---
abstract: 'We study inequalities between general integral moduli of continuity of a function and the tail integral of its Fourier transform. We obtain, in particular, a refinement of a result due to D. B. H. Cline [@Cl] (Theorem \[thm1\] below). We note that our approach does not use a regularly varying comparison function as in [@Cl]. A corollary of Theorem 1.1 deals with the equivalence of the two-sided estimates on the modulus of continuity on one hand, and on the tail of the Fourier transform, on the other (Corollary \[CH3\_s3.lem.1\]). This corollary is applied in the proof of the violation of the so-called entropic area law for a critical system of free fermions in [@GiKl; @Gi2].'
address: 'Department of Mathematics, University of Rochester, Hylan Building, Rochester, NY 14627'
author:
- Dimitri Gioev
date: 'December 15, 2006 and, in revised form, June 15, 2007.'
title: 'Moduli of continuity and average decay of Fourier transforms: two-sided estimates'
---

[The author was supported in part by NSF grant INT–0204308 U.S.–Sweden Collaborative Workshop on PDE’s and Spectral Theory, the Swedish foundation STINT grant PD2001–128 and NSF grant DMS–0550649. The author would like to thank the University of Pennsylvania and the Courant Institute for financial support and hospitality. The author is grateful to Percy Deift for his valuable suggestions which have helped to improve the presentation. ]{}

Introduction and statement of the main results {#s0}
==============================================

A result of this paper (Corollary \[CH3\_s3.lem.1\] below) is applied in the proof of the violation of the so-called entropic area law for a critical system of free fermions, see [@GiKl (6) et seq.], [@GiKl Section on Fractal Boundaries] and [@Gi2 Lemma 2.10]. Corollary \[CH3\_s3.lem.1\] follows from more general results of this paper (Theorems \[thm1\], \[lem4.1\]) which are of independent interest.

It is well-known that the behavior of a modulus of continuity $\omega[f](h)$ of a function $f$ for $|h|$ small is related to the behavior of the Fourier transform $\hat{f}(\xi)$ of $f$ for $|\xi|$ large (precise definitions are given in et seq. below), see e.g. [@St Proposition 5.3.4], [@T Theorem 85]. The main object of our study are inequalities between general averaged moduli of continuity () of $L^p$ functions (defined in  below) and tails of their Fourier transforms (). In [@Cl] several results relevant for our purposes were obtained. Theorem \[thm1\] below gives a lower estimate for a general $L^p$ , $1\leq{}p\leq2$, in terms of the modified tail integral of the improving one of the results in [@Cl] (as in [@Cl], we distinguish between the true and the modified tail integral, as defined in and below). Corollary \[thm2\] gives a two-sided estimate for the in terms of the modified tail integral of the in the case $p=2$. In applications it might be desirable to use the true tail instead of the modified tail that arises naturally in the mentioned inequalities. Theorem \[lem4.1\] gives the best possible power-scale description of the relationship between the true and the modified tails (see Remark \[remstar\]).

Before stating our results we need to introduce some notation and recall two results in [@Cl]. Let $d\in\NN$ and denote by $\|\cdot\|_{p,\RD}$ the standard norm in $L^p(\RD)$, $1\leq p\leq\infty$. The Fourier transform $\hat{f}(\xi):=\int_{\RD} e^{-i\xi\cdot{}x}f(x)\,dx$, $\xi\in\RD$, of a function $f\in{}L^p(\RD)$ for $1\leq p\leq2$ is defined in the standard way (see e.g. [@K Section IV.3]). In the case $2<p\leq\infty$ we consider only the functions $f\in{}L^p(\RD)$ whose transforms belong to $L^\pp(\RD)$, $\pp:=p/(p-1)$. Introduce the difference operator of order $m\in\NN$ acting on functions with domain $\RD$ by [@Ti Section 3.3] $$\label{eqfindiff}
     \D_y^m f(x):=\sum_{k=0}^m (-1)^{m-k}
      \Big(\genfrac{}{}{0pt}{}{m}{k}\Big)\,f(x+ky),\qquad x,y\in\RD,$$ where $\big(\genfrac{}{}{0pt}{}{m}{k}\big)$ denotes the binomial coefficient. Note that the Fourier transform of in $x$ equals $
   \Big(\sum_{k=0}^m (-1)^{m-k}
      \big(\genfrac{}{}{0pt}{}{m}{k}\big)\,e^{iky\cdot\xi}\Big)\,
          \hat{f}(\xi)
  = (e^{iy\cdot\xi}-1)^m\,\hat{f}(\xi),
$ and hence the functions $$\label{eqFpair}
        \D_y^mf(x),\qquad (2i\sin(y\cdot \xi/2))^m\,e^{imy\cdot\xi/2}\,\hat{f}(\xi)$$ form a Fourier pair for $m\in\NN$. Now we define $\D_y^mf(x)$ for any $m>0$, $m\not\in\NN$, as the inverse Fourier transform of the second function in . Let $\SD$ denote the unit sphere in $\RD$ and let $dS$ denote the standard measure on $\SD$. (All the results below with obvious modifications hold if one replaces the standard measure on $\SD$, with a measure $G$ invariant under orthogonal transformations and supported in the unit ball in $\RD$ as in [@Cl].) For $1\leq{}p\leq\infty$ set $$\delta_{p,m}[f](y):=\|\Delta_y^m{}f\|_{p,\RD},\qquad
                          y\in\RD,$$ and define a general averaged (integral) of $f$ as follows. For any $h>0$ in the case $1\leq{}q<\infty$ set $$\label{eq200}
   \omega_{p,m,q}[f](h):=\|\d_{p,m}[f](hy)\|_{q,\SD}
  =  \Big(\int_{\SD} \|\Delta_{hy}^m{}f\|_{p,\RD}^{q}\,dS_y\Big)^{1/q}$$ (where in the case $d=1$ the integral should be interpreted as a sum over $y\in\SS^0=\{\pm1\}$), and in the case $q=\infty$ set $$\omega_{p,m,\infty}[f](h):=\sup_{|y|\leq h}\d_{p,m}[f](y)
            =\sup_{|y|\leq h}\|\Delta_y^m{}f\|_{p,\RD}.$$ The Hölder inequality implies that for any $1\leq q_1\leq q_2\leq\infty$ there exists $c=c(p,q_1,q_2,d)<\infty$ so that $$\label{eq0}
   \omega_{p,m,q_1}[f](h) \leq c\,\omega_{p,m,q_2}[f](h),
          \qquad h>0.$$ Define the tail integral of the for $1\leq\pp<\infty$ and for $\pp=\infty$, respectively, by $$\label{eq00true}
   \psi_{\pp}[\hat{f}](t)
        := \Big(\int_{|\xi|\geq t} |\hat{f}(\xi)|^{\pp}\,d\xi\Big)^{1/\pp},\qquad
   \psi_{\infty}[\hat{f}](t)
        := \sup_{|\xi|\geq t} |\hat{f}(\xi)|,\qquad t>0.$$ Motivated by the results in [@Cl], we wish to compare the $\omega_{p,m,q}[f](1/t)$ and the tail $\psi_{\pp}[\hat{f}](t)$, as $t\ra\infty$. It will be clear from our Theorem \[thm1\] below that the natural choice of $q$ for the purpose of such a comparison is $q=p^\prime$ (see also the discussion preceding Theorem \[Besseltail\]). Note next that it is possible that $\psi_{\pp}[\hat{f}](t)$ is rapidly decreasing, or simply zero, for large $t$ (take e.g., $\hat{f}\in{}C_0^\infty(\RD)$), whereas the modulus of continuity related to the $m$th finite difference of $f$ vanishes generally speaking at the rate $1/t^m$ only. This motivated the author in [@Cl] to introduce the following modified tails: For $1\leq\pp<\infty$ $$\label{eq00}
\begin{aligned}
   \psi_{\pp,m}[\hat{f}](t):&= \Big(\int_{\RD}
   \min\big(1, (|\xi|/t)^{m\pp}\big)
                    \,|\hat{f}(\xi)|^{\pp}\,d\xi \Big)^{1/\pp}\\
           &=\Big( t^{-m\pp}
            \int_{|\xi|\leq t} |\xi|^{m\pp}\,|\hat{f}(\xi)|^{\pp}\,d\xi
          +(\psi_{\pp}[\hat{f}](t))^{\pp}\Big)^{1/\pp},\qquad t>0,
\end{aligned}$$ and for $\pp=\infty$ $$\psi_{\infty,m}[\hat{f}](t):=
\sup_{\xi\in\RD} \Big(\min\big(1,(|\xi|/t)^{m}\big)\,|\hat{f}(\xi)|\Big),\qquad t>0.$$ It might not be immediately obvious why these tails are useful, we give the reason for that in Theorem \[Besseltail\] below. Note that for any $1\leq\pp\leq\infty$, $\psi_{\pp,m}[\hat{f}](t)$ is nonincreasing as $t$ grows, because so is the function $\min(1,r/t)$ for any fixed $r>0$. Note also that in the case $\hat{f}\in{}C_0^\infty(\RD)$, $\hat{f}\not\equiv0$, there exist $\tilde{c}_1=\tilde{c}_1(\hat{f})>0$, $\tilde{c}_2=\tilde{c}_2(\hat{f})>0$ such that $$\label{eqcinfty}
       \tilde{c}_1\,t^{-m}\leq\psi_{\pp,m}[\hat{f}](t)
            \leq \tilde{c}_2\,t^{-m},\qquad t\ra\infty,$$ because $\psi_{\pp}[\hat{f}](t)=0$ for large $t$.

We are ready to state the two results from [@Cl] mentioned above. First [@Cl p. 512], for any $d\in\NN$ in the case $2\leq{}p\leq\infty$, there exists $c_3=c_3(p,d,m)$ such that for for all functions in the set $\{f : f\in{}L^p(\RD)\textrm{ and }\hat{f}\in{}L^{p^\prime}(\RD)\}$, $$\label{eq1}
    \omega_{p,m,\infty}[f](1/t)\leq c_3\,\psi_{\pp,m}[\hat{f}](t),\qquad t>0.$$ Note that in the case $p=2$, holds for all $f\in{}L^2(\RD)$. (The mentioned formula in [@Cl] involves in fact $\psi_{\pp,m}[\hat{f}](2t)$. The formula is then true since $\psi_{\pp,m}[\hat{f}](t)$ is nonincreasing, and suffices for our purposes.) Secondly [@Cl (9)], for any $d\in\NN$ in the case $1\leq{}p\leq2$, for any $a>1$ there exists $c_4=c_4(p,q,d,m,a)$ such that for any $1\leq{}q\leq\infty$ and for all $f\in{}L^p(\RD)$ $$\label{eq2}
       \psi_{\pp}[\hat{f}](t)\leq c_4\,
          \sum_{k=1}^\infty a^m\omega_{p,m,q}[f](1/(a^kt)),\qquad t>0.$$ It would be preferable to have instead of a formula which does not involve an infinite sum, i.e., of the type . It turns out that such a result holds for $d\geq2$ and if $q$ is large enough, at least $q=\pp$.

\[thm1\] For any $d\geq2$ and $1\leq{}p\leq2$ there exists $c=c(p,d,m)$ such that for all $f\in{}L^p(\RD)$ $$\label{eq2p}
    \psi_{\pp,m}[\hat{f}](t) \leq c\,\,\omega_{p,m,\pp}[f](1/t),\qquad t>0.$$

It is explained in Remark \[remdim1\] below why (and even its analog with $\psi_{\pp}[\hat{f}](t)$ in the left-hand side, cf.  below) fails for $d=1$. In view of , one can replace $\pp$ in the right-hand side in with any $q\geq\pp$ (and a different $c$). It would be interesting to know if one could replace $\pp$ in the right-hand side of with $1\leq{}q<\pp$. Note also that for $q\geq\pp$, our Theorem \[thm1\] implies readily all the statements of [@Cl Theorem 2], and is slightly more general since no comparison function $s$ as in [@Cl] is required.

