text:
Pelosi Era Draws to Close as GOP Flips House, Democrats Look for Change
By Anand Shrivastava
(Bloomberg) — Nancy Pelosi's historic run as speaker will come to an end with the Democrats' unexpectedly narrow loss of their US House majority, a power flip that comes as a new generation of leaders clamors to take over the party.
Pelosi, 82, has been the Democratic House leader for almost two decades and is the only woman to be elected speaker, serving in two stretches and under four presidents. She will announce "her future plans" on Thursday, her spokesman, Drew Hammill, tweeted late Wednesday night.
Yet despite her prodigious fundraising prowess and tight control of the caucus, Pelosi's hold on the top spot was challenged even before the midterm elections as centrists and progressives alike agitated for change.
Last week, she was re-elected to her 19th term representing her San Francisco district. On Sunday, Pelosi declined to say whether she would run for caucus leader again in a closed-door party election this month.
"The fact is that any decision to run is about family and also my colleagues. What we want to do is go forward in a very unified way," she said on CNN's "State of the Union."
Pelosi has said that the vicious attack on her husband, Paul Pelosi, in their home by a hammer-wielding assailant on Oct. 28 would be a factor in her plans. But she has also said she'll always have influence over House Democrats, whether or not she's their leader.
Pelosi still has broad support among those Democrats, and the party's better-than-expected performance in the midterms, when it was expected to lose as many as 25 seats, has only served to enhance her prestige.
"Everyone is giving her a wide berth," Representative Rick Larsen of Washington State said Tuesday. "No one works harder than Nancy Pelosi."
President Joe Biden and Senate Majority Leader Chuck Schumer both have encouraged her to stay. Pelosi, in a statement Wednesday night, praised her party for having "defied expectations" in the elections.
"House Democrats will continue to play a leading role in supporting President Biden's agenda — with strong leverage over a scant Republican majority," she said.
Still, the loss of the majority is likely to trigger a broader struggle over control and direction of the party in the House. Majority Leader Steny Hoyer, 83, and Majority Whip James Clyburn, 82, may also come under pressure to step aside, though neither has given an indication they're ready to give up leadership posts.
In a late September meeting with reporters, Hoyer swatted away questions about Pelosi's promise in late 2018 to abide by a term-limit deal with some caucus members to step down from leadership by the end of this term. He said, "We'll see what she does," adding: "I think she will have to answer for herself. I will answer for myself."
The Republican takeover "essentially guarantees a leadership transition," said Joshua Huder, a senior fellow at Georgetown University's governmental affairs institute.
New York Representative Hakeem Jeffries, 52, is widely considered the heir apparent to Pelosi. He has been the chairman of the House Democratic Caucus since 2019, and he has also taken on other leadership roles, most notably as one of seven House managers in the first impeachment trial of President Donald Trump.
Other Democrats expected to seek leadership roles are Massachusetts Representative Katherine Clark, 59, who is the No. 4 House Democrat, and Washington State Representative Pramila Jayapal, 57, the chair of the Congressional Progressive Caucus.
Much of the demand for change has come from the Democrats who got elected in the 2018 midterms and as such were instrumental giving Pelosi a chance to return to the speakership.
"I think we absolutely need a lot of fresh voices in leadership. It's been 20 years," Representative Abigail Spanberger, who was part of that 2018 wave, said at a pre-election campaign event in Fredericksburg, Virginia. "Issues have changed, policies have changes, Congress has changed."
Spanberger, who was re-elected last week, refused to support Pelosi as speaker in 2019 and 2021.
Pelosi has often deflected questions about whether she will keep to the agreement with restive House Democrats to step down from leadership at the end of this congressional term.
"Yes, we need generational change. Of course we do," Pelosi said last month in an interview on MSNBC. "But in some cases there's no substitute for experience."
This is Pelosi's second loss of the House majority as speaker, a position that puts her second in line to the presidency. She first became speaker in 2007 and was re-elected in 2009, but lost the gavel in 2011 after Republicans, propelled by the conservative Tea Party movement, won the House majority in the 2010 election.
Pelosi, a master legislative tactician, was elected speaker again in January 2019 after she had led Democrats back to the majority in the 2018 mid-term elections under Trump. However to secure the role, she had to acquiesce to a term-limit push from factions within her caucus and commit to not seeking the gavel or top House party post beyond the term that ends in January.
Even if Pelosi is entering her final weeks as the top House Democrat, her place in history will go far beyond the barriers she broke as a mother of five first elected to Congress at age 47, then rising through party ranks and eventually ascending to the speakership.
Her career will also be recalled for the often polarizing political battles she waged under four administrations, helping to shape some of the most consequential legislation in modern American history, and in overseeing the two Trump impeachments.
Among her biggest accomplishments was helping to enact the Affordable Care Act with President Barack Obama. She was also instrumental in passing the Dodd-Frank Act to reform Wall Street after the Great Recession, as well as the economic stimulus package to help banks and others recover from that financial crisis.
More recently, she managed her fractious caucus to pass large parts of Biden's economic agenda, including last year's bipartisan infrastructure bill and this summer's Inflation Reduction Act, which included historic investments in climate programs. She also reasserted her longtime hawkish stance on China with a visit this summer to Taiwan despite outrage from Beijing.
Her challenges have included guiding Congress's responses to the coronavirus pandemic and the ongoing investigation into the attack on the Capitol in January 2021.
–With assistance from Mackenzie Hawkins.
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text:
Marcus Breger (* 18. Mai 1905 in Czernowitz, Bukowina, Österreich-Ungarn; † 30. September 1975 in Tucson, Arizona, Vereinigte Staaten) war ein österreichisch-rumänisch-US-amerikanischer Rabbiner und Hochschullehrer.

Leben
Marcus Breger wurde am 18. Mai 1905 in Czernowitz, der traditionellen Hauptstadt der Bukowina, als Sohn von Marcus und Liebe Breger (geborene Druckmann) geboren. Von 1924 bis 1931 studierte er am Jüdisch-Theologischen Seminar in Breslau und besuchte gleichzeitig Studiengänge an der Universität Breslau, an der er im Jahre 1930 promovierte. Bereits im darauffolgenden Jahr legte er seine Rabbinerexamen ab und fungierte von 1931 bis 1932 als Dozent in Berlin. In dieser Zeit war er auch Mitredakteur beim Verfassen der deutschsprachigen Ausgabe der Encyclopaedia Judaica und veröffentlichte weitere Beiträge in Fachzeitschriften in Deutschland. In seiner 1932 veröffentlichten Dissertation schrieb er über Beiträge zur Handelsgeschichte der Juden in Polen während des 17. Jahrhunderts. Nach der Machtergreifung Hitlers emigrierte Breger 1933 nach Italien, ehe er von 1933 bis 1938 als Professor für Talmudwissenschaften und Geschichte am Collegio Rabbinico Convito auf dem von den Italienern besetzten Rhodos tätig war. Gleichzeitig war Breger Oberhaupt des religiösen Gerichts.

1938 wurde er von Italien ausgewiesen und emigrierte im Dezember 1938 bzw. Januar 1939 mit Frau und Kind auf der S.S. Rex, wo er als Marcus Breger-Landesberg in den Passagierlisten erwähnt wird, von Genua über New York City in die Vereinigten Staaten. Dort war er von 1939 bis 1950 als Rabbiner tätig und diente parallel dazu von 1943 bis 1946 als Militärgeistlicher bei der United States Army. Nachdem er 65 Jahre alt geworden war, wurde Breger Rabbiner emeritus der Congregation Anshei Israel in Tucson im US-Bundesstaat Arizona, wo er in den letzten Jahrzehnten bis zu seinem Tod gelebt hatte. Während dieser Zeit war er im Jahr 1971 Professor für Judaistik an der University of Missouri, sowie von 1971 bis 1975 erster Professor für Judaistik an der University of Arizona.

Zeitlebens war Breger Mitglied der Zionist Organization of America (ZOA), des Rabbinical Council of America, der Jewish War Veterans of the United States of America, der Organisation B'nai B'rith, der Military Chaplains Association, der Association of Jewish Chaplains, des Rotary Clubs und des Great Tucson Jewish Community Councils. Er war wiederholt in Zeitungen und im Fernsehen zu sehen. So hatte er im Jahr 1955 einen Auftritt in This Is Your Life mit Ralph Edwards. Am 30. September 1975 starb Breger 70-jährig in Tucson und wurde am Evergreen-Memorial-Park-Friedhof in Tucson beerdigt. Zu seinen Ehren führte The Arizona Center of Judaic Studies der University of Arizona nach seinem Tod das The Rabbi Marcus Breger Memorial Lectureship ein. Zudem betreibt die Universität den Rabbi Marcus Breger Memorial Fund.

Familie
Marcus Breger wurde als Sohn von Jacob und Liebe Breger (geborene Druckmann) geboren. Zu seinen Geschwistern zählen Tony Augenblick und Sol Breger. Im Jahre 1933 heiratete er die jüdische Zahnärztin Bertha Rittenberg (* 1909 in St. Petersburg; † 2001 in Tucson, Arizona). Am 9. Oktober 1936 kam auf Rhodos die gemeinsame Tochter Liba Breger (verheiratete Feuerstein; † 2. Oktober 2013 in Kalifornien), eine spätere Lehrerin, zur Welt. Verschiedenen Quellen zufolge soll er auch noch ein weiteres Kind mit seiner Frau bekommen haben.

Breger wurde mit österreichischer Staatsangehörigkeit geboren, nahm 1918 die rumänische und 1943 die US-amerikanische Staatsangehörigkeit an.

Literatur
 Werner Röder (Hrsg.): Biographisches Handbuch der deutschsprachigen wirtschaftswissenschaftlichen Emigration nach 1933–1945, Band 1: Politik, Wirtschaft, Öffentliches Leben. De Gruyter, Berlin/Boston, 1980, ISBN 978-3-598-11420-5, S. 90
 Susanne Blumesberger, Michael Doppelhofer, Gabriele Mauthe: Handbuch österreichischer Autorinnen und Autoren jüdischer Herkunft 18. bis 20. Jahrhundert. Hrsg. von der Österreichische Nationalbibliothek. Saur, München 2002, ISBN 3-598-11545-8, S. 163–164.

Weblinks

 U Of A Research (englisch)

Einzelnachweise

Österreichischer Emigrant zur Zeit des Nationalsozialismus
Österreichischer Emigrant in den Vereinigten Staaten
Jüdischer Militärseelsorger
Rabbiner (Vereinigte Staaten)
Hochschullehrer (University of Missouri)
Hochschullehrer (University of Arizona)
Person (Cisleithanien)
Rumänischer Emigrant in den Vereinigten Staaten
Österreicher
Rumäne
US-Amerikaner
Geboren 1905
Gestorben 1975
Mann

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

Let $X$ be a random vector in $\R^n$ with density $f$ with respect to the Lebesgue measure. Its Shannon entropy is defined as
$$
h(X)=-\int_{\R^n} f(x)\log f(x)dx.
$$
The entropy power of $X$ is defined as
$$
N(X)=e^{2h(X)/n}.
$$
Shannon's Entropy Power Inequality (i.e., EPI) states that for any two independent random vectors $X$ and $Y$ in $\R^n$ such that
the entropies of $X$, $Y$ and $X+Y$ exist,
\begin{align}\label{eq:epi}
N(X+Y)\geq N(X)+N(Y).
\end{align}
This result was first stated by Shannon \cite{Sha48} with an incomplete proof; the first rigorous proof was provided by Stam \cite{Sta59} using Fisher information and de Bruijn's identity. Various proofs have been found using a number of different approaches, including Young's inequality \cite{Lieb78, DCT91}, restricted Brunn-Minkowski sum \cite{SV00}, minimum mean square estimation (MMSE) \cite{VG06}, and optimal transport \cite{Rio17}. Although the EPI is interesting in its own right, it has further significant impact by connecting several important and active research areas, including functional inequalities on Riemannian manifolds, inequalities in convex geometry, entropic versions of results in additive combinatorics and bounding the capacity of communication channels. We refer to the nice survey by Madiman, Melbourne and Xu \cite{MMX16} for more details.

