text:
\begin{document}

\title[Proof of the HRT conjecture for (2,2) configurations]{Proof of the HRT conjecture for (2,2) configurations}

\author{Ciprian Demeter}
\address{Department of Mathematics, Indiana University, 831 East 3rd St., Bloomington IN 47405}
\email{demeterc@@indiana.edu}

\author{Alexandru Zaharescu}
\address{Department of Mathematics, University of Illinois at Urbana-Champaign, 1409 W. Green Street, Urbana, Illinois 61801-2975}
\email{zaharesc@@math.uiuc.edu}

\keywords{}
\thanks{The  first author is supported by a Sloan Research Fellowship and by NSF Grants DMS-0742740 and 0901208}
\thanks{The  second author is supported by  NSF Grant DMS-0901621}
\thanks{ AMS subject classification: Primary 26A99; Secondary 11K70, 65Q20}
\begin{abstract}

We prove that for any 4 points in a (2-2) configuration, there is no linear dependence between the associated time-frequency translates of any  $L^2(\R)$ function.
\end{abstract}
\maketitle

\section{Introduction}
The following conjecture, known as the HRT conjecture appears in \cite{HRT}. See also \cite{H} for an ample discussion on the subject.

\begin{conjecture}
\label{cc1}
Let $(t_j,\xi_j)_{j=1}^n$ be $n\ge 2$ distinct points in the plane. Then there is no nontrivial $L^2$ function  $f:\R\to \C$ satisfying a  nontrivial linear dependence
$$\sum_{j=1}^nd_if(x+t_j)e^{2\pi i\xi_jx}=0,$$
for a.e. $x\in \R$.
\end{conjecture}

The conjecture  follows trivially when the points $(t_j,\xi_j)_{j=1}^n$ are collinear. The  conjecture was proved  when  $(t_i,\xi_i)_{i=1}^n$ sit on a lattice, \cite{Lin}, using von Neumann algebras techniques. See also \cite{BS}, \cite{DG}, for more elementary alternative arguments. In particular, this is the case with any 3 points. But the question whether the conjecture holds for {\em arbitrary} 4 points is open. Progress on that has been made by the first author in \cite{De2} using a number theoretical approach, and we briefly discuss it below.

We will call an $(2,2)$ configuration, any collection of 4 distinct points in the plane, such that there exist 2 distinct parallel lines each of which containing 2 of the points. One of the results in \cite{De2} is

\begin{theorem}
\label{TT2'}
Conjecture \ref{cc1} holds  for special $(2,2)$ configurations $(0,0),(1,0), (0,\alpha), (1,\beta)$

(a) if
$$\liminf_{n\to\infty}n\log n\min\{\|n\frac{\beta}{\alpha}\|,\|n\frac{\alpha}{\beta}\|\}<\infty$$

(b)  if at least one of $\alpha,\beta$ is rational

In either case, no nontrivial solution $f$ can exist satisfying  minimal decay
$$\lim_{|n|\to\infty\atop{n\in\Z}}|f(x+n)|=0,\;\;a.e.\;x$$
\end{theorem}

In this paper we prove the strongest possible statement about (2,2) configurations, namely

\begin{theorem}
\label{TT2}
Conjecture \ref{cc1} holds  for all $(2,2)$ configurations.
Moreover, when the points sit in a special (2,2) configuration $(0,0),(1,0), (0,\alpha), (1,\beta)$, no nontrivial solution $f$ can exist satisfying  minimal decay
$$\lim_{|n|\to\infty\atop{n\in\Z}}|f(x+n)|=0,\;\;a.e.\;x$$
\end{theorem}

The general approach for proving this theorem is the one developed in \cite{De2}. We first reduce to the case of special configurations, by applying metaplectic transformations.   Then we  turn the hypothetical linear dependence into a recurrence. The contribution from $\beta$ is estimated by using the {\em conjugates trick}. The novelty of our approach here is in the way we treat the contribution coming from the terms containing $\alpha$. In particular, we exploit the Diophantine behavior of $\alpha$ at more than one scale.

\section{Proof of the main theorem}
Define $[x]$, $\{x\}$, $\|x\|$  to be the integer part, the  fractional part and the distance to the nearest integer of $x$.  For two quantities $A$, $B$ that vary, we will denote by $A\lesssim B$ or $A=O(B)$ the fact that $A\le C B$ for some universal constant $C$, independent of $A$ and $B$.  In general, $A\lesssim_p B$ means that the implicit constant is allowed to depend on the parameter $p$. The notation $A\sim_p B$ means that $A\lesssim_p B$ and $B\lesssim _p A$.
 If no parameter is specified, the implicit constants are implicitly understood to depend on the (harmless) fundamental parameters introduced in the proof of Theorem \ref{TT2}. For a set $A\subset \R$, we will denote by $|A|$ its Lebesgue measure, and if the set is finite, $|A|$ will represent its cardinality. Finally, we define
$e(x):=e^{2\pi i x}$.

Let $0<\alpha<1$ be irrational.
Let $\frac{p_k}{N_k}$ be the $k^{th}$ convergent of $\alpha$, so that
\begin{equation}
\label{ecc6}
|\alpha-\frac{p_k}{N_k}|\le \frac1{N_kN_{k+1}},
\end{equation}
and
\begin{equation}
\label{ecc4}
p_{k}N_{k-1}-p_{k-1}N_{k}=(-1)^{k-1}
\end{equation}
Since
$$N_{k}\le N_{k+1},$$there exists an infinite set $E\subset \N$ and a constant $D=D(\alpha)$ such that for each $k\in E$ we have
\begin{equation}
\label{ecc1}
\frac{N_k}{N_{k+1}}\le D  \min_{j\le k}\frac{N_j}{N_{j+1}}.
\end{equation}
Define $\frac1{M_k}:=N_k^2|\alpha-\frac{p_k}{N_k}|$. Of course, $M_k\ge 1$ for each $k$.

The following proposition is the main new ingredient in this paper.

\begin{proposition}
\label{P1}
Let $k\in E$ be odd, and $0<\delta<\frac{1}{100}$. Define $N:=N_k$, $p:=p_k$, $M:=M_k$. Then, for each $x\in [0,1]$ such that \begin{equation}
\label{ecc2}
\min\{\frac{\|x\|}{N},\|x-n\alpha\|,\|x-\frac{n}{N}\|:1\le n\le N\}\ge \frac{\delta}{N}\end{equation} we have
\begin{equation}
\label{ecc36}
\prod_{n=1}^{N}|{e(x)-e(\alpha n)}|\sim_\delta 1.
\end{equation}
\end{proposition}
\begin{remark}
The key thing in \eqref{ecc36} is that the similarity constant does not depend on $N$.
\end{remark}
\begin{proof}
Fix $x$ satisfying \eqref{ecc2}. We will compare $\prod_{n=1}^{N}|{e(x)-e(\alpha n)}|$ to $$\prod_{n=1}^{N}|{e(x)-e(\frac{np}{N})}|=\prod_{n=1}^{N}|{e(x)-e(\frac{n}{N})}|=|e(Nx)-1|\sim_\delta 1,$$
and prove that their ratio is $\sim_\delta 1$.
This is reasonable to expect, since, due to \eqref{ecc6}, we have for each $1\le n\le N$
\begin{equation}
\label{ecc7}
|n\alpha-\frac{np}{N}|\le \frac{1}{N}
\end{equation}
First, let $1\le n_1,n_2,\ldots, n_{200}\le N$ be such that
$$\|x-\frac{n_ip}{N}\|\le \frac{100}N$$
Due to \eqref{ecc2} and \eqref{ecc7}, we get that
\begin{equation}
\label{ecc9}
\delta^3\lesssim \prod_{i=1}^{200}\frac{|{e(x)-e(\alpha n_i)}|}{|{e(x)-e(\frac{n_ip}{N})}|}\lesssim \delta^{-1}.
\end{equation}
Next, we analyze
$$\prod_{n=1\atop{n\not=n_i}}^{N}\frac{|{e(x)-e(\alpha n)}|}{|{e(x)-e(\frac{np}{N})}|}.$$
Note that
$$\frac{|{e(x)-e(\alpha n)}|}{|{e(x)-e(\frac{np}{N})}|}=\left|1+\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}\right|,$$
and that
$$\left|\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}\right|\le \frac{10}{N\|x-\frac{np}{N}\|}<\frac12.$$
Thus,
$$\sum_{n=1\atop{n\not=n_i\atop{\|x-\frac{np}{N}\|\ge \delta}}}^N\frac1{N\|x-\frac{np}{N}\|}\lesssim \sum_{N\delta\le i\le N}\frac{1}{i}\lesssim \log (\delta^{-1}).$$
Using this and the fact that
$$1+x\le e^x,\;\;0<x<1$$
$$e^{-10x}\le 1-x,\;\;0<x<1/2,$$
we get
\begin{equation}
\label{ecc3}
\prod_{n=1\atop{n\not=n_i\atop{\|x-\frac{np}{N}\|\ge \delta}}}^N\left|1+\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}\right|\sim_\delta 1
\end{equation}

Denote by
$$A:=\{1\le n\le N:n\not=n_i,\;\;\|x-\frac{np}{N}\|<\delta\}$$
Using the fact that for $z\in\R$ with $|z|<\frac1{10}$
$$1/2\le\frac{|e(z)-1|}{2\pi|z|}<2,$$we get for each $n\in A$
$$\left|\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}\right|< \frac{\frac{10n}{N^2M}}{\frac{100}{N}}<\frac{1}{10}.$$
It is easy to check that for each $z\in \C$ with $|z|<\frac1{10}$ we have
$$e^{-O(|z|^2)}\le\left|\frac{1+z}{e^z}\right|\le e^{O(|z|^2)}.$$
Apply this inequality to each $z_n:=\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}$. We have seen that $|z_n|\lesssim \frac{1}{N\|x-\frac{np}{N}\|}$, and hence
$$\sum_{n\in A}|z_n|^2\lesssim 1.$$
It follows that
$$\prod_{n\in A}\left|1+\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}\right|\sim \left|e^{\sum_{n\in A}\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}}\right|.$$
Let $\alpha-\frac{p}{N}:=\frac{t}{N^2}$, so $M|t|=1$. Note that since $\|x\|\ge \delta$, it follows that
\begin{equation}
\label{ecc10}
|x-\{\frac{np}{N}\}|<\frac12
\end{equation}
for each  $n\in A$. By invoking Taylor expansions,  \eqref{ecc10}, and  using that
$$|\frac{1}{e(y)-1}-\frac1{2\pi iy}|\lesssim 1$$ for $|y|<\frac12$, we get that
$$\sum_{n\in A}\frac{1-e(\alpha n-\frac{np}{N})}{e(x-\frac{np}{N})-1}=-\sum_{n\in A}\frac{tn}{N^2(x-\{\frac{np}{N}\})}+O(1).$$
We rewrite
$$\sum_{n\in A}\frac{tn}{N^2(x-\{\frac{np}{N}\})}=t\sum_{n=1\atop{\delta\ge|x-\frac{n}{N}|\ge \frac{100}{N}}}^{N}\frac{\frac{n^{*}}{N}}{(Nx-n)},$$
where $n^*:=p^{-1}n$ mod $N$, and $p^{-1}$ is the inverse of $p$ mod $N$.
Our next goal is to prove that
\begin{equation}
\label{ecc12}
\frac{1}{M}\left|\sum_{n=1\atop{\delta\ge|x-\frac{n}{N}|\ge \frac{100}{N}}}^{N}\frac{\frac{n^{*}}{N}}{(Nx-n)}\right|=O(1).
\end{equation}
Since $k$ is odd, it follows from \eqref{ecc4} that $p^{-1}=N_{k-1}$. Let
$$\alpha=\langle a_0,a_1,\ldots\rangle:=a_0+\frac{1}{a_1+\ldots}$$
be the continued fraction expansion of $\alpha$.
We have for each $i\ge 2$
$$p_{i}=a_ip_{i-1}+p_{i-2},$$
$$N_{i}=a_iN_{i-1}+N_{i-2},\,N_0=1,\;N_1=a_1.$$
Due to \eqref{ecc1} we have $a_i\le DM$ for each $i\le k+1$.

Note that $\rho_i:=N_i/N_{i-1}$ satisfies
$$\rho_i=a_i+\frac1\rho_{i-1},\;\rho_1=a_1.$$
Thus,
$$ N/p^{-1}=N_{k}/N_{k-1}=\langle a_k,a_{k-1},\ldots,a_1\rangle.$$
The thing that matters is that all $a_i$ are $O(M)$. Thus, from the recurrence above, the convergents of $N/p^{-1}$, denote them by $M_l/c_l$, have the property that
\begin{equation}
\label{ecc5}
M_{l+1}\lesssim M M_l
\end{equation}
for each $l\le k$ (and similarly for $c_l$, but this will be irrelevant).