In the case $p=2$ the following is an immediate consequence of , and .

\[thm2\] For $d\geq2$ and $p=2$ there exist $c_1,c_2>0$ that depend on $d$ and $m>0$ only, such that for all $f\in{}L^2(\RD)$ and $t>0$ $$\label{eq2p2}
\begin{aligned}
    c_1\,\omega_{2,m,\infty}[f](1/t)\leq \psi_{2,m}[\hat{f}](t)
               &\leq c_2\,\omega_{2,m,2}[f](1/t)\\
   &\leq c_2c(2,2,\infty,d)\,\omega_{2,m,\infty}[f](1/t).
\end{aligned}$$

The estimates , , show that that the modified tail $\psi_{\pp,m}$ is more appropriate than the true tail $\psi_{\pp}$ to be compared with the $\omega_{p,m,\pp}$. The upper estimate in is a Jackson-type inequality, see e.g. [@K Section I.8]. It follows from that for $p=2$ an inequality in the direction opposite to holds: In the case $d\geq2$ for all $f\in{}L^2(\RD)$ $$\label{eq000}
    \omega_{2,m,\infty}[f](1/t)\leq \tilde{c}(2,\infty,2)\,\omega_{2,m,2}[f](1/t),\qquad t>0,$$ and so for any $m>0$ all the moduli $\omega_{2,m,q}$, $2\leq{}q\leq\infty$, are equivalent. It would be interesting to find a direct proof of . We now explain why in the case $d\geq2$ and $1\leq p^\prime<\infty$ the modified $\psi_{\pp,m}$ defined in is a natural quantity to consider. For $d\geq2$ and any $0<\alpha<\infty$ define $$\label{eq_Besselt}
   G_\alpha(|w|):=
         2^\alpha\int_{\SD}\big(1-\cos (y\cdot{}w)\big)^\alpha\,dS_y,
      \qquad w\in\RD.$$ Recalling that the functions in form a Fourier pair and using the Hausdorff–Young inequality (see the proof of Theorem \[thm1\] below and the proof of in [@Cl]) we can compare for $y\in\SD$ fixed and $t>0$, the $L^p$ norm of $\Delta_y^m{}f$, $\delta_{p,m}[f](y)$, and the $L^{p^\prime}$ norm of the function $(2i\sin(y\cdot \xi/(2t)))^m\,e^{imy\cdot\xi/2}\,\hat{f}(\xi)$ (in the case $p>2$ we assume in addition as before that $\hat{f}\in L^{p^\prime}(\RD))$. Raising both quantities to the power $q=p^\prime$ (this explains why the choice $q=p^\prime$ is natural) we can compare the quantities $(\omega_{p,m,p^\prime}[f](t))^{\pp}$ and $$\int_{\RD}
        2^{m\pp/2}
    \bigg(\int_{\SD}\Big(1-\cos\frac{y\cdot\xi}{t}\Big)^{m\pp/2}\,dS_y\bigg)
                      |\hat{f}(\xi)|^{\pp}\,d\xi.$$ With this in mind, for $d\geq2$ and any $\hat{f}\in L^{\pp}(\RD)$, $1\leq p<\infty$, we introduce the [*Bessel tail *]{}of the Fourier transform $$\Psi_{\pp,m}[\hat{f}](t)
          :=
   \bigg(\int_{\RD} G_{m\pp/2}\big({|\xi|}/{t}\big)
                      |\hat{f}(\xi)|^{\pp}\,d\xi\bigg)^{1/\pp}$$ and $\Psi_{\infty,m}[\hat{f}](t):=\psi_{\infty,m}[\hat{f}](t)$. From the above discussion, $\omega_{p,m,p^\prime}[f](t)$ can be compared with the Bessel tail $\Psi_{\pp,m}[\hat{f}](t)$ as in , . The relevance of the modified $\psi_{\pp,m}$ is now apparent from the following

\[Besseltail\] For any $d\geq2$ and any $0<\alpha<\infty$, there exist $C_1,C_2>0$ that depend only on $d$ and $\alpha$ so that $$\label{eqBessel}
      C_1(\min(1,v))^{2\alpha}\leq G_\alpha(v)
          \leq C_2(\min(1,v))^{2\alpha},\qquad v\geq0,$$ and hence for some $\tilde{C}_1,\tilde{C}_2>0$ that depend on $1\leq p<\infty$, $m>0$, $d$ only, $$\tilde{C}_1\Psi_{\pp,m}[\hat{f}](t)
                      \leq \psi_{\pp,m}[\hat{f}](t)
            \leq\tilde{C}_2\Psi_{\pp,m}[\hat{f}](t)$$ for all $\hat{f}\in L^{\pp}(\RD)$ and all $t>0$.

The relation is illustrated for $\alpha=1$, $C_1=\pi/3$, $C_2=6\pi$ in Fig. \[fig\] (note that $G_1(v)=4\pi\big(1-\frac{\sin v}{v})$).

We now describe the relationship between the true and the modified tails, $\psi_\pp$ and $\psi_{\pp,m}$, respectively. From it is clear that for any $1\leq\pp\leq\infty$, $m>0$, and $\hat{f}\in{}L^\pp(\RD)$ $$\label{eqtri}
         \psi_\pp[\hat{f}](t)\leq\psi_{\pp,m}[\hat{f}](t),\qquad t>0.$$ The following statement gives a converse to that is optimal on the power scale (see Remark \[remstar\] below).

\[lem4.1\] Let $d\in\NN$, $m>0$, $\alpha>0$, $1\leq\pp\leq\infty$, $g{}\in{}L^\pp(\RD)$. All the constants below depend on $g$ and are strictly positive and finite.

1\. Let $1\leq\pp<\infty$.

\(i) If $\psi_\pp[g{}](t)\leq c_2(g)\cdot{}t^{-\alpha}$, $t\ra\infty$, then as $t\ra\infty$, $$\psi_{\pp,m}[g{}](t)\leq b_2(g)
      \cdot\begin{cases}
                                      t^{-\alpha},&0<\alpha<m\cr
                                   t^{-\alpha}\,(\log{}t)^{1/\pp},&\alpha=m\cr
                             t^{-m},&\alpha>m.
      \end{cases}$$

\(ii) If $c_1(g)\cdot{}t^{-\alpha}\leq
\psi_{\pp,m}[g{}](t)\leq c_2(g)\cdot{}t^{-\alpha}$, $t\ra\infty$, then as $t\ra\infty$, $$\psi_{\pp}[g{}](t)\geq b_1(g)
      \cdot\begin{cases}
                                      t^{-\alpha},&0<\alpha<m\cr
                                      0,&\alpha\geq m.
      \end{cases}$$

2\. Let $\pp=\infty$.

\(i) If $\psi_\infty[g{}](t)\leq c_2(g)\cdot{}t^{-\alpha}$, $t\ra\infty$, then as $t\ra\infty$, $$\psi_{\infty,m}[g{}](t)\leq b_2(g)
      \cdot\begin{cases}
                                    t^{-\alpha},&0<\alpha< m\cr
                                    t^{-m},&\alpha\geq m.
      \end{cases}$$

\(ii) If $c_1(g)\cdot{}t^{-\alpha}\leq
\psi_{\infty,m}[g{}](t)\leq c_2(g)\cdot{}t^{-\alpha}$, $t\ra\infty$, then as $t\ra\infty$, $$\psi_{\infty}[g{}](t)\geq b_1(g)
      \cdot\begin{cases}
                                      t^{-\alpha},&0<\alpha<m\cr
                                      0,&\alpha\geq m.
      \end{cases}$$

We state finally a result that was applied in a study of the scaling of entanglement entropy for a certain physical system in [@GiKl; @Gi2].

\[CH3\_s3.lem.1\] Let $d\in\NN$. Assume that ${}f\in{}L^2(\RD)$. Then for some $c_1=c_1(f)>0$, $c_2=c_2(f)>0$ and some $\gamma=\gamma(f)\in(0,2)$, $f$ satisfies $$\label{eqneww}
        c_1\,\eps^\gamma\leq
   \int_{\SD}\|{}f(\cdot+\eps{}y)-{}f(\cdot)\|_{2,\RD}^2 \,dS_y\leq c_2\,\eps^\gamma,
             \qquad0\leq\eps\leq1,$$ if and only if there exist $b_1=b_1(f)>0$, $b_2=b_2(f)>0$ such that $$\label{eqnewww}
          b_1\,t^{-\gamma} \leq
                   \int_{|\xi|\geq{}t}
                      |\hat{f}(\xi)|^2\,d\xi
           \leq b_2\,t^{-\gamma},\qquad t\geq1.$$

Note that Corollary \[CH3\_s3.lem.1\] is true in all dimensions: In the proof of Corollary \[CH3\_s3.lem.1\] below, we consider the cases $d\geq2$ and $d=1$ separately. In the former case we employ the general results stated above. In the case $d=1$ we give a direct proof using in particular the ideas in the proofs of [@BCT Lemma 2.10, Lemma 4.2]. The main reason why, in the case $d=1$, Corollary \[CH3\_s3.lem.1\] is true despite the fact that Theorem \[thm1\] fails, is because of the explicit (power-type) form of the estimates in Corollary \[CH3\_s3.lem.1\].

The equivalence of the [*upper *]{}estimates in and in is well-known: it follows follows e.g. from [@MS Lemma 3.3.1], and also from the results obtained in [@Cl]. Note that [@MS Lemma 3.3.1] deals with a Besov space $B_{2,\infty}^{s}(\RD)$, $s>0$, $s\not\in\NN$ (the case $s=\gamma/2\in(0,1)$ is relevant for the upper estimates in Corollary \[CH3\_s3.lem.1\]). In [@BCT], is derived in the case $\gamma=1$ from a more restrictive pointwise condition $$\label{eq_ts}
      c_1\,|y|^\gamma\leq(\d_{2,1}[{}f](y))^2
               \leq c_2\,|y|^\gamma,\qquad|y|\leq1,$$ which in general can not be reversed because $\d_{2,1}[{}f](y)$ can have singular directions.

A simple example of a function that satisfies with $\gamma=1$ is the characteristic function of a compact set with $C^1$ boundary. For any $0<\gamma<1$ there exsits a compact set whose characteristic function satisfies , see [@Gi2 Lemma 2.9].

It turns out that Corollary \[CH3\_s3.lem.1\] fails for $\gamma=2$, see Remark \[remthf\]. Note that for $\gamma>2$ the condition is not satisfied for any $f\in{}C_0^\infty(\RD)$, $f\not\equiv0$, because it involves the finite difference of order $1$.

Theorems \[thm1\], \[Besseltail\], \[lem4.1\] and Corollary \[CH3\_s3.lem.1\] are proved in Section \[sec1.3\].

Proofs and concluding remarks {#sec1.3}
=============================

First consider the case $p=1$, $\pp=\infty$ (in this case the proof below goes through for all $d\in\NN$). We have to prove that for some $c=c(p,d,m)$, $\psi_{\infty,m}[\hat{f}](t)\leq c\,\omega_{1,m,\infty}[f](1/t)$, $t>0$. Recall that $\|\hat{f}\|_{\infty,\RD}\leq\|f\|_{1,\RD}$. Since the functions form a Fourier pair we have $$\label{eq30}
\begin{aligned}
     \omega_{1,m,\infty}[f](1/t) &= \sup_{|y|\leq1/t}\d_{1,m}(y)
             \geq \sup_{|y|\leq1/t} \sup_{\xi\in\RD}
                \big(|2i\sin (y\cdot \xi/2)\big|^m\,|\hat{f}(\xi)|\big)\\
    &= 2^m\,\sup_{\xi\in\RD}\Big( |\hat{f}(\xi)|
            \sup_{|z|\leq1} \big|\sin (z\cdot \xi/(2t))\big|^m \Big).
\end{aligned}$$ It is an easy exercise to prove that for any $\xi\in\RD$ $$\begin{aligned}
      \sup_{|z|\leq1} \big|\sin (z\cdot \xi/(2t))\big| &=
      \begin{cases} 1, &|\xi|\geq |t\pi|\cr
                            \sin(|\xi|/(2t)), &|\xi|< |t\pi|
     \end{cases}\\
   &\geq\frac14\,\min(1,|\xi|/t).
\end{aligned}$$ This together with and proves the result.

Consider now the case $d\geq2$, $1<p\leq2$, $\pp<\infty$. By the Hausdorff–Young inequality $\|\hat{f}\|_{\pp,\RD}\leq(2\pi)^{d/\pp}\|f\|_{p,\RD}$, $1<p\leq2$, and using the fact that is a Fourier pair we get as in [@Cl (4.7)] $$\label{eq300}
\begin{aligned}
   (\d_{p,m}[f](y/t))^{\pp}
    &\geq (2\pi)^{-d}\int_{\RD} \big|2i\sin (y\cdot \xi/(2t))\big|^{m\pp}\,
             |\hat{f}(\xi)|^{\pp}\,d\xi\\
   &= (2\pi)^{-d}\,2^{m\pp/2}\int_{\RD} |\hat{f}(\xi)|^{\pp}
        \,\Big(1-\cos\frac{y\cdot\xi}{t}\Big)^{m\pp/2}\,d\xi.
\end{aligned}$$ Integrating over $y$ and recalling we obtain $$\label{eq3}
\begin{aligned}
    (\omega_{p,m,\pp}[f](1/t))^{\pp}
       &=\int_{\SD}(\d_{p,m}[f](y/t))^{\pp}\,dS_y\\
   &\geq (2\pi)^{-d}\int_{\RD}G_{m\pp/2}(|\xi|/t)\, |\hat{f}(\xi)|^{\pp}\,d\xi.
\end{aligned}$$ Now we need the following elementary result, the proof is given after the end of the present proof.

\[lem1\] Let $d\geq2$ and fix any $0<\alpha<\infty$. The function $G_\alpha(v)$, $v\geq0$, defined in satisfies the following: For any $v_0>0$ there exist $C_\alpha(v_0),c_\alpha(v_0)>0$ such that $$\label{eqlem101}
       {}G_\alpha(v)\geq C_\alpha(v_0) >0,\qquad v\geq v_0$$ and $$\label{eqlem102}
       G_\alpha(v)\geq c_\alpha(v_0)\, v^{2\a}, \qquad 0\leq v\leq v_0.$$

Now we write the integral in as a sum of $\int_{|\xi|\leq{}t}$ and $\int_{|\xi|\geq{}t}$, apply Lemma \[lem1\] with $\alpha=m\pp/2$, $v_0=1$, and set $c_1(p,d,m):=c(m,d)\cdot\min\big( c_\alpha(1),C_\alpha(1))$ to obtain $$\label{eq5}
\begin{aligned}
    (\omega_{p,m,\pp}[f](1/t))^{\pp}
       &\geq c_1(p,d,m)\,
        \Big( \int_{|\xi|\leq{}t}(|\xi|/t)^{m\pp}\,|\hat{f}(\xi)|^{\pp}\,d\xi
          + \int_{|\xi|\geq{}t}|\hat{f}(\xi)|^{\pp}\,d\xi\Big)\\
      &=c_1(p,d,m)\,
             (\psi_{\pp,m}[\hat{f}](t))^{\pp}
\end{aligned}$$ which finishes the proof of Theorem \[thm1\].

