text:
Q: Does HttpClient adds any prefix to a JSON response? I am using Apache HTTPClient 4.5.2 in the following code to fetch the JSON content.
Surprisingly a "?" gets prefixed in the body leading to a malformed JSON.
Code Below:
public class URLDataReader {

     public static void main(String[] args) throws java.lang.Exception {

         URLDataReader obj = new URLDataReader();
         String eventsURL = "http://api.511.org/traffic/events/?api_key=<my-personal-api-key>&format=json";

         String jsonText = null;
         try {
             jsonText = obj.readURLContent(eventsURL, true);
         } catch (java.lang.Exception e) {
             System.out.print("Exception in Reading Data from the URL");
             e.printStackTrace();
         }

         if (jsonText != null) {
             System.out.println("Read JSON Content from URL :" + jsonText + "\n\n\n");
             Gson gson = new GsonBuilder().create();
             EventQueryResponse response = gson.fromJson(jsonText, EventQueryResponse.class);
             System.out.println("JSON read into Java Object:" + response.toString());
         }
     }

     public String readURLContent(String urlToRead, boolean useProxy) throws java.lang.Exception {

         String targetHost = new URL(urlToRead).getHost();

         String inputLine = null;
         StringBuffer result = null;

         String responseJsonBody = null;

         if (useProxy) {
             CredentialsProvider credsProvider = new BasicCredentialsProvider();
             credsProvider.setCredentials(
                 new AuthScope(proxyHost, proxyPort),
                 new UsernamePasswordCredentials(proxyUsername, proxyPassword));

             CloseableHttpClient httpclient = HttpClients.custom().setDefaultCredentialsProvider(credsProvider).build();
             HttpHost target = new HttpHost(targetHost, 80);
             HttpHost proxy = new HttpHost(proxyHost, proxyPort);

             RequestConfig config = RequestConfig.custom().setProxy(proxy).build();

             System.out.println("URL Path:" + new URL(urlToRead).getFile());

             HttpGet httpget = new HttpGet(new URL(urlToRead).getFile());
             httpget.setConfig(config);
             CloseableHttpResponse response = httpclient.execute(target, httpget);

             responseJsonBody = EntityUtils.toString(response.getEntity(), "UTF-8");

             response.close();
             httpclient.close();

         } else {
             System.out.println("Not Implemented Yet!!");
         }

         return responseJsonBody;

     }

     private static String proxyUsername = "username";
     private static String proxyPassword = "password";
     private static String proxyHost = "proxy.host.com";
     private static int proxyPort = 80;

 }

The output I get is
URL Path:/traffic/events/?api_key=<my-personal-ap-key>&format=json
Read JSON Content from URL :?{"events":[{"url":"/traffic/events/511.org/23765","jurisdiction_url":"http://api.511.org/jurisdictions/511.org/","id":"511.org/23765","status":"ACTIVE","headline":"Caltrans : Long-term construction on CA-128 Eastbound and Westbound between Lower Chiles Valley Rd (Napa) and Berryessa Knoxville Rd (Napa). Alternate lanes closed. Expect delays.","event_type":"CONSTRUCTION","severity":"UNKNOWN","created":"2016-08-05T20:20Z","updated":"2016-08-07T17:28Z","geography":{"type":"Point","crs":{"type":"name","properties":{"name":"urn:ogc:def:crs:EPSG::4326"}},"coordinates":[-122.296272,38.503876]},"roads":[{"name":"CA-128 E","from":"Lower Chiles Valley Rd","to":"Berryessa Knoxville Rd","state":"Open"}],"schedule":{"intervals":["2016-08-05T20:20Z/2016-10-01T06:55Z"]}},{"url":"/traffic/events/511.org/24826","jurisdiction_url":"http://api.511.org/jurisdictions/511.org/","id":"511.org/24826","status":"ACTIVE","headline":"CHP : Obstruction on I-580 Eastbound west of Marina Bay Pky (Richmond). Center lanes blocked. Expect delays.","event_type":"INCIDENT","severity":"UNKNOWN","created":"2016-08-08T12:00Z","updated":"2016-08-08T12:01Z","geography":{"type":"Point","crs":{"type":"name","properties":{"name":"urn:ogc:def:crs:EPSG::4326"}},"coordinates":[-122.353792,37.921656]},"roads":[{"name":"I-580 E","from":"Marina Bay Pky","state":"Open"}],"schedule":{"intervals":["2016-08-08T12:00Z/"]}}],"pagination":{"offset":0},"meta":{"url":"/traffic/events/?api_key=65ce84c9-e5c4-43bf-ac16-a672e0b9266f&format=json","up_url":"/","version":"v1"}}

Not sure where the "?" is coming from at the beginning of the JSON response.
Any ideas?