As a generalization of Shannon entropy, R\'enyi entropy is important in quantum information theory, where it can be used as a measure of entanglement. The R\'enyi entropy is also important in ecology and statistics as an index of diversity. For $p\in(0, 1)\cup (1, \infty)$, the R\'enyi entropy of order $p$ (i.e., $p$-R\'enyi entropy) is defined as
$$
h_p(X)=\frac{1}{1-p}\log\int_{\R^n}f(x)^pdx.
$$
Defining by continuity, $h_1(X)$ corresponds to the classical Shannon entropy. By taking limits, we have
$$
h_0(X)=\log|\supp(f)|,
$$
$$
h_\infty(X)=-\log\|f\|_\infty,
$$
where $\supp(f)$ is the support of $f$, and $|\supp(f)|$ is the volume of $\supp(f)$, and $\|f\|_\infty$ is the essential supremum of $f$. The $p$-R\'enyi entropy power is defined by
$$
N_p(X)=e^{2h_p(X)/n}.
$$
Unlike Shannon entropy power, which captures the variances of real-valued Gaussian random variables, the (operational) meaning of R\'enyi entropy power is not clear for the author. It is observed by Bobkov and Chistyakov \cite{BC15} that Shannon's EPI \eqref{eq:epi} does not always hold for the R\'enyi entropy power. So, a more natural definition of the R\'enyi entropy power is necessary. One variant of the R\'enyi entropy power is $N_p^{\alpha}(X)$ with an appropriate exponent $\alpha$ which depends on $p$. Analogous to the EPI \eqref{eq:epi}, one would like the following R\'enyi EPI to hold for independent random vectors
\begin{align}\label{eq:r-epi}
N_p^{\alpha}(X+Y)\geq N_p^{\alpha}(X)+N_p^{\alpha}(Y).
\end{align}
For instance, the Brunn-Minkowski inequality suggests that $\alpha=1/2$ for $p=0$.
For $p>1$, Bobkov and Marsiglietti \cite{BM16} recently showed that \eqref{eq:r-epi} holds for $\alpha=(p+1)/2$. In the next section, we will show that \eqref{eq:r-epi} holds with a smaller exponent.

Related to the R\'enyi EPI is Savar\'e and Toscani's extension \cite{ST14} of Costa's concavity of entropy power theorem \cite{Cos85} from Shannon entropy to R\'enyi entropy. More specifically, let $u_t$ solve the non-linear heat equation
$
\partial_t u_t=\Delta u_t^p
$
in $\R^n$. For $p>1-2/n$, they showed  that $N_p^\alpha(u_t)$ is a concave function of $t>0$, where $\alpha=1+n(p-1)/2$. Another variant of the R\'enyi EPI obtained by Bobkov and Chistyakov \cite{BC15} states that, at the expense of some
universal factor $c$, for $p>1$ and independent random vectors $X_1, \cdots, X_k$,
$$
N_p(X_1+\cdots+X_k)\geq c\sum_{i=1}^kN_p(X_i).
$$
A sharpened version was obtained by Ram and Sason \cite{RS16}.

In general, the EPI can not be reversed. There exist independent real-valued random variables $X, Y$ with finite entropies, but $h(X+Y)=\infty$, see \cite{BC15} for examples. The reversibility of EPI was addressed by Bobkov and Madiman \cite{BM12}, which states that for independent $\kappa$-concave (the definition is given in Section \ref{sec:conjs}) random vectors $X, Y$ in $\R^n$, there exist linear volume preserving maps $T_1, T_2$ such that
$$
N(T_1(X)+T_2(Y))\leq c\cdot(N(X)+N(Y)),
$$
where $c$ is some constant depending on $\kappa$. This can be thought of as the functional lifting of V. Milman's well known reverse Brunn-Minkowski inequality \cite{Mil86}. The following reverse R\'enyi EPI proposed by Ball, Nayar and Tkocz \cite{BNT15} (Shannon entropy version for log-concave random vectors) and Madiman, Melbourne and Xu \cite{MMX16} states that for any symmetric $\kappa$-conave random vector $(X, Y)$ in $\R^2$,
\begin{align}\label{eq:r-r-epi}
N_p^{1/2}(X+Y)\leq N_p^{1/2}(X)+N_p^{1/2}(Y).
\end{align}
Toward this conjecture, it was shown in \cite{BNT15} that \eqref{eq:r-r-epi} holds for $p=1$ in the log-concave case with an exponent $1/10$ instead of $1/2$. Extension in the R\'enyi entropy setting can be found in \cite{MMX16}. We will verify the conjecture in two cases $p=0$ and $p=2$ for log-concave random vectors.

The paper is organized as follows. In Section \ref{sec:r-epi}, we follow the approach of Dembo, Cover and Thomas \cite{DCT91} of employing Young's inequality to derive an improvement of the R\'enyi EPI recently obtained by Bobkov and Marsiglietti \cite{BM16}. Section \ref{sec:conjs} is devoted to the discussion of various formulations of the conjecture \eqref{eq:r-r-epi}. In particular, we will relate the conjecture to the convexity of the $p$-th cross-section bodies introduced by Gardner and Giannopoulos \cite{GG99}. We verify the conjecture \eqref{eq:r-r-epi} in two cases $p=0, 2$ in Section \ref{sec:special cases}. Connections among various $p$-th mean bodies are discussed in Section \ref{sec:convex bodies}.

\section{R\'enyi EPI} \label{sec:r-epi}

The following statement about the R\'enyi EPI is analogous to Lieb's observation of an equivalent formulation of Shannon's EPI \cite{Lieb78}. For the sake of completeness, we include the proof.

\begin{lem}\label{lem:linearization}
Let $p\geq0$ and $\alpha>0$ be some constants. Suppose that $X, Y$ are independent random vectors in $\R^n$ such that $X, Y$ and $X+Y$ have finite $p$-R\'enyi entropies. The following statements are equivalent:
\begin{enumerate}
\item
We have
\begin{align}\label{eq:r-epi-1}
N_p^{\alpha}(X+Y)\geq N_p^{\alpha}(X)+N_p^{\alpha}(Y).
\end{align}

\item
For any $\lambda\in[0, 1]$, we have
\begin{align}\label{eq:renyi-epi-linear}
h_p(\lambda^{\frac{1}{2\alpha}}X+(1-\lambda)^{\frac{1}{2\alpha}}Y)\geq \lambda h_p(X)+(1-\lambda)h_p(Y).
\end{align}
\end{enumerate}
\end{lem}

\begin{proof}
To see that \eqref{eq:r-epi-1} implies \eqref{eq:renyi-epi-linear}, it suffices to assume that $\lambda^{\frac{1}{2\alpha}}X+(1-\lambda)^{\frac{1}{2\alpha}}Y$ has finite entropy. By definition,
\begin{align*}
h_p(\lambda^{\frac{1}{2\alpha}}X+(1-\lambda)^{\frac{1}{2\alpha}}Y) &= \frac{n}{2\alpha}\log N_p^\alpha(\lambda^{\frac{1}{2\alpha}}X+(1-\lambda)^{\frac{1}{2\alpha}}Y)\\
&\geq \frac{n}{2\alpha}\log(N_p^\alpha(\lambda^{\frac{1}{2\alpha}}X)+N_p^\alpha((1-\lambda)^{\frac{1}{2\alpha}}Y)
)\\
&= \frac{n}{2\alpha}\log\left(\lambda N_p^\alpha(X)+(1-\lambda)N_p^\alpha(Y)\right)\\
&\geq \frac{n}{2\alpha}\left(\lambda \log N_p^\alpha(X)+(1-\lambda)\log N_p^\alpha(Y)\right)\\
&= \lambda h_p(X)+(1-\lambda)h_p(Y).
\end{align*}
We use \eqref{eq:r-epi-1} in the first inequality. The second inequality follows from the concavity of the $\log$ function. In the second identity, we use the scaling property that $h_p(aX)=h_p(X)+n\log|a|$. For the reverse, i.e. \eqref{eq:renyi-epi-linear} implies \eqref{eq:r-epi-1}, it is not hard to check that
$$
h_p(\lambda^{-\frac{1}{2\alpha}}X)=h_p((1-\lambda)^{-\frac{1}{2\alpha}}Y),
$$
where
$$
\lambda=\frac{N_p^\alpha(X)}{N_p^\alpha(X)+N_p^\alpha(Y)}.
$$
Then, we have
\begin{align*}
h_p(X+Y) &= h_p(\lambda^{\frac{1}{2\alpha}}\cdot\lambda^{-\frac{1}{2\alpha}}X+(1-\lambda)^{\frac{1}{2\alpha}}\cdot(1-\lambda)^{-\frac{1}{2\alpha}}Y)\\
&\geq h_p(\lambda^{-\frac{1}{2\alpha}}X)\\
&= h_p(X)-\frac{n}{2\alpha}\log \lambda\\
&=\frac{n}{2\alpha}\log(N_p^\alpha(X)+N_p^\alpha(Y)).
\end{align*}
This is equivalent to the desired statement.
\end{proof}

\begin{thm}\label{thm:r-epi}
Let $p>1$. Let $X, Y$ be independent random vectors in $\R^n$ such that $X, Y$ and $X+Y$ have finite $p$-R\'enyi entropies. Then, we have
$$
N_p^{\alpha}(X+Y)\geq N_p^{\alpha}(X)+N_p^{\alpha}(Y),
$$
where
\begin{align*}
\alpha=\left(1+\frac{1}{\log2}\left(\frac{p+1}{p-1}\log\frac{p+1}{2p}+\frac{\log p}{p-1}\right)\right)^{-1}.
\end{align*}
\end{thm}