\[remdim1\] We show now that is not true in the case $d=1$ and $p=2$. More precisely, we show that for $d=1$ there is no $c\in(0,+\infty)$ so that for all $f\in{}L^2(\RR)$ $$\label{eqnn}
    \psi_{2}[\hat{f}](t) \leq c\,\,\omega_{2,1,2}[f](1/t),\qquad t>0.$$ By with $m=1$ using the Parseval formula we find $$\label{eqParseval}
          (\omega_{2,1,2}[f](1/t))^2 = \frac4\pi\int_{\RR}
                   \sin^2\bigg(\frac\xi{2t}\bigg)\,|\hat{f}(\xi)|^2\,d\xi.$$ For $\xi\in\RR$, $t,\tilde{c}>0$, introduce the notation $$H(\xi,t,\tilde{c}) = \sin^2\bigg(\frac\xi{2t}\bigg) - \tilde{c}
              \,\sgn\bigg[\Big(\frac{|\xi|}t-1\Big)_+\bigg]$$ where $\sgn a=1,0,-1$ for $a>0$, $a=0$, $a<0$, respectively, and $(a)_+=\max(0,a)$. If were true for some $c\in(0,+\infty)$ then there would exist $\tilde{c}\in(0,+\infty)$ such that $$         \int_{\RR} H(\xi,t,\tilde{c}) \,|\hat{f}(\xi)|^2\,d\xi \geq0\qquad
       \textrm{for all } f\in L^2(\RR),\quad t>0.$$But $$\begin{aligned}
    \{\, |\hat{f}|^2 \,:\,f\in L^2(\RR)\, \} &=
          \{ \,|f|^2 \,:\,f\in L^2(\RR) \,\}\\
      &=
         \{\, g \,:\,g\in L^1(\RR)\textrm{ and }g\geq0
          \textrm{ almost everywhere (a.e.)} \, \}.
\end{aligned}$$ We have arrived at a contradiction: If $$\int_{\RR} H(\xi,t,\tilde{c}) \,g(\xi)\,d\xi \geq0\qquad
       \textrm{for all } g\in L^1(\RR),\,g\geq0\textrm{ a.e., } t>0,$$ then we must have $H(\xi,t,\tilde{c})\geq0$ for a.e. $\xi$ and all $t>0$, which is clearly false for any choice of $\tilde{c}>0$. This proves the result.

We note that the basic reason for the inapplicability of Theorem \[thm1\] to the case $d=1$ is that for any $v_0>0$ it is not possible to insert a constant function between the graph of $ \sin^2 v$ and the real axis on the interval $[v_0,+\infty)$, cf. Fig. \[fig\]. (In the case $d=1$, $\SS^0=\{\pm1\}$ and so the function $G_\alpha$ in would be given by $2^{2\alpha+1}\big(\sin^2(w/2)\big)^\alpha$, $w\in\RR$, and there is no helpful averaging over $\SS^{d-1}$.)

Recall that $d\geq2$. If $d\geq3$ then integrating over the $d-2$ angles in as in [@GS (II.3.4.2)] we obtain for $v\geq0$ $$\label{eq101z}
    G_\alpha(v) = |\SS^{d-2}|\int_{0}^\pi (1-\cos(v\cos\theta))^\a
 \,\sin^{d-2}\theta\,d\theta.$$ If $d=2$ then and the subsequent formlulae hold true with the convention $|\SS^{0}|=2$.

We show first there is $M_\alpha>0$ such that ${}G_\alpha(v)\geq{}M_\alpha>0$ for sufficiently large $v$. Indeed, in the case $\alpha\geq1$ by the Hölder inequality there is $\tilde{C}_\alpha>0$ so that $$\label{eq101}
\begin{aligned}
   \int_{0}^\pi \big( (1-\cos(v\cos\theta))
 \,&\sin^{(d-2)/\a}\theta\big)^\a\,d\theta \\
         &\geq \tilde{C}_\alpha\,\Big(\int_{0}^\pi (1-\cos(v\cos\theta))
 \,\sin^{(d-2)/\a}\theta\,d\theta \Big)^\a \\
   &\geq \tilde{C}_\alpha\,\Big(\int_{0}^\pi (1-\cos(v\cos\theta))
      \,\sin^{d-2}\theta\,d\theta \Big)^\a
\end{aligned}$$ where the second inequality follows from $$\label{eqeasy}
       w^\beta \geq w,\qquad 0<\beta\leq1,\qquad  0\leq w\leq1$$ with $w=\sin^{d-2}\theta$ and $\beta=1/\a$. In the case $0<\a<1$, in view of with $w=(1-\cos(v\cos\theta))/2$ and $\beta=\a$ we obtain $$\label{eq101p}
\begin{aligned}
   \int_{0}^\pi  (1-\cos(v\cos\theta))^\a
 \,\sin^{d-2}\theta\,d\theta
          &= 2^{\a}\,\int_{0}^\pi
       \Big(\frac{1-\cos(v\cos\theta)}{2}\Big)^\a
            \,\sin^{d-2}\theta\,d\theta \\
        &\geq 2^{\a-1}\,\int_{0}^\pi  (1-\cos(v\cos\theta))
            \,\sin^{d-2}\theta\,d\theta.
\end{aligned}$$ Note that $$\label{eq1019}
\begin{aligned}
     |\SS^{d-2}|\int_{0}^\pi (1-\cos(v\cos\theta))
      \,\sin^{d-2}\theta\,d\theta
  &= \int_{\SD}\Big(1-\cos\big((v,0,\cdots,0)\cdot y\big)\Big)\,dS_y\\
   &=  |\SD|\,\big(1 - 2^s\,\Gamma(s+1)\,
          v^{-s}\,J_s(v)\big)
\end{aligned}$$ where $s=(d-2)/2$, $J_s$ is the Bessel function, and we have used [@GS (II.3.4.2)] and [@GrRy (8.411.4)]. By [@GrRy (8.451)], $v^{-s}J_s(v)=O(v^{-s-1/2})\ra0$, as $v\ra\infty$ for $d\geq2$. Hence the right-hand side of , and also of and , tends to a strictly positive limit, as $v\ra\infty$. Therefore for any fixed $0<\alpha<\infty$, there exist $M_\alpha>0$ and $v_1(\alpha)$ so that ${}G_\alpha(v)\geq{}M_\alpha>0$ for $v\geq{}v_1(\alpha)$.

But ${}G_\alpha(v)$ does not have zeros other than $v=0$. Since $G_\alpha$ is continuous, it is for any $v_0>0$ bounded away from zero on the compact $[v_0,v_1(\alpha)]$. This proves .

Let us now prove for $v_0>0$ small enough. We can rewrite as $$\label{eqlem103}
    G_\alpha(v) = 2^{\alpha}\,|\SS^{d-2}|\int_{0}^\pi
         \sin^{2\alpha}(v\cos\theta)
 \,\sin^{d-2}\theta\,d\theta.$$ Using the elementary estimate $$\label{eqlem104}
         \sin x \geq \frac2\pi\, x,\qquad 0\leq x\leq\frac\pi2,$$ we conclude that $$    G_\alpha(v) \geq v^{2\alpha}\cdot c(\alpha,d) \int_{0}^\pi
         \cos^{2\alpha}\theta
 \,\sin^{d-2}\theta\,d\theta, \qquad 0\leq v\leq \frac\pi2,$$which proves for any $0<v_0\leq\pi/2$. But now if we take any $v_0>\pi/2$ then using for $0\leq v\leq\pi/2$ and the fact that $G_\alpha(v)$ is bounded away from zero for $\pi/2\leq v\leq v_0$ by we can always find $c_\alpha(v_0)>0$ small enough so that holds for $0\leq v\leq v_0$. The proof of Lemma \[lem1\] is complete.

Recall that $d\geq2$. The lower estimate in follows immediately from Lemma \[lem1\] with $v_0=1$. As for the upper estimate, we note first that by the definition , the function $G_\alpha(v)$ is bounded above for $v\geq1$. Next, using the estimate $\sin x\leq x$, $0\leq x\leq1$, in place of , we derive from the upper estimate $$G_\alpha(v) \leq v^{2\alpha}\cdot \tilde{c}(\alpha,d) \int_{0}^\pi
         \cos^{2\alpha}\theta
                    \,\sin^{d-2}\theta\,d\theta, \qquad 0\leq v\leq 1.$$ This proves the upper estimate in . The proof of Theorem \[Besseltail\] is complete.

Recall that $d\in\NN$, $0<\alpha<\infty$, $m>0$.

1\. Consider first the case $1\leq\pp<\infty$.

\(i) Assume $\psi_{\pp}[{}g{}](t)\leq{}c_1(g)\cdot{}t^{-\alpha}$, $t\geq1$. By $$\label{eq201}
 (\psi_{\pp,m}[{}g{}](t))^\pp=   t^{-m\pp}
            \int_{|\xi|\leq t} |\xi|^{m\pp}\,|{}g{}(\xi)|^\pp\,d\xi
          +(\psi_{\pp}[{}g{}](t))^\pp.$$ Then the following gives the result for the case 1(i), all values of $\alpha$: $$\begin{aligned}
   t^{-m\pp} &\int_{|\xi|\leq t} |\xi|^{m\pp}\,|{}g{}(\xi)|^\pp\,d\xi\\
       &\leq t^{-m\pp}\Big(\int_{|\xi|\leq1}|{}g{}(\xi)|^\pp\,d\xi
                    + \sum_{k=0}^{[\log_2t]}
             \int_{2^{k}\leq|\xi|\leq2^{k+1}}|\xi|^{m\pp}\,
                    |{}g{}(\xi)|^\pp\,d\xi\Big)\\
          &\leq t^{-m\pp}\Big(\const(g)+ c_1(g)\sum_{k=0}^{[\log_2t]}
             \big(2^{(m-\alpha)\pp}\big)^k\Big)
\end{aligned}$$ where $[\cdot]$ denotes the integer part of a real number. It is explained in Remark \[remstar\] below why the order in $t$ cannot be improved in the case 1(i) and in all other cases.

\(ii) Assume $c_1(g)\cdot t^{-\alpha}\leq
\psi_{\pp,m}[{}g{}](t)\leq c_2(g)\cdot t^{-\alpha}$, $t\geq1$ ($\pp<\infty$). The example ${}g{}\in{}C_0^\infty(\RD)$, ${}g{}\not\equiv0$, shows that in the case $\alpha\geq{}m$ we can only claim the trivial bound $\psi_\pp[{}g{}](t)\geq0$, $t\ra\infty$. Let now $0<\alpha<m$. We use the idea in the proof of [@BCT Lemma 4.2]. Let $0<B\leq1$ be a number to be chosen later. By the definition $$\label{eqlem120}
    (\psi_{\pp,m}[{}g{}](t))^\pp =
           \Big( \int_{|\xi|\leq Bt} + \int_{|\xi|\geq Bt}
       \Big) \min^{m\pp}(1,|\xi|/t)\,|{}g{}(\xi)|^\pp\,d\xi.$$ We have $$\begin{aligned}
   \int_{|\xi|\leq Bt} &\min^{m\pp}(1,|\xi|/t)\,|{}g{}(\xi)|^\pp\,d\xi
  =  t^{-m\pp} \int_{|\xi|\leq Bt} |\xi|^{m\pp}\,|{}g{}(\xi)|^\pp\,d\xi\\
       & \leq t^{-m\pp} \bigg(\int_{|\xi|\leq B}|\xi|^{m\pp}\,|{}g{}(\xi)|^\pp\,d\xi
   + \sum_{l=1}^{[\log_2 t]+1}
             \int_{2^{l-1}B\leq|\xi|\leq2^{l}B}
                            |\xi|^{m\pp}\,|{}g{}(\xi)|^\pp\,d\xi
    \bigg)\\
  & \leq t^{-m\pp} \bigg(\|g\|_{\pp,\RD}^{\pp}
     + \sum_{l=1}^{[\log_2 t]+1}
            (2^{l}B)^{m\pp}\cdot  c_2^{\pp}(g)\cdot (2^{l-1}B)^{-\alpha\pp}
    \bigg)\\
       &\leq t^{-m\pp}\cdot\const(g)\cdot t^{(m-\alpha)\pp}
             \cdot B^{(m-\alpha)\pp},\qquad t\to\infty.
\end{aligned}$$ Therefore choosing $0<B\leq1$ small enough (recall that $m-\alpha>0$) we obtain that $$\label{eqlem121}
        \int_{|\xi|\leq Bt}\min^{m\pp}(1,|\xi|/t)\,|{}g{}(\xi)|^\pp\,d\xi
             \leq \frac{c_1^{\pp}(g)}2 \cdot t^{-\alpha\pp},\qquad t\to\infty.$$ Since $\psi_{m,\pp}[g](t)\geq c_1^{\pp}(g)\cdot t^{-\alpha\pp}$, $t\to\infty$, together with gives $$\int_{|\xi|\geq Bt}
     \min^{m\pp}(1,|\xi|/t)\,|{}g{}(\xi)|^\pp\,d\xi
      \geq \frac{c_1^{\pp}(g)}2 \cdot t^{-\alpha\pp},\qquad t\to\infty.$$ Setting $s=Bt$ we obtain $$\label{eqlem110prim}
    \int_{|\xi|\geq s}
     \min^{m\pp}(1,B|\xi|/s)\,|{}g{}(\xi)|^\pp\,d\xi
      \geq {c_1(g)}B^{\alpha\pp}\cdot s^{-\alpha\pp},
              \qquad s\to\infty.$$ Noting that $
    1\geq\min(1,B|\xi|/s)
$ we conclude from that $$\psi_{\pp}^\pp[g](t) = \int_{|\xi|\geq s}
     |{}g{}(\xi)|^\pp\,d\xi
               \geq {c_1(g)}B^{\alpha\pp}\cdot s^{-\alpha},\qquad s\to\infty.$$

2\. The case $\pp=\infty$.

