda_j^S{\cal N}f^S_{j}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right]\!. \end{equation} We shall use this fact in the proof of Theorem~\ref{prop2}. More specifically, we will define the constant $\alpha_j^S$ so that \begin{equation}\label{alpha} \alpha_j^S f^S_j(\vec u) = \left[\int_\Omega \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^S(\vec x)}\lambda_j^S{\cal N}f^S_{j}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right]\!. \end{equation} Integrating with respect to $\vec u$ in Equation~(\ref{alpha}), then summing over $j$, we find that $\sum_{j=1}^m \alpha_j^S = 1$. %{\it We still have not proved that $(\vec\lambda^S, \vec f^S)$ exists. %For Theorem~\ref{prop2}, let us merely assume that it does.} \section{Analogue of EL95 Result} %With Theorem~\ref{prop1} proved, we now examine how this theorem, which %describes the behavior of the function $L_\infty(\vec f)$ in the vicinity of %its minimizer $\vec f_S$, may be used. The idea of Eggermont and LaRiccia (1995) is to establish that \[ \| \vec f^* - \vec f_s \|_{L_1} \to 0 \quad\mbox{as $h\to0$}, \] where $\vec f^*$ is the true density, and then that $\vec f^n \to \vec f_S$. This establishes that $\vec f^n \to \vec f^*$ as desired. Rates of convergence are also possible. The first step seems to be the following theorem: %However, in our case, we must treat the whole parameter $(\vec\lambda, \vec f)$ %instead of merely the density $\vec f$. Perhaps it is possible to prove an inequality similar %to~(\ref{ineq}). Let us first re-define $L_\infty$, as follows: %\begin{equation}\label{loglik_infty} %L_{\infty}(\vec\lambda, \vec f) \defn -\int g(\vec x)\log ([{\cal M}_{\vec\lambda} {\cal N}(\vec f)](\vec x))\,d\vec x. %\end{equation} %Also, note as shown in Levine et al (2011) %that $f_{S}$ is the fixed point of the iterative algorithm %\begin{equation}\label{EM} %f^{p+1}_j(\vec u) \propto %\left[\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N}\vec f^p(\vec x)}*{\cal N}f^p_{j}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right]; %\end{equation} %let us therefore define $\alpha_j$ to be the normalizing constant that ensures %\begin{equation}\label{fixedpt} %f^S_j(\vec u) = \alpha_j %\left[\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N}\vec f^S(\vec x)}*{\cal N}f^S_{j}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right]. %\end{equation} \begin{theorem}\label{prop2} Assume that $(\vec\lambda_S, \vec f_S)$ can be defined as the unique (up to label-switching) minimizer of $L_\infty(\vec\lambda, \vec f)$. %For all $(\vec\lambda, \vec f)$, l Let $r_S(\vec x) = g(\vec x)/{\cal M}_{\vec\lambda_S}{\cal N}f_{S}(\vec x)$. Then \begin{eqnarray}\label{newineq} L_\infty(\vec\lambda, \vec f) - L_\infty(\vec\lambda_S, \vec f_S) &\ge & \frac12 \left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- {\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ && +\frac{k^{2}}{8}\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_{j}^{S}}{\lambda_{j}^{S}} \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)} + \sum_{j=1}^m \alpha_j^S \left( 1 - \frac{\lambda_j }{\lambda_j^S} \right).