\begin{proof}
Let $0<\lambda<1$. Let $f$ and $g$ be the densities of $\lambda^{\frac{1}{2\alpha}}X$ and $(1-\lambda)^{\frac{1}{2\alpha}}Y$, respectively. Then, $X$ and $Y$ have densities $f_{\lambda, \alpha}(x)=\lambda^{\frac{n}{2\alpha}}f(\lambda^{\frac{1}{2\alpha}}x)$
and $g_{1-\lambda, \alpha}(x)=(1-\lambda)^{\frac{n}{2\alpha}}g((1-\lambda)^{\frac{1}{2\alpha}}x)$, respectively. By Lemma \ref{lem:linearization}, it suffices to prove \eqref{eq:renyi-epi-linear}, which is equivalent to
\begin{align*}
\|f\ast g\|_p^{-p'} &\geq \|f_{\lambda, \alpha}\|_p^{-\lambda p'}\|g_{1-\lambda, \alpha}\|_p^{-(1-\lambda) p'}\\
&= (\lambda^{\lambda}(1-\lambda)^{1-\lambda})^{-\frac{n}{2\alpha}}\|f\|_p^{-\lambda p'}\|g\|_p^{-(1-\lambda)p'},
\end{align*}
where $p'$ is the H\"{o}lder conjugate of $p$, i.e., $1/p+1/p'=1$.
For $p>1$, we have $p'>1$. The above inequality is equivalent to
\begin{align}\label{eq:equi-1}
\|f\ast g\|_p\leq(\lambda^\lambda(1-\lambda)^{1-\lambda})^{\frac{n}{2\alpha p'}}\|f\|_p^{\lambda }\|g\|_p^{1-\lambda}.
\end{align}
Using the log-convexity of $L^s$ norm, for $1\leq q\leq p$, we have
$$
\|f\|_q\leq \|f\|_1^{\theta_q}\|f\|_p^{1-\theta_q}=\|f\|_p^{1-\theta_q},
$$
where $\theta_q=1-p'/q'$ is the solution of $1/q=\theta_q+(1-\theta_q)/p$. (It also easily follows from H\"{o}lder's inequality). Similarly,  for $1\leq r\leq p$, we have
$$
\|f\|_r\leq\|f\|_p^{1-\theta_r},
$$
where $\theta_r=1-p'/r'$ is the solution of $1/r=\theta_r+(1-\theta_r)/p$.
Then, the estimate \eqref{eq:equi-1} holds if we have
\begin{align}\label{eq:equi-2}
\|f\ast g\|_p\leq(\lambda^\lambda(1-\lambda)^{1-\lambda})^{\frac{n}{2\alpha p'}}\|f\|_q^{\frac{\lambda}{1-\theta_q}}\|g\|_r^{\frac{1-\lambda}{1-\theta_r}}.
\end{align}
Given $p>1$ and $0<\lambda<1$, we can define $q, r$ such that $\lambda=p'/q'$ and $1-\lambda=p'/r'$. It is easy to check that $1+1/p=1/q+1/r$. Young's inequality says
$$
\|f\ast g\|_p\leq \left(\frac{c_qc_r}{c_p}\right)^{n/2}\|f\|_q\|g\|_r,
$$
where $c_s=s^{1/s}(s')^{-1/s'}$. Hence, the estimate \eqref{eq:equi-2} holds if we have
$$
(\lambda^\lambda(1-\lambda)^{1-\lambda})^{\frac{n}{2\alpha p'}}\geq\left(\frac{c_qc_r}{c_p}\right)^{n/2}.
$$
Particularly, we can set
$$
\alpha=\sup_{q, r}\frac{H(\lambda)}{p'\log\left(\frac{c_p}{c_qc_r}\right)},
$$
where $H(\lambda)$ is the Shannon entropy of Bernoulli($\lambda$), and the supremum is taken over all $q, r$ such that $1/q+1/r=1+1/p$. Using $c_s=s^{1/s}(s')^{-1/s'}$, we have
\begin{align*}
p'\log\left(\frac{c_p}{c_qc_r}\right) &= p'\left(\frac{\log p}{p}-\frac{\log q}{q}-\frac{\log r}{r}\right)-\left(\log p'-\frac{p'}{q'}\log q'-\frac{p'}{r'}\log r'\right)\\
&= p'\left(\frac{\log p}{p}-\frac{\log q}{q}-\frac{\log r}{r}\right)-\left(\frac{p'}{q'}\log \frac{p'}{q'}+\frac{p'}{r'}\log \frac{p'}{r'}\right)\\
&= H(\lambda)-A(\lambda),
\end{align*}
where
$$
A(\lambda)=p'\left(\frac{1}{p}\log p-\frac{1}{q}\log q-\frac{1}{r}\log r\right),
$$
which can be rewritten as
$$
p'\left(\left(1-\frac{1}{p'}\right)\log \left(1-\frac{1}{p'}\right)-\left(1-\frac{\lambda}{p'}\right)\log \left(1-\frac{\lambda}{p'}\right)-\left(1-\frac{1-\lambda}{p'}\right)\log \left(1-\frac{1-\lambda}{p'}\right)\right).
$$
Thus, we need to solve the optimization problem
$$
\alpha=\left(1-\sup_{0\leq \lambda\leq 1}\frac{A(\lambda)}{H(\lambda)}\right)^{-1}.
$$
The theorem follows from the fact that $A(\lambda)/H(\lambda)$ achieves the maximum at $\lambda=1/2$.
Notice that $H(\lambda)$ and $A(\lambda)$ are symmetric about $\lambda=1/2$. It suffices to show that $A(\lambda)/H(\lambda)$ is increasing for $0\leq\lambda\leq1/2$. The latter statement is equivalent to
\begin{align}\label{eq:increase1}
\frac{A'(\lambda)}{H'(\lambda)}\geq\frac{A(\lambda)}{H(\lambda)}, ~0\leq\lambda\leq1/2,
\end{align}
where $A'(\lambda)=\log\frac{p'-\lambda}{p'-(1-\lambda)}$, and $H'(\lambda)=\log\frac{1-\lambda}{\lambda}$.
Notice that $A(0)=H(0)=0$. We have
$$
\frac{A(\lambda)}{H(\lambda)}=\frac{\int_0^\lambda A'(s)ds}{\int_0^\lambda H'(s)ds}=\lim_{m\to\infty}\frac{\sum_{k=1}^m A'(k\lambda/m)}{\sum_{k=1}^m H'(k\lambda/m)}.
$$
The statement \eqref{eq:increase1} holds, if we can show that $A'(\lambda)/H'(\lambda)$ is increasing for $0\leq\lambda\leq 1/2$. This is equivalent to
\begin{align}\label{eq:increase2}
\frac{A''(\lambda)}{H''(\lambda)}\leq\frac{A'(\lambda)}{H'(\lambda)}, ~0\leq\lambda\leq1/2,
\end{align}
where $A''(\lambda)=\frac{1-2p'}{(p'-\lambda)(p'-(1-\lambda))}$ and $H''(\lambda)=-\frac{1}{\lambda(1-\lambda)}$. Notice that $A'(1/2)=H'(1/2)=0$. We have
$$
\frac{A'(\lambda)}{H'(\lambda)}=\frac{\int_\lambda^{1/2} A''(s)ds}{\int_\lambda^{1/2} H''(s)ds}=\lim_{m\to\infty}\frac{\sum_{k=0}^{m-1}A''(\lambda+k(1/2-\lambda)/m)}{\sum_{k=0}^{m-1}H''(\lambda+k(1/2-\lambda)/m)}.
$$
The statement \eqref{eq:increase2} holds, if we can show that $A''(\lambda)/H''(\lambda)$ is increasing for $0\leq\lambda\leq 1/2$. One can check that
$$
\frac{A''(\lambda)}{H''(\lambda)}=(2p'-1)\left(1+\frac{p'(p'-1)}{\lambda(1-\lambda)}\right)^{-1},
$$
which is increasing for $0\leq\lambda\leq 1/2$ (recall that $p'>1$). Hence, the theorem follows.
\end{proof}

\begin{rmk}
The exponent $\alpha$ in the theorem is strictly smaller than $(p+1)/2$ obtained in \cite{BM16}. For large $p$, we have $\alpha\approx \frac{p-1}{\log_2 p}$, which, in view of the lower bound $\frac{p-1}{2\log_2\left(\frac{p+1}{2}\right)}$ obtained in  \cite{BM16}, is asymptotically optimal up to a multiplicative constant. As $p\to1$, we can recover the classical EPI.
Costa, Hero and Vignat \cite{CHV03} proved that, under covariance constraints,  maximizers of R\'enyi entropies are generalized Gaussian distributions, indexed by a parameter $-\infty<\beta\leq \frac{2}{n+1}$, of the form
$$
A_\beta\left(1-\frac{\beta}{2}\|x\|^2\right)_+^{\frac{1}{\beta}-\frac{n}{2}-1}.
$$
It is not clear whether the sharp exponent can be obtained from the generalized Gaussian case. As $p\to0$, we have $\alpha\to 1/2$, which captures the classical Brunn-Minkowski inequality. But it is not clear if Theorem \ref{thm:r-epi} holds for $0<p<1$, in which case we might need a reverse H\"{o}lder argument.
\end{rmk}

\section{Reverse R\'enyi EPI} \label{sec:r-r-epi}

This section is devoted to the reversibility of the R\'enyi EPI. In general, the R\'enyi EPI can not be reversed, even in the Shannon case. Variants of the reverse R\'enyi EPI might hold for certain distributions.

\subsection{Discussion of conjectures} \label{sec:conjs}
Recall that a probability measure $\mu$ on $\R^n$ is called $\kappa$-concave if the following Brunn-Minkowski type inequality
$$
\mu(\lambda A+(1-\lambda)B)\geq (\lambda \mu(A)^\kappa+(1-\lambda)\mu(B)^\kappa)^{1/\kappa}
$$
holds for any $0\leq \lambda\leq 1$ and any non-empty Boreal sets $A, B\subseteq\R^n$. Defining by continuity, for $\kappa=0$, we have
$$
\log\mu(\lambda A+(1-\lambda)B)\geq \lambda \log\mu(A)+(1-\lambda)\log\mu(B),
$$
which characterizes the class of log-concave measures. For $\kappa=-\infty$, we obtain the largest class, whose members are called convex measures. A $\kappa$-concave measure is absolutely continuous with respect to the Lebesgue measure if $\kappa\leq 1/n$, and its density $f$ satisfies the so-called $s$-concave property
$$
f(\lambda x+(1-\lambda)y)\geq(\lambda f(x)^s+(1-\lambda)f(y)^s)^{1/s}
$$
for all $x, y\in\R^n$ such that $f(x)f(y)\neq0$ and $s=\kappa/(1-n\kappa)$. We refer to the seminal work of Borell \cite{Bor75} for the hierarchical properties of such measures.

The following conjecture was raised by Ball, Nayar and Tkocz \cite{BNT15} (Shannon entropy version for log-concave measures) and Madiman, Melbourne and Xu \cite{MMX16} (R\'enyi entropy version for general $\kappa$-concave measures).

\begin{conj}\label{conj:1}
Let $p\in[0, \infty]$. Let $\kappa\in[-\infty, 1/n]$, and let $X$ be a $\kappa$-concave random vector in $\R^n$.  The function
$$
v\mapsto
\begin{cases}
N^{1/2}_p(v\cdot X), & v\neq0\\
0, & v=0
\end{cases}
$$
defines a norm of $v\in\R^n$.
\end{conj}

It is easy to check the homogeneity. The veracity of the conjecture really depends on proving the triangle inequality
$$
N^{1/2}_p((u+v)\cdot X)\leq N^{1/2}_p(u\cdot X)+N^{1/2}_p(v\cdot X),
$$
which essentially depends on the marginal distribution of $X$ on the plane spanned by $u, v$. We can apply a linear transformation to the plane such that $u, v$ end up perpendicular without changing the convexity of the marginal distribution. Hence, the conjecture is equivalent to that
\begin{align}\label{eq:r-r-epi-1}
N_p^{1/2}(X+Y)\leq N_p^{1/2}(X)+N_p^{1/2}(Y)
\end{align}
holds for any symmetric $\kappa$-concave random vector $(X, Y)$ in $\R^2$. Similar to Lemma \ref{lem:linearization}, we have the following linearized version of \eqref{eq:r-r-epi-1}.  Let $(X, Y)$ be a symmetric $\kappa$-concave random vector in $\R^2$, and assume that $X, Y$ have equal $p$-R\'enyi entropy. For any $\lambda\in[0, 1]$, we have
\begin{align}\label{eq:r-epi-linear}
h_p(\lambda X+(1-\lambda)Y)\leq h_p(X).
\end{align}
It is not hard to check that equality holds for $p=0$.
The statement is not true without the equal entropy assumption. Dembo, Cover and Thomas \cite{DCT91} proved that for any independent random vectors $X$ and $Y$ in $\R^n$,
$$
h_0(\lambda X+(1-\lambda)Y)\geq \lambda h_0(X)+(1-\lambda)h_0(Y).
$$
As is observed in \cite{BNT15}, the statement \eqref{eq:r-r-epi-1} holds for $p=1$ in the Shannon case when $(X, Y)$ is log-concave and one marginal has the same law as the other one rescaled, say $Y\sim tX$ for some $t>0$.  The essential case $t=1$ was first observed by Cover and Zhang \cite{CZ94}.