\(i) Assume $\psi_\infty[{}g{}](t)\leq{}c_1(g)\cdot t^{-\alpha}$, $t\geq1$. Note that $$\label{eqggg}
        \psi_{\infty,m}[{}g{}](t)
              = \max\Big(t^{-m}\sup_{|\xi|\leq t} |\xi|^m|{}g{}(\xi)|,
                          \,\sup_{|\xi|\geq t} |{}g{}(\xi)|\Big).$$ Clearly $$\label{eqgg}
       |{}g{}(\xi)|\leq \sup_{|\eta|\geq|\xi|} |{}g{}(\eta)|
             \leq c_1(g)\cdot |\xi|^{-\alpha},\qquad|\xi|\geq1.$$ Therefore $$\begin{aligned}
    \sup_{|\xi|\leq t} |\xi|^m&|{}g{}(\xi)|
        =\max\big(
             \sup_{|\xi|\leq 1} |\xi|^m|{}g{}(\xi)|,
                \sup_{1\leq|\xi|\leq t} |\xi|^m|{}g{}(\xi)|\big)\\
       &\leq \max\Big(c({}g{}),\,
               c_1({}g{})\cdot \sup_{1\leq|\xi|\leq{}t}|\xi|^{m-\alpha}\Big)
          =\begin{cases}C({}g{}),&0<\alpha<m\cr
                        C_1({}g{})\cdot{}t^{m-\a},&\a\geq m.
             \end{cases}
\end{aligned}$$ This together with proves the case 2(i).

\(ii) Let $c_1(g)\cdot t^{-\alpha}\leq
\psi_{\infty,m}[{}g{}](t)\leq c_2(g)\cdot t^{-\alpha}$, $t\to\infty$. Again if $\alpha\geq{}m$ then the example of ${}g{}\in{}C_0^\infty(\RD)$, $g\not\equiv0$, shows that generally speaking only the trivial bound $\psi_{\infty}[{}g{}](t)\geq0$ holds for large $t$. Let now $0<\alpha<m$. Let $0<B\leq1$ be a number to be chosen later. We have $$\begin{aligned}
   \psi_{\infty,m}[{}g{}](t)
              &= \sup_{\xi\in\RD}
         \min^m(1,|\xi|/t)\,|g(\xi)|\\
   &=\max\Big( \sup_{|\xi|\leq Bt}
         \min^m(1,|\xi|/t)\,|g(\xi)|,\,\sup_{|\xi|\geq Bt}
         \min^m(1,|\xi|/t)\,|g(\xi)|
         \Big).
\end{aligned}$$ Now $$\begin{aligned}
      \sup_{|\xi|\leq Bt}
         &\min^m(1,|\xi|/t)\,|g(\xi)|
   =
              t^{-m} \sup_{|\xi|\leq Bt} |\xi|^m\,|g(\xi)|\\
    &\leq    t^{-m} \max\bigg( \sup_{0\leq|\xi|\leq B} |\xi|^m\,|g(\xi)|,
         \max_{l=1,\cdots,[\log_2t]+1}
          \sup_{2^{l-1}B\leq|\xi|\leq 2^lB} |\xi|^m\,|g(\xi)|\bigg)\\
    &\leq    t^{-m} \max\bigg( B^m\|g\|_{\infty,\RD},
         \max_{l=1,\cdots,[\log_2t]+1}
          2^{lm} B^m\cdot c_2(g)\cdot (2^{l-1}B)^{-\alpha}\bigg)\\
&\leq \const(g)\cdot B^{m-\alpha}\cdot t^{-\alpha},\qquad t\to\infty.
\end{aligned}$$ Choosing $B$ small enough (note that $m-\alpha>0$) we obtain $${c_1(g)}\cdot t^{-\alpha} \leq
     \psi_{\infty,m}[{}g{}](t)
         \leq \max\Big( \frac{c_1(g)}2\cdot t^{-\alpha},\,\sup_{|\xi|\geq Bt}
         \min^m(1,|\xi|/t)\,|g(\xi)|
         \Big),\qquad t\to \infty,$$ which implies that $\sup_{|\xi|\geq Bt}
         \min^m(1,|\xi|/t)\,|g(\xi)| \geq {c_1(g)}\cdot t^{-\alpha}$. Replacing $s=Bt$ we obtain $$\label{eqlem110}
     \sup_{|\xi|\geq s}
         \min^m(1,B|\xi|/s)\,|g(\xi)| \geq {c_1(g)}B^\alpha\cdot s^{-\alpha},
              \qquad s\to\infty.$$ Noting that $
    1\geq\min(1,B|\xi|/s)
$ we conclude from that $$\psi_\infty[g](t) = \sup_{|\xi|\geq s} |g(\xi)|
      \geq {c_1(g)}B^\alpha\cdot s^{-\alpha},\qquad s\to\infty.$$ This completes the proof of Theorem \[lem4.1\].

\[remstar\] Let us explain why the estimates in Theorem \[lem4.1\] have the best possible order in $t$. In view of , the estimates 1(i) and 2(i) for $0<\a<m$ can not be improved. The example of $g\in{}C_0^\infty(\RD)$, $g\not\equiv0$, in view of the lower estimate in shows that estimates 1(i) for $\a>m$ and 1(ii), 2(i), 2(ii) for $\a\geq{}m$ can not be improved. The example of $g\in{}L^\pp(\RD)$, $1\leq\pp<\infty$, such that $g(\xi)=|\xi|^{-(d/\pp)-\a}$ for $|\xi|\geq1$ and $g$ smooth for $|\xi|<1$ shows that the estimate 1(ii) for $0<\a<m$ can not be improved (note that the mentioned $g$ satisfies $\psi_{\pp,m}[g](t)]\leq
c_1(g)\cdot{}t^{-\a}$, $t\ra\infty$, and also $\psi_\pp[g](t)=
\big(|\SD|/(\a\pp)\big)^{1/\pp}\cdot{}t^{-\a}$, $t\geq1$). Choosing $g\in{}L^\infty(\RD)$ such that $g(\xi)=|\xi|^{-\a}$ for $|\xi|\geq1$ and $g$ smooth for $|\xi|<1$ shows that the estimate 2(ii) for $0<\a<m$ can not be improved (for this $g$ we have $\psi_\infty[g](t)=t^{-\a}$, $t\geq1$). Finally, the example ${}g{}\in{}L^\pp(\RD)$, $1\leq\pp<\infty$, such that ${}g{}(\xi)=|\xi|^{-(d/\pp)-m}$, $|\xi|\geq1$, and $g$ smooth for $|\xi|<1$ shows that 1(i) for $\a=m$ can not be improved. Indeed for this $g$, $\psi_\pp[{}g{}](t)=\big(|\SD|/(m\pp)\big)^{1/\pp}\cdot t^{-m}$, $t\geq1$, but $\psi_{\pp,m}[{}g{}](t)\geq
|\SD|^{1/\pp}\cdot t^{-m}(\log{}t)^{1/\pp}$, $t\geq1$.

[*The case $d\geq2$. *]{}The result follows readily from Corollary \[thm2\] combined with Theorem \[lem4.1\] for $p=\pp=2$, $\a=\gamma/2$, $m=1$, $g=\hat{f}$ with $\eps=1/t$. Note that the quantitity in the middle in is $(\omega_{2,1,2}[f](\eps))^2$. The integral in is the true tail integral of the of $f$, $\psi_2[\hat{f}](t)$. By Corollary \[thm2\], the exist $C_1,C_2>0$ that depend on $d$ only so that for all $f\in{}L^2(\RD)$ $$\label{eqappr1}
     C_1\,\omega_{2,1,2}[f](1/t)
       \leq \psi_{2,1}[\hat{f}](t)\leq C_2\,\omega_{2,1,2}[f](1/t),\qquad t>0.$$ Set $\alpha=\gamma/2$. Note that since $0<\gamma<2$ we have $0<\a=\gamma/2<m=1$, and the reult follows from Theorem \[lem4.1\], cases 1 (i) and (ii), applied to the comparison function $t^{-\alpha}=t^{-{\gamma/2}}$, $t\geq1$.

[*The case $d=1$. *]{}We prove first that implies . This follows from a straightforward modification of the proof of [@BCT Lemma 4.2]. We prefer to give the details for the convenience of the reader. Below, $c$ will denote a positive constant whose precise value may change from equation to equation and may depend on $f$ but which is independent of $\eps$ and $t$. By $$\begin{aligned}
              (\omega_{2,1,2}[f](\eps))^2 = \frac4\pi\int_{\RR}
                   \sin^2\bigg(\frac{\eps\xi}{2}\bigg)\,|\hat{f}(\xi)|^2\,d\xi
         \geq c \int_{{\pi}/(2\eps)
                       \leq|\xi|\leq2{\pi}/(2\eps)}
                   |\hat{f}(\xi)|^2\,d\xi.
\end{aligned}$$ Setting $\tilde{\eps}=2\eps/\pi$ and denoting $\tilde{\eps}$ by $\eps$ again, we find from the upper inequality in $$\begin{aligned}
           c_2\cdot(\pi\eps/2)^\gamma
        \geq   \omega_{2,1,2}[f](\pi\eps/2)
           \geq c \int_{1/\eps
                       \leq|\xi|\leq2/\eps}
                   |\hat{f}(\xi)|^2\,d\xi
\end{aligned}$$ which after setting $t=1/\eps$ implies $$\label{eqdyadic}
         \int_{t
                       \leq|\xi|\leq2t}
                   |\hat{f}(\xi)|^2\,d\xi \leq c\cdot t^{-\gamma},\qquad t\geq1.$$ Using the latter estimate and representing $$        \int_{|\xi|\geq t}|\hat{f}(\xi)|^2\,d\xi
          =\sum_{j=0}^\infty
                 \int_{2^j t\leq|\xi|\leq2^{j+1} t} |\hat{f}(\xi)|^2\,d\xi,$$we prove the upper inequality in . In order to prove the lower inquality in , we note first that implies $$\label{eq1dyadic}
         \int_{|\xi|\leq r}
                  |\xi|^2\, |\hat{f}(\xi)|^2\,d\xi \leq c_3
                          \cdot r^{2-\gamma},\qquad r\geq1.$$ Indeed using the upper inequality in we obtain $$\begin{aligned}
      \int_{|\xi|\leq r}
                  |\xi|^2\, |\hat{f}(\xi)|^2\,d\xi
          &\leq  \int_{|\xi|\leq 1}
                  |\xi|^2\, |\hat{f}(\xi)|^2\,d\xi
          +\sum_{j=0}^{[\log_2r]+1}
                 \int_{2^j \leq|\xi|\leq2^{j+1} }  |\xi|^2\, |\hat{f}(\xi)|^2\,d\xi\\
         &\leq\const
          +c\sum_{j=0}^{[\log_2r]+1}
                (2^{j+1})^2 \cdot b_2\cdot (2^{j})^{-\gamma}\\
       &\leq c^\prime\cdot r^{2-\gamma}
\end{aligned}$$ (recall that $\gamma<2$). Next, let $\alpha,\beta>0$ be two numbers to be chosen later. From the lower inequality in and using $|\sin x|\leq x$ and $|\sin x|\leq1$ for $x\geq0$, we obtain $$\begin{aligned}
    c_1\cdot \eps^\gamma&\leq
           (\omega_{2,1,2}[f](\eps))^2\\
          &= \frac4\pi\int_{\RR}
                   \sin^2\bigg(\frac{\eps\xi}{2}\bigg)\,|\hat{f}(\xi)|^2\,d\xi\\
          &\leq c \bigg(\eps^2\int_{|\xi|\leq\alpha/\eps}
                   |\xi|^2\,|\hat{f}(\xi)|^2\,d\xi
             +\int_{\alpha/\eps\leq|\xi|\leq\beta/\eps}
                   |\hat{f}(\xi)|^2\,d\xi
               +\int_{|\xi|\geq\beta/\eps}
                   |\hat{f}(\xi)|^2\,d\xi\bigg).
\end{aligned}$$ Using and the upper inequality in we find $$\int_{\alpha/\eps\leq|\xi|\leq\beta/\eps}
                   |\hat{f}(\xi)|^2\,d\xi
        \geq c_1c^{-1}\cdot \eps^\gamma
          -\beta^{-\gamma}b_2\cdot \eps^\gamma
          -\alpha^{\gamma}c_3\cdot \eps^\gamma$$ which after choosing $\alpha>0$ small enough and $\beta>0$ large enough gives $$\int_{|\xi|\geq\alpha/\eps}
                   |\hat{f}(\xi)|^2\,d\xi
        \geq\int_{\alpha/\eps\leq|\xi|\leq\beta/\eps}
                   |\hat{f}(\xi)|^2\,d\xi
        \geq c\cdot\eps^\gamma,\qquad 0<\eps\leq1.$$ Setting $t=\alpha/\eps$ we prove the lower inequality in .