\nonumber \end{eqnarray} \end{theorem} {\bf Proof} As a first step, \begin{eqnarray*} &&\left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- {\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ &=&\left( \int_{\Sigma} \sqrt{r_{S}(\vec x)}\sqrt{ \left [ ({\cal M}_{\vec\lambda}{\cal N}\vec f- {\cal M}_{\vec\lambda_S} {\cal N}\vec f_{S})(\vec x) \right ]^{2}}\,d\vec x\right)^{2}\\ &\le & \Bigg[ \int_{\Sigma} \sqrt{r_{S}(\vec x)}\sqrt{\left( \frac{4}{3}[{\cal M}_{\vec\lambda}{\cal N}f](\vec x) +\frac{2}{3}[{\cal M}_{\vec\lambda_S}{\cal N}f_{S}](\vec x)\right)} \\ &&\times \sqrt{\left({\cal M}_{\vec\lambda_S}{\cal N}f_{S}(\vec x) \log \frac{ [{\cal M}_{\vec\lambda_S}{\cal N}f_{S}](\vec x)}{[{\cal M}_{\vec\lambda} {\cal N}f](\vec x)}+{\cal M}_{\vec\lambda}{\cal N}f(\vec x) - {\cal M}_{\vec\lambda_S}{\cal N}f_{S}(\vec x)\right)}\,d\vec x \Bigg ]^{2} \end{eqnarray*} using the following inequality of Kemperman (1967), which is also used by Eggermont and LaRiccia (1995), though the latter state it incorrectly: \[ (u-v)^{2}\le \left(\frac{2}{3}u+\frac{4}{3}v\right)\left(u\log \frac{u}{v}+v-u\right). \] Continuing, the Cauchy-Schwartz inequality together with the fact that ${\cal N}\phi(\vec x) \le {\cal S}\phi(\vec x)$ allows us to bound the above expression above by \begin{eqnarray*} && \int \left( \frac{2}{3}[{\cal M}_{\vec\lambda}{\cal S}f](\vec x) +\frac{4}{3}[{\cal M}_{\vec\lambda_S}{\cal S}f_{S}](\vec x)\right)\, d\vec x \\ && \times \int \left[g(\vec x)\log \frac{[{\cal M}_{\vec\lambda_S}{\cal N}f_{S}](\vec x)}{[{\cal M}_{\vec\lambda}{\cal N}f](\vec x)}+r_{S}(\vec x)\left[{\cal M}_{\vec\lambda}{\cal N}f(\vec x) - {\cal M}_{\vec\lambda_S}{\cal N}f_{S}(\vec x)\right]\right]\,d\vec x \\ &= & 2 \int \left[g(\vec x)\log \frac{[{\cal M}_{\vec\lambda_S}{\cal N}f_{S}](\vec x)}{[{\cal M}_{\vec\lambda}{\cal N}f](\vec x)}+ r_{S}(\vec x){\cal M}_{\vec\lambda}{\cal N}f(\vec x) - g(\vec x) \right]\,d\vec x \\ & = &2 \left[ L_{\infty}(\vec\lambda, \vec f)-L_{\infty}(\vec\lambda_S, \vec f_{S}) \right] +2\int r_{S}(\vec x){\cal M}_{\vec\lambda}{\cal N}f(\vec x) \,d\vec x - 2, \end{eqnarray*} where the penultimate equality uses Fubini's theorem and the last uses the definition of $L_{\infty}$. We continue by writing \begin{eqnarray*} \int r_{S}(\vec x){\cal M}_{\vec\lambda}{\cal N}\vec f(\vec x) &=&\sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N}\vec f^{S}(\vec x)}{\cal N} f_{j}^{S}(\vec x)\left[{\cal N}\left(\frac{f_{j}(\vec x)}{f_{j}^{S}(\vec x)}\right) \right]\,d\vec x \\ &=&\sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N}\vec f^{S}(\vec x)}{\cal N} f_{j}^{S}(\vec x)\left[{\cal N}\left(\frac{f_{j}(\vec x)}{f_{j}^{S}(\vec x)}\right)-S\left(\frac{f_{j}(\vec x)}{f_{j}^{S}(\vec x)}\right)\right]\,d\vec x\\ && + \sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N}\vec f^{S}(\vec x)}{\cal N} f_{j}^{S}(\vec x)S\left(\frac{f_{j}(\vec x)}{f_{j}^{S}(\vec x)}\right)\,d\vec x. \end{eqnarray*} The last term above can be rewritten using Fubini's theorem as \begin{equation} \int\, d\vec u \sum_{j=1}^{m}\lambda_{j}\frac{f_{j}(\vec u)}{f_{j}^{S}(\vec u)}\left[\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^{S}(\vec