redpajama_set_name: RedPajamaStackExchange


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text: 44 | M R EP O RT O R I G I NAT I O N S E R V I C I N G DATA G O V E R N M E N T S E C O N DA R Y M A R K E T THE LATEST ORIGINATION The Far-Reaching Implications of California's Affordability Crisis The high cost of living is causing more than 60% of millennials to look elsewhere. A nalysis from the San Francisco Chronicle states that California's housing crisis is not just impacting homeowners, but the business community as well. "We sold our 1,310-square-foot house on a 10,000-square-foot lot for $335,000. In Oregon, we purchased a 1,408-square-foot home on two-thirds of an acre for $168,000. More houses, more property, no sales tax, much lower gas taxes, similar property tax rates. The decision was a no- brainer," said Nina Kingsley-Keller in the Chronicle's article. The author of the article, Ken Monroe, Chairman of the Family Business Association of California, said data shows 53% of Californians are considering moving because of the high cost of living. Also, 63% of millennials want to leave. "If they leave, who will be left in the workforce when the baby boomers retire?" Monroe asks. Monroe continued by saying lawmakers "must come up with real-world solutions to clear the way to build the millions of new homes and apartments our state desperately needs." "Lawmakers must realize that the government isn't going to build the 3.5 million housing units the governor has set as his goal," Monroe said. "The private sector—and their thousands of employees—are going to build them. What government officials can do is recognize that their actions over the past 40 years, no matter how well-intended, are a major reason for our shortage." He added that legislators must stop adding new mandates that increase costs. The California Environmental Quality Act requires all new homes to have solar panels, which could add between $9,000 and $10,000 to the cost of the home. Zillow previously reported that the total value of U.S. homes reached $33.6 trillion in 2019. California made up the highest share of the nation's housing value at 21.2% and 12% of its population. Texas is the next most populated state at 8.8% but accounts for just 5.9% of housing value. The home values of New York ($2.7 trillion); Florida ($2 trillion); Texas ($2 trillion); and Washington ($1 trillion) combined to exceed California's $7.1 trillion in home value. Three metros reached the trillion-dollar mark—New York ($3.2 trillion); Los Angeles ($2.5 trillion); and San Francisco ($1.6 trillion). Los Angeles was the only market to add more than a trillion dollars of housing during the decade, gaining $1.1 trillion. California was home to three of the five metros that gained the most value: San Francisco ($827 billion); New York ($657 billion); San Jose, California ($360 billion); and Seattle, Washington ($356 billion). The Gen Z Homebuyer Landscape The post-millennial generation led the way for first-time buyers during Q4. W hile the National Association of Homebuilder's latest Housing Trends Report found the share of adults looking to buy a home over the next year fell to 11% in Q 4 2019, the number of prospective first-time buyers rose to 63%. The share of first-time buy- ers rose from 53% in Q4 2018. Prospective adults looking to buy a home fell from 13% during the same quarter last year. This drop marks the fifth-consecutive year-over- year decline in the share of prospective adults looking to buy a home. The NAHB said this trend is likely spurred by low levels of housing inventory. The share of prospective homebuyers for adults has consistently fallen since Q 4 2017 when the share was 24%. The most recent figures are the second-consecutive quarter of increases for first-time buyers and is an increase from Q 4 2018's 53%. Millennials are most likely to purchase a home within the next year at 19%. The millennial genera- tion was followed by Gen Z (13%) and Gen X (12%). Just 5% of baby boomers are planning to purchase a home. While millennials are more likely to buy a home, it is Gen Z consumers who are most likely to be first-time buyers at 88%. Millennials were close behind at 78%. The share of first-time buyers drops significantly for Gen Z buy- ers 57%) and baby boomers (20%). The NAHB states that more than 60% of prospective buyers in every region are buying a home for the first time. The Northeast led the nation with 65% of first-time buyers plan- ning to buy a home. The South region followed with 64%, the West had 62%, and the Midwest had 61%. The South and the West regions led the nation in potential home sales for adults at 12%. Ten percent of adults plan to buy a home in the Northeast and 9% in the Midwest. Forty-percent of first-time buy- ers are looking to buy an existing, while 19% prefer a newly-built home. The remaining 41% would purchase both options. While millennials are more likely to buy a home, it is Gen Z consumers who are most likely to be first- time buyers

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: send data from s3 to postres rds with glue I'm trying to create an automated pipeline with aws. I'm able to get my csv file into my s3 bucket and that automatically triggers a lambda function to send the csv to my glue job. The glue job then turns the csv into a dataframe with pyspark. you cannot use psycopg2, pandas or sqlalchemy, or else glue will give an error saying the module doesn't exist. I have a postgres rds setup in aws rds. This is what i have so far
import sys
from pyspark.context import SparkContext
from pyspark.sql.functions import *
from pyspark.sql.types import *
from pyspark.sql import SparkSession
from setuptools import setup
from sqlalchemy import create_engine
spark = SparkSession.builder.getOrCreate()
args = getResolvedOptions(sys.argv, ["VAL1", "VAL2"])
file_name = args['VAL1']
bucket_name = args["VAL2"]
file_path = "s3a://{}/{}".format(bucket_name, file_name)
df = spark.read.csv(file_path, sep=',', inferSchema=True, header=True)
df.drop("index")
url = "my rds endpoint link"

i have tried almost a dozen solutions before asking on stackoverflow. So any help would be amazing

A: I used this df.write approach before.  Starting where you left off with your pyspark dataframe
jdbc_url = 'jdbc:postgresql://<instance_name>.xxxxxxxxx.us-west-2.rds.amazonaws.com:5432/<db_name>'

(df.write.format('jdbc').option('url', 'jdbc_url')
                        .option('user', 'myUsername')
                        .option('password', 'myPassword')
                        .option('dbtable', 'myTable')
                        .option('driver', 'org.postgresql.Driver')
                         .mode('append').save())


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As you can see, the pros definitely outweigh the cons. A few of the cons are rather nitpicky, like the URLs being cluttered, but that is more of a personal preference than anything. I'm aware that there are many other landing page builders that we'll use, such as Thrive Architect from Thrive Themes. For me , I've found the funnel measure integration for a few of the easiest funnel applications to design however.
Using TA, you are going to want to design the webpage, then connect to the following pages in the sequence and it is done inside WordPress. While I love Thrive and their subjects, I really do believe ClickFunnels is inherently simpler to use than TA, but it comes in a monthly price.
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WhatsApp recently released a feature which enables you to delete messages for everyone in a conversation within 7 minutes. It does come in handy when you are recalling wrongly sent texts but it has been discovered that recalled texts don't completely disappear from the receiver's mobile phone.
First spotted by a Spanish publication Android Jefe, this individual bug has been related to the notifications we see on receiving a new WhatsApp message in Android smartphones. It has been found that the deleted message's notification still exists on the device even after the message is recalled by the sender. This is because the sent messages are stored in the notification log of the Android system.
You can see in the screenshots attached above that the message has been deleted from within WhatsApp, but it still does appear in the notification log on the right. This means every time you receive a message, whose notification is shown, a copy of the same gets stored in the notification log. This log records everything in an Android session, until the device is restarted.
You can access the notification log via two methods. If you have stock Android running on your device, you can easily create a notification log shortcut on your homescreen. You can even install a third party app, such as Notification History, which lets you view your device's notification log.
This is a huge security loophole in the implementation of the recall feature and the deleted texts can easily be read by someone, who has access to your Android device. They can be read by hackers as well, if your device's security has been compromised by some intrusive malware. The WhatsApp notifications even store sensitive information and it's a huge lapse in security, which may not be fixable because WhatsApp doesn't recall notifications. What do you think of this security loophole? Let us know in the comments down below.


text:
\section{Introduction} \label{Intro}

Let $\Omega$ be a bounded domain in $\mathbb{C}^n$ and $A^{2}(\Omega)$ denote the Bergman space
of square-integrable holomorphic functions on $\Omega.$ The Bergman projection on
$\Omega$ is the orthogonal projection from $L^2(\Omega)$ onto $A^{2}(\Omega).$

The Bergman projection is known to be regular, in the sense that it maps $W^s$ to
$W^s$ for all $s\geq 0$ where $W^s$ denotes the Sobolev space of order $s,$ on a
large class of smooth bounded pseudoconvex domains (throughout this paper a domain is
smooth if its boundary is a smooth manifold). Regularity is, usually,
established through the $\overline{\partial}$-Neumann problem, the solution operator for the
complex Laplacian $\Box=\overline{\partial}\dbar^*+\overline{\partial}^*\overline{\partial}$ on square integrable
$(0,1)$-forms. For more information on this matter we refer the  reader to
\cite{BoasStraube99,StraubeBook} and the references therein.