An interesting related problem one would like to study is the maximal entropy increment $h(X+Y)-h(X)$, where $Y$ is an independent copy of $X$. Abbe \cite{Abb} conjectured that exponential distribution is the maximizer among log-concave measures for given covariance matrix. This might be related to the minimal entropy problem for log-concave measures with fixed covariance matrix. The latter is the entropy version of Bourgain's slicing problem \cite{Bou86} formulated by Bobkov and Madiman \cite{BM11-1}. The lower bound of the entropy jump $h(X+Y)-h(X)$
has been studied in \cite{BBN03, ABBN04, BN12} under the assumption that $X$ satisfies the Poincar\'e inequality.

The following statement is stronger than Conjecture \ref{conj:1}.
\begin{conj}
Let $(X, Y)$ be a symmetric $\kappa$-concave random vector in $\R^2$, and assume that $X, Y$ have equal $p$-R\'enyi entropy. The function
$$
\lambda\mapsto h_p(\lambda X+(1-\lambda)Y)
$$
is convex for $\lambda\in[0, 1]$.
\end{conj}

Before giving another geometric formulation of Conjecture \ref{conj:1}, we introduce some notations commonly used in convex geometry. Let $K\subset\R^n$ be a compact set with the origin in its interior. For $v\in{\mathbb S}^{n-1}$, the radial function $\rho_K(v)$ is defined as
$$
\rho_K(v)=\sup\{r\geq0: rv\in K\}.
$$
For a symmetric convex body $K$, it is not hard to see that $\rho_K(v)=\|v\|_{K}^{-1}$, where $\|v\|_{K}$ is the norm of $v$ for which $K$ is the unit ball.

Conjecture \ref{conj:1} is true when $p=\infty$ and $X$ is uniformly distributed over a symmetric convex body $K$. This is known as Buseman's theorem \cite{Bus49},
and the unit ball is called the intersection body, $I(K)$, of $K$.
Generalizations of Buseman's theorem hold for symmetric log-concave measures \cite{Ball88} and more general convex measures \cite{MMX16}.

As a generalization of intersection body, the $p$-th cross-section body of a convex body was introduced by Gardner and Giannopoulos \cite{GG99}. Its extension for probability measures is given below. We will see that that the $p$-th cross-section body of a random vector (its density) agrees with the unit ball associated to the norm claimed in Conjecture \ref{conj:1}.

\begin{defn}
Let $f$ be a probability density on $\R^n$. The $p$-th cross-section body $C_p(f)$ is the set with the radial function
\begin{align}\label{eq:cpf-radial}
\rho_{C_p(f)}(v)=\left(\int_{\R^n}\left(\int_{x+v^\perp}f(y)dy\right)^pf(x)dx\right)^{1/p},~v\in{\mathbb S}^{n-1},
\end{align}
where $v^\perp$ is the subspace perpendicular to $v$.
\end{defn}

It is not hard to see that the radial function can be rewritten as
$$
\rho_{C_pf}(v)=\left(\int_{\R}\left(\int_{x\cdot v=t}f(x)dx\right)^{p+1}dt\right)^{1/p}.
$$
For $v\in\R^n$, the density of $v\cdot X$ is
$$
|v|^{-1}\int_{v\cdot x=t}f(x)dx.
$$
Therefore, we have
$$
\rho_{C_pf}(v)=N^{-1/2}_{p+1}(v\cdot X).
$$
Hence, we can rephrase Conjecture \ref{conj:1} in the following way.
\begin{conj}\label{conj:2}
Let $p\in[-1, \infty]$. Let $s\in[-1/n, \infty]$, and let $f$ be a symmetric $s$-concave probability density on $\R^n$. Then, the $p$-th corss-section body $C_p(f)$ is a symmetric convex body.
\end{conj}

We introduce other types of $p$-th mean bodies, which are closely related to each other. Their relationships will be discussed in Section \ref{sec:convex bodies}. The following is a natural extension of the intersection body of a convex body to a probability density $f$.

\begin{defn}
Let $f$ be a probability density on $\R^n$. The intersection body $I(f)$ is the set with the radial function
\begin{align*}
\rho_{I(f)}(v)=\int_{v^\perp}f(x)dx,~v\in{\mathbb S}^{n-1}.
\end{align*}
\end{defn}

The following $p$-th radial mean body of a probability density function reduces to the $p$-th radial mean body of a convex body introduced by Gardner and Zhang \cite{GZ98} with some normalizing constant if the density is taken to be uniform over that convex body.
\begin{defn}
Let $f$ be a probability density on $\R^n$. The $p$-th radial mean body $R_p(f)$ is the set with the radial function
\begin{align}\label{eq:rpf-radial}
\rho_{R_p(f)}(v)=\left(\int_{\R^n}f(x)\int_{\R_+}r^{p-1}f(x+rv)drdx\right)^{1/p},~v\in{\mathbb S}^{n-1}.
\end{align}
\end{defn}

\begin{defn}
Let $f$ be a probability density function on $\R^n$. Ball's $p$-th mean body is defined as the set with radial function
\begin{align*}
\rho_{B_p(f)}(v)=\left(\int_{\R_+}r^{p-1}f(rv)dr\right)^{1/p}, ~v\in{\mathbb S}^{n-1}.
\end{align*}
\end{defn}

\begin{defn}
Let $f$ be a probability density on $\R^n$. The polar $p$-th centroid body $\Gamma_p^\circ (f)$ is the set with the radial function
\begin{align*}
\rho_{\Gamma_p^\circ (f)}(v)=\left(\int_{\R^n}|v\cdot x|^pf(x)dx\right)^{-1/p},~v\in{\mathbb S}^{n-1}.
\end{align*}
where $v\cdot x$ is the inner product of $v$ and $x$.
\end{defn}

We let $Z_p(f)$ be the dilation of $\Gamma_p^\circ (f)$ defined as
\begin{align}\label{eq:dilated}
Z_p(f)=\left(\frac{2}{p+1}\right)^{1/p}\Gamma_p^\circ (f).
\end{align}

\subsection{$C_{-1}(f)$ and $C_1(f)$}\label{sec:special cases}

In the following, we verify Conjecture \ref{conj:2} in two special cases $p=-1, 1$ (log-concave densities when $p=1$), equivalently Conjecture \ref{conj:1} for $p=0, 2$. As we mentioned, it is easy to see that \eqref{eq:r-epi-linear} holds for $p=0$. Equivalently, Conjecture \ref{conj:1} holds for $p=0$, and Conjecture \ref{conj:2} holds for $p=-1$. We include this as a theorem to show the relation between $C_{-1}(f)$ and other geometric bodies.

\begin{thm}
Let $f$ be a probability density supported on a symmetric convex body $K\subset\R^n$. We have
$$
C_{-1}(f)=(2K)^\circ=(R_\infty (f))^\circ.
$$
\end{thm}

\begin{proof}
In this case, we have
$$
N_0^{1/2}(v\cdot X)=|\text{Range}(v\cdot X)|=2\sup_{x\in K}v\cdot x=2h_K(v)=h_{2K}(v).
$$
Since $K$ is a symmetric convex body, the dilated set $2K=\{2x: x\in K\}=K\pm K$ is a symmetric convex body as well. Therefore, the support function $h_{2K}(v)$ is a homogeneous convex function, and it defines a norm on $\R^n$. Then, we have
$$
\rho_{C_{-1}(f)}(v)=h_{2K}(v)^{-1},~v\in{\mathbb S}^{n-1},
$$
which implies that
$$
C_{-1}(f)=(2K)^\circ.
$$
To see the 2nd identity in the theorem, we let $p\to\infty$ in the definition \eqref{eq:rpf-radial}. We have
\begin{align*}
\rho_{R_\infty (f)}(v) &= \sup\{r>0: \text{there is $x\in K$ such that}~ x+rv\in K\}\\
&= \sup\{r>0: rv\in K-K\}\\
&= \rho_{2K}(v).
\end{align*}
The first identity follows from the principle of
the largest term. This is equivalent to the desired statement.
\end{proof}

\begin{rmk}
If we allow the norm $\|v\|=N_0^{1/2}(v\cdot X)$ to be infinite, the support of $f$ only need to be symmetric and convex, not necessary bounded.
\end{rmk}

\begin{thm}[Ball's Theorem \cite{Ball88}]
Let $p>0$ and let $f$ be a symmetric log-concave probability density on $\R^n$. Then
$$
\rho(v)=\left(\int_{\R_+}r^{p-1}f(rv)dr\right)^{1/p},~v\in{\mathbb S}^{n-1}
$$
is the radial function of a symmetric convex body in $\R^n$.
\end{thm}

\begin{thm}\label{thm:c1f}
Let $p>0$, and let $f$ be a log-concave (not necessary symmetric) probability density on $\R^n$. Then, $C_1(f)$ and $R_p(f)$ are symmetric convex bodies. Furthermore, we have
$$
C_1(f)=I(\hat{f})=(n-1)I(R_{n-1}(f)),
$$
$$
R_p(f)=B_p({\hat{f}}),
$$
where
$$
\hat{f}(x)=\int_{\R^n}f(y)f(x+y)dy.
$$
\end{thm}

\begin{proof}
By the definition of $C_1(f)$ in \eqref{eq:cpf-radial}, for $v\in{\mathbb S}^{n-1}$, we have
\begin{align*}
\rho_{C_1(f)}(v) &= \int_{\R^n}f(x)\int_{x+v^\perp}f(y)dydx\\
&= \int_{\R^n}f(x)\int_{v^\perp}f(x+y)dydx\\
&=\int_{v^\perp}\int_{\R^n}f(x)f(x+y)dxdy\\
&= \rho_{I(\hat{f})}(v).
\end{align*}
To see the second identity, we rewrite $\rho_{C_1(f)}(v)$ in the spherical polar coordinates as
\begin{align*}
\rho_{C_1(f)}(v) &= \int_{\R^n}f(x)\int_{{\mathbb S}^{n-1}\cap v^\perp}\int_{\R_+}r^{n-2}f(x+ru)drdudx\\
&=\int_{{\mathbb S}^{n-1}\cap v^\perp}\int_{\R^n}f(x)\int_{\R_+}r^{n-2}f(x+ru)drdxdu.
\end{align*}
The 2nd equation follows from Fubini's theorem. Using the definition of $R_{n-1}(f)$ in \eqref{eq:rpf-radial}, we have
\begin{align*}
\rho_{C_1(f)}(v) &= \int_{{\mathbb S}^{n-1}\cap v^\perp}\rho_{R_{n-1}(f)}(u)^{n-1}du\\
&= (n-1)\int_{{\mathbb S}^{n-1}\cap v^\perp}\int_0^{\rho_{R_{n-1}(f)}(u)}r^{n-2}drdu\\
&= (n-1)|R_{n-1}(f)\cap v^\perp|\\
&= (n-1)\rho_{I(R_{n-1}(f))}(v).
\end{align*}
By the definition of $R_p(f)$ in \eqref{eq:rpf-radial}, we have
\begin{align*}
\rho_{R_p(f)}(v) &= \left(\int_{\R^n}f(x)\int_{\R_+}r^{p-1}f(x+rv)drdx\right)^{1/p}\\
&= \left(\int_{\R_+}r^{p-1}\int_{\R^n}f(x)f(x+rv)dxdr\right)^{1/p}\\
&= \left(\int_{\R_+}r^{p-1}\hat{f}(rv)dr\right)^{1/p}\\
&= \rho_{B_p(\hat{f})}(v).
\end{align*}
Notice that $\hat{f}$ is a symmetric log-concave function, since it is the density of the difference of two independent random vectors with the same density $f$. Ball's theorem implies the convexity of $B_p(\hat{f})$, i.e., $R_p(f)$, and $C_1(f)$, i.e., $I(\hat{f})$.
\end{proof}

\begin{rmk}
The theorem can be thought of as the functional version of Theorem 5.2 in \cite{GG99} and Theorem 4.3 in \cite{GZ98}. The proofs given here are much simpler. Generalization of Ball's theorem do hold for convex measures \cite{Bob10}. But it is not clear for us if the proof can be extended to general convex measures, since the convolution of two $s$-concave densities is not necessary $s'$-concave for some $s'$.
\end{rmk}

\section{Discussion of $p$-th mean bodies} \label{sec:convex bodies}

The various $p$-th mean bodies introduced in the previous section are closely related to each other. We take $f=1_K$, where $K$ is a convex body. It is known that $R_\infty(K)$ is the difference body of $K$, and that $R_{-1}(K)$ is the polar projection body of $K$. So, the $p$-th radial mean body, $R_p(K)$, forms a spectrum connecting the difference body and polar projection body. Similar property holds for the dilated polar $p$-th centroid body defined in \eqref{eq:dilated}, $Z_p(K)$, which connects the intersection body of $K$ when $p=-1$ and the polar body of $K$ when $p=\infty$.