It is known that the $l^{th}$ convergent of $p^{-1}/N$ will equal $\frac{c_{l-1}}{M_{l-1}}$, and that the last convergent will equal $p^{-1}/N$. Choose $l_0$ such that
$\frac{N\delta}{M^{3/2}}\lesssim M_{l_0}^{3/2}<{N}\delta$. This is possible due to \eqref{ecc5}. Reasoning as before, we get
$$\frac{1}{M}|\sum_{n=1\atop{|x-\frac{n}{N}|\ge \frac{M_{l_0}^{3/2}}{N}}}^{N}\frac{\frac{n^{*}}{N}}{(Nx-n)}|\lesssim \frac{1}{M}\sum_{N\ge i\gtrsim M_{l_0}^{3/2}}\frac1i\lesssim \frac{\log M+\log(\delta^{-1})}{M}\lesssim_\delta 1.$$

Next, we observe that the remaining part of the sum can be written as
$$\frac{1}{M}\sum_{|j|<M_{l_0}^{3/2}}\frac{\{u+\frac{N_{k-1}j}{N}\}}{j}+O(1),$$
where $u$ is a number whose value is completely irrelevant.

Note that if, say,  $M^5>N$ then the sum above is trivially bounded by $\frac{1}{M}\sum_{|j|<M^5}\frac1{|j|}=O(1)$, and we are fine. Otherwise, we can choose $l_1<l_0$ such that $M^4\lesssim M_{l_1}<M^5$.  The sum above restricted to $|j|\le M_{l_1}^{3/2}$ is trivially $O(1)$.

For $l_1\le l\le l_0-1$ and $M_l^{3/2}\le |j|\le M_{l+1}^{3/2}$, we use that
$$|\frac{N_{k-1}}{N}-\frac{c_l}{M_l}|\le \frac {1}{M_lM_{l+1}},$$
and thus by \eqref{ecc5}
$$|\frac{N_{k-1}j}{N}-\frac{c_lj}{M_l}|\le \frac {M_{l+1}^{3/2}}{M_lM_{l+1}}\lesssim M^{1/2}M_l^{-1/2}.$$
Define
$$C_l:=\{M_l^{3/2}\le |j|\le M_{l+1}^{3/2}: \|u+\frac{c_lj}{M_l}\|\gtrsim M^{1/2}M_l^{-1/2}\}.$$
It follows that
$$|\{M_l^{3/2}\le |j|\le M_{l+1}^{3/2}\}\setminus C_l|\le |\{|j|\le M_{l+1}^{3/2}:\|u+\frac{c_lj}{M_l}\|\lesssim M^{1/2}M_l^{-1/2}\}|$$$$\lesssim M_{l+1}^{3/2}M^{1/2}M_l^{-1/2},$$ and that for each $j\in C_l$
$$|\{u+\frac{N_{k-1}j}{N}\}-\{u+\frac{c_lj}{M_l}\}|\lesssim M^{1/2}M_l^{-1/2}.$$

So we have the following estimate for the error term corresponding to  some $l$
$$\left|\sum_{M_l^{3/2}<|j|<M_{l+1}^{3/2}}\frac{\{u+\frac{N_{k-1}j}{N}\}}{j}-\sum_{M_l^{3/2}<|j|<M_{l+1}^{3/2}}\frac{\{u+\frac{c_lj}{M_l}\}}{j}\right|$$$$\lesssim M^{1/2}M_l^{-1/2}\sum_{j\in C_l}\frac{1}{|j|}+\sum_{M_l^{3/2}\le |j|\le M_{l+1}^{3/2}\atop{j\notin C_l}}\frac{1}{|j|}\lesssim M^2M_l^{-1/2}.$$
Since for each $i$
\begin{equation}
\label{ecc11}
M_{i}\ge M_{i-1}+M_{i-2}\ge 2M_{i-2},
\end{equation}
and since $M_{l_1}\gtrsim M^4$ it follows that the sum of all error terms is bounded by
$$\sum_{l_1\le l}\frac{M^2}{M_l^{1/2}}\lesssim 1$$
as desired. But
$$\sum_{M_l^{3/2}<|j|<M_{l+1}^{3/2}}\frac{\{u+\frac{c_lj}{M_l}\}}{j}=\sum_{r=1}^{M_l}\{u+\frac{c_lr}{M_l}\}\sum_{M_l^{3/2}<|j|<M_{l+1}^{3/2}\atop{j=r \text{ mod }M_l}}\frac{1}{j},$$
and this is $O(\frac{1}{M_l^{1/2}})$, since actually
$$\sup_{P>M_l^{3/2}}|\sum_{M_l^{3/2}<|j|\le P\atop{j=r \text{ mod }M_l}}\frac{1}{j}|=O(\frac{1}{M_l^{3/2}})$$
for each $r$. Summing over $l\ge l_1$ we get using \eqref{ecc11}
$$\sum_{l_0-1\ge l\ge l_1}\left|\sum_{M_l^{3/2}<|k|<M_{l+1}^{3/2}}\frac{\{u+\frac{c_lk}{M_l}\}}{k}\right|\lesssim 1.$$
By putting everything together we conclude that \eqref{ecc12} holds.
\end{proof}

An immediate consequence which only requires trivial modifications is the following.

\begin{corollary}
\label{Coro:745789760598}
Let $A,B\in\C$ with $|A|=|B|=1$. Let also $\alpha$ and $N$ be as in Proposition \ref{P1}. Define $$P(x)=A+Be(\alpha x).$$
Then for each $0<\epsilon<1$ there exist $c_1(\epsilon,A,B,\alpha),c_2(\epsilon,A,B,\alpha)>0$ and a set $P(A,B,\epsilon,\alpha,N)\subset [0,1]$ with measure at least $1-\epsilon$ such that for each $y\in P(A,B,\epsilon,\alpha,N)$
$$c_2(\epsilon,A,B,\alpha)\ge \prod_{n=-N}^{-1}|P(y+n)|\ge c_1(\epsilon,A,B,\alpha)$$
$$c_2(\epsilon,A,B,\alpha)\ge\prod_{n=0}^{N-1}|P(y+n)|\ge c_1(\epsilon,A,B,\alpha).$$

\end{corollary}
The relevance of this result for later applications is that while the sets $P$ are allowed to depend on $N$, the constants $c_1,c_2$ do not depend on $N$.

We can now begin the proof of Theorem \ref{TT2}. By applying the area preserving affine transformations -also called {\em metaplectic transforms}- of the plane (such as translations, rotations, shears, and area one rescalings), it suffices to rule out minimal decay \eqref{eeeeeeeeeeeeeeeeeeeeeee3} for special configurations. See Section 2 in \cite{HRT} for a discussion on this.

Assume for contradiction that there exists a measurable function  $f:\R\to \C$, some $d\in (0,\infty)$ and some $S\subset [0,1]$ with positive measure such that
\begin{equation}
\label{eeeeeeeeeeeeeeeeeeeeeee2}
d<|f(x)|<\infty\;\;\text{for each }x\in S,
\end{equation}
\begin{equation}
\label{eeeeeeeeeeeeeeeeeeeeeee3}
\lim_{|n|\to\infty\atop_{n\in\Z}}f(x+n)=0,
\end{equation}
and
$$f(x+1)(A+Be(\alpha x))=f(x)(E+Fe(\beta x)),$$
for a.e. $x$, for some fixed $A,B,E,F\in\C$, $\alpha,\beta\in\R$, none of them zero. We can also assume $\alpha$ and $\beta$ to be irrational, since the rational case was treated in \cite{De2}. The same metaplectic transforms allow us to assume $0<\alpha<1$.  By re-normalizing, we can trivially assume $E=1$. Let
$$P(x)=A+Be(\alpha x),\;\;\;Q(x)=1+Fe(\beta x).$$
Also, the argument from \cite{De2} shows that the worst case scenario (and the only one that needs to be considered here) is when $|B|=|A|$. Equivalently, $P$ will have zeros. We comment on this in the end of the argument.

By making $S$ a bit smaller, we can also assume that $S+\Z$ contains no zeros of $P$ and $Q$.

Note that by Egoroff's Theorem, \eqref{eeeeeeeeeeeeeeeeeeeeeee3} will allow us to assume (by making $S$ a bit smaller if necessary) that
\begin{equation}
\label{eeeeyurie848eeeee356frsd332}
\lim_{|n|\to\infty\atop_{n\in\Z}}f(x+n)=0,
\end{equation}
uniformly on $S$.

The parameters $D,\alpha,\beta,A,B,F,\epsilon_1,\epsilon_2,\epsilon_3, c_1,c_2,d,m,\gamma$ (some of which are introduced below) will be referred to as {\em fundamental parameters}. They will stay fixed throughout the argument, and in particular will not vary with $N$.

Let us first see how to deal with the contribution coming from the polynomials $Q$. This is done via the {\em conjugates trick} introduced in \cite{De2}. More precisely, let $F=e(\theta)$. Since $S$ has positive measure, it follows that $1_S*1_S$ is continuous and that there exists  an interval $I\subset [0,2]$ and $\epsilon_1>0$ such that
\begin{equation}
\label{ecc14}
1_S*1_S(w)>\epsilon_1
\end{equation}
for each $w\in I$. We can assume without any loss of generality that $I\subset [0,1]$.  There exists $n'\in \N$ large enough such that $m:=[-\frac{2\theta}{\beta}+n'\beta^{-1}]>0$ and $\gamma:=\{-\frac{2\theta}{\beta}+n'\beta^{-1}\}\in I$.  It follows from \eqref{ecc14} that the set $S':=\{x\in S: \gamma-x\in S\}$ has measure at least $\epsilon_1$. The point of this selection is that for each $n\in\Z$, and each $y:=-x-\frac{2\theta}{\beta}+n'\beta^{-1}$, the numbers
$1+Fe(\beta y-n\beta )$ and $1+Fe(\beta x+n\beta)$ are complex conjugates and thus, for each $L\ge 1$, and each $x\in\R$
\begin{equation}
\label{ecc26}
\prod_{n=-L}^{-1}|Q(\gamma-x+n)|=\prod_{n=m+1}^{L+m+1}|Q(x+n)|.
\end{equation}
Let $S''$ be a subset of $S'$ of measure at least $\epsilon_1/2$, and let  $\epsilon_2>0$ depending only on the fundamental parameters $\beta,F$ and $m$ such that
\begin{equation}
\label{ecc16}
\prod_{n=0}^{m}|Q(x+n)|\ge \epsilon_2
\end{equation}
for each $x\in S''$. Let $N$ be as in Corollary \ref{Coro:745789760598}. Let $\epsilon_3>0$ be small enough (depending only on $\epsilon_1$, in particular not depending on $N$) such that the set
$$S(N):=S''\cap\{x\in P(A,B,\epsilon_3,\alpha,N)\}\cap\{x:\gamma-x\in P(A,B,\epsilon_3,\alpha,N)\},$$
has positive measure, and thus is non-empty.  For each $N$ as above, choose a point $x_N\in S(N)$. Let $z_N:=\gamma-x_N$. The recurrence along the orbits of $x_N$ and $z_N$ implies that
$$|f(x_N+N+m+2)|=|f(x_N)|\frac{\prod_{n=0}^{N+m+1}|Q(x_N+n)|}{\prod_{n=0}^{N+m+1}|P(x_N+n)|}$$
$$|f(z_N-N)|=|f(z_N)|\frac{\prod_{-N}^{n=-1}|P(z_N+n)|}{\prod_{-N}^{n=-1}|Q(z_N+n)|}.$$
Multiply these equalities. Using the fact that $x_N,z_N$ are in $S$, \eqref{eeeeeeeeeeeeeeeeeeeeeee2}, \eqref{ecc26} with $x:=x_N$ and $L:=N$, \eqref{ecc16} with $x:=x_N$, Corollary \ref{Coro:745789760598} and the fact that
$$\prod_{n=N}^{N+m+1}|P(x_N+n)|\le (2|A|)^{m+2},$$
it follows that
$$|f(x_N+N+m+2)||f(z_N-N)|\ge \frac{d^2\epsilon_2c_1(\epsilon_3,A,B,\alpha)}{(2|A|)^{m+2}c_2(\epsilon_3,A,B,\alpha)}.$$
The important thing is that the constant on the right depends only on the fundamental parameters, and not on $N$. By letting $N\to\infty$, this will contradict the uniformity assumption \eqref{eeeeyurie848eeeee356frsd332}. This ends the proof of Theorem \ref{TT2}, under the assumption that $|A|=|B|$.