Irregularity of the Bergman projection is not understood nearly as well as
regularity. The story of irregularity goes back to the discovery of the
worm  domains in $\mathbb{C}^2$ by Diederich and Forn\ae{}ss \cite{DiederichFornaess77}.
Worm domains were constructed to show that  the closure of some smooth
bounded pseudoconvex domains may not have Stein neighborhood bases (a compact
set $K\subset \mathbb{C}^n$ is said to have a Stein neighborhood basis if for every open
set $U$ containing  $K$ there exists a pseudoconvex domain $V$ such that
$K\subset V \subset U$). Indeed, Diederich and Forn\ae{}ss in
\cite{DiederichFornaess77} showed that the closure a worm domain does not
have a Stein neighborhood basis if the total winding is bigger than or equal to
$\pi.$ It turned out that worm domains are also counter-examples for regularity
of the Bergman projection. In 1991,  Kiselman \cite{Kiselman91} showed that the
Bergman projection does not satisfy Bell's condition R on nonsmooth  worm
domains (a domain $\Omega$ satisfies Bell's condition
R if  the Bergman projection maps $C^\infty(\overline\Omega)$ to
$C^\infty(\overline\Omega)$). In 1992,  the first author
\cite{Barrett92} showed that the Bergman projection on a smooth worm domain   does not map
$W^s$ into $W^s$ if $s\geq \pi /(\text{total winding}).$  On the other hand, Boas
and Straube \cite{BoasStraube92} showed that the Bergman projection maps $W^k$
into $W^k$ if $k\leq \pi /(2\times \text{total winding})$ and $k$ is a positive
integer or $k=1/2.$ Finally, in 1996 Christ \cite{Christ96} showed that the
Bergman projections on smooth worm domains, with any positive
winding, do not satisfy Bell's condition R. Recently, Krantz and Peloso
\cite{KrantzPeloso08a,KrantzPeloso08b} studied the asymptotics
for the Bergman kernel on the model domains in $\mathbb{C}^2$ and derived $L^p$
(ir)regularity for the Bergman projection on worm domains in $\mathbb{C}^2.$

In this note we will construct smooth bounded pseudoconvex domains
$\Omega_{\alpha\beta}\subset \mathbb{C}^n$ that are  higher dimensional generalizations of
the worm domains in $\mathbb{C}^2$ and  study the irregularity of the Bergman projection
on  these domains on $L^p$ Sobolev spaces for $1\leq p<\infty.$ We will use
 the method developed by the first author in \cite{Barrett92} to show that
irregularity on $L^2$ Sobolev spaces depends only on the total winding whereas the
irregularity on $L^p$ spaces with $p\neq 2$ depend on the total winding as well as the
dimension $n$.

The two parameters $\alpha$ and $\beta$ in $\Omega_{\alpha\beta}$ represent the speed of
the winding and the thickness of the annulus, respectively. Both parameters play a role in
the proof of Theorem \ref{Theorem}, but we find it interesting to note that the actual
results depend only on the total winding whether this is achieved by fast winding along a
thin annulus or slow winding along a thick annulus.

The domains  $\Omega_{\alpha\beta} \subset \mathbb{C}^n, n\geq 3,$ are defined by
\begin{equation*}\label{Domain}
\Omega_{\alpha\beta}=\left\{(z_{1}, z',z_{n})\in \mathbb{C}^{n}: r(z_1,z',z_n) <0\right\}
\end{equation*}
with
\[r(z_1,z',z_n) =\left|z_{1}-e^{2i\alpha \ln|z_{n}|}
\right|^{2}+|z'|^{2}-1 +\sigma(|z_n|^2-\beta^2)+ \sigma(1-|z_n|^2);\]
here  $z'=(z_{2},\ldots,z_{n-1}), |z'|^{2}=|z_{2}|^{2}+\cdots+|z_{n-1}|^{2}$,
the constants $\alpha>0, \beta>1,$  and
$\sigma(t)=Me^{-1/t}$ for $t> 0,\,\sigma(t)=0$ for $t\leq 0$ for some $M>0$.

In section \ref{levi} below we show that $\Omega_{\alpha\beta}$
is smooth bounded pseudoconvex when $M$ is sufficiently large.
The main result of this paper is the following theorem.

\begin{theorem}\label{Theorem}
The Bergman projection for $\Omega_{\alpha\beta}$ does not map
$W^{p,s}\left(\Omega_{\alpha\beta}\right)$ into
$W^{p,s}\left(\Omega_{\alpha\beta}\right)$ where $1\leq p<\infty$ and
$s\ge \frac{\pi}{2\alpha\ln\beta}+n\left(\frac{1}{p}-\frac{1}{2}\right).$
\end{theorem}
Here $W^{p,s}\left(\Omega_{\alpha\beta}\right)$ is the Sobolev space of order
$s$ with exponent $p$ and when
$W^{p,s}\left(\Omega_{\alpha\beta}\right)\not\subset L^2\left(\Omega_{\alpha\beta}\right)$
we mean that the $W^{p,s}$ bounds do not hold for the Bergman projection on
$W^{p,s}\left(\Omega_{\alpha\beta}\right)\cap L^2\left(\Omega_{\alpha\beta}\right).$ The
denominator $2\alpha\ln\beta$ appearing above may be interpreted as the
total amount of winding along the annulus $1<|z_{n}|<\beta$
(see \eqref{LimitDomain} below).

If we choose $p=2$ then  the amount of irregularity provided by a fixed amount
of winding is independent of the dimension.

\begin{corollary}\label{corollary1}
The Bergman projection for $\Omega_{\alpha\beta}$ does not map
$W^{2,s}\left(\Omega_{\alpha\beta}\right)$ to
$W^{2,s}\left(\Omega_{\alpha\beta}\right)$ when $s\geq
\frac{\pi}{2\alpha\ln\beta}$.
\end{corollary}

\begin{remark}
Assume that the Bergman projection $P_U$ of a domain $U$ bounded on $L^p(U)$
where $p> 2.$ Then the duality and self-adjointness of the Bergman
projection imply that $P_U$ is also bounded on $L^q(U)$
where $\frac{1}{p}+\frac{1}{q}=1.$ Furthermore, interpolation implies that
$P_U$ is bounded on $L^r$ for all $r\in [q,p].$
\end{remark}
Therefore, when $s=0$ and $n\alpha\ln\beta>\pi,$ the previous remark and Theorem
\ref{Theorem}  imply the following corollary.

\begin{corollary}\label{corollary2}
The Bergman projection for $\Omega_{\alpha\beta}$ does not map
$L^p\left(\Omega_{\alpha\beta}\right)$ to
$L^p\left(\Omega_{\alpha\beta}\right)$ when
$0<\frac{1}{p}\le\frac{1}{2}-\frac{\pi}{2n\alpha\ln\beta}$ or
$\frac{1}{2}+\frac{\pi}{2n\alpha\ln\beta}\leq \frac{1}{p}<1$.
\end{corollary}

Theorem \ref{Theorem} is proved in section \ref{ProveThm} below.  The proof is
based on model domain asymptotics developed in section \ref{Model}.