We now derive from . Using the upper inequality in and employing and Theorem \[lem4.1\] (recall that in our case $p=p^\prime=2$, $m=1$, $\alpha=\gamma/2\in(0,1)$) $$     (\psi_{2}[\hat{f}](t))^2
              \leq (\psi_{2,1}[\hat{f}](t))^2
               \leq c\cdot t^{-\gamma}.$$ Combining this with (that holds for $d=1$) and using we obtain $$(\omega_{2,1,2}[f](1/t))^2 \leq c \cdot(\omega_{2,1,\infty}[f](1/t))^2
         \leq c^\prime \cdot(\psi_{2,1}[\hat{f}](t))^2
           \leq c^{\prime\prime}\cdot t^{-\gamma}$$ which proves the upper estimate in . It remains to prove the lower estimate in . Note that the two-sided estimate implies that for $A>1$ large enough $$\bigg( \int_{|\xi|\geq t} -
                     \int_{|\xi|\geq At} \bigg)\,|\hat{f}(\xi)|^2\,d\xi
          =  \int_{t\leq|\xi|\leq At} |\hat{f}(\xi)|^2\,d\xi \geq c\cdot t^{-\gamma}$$ or after setting $\eps=\pi/(2t)$ $$\label{eq2dyadic}
   \int_{\pi/(2\eps)\leq|\xi|\leq A\pi/(2\eps)}
               |\hat{f}(\xi)|^2\,d\xi \geq c\cdot \eps^{\gamma},
                     \qquad 0<\eps\leq1.$$ On the other hand again using $$\begin{aligned}
              (\omega_{2,1,2}[f](\eps))^2 = \frac4\pi\int_{\RR}
                   \sin^2\bigg(\frac{\eps\xi}{2}\bigg)\,|\hat{f}(\xi)|^2\,d\xi
         \geq c \int_{{\pi}/(2\eps)
                       \leq|\xi|\leq3{\pi}/(2\eps)}
                   |\hat{f}(\xi)|^2\,d\xi
\end{aligned}$$ and replacing $\eps$ with $\eps/3^j$, $j\in\NN$, we find $$(\omega_{2,1,2}[f](3^{-j}\eps))^2
           \geq c \int_{3^j{\pi}/(2\eps)
                       \leq|\xi|\leq3^{j+1}{\pi}/(2\eps)}
                   |\hat{f}(\xi)|^2\,d\xi,\qquad j=0,1,2,\cdots.$$ Choosing $N=[\log_3 A]$ we then obtain $$\begin{aligned}
          \sum_{j=0}^N (\omega_{2,1,2}[f](3^{-j}\eps))^2
                 &\geq c\cdot
           \sum_{j=0}^N  \int_{3^j{\pi}/(2\eps)
                       \leq|\xi|\leq3^{j+1}{\pi}/(2\eps)}
                   |\hat{f}(\xi)|^2\,d\xi \\
       & \geq c\cdot
           \sum_{j=0}^N  \int_{{\pi}/(2\eps)
                       \leq|\xi|\leq A{\pi}/(2\eps)}
                   |\hat{f}(\xi)|^2\,d\xi \\
        &\geq c^\prime\cdot \eps^\gamma,\qquad 0<\eps\leq1,
\end{aligned}$$ where we have used . Therefore $$\liminf_{\eps\to0}\eps^{-\gamma}
              \sum_{j=0}^N \big(\omega_{2,1,2}[f](3^{-j}\eps)\big)^2\geq c>0$$ and hence as $N\in\NN$ is fixed, there exists at least one $J\in\{1,\cdots,N\}$ such that $$\tilde{c}:=
       \liminf_{\eps\to0} \eps^{-\gamma}
                   \big(\omega_{2,1,2}[f](3^{-J}\eps)\big)^2 > 0.$$ Then $$\big(\omega_{2,1,2}[f](3^{-J}\eps)\big)^2\geq \tilde{c}\cdot \eps^\gamma,
                 \qquad 0<\eps\leq1,$$ and denoting $3^{-J}\eps$ by $\eps$ we finish the proof of the lower inequality in .

We note finally that the proof of Corollary \[CH3\_s3.lem.1\] for $d=1$ can be modified to give an alternative proof of Corollary \[CH3\_s3.lem.1\] also for all $d\geq2$ from scratch (in this connection, see an explanation of an argument from [@BCT] given in the proof of [@thesis Lemma 3.4.1]).

\[remthf\] After the above general discussion it is not difficult to understand why Corollary \[CH3\_s3.lem.1\] fails for $\gamma=2$. Recall that in , $p=\pp=2$ and $m=1$. Let $\hat{f}\in{}L^2(\RD)$ be defined by $\hat{f}(\xi):=|\xi|^{-(d/2)-1}$ for $|\xi|\geq1$ and smooth for $|\xi|<1$. Let $f\in{}L^2(\RD)$ be the inverse of this $\hat{f}$. It is easy to check that for some $\tilde{b}_1,\tilde{b}_2>0$ that depend on $\hat{f}$ $$\label{eqtss}
     \tilde{b}_1\,t^{-1}(\log{t})^{1/2} \leq\psi_{2,1}[\hat{f}](t)
           \leq \tilde{b}_2\,t^{-1}(\log{t})^{1/2},\qquad t\geq2$$ whereas for certain $\tilde{c}_1,\tilde{c}_2>0$ that depend on $\hat{f}$ $$\tilde{c}_1\,t^{-1} \leq\psi_{2}[\hat{f}](t)
           \leq \tilde{c}_2\,t^{-1},\qquad t\geq2.$$ Note that by and for $c_1,c_2>0$ that depend on $f$ $$c_1\,t^{-1}(\log{t})^{1/2} \leq\omega_{2,1,2}[f](1/t)
           \leq c_2\,t^{-1}(\log{t})^{1/2},\qquad t\geq2.$$ This shows that Corollary \[CH3\_s3.lem.1\] fails for $\gamma=2$. It is only true in the case $\gamma=2$ that the upper estimate in implies the upper estimate in (simply because for any $f\in{}L^2(\RD)$ and all $t>0$, $\psi_2[\hat{f}](t)\leq{}\psi_{2,1}[\hat{f}](t)\leq{}c(2,d,1)\,\omega_{2,1,2}[f](1/t)$ in view of and ). The fact that the lower estimate in need not hold is shown by considering the example of $\hat{f}\in{}C_0^\infty(\RD)$, $\hat{f}\not\equiv0$ (for which $\psi_2[\hat{f}](t)=0$ identically for large $t$). Finally, the first example of this remark shows the upper estimate in for $\gamma=2$ need not follow even from a two-sided estimate in .

[12]{} [BCT]{} L. Brandolini, L. Colzani and G. Travaglini, *Average decay of Fourier transforms and integer points in polyhedra.* Ark. Mat. **35** (1997), 253–275. [Cl]{} D. B. H. Cline, *Regularly varying rates of decrease for moduli of continuity and Fourier transforms of functions on $\RD$.* J. Math. Anal. Appl. **159** (1991), 507–519. [GS]{} I. M. Gel’fand and G. E. Shilov, *Generalized Functions*, vol. 1, Academic Press, New York–London, 1964. D. Gioev and I. Klich, *Entanglement entropy of fermions in any dimension and the Widom conjecture.* Phys. Rev. Lett. **96** (2006), 100503, 4 pp. [Gi2]{} D. Gioev, [*Szegö limit theorem for operators with discontinuous symbols and applications to entanglement entropy.*]{} Int. Math. Res. Not. **2006**, Art. ID 95181, 23 pp. D. Gioev, *Generalizations of Szegö Limit Theorem: Higher Order Terms and Discontinuous Symbols*, Ph.D. Thesis, Royal Institute of Technology, Stockholm, 2001. [www.math.rochester.edu/people/faculty/gioev/thesis.pdf]{} [GrRy]{} I. S. Gradshteyn and I. M. Ryzhik, *Tables of Integrals, Series, and Products*, 4th ed., Academic Press, New York–London–Toronto, 1980. Y. Katznelson, *An Introduction to Harmonic Analysis*, 3rd Ed., Cambridge University Press, Cambridge, 2004. [MS]{} V. G. Maz’ya and T. O. Shaposhnikova, *Theory of Multipliers in Spaces of Differentiable Functions*, Pitman Publishing, Boston, 1985. [St]{} E. Stein, *Singular Integrals and Differentiability Properties of Functions*, Princeton University Press, Princeton, 1970. [Ti]{} A. F. Timan, *Theory of Approximation of Functions of a Real Variable*, Pergamon Press, Oxford–New York, 1963. [T]{} E. C. Titchmarsh, *Introduction to the Theory of Fourier Integrals*, 2nd ed., Oxford University Press, London–New York, 1948.


Q:

Functorial type constructors in System F

I have come across the claim that all basic data types in System F, such as Bool, Nat, and List(U), can be expressed in the form $\forall \alpha (((T\alpha \rightarrow \alpha) \rightarrow \alpha)$, where $T(\alpha)$ is a definable functorial type constructor.
Can someone explain why this is true or refer me to a proof? Because I have not found a proof for this claim yet.

A:

The usual (categorical) characterization of inductive data types is as initial $F$-algebras for a functor $F$.  An $F$-algebra is a pair of a type, $A$, and a function $a : FA \to A$.  A homomorphism of $F$-algebras, $\varphi : (A,a) \to (B,b)$ is a function $\varphi : A \to B$ which satisfies $\varphi\circ a = b\circ F\varphi$.  An $F$-algebra, $(\mu F,\mathsf{in}_F)$, is initial if there exists a unique homomorphism from it to any other $F$-algebra.  This means we have a function which takes an $F$-algebra $(A,a)$ and produces a function $\mathsf{fold}_A(a):\mu F\to A$ which satisfies $$\mathsf{fold}_A(a)\circ\mathsf{in}_F=a\circ F(\mathsf{fold}_A(a))$$
The type of $\mathsf{fold}$ in System F would be $\forall \alpha.(F\alpha\to\alpha)\to\mu F\to\alpha$.  The free theorem for this (as calculated by this) specializes to
$$\forall \alpha,\beta. \forall \varphi : \alpha\to \beta.
   \forall a : F \alpha \to \alpha.
   \forall b : F \beta \to \beta.\\
    \varphi \circ a = b \circ F \varphi \implies \varphi \circ \mathsf{fold}_{\alpha}(a) = \mathsf{fold}_\beta(b)$$
which further specializes to
$$\forall \beta.
   \forall b : F \beta \to \beta.
    b\circ \mathsf{fold}_{F\beta}(Fb) = \mathsf{fold}_\beta(b)$$
and
$$\forall \beta. \forall \varphi : \mu F\to \beta.
   \forall b : F \beta \to \beta.\\
    \varphi \circ \mathsf{in}_F = b \circ F \varphi \implies \varphi \circ \mathsf{fold}_{\mu F}(\mathsf{in}_F) = \mathsf{fold}_\beta(b)$$
If we can prove $\mathsf{fold}_{\mu F}(\mathsf{in}_F) = id_{\mu F}$, then this would mean that $(\mu F,\mathsf{in}_F)$ was an initial $F$-algebra as the above equation says: "any $F$-algebra homomorphism $\varphi : (\mu F, \mathsf{in}_F)\to(B,\beta)$ is $\mathsf{fold}_\beta(b)$ and thus exists and is unique". Of course to do this we need to actually give a definition of $\mu F$, $\mathsf{fold}$, and $\mathsf{in}_F$.
By noting that $\forall \alpha.(F\alpha\to\alpha)\to\mu F\to\alpha$ is isomorphic to $\mu F \to \forall \alpha.(F\alpha\to\alpha)\to\alpha$, there's an obvious choice for both $\mathsf{fold}$ and $\mu F$: simply set $\mu F = \forall \alpha.(F\alpha\to\alpha)\to\alpha$ and $\mathsf{fold}$ to be the inverse image of $id$ with respect to that isomorphism.  Explicitly, $$\mathsf{fold} = \Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.\lambda g\!:\!\mu F.g[\alpha]a$$ All that remains is to define $\mathsf{in}_F$ and show that $\mathsf{fold}_{\mu F}(\mathsf{in}_F) = id_{\mu F}$.  The easiest way to define $\mathsf{in}_F$ is to just follow the types: $\mathsf{in}_F : F\mu F \to \forall \alpha.(F\alpha\to\alpha) \to \alpha$ so $$\mathsf{in}_F = \lambda x\!:\!F\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.a(F(\mathsf{fold}_{\alpha}(a))x)$$
This definition is particularly convenient as you may have noticed that while I've shown that if there's a homomorphism $\varphi : (\mu F,\mathsf{in}_F)\to(A,a)$ then it is $\mathsf{fold}_A(a)$, and I've shown that there is a function $\mathsf{fold}_A(a):\mu F\to A$, I haven't shown that $\mathsf{fold}_A(a)$ is a homomorphism. That is, I haven't shown $\mathsf{fold}_A(a)\circ\mathsf{in}_F = a\circ F(\mathsf{fold}_A(a))$, but with the above definitions of $\mathsf{fold}$ and $\mathsf{in}_F$ this is easily shown by normalizing. To finish, $$\begin{align}
\mathsf{fold}_{\mu F}(\mathsf{in}_F)
& = \mathsf{in}_F\circ\mathsf{fold}_{F\mu F}(F(\mathsf{in}_F)) \\
& = \lambda g\!:\!\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.(a\circ F(\mathsf{fold}_{\alpha}(a)) \circ \mathsf{fold}_{F\mu F}(F(\mathsf{in}_F)))g \\
& = \lambda g\!:\!\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.(\mathsf{fold}_\alpha(a)\circ \mathsf{in}_F \circ \mathsf{fold}_{F\mu F}(F(\mathsf{in}_F)))g \\
& = \lambda g\!:\!\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.(\mathsf{fold}_\alpha(a)\circ \mathsf{fold}_{\mu F}(\mathsf{in}_F))g \\
& = \lambda g\!:\!\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.\mathsf{fold}_\alpha(a)g \\
& = \lambda g\!:\!\mu F.\Lambda\alpha.\lambda a\!:\!F\alpha\to\alpha.g[\alpha]a \\
& = \lambda g\!:\!\mu F.g
\end{align}$$


Those that invested in inXile’s Wasteland 3 through its Fig crowdfunding campaign will receive a 132 percent return on their investment, rather than the performance-based payout outlined in Fig’s usual equity crowdfunding program.