If $|A|\not=|B|$, then things are much easier, and have already been addressed in \cite{De2}. We briefly recap the argument. By invoking Riemann sums and the fact that the derivative of $\phi(x):=\ln|A+Be(x)|$ satisfies
$$\inf_{x\in [0,1]}|\phi'(x)|\gtrsim_{A,B} 1,$$
we get that
$$|\sum_{n=0}^{N-1}\ln |P(x+n)|-N\int_{0}^1\phi|\lesssim_{A,B} 1$$
$$|\sum_{n=-N}^{-1}\ln |P(x+n)|-N\int_{0}^1\phi|\lesssim_{A,B} 1,$$
for each $x\in [0,1]$ and each $N$ such that
$$N\|N\alpha\|\le 1.$$
In particular,

$$|\sum_{n=0}^{N-1}\ln |P(x_N+n)|-\sum_{n=-N}^{-1}\ln |P(z_N+n)||\lesssim_{A,B} 1$$
and thus
$$\frac{\prod_{-N}^{n=-1}|P(z_N+n)|}{\prod_{n=0}^{N-1}|P(x_N+n)|}\sim_{A,B} 1,.$$
This will replace Corollary \ref{Coro:745789760598} in the argument above. Everything else will be the same.

config: arxiv


TITLE: How should I do this? Euler's Method?
QUESTION [2 upvotes]: I am approximating a solution to a first order LODE using Euler's method. I made two tables, one using a step size of .01 and another using .05 ( I had to start at x=0 and end at x=1). I am not understanding the directions for the second part of my assignment:
It states that the order of numerical methods (like Euler's) is based upon the bound for the cummulative error; i.e. for the cummulative error at, say x=2, is bounded by $Ch^n$, where $C$ is a generally unknown constant and $n$ is the order.
For Euler's method, plot the points:
$$(0.1,  \phi(1)-y_{10}),$$ $$(.05, \phi(1)-y_{20}),$$ $$(.025, \phi(1)-y_{40}),$$ $$(.0125, \phi(1)-y_{80}),$$ $$(.00625, \phi(1)-y_{160})$$
And then fit a line to the above data of the form $Ch$. I don't understand, am I supposed to plot these using a step size of .1 or .05? Or am I supposed to use another step size?
Any clarification is appreciated.
Thanks
Edit:
The LODE I am given is $y'=x+2y, y(0)=1$ and the exact solution I found was $\phi(x)=\frac{1}{4}(-2x+5e^{2x}-1)$.

REPLY [2 votes]: From the way I see it, if $\phi$ is the exact solution, the teacher probably wants you to compare the errors at $x=1$ for different step sizes.
Take for example the one before the last $$(0.0125, \phi(1)-y_{80})$$
the step size to use here is $h$ such that $n=80$ and $0+nh=1\implies h=1/80=0.0125$ (I didn't see that coming).
The first number in the bracket is the step size to use for each point. For each point evaluate the approximated value of $\phi(1)\approx y_n$ Then ploy the graph of points $(h_n,\epsilon_n)$ where $\epsilon_n=\phi(1)-y_n$ and $h_n$ is the equivalent step size.


TITLE: Find $R$ such that $x>R$ implies $f(x) > 20$.
QUESTION [0 upvotes]: Let f$(z) = z^3 + iz^2 + 8z + 3$.
Find some $R > 0$ so that for all $z$ with $|z|>R, |f(z)| > 20$.
My approach was to find the point where $f(z) = 20$ since as $z$ increases, $f$ increases. However, this point is a complex number (which I also don't know how to calculate).
I feel like I am doing something wrong?

REPLY [0 votes]: For $|z|>24$,
$|f(z)/z^3|=|1+(iz^2+8z+3)/z^3|\ge 1-|(iz^2+8z+3)/z^3| \ge 1-(1/|z|+8/|z^2|+3/|z^3|) \ge 1-12/|z|\ge 1/2$.


text:
\begin{document}

\maketitle

\begin{abstract}
Consider the Navier-Stokes flow past a rotating obstacle with
a general time-dependent angular velocity and a time-dependent
outflow condition at infinity. After rewriting the problem on a fixed domain,
one obtains a non-autonomous system of equations with unbounded drift terms.
It is shown that the solution to a model problem in the whole space case $\R^d$
is governed by a strongly continuous evolution
system on $L^p_\sigma(\R^d)$ for $1<p<\infty$. The strategy is to
derive a representation formula, similar to the one known in the case
of non-autonomous Ornstein-Uhlenbeck equations. This explicit formula allows to prove $L^p$-$L^q$ estimates
and gradient estimates for the evolution system. These results are key ingredients to
obtain (local) mild solutions to the full nonlinear problem by a version of Kato's iteration
scheme.

\end{abstract}

\section{Introduction and main result}

In this paper we consider  a model problem in $\R^d$ for the flow of an incompressible, viscous fluid
past a rotating obstacle with an additional time-dependent outflow condition at
infinity. The equations describing this problem are the Navier-Stokes
equations in an exterior domain varying in time with an additional
condition for the velocity field at infinity.

In order to motivate our model problem, let $\mathcal O \subset \R^d$ be a compact obstacle with
smooth boundary, let $\Omega:=\R^d \setminus \mathcal O$ be the
exterior of the obstacle and

let $m \in C([0,\infty); \R^{d\times d})$ be a
continuous matrix-valued function.

Then, the exterior of the rotated obstacle at time $t>0$ is represented by $\Omega(t):=
Q(t)\Omega$ where $Q(t)$ solves the ordinary differential equation
\begin{equation}
\left\{\begin{array}{rcll}
\partial_t Q(t) &=& m(t)Q(t),& t>0,\\[0.2cm]
Q(0)&=&\mathrm{Id}.
\end{array}\right.
\end{equation}\normalsize
With a prescribed velocity field $v_\infty\in C^1([0,\infty);\R^d)$
at infinity, the equations for the fluid on the time-dependent
domain $\Omega(t)$ with no-slip boundary condition
take the form
\begin{align}\label{eq:NS_2}
v_t-\Delta v +v\cdot \nabla v+\nabla \q&=0&\quad\quad\quad\mbox{in $\Omega(t)\times (0,\infty) $,}\notag\\
\div v&=0&\quad\quad\quad\mbox{in $\Omega(t)\times (0,\infty) $,}\notag\\
v(t,y)&=m(t)y&\quad\quad\quad\mbox{ on $\partial\Omega(t)\times (0,\infty) $,}\\
\lim_{|y|\to \infty} v(t,y)&=v_\infty(t)&\quad\quad\quad\mbox{ \mbox{for} $t\in(0,\infty) $,}\notag\\
v(0,y)&=u_0(y)&\quad\quad\quad\mbox{in $\Omega$}\notag,
\end{align}
where $v$ and $\q$ are the unknown velocity field and the pressure of
the fluid, respectively.

The disadvantage of this description is the
variability of the domain $\Omega(t)$, and the fact that the
equations do not fit into the $L^p$-setting, due the velocity
condition at infinity. Assume for the time beeing that $m(t)$ is skew
symmetric for $t>0$; this implies that for all $t>0$ the matrix $Q(t)$ is orthogonal. Then, by setting
\begin{equation}
x=Q(t)^{\mathrm T}y, \quad u(t,x)=Q(t)^{\mathrm T}
(v(t,y)-v_\infty(t)), \quad \p(t,x)=\q(t,y),
\end{equation}
the above equations can be transformed to the reference domain
$\Omega$ and the new velocity field $u$ vanishes at infinity. Then
\eqref{eq:NS_2} is equivalent to the following system of equations

\begin{align}\label{eq:NS_3}
	\left.
	\begin{array}{l}
u_t- \Delta u - \mathcal M(t)x \cdot \nabla u + \mathcal M(t)u  \\[0.1cm]
\quad+ Q(t)^{\mathrm T}v_{\infty}(t)\cdot \nabla u
-Q(t)^{\mathrm T}\partial_t v_\infty(t)\\[0.1cm]
\quad +u\cdot \nabla u
+\nabla \p
\end{array}\right\}&=0&\;\mbox{in   $\Omega \times (0,\infty)$,}\notag\\
\div u&=0&\quad\mbox{in $ \Omega \times (0,\infty) $,}\notag\\
u(t,x)&= \mathcal M(t)x-Q(t)^{\mathrm T}v_\infty(t)&\;\mbox{on $\partial\Omega \times (0,\infty)$},\\
\lim_{|x|\to \infty} u(t,x)&= 0&\;\mbox{\mbox{for} $t\in(0,\infty) $,}\notag\\
u(0,x)&=u_0(x)&\;\mbox{in $\Omega$},\notag
\end{align}
\normalsize
where $\mathcal M(t):=Q(t)^{\mathrm T}m(t)Q(t)$. The  main difficulty in dealing with this problem arises since the
term $\mathcal M(t)x \cdot \nabla$ has unbounded coefficients. In particular, the
lower order terms cannot be treated by classical perturbation theory for the
Stokes operator.

Note that even if we assume that $m(t)\equiv m$ is independent of
time (this implies that also $\mathcal M(t)\equiv \mathcal
M$ is independent of time), equation (\ref{eq:NS_3}) is still non-autonomous due to the
time-dependent first order term $Q(t)^{\mathrm T}v_{\infty}\cdot
\nabla$ (except in some special cases discussed below).

However, by using localization techniques similar to \cite{GHH06}, this problem is
finally reduced to a model problem in $\R^d$ and a model problem in a bounded domain.
Since $Q(t)\partial_t v_\infty(t)\equiv F(t)$, $t>0$, i.e. it is
constant in space, we may put this term in the pressure $\p$.
Hence, in this paper we discuss the following linearized model problem
in $\R^d$\vspace{0.2cm}
\begin{align}
	u_t-\Delta
u-\left(M(t)x + f(t)\right)\cdot\nabla u +M(t)u + \nabla \p &=0&\quad\mbox{in $\R^d \times (0,\infty)$,}\notag\\
\div u&=0& \quad\mbox{in $\R^d \times (0,\infty)$,}\label{eq:NS_1}\\
u(0)&=u_0&\quad\mbox{in $\R^d$,}\notag\vspace{0.2cm}
\end{align}
where we allow general coefficients $M\in C([0,\infty); \R^{d\times d})$ and $f\in C([0,\infty); \R^{d})$. If we set $M(t):=Q(t)^{\mathrm T}m(t)Q(t)$ and
$f(t):=-Q(t)^{\mathrm T}v_\infty(t)$ then we obtain the linearization of equation \eqref{eq:NS_3} with $\Omega=\R^d$.
Such a model problem also arises in the analysis of a rotating body with translational velocity
$-v_\infty(t)$, see \cite{Far05}.

Existence and uniqueness of a mild solution of an autonomous variant of problem \eqref{eq:NS_2}
{\em without} an outflow condition, i.e.  $v_\infty\equiv0$,
and $m(t)\equiv m$, was investigated in quite a few papers, see
\cite{His99a}, \cite{His99b}, \cite{GHH06} and \cite{HS05}. Hishida was even able to deal
with a time dependent rotation in \cite{His01}, however only for angular velocities of a special form.

For the problem including an additional outflow condition at infinity,
there are only a few results.
Indeed, in the special case, where $m(t)x=\omega(t) \times x$ and
$\omega:[0,\infty) \rightarrow \R^3$ is the angular velocity of the
obstacle and $v_\infty:[0,\infty)\to\R^3$ a time-dependent outflow velocity,
Borchers \cite{Bor92} constructed weak non-stationary solutions for the equations (\ref{eq:NS_3}).
Moreover, Shibata \cite{Shi08} studied
the special case where $m(t)\equiv m$, $v_\infty(t)=v_\infty$ and
$mv_\infty=0$. The condition $mv_\infty=0$,  i.e.
$Q(t)^{\mathrm T}v_\infty= k v_\infty$ for $k\in \{-1,1\}$, ensures that
\eqref{eq:NS_3} is still an autonomous equation and the
solution of \eqref{eq:NS_3} is governed by a $C_0$-semigroup which is
{\em not} analytic.
The physical meaning of the additional condition $mv_\infty=0$ is that the outflow
direction of the fluid is parallel to the axis of rotation of the
obstacle. The stationary problem of this latter situation was
analysed in \cite{Far05}.

The assumption $mv_\infty=0$ was recentely relaxed by the second author in
\cite{Han10}. Indeed, he was able to deal with the model problem in $\R^d$ where
$m(t)v_\infty\neq 0$ and $v_\infty(t)\equiv v_\infty$.
However he assumes that $m(t)$ and $m(s)$ commute for all
$t,s>0$ which can physically be interpreted by the fact that
the axis of rotation is fixed.

The aim of this work is to remove the latter additional condition, i.e.
$m(t)$ and $m(s)$ need not to commute and $v_\infty$ may be time-dependent.