\section{Geometry of the Worm Domains} \label{levi}

\begin{proposition}
The domain $\Omega_{\alpha\beta}$ is smooth bounded and pseudoconvex whenever
$M$ is sufficiently large.
\end{proposition}

\begin{proof}
Start by requiring $M>e^2$.  Then  $\Omega \subset \{z\in \mathbb{C}^n:
|z_1|<3,|z'|<2,1/2 <|z_n|<\sqrt{\beta^2+1/2}\}$. Then $\Omega$ is bounded.
Also, by considering $z_1$-, $z'$-, and $z_n$-derivatives in order it is easy to
check that the gradient of $r(z)$ does not vanish on $\{z\in \mathbb{C}^n:r(z)=0\}$, so
$\Omega$ has smooth boundary.

It remains to show that $\Omega_{\alpha\beta}$ is pseudoconvex. It suffices to
check this locally.  We focus on the case $|z_n|\geq (1+\beta)/2$, the case
$|z_n|\leq (1+\beta)/2$ being similar.

Multiplying $r(z)$ by $e^{\Arg(z^{2\alpha}_{n})}$
we obtain the new defining function
\begin{equation*}
r_1(z)=r_2(z)-2\,\re\left(z_1 z_n^{-2\alpha i}\right)
\end{equation*}
where
\[r_2(z) = \left( |z_1|^2 + |z'|^2+\lambda(z_n)\right) e^{\Arg(z^{2\alpha}_{n})}
\text{ and } \lambda(z_n)=\sigma\left(|z_n|^2-\beta^2\right).\]
Since $2\,\re\left(z_1 z_n^{-2\alpha i}\right)$ is pluriharmonic it will
suffice now to show that $r_2$ is plurisubharmonic.
To simplify the notation let $A(z)=|z_1|^2+|z'|^2+\lambda(z_n)$ and
$B(z)=\Arg(z^{2\alpha}_{n}).$ Let $W=\sum_{j=1}^nw_j \partial/\partial z_j$ with
$w_j$ constant. In the following calculations  $H_{f}(W)$ denote the complex
Hessian of $f$ in the direction $W.$ Then $W(r_2)=e^B(W(A)+AW(B))$ and
Cauchy-Schwarz inequality implies that
\[-2\re\left( \overline w_n B_{\overline
z_n}\sum_{j=1}^{n-1} w_j\overline z_j \right)\leq
\sum_{j=1}^{n-1}|w_j|^2+|\overline w_nB_{\overline
z_n}|^2 \sum_{j=1}^{n-1}|z_j|^2.\]
Using the above inequality in the second line below we get
\begin{align*}
H_{r_2}(W)
&=e^B(H_A(W)+2\re(W(A)\overline{W}(B))+A|W(B)|^2+A H_B(W)) \\
&\geq |w_n|^2e^B(\lambda_{z_n\overline  z_n}+2\re(\lambda_{z_n}B_{\overline
z_n})+\lambda |B_{\overline  z_n}|^2).
\end{align*}
One can check that
$\lambda_{z_n}(z_n)=\overline{z}_n \sigma'(|z_n|^2-\beta^2),|B_{\overline
z_n}|=\frac{\alpha}{|z_n|},$ and
\begin{equation*}
\lambda_{z_n\overline{z}_n}(z_{n})= |z_n|^2\sigma''(|z_n|^2-\beta^2)
+\sigma'(|z_n|^2-\beta^2).
\end{equation*}
We note that since $\lambda
(z_n)=\lambda_{z_n}(z_n)=\lambda_{z_n\overline z_n}(z_n)=0$ for
$|z_n|\leq \beta,$ without loss of generality  we can assume that
$|z_n|>\beta.$ Using  the fact that $\beta<|z_n|<\sqrt{\beta^2+1/2}$ and
$t=|z_n|^2-\beta^2$ on the third line below we get
\begin{align*}
\lambda_{z_n\overline z_n}
+2\re(\lambda_{z_n}B_{\overline  z_n})+\lambda |B_{\overline  z_n}|^2 \geq &
\lambda_{z_n\overline z_n}
-\frac{2\alpha|\lambda_{z_n}|}{|z_n|}\\
\geq &|z_n|^2\sigma''(|z_n|^2-\beta^2)
+(1-2\alpha)\sigma'(|z_n|^2-\beta^2) \\
=& Me^{-1/t} \left(
\frac{\beta^2+t}{t^4}-\frac{2(\beta^2+t)}{t^3}+\frac{1-2\alpha}{t^2}
\right)\\
=&\frac{M(\beta^2+t)e^{-1/t}}{t^4} \Big( 1-2t+\frac{(1-2\alpha)t^2}{\beta^2+t}
\Big)
\end{align*}
We can choose $M$ sufficiently large so that $z\in \Omega_{\alpha\beta}\cap
\{z\in\mathbb{C}^n:|z_n|\geq \beta\}$  implies that $t$ is sufficiently small. In return, this
implies that
\[1-2 t+\frac{(1-2\alpha)t^2}{\beta^2+t} > 0.\]
The last inequality above implies that
$\lambda_{z_n\overline z_n} +2\re(\lambda_{z_n}B_{\overline  z_n})+\lambda
|B_{\overline  z_n}|^2 \geq 0$ for
$z\in \Omega_{\alpha\beta}$ such that $|z_n|\geq (1+\beta)/2.$
Hence, the domain $\Omega_{\alpha\beta}$ is pseudoconvex for sufficiently large $M.$
\end{proof}

\begin{remark}
A similar calculation shows that the set of weakly pseudoconvex points in the
boundary is the set $\{(0,\ldots,0,z_n)\in \mathbb{C}^n:1\leq |z_n| \leq \beta \}.$
\end{remark}

\begin{remark}
We note that regularity of the $\overline{\partial}$-Neumann operator is closely connected
to regularity of the Bergman projection \cite{BoasStraube90}. In particular,
if the $\overline{\partial}$-Neumann operator of a smooth bounded pseudoconvex domain is
globally regular then  the Bergman projection satisfies Bell's condition R.
One can show that on the set $\{(0,\ldots,0,z_{n})\in \mathbb{C}^n:1\leq |z_n| \leq
\beta \}$ the Levi form of $r$ has only one vanishing eigenvalue as the Levi
form has positive eigenvalues in the direction transversal to $z_{n}$-axis. In
this case Theorem 1 in \cite{SahutogluStraube06} applies and it implies that the
$\overline{\partial}$-Neumann operator is not compact on $(0,1)$-forms (compactness of the
$\overline{\partial}$-Neumann operator implies that it is globally regular
\cite{KohnNirenberg65}). However, to show irregularity of the Bergman
projection in Sobolev scale one needs to work harder.
\end{remark}

\section{Model Domains} \label{Model}

In this section we are going to define a family of simplified model domains and
calculate the asymptotics for the Bergman kernels of these model domains. We
use a modified version of the method developed by the first author
in \cite{Barrett92}.