That payout will be one lump sum rather than a series of returns following the game’s launch, though it makes for a higher dividend rate than the one originally listed in the Wasteland 3 campaign. The reason for this sudden switch, Fig explains in an email, is that Microsoft has purchased Fig's publishing rights for the game for nearly $2.3 million.

If this all sounds a little familiar, it’s because nearly the exact same thing happened with the Fig-funded game Psychonauts 2 this week, though the percentages and payouts differ just slightly.

Both inExile and Psychonauts 2 developer Double Fine were acquired by Microsoft within the past year, and it seems the company has now sat down with Fig to settle some of the details surrounding the crowdfunding campaigns that helped the games raise millions.

The Wasteland 3 deal sees investors receiving a dividend rate of 85 percent (as opposed to the campaign’s original 70 percent rate), and results in a payout of $1,320.75 per $1,000 share.

“As you may already know, inXile joined the Microsoft family late last year - we are excited for them for this next stage in their journey of making unique and creative games and can't wait to see what they do next,” explains the email. “As a part of this transition, Microsoft has also recently purchased Fig's publishing rights for Wasteland 3, for a price of $2,295,000 which will provide a positive return to all investors in Fig Game Shares.”

Both Wasteland 3 and Psychonauts 2 are among the largest campaigns Fig has hosted to date, with both raising over $3 million before funding closed. In the case of Wasteland 3, the game raised $3.12 million of its $2.75 million goal. Around $2.25 million of that total came from the sale of Fig Shares.


REVIEW: Poor Lou finds himself in the middle of his own ASSAULT ON PRECINCT 13 in this week's amazing episode of Fargo. It's incredible how Noah Hawley's show manages to outdo itself each and every week, and after last week's tense, bloody installment I wasn't at all sure this week's episode would be able to measure up. I shouldn't have under-estimated Hawley, as this week's episode might just be the best so far.

While these kinds of stand-off, bottle episodes are nothing new (the PRECINCT 13 model far predates that film – going back to the classic westerns that influenced John Carpenter), what is unique about the way Hawley's show handled this formula. Specifically, how he lets the feeling of dread mount throughout the episode but then manages to have it come to an unexpectedly non-violent conclusion, which the audience will no doubt find a relief, as Lou, Ted Danson's Sherrif and the rest of the Luverne PD are such a likable crew you don't want to see any of them buy the farm.

While the episode comes to a peaceful end, Lou's dealing with the Gerhardt clan will no doubt have a violent end, especially once Jeffrey Donovan's psycho Dodd gets involved, with him being sidelined by a bout with Kirsten Dunst's surprisingly capable Peggy.

There's so much to enjoy in this episode, with Ted Danson and Nick Offerman getting loads of screen-time this week. While Danson's sheriff always seemed a tad simple, he displays some of the sharpness his grand-daughter Molly possesses in season one, with him having a great exchange with Peggy where he tries to fathom why she didn't just come clean the moment she hit Rye, and clues her-in on the grim fate that now awaits her and Ed.

Offerman fares even better. While his whiskered, big-talking Carl has seemed like a bit of a joke so far, here he's revealed to be anything but, with him being roused from a drunken night out to work as Ed's attorney. His unexpectedly silver-tongued knowledge of the law comes in handy with him able to talk-out a peaceful solution with the Gerhardts, giving him something to brag about at the next lodge meeting (an end-credits treat).

As always, Fargo just keeps getting better and better. This was one hell of a good episode – maybe the best so far, and amazingly only one person died this week (a Gerhardt shooter accidentally plugged by Dodd). Oh – and as a treat for Coen Bros fans, we get cool seventies version of “A Man of Constant Sorrow” - which was famously featured in O BROTHER WHERE ART THOU. The perfect end to a perfect episode.


Purpose:
Titmus stereoacuity testing is used to estimate visual acuity (VA) in the diagnosis of non-organic visual loss (NOVL), but only predicts mean VA and doesn’t account for normal inter-subject variability. These predictions were derived from optical degradation of VA in normal subjects and may not account for the variability seen in patients with neuro-ophthalmic pathologies included in the differential diagnosis of NOVL. The purpose of this study was to evaluate the relationship between Titmus stereoacuity and minimal visual acuity based on a real-world testing environment.

Methods:
All patients presenting to our service between 4/25/2014 and 6/26/2014 underwent routine neuro-ophthalmic examination, including Titmus stereoacuity measurements. A compound Bayesian logit-lognormal model accounting for heteroskedasticity was used to determine 95% and 99% prediction intervals of the worse eye’s near visual acuity (VA) based on stereoacuity. LogMAR acuity and log stereoacuity were analyzed.

Conclusions:
When fully accounting for individual variation and the full spectrum of neuro-ophthalmic diseases affecting VA, stereoacuity remains associated with VA, but commonly-used VA estimates based on stereoacuity overestimate VA. Our results more accurately predict minimum VA from Titmus stereoacuity and should be preferentially used when evaluating patients with suspected non-organic visual loss. We demonstrate that Titmus stereoacuity cannot establish definitively normal VA, and therefore can only suggest, but not establish, the diagnosis of NOVL.


590 So.2d 331 (1991)
Carl FREMAN
v.
CITY OF MOBILE and the Personnel Board of Mobile County.
2900492.
Court of Civil Appeals of Alabama.
November 15, 1991.
*332 Richard R. Williams, Mobile, for appellant.
Richard T. Dorman, Patricia Ponder and William R. Lauten, Mobile, for appellee City of Mobile.
Mylan R. Engel and Edgar P. Walsh, Mobile, for appellee Mobile County Personnel Bd.
RUSSELL, Judge.
Carl Freman appeals from the trial court's order which, pursuant to § 34, Act No. 470, Local Acts of Alabama 1939, sustained the findings of the Personnel Board of Mobile County (board). The board upheld the dismissal of Freman from the City of Mobile Police Department. We affirm.
The dispositive issue is whether the findings of the board are supported by substantial evidence.
We note at the outset that the trial court's review of the board's findings is limited to the record made before the board and to the questions of law presented and that substantial evidence to support the board's findings requires that court's affirmance of the board's judgment. City of Mobile v. Seals, 471 So.2d 431 (Ala.Civ. App.1985). The trial court may not substitute its judgment for that of the board. Id. Further, the trial court's standard of review applies to this court as well. Id.
Substantial evidence is that which would be more than a mere scintilla of the evidence and less than a preponderance of the evidence. Ex parte Morris, 263 Ala. 664, 83 So.2d 717 (1955). It is such relevant evidence as might be accepted as adequate to support a conclusion by reasonable minds. Ex parte Morris, 263 Ala. 664, 83 So.2d 717; Seals, 471 So.2d 431.
Freman was terminated due to numerous alleged acts of perverse, abusive, and harassing sexual behavior, both on and off the job. The charges stemmed from actions with his former wife, two of his daughters, co-employees, and a prostitute.
Freman contends that there was not substantial legal evidence to support the board's decision because, he claims, the trial court eliminated all charges relating to "kinky sex" and "sexual affairs" and found substantial evidence based on child abuse and gun play, which allegedly occurred at home. He argues that he cannot be dismissed for conduct unbecoming an officer if the unbecoming conduct occurred while he was off duty.
However, although the trial court told Freman's attorney at the hearing that he did not want the attorney to talk about kinky sex because he was more concerned about the use of guns, the trial court's review of the evidence was limited to the record made before the board. Therefore, the evidence concerning those issues was already before the trial court. In addition, the trial court stated at the conclusion of the hearing as follows: "I do not find any particular ground that the Personnel Board used to be entirely invalid and I am not making any effort to rule one way or another on the validity or invalidity of particular grounds. But I simply say that I see *333 a number of grounds that are patently valid and patently supported by a super abundance of evidence." Therefore, we find that the trial court did not eliminate any of the charges considered by the board.
Furthermore, even if the issues concerning Freman's on-the-job behavior had been eliminated, off-duty conduct can support a charge of behavior unbecoming a police officer. See Guthrie v. Civil Service Board of Jasper, 342 So.2d 372 (Ala. Civ.App.1977).
After reviewing the record, we find that there was substantial evidence to support the findings of the board and further find that this case is due to be affirmed.
AFFIRMED.
ROBERTSON, P.J., and THIGPEN, J., concur.


HONG KONG (Reuters) - A U.S. congressional panel has warned of an “alarming” rise in China’s interference in Hong Kong, noting fears over the former British colony’s continued role as a global financial hub.

In its annual report to Congress on Wednesday, the bipartisan U.S.-China Economic and Security Review Commission highlights the “chilling” abduction and detention of five booksellers based in Hong Kong as well as pressure on media and academic freedoms.

Britain handed Hong Kong back to China in 1997 under a “one country, two systems” agreement that ensures its freedoms, independent legal system and wide-ranging autonomy remain intact.

The commission, in a detailed 33-page section, urges a fresh probe from the State Department into Hong Kong’s autonomy and freedoms, as well as continued congressional oversight.

“Hong Kong’s traditional standing as a global financial hub has significant economic implications for the United States, as U.S. trade and investment ties with Hong Kong are substantial,” the report notes.

It says the booksellers’ detentions - including two foreign nationals and one who was abducted inside Hong Kong - broadened domestic fears of mainland encroachment and sparked a record turnout in September’s legislative election.

“This incident has threatened the maintenance of the ‘one country, two systems’ framework and led some observers to question Hong Kong’s status as a leading global financial hub,” the report warns.

“The election took place against the backdrop of an alarming rise in mainland interference in Hong Kong.”

U.S. senators Marco Rubio and Tom Cotton also introduced a bill that would freeze U.S.-based assets and ban U.S. entry of those responsible for the “surveillance, abduction, detention, or forced confessions of certain booksellers and journalists in Hong Kong”, according to a statement on Rubio’s senate website.

In Beijing, Chinese Foreign Ministry spokesman Geng Shuang said Hong Kong was an internal matter for China and no foreign country had the right to interfere.

“In order to try to split the country, some forces have openly sought foreign support,” Geng told a daily news briefing.

“Those who attempt to use foreign forces to achieve their own political goals will not succeed.”

The report comes amid deepening concerns in Beijing over a fledgling independence movement in the city. The High Court this week backed a government demand to bar two recently elected lawmakers, who insulted China when taking their oath of office, from the legislature.

In a statement after the report’s release, the Hong Kong government said the “one country, two systems” principle was being implemented successfully, as was the city’s role as a global commercial hub. It also urged foreign parties not to interfere.

The booksellers were involved in the production and sale of gossipy political titles banned in mainland China but freely available in Hong Kong. Their plight fueled Western governments’ concerns and sparked formal diplomatic protests.

One of the five men, Swedish passport holder Gui Minhai, who disappeared from the Thai resort of Pattaya last October, is the only one still in detention in China.

Another, Lam Wing-kee, returned to Hong Kong in June, saying he had been held captive by Chinese agents for eight months.

The report places the worsening climate in the city in the context of China’s broader disregard for international legal agreements and norms on issues such as the South China Sea and Taiwan.


A new study documents structural differences in the brains of preschool-age children related to screen-based media use.

The study, published in JAMA Pediatrics, shows that children who have more screen time have lower structural integrity of white matter tracts in parts of the brain that support language and other emergent literacy skills. These skills include imagery and executive function -- the process involving mental control and self-regulation. These children also have lower scores on language and literacy measures.

The Cincinnati Children's Hospital Medical Center study assessed screen time in terms of American Academy of Pediatrics (AAP) recommendations. The AAP recommendations not only take into account time spent in front of screens but also access to screens, including portable devices; content; and who children are with and how they interact when they are looking at screens.

"This study raises questions as to whether at least some aspects of screen-based media use in early childhood may provide sub-optimal stimulation during this rapid, formative state of brain development," says John Hutton, MD, director of the, Reading & Literacy Discovery Center at Cincinnati Children's and lead author of the study. "While we can't yet determine whether screen time causes these structural changes or implies long-term neurodevelopmental risks, these findings warrant further study to understand what they mean and how to set appropriate limits on technology use."

Among the AAP recommendations:

For children younger than 18 months, avoid use of screen media other than video-chatting. Parents of children 18 to 24 months of age who want to introduce digital media should choose high-quality programming, and watch it with their children to help them understand what they're seeing.

For children ages 2 to 5 years, limit screen use to 1 hour per day of high-quality programs. Parents should co-view media with children to help them understand what they are seeing and apply it to the world around them.

Designate media-free times together, such as dinner or driving, as well as media-free locations at home, such as bedrooms.

Dr. Hutton's study involved 47 healthy children -- 27 girls and 20 boys -- between 3 and 5 years old, and their parents. The children completed standard cognitive tests followed by diffusion tensor MRI, which provides estimates of white matter integrity in the brain. The researchers administered to parents a 15-item screening tool, the ScreenQ, which reflects AAP screen-based media recommendations. ScreenQ scores were then statistically associated with cognitive test scores and the MRI measures, controlling for age, gender and household income.

Among the key findings:

Higher ScreenQ scores were significantly associated with lower expressive language, the ability to rapidly name objects (processing speed) and emergent literacy skills.