As usual the Helmholtz projection $\mathbb P$ allows us to rewrite \eqref{eq:NS_1} as an abstract Cauchy problem in
$L^p_\sigma(\R^d)$,
where $L^p_\sigma(\R^d)$ denotes  the space of all
solenoidal vector fields in $L^p(\R)^d$:
\begin{equation}\label{eq:NS_5}
\begin{array}{rclll}
u'(t)- A(t)u(t) &=&0, & t>0,  \\[0.15cm]
u(0)&=&u_0.&
\end{array}
\end{equation}
Here:
\begin{align*}
	A(t)u&:=\mathbb P\left(\Delta u+\left(M(t)x + f(t)\right)\cdot\nabla u
	+M(t)u\right)\\
	D(A(t))&:=\{ u\in W^{2,p}(\R^d)^d\cap L^p_\sigma(\R^d): M(t)x\cdot
	\nabla u\in L^p(\R^d)^d\}.
\end{align*}
Note that it immediately follows from \cite{HS05} that for fixed $t>0$, the operator $A(t)$ is the
generater of a $C_0$-semigroup, which is not analytic.
The fact that the semigroup is not analytic prevents us from employing
standard generation results for evolution systems, see \cite[Chapter
5]{Paz83} and references therein. For the same reason, $L^p$-$L^q$
estimates and gradient estimates don't follow from standard arguments.

Therefore, we first derive a representation formula for the solution of
\eqref{eq:NS_1}. In order to derive this representation formula
we transform  \eqref{eq:NS_1} to a non-autonomous heat equation
which can be explicitly solved, see Section~\ref{sec:cov}.
It turns out that the transformation to a non-autonomous heat equation
is crucial to deal with our problem in this generality since the
different transformation used in \cite{Han10} caused the additional
assumption that $M(t)$ and $M(s)$ commute for all $t,s>0$.

In the following we denote by $\{U(t,s)\}_{t,s\geq 0}$ the evolution system on $\R^d$ generated by the family of matrices $\{-M(t)\}_{t\geq 0}$, i.e.
\begin{equation}
\left\{\begin{array}{lcl}
\partial_t U(t,s) & = & -M(t)U(t,s),\\[0.15cm]
U(s,s) & = & \mathrm{Id}.
\end{array}\right.
\vspace{0.1cm}
\label{eq:evol}
\end{equation}
Note that $\partial_s U(t,s)  =  U(t,s)M(s)$.

We are now ready to present our main result.

\begin{theorem}\label{thm:main}
	Let $1<p<\infty$, $M\in C([0,\infty);\R^{d\times d})$ and $f\in
	C([0,\infty);\R^d)$. The the solution of
\eqref{eq:NS_5} is governed by a strongly continuous evolution system
$\{T(t,s)\}_{t\geq s\geq0}\subset \cL(L^p_\sigma(\R^d)^d)$.
Moreover, the evolution system $\{T(t,s)\}_{t\geq s\geq0}$ admits the
following properties:
\begin{enumerate}
	\item[(a)] For $T_0>0$ set
	$
		M_{T_0}:=\sup\{\|U(t,s)\|:t,s\in[0,T_0]\}.
	$
	Then for $1<p<\infty$ and $p\leq q \leq \infty$ there exists
	$C:=C(M_{T_0},d)>0$ such that for $u\in L_\sigma^p(\R^d)$
\begin{align}
	\|T(t,s)u\|_{L^q_\sigma(\R^d)}
	&\leq C
(t-s)^{-\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|u\|_{L^p_\sigma(\R^d)},&
 0\leq s<t<T_0,\label{eq:LpLqsmoothing}\\
\|\nabla T(t,s)u\|_{L^q(\R^d)}
&\leq C
(t-s)^{-\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)-\frac{1}{2}}\|u\|_{L^p_\sigma(\R^d)},
&  0\leq s<t<T_0. \label{eq:GradientEstimate}
\end{align}
In particular, if the evolution system $\{U(t,s)\}_{s,t\geq 0}$ is uniformly
bounded, i.e. $M_{T_0}\leq M$, for some $M>0$ and all $T_0>0$, we may set
$T_0=\infty$.
\item[(b)]
For $1<p<q<\infty$, $s\geq0$ and $u\in L^p_{\sigma}(\R^d)$ we have
\begin{align*}\label{eq:behavior_ts_1}
	\lim\limits_{t\to s,\ t> s}(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q\right)} \|T(t,s)u\|_{L^q_{\sigma}(\R^d)}
	=0 \mbox{ and }
	\lim\limits_{t\to s,\ t> s}
(t-s)^{\frac 1 2 } \|\nabla T(t,s)u\|_{L^p(\R^d)}=0.
\end{align*}
\end{enumerate}
\end{theorem}

Next we consider the nonlinear problem
\begin{equation}\label{eq:NS_abstract}
\begin{array}{rclll}
u'(t)- A(t)u(t) + \mathbb{P}( (u(t) \cdot \nabla) u(t) )&=&0, & t>0,  \\[0.15cm]
u(0)&=&u_0,&
\end{array}
\end{equation}

with initial value $u_0 \in L^p_{\sigma}(\R^d)$.

For given $0<T_0\leq \infty$,
we call a function $u\in C([0,T_0);L^p_{\sigma}(\R^d))$ a
\textit{mild solution} of (\ref{eq:NS_abstract})  if $u$ satisfies
the integral equation
\begin{equation}\label{eq:Duhamel}
u(t) = T(t,0)u_0 - \int_0^t T(t,s) \mathbb P((u(s) \cdot \nabla)
u(s)) \d s, \quad t>0,
\end{equation}
in $L^p_{\sigma}(\R^d)$.
By adjusting Kato's iteration scheme (see \cite{Kat84}) to our
situation the existence of a unique (local) mild solution follows, cf.
\cite{Han10} for details.

\begin{corollary}\label{cor:Kato}
Let $2\leq d\leq p \leq q < \infty$, $M\in C([0,\infty);\R^{d\times d})$, $f\in
	C([0,\infty);\R^d)$ and $u_0\in
L^p_{\sigma}(\R^d)$. Then there exists $T_0>0$ and a unique mild
solution $u\in C([0,T_0);L^p_{\sigma}(\R^d))$ of
(\ref{eq:NS_abstract}), which has the properties
\begin{equation}\label{eq:mild_solution_prop1}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}u(t) \in
C([0,T_0);L^q_{\sigma}(\R^d)),
\end{equation}
\begin{equation}\label{eq:mild_solution_prop2}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)+\frac{1}{2}}\nabla
u(t) \in C([0,T_0);L^q(\R^d)^{d\times d}).
\end{equation}
If $p<q$, then in addition
\begin{equation}\label{eq:mild_solution_prop3}
t^{\frac{d}{2}\left(\frac{1}{p}-\frac{1}{q}\right)}\|u(t)\|_{L^q(\R^d)} +
t^{\frac{1}{2}}\|\nabla u(t)\|_{L^p(\R^d)} \rightarrow 0 \qquad \mbox{as\;} t
\rightarrow 0.
\end{equation}
Moreover, in the case $d=p$ we may set $T_0 =+\infty$
provided
$\|u_0\|_{L^d(\R^d)}$ is small enough and $\{U(t,s)\}_{s,t\geq 0}$ is
uniformly
bounded.

\end{corollary}

\begin{remark}
	In particular, $\{U(t,s)\}_{s,t\geq 0}$ is
	uniformly bounded if $M(t)$ is skew symmetric for all $t>0$.
\end{remark}

\section{Proof of Theorem~\ref{thm:main}}

Let $M$ be as in  Theorem~\ref{thm:main},
and let $\{U(t,s)\}_{s,t\geq0}$  be the evolution system on $\R^d$ that satisfies \eqref{eq:evol}.
We consider the system of parabolic equations of the form
\begin{equation}\label{eq:1}
\left\{\begin{array}{rcll}
\partial_t u(t,x)  -\mathcal A(t) u(t,x)&=&0, & \; t> s,\  x\in \R^d,\\[0.15cm]
u(s,x) &=& \varphi(x),& \;x\in \R^d,
\end{array}\right.\end{equation}
for $s\geq 0$ fixed, initial value $\varphi\in L^p(\R^d)^d$ and some $p\in(1,\infty)$. Here the family of operators $\mathcal A(t)$ is of the form
$$
\mathcal A(t)u(x) := \Big( \Delta u_i(t,x) + \langle M(t)x + f(t) , \nabla
u_i(t,x) \rangle \Big)_{i=1}^{d} - M(t)u(t,x), \quad t>0,\; x\in\R^d.
$$

 As in \cite[Lemma
3.2]{GL08a} or \cite{Han10}, we first develop an explicit representation
formula. To be more precise,
 we show in Section~\ref{sec:cov} that for $p\in(1,\infty)$ and $\varphi\in L^p(\R^d)^d$
 the solution $u$ to (\ref{eq:1}) is governed by a strongly continuous evolution system
 $\{\tilde T(t,s)\}_{t\geq s}\subset \cL(L^p(\R^d)^d)$ which is explicitly given by

 \begin{align}
u(t,x):=& (\tilde T(t,s)\varphi)(x):=(k(t,s,\cdot)*\varphi)(U(s,t)x+g(t,s)),

\quad t>s,\ x\in\R^d,
\label{eq:solformul}
\end{align}
where
 \begin{align}
	k(t,s,x)&:= \frac{1}{(4\pi)^{d/2} (\det Q_{t,s})^{1/2}} U(t,s)
\mathrm e^{-\frac 1 4 \langle
Q_{t,s}^{-1}x,x\rangle}  \d y,\quad t>s\geq 0,\ x\in\R^d,\label{eq:kern}
\\
g(t,s)&:= \int_s^t U(s,r) f(r) \d r,\quad
Q_{t,s}: = \int_s^t U(s,r)U^*(s,r) \d r, \quad t\geq s\geq0.
\nonumber
\end{align}

Similar to \cite{DPL07} one can show that
for $\varphi\in
C_c^\infty(\R^d)^d$ the solution $u$ of \eqref{eq:1} given by \eqref{eq:solformul}  is a classical
solution.

A simple calculation shows that $\div \tilde T(t,s)\varphi=0$ for $\varphi\in
C_{c,\sigma}^\infty(\R^d)$ and $t\geq s\geq0$. Hence, the restriction
$T(t,s):=\tilde T(t,s)|_{L^p_\sigma(\R^d)}$ is an evolution system on
$L^p_\sigma(\R^d)$. In particular, $u(t):=T(t,0)u_0$ is a solution to
\eqref{eq:NS_5}.

By similar arguments as in the proofs of \cite[Lemma
3.2]{GL08a} or \cite[Lemma 2.4]{Han10}, for $T_0>0$ there exists $C:=C(d,M_{T_0})>0$ (see
Theorem~\ref{thm:main} for the definition of $M_{T_0}$)
such that
\begin{align}
	\begin{split}
\|Q_{t,s}^{-\frac 1 2}\| \leq C (t-s)^{-\frac 1 2}, \quad 0\leq s<t<T_0,\\
(\det Q_{t,s})^{\frac 1 2} \geq
C(t-s)^{\frac d 2},  \quad 0\leq s<t<T_0.
\end{split}
\label{eq:estimateQ}
\end{align}
Moreover, if $M_{T_0}$ is uniformly bounded in $T_0$ we may write
$T_0=\infty$ in \eqref{eq:estimateQ}.

\begin{proof}[Proof of Theorem~\ref{thm:main}]
We start by showing the estimate (\ref{eq:LpLqsmoothing}). Let $T_0>0$.
By the change
of variables $\xi=U(s,t)x$ and by Young's inequality we obtain
\begin{align*}
 \|T(t,s)u\|_{L_{\sigma}^q(\R^d)}
 \leq |\det U(s,t)|^{\frac1q}\|k(t,s,\cdot)\|_{L^r(\R^d)}
\|u\|_{L_{\sigma}^p(\R^d)},\quad t> s\geq0,

\end{align*}
where $1<r<\infty$ with $\frac 1 p + \frac 1 r = 1 + \frac 1 q$.
Further, by the change of variable $y= Q_{t,s}^{1/2} z$ we obtain
\begin{align*}
	\|k(t,s,\cdot)\|_{L^r(\R^d)}^r

&= \|U(t,s)\|\int_{\R^d}
\left(\frac1{(4\pi)^\frac d2}
\e^{-\frac{ |z|^2}{4}}\right)^r (\det Q_{t,s})^{\frac {1-r}2} \d
z\\
&\leq C \|U(t,s)\|(\det Q_{t,s})^{\frac{1-r}2},\quad t\geq s\geq0,

\end{align*}
for some $C>0$. Now \eqref{eq:estimateQ}
yields \eqref{eq:LpLqsmoothing}.

To prove the gradient estimate (\ref{eq:GradientEstimate}), we first
observe that

\begin{align*}
	\nabla
	T(t,s)u(x)=\int\limits_{\R^d}u(U(s,t)x+g(t,s)k(t,s,y)\left(U^T(s,t)Q^{-1}_{s,t}y\right)^T\d
	y,\quad t> s\geq0,\ x\in\R^d.
\end{align*}
Now, \eqref{eq:GradientEstimate} follows similarly as above.