For $\lambda>0$ let
\begin{align*}
\tau_{\lambda}(z_{1},z',z_{n})&=(2\lambda^{2}z_{1},\lambda z',z_{n}),\\
r_{\lambda}&=\lambda^{2} r \circ \tau_{\lambda}^{-1},\\
D_{\lambda}&=\tau_{\lambda}(\Omega_{\alpha\beta}).
\end{align*}
Then for $1\le|z_{n}|\le\beta$ we have $r_{\lambda}\searrow r_{\infty}$ as
$\lambda\to \infty$ where
\[r_{\infty}(z_{1},z',z_{n})=|z'|^{2}-\re\left(z_{1}e^{
-2\alpha i\ln|z_{n}|}\right);\]
for $|z_n|$ outside this range we have $r_{\lambda}\to\infty$.  It follows that
the $D_{\lambda}$ converge in an appropriate sense to the limit domain
\begin{equation}\label{LimitDomain}
D=D_{\alpha\beta}= \left\{(z_{1},z',z_{n})\in \mathbb{C}^{n}:\re\left(
z_{1}e^{-2\alpha
i\ln|z_{n}|}\right) >|z'|^{2},1<|z_{n}|<\beta\right\},
\end{equation}
the limit being increasing over the annulus $1\le|z_{n}|\le\beta$.

Bergman projection $P$ of $D$ is defined by $Pf(z)=\int_D K(z,w) f(w)\,dV(w)$
where $f\in L^{2}(D)$ and  $K:D\times D\to \mathbb{C},$ is the Bergman kernel
characterized by the following conditions
\begin{itemize}
 \item [i.] $K(z,w)\in A^2(D)$ for fixed $w\in D$,
\item[ii.] $K(w,z)=\overline{K(z,w)}$,
\item[iii.] $\int_D K(z,w) f(w)\,dV(w)=f(z)$ for $f\in A^{2}(D)$.
\end{itemize}
If $f_1, f_2,\dots$ is an orthonormal basis for $A^{2}(D)$ then we
have $K(z,w)=\sum_j f_j(z)\overline{f_j(w)}.$

To study the Bergman kernel of $D$ we begin by performing a Fourier
decomposition. We define
\begin{equation}\label{projection}
(P_{Jk}f)(z_{1},z',z_{n})=\frac{1}{2^{n-1}\pi^{n-1}}\int\limits_{[-\pi,\pi]^{n-1
}}f(z_{1},e^{iS}z',e^{it}z_{n})e^{-iJ S}e^{-ikt}dS\,dt,
\end{equation}
where
\begin{align*}
e^{iS}&=(e^{is_{1}},\ldots,e^{is_{n-2}}),\\
S&=(s_{1},\ldots,s_{n-2})\in [-\pi,\pi]^{n-2},\\
J&=(j_{1},\ldots,j_{n-2})\in \mathbb{N}^{n-2},\\
k&\in \mathbb{Z},\\
J S&=j_{1}s_{1}+\cdots+j_{n-2}s_{n-2},\\
dS&=ds_{1}\cdots ds_{n-2}.
\end{align*}
Let us define the mapping
$\rho_{St}(z_{1},z',z_{n})=(z_{1},e^{iS}z',e^{it}z_{n}).$
Then $P_{Jk}$ is the orthogonal projection from $A^{2}(D)$ onto
\[A_{Jk}^{2}(D)=\{f\in A^{2}(D): f\circ \rho_{St}=e^{iJS}e^{ikt}f \text{ for all
} S,t\}.\]
Therefore the Bergman space $A^{2}(D)$  can be written as an orthogonal sum
\[A^{2}(D)=\underset{\begin{subarray}{c}
 J\in \mathbb{N}^{n-2},\,k\in\mathbb{Z} \end{subarray}}{\bigoplus} A_{Jk}^{2}(D)\]
and the Bergman kernel $K(z,w)$ for $D$ satisfies
\[K(z,w)=\sum\limits_{J\in \mathbb{N}^{n-2},\,k\in\mathbb{Z}}K_{Jk}(z,w)\]
where $K_{Jk}(z,w)$ is the kernel for $A_{Jk}^{2}(D).$

One can show that for $f\in A_{Jk}^{2}(D)$ the function
$f(z_{1},z',z_{n})z_{2}^{-j_1}\cdots z_{n-1}^{-j_{n-2}}z_{n}^{-k}$ is  a
function that is locally independent of $(z',z_n)$.  We notate such functions as
functions of $z_1$, where it is understood that $z_1$ ranges over the Riemann
domain described by $-\pi/2<\Arg z_1<2\alpha\ln\beta+\pi/2$.

Let $|J|=j_{1}+\cdots+j_{n-2}.$  Then a square integrable holomorphic function
$f$ on $D$ can be written as
\[ f(z)=\sum_{J\in \mathbb{N}^{n-2},\,k\in \mathbb{Z} }F_{Jk}(z)\,\]
where
\[F_{Jk}(z_{1},z',z_{n})=z_{1}^{-\frac{|J|+n}{2}}f_{Jk}(z_{1})z'^{J}z_{n}^{k}\]
and the sum converges locally uniformly.