x)}{\cal N}f_{j}^{S}(\vec x)K_{h}(\vec x-\vec u)\,d\vec x\right], \end{equation} which equals \[ \sum_{j=1}^m \frac{\lambda_j\alpha_j^S}{\lambda_j^S } \] by equation~(\ref{alpha}). Finally, we consider the term \begin{equation}\label{lastterm} \sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N} \vec f^{S}(\vec x)}{\cal N} f_{j}^{S}(\vec x)\left[{\cal N} \left(\frac{f_{j}(\vec x)}{f_{j}^{S}(\vec x)}\right)-S\left(\frac{f_{j}(\vec x)} {f_{j}^{S}(\vec x)}\right)\right]\,d\vec x. \end{equation} Following Eggermont and LaRiccia (1995), denote $\phi_j=\sqrt{f_j/f_j^S}$. Then, since $N(\psi)\le \{S(\sqrt{\psi})\}^{2}$ for any $\psi$, the term in square brackets in~(\ref{lastterm}) is bounded above by $(S\phi_{j})^{2} - S(\phi_{j}^{2})$. Putting all of this together, we obtain \begin{eqnarray*} &&\frac12 \left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- {\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ & \le & L_\infty(\vec\lambda, \vec f) - L_\infty(\vec\lambda_S, \vec f_S) - 1 + \sum_{j=1}^m \frac{\lambda_j \alpha_j^S}{\lambda_j^S } \\ &-& \sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda_S}{\cal N} \vec f^{S}(\vec x)}{\cal N} f_{j}^{S}(\vec x) \left ( [S(\phi^{2}_{j})](\vec x)-\{[S\phi_{j} ](\vec x)\}^{2} \, \right ) d\vec x. \end{eqnarray*} In order to continue, one needs to use the Lagrange Identity: \[ [S(\phi^{2}_{j})](\vec x)-\{[S\phi_{j}](\vec x)\}^{2}=\int s_{h}(\vec x,\vec y)s_{h}(\vec x,\vec z)(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\,d\vec y\,d\vec z \] Due to the Lagrange identity, we now have \begin{eqnarray*} &&\sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^{S}(\vec x)}*{\cal N} f_{j}^{S}(\vec x) \left\{S(\phi_{j}^{2})-(S\phi_{j})^{2}\right\}\,d\vec x\\ &=&\sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^{S}(\vec x)}{\cal N}f_{j}^{S}(\vec x) \left[\int s_{h}(\vec x,\vec y)s_{h}(\vec x,\vec z)(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\,d\vec y\,d\vec z\right]\,d\vec x\\ &=&\sum_{j=1}^{m}\lambda_{j}\int_{\Omega\times \Omega}(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\left[\int_{\Omega}\frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^{S}(\vec x)}{\cal N}f_{j}^{S}(\vec x)s_{h}(\vec x,\vec y)s_{h}(\vec x,\vec z)\,d\vec x\right] d\vec y d\vec z\\ &\ge & \sum_{j=1}^{m}\lambda_{j}\int_{\Omega \times \Omega}(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\frac{k^{2}}{\vert \Omega \vert}\left[\int_{\Omega}\frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}f^{S}(\vec x)}{\cal N}f_{j}^{S}(\vec x)s_{h}(\vec x,\vec y)\,d\vec x\right]d\vec y d\vec z\\ &=&\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_j^S}{\lambda_j^S}\int_{\Omega\times\Omega}(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\frac{k^{2}}{\vert \Omega\vert}f_{j}^{S}(\vec y)\,d{\vec y}\, d\vec z, \end{eqnarray*} where the last step follows from~(\ref{fixedpt}) as before. Replacing the outermost integral (in $\vec z$) with a minimum over all possible vectors $\mu \in R^{r},$ one obtains \begin{eqnarray*} &&\int_{\Omega\times\Omega}(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\frac{k^{2}}{\vert \Omega\vert}f_{j}^{S}(\vec y)\,d{\vec y}\, d\vec z \\ %&\ge &\frac{k^{2}}{\vert \Omega\vert} \int_{\Omega}\min_{\vec z} \int_{\Omega} (\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}f_{j}^{S}(\vec y)\,d{\vec y}\, d\vec z\\ &\ge & k^{2} \min_{\vec \mu} \int_{\Omega} (\phi_{j}(\vec y)-\mu)^{2}f_{j}^{S}(\vec y)\,d{\vec y}\\ &=& k^{2}\int_{\Omega} (\phi_{j}(\vec y)-\nu)^{2}f_{j}^{S}(\vec y)\,d{\vec y}, \end{eqnarray*} where the minimum is clearly achieved when $\mu\equiv \nu= \int_{\Omega}\phi_{j}(\vec y)f_{j}^{S}(\vec y)\,d\vec y=\int_{\Omega} \sqrt{f_{j}(\vec y)}\sqrt{f_{j}^{S}(\vec y)}\,d\vec y.$ Plugging in the true values of $\phi_{j}={\sqrt{f_{j}}}/{\sqrt{f_{j}^{S}}}$ and $ \nu$, we find that \begin{eqnarray*} &&\int_{\Omega\times\Omega}(\phi_{j}(\vec y)-\phi_{j}(\vec z))^{2}\frac{k^{2}}{\vert \Omega\vert}f_{j}^{S}(\vec y)\,d{\vec y}\, d\vec z \\ &\ge &k^{2}\int_{\Omega}\left(\sqrt{f_{j}(\vec y)}-\nu \sqrt{f_{j}^{S}(\vec y)}\right)^{2}\,d\vec y=k^{2}%\sum_{j=1}^{m}\lambda_{j} \left\|\sqrt{f_{j}}-\nu\sqrt {f_{j}^{S}}\right\|^{2}_{L^{2}(\Omega)} \end{eqnarray*} Due to the above, $\nu \sqrt{f_{j}^{S}}$ can be viewed as the Euclidean projection (in the space $L_{2}(\Omega)$) of $\sqrt{f_{j}}$ onto the linear space spanned by $\sqrt{f_{j}^{S}};$ such a space is, essentially, a straight line in $L_{2}(\Omega).$ Therefore, on one hand, \begin{equation*} \left\| \sqrt{f_{j}}-\nu\sqrt {f_{j}^{S}} \right\| ^{2}_{L^{2}(\Omega)}\le \left\| \sqrt{f_{j}}-\sqrt {f_{j}^{S}}\right \|^{2}_{L^{2}(\Omega)} \end{equation*} for any $j=1,\ldots,m$. On the other hand, the constant $0< \nu=\int_{\Omega}\sqrt{f_{j}f_{j}^{S}}<1$ due to Cauchy-Schwarz inequality and the fact that both $f_{j}$ and $f_{j}^{S}$ are true densities. Therefore, \[ \left \| \sqrt{f_{j}}-\nu \sqrt {f_{j}^{S}} \right\|^{2}_{L^{2}(\Omega)}=1-\nu^{2}\ge 1-\nu=\frac{1}{2} \left\|\sqrt f_{j}-\sqrt {f_{j}^{S}} \right \|^{2}_{L^{2}(\Omega)} \] and we can now say that \[ \frac{1}{2} \left\|\sqrt{f_{j}}-\sqrt {f_{j}^{S}} \right\|^{2}_{L^{2}(\Omega)}\le \left\|\sqrt{f_{j}}-\nu\sqrt {f_{j}^{S}} \right\|^{2}_{L^{2}(\Omega)}. \] This, in turn, means that \begin{eqnarray*} &&\sum_{j=1}^{m}\lambda_{j}\int \frac{g(\vec x)}{{\cal M}_{\vec\lambda^S}{\cal N}\vec f^{S}(\vec x)}*{\cal N} f_{j}^{S}(\vec x) \left\{S(\phi_{j}^{2})-(S\phi_{j})^{2}\right\}\,d\vec x\\ &\ge& \frac{k^{2}}{2}\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_j^S}{\lambda_j^S} \left\|\sqrt{f_{j}}-\sqrt{f_{j}^{S}} \right\|^{2}_{L^{2}(\Omega)}\\ &\ge & \frac{k^{2}}{8}\sum_{j=1}^{m} \frac{\lambda_{j}\alpha_j^S}{\lambda_j^S} \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)}; \end{eqnarray*} the last inequality follows from the obvious fact that \[ \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)}\le 4 \left\|\sqrt {f_{j}}-\sqrt{f_{j}^{S}} \right\|^{2}_{L^{2}(\Omega)}. \] We thus obtain %\begin{eqnarray*} %&&\frac12 \left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- %{\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ %& \le & %L_\infty(\vec\lambda, \vec f) - L_\infty(\vec\lambda_S, \vec f_S)-1 %+\sum_{j=1}^m \frac{\lambda_j \alpha_j^S}{\lambda_j^S } %\left\{ 1 - \frac{k^{2}}{8} \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)} \right \}. %\end{eqnarray*} %{\bf It is more instructive to rewrite the above inequality as \begin{eqnarray*} L_\infty(\vec\lambda, \vec f) - L_\infty(\vec\lambda_S, \vec f_S) &\ge & \frac12 \left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- {\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ && +\frac{k^{2}}{8}\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_{j}^{S}}{\lambda_{j}^{S}} \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)} + \sum_{j=1}^m \alpha_j^S \left( 1 - \frac{\lambda_j }{\lambda_j^S} \right). \end{eqnarray*} %{\it %I believe I have followed all of the algebra correctly. At this point, note that %if $\vec\lambda=\vec\lambda^s$, things simplify considerably: %\begin{eqnarray*} %&&\frac12 \left( \int_{\Sigma} \sqrt{r_{S}(\vec x)} \left| ({\cal M}_{\vec\lambda}{\cal N}\vec f- %{\cal M}_{\vec\lambda_S}{\cal N}\vec f_{S})(\vec x) \right |\,d\vec x\right)^{2} \\ %& \le & %L_\infty(\vec\lambda, \vec f) - L_\infty(\vec\lambda_S, \vec f_S) %-\frac{k^2}{8}\sum_{j=1}^m %\alpha_j^S \left\|f_{j}-f_{j}^{S} \right\|^{2}_{L^{1}(\Omega)}, %\end{eqnarray*} %where once again we must remember that $\sum_j\alpha_j^S=1$. %Given this fact, it should be possible to express the final inequality using %some sort of bound on the distance between $\vec\lambda$ and $\vec\lambda^S$. %} \section{Convergence of $\vec f_{n}$ to $\vec f_{S}$} Consider two different sets of parameters $\theta_{1}=(\vec \lambda^{1},\vec f^{1})$ and $\theta_{2}=(\vec \lambda^{2},\vec f^{2}).$ Recall that $L_{n}(\vec \lambda, \vec f)=-\sum_{i=1}^{n}\log\{({\cal M_{\vec \lambda}}{\cal N}\vec f )(\vec x_{i})\}.$ Let $(\vec \lambda_{n},\vec f_{n})$ be a minimizer of $L_{n}(\vec \lambda,\vec f).$ Define $L_{n}(\vec \lambda_{1},\vec f^{1},\vec \lambda^{2},\vec f^{2})=L_{n}(\vec \lambda^{1}, \vec f^{1})-L_{n}(\vec \lambda^{2},\vec f^{2})$ and likewise for $L_{\infty}(\vec \lambda_{1},\vec f^{1},\vec \lambda^{2},\vec f^{2}).$ More explicitly, \begin{equation}\label{4argfunc1} L_{n}(\vec \lambda^{1},\vec f^{1},\vec \lambda^{2},\vec f^{2})=\sum_{k=1}^{n}\log{ \frac{\sum_{j=1}^{m}\lambda^{1}_{j}{\cal N}\vec f^{1}_{j}(\vec x_{k})}{\sum_{j=1}^{m}\lambda^{2}_{j}{\cal N}\vec f^{2}_{j}(\vec x_{k})}} \end{equation} and \begin{equation}\label{4argfunc2} L_{\infty}(\vec \lambda_{1},\vec f_{1},\vec \lambda_{2},\vec f_{2})= \int g(\vec x)\log{ \frac{\sum_{j=1}^{m}\lambda^{1}_{j}{\cal N}\vec f^{1}_{j}(\vec x)}{\sum_{j=1}^{m}\lambda^{2}_{j}{\cal N}\vec f^{2}_{j}(\vec x)}\,d\vec x}. \end{equation} Therefore, the difference \begin{equation} L_{n}(\vec \lambda_{1},\vec f_{1},\vec \lambda_{2},\vec f_{2})-L_{\infty}(\vec \lambda_{1},\vec f_{1},\vec \lambda_{2},\vec f_{2})= \int_{\Sigma} \Phi(\vec x)[dG_{n}(\vec x)-dG(\vec x)] \end{equation} where \[ \Phi(\vec x)=\log{ \frac{\sum_{j=1}^{m}\lambda^{1}_{j}{\cal N}\vec f^{1}_{j}(\vec x)}{\sum_{j=1}^{m}\lambda^{2}_{j}{\cal N}\vec f^{2}_{j}(\vec x)}} \] $G_{n}(\vec x)$ is an empirical sampling distribution function and $G(\vec x)$ is the distribution function corresponding to the target density $g(\vec x).