Since \eqref{eq:1} is uniquely solvable for $\varphi\in C_c^\infty(\R^d)^d$, see Section~\ref{sec:cov}, the
law of evolution is valid, i.e.
\begin{equation}\label{eq:law of evolution}
\tilde T(t,s)\varphi = \tilde T(t,r)\tilde T(r,s)\varphi,
\end{equation}
holds for $ 0\leq s \leq r \leq t $ and every $\varphi \in C_{c}^\infty(\R^d)^d$.
The density of $C_{c}^\infty(\R^d)^d$ in $L^p(\R^d)^d$ yields that (\ref{eq:law of evolution}) even holds for all $\varphi\in L^p(\R^d)^d$.

In order to prove the strong continuity of the map $(t,s)\mapsto \tilde T (t,s)$ on $0\leq s \leq t $ we apply the change of the variables $y=Q_{t,s}^{1/2}z$, to see that
$$
\tilde T(t,s)\varphi(x)= \frac{1}{(4\pi)^{\frac d 2}}U(t,s)\cdot \int_{\R^d}
\varphi(U(s,t)x+g(t,s)-Q_{t,s}^{\frac 1 2}z) \e^{-\frac{|z|^2}{4}} \d z
$$
holds. For $t>s$ fixed, we pick two sequences $(t_n)_{n\in \N}$ and $(s_n)_{n\in \N}$ such that $t_n \geq s_n $ holds for every $n\in \N$ and $(t_n,s_n)\to (t,s)$ as $n \to \infty$. For every $\varphi \in C_c^{\infty}(\R^d)^d$ and every
$x\in\R^d$ we now obtain
$$
\varphi(U(s_n,t_n)x+g(t_n,s_n)-Q_{t_n,s_n}^{\frac 1 2}z)
\rightarrow \varphi(U(s,t)x+g(t,s)-Q_{t,s}^{\frac 1 2}z)
$$
as $n\rightarrow \infty$. Lebegue's theorem now yields
$\tilde T(t_n,s_n)\varphi \rightarrow\tilde T(t,s)\varphi$
as $n\rightarrow \infty$ for every $\varphi\in
C_c^{\infty}(\R^d)^d$. The density of $C_c^{\infty}(\R^d)^d$ in
$L^p(\R^d)^d$ implies the strong continuity.

In order to prove Theorem~\ref{thm:main}(b)
let $u\in L^p_\sigma(\R^d)$, $t-s \leq 1$ and choose $(u_n)_{n\in\N}
\subset C_{c,\sigma}^{\infty}(\R^d) \subset L_{\sigma}^p(\R^d)$,
such that $\lim_{n\to\infty}\|u-u_n\|_{L^p(\R^d)}=0$. The triangle inequality
together with the $L^p$-$L^q$ estimates (\ref{eq:LpLqsmoothing})
imply that there exist constants $C_1, C_2>0$ such that
\begin{align*}
&(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q \right)}
\|T(t,s)u\|_{L_{\sigma}^q(\R^d)}\\
&\qquad \leq (t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q \right)}
\|T(t,s)(u-u_n)\|_{L_{\sigma}^q(\R^d)} + (t-s)^{\frac d 2 \left(\frac 1 p - \frac 1
q
\right)} \|T(t,s)u_n\|_{L_{\sigma}^q(\R^d)}\\
&\qquad \leq  C_1 \|u-u_n\|_{L_{\sigma}^p(\R^d)} + C_2 (t-s)^{\frac d 2 \left(\frac 1 p -
\frac 1 q \right)} \|u_n\|_{L_{\sigma}^q(\R^d)},\quad 0\leq t-s\leq 1,\
n\in\N.
\end{align*}
Hence,
$
	\lim\limits_{t\to s}(t-s)^{\frac d 2 \left(\frac 1 p - \frac 1 q
	\right)} \|T(t,s)u\|_{L_{\sigma}^q(\R^d)}=0
$
by letting first $t\rightarrow s$ and then $n\rightarrow \infty$.
The second assertion in Theorem~\ref{thm:main}(b) is proved in a similar
way.
\end{proof}

\section{Representation Formula}
\label{sec:cov}

In this section the representation formula (\ref{eq:solformul}) is derived. The general idea is to do a coordinate transformation in order to eliminate the unbounded drift and the zero order term of the operator $\mathcal A(t)$. For this purpose we set
\begin{equation*}
z := U(s,t)x +g(t,s),
\end{equation*}
where
\begin{equation*}
g(t,s):= \int_s^t U(s,r) f(r) \d r,
\end{equation*}
and we look for a solution $u$ of (\ref{eq:1}) with initial value  $\varphi\in C_c^{\infty}(\R^d)^d$ in the form
\begin{equation}\label{eq:coord_transf}
u(t,x)= U(t,s) w(t,U(s,t)x+g(t,s)).
\end{equation}

By recalling \eqref{eq:evol} we obtain from a straightforward computation that
\begin{align*}
\partial_t u(t,x) &= -M(t) U(t,s)w(t,z) + U(t,s) \Big(\langle U(s,t) M(t) x + U(s,t) f(t), \nabla w_i(t,z) \rangle\Big)_{i=1}^d \\
&\qquad \qquad + U(t,s)\partial_t w (t,z),
\end{align*}\normalsize

holds. Moreover, we can write equation \eqref{eq:coord_transf} component-wise as
\begin{equation*}
u_i(t,x)= \sum_{j=1}^d U_{ij}(t,s) w_j(t,U(s,t)x+g(t,s)), \qquad \mbox{for}\; i=1,\ldots,d,
\end{equation*}
and thus for the spatial derivatives of $u$ we obtain
\begin{align*}
\nabla u_i(t,x) &= \sum_{j=1}^d U_{ij}(t,s) U^*(s,t) \nabla w_j(t,z),\\[0.1cm]
\nabla^2 u_i(t,x) &= \sum_{j=1}^d U_{ij}(t,s) U^*(s,t) \nabla^2 w_j(t,z)U(s,t).\vspace{0.2cm}
\end{align*}

In particular, the drift term can be written as
\begin{align*}
\langle M(t)x + f(t), \nabla u_i(t,x) \rangle &=
  \sum_{j=1}^d U_{ij}(t,s) \langle U(s,t) M(t)x + U(s,t)f(t) , \nabla w_j(t,z) \rangle.
\end{align*}\normalsize

Thus, the function $u$ solves problem (\ref{eq:1}) if and only if for every $i=1,\ldots,d$, the function $w_i : \R^d \rightarrow \R$ is a solution to\vspace{0.1cm}
\begin{equation}\label{eq:2}
\left\{\begin{array}{lcll}
\partial_t w_i(t,z)  &=&\mathrm{Tr} [U(s,t) U^*(s,t) \nabla^2 w_i(t,z)], &  \; t> s, z\in \R^d,\\[0.15cm]
w_i(s,z) &=& \varphi_i(z),&  \;z\in \R^d.
\end{array}\right.\vspace{0.1cm}\end{equation}\normalsize

By our transformation we now obtained an uncoupled system of parabolic equations with coefficients only depending on $t$. More
precisely, for $i=1,\ldots,d$, the equation \eqref{eq:2} is a non-autonomous heat equation. It is well known that such a problem can be
uniquely solved (cf. \cite[Proposition 2.1]{DPL07}) and that for every $\varphi_i \in C_{c}^\infty(\R^d)$ its unique solution is explicitly given
by the formula
\begin{equation}
w_i(t,z) = \frac{1}{(4\pi)^{\frac d 2} (\det Q_{t,s})^{\frac 1 2}} \int_{\R^d} \varphi_i(z-y)\mathrm e^{-\frac 1 4 \langle Q_{t,s}^{-1}y,y\rangle}  \d y,
\end{equation}
where
\begin{equation}
Q_{t,s} = \int_s^t U(s,r)U^*(s,r) \d r.
\end{equation}

Now, via \eqref{eq:coord_transf}, the unique solution to our original problem (\ref{eq:1}) is given by the representation formula

 \begin{equation}\label{eq:OUsol}
u(t,x)= (k(t,s,\cdot)\ast u)(U(s,t)x+g(t,s)),
\end{equation}
where the kernel $k(t,s,x)$ is defined in \eqref{eq:kern}.

Note that the right hand side of (\ref{eq:OUsol}) is even well defined
for each $L^p(\R^d)^d$-function $\varphi$. Thus, this explicit formula can be used to define an evolution system on $L^p(\R^d)^d$ in the following way. For $\varphi \in L^p(\R^d)^d$ we set
\begin{equation*}
\tilde T(t,s)\varphi := \left\{\begin{array}{cl}
\varphi & \mbox{for}\; t=s,\\[0.2cm]
(k(t,s,x)\ast \varphi)(U(s,t)x+g(t,s))&\mbox{for}\; t>s.
\end{array}\right.
\end{equation*}

Since problem (\ref{eq:2})  is uniquely solvable it follows via (\ref{eq:coord_transf}) that $\tilde T(t,s)\varphi$ is the unique solution of (\ref{eq:1}) for initial value $\varphi \in C_{c}^\infty(\R^d)^d$.

\bibliographystyle{amsalpha-ag4}
\bibliography{ref}

\end{document}


TITLE: Euclidean Domain and algorithm questions
QUESTION [1 upvotes]: If $R$ is a Euclidean Domain. Describe an algorithm for computing the greatest common divisor of two non-zero elements $a$ and $b$ of $R$.
Would this just be the Euclidean algorithm?
Also why does the algorithm terminate and return to the correct answer?

REPLY [0 votes]: Euclidean Domain
The definition of a Euclidean Domain, $R$, asserts that you have a Euclidean function $N:R\setminus\{0\}\rightarrow\mathbb N$ so that the algorithm will finish in finitely many steps. This function must satisfy that for $x\in R$ and $d\in R\setminus\{0\}$ there exists $q,r\in R$ such that
$$
x=qd+r
$$
where either $N(r)<N(d)$ or $r=0$.
Euclidean Algorithm
To see how this ensures the algorithm to terminate let us say we are to find $\gcd(a,b)$ for some $a,b\in R$ with $N(a)>N(b)$ just write
$$
a=q_1 b+r_1\\
b=q_2 r_1+r_2\\
r_1=q_3 r_2+r_3\\
\vdots\\
r_{i-2}=q_i r_{i-1}+r_i
$$
using the properties of the Euclidean function in each step so that
$$
N(b)>N(r_1)>N(r_2)>...>N(r_i)>N(r_{i+1})>...
$$
Since the sequence given above is stricly decreasing in $\mathbb N$ by each step in the algorithm we will eventually run out of natural numbers so that we are forced to have $r_n=0$ for some $r_n$ after at most $N(b)$ steps. Then (since a Euclidean Domain is also a Unique Factorization Domain) $\gcd(a,b)=r_{n-1}$ is the 'unique' greatest common divisor of $a$ and $b$ up to multiplication by a unit.


TITLE: Expected number of rolls for an unfair die to get all possibile values at least once
QUESTION [7 upvotes]: Suppose that we have a 6-sided unfair dice, where rolling a 1 is twice as likely as rolling any other number, and the other numbers have the same likelihood. What is the expected number of rolls to get each value at least once?
Thus, $$p(1) = 2/7\qquad p(2) = p(3) = \cdots = p(6) = 1/7$$
I understand how to approach this when the probabilities are the same, as it's just the Coupon collector's problem, but throwing in a non-uniform probability distribution throws me off. I know there is a general solution for this problem, but I can't seem to get any intuition as to why it's the case.

REPLY [7 votes]: Consider the die as a $7$-sided die with two $1$s and wait for all sides to appear. With probability $\frac27$, you unnecessarily waited for the second $1$ for an expected number of $7$ rolls at the end. Correcting for that yields
$$7H_7-\frac27\cdot7=\frac{323}{20}\;.$$


TITLE: Showing that $\text{Var}(X_t) = t$ if $(X_t)_{t > 0}$ is a gaussian process
QUESTION [0 upvotes]: Let $(X_t)_{t > 0}$ be a stochastic process such that

$(X_t)_{t > 0}$ is a gaussian process,
it has continuous trajectories,
$\mathbb{E}(X_t) = 0$,
$\text{Cov}(X_t, X_s) = \min \{t, s \}$.

I would like to show that $(X_t)_{t > 0}$ is a Brownian motion. Continuity is obvious, $X_0 = 0$ also. I'm stuck with showing that $X_t - X_s \sim N(0, t-s), t > s$. Of course it's easy to show that $\mathbb{E}(X_t - X_s) = 0$ but I have some problems in computing the variance.
$$\text{Var}(X_t - X_s) = \text{Var}(X_t) + \text{Var}(X_s) - 2\text{Cor}(X_t, X_s) = \text{Var}(X_t) + \text{Var}(X_s) - 2s.$$
How can we show that $\text{Var}(X_t) = t$?