The next two propositions are functional liftings of Lemma 5.1 and Proposition 3.1 in \cite{GG99}, which could be used to give an alternative proof of the first part of Theorem \ref{thm:c1f}.

\begin{prop}\label{prop:Z-R}
Let $p\geq-1$ be non-zero. Let $f$ be a probability density on $\R^n$. For $v\in{\mathbb S}^{n-1}$, we have
$$
\rho_{Z_p(R_{n+p}(f))}(v)=\left(\frac{1}{n+p}\int_{\R^n}f(x)\rho_{Z_p(f_x)}(v)^{-p}dx\right)^{-1/p},
$$
where $f_x(y)=f(x+y)$.
\end{prop}

\begin{proof}
\begin{align*}
\rho_{Z_p(R_{n+p}(f))}(v)^{-p} &= \frac{p+1}{2}\int_{R_{n+p}(f)}|v\cdot x|^pdx\\
&=\frac{p+1}{2}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\int_0^{\rho_{R_{n+p}(f)}(u)}r^{n+p-1}drdu\\
&=\frac{p+1}{2(n+p)}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\rho_{R_{n+p}(f)}(u)^{n+p}du\\
&= \frac{p+1}{2(n+p)}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\int_{\R^n}f(x)\int_{\R_+}r^{n+p-1}f_x(ru)drdxdu\\
&= \frac{p+1}{2(n+p)}\int_{\R^n}f(x)\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\int_{\R_+}r^{n+p-1}f_x(ru)drdudx\\
&= \frac{p+1}{2(n+p)}\int_{\R^n}\left(\int_{\R^n}|v\cdot y|^pf_x(y)dy\right)f(x)dx\\
&=\frac{1}{n+p}\int_{\R^n}f(x)\rho_{Z_p(f_x)}(v)^{-p}dx.
\end{align*}
\end{proof}

For $p>-1$, the $p$-cosine transform $T_pf$ of a continuous function $f$ on ${\mathbb S}^{n-1}$ is defined as
$$
T_pf(v)=\int_{{\mathbb S}^{n-1}}|u\cdot v|^pf(u)du,~ v\in{\mathbb S}^{n-1}.
$$
The spherical Radon transform $\mathcal{R}f$ is defined as
$$
\mathcal{R}f(v)=\int_{{\mathbb S}^{n-1}\cap v^\perp}f(u)du,~ v\in{\mathbb S}^{n-1}.
$$
The following connection between $p$-cosine transform and Radon transform is known, see \cite{Kol97, GZ99}.
\begin{align}\label{eq:T-R}
\lim_{p\to-1+}\frac{p+1}{2}T_pf=\mathcal{R}f.
\end{align}
\begin{prop} \label{prop:Z-I}
Let $f$ be a probability density on $\R^n$. For $v\in{\mathbb S}^{n-1}$, we have
$$
\lim_{p\to-1+}\rho_{Z_p(f)}(v)^{-p}=\rho_{I(f)}(v).
$$
In other words, we have
$$
Z_{-1}(f)=I(f).
$$
\end{prop}

\begin{proof}
By definition, we have
\begin{align*}
\rho_{Z_p(f)}(v)^{-p} &= \frac{p+1}{2}\int_{\R^n}|v\cdot x|^pf(x)dx\\
&=\frac{p+1}{2}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\int_{\R_+}r^{n+p-1}f(ru)drdu.\\
\end{align*}
Using identity \eqref{eq:T-R}, we have
\begin{align*}
\lim_{p\to-1+}\rho_{Z_p(f)}(v)^{-p} &= \int_{{\mathbb S}^{n-1}\cap v^\perp}\int_{\R_+}r^{n-2}f(ru)drdu\\
&=\int_{v^\perp}f(x)dx\\
&=\rho_{I(f)}(v).
\end{align*}
\end{proof}

In the following, we apply these properties to give another proof of the first statement of Theorem \ref{thm:c1f}.
\begin{proof}
\begin{align*}
\rho_{C_1(f)}(v) &= \int_{\R^n}f(x)\int_{x+v^\perp}f(y)dydx\\
&=\int_{\R^n}f(x)\int_{v^\perp}f(x+y)dydx\\
&=\int_{\R^n}f(x)\rho_{I(f_x)}(v)dx\\
&=\int_{\R^n}f(x)\rho_{Z_{-1}(f_x)}(v)dx\\
&=(n-1)\rho_{Z_{-1}(R_{n-1}(f))}(v)\\
&=(n-1)\rho_{I(R_{n-1}(f))}(v).
\end{align*}
We use Proposition \ref{prop:Z-I} in the last equation and the 3rd last equation. The 2nd last identity follows from Proposition \ref{prop:Z-R}.
\end{proof}
Using a similar argument, we can prove the following result.
\begin{prop}
Let $f$ be a probability density on $\R^n$. For $v\in{\mathbb S}^{n-1}$,  we have
$$
\lim_{p\to-1+}\frac{p+1}{2(n-1)}T_p\rho_{C_{n-1}(f)}^{n-1}(v)=\rho_{I(C_{n-1}(f))}(v).
$$
\end{prop}

\begin{proof}
For $u\in{\mathbb S}^{n-1}$, we have
\begin{align*}
\rho_{I(C_{n-1}(f))}(v) &=|C_{n-1}f\cap v^\perp| \nonumber\\
&=\int_{C_{n-1}(f)\cap v^\perp}1dx \nonumber\\
&=\int_{{\mathbb S}^{n-1}\cap v^\perp}\int_0^{\rho_{C_{n-1}f}(u)}r^{n-2}drdu \nonumber\\
&=\frac{1}{n-1}\int_{{\mathbb S}^{n-1}\cap v^\perp}\rho_{C_{n-1}f}(u)^{n-1}du \nonumber\\
&=\frac{1}{n-1}\int_{{\mathbb S}^{n-1}\cap v^\perp}\int_{\R^n}f(x)\left(\int_{x+u^\perp}f(y)dy\right)^{n-1}dxdu \nonumber\\
&=\frac{1}{n-1}\int_{\R^n}f(x)\int_{{\mathbb S}^{n-1}\cap v^\perp}\left(\int_{x+u^\perp}f(y)dy\right)^{n-1}dudx.
\end{align*}
Using the connection \eqref{eq:T-R}, we have
\begin{eqnarray*}
\rho_{I(C_{n-1}(f))}(v) &=& \lim_{p\to-1+}\frac{p+1}{2(n-1)}\int_{\R^n}f(x)\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\left(\int_{x+u^\perp}f(y)dy\right)^{n-1}dudx\\
&=& \lim_{p\to-1+}\frac{p+1}{2(n-1)}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\int_{\R^{n}}f(x)\left(\int_{x+u^\perp}f(y)dy\right)^{n-1}dxdu\\
&=& \lim_{p\to-1+}\frac{p+1}{2(n-1)}\int_{{\mathbb S}^{n-1}}|u\cdot v|^p\rho_{C_{n-1}(f)}^{n-1}(u)du\\
&=&\lim_{p\to-1+}\frac{p+1}{2(n-1)}T_p\rho_{C_{n-1}(f)}^{n-1}(v).
\end{eqnarray*}

\end{proof}

\section*{Acknowledgement}
The author is grateful to Mokshay Madiman and Muriel M\'edard for helpful discussions. The author also would like to thank the anonymous referee for a thorough reading of the paper, and valuable suggestions which make the paper much improved.

redpajama_set_name: RedPajamaArXiv


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text:
Q: Creating NRG Multiband TIF from single bands using QGIS? I'm attempting to take 3 single bands (NIR, Red, and Green) and turn them into a single multiband tif image.
Someone working on the images was able to take this image which I merged...

And then using QGIS remove the transparency bands and turn it into this image...

Unfortuantely, the person that was able to complete this is not able to help us get to the final product, so I've been trying to find out how to merge single bands into the final product and have it come our looking like a normal CIR image.
I have each band processed separately without the transparency (using Pix4D) and have tried merging those, but still continue to get an image that looks nothing like the second image above. I need to be able to get these single bands into the multiband in NRG order.
EDIT: I followed Lukes advice below and did the following:
Raster -> Miscellaneous -> Build Virtual Raster, used visible raster layers (to make sure they were in order with NIR at top of layers, then Red, then Green.) Then set the output file and used separate to make sure the layers would stay as separate bands.
When I hit Ok, this error came up.

Pretty sure that was unrelated, but then the final image it created looked like this:

Then I did this path:
Raster menu -> Conversion -> Translate (Convert Format)...
Which was set up like this:

Then when I hit Ok, it came up with an image that looks like this: (The error didn't happen this time. Sweet!)

Still don't understand why my image looks so different from the one he created. I'm testing the upload now.
EDIT 2:
Here are the properties for the correct image that the other guy was able to create :

And here are the properties to my image after I've made it:

Why do my properties have a "Gray" Band at the top? Each of them should be coming through as color bands since it's NIR, Red, and Green, right?
EDIT 3 : Here is the txt of the VRT that I've created with my single bands.
<VRTDataset rasterXSize="2720" rasterYSize="2592">
  <SRS>PROJCS["WGS 84 / UTM zone 11N",GEOGCS["WGS 84",DATUM["WGS_1984",SPHEROID["WGS 84",6378137,298.257223563,AUTHORITY["EPSG","7030"]],AUTHORITY["EPSG","6326"]],PRIMEM["Greenwich",0,AUTHORITY["EPSG","8901"]],UNIT["degree",0.0174532925199433,AUTHORITY["EPSG","9122"]],AUTHORITY["EPSG","4326"]],PROJECTION["Transverse_Mercator"],PARAMETER["latitude_of_origin",0],PARAMETER["central_meridian",-117],PARAMETER["scale_factor",0.9996],PARAMETER["false_easting",500000],PARAMETER["false_northing",0],UNIT["metre",1,AUTHORITY["EPSG","9001"]],AXIS["Easting",EAST],AXIS["Northing",NORTH],AUTHORITY["EPSG","32611"]]</SRS>
  <GeoTransform> 3.5246044199000002e+005, 4.3333333333333335e-001, 0.0000000000000000e+000, 5.1834971052700002e+006, 0.0000000000000000e+000,-4.3333333333333335e-001</GeoTransform>
  <VRTRasterBand dataType="UInt16" band="1">
    <NoDataValue>-10000</NoDataValue>
    <ComplexSource>
      <SourceFilename relativeToVRT="0">A:/UpAngle eBee Flights/Crop Production Services/9.13.16 CPS Connell Johnson/Sequioa Images/Single Band Try/Merged 30cm/NO transparency/cps 6-14 nir_mosaic_nir.tif</SourceFilename>
      <SourceBand>1</SourceBand>
      <SourceProperties RasterXSize="3756" RasterYSize="3711" DataType="UInt16" BlockXSize="256" BlockYSize="256" />
      <SrcRect xOff="0" yOff="0" xSize="3756" ySize="3711" />
      <DstRect xOff="8.91590769237015" yOff="0" xSize="2600.30769230769" ySize="2569.15384615385" />
      <NODATA>-10000</NODATA>
    </ComplexSource>
  </VRTRasterBand>
  <VRTRasterBand dataType="UInt16" band="2">
    <NoDataValue>-10000</NoDataValue>
    <ComplexSource>
      <SourceFilename relativeToVRT="0">A:/UpAngle eBee Flights/Crop Production Services/9.13.16 CPS Connell Johnson/Sequioa Images/Single Band Try/Merged 30cm/NO transparency/cps 6-14 red_mosaic_red.tif</SourceFilename>
      <SourceBand>1</SourceBand>
      <SourceProperties RasterXSize="2357" RasterYSize="2239" DataType="UInt16" BlockXSize="256" BlockYSize="256" />
      <SrcRect xOff="0" yOff="0" xSize="2357" ySize="2239" />
      <DstRect xOff="0" yOff="8.97618461543551" xSize="2719.61538461538" ySize="2583.46153846154" />
      <NODATA>-10000</NODATA>
    </ComplexSource>
  </VRTRasterBand>
  <VRTRasterBand dataType="UInt16" band="3">
    <NoDataValue>-10000</NoDataValue>
    <ComplexSource>
      <SourceFilename relativeToVRT="0">A:/UpAngle eBee Flights/Crop Production Services/9.13.16 CPS Connell Johnson/Sequioa Images/Single Band Try/Merged 30cm/NO transparency/cps 6-14 green_mosaic_green.tif</SourceFilename>
      <SourceBand>1</SourceBand>
      <SourceProperties RasterXSize="2338" RasterYSize="2204" DataType="UInt16" BlockXSize="256" BlockYSize="256" />
      <SrcRect xOff="0" yOff="0" xSize="2338" ySize="2204" />
      <DstRect xOff="0.423092307666173" yOff="17.198976921634" xSize="2697.69230769231" ySize="2543.07692307692" />
      <NODATA>-10000</NODATA>
    </ComplexSource>
  </VRTRasterBand>
</VRTDataset>

A: In QGIS:

*

*Raster menu -> Miscellaneous -> Build Virtual Raster (Catalog)...

*Ensure the "Separate" option is ticked so each raster is put into a separate band

*Raster menu -> Conversion -> Translate (Convert Format)...

*Output to GeoTIFF

Command-line GDAL:

*

*gdalbuildvrt -separate NRG.vrt NIR.tif RED.tif GREEN.tif

*gdal_translate NRG.vrt NRG.tif
Note there are a few other options you could play around with, to set nodata values, compress the output, define projection etc.
Here's some screenshots. Note: output VRT and tif look the same as the original RGB tif. Also note that the order of the input rasters is  important when building the VRT.

redpajama_set_name: RedPajamaStackExchange


text:
Local League Directory
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Issues: Charter Schools
The League of Women Voters of Illinois affirmed an update to our Charter Schools Position during the 2015 State Convention.
Since the original position was adopted in 2001, there have been many new developments that were not addressed. A state-wide study was conducted during the 2013 - 2015 biennium. Following is resulting position to be published in the 2015 - 2017 issue of Where We Stand:
Updated Charter School Position Proposed March, 2015
Accepted by League of Women Voters Board of Directors March 14, 2015
Affirmed by Delegates to the 2015 LWVIL Convention
The League of Women Voters of Illinois supports accountability, transparency, and equity in the use of public funds for education, including charter schools.
The League of Women Voters of Illinois believes that
Adequate and equitable funding is needed for all public schools, including charters. The financial impact on traditional public schools must be evaluated before charters are granted.
Charter school authorizers should evaluate the performance of existing charters before granting new charters or approving additional campuses.
Existing charter schools should be evaluated in incremental periods not to exceed five years (and preferably more frequently) for purposes of charter renewal.
Charters should be revoked or not renewed if a school repeatedly or egregiously
o Failed to meet content standards, pupil performance standards, or school climate standards identified in the charter contract.
o Failed to meet generally accepted standards of fiscal management.
o Violated any provision of state or federal law from which the charter school was not exempt.
o Discriminated against students based on characteristics such as family economic circumstances, disability, race, creed, color, national origin, religion, ancestry, or need for special education services.
o Used curriculum or instruction that failed to maintain the separation of church and state.
o Failed to admit prospective students periodically throughout the year as space became available.
o Failed to enforce truancy laws.
Employees should have the right to organize unions at charter schools.
The authority for approving and renewing charters should reside exclusively in the local school board.
Local voters should be able through referendum to request their school board to establish a charter school.
There should be adequate provisions for education of the public and participation in the authorizer's decision-making process.
If the Illinois State Board of Education (ISBE) or a body appointed by ISBE, such as the State Charter School Commission, has the power to reverse a decision by a local school board, local voters should be able through referendum to override the decision of ISBE or the appointed commission.
There needs to be good communication among all public schools regarding effective innovations.
LWVIL has identified the following areas of concern about charter schools:
1. Financial impact on school districts. For example,
Charters can place financial pressure on districts by drawing money out of the traditional public schools.
When a student transfers between traditional public and charter schools, funding for that pupil should be prorated and transferred in a timely fashion.
All property, equipment, and materials purchased by a charter school with public dollars should be surrendered to the local district if the school closes for any reason.
2. Privatization and profit-making. The LWVIL opposes the presence of private, for-profit companies in the governance of public education. Subcontracting most of the management of a school to a for-profit organization should be prohibited.
3. Lack of transparency. Like other public schools, charter schools should be subject to disclosure of
Potential conflicts of interest of members of the governing board
Salaries and benefit packages
The identity of the owner of any property and buildings
Admission processes
4. Virtual education (provided by computers off-site, with students spending little if any time in a brick-and-mortar school). Virtual education is not appropriate for elementary-aged students except in rare circumstances. Funding for virtual schools should be based on actual costs, rather than the same level of funding provided to other charter schools.
For information about the League's position on schools, please see the Schools issue page.
To contact the issues specialist: issues@lwvil.org.
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Phone: 312-939-5935 | Fax: 312-939-6887 | info@lwvil.org | Contact the Webmaster | Privacy Policy
The League of Women Voters of Illinois does not endorse the contents of any web sites to which it links; such Internet addresses contain information not posted by the League and are included to provide convenient access to additional information. The League does not endorse or control these external Internet addresses nor guarantee the accuracy of such addresses.

redpajama_set_name: RedPajamaCommonCrawl


Your search for 'Naumann' returned 46 results.
Werner Naumann–Joseph Goebbels' chosen successor as head of the Nazi propaganda ministry–was arrested plotting a coup with other former high-ranking Nazis in 1953. The investigation revealed that Naumann and his associates were acting at the direction of a Nazi government-in-exile based in Madrid. With the support of German industry and finance, this Underground Reich operated all over the world, executing conspiracies in foreign countries on behalf of German cartels.
That struggle is framed against a larger political dynamic embracing advocacy of the elimination of formal national borders in Europe in favor of "regionalist plans." Just such regionalist advocacy was the focal point of a prominent article (with accompanying maps of the projected realignment) in Die Zeit, a major German weekly.
The two Twitter accounts that appear to account for nearly a third of all Twitter traffic with the #Catalonia hashtag, in reference to the Catalonian secession movement belong to Julian Assange and Edward Snowden.
Of more than passing interest, under the circumstances, is the Twitter effort by both Julian Assange and Edward Snowden on behalf of Catalonian independence.
As seen in many past programs and posts, Snowden and Assange are as far to the right as it is possible to be.
Snowden and Assange's ostensibly "liberating" doctrines, if put into effect, would leave citizenry at the mercy of unfettered economic will, exercised by corporations and their associated elites.
Snowden specifically appears to be advocating that no secession movement anywhere ever can be rejected by the government under the premise that self-determination is a human right, viewing this as a "natural law" issue.
In that context, the right to secede is championed by the Libertarian far-right, all the way down to the right to individuals to secede from all government. As this piece from Libertarian David S. D'Amato demonstrates, extending the right to secede down to the individual facilitates the implementation of an anarcho-capitalist society with no government at all, as seen by figures like Murray Rothbard. This is envisioned as an excellent wayof achieving an anarcho-capitalist utopia.
The Snowden/Assange pro-secessionist movement should also be seen against the background of the Neo-Confederate movement, championed by Ron Paul and the Ludwig Von Mises Institute.
Following capture of 13 percent of the vote in Germany's federal elections on Sunday by the Alternative For Germany (AfD), Alexander Gauland, the AfD leader, provoked outrage after suggesting that Germans should no longer be reproached with the Nazi past.
This type of behavior apparently motivated AfD leader Frauke Petry to leave the party, just hours after the election over its extremism.
1.-Review of Dorothy Thompson's 1941 article about what a Nazi victory in Europe would look like–a scenario that bears considerable resemblance to the regionalization plan discussed above.
2.-Discussion of the potential fortunes of Austria's Freedom Party, formed in 1956 as a vehicle for the re-introduction of Austrian Third Reich alumni into that nation's political process.
The entry point to our exploration of Julius Evola is top Trump adviser and first tier NSC member Steve Bannon. Evola is a key influence on Bannon. Evola was an early occult fascist, with strong connections with Mussolini's Italy. Eventually Evola established strong, lasting connections with the Nazi SS, both operationally and ideologically.
Evola has also influenced Alexander Dugin, a prominent Russian ideologue and politician.
The broadcast recaps FTR #233, which details Evola's work for the SS and Kevin Coogan's theory that Evola was involved with an SS occult network incorporating important people and institutions in both the West and behind the so-called "Iron Curtain." Later in the program, we further develop the story of Alexander Dugin, a Russian "Alt-right" thinker and politician prominent in the Russian government. As mentioned above, Dugin, like Bannon, has been influenced by Evola.
Like SS chief Himmler, Evola saw the SS as the successors to the Kshatriya class (the Hindu warrior caste.) Seeing Germany and Europe as succumbing to "barbarian invasion," Evola saw a pagan, anti-Christian mysticism as necessarily antithetical to the Judeo-Christian culture which, he felt, had led the West to decline before the "Bolshevik hordes" of the Soviet Union and the "chewing gum imperialism" of the United States.
Kevin felt that this organization (reflecting the ideological stance of an element of the Waffen SS) would be pan-European in scope and orientation, and not necessarily entirely chauvinistic from a Nordic or Germanic racial and national standpoint. Nourished by bank accounts secreted abroad, this hypothetical organization functions in an underground fashion. (The funds that nourished this institution would necessarily have derived from the Bormann Organization.) The Order appears to have established ostensibly friendly relations with the West.
This organization may very well have begun working with the U.S. intelligence apparat after the war, as evidenced by, among other things, the collaboration between post-war SS elements and the CIA. Coogan hypothesizes that CIA director Allen Dulles may have played a primary role in such an accord.
Another influence on a Dulles/Order collaborative relationship may have been psychologist Carl Jung, who was connected to Dulles and to the Third Reich.
Significantly, the Order appears to have overlapped, and also worked with, elements of the East Bloc, including former Soviet and East German national security officials. The organization also maintained contacts with "anti-imperialist," Third World liberation movements.
Steve Bannon's discussion of Alexander Dugin gains significance in this context.
The Order appears to have exploited its contacts within both East and West blocs to further its own fascistic and elitist agenda, playing both sides against the middle during the Cold War.
The Dugin/Evola affiliation and the Bannon/Evola affiliation may be significant in that context.
Adbusters magazine's publicizing of Alexander Dugin–we review the fact that Adbusters appears to have played a key role in jumpstarting the "Occupy" movement.
Keyed by the victory of a Crimean Tatar singer in the Eurovision song contest–she sang about their deportation by Stalin in 1944–the Crimean Tatars are in the political spotlight again.
Agitating and conspiring alongside OUN/B fascist heirs Pravy Sektor in Ukraine, the Crimean Tatars have become something of a cause celebre in the course of collaborating to destabilize Crimea and Russia. What has been eclipsed by the current controversy is the long history of Crimean Tatar collaboration with the Third Reich, initially, and then Western intelligence.
Following military defeat in the Second World War, the Crimean Tatars and their primary sponsor in the Ostministerium–Gerhard von Mende–continued their work apace under the auspices of Frank Wisner and Allen Dulles's CIA and the Gehlen organization, both in its CIA incarnation and after it's incorporation as the BND, the intelligence service of the Federal Republic of Germany. The Crimean Tatar/Ukrainian fascist collaboration has continued into the post-Maidan period.
Program Highlights Include: Review of Crimean Tatar/Pravy Sektor blockading of road traffic into Crimea; review of Crimean Tatar/Pravy Sektor collaboration in the sabotage of Crimea's electric grid; contemporary German political liaison with dissident Crimean Tatars; Turkey's ongoing support for the Crimean Tatars, part of Erdogan's "neo-Ottoman"/Pan-Turkic policy; von Mende's use of Third Reich collaborator Edige Kirimal during the Cold War; review of the UNA-UNSO's collaboration with Chechens and other Caucasian Islamists following the breakup of the Soviet Union.
As Americans process the Orlando nightclub shooting, we highlight the Brussels attacks by ISIS devotees. Against the background of deep and longstanding Muslim Brotherhood infiltration of Belgian and European civil society and political infrastructure, we once again underscore that elements of Western intelligence, including elements of CIA, continue to use jihadist groups as proxy warriors and armed heralds of corporate economics.
In addition to the Caucasus, where Chechen jihadis have emigrated to Iraq and Syria and have formed a major foundational element of ISIS, Syria has been a major theater of operations for U.S., Saudi, Turkish and Qatari-backed Islamist fighters. Perhaps the most terrifying aspect of the Brussels attacks is the fact that the assaults appear to have been a "plan B," put into effect when other plans to attack Belgian nuclear facilities fell through. Belgian nuclear facilities are poorly secured and have been infiltrated by ISIS-linked individuals.
In addition to video footage surveilling personnel from Belgian nuclear facilities, a security guard at one of them was found murdered and his security badge had been stolen. This comes as King Abdullah of Jordan has charged that Turkey is deliberately sending ISIS fighters to Europe, where their actions have precipitated the kid-glove treatment the EU has afforded Erdogan and his burgeoning Islamic fascist government's suppression of journalistic and political freedom.
This program is set against the background of our recent discussions with Peter Levenda about the profound Nazi presence in Indonesia before, during and after World War II. The bloody 1965 coup against Sukarno is known to have been largely engineered by the CIA. We now learn that the BND–the German foreign intelligence service and the final incarnation of the Gehlen spy outfit–played a major role in the coup, as well. The apparent BND point man for the Federal Republic's role in the coup was Rudolf Oebsger-Roder, like so many Gehlen org officers, a veteran of the SS. The second half of the broadcast consists of a reading of a section of Terence McKenna's book "True Hallucinations." McKenna allegedly came across a German mining engineer working for what may well have been the post-war Bormann capital network.