$ The reason we are concerned with this is because the behavior of the functional $L_{\infty}$ in the vicinity of its minimizer can be described as \begin{align}\label{diff} &L_{\infty}(\vec \lambda_{n},\vec f_{n})-L_{\infty}(\vec \lambda_{S},\vec f_{S})=-L_{\infty}(\vec \lambda_{S},\vec f_{S},\vec \lambda_{n},\vec f_{n})\\\nonumber &\le L_{n}(\vec \lambda_{S},\vec f_{S},\vec \lambda_{n},\vec f_{n})-L_{\infty}(\vec \lambda_{S},\vec f_{S},\vec \lambda_{n},\vec f_{n})\\\nonumber &=\int_{\Sigma} \Phi(\vec x)[dG_{n}(\vec x)-dG(\vec x)] \end{align} using definitions \eqref{4argfunc1} and \eqref{4argfunc2}. In the above, \[ \Phi(\vec x)=\log{ \frac{\sum_{j=1}^{m}\lambda^{S}_{j}{\cal N}\vec f^{S}_{j}(\vec x)}{\sum_{j=1}^{m}\lambda^{n}_{j}{\cal N}\vec f_{j}^{n}(\vec x)}}. \] \hrule {\em New (May 11, 2012):} Combining Theorem~1 with inequality~(\ref{diff}) gives \begin{eqnarray*} && K\sum_{j=1}^m \frac{\alpha^S_j\lambda^n_j}{\lambda^S_j} \| f_j^n - f_j^S \|_1 + \frac12 \left( \int \sqrt{r_S(\vec x)} \left| {\cal M}_{\lambda^n} {\cal N} \vec f^n (\vec x) - {\cal M}_{\lambda^S} {\cal N} \vec f^S (\vec x) \right| \, d\vec x \right)^2 \\ &\le & L_\infty(\vec\lambda^n, \vec f^n) - L_\infty(\vec\lambda^S, \vec f^S) + \frac{\max_j \alpha_j^S}{\min_j \lambda_j^S} \| \vec\lambda^n - \vec\lambda^S \|_1 \\ &\le & \int \log \frac{ {\cal M}_{\lambda^S} {\cal N} \vec f^S (\vec x) } {{\cal M}_{\lambda^n} {\cal N} \vec f^n (\vec x)} \left[ dG_n (\vec x) - dG (\vec x) \right] + \frac{\max_j \alpha_j^S}{\min_j \lambda_j^S} \| \vec\lambda^n - \vec\lambda^S \|_1 \\ &\le& \left( 2+ \frac{\max_j \alpha_j^S}{\min_j \lambda_j^S} \right) \| \vec\lambda^n - \vec\lambda^S \|_1 + \int | {\cal N}\vec f^n(\vec x) - {\cal N} \vec f^S(\vec x) | \left[ dG_n(\vec x) - dG(\vec x) \right] \end{eqnarray*} as long as it is possible to show that \[ | \Phi(\vec x) | \le \| \vec\lambda^n - \vec\lambda^S \|_1 + | {\cal N} \vec f^n (\vec x) - {\cal N} \vec f^S (\vec x) |. \] {\em Michael, I believe this latter fact is what you have shown, correct?} \hrule Our next step will be to analyze the function $\Phi(\vec x)$ in detail; note that the $\sum_{j=1}^{m}\lambda^{S}_{j}{\cal N}\vec f^{S}_{j}(\vec x)$ can be thought of as a trace of the positive definite diagonal matrix with a typical element $\lambda^{S}_{j}{\cal N}\vec f^{S}_{j}(\vec x)$ and the same is true of the denominator in the definition of $\Phi(\vec x).$ Since for any two positive definite square matrices $\frac{ tr(B)}{tr (A)} \le tr (A^{-1}B)$ (see, e.g. DasGupta (2008)) we have \begin{align} &\Phi(\vec x) \le \log \left( \sum_{j=1}^{m}\frac{\lambda^{S}_{j}}{\lambda^{n}_{j}}\exp\left(\int_{\Omega}K_{h}(\vec x -\vec u)[\log f^{S}_{j}(\vec x)-\log f^{n}_{j}(\vec x)]\right)\right)\\ &= \log \left( \sum_{j=1}^{m}\frac{\lambda^{S}_{j}}{\lambda^{n}_{j}}{\cal N}\left\{\frac{f_{j}^{S}(\vec x)}{f_{j}^{n}(\vec x)}\right\}\right)\equiv \Psi(\vec x) \end{align} %First, using Jensen's inequality one can show that $Nf_{S}^{j}(\vec x)=\exp(\int K_{h}(\vec x-\vec u)\log f_{S}^{j}(\vec u)\,d\vec u\le \int %K_{h}(\vec x-\vec u)f_{S}^{j}(\vec u)\,d\vec u\le \max_{\vec x\in \Omega} K_{h}(\vec x)=K_{1}.