REPLY [3 votes]: $var(X_t)$ is same as $cov(X_t,X_t)$ so it is $\min \{t,t\}=t$.


text:
\begin{document}

\begin{center}

\bf \Large

Properization:

Constructing Proper Scoring Rules via Bayes Acts

\end{center}

\centerline{\bf Jonas R. Brehmer}
\centerline{University of Mannheim, Mannheim, Germany}

\bigskip

\centerline{\bf Tilmann Gneiting}
\centerline{Heidelberg Institute for Theoretical Studies, Heidelberg, Germany}
\centerline{Karlsruhe Institute of Technology, Karlsruhe, Germany}

\bigskip

\centerline{August 16, 2018}

\medskip

\begin{abstract}

Scoring rules serve to quantify predictive performance.  A scoring
rule is proper if truth telling is an optimal strategy in expectation.
Subject to customary regularity conditions, every scoring rule can be
made proper, by applying a special case of the Bayes act construction
studied by \citet{GruenDawid2004} and \citet{Dawid2007}, to which we
refer as properization.  We discuss examples from the recent
literature and apply the construction to create new types, and
reinterpret existing forms, of proper scoring rules and consistent
scoring functions. In an abstract setting, we formulate sufficient
conditions under which Bayes acts exist and scoring rules can be made
proper.

\smallskip
\noindent
{{\em Key words and phrases}.  Bayes act; consistent scoring function;
  forecast evaluation; misclassification error; proper scoring rule.}

\end{abstract}

\section{Introduction}  \label{sec:introduction}

Let $\cB$ be a $\sigma$-algebra of subsets of a general sample space
$\Omega$.  Let $\cP$ be a convex class of probability measures on
$(\Omega, \cB)$.  A {\em scoring rule}\/ is any extended real-valued
function $\myS$ on $\cP \times \Omega$ such that
\[
\myS(P,Q) = \int \myS(P,\omega) \dd Q(\omega)
\]
is well-defined for $P, Q \in \cP$.  The scoring rule $\myS$ is {\em
  proper}\/ relative to $\cP$ if
\begin{equation}  \label{eq:proper}
\myS(Q,Q) \leq \myS(P,Q) \quad \mbox{for all} \quad P, Q \in \cP.
\end{equation}
In words, we take scoring rules to be negatively oriented penalties
that a forecaster wishes to minimize.  If she believes that a future
quantity or event has distribution $Q$, and the penalty for quoting
the predictive distribution $P$ when $\omega$ realizes is
$\myS(P,\omega)$, then~\eqref{eq:proper} implies that quoting $P = Q$
is an optimal strategy in expectation.  The scoring rule is {\em
  strictly proper} if~\eqref{eq:proper} holds with equality only if $P
= Q$.  For recent reviews of the theory and application of proper
scoring rules see \citet{Dawid2007}, \citet{GneitRaft2007},
\citet{DawidMusio2014}, and \citet{GneitKatz2014}.

The intent of this note is to draw attention to the simple fact that,
subject to customary regularity conditions, any scoring rule can be
{\em properized}, in the sense that it can be modified in a
straightforward way to yield a proper scoring rule, so that truth
telling becomes an optimal strategy.  Implicitly, this construction
has recently been used by various authors in various types of
applications; see, e.g., \citet{Diks2011}, \citet{Christetal2014} and
\citet{HolzKlar2018}.

\begin{theorem}[properization]  \label{th:properization}
Let\/ $\myS$ be a scoring rule.  Suppose that for every\/ $P \in \cP$
there is a probability distribution\/ $P^* \in \cP$ such that
\begin{equation}  \label{eq:Bayes}
\myS(P^*,P) \leq \myS(Q,P) \quad \mbox{for all} \quad Q \in \cP.
\end{equation}
Then the function
\begin{equation}  \label{eq:S*}
\myS^* : \cP \times \Omega \to \Real, \quad
(P,\omega) \mapsto \myS^*(P,\omega) = \myS(P^*,\omega),
\end{equation}
is a proper scoring rule.
\end{theorem}

Here and in what follows we denote the real line by $\real$ and the
extended real line by $\Real := \real \cup \{ - \infty, \infty \}$.
Any probability measure $P^*$ with the property~\eqref{eq:Bayes} is
commonly called {\em Bayes act}; for the existence of Bayes acts,
see Section~\ref{sec:existence}.  In case there are multiple
minimizers of the expected score $Q \mapsto \myS(Q, P)$, the function
$\myS^*$ is well-defined by using a mapping $P \mapsto P^*$ that
chooses a $P^*$ out of the set of minimizers.  If $\myS$ is proper and
$P^* = P$, then $\myS^* = \myS$, so the proper scoring rules are fixed
points under the properization operator.

Importantly, Theorem~\ref{th:properization} is a special case of a
general and powerful construction studied in detail by
\citet{GruenDawid2004} and \citet{Dawid2007}.  Specifically, given
some action space $\cA$ and a loss function $\myL : \cA \times \Omega
\to \Real$, suppose that for each $P \in \cP$ there is a Bayes act
$a_P \in \cA$, such that
\[
\int \myL(a_P,\omega) \dd P(\omega) \leq \int \myL(a, \omega) \dd P(\omega)
\quad \mbox{for all} \quad a \in \cA.
\]
Then the function
\[
\myS^* : \cP \times \Omega \to \Real, \quad
(P,\omega) \mapsto \myS^*(P,\omega) = \myL(a_P,\omega),
\]
is a proper scoring rule.  Note the natural connection to decision-
and utility-based scoring approaches, where the quality of a forecast
is judged by the monetary utility of the induced acts and decisions
\citep{GranPesa2000, GranMach2006, Ehmetal2016}.

In the remainder of the paper we focus on the above special case in
which the action domain $\cA$ is the class $\cP$.  In
Section~\ref{sec:examples} we identify scattered results in the
literature as prominent special cases of properization
(Examples~\ref{ex:binary}--\ref{ex:PMC}), and we use
Theorem~\ref{th:properization} to construct new proper scoring rules
from improper ones (Examples~\ref{ex:CRPS}--\ref{ex:probScore}).
Section~\ref{sec:existence} gives sufficient conditions for the
existence of Bayes acts and Section~\ref{sec:discussion} contains a
brief discussion. All proofs and technical details are moved to the
\hyperref[sec:appendix]{Appendix}.

\section{Examples}  \label{sec:examples}

This section starts with an example in which we review the ubiquitous
misclassification error from the perspective of properization. We go
on to demonstrate how Theorem~\ref{th:properization} has been used
implicitly to construct proper scoring rules in econometric,
meteorological, and statistical strands of literature.  The notion of
properization simplifies and shortens the respective proofs of
propriety, makes them much more transparent, and puts the scattered
examples into a unifying and principled joint framework.  Further
examples show other facets of properization: The scoring rules
constructed in Example~\ref{ex:CRPS} are original, and the discussion
in Example~\ref{ex:CRPS2} illustrates a connection to the practical
problem of the treatment of observational uncertainty in forecast
evaluation. Finally, Example~\ref{ex:probScore} includes an instance
of a situation in which properization fails.

\begin{example}  \label{ex:binary}
Consider probability forecasts of a binary event, where $\Omega =
\lbrace 0,1 \rbrace$ and $\cP$ is the class of the Bernoulli measures.
We identify any $P \in \cP$ with the probability $p = P(\lbrace 1
\rbrace) \in [0,1]$ and consider the scoring rules
\[
\myS_1(P,\omega) := 1 - p \omega - (1-p)(1-\omega)
\quad \text{ and } \quad
\myS_2(P,\omega) := |p-\omega|.
\]
The scoring rule $\myS_1$ corresponds to the mean probability rate
(MPR) in machine learning \citep[p.~30]{Ferrietal2009}.  The scoring
rule $\myS_2$ was first considered by \citet{Dawid1986}. It agrees
with the special case $c_1 = c_2$ in Section~4.2 of
\citet{Parry2016} and corresponds to the mean absolute error (MAE)
as discussed by \citet[p.~30]{Ferrietal2009}.\footnote{As noted by
  \citet{Parry2016}, the improper score $\myS_2$ shares its (concave)
  expected score function $P \mapsto \myS_2(P,P)$ with the proper
  Brier score.  This illustrates the importance of the second
  condition in Theorem 1 of \citet{GneitRaft2007}: For a scoring rule
  $\myS$ the (strict) concavity of the expected score function $ G(P)
  := \myS(P,P)$ is equivalent to the (strict) propriety of $\myS$ only
  if, furthermore, $- \myS(P,\cdot)$ is a subtangent of $- G$ at $P$.}
Both $\myS_1$ and $\myS_2$ are improper with common Bayes act
\[
\textstyle
p^* = \one{p \geq \frac{1}{2}} \in \lbrace 0,1 \rbrace,
\]
and with the same properized score given by the zero-one rule
\[
\myS^*(P,\omega) = \left\lbrace
\begin{array}{cl}
0, & p^* = \omega, \\
1, & \text{otherwise.}
\end{array}
\right.
\]
A case-averaged zero-one score is typically referred to as {\em
  misclassification rate}\/ or {\em misclassification error};
undoubtedly, this is the most popular and most frequently used
performance measure in binary classification.  While the scoring rule
$\myS^*$ is proper it fails to be strictly proper
(\citealp[Example 4]{GneitRaft2007};
\citealp[Section 4.3]{Parry2016}). Consequently, misclassification
error has serious limitations as
a performance measure, as persuasively argued by
\citet[p.~258]{Harrell2015}, among others.  Nevertheless, the scoring
rule $\myS^*$ is proper, contrary to recent claims of impropriety in
the blogosphere.\footnote{See, e.g.,
  \url{http://www.fharrell.com/post/class-damage/} and
  \url{http://www.fharrell.com/post/classification/}.}
\end{example}

For the remainder of the section, let $\Omega = \real$ and let $\cB$
be the Borel $\sigma$-algebra.  We let $\cL$ be the class of Borel
measures $P$ with a Lebesgue density, $p$.  Furthermore, we write
$\cP_k$ for the measures with finite $k$-th moment and $\cP_k^+$ for
the subclasses when Dirac measures are excluded.  Whenever it
simplifies notation, we identify $P$ with its cumulative distribution
function $x \mapsto P((-\infty, x])$.

\begin{example}  \label{ex:weighted}
Let $\myS_0$ be a proper scoring rule on some subclass $\cP$ of $\cL$
and let\/ $w$ be a nonnegative weight function such that\/ $0 < \int
w(z) \, p(z) \dd z < \infty$ for $p \in \cP$.  Let
\[
\myS : \cP \times \real \to \real, \quad (P,y) \mapsto \myS(P,y) = w(y) \, \myS_0(P,y);
\]
this score is improper unless the weight function is constant.
Indeed, by Theorem~1 of \citet{GneitRanjan2011}, the Bayes act $P^*$
under $\myS$ has density
\[
p^*(y) = \frac{w(y) \, p(y)}{\int w(z) \, p(z) \dd z}.
\]
From this we see that the key statement in Theorem~1 of
\citet{HolzKlar2018} constitutes a special case of Theorem
\ref{th:properization}. In the further special case in which $\myS_0$
is the logarithmic score, the properized score~\eqref{eq:S*} recovers
the conditional likelihood score of \cite{Diks2011} up to equivalence,
as noted in Example 1 of \citet{HolzKlar2018}.  For analogous results
for consistent scoring functions see Theorem~5 of \citet{Gneit2011}
and Example~2 of \citet{HolzKlar2018}.
\end{example}

\begin{example}  \label{ex:error.spread}
For a probability measure $P \in \cP_4$, let $\mu_P$, $\sigma^2_P$,
and $\gamma_P$ denote its mean, variance, and centered third moment.
Let
\[
\myS(P,y) = \left( \sigma_P^2 - (y - \mu_P)^2 \right)^2
\]
be the `trial score' in equation~(16) of \citet{Christetal2014}. As
\citet{Christetal2014} show in their Appendix A, any Bayes act
$P^*$ under $\myS$ has mean $ \mu_P + \tfrac{1}{2}
\tfrac{\gamma_P}{\sigma_P^2}$ and variance
\[
\sigma_P^2 \left( 1 + \frac{1}{4} \frac{\gamma_P^2}{\sigma_P^6} \right),
\]
so properization yields the spread-error score,
\[
\myS^*(P,y) = \left( \sigma_P^2 - \left( y - \mu_P \right)^2
+ \left( y - \mu_P \right) \frac{\gamma_P}{\sigma_P^2} \right)^2 ,
\]
which is proper relative to the class $\cP_4^+$.  Hence the
construction of the spread-error score in \citet{Christetal2014}
constitutes another special case of Theorem~\ref{th:properization}.
\end{example}

\begin{example}  \label{ex:PMC}
The predictive model choice criterion of \citet{LaudIbra1995} and
\citet{GelfGhosh1998} uses the scoring rule $\myS(P,y) = \left( y -
\mu_P \right)^2 + \sigma_P^2$, where $\mu_P$ and $\sigma_P^2$ denote
the mean and the variance of a distribution $P \in \cP_2$,
respectively.  As pointed out by \citet{GneitRaft2007}, this score
fails to be proper.  Specifically, any Bayes act $P^*$ under $\myS$
has mean $ \mu_P$ and vanishing variance, so properization yields the
ubiquitous squared error, $\myS^*(P,y) = \left( y - \mu_P \right)^2$.
\end{example}

The original scoring rules of Examples~\ref{ex:error.spread}
and~\ref{ex:PMC} can be interpreted as functions $ \myS : \cA \times
\Omega \rightarrow \real$ in the Bayes act setting of
\citet{GruenDawid2004} and \citet{Dawid2007}, where the action space
$\cA$ is given by $\real \times [0, \infty)$.  Hence, the
  properization method can be interpreted as an application of
  Theorem~3 of \citet{Gneit2011} to consistent scoring functions for
  elicitable two-dimensional functionals, as discussed by
  \citet{FissZieg2016}.