Now we will calculate the $L^{2}$-norm of $F_{Jk}$ on $D.$ Let
$z_{1}=r_{1}e^{i\theta_{1}}, r_{j}=|z_{j}|$ for $j=1,\ldots n
,r'=\sqrt{r_{2}^{2}+\cdots+r_{n-1}^{2}},s=\ln|z_{n}|^{2}.$ Then $D$ is described
by the inequalities
\begin{align*}
0&< r_{1}<\infty,\\
0&<s<2\ln\beta,\\
 |\theta_{1}-\alpha s|&<\pi/2,\\
 0&\leq r'<\sqrt{r_{1}\cos(\theta_{1}-\alpha s)}.
\end{align*}
We have
\begin{eqnarray} \displaystyle
\nonumber  \|F_{Jk}\|^{2}_{D}&=&
\int\limits_{D}
 |f_{Jk}(r_{1}e^{i\theta_{1}})|^{2}r_{1}^{-|J|-n+1}r_2^{2j_2+1}\cdots
r_{n-1}^{2j_{n-2}+1} r_{n}^{2k+1}d\theta_{1}\cdots d\theta_{n} dr_{1}\cdots
dr_{n} \\
 \nonumber &=&
 C_{nJ} \int\limits_{\substack{0< r_{1}<\infty \\
 |\theta_{1}-\alpha s|<\pi/2 \\
  0<s<2\ln \beta}}
   |f_{Jk}(r_{1}e^{i\theta_{1}})|^{2}\cos^{|J|+n-2}(\theta_{1}-\alpha
s)e^{s(k+1)}r_{1}^{-1} d\theta_{1} dr_{1} ds  \\
 &=& \label{IntOnGamma}
\int\limits_{\substack{
 0< |z_{1}|<\infty \\
 -\pi /2<\arg(z_{1})<2\alpha \ln \beta +\pi /2 }}
 |f_{Jk}(z_{1})|^{2}W_{Jk}(\theta_{1})|z_{1}|^{-2}\,dV(z_{1})
 \end{eqnarray}
 where  $C_{nJ}$ is a positive constant,
\[W_{Jk}(\theta_{1})=C_{nJ}\int_{-\infty}^{\infty}\cos^{|J|+n-2}(\theta_{1}
-\alpha t)\chi_{\pi/2}(\theta_{1}-\alpha t)e^{t(k+1)}\chi_{\ln
\beta}(t-\ln\beta)\,dt, \]
and $\chi_a(t)$ is the characteristic function of the interval $[-a,a]$ for
$a>0.$  (The positivity of   $C_{nJ}$ follows from the fact that we are only
integrating over positive values of $r_j$.)

Let us use a change of coordinates $z=\ln z_1$ in the last integral to obtain
\begin{align}\label{ChangeCoord}
\|F_{Jk}\|^{2}_{D}&= \int\limits_{\substack{
 -\infty< x<\infty \notag\\
 -\pi /2<y<2\alpha \ln \beta +\pi /2 }}|f_{Jk}(e^z)|^{2}W_{Jk}(y)\,dV(z) \\
 &= \int\limits_{\substack{-\infty< x<\infty \\
 -\pi /2<y<2\alpha \ln \beta +\pi /2 }}|\widetilde
f_{Jk}(z)|^{2}W_{Jk}(y)\,dV(z)
 \end{align}
where $z=x+iy$ and  $\widetilde f_{Jk}(z)=f_{Jk}(e^{z}).$
Then $\widetilde f_{Jk}$ is a square integrable holomorphic function on
$S_{\alpha\beta}=\{z\in \mathbb{C}:-\pi/2<\im(z)<\pi /2+2\alpha \ln \beta \}$ with
weight $W_{Jk}.$ Furthermore, the Bergman kernel $K_{Jk}$ for $A_{Jk}^{2}(D)$
can be calculated as
\begin{equation}\label{EqnKernelTransfrom}
K_{Jk}(z,w)=K_{Jk}^{\alpha\beta} (\ln z_{1},\ln
w_{1})\frac{z'^{J}z^{k}_{n}\overline  w'^{J}\overline
w^{k}_{n}}{z_{1}^{\frac{|J|+n}{2}}\overline  w^{\frac{|J|+n}{2}}_{1}}
\end{equation}
where $K_{Jk}^{\alpha\beta}$ is the Bergman kernel on $S_{\alpha\beta}$ with the
weight $W_{Jk}.$
(One way to see this is to note that \eqref{ChangeCoord} allows us to convert an
orthonormal basis for the Bergman space on $S_{\alpha\beta}$ with  weight
$W_{Jk}$ to an orthonormal basis for $A_{Jk}^{2}$.)

Let $\mathcal{F}(f)$ denote the Fourier transform of $f$; thus
$\mathcal{F}(f)(\xi)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\xi t}dt$ and
$\mathcal{F}^{-1}(f)(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(\xi)e^{i\xi
t}d\xi$.

\begin{proposition}\label{JkKernel}
$K^{\alpha\beta}_{Jk}$ is given by the integral
\begin{equation}\label{IntFormula}
K^{\alpha\beta}_{Jk}(z,w)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} \frac{ e^{i(z-\overline
 w)\xi}}{\mathcal{F}(W_{Jk})(-2i\xi)}d\xi.
\end{equation}
\end{proposition}

\begin{proof}
See \cite{Barrett92} and \cite[Lemma 6.5.1]{ShawBook}.
\end{proof}

Note also that $-\pi<\text{Im}(z-\overline  w)<\pi+4\alpha\ln\beta $ for $
z,w\in S_{\alpha\beta}.$

\begin{proposition}\label{WeightTransform}
The Fourier transform of $W_{Jk}$ is given by
\begin{equation}\label{WeightTransformFormula}
\mathcal{F}(W_{Jk})(\xi)
=D_{nJ}e^{-\frac{i\xi\pi}{2}}\frac{E_{Jk}(\xi)}{
(\xi+|J|+n-2)(\xi+|J|+n-4)\ldots(\xi-|J|-n+2)}
\end{equation}
 where
\[E_{Jk}(\xi)=\left(e^{i\xi\pi}-(-1)^{|J|+n}\right)\left(\frac{e^{
2(k+1-i\alpha\xi)\ln\beta}-1}{k+1-i\alpha\xi}\right). \]
\end{proposition}

We postpone the proof of this Proposition.

To apply residue methods to \eqref{IntFormula} we need to find the zeros of
$\mathcal{F}(W_{Jk})(-2i\xi)$. Let us denote the set $\{s\in \mathbb{Z}:-m\leq s\leq m\}$
by $\mathbb{I}(m).$ From Proposition \ref{WeightTransform} we see that if
$|J|+n$ is even  then the zeros of $\mathcal{F}(W_{Jk})(-2i\xi)$ are located at
\[\left\{ mi:m\in
\mathbb{Z}\setminus\mathbb{I}\left(\frac{|J|+n-2}{2}\right)\right\} \bigcup
\left\{\frac{m\pi i}{2\alpha\ln\beta}+\frac{k+1}{2\alpha
}:m\in \mathbb{Z}\setminus\{0\}\right\} \]
and in case  $|J|+n$ is odd they are located at
\begin{multline*}
\left\{ mi+\frac{ i}{2}:m\in
\mathbb{Z}\setminus\left(\mathbb{I}\left(\frac{|J|+n-3}{2}\right)\cup\{
-(|J|+n-1)/2\}\right)\right\} \\
\bigcup \left\{\frac{m\pi i}{2\alpha\ln\beta}+\frac{k+1}{2\alpha
}:m\in \mathbb{Z}\setminus\{0\}\right\}.
\end{multline*}

 For simplicity we focus now on the case  $J=0,\, k=-2$; note that this
guarantees that the zeros enumerated above are simple  (see Remark \ref{kChoice}
below).