This (free) extension implements a rain animation inspired by the movie "The Matrix".
The embedded script is partly based on a codepen by Ebram Marzouk: https://codepen.io/P3R0/pen/MwgoKv.
The animation can be rendered on the page background or in the specified box.


text:
Posts in 'Constitutional Carry'
2017 is Best Chance Yet for Permitless "Constitutional" Carry in Indiana
Indiana has always been a good candidate for permitless carry. It was one of the first states to have a shall issue permit. That was because of a judicial ruling on the Indiana Constitution, not legislative action.
TX: Constitutional Carry Bill HB 375 Filed
"It's time in Texas to restore our Second Amendment rights to their originally intended level," said state Rep. Jonathan Stickland, a Fort Worth-area Republican who last year spearheaded the successful open-carry bill.
Silly Carry Permit Requirements
Carry permits do not have to be expensive and complex. The New Hampshire permit has been fairly easy to obtain for many years, if you already had a permit from your state of residence. During the last few months I have renewed a couple of carry permits. It is not intensely difficult, but when you have better things to do, it can be irritating.
One Week Since Trump Won…And With Him America
I think Trump will be good for America for a number of reasons, but when it comes to the 2nd Amendment, I truly believe that the election of Donald Trump over Hillary Clinton has all but saved it from extinction. Or rather, saved patriots from having to literally fight for the right.
Rhode Island Supreme Court Chides Police Chief for Failure to Issue Carry Permits
East Providence Police Chief Christopher J. Parella is in trouble with the Supreme Court of Rhode Island. Again. He failed to issue licenses to carry handguns to three more applicants, without explaining any reason for doing so.
State Police Illegally Ties Drivers License To Carry Permits Violating the Law
State Police are actively breaking the law by having a persons LTCF disclosed to them whenever they pull over a driver. You see, a recent update to the NCIC was done that now couples the information. Only one problem…such a move is completely illegal.
Welcome To Constitutional Carry Missouri, Governor's Veto OVERRIDDEN
Missouri becomes the 10th Constitutional Carry state in the nation, meaning that 20% of the states in America no longer require the extortion of money from and the hat in hand begging of citizens in order to exercise their 2nd Amendment rights.
Idaho Set To Become 9th Constitutional Carry State
On the heels of West Virginia's override of gun control Governor Earl Ray Tomblin's veto of Constitutional Carry, the Idaho Senate has overwhelmingly passed their own permit-less carry bill by a vote of 27-8.
West Virginia Overrides Gun Control Governor's Veto For BIG Win
The lawmakers of West Virginia stood their ground in the face of Gun Control Governor Earl Ray Tomblin's veto attempt to ground the 2nd Amendment under heel.
Constitutional Carry: New Hampshire Falters While Maine Rises
In an update from earlier articles I have written about the 2 New England states bucking the gun control trends from nearby states of Massachusetts, Connecticut and New York, Maine and New Hampshire have both passed legislation allowing their citizens to fully exercise their right to keep and bear arms without the bureaucracy, cost, or invasion of privacy and human dignity that goes along with begging the government to do so. Unfortunately, New Hampshire's leading mouth breather, Gov. Maggie Hassan has gone against the will of the legislature and the will of the people by vetoing the Constitutional Carry measure. Hassan gave the standard two faced talk from both sides of her mouth by stating that she supports the 2nd Amendment but by eliminating the need for people to beg the government to exercise their rights it would "compromise the public safety of our citizens." No…the only thing it...
Maine Poised To Join States With Constitutional Carry
In big news the Maine Senate has passed legislation for Constitutional Carry by a vote of 21-14. In a surprising twist, the vote did not fall completely on party lines, as three Democrats voted in favor of liberty while 2 Republicans turned their back on the Constitution and on the rights of the people. The bill now heads to the state Legislature which is expected to pass it when it is taken up within the next couple of days. The bill, L.D. 652, will Maine the seventh state that allow individuals to carry a concealed handgun without a permit. It has 96 co-sponsors, more than half the members of the Legislature, and the blessing of some members of the Republican and Democratic leadership. It will also be the first New England state that passes constitutional carry. Vermont doesn't count because the Green Mountain State has ALWAYS respected the rights of...
The Constitutional Carry Tide Keeps Rolling
Despite one of the most anti gun Presidents in our history at the helm, more and more states are poised to do away with the permit system for its citizens to carry firearms. Currently 10% of states have laws enacting Constitutional Carry; the carrying of firearms without first coming to the government, hat in hand to ASK if its ok. That number could more than double by the end of the year with states currently debating and others on the precipice of introducing legislation constitutional carry laws. From Godfather Politics: Just when Barack Obama and his liberal tyrant friends have been pushing so hard to enact stricter anti-gun and Second Amendment defying laws, some states are looking to pass laws that would not require a special permit to carry a concealed weapon. Currently, Alaska, Arizona, Arkansas, Vermont and Wyoming do not require any special permit to carry a concealed weapon....
Wyoming Moves To Ban Criminal Empowerment Areas aka "Gun Free Zones"
Between states like Indiana and New Hampshire pushing for Constitutional Carry and court victories in places like California that are making May-Issue states an endangered species, so far this is a good year for the 2nd Amendment and common sense. Common sense is poised for another big win as in Wyoming, a sensible thinking state rep by the name of Allen Jaggi introduced House Bill 114. A bill which passed the Wyoming House of Representatives without any watering down amendments by an overwhelming vote of 42-17. What does HB 114 do? It outlaws the state of Wyoming from enforcing the fatal fallacy of Gun Free Zones. From the bill itself: (a) This section shall be known and may be cited as the "Wyoming Repeal Gun Free Zones Act." (b) Persons holding a valid concealed carry permit 2 issued by the state of Wyoming under W.S. 6-8-104(a)(ii) may carry a concealed...
New Hampshire Seeks To Empower Citizens By Equal Firepower Measures
I am appreciating New Hampshire more and more as of late. The Live Free Or Die state is really living up to its motto. As you may have read about last week, NH is on the verge of becoming likes its neighbor Vermont and doing away with its firearm carrying licensing system to become a Constitutional Carry state. On the heels of that legislation is another bill, brought forth by State Representative J.H Hoell, that would bar the police from becoming a paramilitary force without consent of the people. The proposed legislation is now in committee. If it's passed and signed into law by the governor, state and municipal agencies in New Hampshire will be barred from buying or even accepting free offers of "military style equipment" for police use, except with the approval of the assembled citizenry at a public town meeting. That prohibition would include not just MRAPS...
Live Free Or Die: New Hampshire On Verge Of Enacting Constitutional Carry
The Granite state is poised to crack under the pressure of inalienable rights as legislation has been introduced to make New Hampshire the seventh Constitutional Carry state in the nation. The legislation I refer to is Senate Bill 116 that was recently introduced by Republican Senators Jeb Bradley and David Boutin. The bill is rather direct and concise in its objective and intent. This bill: I. Increases the length of time for which a license to carry a pistol or revolver is valid. II. Allows a person to carry a loaded, concealed pistol or revolver without a license unless such person is otherwise prohibited by New Hampshire statute. III. Requires the director of the division of state police to negotiate and enter into agreements with other jurisdictions to recognize in those jurisdictions the validity of the license to carry issued in this state. IV. Repeals the requirement to obtain...
Indiana Takes First Step To Become Constitutional Carry State
Constitutional Carry is the precedent currently set in 5 states which dictates that a law abiding citizen needs no permission or permit from the government to exercise their God given right to keep and bear arms. Alaska, Wyoming, Vermont, Arizona and Arkansas may have a new state joining their ranks if State Rep. Jim Lucas of Indiana has anything to say about it. Lucas has introduced House Bill 1144 which would strike down the permitting scheme currently in place and would repeal whole chunks of the nearly 30 pages of laws dealing with carrying firearms. Most important among them is the "poll tax" that requires law abiding citizens to pay for the ability to exercise their rights. Colloquially known as purchasing a carry permit. If this bill passes it will take effect on July 1, 2015. Lucas puts it rather succinctly as to whom this bill is directed at: "This...
2nd Amendment Infringements Endanger US Troops and Their Families
When you add up the number of military personnel, including National Guard, Reservists and the Coast Guard, the United States has about 2.2 million service members who have volunteered to put their lives on the lines to defend this nation of ours. Yet a great deal of these troops have been betrayed by their state governments and denied the protection that the 2nd Amendment enumerates. Even by the MOST LIBERAL ANTI GUN interpretation of the 2nd Amendment, these 2.2 million people (if no one else) should not have their 2nd Amendment right infringed. If you are a service member in Vermont or Alaska or Arizona then you have nothing to worry about because your state recognizes the 2nd Amendment rights of all it's citizens. Yet if you live in New York, Maryland, San Francisco and other such dens of iniquity you are pretty much out of luck, just like...
Gun Grabber Heads Implode as DC Ban Ruled Unconstitutional – All law abiding can now carry
Shockingly over the weekend a Judge ruled that "the right to keep and bear arms" ACTUALLY included bearing arms. In Palmer v District of Columbia, Federal Judge Frederick Scullin Jr. ruled that the Second Amendment existed OUTSIDE a person's home. With a decision so obvious I do wonder why it took 5 years to issue this decision. Perhaps the five year wait had to do with Judge Scullin delving DEEPLY into both Heller v DC and MacDonald v Chicago to issue this resounding and total conclusion: "there is no longer any basis on which this court can conclude that the District of Columbia's total ban on the public carrying of ready-to-use handguns outside the home is constitutional under any level of scrutiny." That wording is important. According to the correct ruling of Judge Scullin, the right to bear arms is just as paramount as the right to keep them....
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Neutral Bay, New South Wales : Distributed by Warner Bros. Entertainment, 2014.
1 DVD-video (approximately 87 min.) : sound, colour ; 12 cm.
In English. Subtitles in English for the hearing impaired (SDH). English audio description.
Cast: Sandra Bullock, George Clooney, Ed Harris.