Higher ScreenQ scores were associated with lower brain white matter integrity, which affects organization and myelination -- the process of forming a myelin sheath around a nerve to allow nerve impulses to move more quickly -- in tracts involving language executive function and other literacy skills.

"Screen-based media use is prevalent and increasing in home, childcare and school settings at ever younger ages," says Dr. Hutton. "These findings highlight the need to understand effects of screen time on the brain, particularly during stages of dynamic brain development in early childhood, so that providers, policymakers and parents can set healthy limits."

The study was funded by a Procter Scholar Award from the Cincinnati Children's Research Foundation. The researchers report no financial relationships of conflicts of interest in regard to the study.


Some of the chemicals are carcinogenic and incredibly dangerous when used improperly. It should also be noted that these are expensive and often difficult to obtain chemicals, best bought in bulk. They are not conducive to only making one or two glow sticks. At the very least, you should have:

This is your base and makes up the majority of the liquid in your glow sticks. It will hold the chemicals that actually glow and amplify them. Start with 10mL of DP, but know that you can double or half this for bigger or smaller sticks. This looks like water, but the chemical reaction necessary for glowing won't work in water

Youuse normal or additive dyes; make sure you're usingdyes. These dyes will not be the same color un-mixed as they are when they glow, so trust the guide given below to make your colors. There are a variety of different dyes you can use, depending on the color you want:

TCPO stands for bis (2,4,6-trichlorophenyl) oxalate, which you may need to know to purchase it. It's quite expensive to buy, but you can make it fairly cheaply if , again, you're experienced and competent around chemicals. Otherwise, making it is not advised.

Add 100 mg of sodium acetate to the mixture. If you don't have sodium acetate, an even mixture of sodium bicarbonate (baking soda) and sodium salicylic can effectively be used instead. When added, put the lid on and shake the bottle well.

It's important to do this step last, as it creates the chemical reaction. Put the lid back on, shake it up well, and turn off the lights. You should have a quite impressive stick/tube/container glowing ferociously in front of you.

7

Add larger quantities of TCPO and Sodium Acetate to make the reaction go on for longer. If you so desire, mess around with the recipe to see what warrants the best results. This reaction works because TCPO and Sodium Acetate release energy when combined, as they start to decay. This energy is picked up by the florescent dyes, which can convert energy into light. More of each leads to more energy and a longer reaction.


Predicted multiple selected reaction monitoring to screen activated drug-mediated modifications on human serum albumin.
Metabolic activation of drugs frequently generates electrophilic products that may undergo covalent binding to biological macromolecules, such as proteins and DNA. The resulting covalent adducts are of considerable concern in drug discovery and development. Several strategies for assessing the potential risks of candidate drugs have been reported. Of these, glutathione trapping is the most commonly used method together with mass spectrometry. Furthermore, drug-mediated protein modifications have been studied using serum albumin and CYP enzymes to clarify target amino acids and mechanism-based inhibition, respectively. In this article, we introduce a practical way to screen drug-mediated protein modifications. The method, referred to as "predicted multiple selected reaction monitoring," is based on the selected reaction monitoring (SRM) strategy, but targets all possible chemically modified tryptic peptides. The creation of SRM lists may require patience; however, this strategy could facilitate more sensitive screening compared with the common strategy of data-dependent product ion scanning. Ketoprofen-N-hydroxysuccinimidyl ester (equivalent to glucuronide) and N-acetyl-p-benzoquinone imine (NAPQI) were allowed to react with human serum albumin as a model experiment. Using this strategy, 11 ketoprofen-adduction sites (at Lys(137, 195, 199, 212, 351, 402, 432, 436, 525, 536, and 541)) and 1 NAPQI-adduction site (at Cys(34)) were easily identified.


Q:

pseudcode for minmax algorithm

I want to get the pseudocode for minmax algorithm. I have to make 2 functions, def maxAgent(gameState, depth) and minAgent. Is there any body who has the right and easy pseudocode for it.

A:

Two players, A and B, take turns to play.
We are given a scoring function f that evaluates a given board position, P.  Larger values of f(P) are better for A and worse for B (i.e., f(P) is an estimate of how "good" P is for A without doing any further lookahead).
Consider a board position P.
If P is a leaf node (i.e., P is a winning position or we have looked as far ahead as we want to) then we return f(P) as the score for this node.
Otherwise P is not a leaf node and has children C1, ..., Cn.  We recursively compute the scores for the children, giving S1, ..., Sn.
If A plays at P then the score for P is max{S1, ..., Sn} since A will always play to maximise his advantage.
If B plays at P then the score for P is min{S1, ..., Sn} since B will always play to minimise A's advantage.
That should be enough to turn into code.
Once you've done that, have a look at alpha-beta pruning, which should (drastically) reduce the amount of search you need to do.  Alpha-beta pruning is based on the idea that if A deduces that B can play to force A's maximum advantage to be M, then there is no point in considering any subtree whose score is greater than M since B will never allow A that option!


Erin Markey

Erin Markey (born 1981) is an American writer, comedian, and performance artist. Markey's work combines elements of cabaret theater and comedy and often incorporates stories of their childhood in the Midwest. The New York Times has described Markey as having "a cult following as an alt-cabaret star with swaggering confidence and off-kilter sense of humor."

Biography

Markey was born in Michigan in 1981. They were raised Catholic. Their father was a marketer in telecommunications and their mother was a medical assistant. In high school, Markey flirted with being a "Bible Belt Christian," due to the influence of friends and their sense that " it felt like really impassioned, relatively speaking, to Catholicism." In high school, Markey informed their parents that they wanted to become a performer, rather than a veterinarian, which made their mother "really mad." Markey studied performance in college at the University of Michigan, taking a class taught by well-known performance artist Holly Hughes. After college, Markey worked for five months at a strip club, an experience that, according to a New York Times profile, was "helpful in developing her [sic] confidence and stage presence." Markey drew upon this experience in several performance works.

Markey has described, in an interview published in Bomb, their self-imposed separation from their family after they came out as gay, a period they described as a "wonderful break of me being able to just develop artistically outside the bounds of me feeling as accountable to them." In the same interview, Markey described their misgivings and anxieties about their family seeing their work, which draws upon their family life. Markey explained: "It’s not exactly dirty laundry, but just things that feel private are like, being made into a show where tons of people are watching it."

Live Performance Work

In 2010, Markey's show "Puppy Love: A Stripper's Tail" was produced at Performance Space 122 in New York City. Based on Markey's experiences working as a stripper after college, "Puppy Love" was praised as "a must-see tale/tail for all you ladies and gents (but mostly ladies) who work in the gray area between theater and erotic arts in New York. It’s also worth seeing if you don’t work in that lovely, glistening niche of the alternative entertainment world." Michelle Tea praised "Puppy Love," calling it

a sex work narrative, but in the form of a musical. The problem with most sex work narratives is that they are not musicals ... they don’t innovate on a story that we’ve now heard a lot, a story that doesn’t really change up all that much. But Erin brings into it her [sic] voices, her [sic] singular artistic voice of course, but also her [sic] singing voice which is really good. Like, she [sic] could go on American Idol or something.

In 2013, Markey starred in Progressive Theatre Workshop's "God Hates This Show: Shirley Phelps-Roper in Concert, Live From Hell," a musical written and directed by John J. Caswell, Jr. satirizing Westboro Baptist Church. Markey played Shirley Phelps-Roper, daughter of Fred Phelps, the founder of Westboro Baptist Church. In the play, Phelps-Roper has died and gone to hell, and sings songs expressing the hateful beliefs of the church.  As one account described their performance, "[b]acked by a full rock band, Markey as Phelps-Roper sing such 'hits' as 'Death in Your Window,' a parody of Melissa Etheridge’s 'Come to My Window,' and 'Stinky Freak,' known in non-helllish circles as 'Super Freak.' She’s [sic] accompanied by a trio of backup singers and dancers in full-on cabaret mode."

In 2013-14, Markey's show "Erin Markey" ran at Joe's Pub. Jarrett Earnest, writing about the show, observed that

Erin Markey’s energy, her [sic] sensibility, is so singularly constructed for herself [sic] that it couldn’t be transferred to another person, maybe that is one thing people mean in calling her [sic] a performance artist. She [sic] plays "Erin Markey" as an outsized persona through autobiographical sketches interlaced with musical numbers—a tried and true nightclub strategy. The songs, usually contemporary pop songs, are performed with an intensity that unravels them.

Asked about their work in "Erin Markey," Markey explained:

 It’s funny to be making this show right now where I’m supposed to be Erin Markey the whole time because I feel like I work best when I’m in character. That’s when I’m most honest. So that’s not difficult for me to go back and forth between characters either, because you know different situations call for different folks. Not that that’s made so explicit in my life, in my practical life I mean. I do think we’re going through this time where identity is not as … people aren’t as interested in being one singular thing anymore. I just had trouble writing a bio recently because I was like, "I don’t want to be called a performance artist. I don’t wanna be called a comedian, I don’t want to be called an actor, or a singer or a cabaret artist," because none of those things seem to really describe who I am.

In 2015, Markey's play "Deleted Scenes from Fun Home," based on the Broadway adaptation Alison Bechdel's graphic novel Fun Home, was produced at the Duplex Cabaret Theatre in New York City. A one-person show written by and starring Markey, the show purported to present deleted scenes from the Broadway musical Fun Home. Asked about the inspiration for "Deleted Scenes from Fun Home," Markey described how they became obsessed with the score to the Broadway musical Fun Home, saying that "for largely unnameable or unknowable reasons, I was activated in a strange way that I had no control over." Markey went on to explain the appeal that the Broadway musical Fun Home had for them:

The fact that I was having this sensation was a huge revelation to me—or really, a prior absence of these feelings was the revelation. The fact that the story of a butch lesbian was being supported on a Broadway budget level was a huge part of it. The money. Or maybe it was the fact that Alison as a character and as a woman was not operating in the traditional economies of desire and that there was a huge budget for that. Or maybe it was the very particular sound of conservatory-trained voices (and all the gender/access stuff that sound comes with) singing some actual thoughts I’ve had in the context of a Broadway musical, which is a pretty standardized form (though Fun Home took some major liberties) that I generally don’t feel compelled to deeply emotionally connect to as an artist, or so I thought.

On January 13, 2016, Markey's show "A Ride on the Irish Cream," which was funded in part by an Indiegogo campaign, premiered at the Abrons Arts Center of the Henry Street Settlement. The show has been described as

a musical anchored inside the memory of a Michigan backyard on the bank of the Kawkawlin River. A live band and original score become the space for the thrills and terrors of a relationship between Reagan (Markey), a vainglorious self-made girl, and Irish Cream (Becca Blackwell), her family's pontoon boat/horse. They are in love, but when their relationship is tested by dust ruffles, sex for money, severe T-storms, and a secret cellar, the only way to stay together is to remember all the parts of themselves their bodies tried to forget.

Markey has stated that they consider alt-cabaret performer Bridget Everett a "possible career model" for her. Describing their troubles writing an accurate description of their career, Markey said that the "most truthful" description of what they do is "Erin Markey makes stuff for stage and video that has music in it."

Television Work

Markey has appeared in the Season 3 of the HBO series High Maintenance, playing the character Cori. The Atlantic described their performance as a "lived-in depiction of blue-collar crunchiness."

References

Additional sources

External links
 Erin Markey's website
 Erin Markey's Youtube channel
 Erin Markey's Twitter
 I Deserve This, a blog by Erin Markey and Lorelei Lee

Category:1981 births
Category:Living people
Category:American performance artists
Category:Performance art in New York City
Category:University of Michigan alumni
Category:21st-century American artists
Category:21st-century women artists
Category:People from Brooklyn


436 S.W.2d 234 (1968)
Marjorie Saunders WATTENBURGER et al., Appellants,
v.
Frank MORRIS, Jr., et al., Ex. and Tr., Appellees.
No. 16971.
Court of Civil Appeals of Texas, Fort Worth.
December 6, 1968.
Rehearing Denied January 17, 1969.
*236 Stark & Underwood, and Richard S. Stark, Gainesville, for appellants.
John Atchison, and Ray Winder, Gainesville, for appellees Frank Morris, Jr., and Claude Jones, as executors and trustees.
Sullivant & Harris, Gainesville, McDonald, Sanders, Wynn, Ginsburg, Phillips & Maddox, and Atwood McDonald, Fort Worth, for appellees L. H. Saunders, Jr., Mary Jo Bullard and husband Ralph Bullard, Marie Saunders, Nancy Saunders and Vera Gertrude Saunders.