Let $\nu_{\alpha\beta}=\frac{\pi}{2\alpha\ln\beta}$ and
$\mu_{\alpha}=\frac{1}{2\alpha }>0$.

\begin{proposition} \label{Kernel}
The  kernels $K_{0,-2}$ satisfy
\begin{multline}\label{BergmanKernel}
K_{0,-2}(z,w)= \sum_{\ell=0}^{[\nu_{\alpha\beta}-n/2]}
C_\ell z_{1}^{\ell}\overline w_{1}^{-\ell-n}
z_n^{-2}\overline  w_n^{-2}
+Cz_{1}^{\nu_{\alpha\beta}-n/2-i\mu_{\alpha} }\overline
w_1^{-\nu_{\alpha\beta}-n/2+i\mu_{\alpha}} z_n^{-2}\overline  w_n^{-2}
+\rem(z,w)
\end{multline}
 where $\varepsilon>0$,  the constants $C$ and $C_\ell$ are nonzero  and  the
remainder term $\rem(z,w)$ satisfies
 \begin{equation*}
\left(\frac{\partial}{\partial z_{1}}\right)^{m}\rem(z,w) =
O\left(z_1^{\nu_{\alpha\beta}-n/2+\varepsilon-m}
\overline w_1^{-\nu_{\alpha\beta}-n/2-\varepsilon}\right)
\end{equation*}
uniformly on closed subannuli of $1<|z_{n}|<\beta$.
\end{proposition}

\begin{proof}
We apply the residue theorem to the integral in  \eqref{IntFormula} along the
strip $-\nu_{\alpha\beta}-\varepsilon\le \im\xi\le 0$ to obtain
\begin{equation*}
K^{\alpha\beta}_{0,-2}(z,w) =
\sum_{\ell=0}^{[\nu_{\alpha\beta}-n/2]} C_\ell e^{(\ell+\frac{n}{2})( z-
\overline w)} + C e^{(\nu_{\alpha\beta}-i\mu_{\alpha})( z- \overline w)} +
\widetilde\rem(z,w)
\end{equation*}
for non-zero $C, C_\ell$, where $\widetilde\rem(z,w)$ and all of its derivatives
are $O\left( e^{(\nu_{\alpha\beta}+\varepsilon)( z-\overline w)}\right)$
on closed substrips of $S_{\alpha\beta}$.

Plugging this into \eqref{EqnKernelTransfrom} we obtain \eqref{BergmanKernel}.
\end{proof}

\begin{remark}\label{kChoice}
We have focused on the case $J=0,\, k=-2$   because this is the simplest choice
which avoids possible problems with double poles.  Analogous formulae hold for
other values of $k$ in the  absence of double poles.  When double poles do occur
they contribute factors of $\ln(z_1-\overline w_1)$.
\end{remark}

\begin{lemma}\label{SumEval}
$\displaystyle{ \sum_{s=0}^{j}\binom{j}{s} \frac{(-1)^{s}}{\xi+\alpha(j-2s)}
= \frac{(-2\alpha)^j j!} {(\xi+\alpha j)(\xi+\alpha(j-2))\cdots(\xi-\alpha
j)}}.$
\end{lemma}
\begin{proof}
The statement is true for $j=0$.

Working inductively and recalling that
$\binom{j}{s}=\binom{j-1}{s-1}+\binom{j-1}{s}$ we have
\begin{align*}
\sum_{s=0}^{j}\binom{j}{s}\frac{(-1)^{s}}{\xi+\alpha(j-2s)}
&=
\sum_{s=0}^{j-1}\binom{j-1}{s}\frac{(-1)^{s}}{\xi+\alpha(j-2s)}
+
\sum_{s=1}^{j}\binom{j-1}{s-1}\frac{(-1)^{s}}{\xi+\alpha(j-2s)}
\\
&=
\frac{(-2\alpha)^{j-1} (j-1)!}
{(\xi+\alpha j)(\xi+\alpha(j-2))\cdots(\xi+\alpha(-j+2))}\\
&\qquad -\frac{(-2\alpha)^{j-1} (j-1)!}
{(\xi+\alpha(j-2))(\xi+\alpha(j-4))\cdots(\xi-\alpha j)}\\
&=
\frac{(-2\alpha)^{j-1} (j-1)!}
{(\xi+\alpha(j-2))\cdots(\xi+\alpha(-j+2))}
\left(
\frac{1}{\xi+\alpha j}-\frac{1}{\xi-\alpha j}
\right)\\
&=\frac{(-2\alpha)^j j!}
{(\xi+\alpha j)(\xi+\alpha(j-2))\cdots(\xi-\alpha j)}.
\end{align*}
\end{proof}
\begin{proof}[Proof of Proposition \ref{WeightTransform}]
Write
\[W_{Jk}(y)= C_{nJ} \Big( W_{Jk1} * W_{Jk2}\Big) (y/\alpha)\]
for $ -\pi /2<y<\pi/2+2\alpha \ln \beta$ where $f*g$ denotes the convolution of
$f$ and $g$ and
\begin{align*}
W_{Jk1}(t)&=\cos^{|J|+n-2}(\alpha t)\chi_{\pi/2}(\alpha t), \\
 W_{Jk2}(t)&=e^{t(k+1)}\chi_{\ln \beta}(t-\ln\beta).
\end{align*}
To calculate the Fourier transform of $W_{Jk}$ we first calculate
\[\cos^{j}(t)=\frac{1}{2^{j}}\sum_{s=0}^{j}\binom{j}{s}e^{i(2s-j)t}.\]
One can calculate that
\[\mathcal{F}(\cos^{j}(t)\chi_{\pi/2}(t))(\xi)=\frac{1}{i\sqrt{2\pi}2^{j-1}}
\sum_{s=0}^{j}\binom{j}{s}\frac{\left(e^{\frac{i(\xi+j-2s)\pi}{2}}-e^{-\frac{i
(\xi+j-2s)\pi}{ 2 }}\right)}{2(\xi+j-2s)}.\]
 Lemma \ref{SumEval} implies that
\begin{align*}
\mathcal{F}(\cos^{j}(\alpha t)\chi_{\pi /2}(\alpha t))(\xi)
&= \frac{1}{\alpha}\mathcal{F}(\cos^{j}( t)\chi_{\pi /2}(t))(\xi/\alpha)
\\
&=\frac{i^{j-1}\Big(e^{\frac{i\xi\pi}{2\alpha}}-(-1)^{j}e^{-\frac{i\xi\pi}{
2\alpha}}\Big)}{\sqrt{2\pi}2^{j}}\sum_{s=0}^{j}\binom{j}{s}\frac{(-1)^{s}}{
\xi+\alpha (j-2s)}\\
&=\frac{(-\alpha i)^j j!
\Big(e^{\frac{i\xi\pi}{2\alpha}}-(-1)^{j}e^{-\frac{i\xi\pi}{2\alpha}}\Big)}
{i\sqrt{2\pi}{(\xi+\alpha j)(\xi+\alpha(j-2))\cdots(\xi-\alpha j)}}.
\end{align*}