$ This allows us to conclude that %$\sum_{j=1}^{m}\lambda^{j}_{S}{\cal N}f^{j}_{S}\le K_{1}.$ %Note, first, that it can be bounded from the above as %\begin{align}\label{phi} %\Phi(\vec x)&\le \log\left\{K*\left(\sum_{j=1}^{m}[\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec %x)]\right)+1\right\}\\ %&=K*\left(\sum_{j=1}^{m}[\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec x)]\right)+o()\equiv \Psi(\vec x) %\end{align} %for some positive constant $K.$ %On the other hand, for any density function $f_{j}\in B$ (see Levine, Hunter and Chauveau (2011)), $f_{j}\ge \inf K_{h}(\vec x)\ge K_{2};$ therefore, %${\cal N}f_{j}^{S}(\vec x)\ge \exp[\int K_{h}(\vec x-\vec u)\log K_{2}\,d\vec u]\ge K_{2}>0.$ In the limit, we can also say that ${\cal N}f_{j}^{S}$ %is bounded away from zero as well. Since we assume that the weights $\lambda_{j}^{S}$ are bounded away from zero, we have that %$\sum_{j=1}^{m}\lambda_{j}^{S}{\cal N}f_{j}^{S}$ is bounded away from zero; thus, we can bound $\Phi(\vec x)$ from above regardless of the sign of %the denominator. Thus, \begin{align}\label{int_by_parts} &L_{n}(\vec \lambda_{1},\vec f_{1},\vec \lambda_{2},\vec f_{2})-L_{\infty}(\vec \lambda_{1},\vec f_{1},\vec \lambda_{2},\vec f_{2})\le \int_{\Sigma}\Psi(\vec x)[dG_{n}(\vec x)-dG(\vec x)]\nonumber\\ &=(-1)^{r}\int_{\Sigma}D\Psi(\vec x)[G_{n}(\vec x)-G(\vec x)]\,d\vec x \end{align} where $D\Psi(\vec x)=\frac{{\partial }^{r}}{{\partial }x_{1}{\partial }x_{2}\ldots {\partial }x^{r}}\Psi(\vec x)$ with $r$ being the dimensionality of the space $\Sigma.$ The reason for using integration by parts in \eqref{int_by_parts} is that it is easy to characterize the rate of convergence of $G_{n}(\vec x)$ to $G(\vec x)$ using the empirical process theory. Indeed, due to a well known estimate of Shorack and Wellner, $||G_{n}(\vec x)-G(\vec x)||_{L_{1}(\Sigma)} =O\left( \sqrt{\frac{\log\log n}{n}}\right)\equiv LL(n).$ Therefore, it seems advantageous to present the right-hand side of \eqref{diff} in the form \eqref{int_by_parts}. Now, note that differentiation of ${\cal N}\left\{\frac{f_{j}^{S}(\vec x)}{f_{j}^{n}(\vec x)}\right\}$ w.r.t. any coordinate $x_{k},$ $k=1,\ldots,r$ produces \[ \frac{\partial }{\partial x^{k}}{\cal N}\left\{\frac{f_{j}^{S}(\vec x)}{f_{j}^{n}(\vec x)}\right\}= {\cal N}\left\{\frac{f_{j}^{S}(\vec x)}{f_{j}^{n}(\vec x)}\right\}\int \frac{\partial }{\partial x^{k}}K_{h}(\vec x-\vec u)[\log f_{j}^{S}(\vec u)-\log f_{j}^{n}(\vec u)]\, d\vec u. \] If we assume that the $k$th partial derivative of the kernel $K(\cdot)$ is continuous on the compact set $\Omega$, it is easy to see that the above does not exceed $C_{1}{\cal N}\left\{\frac{f_{j}^{S}(\vec x)}{f_{j}^{n}(\vec x)}\right\}$ for some positive constant $C_{1}.