OPINION
MASSEY, Chief Justice.
Owen Saunders, Sr., often to be referred to as the testator, died testate on August 29, 1960. By an original will executed August, 21, 1944, he had made material provisions for the benefit of his three sons, Owen T. Saunders, Blackwood Saunders, and L. H. Saunders. Concerning the provisions therein made, particularly as applied to his reasons for burdening the inheritance of his three sons by placing the real estate in trust for them for 25 years after the date of his death, and the money in trust for them for 20 years, he stated that he was "persuaded that it is wise to so limit the disposition thereof as to assure them and each of them a competency in their old age, realizing that those who become the beneficiaries of property without having earned it usually do not appreciate its real value and are less likely to manage it in a wise and prudent manner."
Alleged date of Owen T. Saunders' death was November 29, 1951. He was *237 one of the three sons of the testator, for whose benefit the testator had left real property and money in trust by will. August 24, 1954 the testator executed a codicil, one reason therefor being that Owen T. Saunders was dead, leaving three children surviving. They were Marjorie Saunders, Joyce Saunders, and Owen T. Saunders, Jr.
The material language of the August 24, 1954 codicil read:
"1. Since the making of my said will (reference made was to original will) certain conditions have changed and it is my desire, in lieu of all bequests, devise or other benefits whatsoever which my grandchildren, Owen Thomas Saunders, Margie Saunders and Joyce Saunders, would or could have under my said will and testament, to bequeath to my grandchildren, Owen Thomas Saunders, Margie Saunders and Joyce Saunders, being the children of my deceased son, Owen Saunders, Two Thousand Dollars each."
The executors named by Owen Saunders' will were Frank Morris, Jr., et al. They were also named trustees to hold, manage and control the realty for 25 years and the money for 20 years.
As Morris, et al. interpreted the will they deemed it their duty as trustees to manage the estates of which they acted as trustees for the periods prescribed, for the use and benefit of Blackwood and L. H. Saunders (Owen T. Saunders having predeceased the testator, and the codicil aforementioned having been thereafter executed). They deemed it their duty, as executors of the will, to pay to each of the grandchildren who were the children of Owen T. Saunders (who had died before his father, the testator) the $2,000.00 prescribed in the codicil previously mentioned and from which we have quoted. After qualifying before the Probate Court they did pay $2,000.00 to each of said grandchildren in the latter part of the year 1960.
Before making payment of the several $2,000.00 bequests the executors of the will required the execution and delivery by each grandchild of a purported release. Each such read substantially, as follows:
"Now comes (beneficiary named) and acknowledges receipt of the sum of $2,000.00 in full payment of the bequest to the said (beneficiary named) contained in the last will of Owen Saunders, Sr., deceased, and in consideration thereof hereby releases said estate from any and all claims of every kind whatsoever."
It is not a matter of dispute in the case and may be accepted as factthat each grandchild, Owen Thomas Saunders, Marjorie Saunders Wattenburger and Joyce Saunders Lankford (together with the husbands of Marjorie and Joyce) were "on notice" of the testator's provisions by the will and codicils thereto as of the times each executed the purported release form.
A little less than five (5) years after the testator's death, his son, L. H. Saunders, died intestateas of May 7, 1965, leaving as his only heirs his widow, Marie, and his two children, L. H. Saunders, Jr. and Mary Jo Saunders Bullard.
A little more than five (5) years after the testator's death, his son, Blackwood Saunders, died testateas of November 24, 1965, leaving as his only heirs his widow, Gertrude, and his only child, Nancy Saunders.
Though the original pleadings filed November 18, 1966, in the District Court of Cooke Countydid not present the executor's/trustee's petition for declaratory judgment in the regular form we have no doubt but that the state of all the pleadings in the case properly invoked the jurisdiction of the court to render the judgment it did render as of trial date beginning February 19, 1968. We so hold.
By the judgment the court, among other things found and declared:
That the second codicil to the will revoked the provisions of the original will in favor of Owen T. Saunders and that the widow and children of L. H. Saunders, deceased, and the widow and child of Blackwood *238 Saunders, deceased, are the only persons parties to the suitwho have any right, title, or interest in or any claim against the estate of Owen Saunders, Sr.; that the real estate trust and the money trust created by the will of said testator terminated on November 24, 1965 when Blackwood Saunders died, he being the last surviving son of the testator; that the property at that time remaining, and in trust, vested in the widows and children of L. H. and Blackwood Saunders, both deceased, subject to the winding up and closing of the trusts; and that in the winding up and closing of the trusts the trustees have authority to sell and convey real estate and perform all reasonably necessary functions and are allowed a reasonable time within which to act. Further declarations concerned amounts from the trusts to which the trustees were entitled as fees and for expenses, including legal fees, necessarily incident to suit and for the management and handling of the trust estates, plus amounts awarded as attorneys' fees to pay named guardians ad litem appointed by the court in behalf of interested minor children who were made parties to the suit.
To said judgment and decree an appeal was taken by the three children of Owen T. Saunders, deceased, and/or the children of his children, as lineal descendants and heirs entitled to share in the estate of Owen Saunders, Sr. who died August 29, 1960.
We affirm the judgment.
Three important questions are encountered in connection with the determination of whether the lineal descendants and heirs of Owen T. Saunders, who died in 1951 prior to the death of Owen Saunders, Sr. on August 29, 1960:
(1) Whether they were eliminated as being entitled to anything under the will in question other than the $6,000.00 ($2,000.00 for each of the three children of Owen T. Saunders) because of language in the codicil of August 24, 1954. We have concluded, and therefore hold, that the language of the codicil revoked the bequests contained in the original will (of August 21, 1944) in favor of Owen T. Saunders.
(2) In connection therewith it is of importance that consideration be given to the question of whether the language of the codicil in question be held to be mandatory rather than precatory. We have concluded, and therefore hold, that the language should be construed as mandatory.
(3) But, in the event we err in construing the language to be mandatoryand assuming such to have been precatory, whether the children of Owen T. Saunders, deceased, are precluded from claiming under the original will (of August 21, 1944), either (a) under the theory that in accepting $2,000.00 each (as provided by the codicil of August 24, 1954), and in executing the instruments of release, they released any claims they otherwise might have had under the will, and/or (b) whether they are now precluded from claiming under the will by reason of the doctrine of election of rights and remedies. We have concluded, and therefore hold,in the assumption that the language of the codicil be properly treated as precatory merely (contrary to our belief)that they are now precluded from claiming under the will by reason of the doctrine of election of rights and remedies, but should not be held bound under the theory that they had released and settled any claim to which they were otherwise entitled merely because they had executed instruments purporting to be a release.
Only in the event of error on our part under the foregoing matters, upon which we have stated our holdings, would points of error presented in connection with the trusts and termination thereof become important. If we have correctly decided that the complainants have no further interest in or right to take anything remaining of the estate of the testator they would not be interested in the duration of any trust affecting it nor could they be heard to complain upon the matter. Should we have erred thereupon, and should it be ultimately held that the children *239 and lineal descendants of Owen T. Saunders, deceased, are entitled to a proportionate interest in the estate, we are of the opinion, and therefore hold, that the trusts terminated at the times and subject to the rights of the trustees in every respect as determined by the trial court.
Where there are codicils to be taken into consideration in a case requiring will construction it is to be presumed in law that the testator has re-examined the testamentary disposition in the original will before executing the codicil; the will and codicil are to be regarded as one instrument, speaking from the date of the codicil; and where it is practically possible to construe the codicil so as to make it consistent with the purposes declared in the body of the will such should be done, it being proper to hold that provision of the codicil would override a will provision and prevail over it only when the provisions can not be reconciled. Boyd v. Frost Nat. Bank, 145 Tex. 206, 196 S.W.2d 497, 168 A.L.R. 1326 (1946); Hinson v. Hinson, 154 Tex. 561, 280 S.W.2d 731 (1955); 61 Tex.Jur.2d, p. 249, "Wills", Sec. 127, "Construction of codicils".
The first concern of the court in construing a will is to ascertain the intent of the testator, and the second is to give effect to that intention if not unlawful. A construction of a testamentary provision that has the effect of defeating or thwarting the intention and purpose of the testator, as expressed by the will as a whole, should be avoided. 61 Tex.Jur.2d, p. 266, "Wills", Sec. 145, "Duty of court", and p. 292, Sec. 165, "Defeat of testator's intent".
In will construction form should be subordinated to substance and if the testator's intention can be ascertained, from a reading of the whole will, such intention should be given effect, however informal the language. Where, from the context of the will and surrounding circumstances (application to the physical facts) the testator's intention is manifest, though somewhat obscured by an inaccurate mode of expression, the language will be subordinated to the intention. In order to carry out the intention of the testator, as expressed by the whole will, the court will, so far as possible, depart from the strict words, and read a word or phrase in a sense different from that which is ordinarily attributed thereto, and for such purpose may change the language of the will, such as by rejecting superfluous words or restricting them in their application. This rule, of course, only applies where it is necessary in order to effectuate the testator's intention, clearly apparent from the will. The whole scheme of the will should be kept in mind and the will construed from its four corners so as to give effect to the clearly expressed intention of the testator. McMullen v. Sims, 37 S.W.2d 141 (Tex.Com.App., 1931).
The presentation on appeal, made in connection with the contention that the codicil's provision is precatory merely is based upon the use of the word "desire" in the language we have earlier quoted from the codicil of August 24, 1954. We have once stated our opinion and holding that the whole will, including this codicil, should be construed in a manner which would necessitate treating the questioned provision of the codicil as mandatory.
In determining whether particular words are to be construed as precatory or mandatory, the court will look to the expressed intent of the testator, as found from the context of the will and surrounding circumstances; and words which are precatory in their ordinary meaning will nevertheless be construed as mandatory when it is evident that such was the testator's intent. Where words of recommendation, request, desire, and the like are used in direct reference to the disposition of the testator's own property and show a clear intent to make such disposition without the intervention of any act by the first donee, they are ordinarily regarded as imperative and testamentary rather than precatory. Further, distinctions have been drawn between *240 expressions of desire directed to beneficiaries, to executors, and to trustees. A wish directed to a beneficiary is generally regarded as precatory, unless the words clearly express the testator's intention to the contrary; but where the words are addressed to an executor, they are more often regarded as mandatory, or at least prima facie mandatory. 95 C.J.S. "Wills" § 602, p. 801, "(General Principles of ConstructionLanguage of Will) Advisory and Precatory Words, (b) Whether Precatory or Mandatory."
We are referred to two authoritative Texas cases in which language contended to be precatory merely was held to be mandatory in each instance because of and upon its consideration along with the entire will. Bergin v. Bergin, 159 Tex. 83, 315 S.W.2d 943 (1958); Levin v. Fisch, 404 S.W.2d 889 (Eastland, Tex.Civ.App., 1966, ref. n. r. e.).
In our opinion no one could reasonably believe that the testator had any intention other than to make the codicil an effective part of his will, despite his use of the word "desire" therein. If such was his intent it necessarily follows that the questioned provision of the codicil should be deemed mandatory. The testator decided, in view of the death of his son, Owen T. Saunders, to give to his grandchildren, who were the children of the deceased son, the sum of $2,000.00 each, in cash, in lieu of leaving them beneficiaries of the trusts.
In said connection we hold that Texas' Anti-Lapse Statute, V.A.T.S., Probate Code, Sec. 68, "Prior Death of Legatee", was inapplicable. By force of Section 68 the children of Owen T. Saunders, deceased, would have stood to inherit under the will to the same extent and by the same manner as would their father, had the testator not executed the codicil after Owen T. Saunders died. It is to be presumed that the testator was aware of the law in this respect, and that his action in executing the codicil was in derogation of rights otherwise existent in his grandchildren but for such action on his part. By the codicil each grandchild was given $2,000.00 "in lieu of all bequests, devise or other benefits whatsoever which my grandchildren, Owen Thomas Saunders, Margie Saunders and Joyce Saunders, would or could have under my said will and testament."
We hold that the purported releases that the three children of Owen T. Saunders were required to execute as a condition of their individual rights to receive $2,000.00 in cash upon the death of the testator, and under the provisions of the codicil, were inoperative and without effect as a release of any other right they might have because there was no consideration to support same. However, we hold as heretofore indicatedthat they may not take under the will as it is to be considered apart from and with an absence of provisions thereof as embodied in the codicil because they must be held to have "elected" to take under the codicil "in lieu of" taking under the original will. Where an election is once made by a party bound to elect, with full knowledge of all the facts, his election binds him and those who claim under him though made in ignorance of the law. See cases under 72 A.L.R. 1135 and 28 A.L.R.2d 116, Anno.: "Estoppel to contest will or attack its validity"; 21 Tex.Jur.2d, p. 196 et seq., "Election of Remedies", all sections; Calvert v. Fort Worth National Bank, 348 S.W.2d 19 (Austin, Tex.Civ.App., 1961, affirmed at 163 Tex. 405, 356 S.W.2d 918 (1962)).
The foregoing constitutes a discussion, with holdings indicated, of what we consider to be the material and controlling questions determinative of the appeal. No worthwhile purpose would be served by lengthening the opinion. All points of error have been severally considered and are overruled.
Judgment is affirmed.


Posted in: BlackBerry

Today, at the BlackBerry 10 launch event, the BlackBerry’s CEO announced Alicia Keys will be company’s Global Creative Director. Now I will try to shed some light on what the singer’s new job actually means.

According to the press release, she will be working closely with developers, content creators, carriers and retailers to promote, innovate and expand the capabilities of the new BlackBerry 10 OS and devices.

The lovely Miss Alicia Keys will be in charge of the Keep Moving project and will be working along with the director Robert Rodriguez and the author Neil Gaiman. Here is the promo video.

As far as I understand, she’ll be in charge of the promotional campaign of the new platform and will be responsible for some fresh ideas that will hopefully make it into the next versions of the BlackBerry OS.

This all sounds great, but is it really going to help the BlackBerry’s future? I like Alicia Keys as a musician and I like the new BlackBerry Z10, but this just seems like bag of money thrown away. Only time will tell if this was actually a smart move by BlackBerry, I guess.

Source


package com.webank.cmdb.dto;

public class EnumPair {
    private int codeId;
    private String code;
    private String name;

    public EnumPair() {
    }

    public EnumPair(int codeId, String code, String name) {
        this.codeId = codeId;
        this.code = code;
        this.name = name;
    }

    public String getCode() {
        return code;
    }

    public void setCode(String code) {
        this.code = code;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public int getCodeId() {
        return codeId;
    }

    public void setCodeId(int codeId) {
        this.codeId = codeId;
    }
}