Detailed arguments and calculations for the subsequent examples are
deferred to the \hyperref[sec:appendix]{Appendix}.

\begin{example}  \label{ex:CRPS}
For $\alpha > 0$ consider the scoring rule
\[
\myS_\alpha (P,y) = \int \left| P(x) - \one{y   \le   x} \right|^\alpha \dd x,
\]
where $P$ is identified with its cumulative distribution function
(CDF).  For $\alpha = 2$ this is the well known proper continuous
ranked probability score (CRPS), as reviewed in Section 4.2 of
\citet{GneitRaft2007}.  For $\alpha = 1$ the score $\myS_\alpha$ was
proposed by \citet{Muelleretal2005}, and \citet{ZamoNaveau2018} show
in their Appendix A that for discrete distributions every Dirac
measure in a median of $P$ is a Bayes act.  The same holds true for
general distributions and for all $\alpha \in (0,1]$.  If $\alpha >
1$, the Bayes act $P^*$ under $\myS_\alpha$ is given by
\begin{equation}  \label{eq:CRPS_alpha_P*}
P^* (x) = \left(
1 + \left( \frac{1 - P(x)}{P(x)} \right)^{1/(\alpha - 1)} \right)^{-1} \one{P(x) > 0},
\end{equation}
and all in all we see that properization of $\myS_\alpha$ works for
any $\alpha > 0$.

Moreover, in the case $\alpha >1$ the mapping $P \mapsto P^*$ is even
injective. Consequently, if the class $\cP$ is such that $P^* \in \cP$
and $\myS_\alpha(P^*,P)$ is finite for $P \in \cP$, the properized
score~\eqref{eq:S*} is even strictly proper relative to $\cP$.  If
$\alpha \in (1,2]$, this can be ensured by restricting $\myS_\alpha$
to the class $\cP_1$. For $\alpha > 2$ the class $\cP_{\mathrm c}$ of
the Borel measures with compact support is a suitable choice.
\end{example}

\begin{example}  \label{ex:CRPS2}
\citet[p.~58]{FriedThor2012} propose a modification of the CRPS that
aims to account for observational error in forecast evaluation.
Specifically, they consider the scoring rule
\[
\myS_\Phi (P, y) = \int \left| P(x) - \Phi(x-y) \right|^2 \dd x,
\]
where $\Phi \in \cP_1^+$ represents additive observation error.  This
scoring rule fails to be proper, as for probability measures $P, Q \in
\cP_1$ we have
\begin{equation}  \label{eq:NoisyCRPS}
\myS_\Phi (P, Q) = \crps (P, Q * \Phi) - \crps (\Phi, \Phi),
\end{equation}
where $*$ denotes the convolution operator.  Due to the strict
propriety of the CRPS relative to the class $\cP_1$, the unique Bayes
act under $\myS_\Phi$ is given by $P^* = P * \Phi$.
Theorem~\ref{th:properization} now gives the scoring rule
$\myS(P,y) := \myS_\Phi (P^*, y)$, which is proper relative to
$\cP_1$.

In order to account for noisy observational data in forecast
evaluation, equation~\eqref{eq:NoisyCRPS} suggests using the scoring
rule $\myS (P,y) := \crps (P^*, y)$ if the noise is independent,
additive, and has distribution $\Phi$. This corresponds to predicting
hypothetical true values, to which noise is added before they are
compared to observations. The drawbacks of this approach and
alternative techniques are discussed by~\citet{Ferro2017}. The
associated issues in forecast evaluation remain challenges to the
scientific community at large; see, e.g., \cite{Ebertetal2013}
and \cite{Ferro2017}.

\end{example}

\begin{example}  \label{ex:probScore}
Let $\myS$ be a scoring rule, and let $\Phi \in \cL$ be a distribution
with Lebesgue density $\varphi$.  Suppose $\cP$ is a class of
distributions such that $P * \Phi \in \cP$ for $P \in \cP$.  For $P
\in \cP$ define
\[
\myS^\varphi (P, y) := \int \varphi (x-y) \, \myS (P, x) \dd x,
\]
which is again a scoring rule.  If $\myS$ is proper, a Bayes act
under $\myS^\varphi$ is given by $P^* = P * \Phi$, since $\myS^\varphi
(P, Q) = \myS(P, Q * \Phi )$ for $Q \in \cP$, and if $\myS$ is
strictly proper, the Bayes act is unique. Properization now gives the
proper scoring rule $\myS(P,y) := \myS^\varphi (P^*, y)$. An
interesting special case emerges when substituting the CRPS for
$\myS$. This leads to
\begin{equation}  \label{eq:NoisyCRPS2}
\crps^\varphi (P,y) = \myS_\Phi (P,y ) + \crps (\Phi, \Phi),
\end{equation}
where $\myS_\Phi$ is the scoring rule in the previous example.  For
another special case, let $c > 0$ and $P \in \cL$, to yield
\[
\mathrm{PS}_c(P, y) := - \int_{y-c}^{y+c} p(x) \dd x,
\]
which recovers the {\em probability score}\/ of
\citet{Wilsonetal1999}.  We have that $\mathrm{PS}_c = 2c \,
\mathrm{LinS}^{\varphi_c}$, where $ \mathrm{LinS}(P,y) := - p(y)$ is
the improper linear score and $\varphi_c$ is a uniform density on
$[-c,c]$.  Properization is not feasible relative to sufficiently rich
classes $\cP$, as Bayes acts fail to exist under both the linear
score and the probability score.  For details, see the
\hyperref[sec:appendix]{Appendix}.
\end{example}

\section{Existence of Bayes acts}   \label{sec:existence}

In Example~\ref{ex:probScore} we presented a scoring rule that cannot
be properized, due to the non-existence of Bayes acts. This section
addresses the question under which conditions on $\myS$ and $\cP$ a
minimum of the expected score function exists. To illustrate the
ideas, we start with a further example.

\begin{example}  \label{ex:normalized}
Using the notation of Example~\ref{ex:error.spread}, consider the
normalized squared error,
\begin{equation*}
\myS(P,y) = \frac{(y - \mu_P)^2}{\sigma_P^2},
\end{equation*}
as a scoring rule on the classes $\cP_{2,m}$ of the Borel measures
with variance at most $m$, and $\cP_2 = \cup_{m > 0} \cP_{2,m}$,
respectively.  Relative to $\cP_{2,m}$ any Bayes act $P^*$ under
$\myS$ has mean $\mu_P$ and variance $m$, so properization yields
(non-normalized) squared error up to equivalence.  Relative to $\cP_2$
however, there is no Bayes act, since increasing the variance will
always lead to a smaller expected score.
\end{example}

We now turn to a general perspective and discuss sufficient
conditions for the existence of Bayes acts. At first, consider a
finite probability space $\Omega = \lbrace \omega_1, \ldots, \omega_k
\rbrace$. In this situation, geometrical arguments yield  sufficient
conditions. In particular, a Bayes act under $\myS$ exists if the
\textit{risk set}
\[
\mathcal{S} := \lbrace (x_1, \ldots, x_k) \mid \exists \, P \in
\mathcal{P} : x_j = \myS (P, \omega_j) , \, j=1,\ldots, k \rbrace
\subset \mathbb{R}^k
\]
is closed from below and bounded from below; see Theorem~1 in
Chapter~2.5 of \citet{Ferguson1967}. Extending this result to a general
sample space $\Omega$ is non-trivial since in this case $\mathcal{S}$
can be a subset of an infinite-dimensional vector space. In the
following we employ well-known concepts of functional analysis in order
to discuss two possible extensions. All proofs are deferred to the
\hyperref[sec:appendix]{Appendix}.

Let $\cP$ be a set of probability measures on a general probability
space $\Omega$ and let $\cA$ be a topological space. We return to the
setting of Section~\ref{sec:introduction} and consider functions
$\myS: \cA \times \Omega \rightarrow \real$. This makes the results
more general and easier to apply in situations where the scoring rule
depends on $P$ only via some finite number of parameters. Concerning
the latter point, note that the normalized squared error of
Example~\ref{ex:normalized} can be written as a composition of the
mapping $P \mapsto (\mu_P, \sigma^2_P)$ and the function $s(x_1,x_2,y)
:= (y -x_1)^2 / x_2$, with $s$ being defined on $\real \times (0,
\infty) \times \real$. Consequently, the expected normalized squared
error attains its minimum if the expected score of $s$ attains its
minimum. Note that such a decomposition of the scoring rule is
possible for Examples~\ref{ex:error.spread} and~\ref{ex:PMC} as well,
as alluded to in the comments that succeed these examples.

We impose the following integrability assumption on $\myS$.

\begin{definition}
The mapping $\myS: \cA \times \Omega \rightarrow \real$ is
\textit{uniformly bounded from below} if there exists a function
$g: \Omega \rightarrow \real$ which is integrable with respect to any
$P \in \cP$ and such that $\myS (a,\cdot) \geq g$ holds for all
$a \in \cA$.
\end{definition}

Our first result is similar to Theorem~2 in Chapter~2.9 of
\citet{Ferguson1967}, which proves the existence of minimax decision
rules.

\begin{theorem}  \label{th:existence1}
Suppose $\myS$ is lower semicontinuous in its first component and
uniformly bounded from below. If\/ $\cA$ is compact, then the function
$a \mapsto \myS(a,P)$ attains its minimum for any $P \in \cP$.
\end{theorem}

This theorem can be used to prove the existence of a Bayes act
for a given scoring rule. However, it is not applicable to
Example~\ref{ex:normalized}. To see this, recall the decomposition of
$\myS$ mentioned above and note that restricting $\myS$ to $\cP_{2,m}$
corresponds to restricting $s$ to $\real \times (0,m]$. The latter set
is not a compact space and neither is its closure. Consequently, we aim
to dispense with the compactness assumption used in
Theorem~\ref{th:existence1}.

To do so, we need additional concepts from functional analysis. Let
$\cX$ be a real normed vector space.  Recall that a function $h : \cX
\rightarrow \real$ is called \textit{coercive} if for any sequence
$(x_n)_{n \in \mathbb{N}} \subset \cX$ the implication
\begin{equation*}
\lim_{n \rightarrow \infty} \Vert x_n \Vert = \infty  \quad \Rightarrow \quad \lim_{n \rightarrow \infty} h(x_n) = \infty
\end{equation*}
holds true, see, e.g.~Definition~III.5.7 in~\citet{Werner2018}. By
\textit{weak topology} on $\cX$, we mean the weakest topology such
that all real-valued linear mappings on $\cX$ are continuous; see,
e.g.~Chapters~2.13 and~6.5 in~\citet{AlipBord2006}. The space $\cX$ is
called a \textit{reflexive Banach space} if it is complete and the
canonical embedding of $\cX$ into its bidual space is surjective; see,
e.g.~Chapter~III.3 in~\citet{Werner2018} or Chapter~6.3
in~\citet{AlipBord2006}. Combining these concepts, we obtain a
complement to Theorem~\ref{th:existence1}.

\begin{theorem}  \label{th:existence2}
Let $\cA$ be a weakly closed subset of a reflexive Banach space.
Moreover, suppose $\myS$ is weakly lower semicontinuous in its first
component and uniformly bounded from below. If the function $a \mapsto
\myS(a,P)$ is coercive, then it attains its minimum.
\end{theorem}

This result yields the existence of Bayes acts as long as the
integrated scoring rule is coercive for any $P \in \cP$, where $\cP$
is a reflexive Banach space. To conclude this section, we connect
Theorem~\ref{th:existence2} to Example~\ref{ex:normalized}: The
function $s(\cdot, \cdot, y)$ from the decomposition of $\myS$
mentioned above is defined on $\real \times (0, \infty)$, which is a
subset of the reflexive Banach space $\real^2$. Moreover, $s$ is
bounded from below by zero and continuous in its first component. As
mentioned above, restricting the class $\cP_2$ to $\cP_{2,m}$
corresponds to restricting the domain of $s$ to $\real \times (0, m]$
and in this situation, integrating $s$ with respect to $y$ gives a
coercive function. Consequently, Theorem~\ref{th:existence2} can be
used to show that $\myS$ can be properized if restricted to
$\cP_{2,m}$.