$ ********************To be continued*********** %One can easily show that the same is true for the $r$ th order derivative of ${\cal N}f_{j}^{n}(\vec x)$ as well. On the other hand, given that $\log f_{S}$ is integrable on the compact set $\Omega,$ one can also show that the $r$th order partial derivative of ${\cal N}f_{j}^{S}(\vec x)$ is bounded from below. Therefore, there exists a constant $C$ that does not depend on $n$ and $h$ such that \[ D\Psi(\vec x) \le C[\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec x); \] thus, \[ L_{\infty}(\vec \lambda_{n},\vec f_{n})-L_{\infty}(\vec \lambda_{S},\vec f_{S})=C\int [\sum_{j=1}^{m}(\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x))]\,d\vec x*LL(n) \] Now, define a set $\Omega_{j}^{+}=\{\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)\ge \lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec x)\}$ and $\Omega^{-}_{j}=\Omega\setminus\Omega_{j}^{+}.$ Then, \begin{align*} &\int_{\Omega_{j}^{+}}[\sum_{j=1}^{m}(\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x))]\\ &=\int_{\Omega_{j}^{+}}[\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{n}Sf_{j}^{n}(\vec x)]\,d\vec x+\int_{\Omega_{j}^{+}}\lambda_{j}^{n}S[f_{j}^{n}(\vec x)-f_{j}^{S}(\vec x)]\,d\vec x+\int_{\Omega_{j}^{+}} [\lambda_{j}^{n}Sf_{j}^{n}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}\vec (x)]\,d\vec x \end{align*} The first integral out of three above is negative since ${\cal N}f_{j}^{n}(\vec x)\le Sf_{j}^{n}(\vec x);$ for the same reason the last integral is positive. If we define \[ \Lambda f_{j}^{S}=h^{-2}\int_{\Omega}[\lambda_{j}^{n}Sf_{j}^{S}-\lambda_{j}^{S}{\cal N}f_{j}^{S}](\vec x)\,d\vec x \] as a linear operator with $\sup \Lambda f_{j}^{S}\le M$ for some nonnegative constant $M,$ we have \[ \int_{\Omega_{j}^{+}}[\sum_{j=1}^{m}(\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x))]\le ||S( \vec f^{n}-\vec f_{S})||_{L_{1}(\Omega)}+h^{2}M \] In exactly the same way, one can show that \begin{align*} &\int_{\Omega_{j}^{+}}[\sum_{j=1}^{m}(\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x))]\\ &=||S(\vec f^{n}-\vec f_{S})||_{L_{1}(\Omega)}+h^{2}M+\int \sqrt{r_{S}(\vec x)}[\lambda_{j}^{n}{\cal N}f_{j}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec x)]\,d\vec x \end{align*} Thus, we can conclude that \begin{align*} &L_{\infty}(\vec \lambda_{n},\vec f_{n})-L_{\infty}(\vec \lambda_{S},\vec f_{S})\le C*LL(n)\sum_{j=1}^{m}||f_{j}^{n}-f_{j}^{S}||_{L^{1}(\Omega)}+Mh^{2}\\ &+\int \sqrt{r_{S}}(\vec x)\{\lambda_{j}^{n}{\cal N}f_{j}^{n}(\vec x)-\lambda_{j}^{s}{\cal N}f_{j}^{S}(\vec x)\}\,d\vec x \end{align*} Let $E\equiv \int \sqrt{r_{S}(\vec x)}\sum_{j=1}^{m}[\lambda_{j}^{n}{\cal N}f_{j}(\vec x)-\lambda_{j}^{S}{\cal N}f_{j}^{S}(\vec x)]\,d\vec x$ and $e_{j}=||f_{j}^{n}-f_{j}^{S}||_{L_{1}(\Omega)}$for brevity. Finally, using inequality \eqref{newineq}, one obtains that \begin{align*} &\frac{1}{2}E^{2}+1-\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_{j}^{S}}{\lambda_{j}^{S}}+\frac{k^{2}}{8}\sum_{j=1}^{m}\frac{\lambda_{j}\alpha_{j}^{S}}{\lambda_{j}^{S}}e_{j}^{2}\\ &\le C*LL(n)\sum_{j=1}^{m}[e_{j}+Mh^{2}+E] \end{align*} The terms with $E$ can be completed to a full square on the left hand side and the resulting full square dropped due to its positivity, thus obtaining an expression for the rate of convergence for $f_{j}^{n}$ to $f_{j}^{S}$ if something preliminary is known about convergence of Euclidean parameters. \end{document}