No, we're not humble about it. The two-day TrawlerFest seminar "Everything You Need To Know About Diesel Engines" is one of the best preparatory courses of its kind anywhere. The instructors, author Nigel Calder and boatyard owner Steve Zimmerman, are knowledgeable, articulate and bring a welcome sense of humor to their presentations.
The next TrawlerFest is in Stuart, Florida, in March, followed by Bremerton, Washington, in May. The classes are limited to 36 students and usually sell out. Visit www.trawlerfest.com.
But not everyone has the wherewithall to fly from whereever they live to attend.
We have a solution for you. It's a good solution, albeit a partial one. Boaters University, operated by our parent company, offers an online class called "Marine Diesel Maintenance & Troubleshooting," taught by none other than Steve Zimmerman with help from our TrawlerFest partners at Whiticar Boatworks of Stuart, Florida.
Let Steve tell you what it's all about.


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Q: Dependency problems after update nuget packages of a VISX project I'm creating a VISX project. It runs normally, but after I update all it's nuget packages, it start to show errors when I try to run it in the Visual Studio Experimental Instance.
I tried to revert the packages, but it seems that some packages versions are no longer available for download.
The main reason is that the VISX can't load the Microsoft.VisualStudio.Threading assembly after the dependecies are updated.
I tried adding the assembly to the package's Assets too. The VISX can load it now, but can't find the following method in the AsyncPackage class, like it would be loading the wrong assembly version.
JoinableTaskFactory.SwitchToMainThreadAsync(cancellationToken);

There's also the same error (assembly not found) for the Newtonsoft.Json dependency.
Can someone help me with this?

I zipped everything and upload in my onedrive: https://1drv.ms/u/s!AmhLdtS8gKjrl71Vmnp85m4ZPUILgw
I included all packages, since there's a chance that who download the project could not execute it because the packages could not be downloaded from nuget.org.
To simulate the error, just update the dependencies.

Update 1
As suggested, I was able to redo the entire project using "package reference" to target dependencies.
The visx builds and work, but the problem with the dependency for Newtonsoft.Json still occurs.
Also, I tried adding another dependency from a nuget package that I created, and the same error occurs too.
I wrote a piece of code that uses JsonConvert just to make use of the dependency.
When I execute the "TestCommand" from VS Experimental instance, them the error popups.

Link for the second project in my onedrive:
https://1drv.ms/u/s!AmhLdtS8gKjrl71ZCd1WIYNDTZaSCw


The Beatles were back in Liverpool today in 1964 for the hometown premier of A Hard Day's Night, which had opened in London 4 days earlier. The Liverpudlian Lord Mayor saw fit to honor the band with a special ceremony at the town hall, and as some 200,000 fans gathered in the streets below with the band on a balcony, John Lennon was struck by the Nüremberg-Rally like setting and jokingly gave a few Nazi "Seig-Heil" salutes, which not everyone found funny.
Eric Clapton announced today in 1968 that Cream would break up after a farewell American tour (Clapton was huge at home in England, but had left The Yardbirds before For Your Love became a hit, and Cream had made him a superstar in the U.S.), and two shows at London's Royal Albert Hall. Jack Bruce and Ginger Baker had been famous for fighting in the Graham Bond Organization, and had promised Eric they'd get along, but of course they didn't and Cream had become unbearable. Clapton had invited keyboard player Steve Winwood to join the group in hopes he'd be a calming influence, but the two decided to form a new band after the breakup, and had no intention of using Baker in Blind Faith, but somehow he found out about the first drummerless practice session, and showed up anyway, and neither had the heart to tell him he couldn't be in the band.
The Nice were told they were "banned for life" from the Royal Albert Hall today in 1968 after organist Keith Emerson had burned an American flag on stage in protest of the Vietnam War while they played a rocking rendition of Leonard Bernstein's America from West Side Story, which Keith had described as "the world's first instrumental protest song" at their show there in late June. The band would break up not long after, and Emerson started Emerson Lake and Palmer two years later.
Today in 1969 was the funeral of The Rolling Stones founding guitarist Brian Jones at the Hatherly Road Parish Church in his hometown of Cheltenham. Bandmates Bill Wyman and Charlie Watts attended, but Mick Jagger was in Australia filming Ned Kelly, and Keith Richards was shacking up with Brian's old girlfriend Anita Pallenberg, and thought it best not to attend. Brian was buried at 12 feet deep to prevent morbid fans from digging him up.
Harry Nilsson released his 8th album Son of Schmillson today in 1972. It listed some interesting backup musicians: George Harrysong on guitar (George Harrison), Richie Snare on drums (Ringo Starr), and a relatively unknown guitarist named Peter Frampton.
Chuck Berry was sentenced to 5 months in prison today in 1979 after being found guilty of tax evasion. Chuck always demanded to be paid in cash, and always used a back up band of local musicians he figured should be honored to play with Chuck Berry, and never paid any of them.
The Who's bass player John Entwistle was buried in his hometown of Gloucestershire today in 2002, a week after dying of a heart attack in his hotel room at the Las Vegas Hard Rock Hotel and Casino after retiring for the evening with Hookers and Blow. The funeral at a 12th century church was attended by over 200, including former bandmates Roger Daltrey and Pete Townsend, who later remarked that he'd known John since they started the band, but only found out at the funeral that he'd been a Freemason.
Elf, Rainbow, Black Sabbath, and Dio singer Ronnie James Dio would be 76 if he hadn't died of cancer at 67.
Moby Grape guitarist and Seattle area resident Jerry Miller is 75, and still plays for the bikers most every Taco Thursday at Uncle Sam's Bar and Grill in Spanaway.
Triumph guitarist Rik Emmett is 65.
The Runaways drummer Sandy West would be 59 if she hadn't died of cancer at 47.


This year has definitely been a good year for my zucchini. I have steadily been harvesting at least one a day. I refuse to let any go to waste so I have cooking, baking and freezing a whole of zucchini-centric dishes.
I've seen lots of folks passing around recipes pairing zucchini with chocolate. Now that is something I hadn't tried yet. So, I decided to modify my zucchini bread recipe by adding in some cocoa powder and messing around with a few other things.
The result: Not bad, not bad at all. The bread came out moist and chocolaty. The zucchini taste was subtle. Renato gave it a thumbs up so I figured it was worthy to share with all of you.
Preheat oven to 350 degrees Fahrenheit. Lightly grease two loaf pans.
In a large bowl beat the eggs. Beat in the sugar and coconut oil. Add the cocoa, vanilla, zucchini and stir well.
In another large bowl sift the flour, baking soda, baking powder, salt, cinnamon, nutmeg, and cloves.
Pour the batter equally into the two loaf pans. Bake in the oven for 50-60 minutes. Rotating the pans halfway through the cooking time.
Remove the pans from the oven and let cool.
Eat now or save for later. Once fully cooled you can freeze the loaves. Wrap each one in aluminum foil and place inside a plastic freezer bag with as much of the air pressed out as possible.
So, what are you making with your abundance of zucchini this year?
oh, yum. we have about 43923432 zucchini i need to do something with, too!
I hear you. I'm trying to get creative in how to use them all. I just tried a flourless zucchini brownie recipe. It needs a few tweaks but it was surprisingly good.
Yum! I do not like zucchini (over did it as a child), but I love zucchini bread. I'll have to give this a try!
This recipe is especially good at "disguising" the zucchini.
My theory was that I'd use those spices in something like pumpkin bread so they would theoretical work with summer squashes too.
My wife is from Maine and her family has a pretty good Zucchini bread recipe, but I bet the cocoa would make it taste even better!
Glad you're going to give it a try.
This sounds great...can't wait to try it!
Zucchini is one of those plants that seems to always give way more veggies then I'd planned for - I hadn't tried it with chocolate, but it sounds like a great idea!


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Pardais es una freguesia portuguesa del concelho de Vila Viçosa, con 17,83 km² de superficie y  habitantes (2001). Su densidad de población es de 31,3 hab/km².

Enlaces externos

  (en portugués)
 Gobierno civil e información administrativa de Vila Viçosa (en portugués)

Freguesias de Vila Viçosa


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The Sounds of...
Every environment has its typical background noise. Every country has its own language. Check-out the atmosphere in a number of selected sites in the Czech Republic and compare it to similar sites in countries where Radio Prague has media partners.
Religion in the Czech Republic
The Czech Republic is one of the most atheistic countries in the world. The number of believers has consistently been dropping and has halved in the course of the last two decades. Only a fifth of Czech citizens claim to belong to a faith. Only a tenth of believers go to church regularly, at least once a month. The Roman Catholic Church has the largest number of followers.
The majority of Swiss people are Christians, most of them Roman Catholics. A 2016 report showed that about 70% of Swiss visited a place of worship at most five times a year. The number of believers has dropped in the past few years. In October 2018, Arlesheim Cathedral community welcomed their new pastor Sylvester Ihuoma in a solemn mass. The cathedral was built in the 1680s and is known for its impressive late Baroque architecture.
Mass at the Shrine of Kateri Tekakwitha in Kahnawake ( Canada)
Kateri Tekakwitha (1656-1680), a Mohawk woman from Kahnawake, an Indigenous community near Montreal, was canonized by the Catholic Church in 2012. On October 21, 2018, a special mass was celebrated for the 6th anniversary of her canonization at the St. Francis Xavier Mission, in Kahnawake, which also houses her shrine. The choir sings in Mohawk and some of the prayers are also in that language.
86.5% of the Romanians have declared themselves Orthodox Christians, in the latest census. What's the sound of an Orthodox mass in Romania, in a centuries-old monastery with World Heritage frescoes in the region of Bukovina, in a district church, in the Patriarchal Cathedral in Bucharest or in a small wooden church in Maramures County?
Sunday mass in Warsaw (Poland)
About 93 percent of Poles declare themselves to be Catholics. But even in Poland is the number of church service attendees falling. By 2016, the figure had dropped to 36.7 percent. Despite the fall, the Sunday mass remains for many Poles a very important part of life.
This year was the first time since 1942 that Slavia Prague football club secured the double, meaning lifting both the league title…
The sounds of international airports
Every environment has its typical background noise. Check-out the atmosphere of Prague's Václav Havel International Airport in the…
Church and social historian Tomáš Petráček on Czech secularism, Slavic paganism, and Easter as a celebration of life
In a wide-ranging interview at the start of Holy Week, leading up to Easter, the Roman Catholic priest Tomáš Petráček – a leading church and…