\section{Discussion}   \label{sec:discussion}

In this article we have introduced the concept of properization, which
is rooted in the Bayes act construction of \citet{GruenDawid2004} and
\citet{Dawid2007}, and we have drawn attention to its widespread
implicit use in the transdisciplinary literature on proper scoring
rules, where our unified approach yields simplified, shorter, and
considerably more instructive and transparent proofs than extant
methods.  Moreover, using new examples, we have demonstrated the power
of the properization approach in the creation of new proper scoring
rules from existing improper ones.

Since the central element in the construction of a properized score is
a Bayes act, we have discussed conditions on the scoring rule $\myS$
and the class $\cP$ that guarantee its existence.  Undoubtedly, there
are alternative paths to existence results in the spirit of
Theorems~\ref{th:existence1} and~\ref{th:existence2}, and the
derivation of sufficient conditions in alternative situations is an
interesting open problem.  Furthermore, we have not explored necessary
conditions for the existence of Bayes acts in this work. Their
derivation and the refinement of sufficient conditions on $\myS$ and
$\cP$ remain challenges that we leave for future work.

\section*{Appendix: Proofs} \label{sec:appendix}
\addcontentsline{toc}{section}{Appendix: Proofs}

Here we present detailed arguments for the technical claims in
Examples~\ref{ex:CRPS}, \ref{ex:CRPS2}, and~\ref{ex:probScore}
as well as the proofs of Theorems~\ref{th:existence1}
and~\ref{th:existence2}.

\subsection*{Details for Example~\ref{ex:CRPS}}
\addcontentsline{toc}{subsection}{Details for Example~\ref{ex:CRPS}}

We fix some distribution $P$ and start with the case $\alpha > 1$. An
application of Fubini's theorem gives
\begin{equation} \label{eq:Ex4SQP}
\myS_\alpha (Q,P) = \int \int \vert Q(x) - \one{y \leq x} \vert^\alpha  \dd P(y) \dd x .
\end{equation}
Given $x \in \real$ we seek the value $Q(x) \in [0,1]$ that minimizes
the inner integral in~\eqref{eq:Ex4SQP}.  If $x$ is such that $P(x)
\in \lbrace 0, 1 \rbrace$, the equality $\one{y \leq x} = P(x)$ holds
for $P$-almost all $y$, hence $Q(x)= P(x)$ is the unique minimizer.
If $x$ satisfies $P(x) \in (0,1)$, define the function
\[
g_{x,P}(q) := \int \vert q - \one{y \leq x} \vert^\alpha \dd P(y)
= (1-P(x)) q^\alpha + P(x) (1-q)^\alpha,
\]
which is strictly convex in $q \in (0,1)$ with derivative
\[
g_{x,P}'(q) = \alpha (1- P(x)) q^{\alpha - 1} - \alpha P(x) (1- q)^{\alpha - 1}
\]
and a unique minimum at $q = q_{x,P}^* \in (0,1)$.  As a consequence,
the minimizing value $Q(x)$ is given by
\begin{equation*}
Q(x) = q_{x,P}^* = \left( 1 + \left( \frac{1-P(x)}{P(x)} \right)^{1/(\alpha - 1)} \right)^{-1}.
\end{equation*}
The function $Q$ defined by the minimizers $Q(x)$, $x \in \real$ is a
minimizer of $\myS_\alpha ( \cdot ,P)$ and if $\myS_\alpha (Q, P)$ is
finite, it is unique Lebesgue almost surely.  Since $\alpha >1$ the
function $Q$ has the properties of a distribution function and hence
$P^*$ defined by~\eqref{eq:CRPS_alpha_P*} is a Bayes act for
$P$. Moreover, equation~\eqref{eq:CRPS_alpha_P*} shows that the
relation between $P$ and $P^*$ is one-to-one.

It remains to be checked under which conditions the properization of
$\myS_\alpha$ is not only proper but strictly proper.  The
representation~\eqref{eq:CRPS_alpha_P*} along with two Taylor
expansions implies that $P^*$ behaves like $P^{1/(\alpha -1)}$ in the
tails. This has two consequences. At first, the above arguments show
that for $\myS_\alpha (P^*, P)$ to be finite $x \mapsto g_{x,P}
(P^*(x))$ has to be integrable with respect to Lebesgue measure.
Hence, the tail behavior of $P^*$ and the inequality $\alpha/(\alpha
- 1) > 1$ for $\alpha > 1$ show that $\myS_\alpha (P^*, P)$ is finite
for $P \in \cP_1$.  Second, $P^*$ has a lighter tail than $P$ for
$\alpha \in (1,2)$ and a heavier tail for $\alpha > 2$.  In the latter
case $P \in \cP_1$ does not necessarily imply $P^* \in \cP_1$.  Hence,
without additional assumptions, strict propriety of the properized
score~\eqref{eq:S*} can only be ensured relative to $\cP_\mathrm{c}$
for $\alpha > 2$ and relative to the class $\cP_1$ for $\alpha \in (1,
2]$.

We now turn to $\alpha \in (0,1)$.  In this case, the function
$g_{x,P}$ is strictly concave, and its unique minimum is at $q = 0$
for $P(x) < \frac{1}{2}$ and at $q = 1$ for $P(x) > \frac{1}{2}$.  If
$P(x) = \frac{1}{2}$, then both $0$ and $1$ are minima.  Arguing as
above, every Bayes act $P^*$ is a Dirac measure in a median of $P$.

Finally, $\alpha = 1$ implies that $g_{x,P}$ is linear, thus, as for
$\alpha \in (0,1)$, every Dirac measure in a median of $P$ is a Bayes
rule.  The only difference to the case $\alpha \in (0,1)$ is that if
there is more than one median, there are Bayes acts other than Dirac
measures, since $g_{x,P}$ is constant for all $x$ satisfying $P(x) =
\frac{1}{2}$.

\subsection*{Details for Example~\ref{ex:CRPS2}}
\addcontentsline{toc}{subsection}{Details for Example~\ref{ex:CRPS2}}

Let $P, Q$ and $\Phi$ be distribution functions.  By the
definition of the convolution operator
\begin{equation*}
\int \one{y \leq x}  \dd (Q * \Phi) (y) = \int \Phi (x-y)  \dd Q(y)
\end{equation*}
holds for $x \in \real$.  Using this identity and Fubini's theorem
leads to
\begin{align*}
\myS_\Phi (P,Q)
& = \int \! \int \left( P(x)^2 - 2 P(x) \Phi(x-y) + \Phi(x-y)^2 \right) \dd Q(y) \dd x \\
& = \int \! \int \left( P(x)^2 - 2 P(x) \one{y \leq x} + \one{y \leq x} \right) \dd (Q * \Phi)(y) \dd x \\
& \quad + \int \! \int \Phi(x-y) (\Phi(x-y) - 1) \dd Q(y) \dd x \\
& = \int \! \int (P(x) - \one{y \leq x})^2  \dd x \dd (Q * \Phi)(y)
  - \int \Phi (x) (1- \Phi (x))  \dd x,
\end{align*}
which verifies equality in~\eqref{eq:NoisyCRPS}.  Moreover, the strict
propriety of the CRPS relative to the class $\cP_1$ gives $\myS_\Phi
(P, Q) < \infty$ for $P, Q, \Phi \in \cP_1$, thereby demonstrating
that the Bayes act is unique in this situation.

\subsection*{Details for Example~\ref{ex:probScore}}
\addcontentsline{toc}{subsection}{Details for Example~\ref{ex:probScore}}

For distributions $P, Q \in \cP$ and $c > 0$, the Fubini-Tonelli
theorem and the definition of the convolution operator give
\begin{align*}
\myS^\varphi (P,Q) &= - \int \int \varphi (x-y) \myS(P,x) \dd Q(y) \dd x \\
&= \int \int \varphi (x-y) \dd Q(y) \, \myS(P,x) \dd x = \myS (P, Q * \Phi),
\end{align*}
so the stated (unique) Bayes act under $\myS^\varphi$ follows from
the (strict) propriety of $\myS$.  Proceeding as in the details for
Example~\ref{ex:CRPS2} we verify identity~\eqref{eq:NoisyCRPS2}.

For $P \in \cL$ the same calculations as above show that the
probability score satisfies
\[
\mathrm{PS}_c(P,Q) =
2c \int \frac{Q(x + c) - Q(x - c)}{2c} \: \mathrm{LinS}(P,x) \dd x,
\]
where $\mathrm{LinS}(P,y) = - p(y)$ is the linear score.
Consequently, to demonstrate that Theorem~\ref{th:properization} is
neither applicable to $\mathrm{PS}_c$ nor to $\mathrm{LinS}$, it
suffices to show that there is a distribution $Q$ such that $P \mapsto
\mathrm{LinS}(P,Q)$ does not have a minimizer.  We use an argument
that generalizes the construction in Section 4.1 of
\citet{GneitRaft2007} who show that $\mathrm{LinS}$ is improper.  Let
$q$ be a density, symmetric around zero and strictly increasing on
$(-\infty, 0)$.  Let $\epsilon > 0$ and define the interval $I_k :=
((2k - 1) \epsilon, (2k + 1) \epsilon]$ for $k \in \mathbb{Z}$.
  Suppose $p$ is a density with positive mass on some interval $I_k$
  for $k \neq 0$.  Due to the properties of $q$, the score
  $\mathrm{LinS}(P,Q)$ can be reduced by substituting the density
defined by
\[
\tilde{p}(x) := p(x) - \one{x \in I_k} \, p(x) + \one{x + 2k \epsilon
  \in I_k} \, p(x + 2k \epsilon)
\]
for $p$, i.e., by shifting all probability mass from $I_k$ to the
modal interval $I_0$.  Repeating this argument for any $\epsilon > 0$
shows that no density $p$ can be a minimizer of the expected score
$\mathrm{LinS}(P,Q)$.  Note that the assumptions on $q$ are stronger
than necessary in order to facilitate the argument.  They can be
relaxed at the cost of a more elaborate proof.

\subsection*{Proof of Theorem~\ref{th:existence1}}
\addcontentsline{toc}{subsection}{Proof of Theorem~\ref{th:existence1}}

Let $(a_n)_{n \in \mathbb{N}} \subset \cA$ be a sequence with $ a :=
\lim_{n \rightarrow \infty} a_n$. Since $\myS$ is lower semicontinuous
in its first component and uniformly bounded from below by $g$, Fatou's
lemma gives
\begin{equation*}
\liminf_{n \rightarrow \infty} \int \myS (a_n, \omega) \dd P(\omega)
\geq \int \liminf_{n \rightarrow \infty} \myS (a_n,\omega) \dd
P(\omega) \geq \myS(a,P)
\end{equation*}
for any $P \in \cP$. Hence, $a \mapsto \myS(a, P)$ is a lower
semicontinuous function for any $P \in \cP$ and due to the assumed
compactness of $\cA$, the result now follows from Theorem~2.43
in~\citet{AlipBord2006}.

\subsection*{Proof of Theorem~\ref{th:existence2}}
\addcontentsline{toc}{subsection}{Proof of Theorem~\ref{th:existence2}}

The same arguments as in the proof of Theorem~\ref{th:existence1}
show that $a \mapsto \myS(a, P)$ is a weakly lower semicontinuous
function for any $P \in \cP$. If $P \in \cP$ is such that this function
is also coercive, then proceeding as in the proof of Satz~III.5.8
in~\citet{Werner2018} gives a weakly convergent sequence
$(a_n)_{n \in \mathbb{N}} \subset \cA$ with $\lim_{n \rightarrow
\infty} \myS( a_n, P) = \inf_{a \in \cA} \myS(a, P)$. Since $\cA$ is
weakly closed by assumption, it contains the weak limit  $a^*$ of the
sequence $(a_n)_{n \in \mathbb{N}}$ and hence weak lower semicontinuity
implies that $a \mapsto \myS(a, P)$ attains its minimum at $a^* \in
\cA$.

\section*{Acknowledgments}
\addcontentsline{toc}{section}{Acknowledgments}

Tilmann Gneiting is grateful for funding by the Klaus Tschira
Foundation and by the European Union Seventh Framework Programme under
grant agreement 290976.  Part of his research leading to these results
has been done within subproject C7 ``Statistical postprocessing and
stochastic physics for ensemble predictions'' of the Transregional
Collaborative Research Center SFB / TRR 165 ``Waves to Weather''
(\url{www.wavestoweather.de}) funded by the German Research Foundation
(DFG).  Jonas Brehmer gratefully acknowledges support by DFG through
Research Training Group RTG 1953.  We thank Tobias Fissler and Matthew
Parry for instructive discussions.