We also need to find the Fourier transform of $e^{kt}\chi_{a}(t-a)$:
 \[\mathcal{F}(e^{kt}\chi_{a}(t-a))(\xi)= \frac{1}{\sqrt{2\pi}}\frac{e^{2a(k-i\xi)}-1}{
k-i\xi}.\]
Using $\mathcal{F}(f*g)=\sqrt{2\pi}\mathcal{F}(f)\mathcal{F}(g)$ we find that the Fourier transform of
$W_{Jk}$ is given by \eqref{WeightTransformFormula}.
\end{proof}

\section{Proof of Theorem \ref{Theorem}}\label{ProveThm}

The proof of Theorem \ref{Theorem} follows immediately from
Lemmas \ref{SupposeContrary}  and \ref{Bad-f} below.

\begin{lemma}
If $P$ is continuous on $W^{p,s}(\Omega_{\alpha\beta})$ then
\begin{equation}\label{Estimate}
\left\| |r_{\lambda}|^{t} \left(\frac{\partial}{\partial z_1}\right)^m
P_{\lambda}f\right\|_{L^{p}(D_{\lambda})} \leq C  \left\|
f\right\|_{W^{p,s}(D_{\lambda})}
\end{equation}
where $m$ is a nonnegative integer, $0\leq t<1$ such that $m=s+t$ and  the
constant $C$ is independent of $\lambda$ and $f.$
\end{lemma}
\begin{proof}
Assume that $P$ maps $W^{p,s}(\Omega_{\alpha\beta})$ onto itself continuously
and let $T_{\lambda}f=f\circ \tau_{\lambda}.$ Then one can check that
\[\left\| \left(\frac{\partial}{\partial z}\right)^{P}
\left(\frac{\partial}{\partial \overline  z}\right)^{Q}T_{\lambda}f
\right\|_{L^{p}(\Omega_{\alpha\beta})}=2^{p_{1}+q_{1}-2/p}\lambda^{2p_{1}
+2q_{1}+|P'|+|Q'|-2n/p} \left\| \left(\frac{\partial}{\partial z}\right)^{P}
\left(\frac{\partial}{\partial \overline  z}\right)^{Q}f
\right\|_{L^{p}(D_{\lambda})} \]
where
$P=(p_{1},\ldots,p_{n}),Q=(q_{1},\ldots,q_{n}),P'=(p_{2},\ldots,p_{n-1}),Q'=(q_{
2},\ldots,q_{n-1}),|P'|=p_{1}+\cdots+p_{n-1},$ and $|Q'|=q_{1}+\cdots+q_{n-1}.$
Therefore we have
\[\| T_{\lambda}f \|_{W^{p,k}(\Omega_{\alpha\beta})}\leq
2^{k-2/p}\lambda^{2k-2n/p}\| f \|_{W^{p,k}(D_{\lambda})} .\]
By interpolation we also have
$\| T_{\lambda}f \|_{W^{p,s}(\Omega_{\alpha\beta})}\leq
2^{s-2/p}\lambda^{2s-2n/p}\| f \|_{W^{p,s}(D_{\lambda})}$ for all $s>0$.

Let $s=m-t$ where $m$ is a nonnegative integer and $0\leq t<1$. We have
\begin{equation}\label{HoloSob}
\left\| |r|^{t} \left(\frac{\partial}{\partial z_{1}}\right)^{m}
f\right\|_{L^{p}(\Omega_{\alpha\beta})} \le  C_1
\left\|f\right\|_{W^{p,s}(\Omega_{\alpha\beta})}
\end{equation}
for $f$ holomorphic on $\Omega_{\alpha\beta}$ (see, for example,
\cite{LIgocka87}).

Let $P_{\lambda}$ be the Bergman projection for $D_{\lambda}$. Then
$P_{\lambda}=T_{\lambda}^{-1}PT_{\lambda}$ and
\begin{eqnarray*}
\left\| |r_{\lambda}|^{t} \left(\frac{\partial}{\partial z_1}\right)^m
P_{\lambda}f\right\|_{L^{p}(D_{\lambda})}
 &=& \left\| |r_{\lambda}|^{t} \left(\frac{\partial}{\partial z_1}\right)^{m}
T_{\lambda}^{-1}PT_{\lambda}f\right\|_{L^{p}(D_{\lambda})}\\
&=&2^{2/p-m}\lambda^{2t+2n/p-2m}\left\| |r|^{t} \left(\frac{\partial}{\partial
z_{1}}\right)^{m} PT_{\lambda}f\right\|_{L^{p}(\Omega_{\alpha\beta})}\\
&\leq& C_2 \lambda^{2t+2n/p-2m}\left\|
PT_{\lambda}f\right\|_{W^{p,s}(\Omega_{\alpha\beta})}\\
&\leq& C_3  \lambda^{2n/p-2s}\left\|
T_{\lambda}f\right\|_{W^{p,s}(\Omega_{\alpha\beta})}\\
&\leq& C_4 \left\| f\right\|_{W^{p,s}(D_{\lambda})}
\end{eqnarray*}
where the constants are independent of $\lambda$.
\end{proof}
\begin{lemma} \label{SupposeContrary}
If the estimate \eqref{Estimate} holds on $D_{\lambda}$ then
$$\left\| |r_{\infty} |^{t} \left(\frac{\partial}{\partial z_1}\right)^m
P_{\infty}f\right\|_{L^{p}(D)} \leq C  \left\| f\right\|_{W^{p,s}(D)} $$
where $P_{\infty}$ is the Bergman projection on $D$ and the constant $C$ is
independent of $f.$
\end{lemma}
The above lemma can be proved like Lemma 1 in \cite{Barrett92}.
\begin{lemma} \label{Bad-f}
Let $s\geq \nu_{\alpha\beta}+n\left(\frac{1}{p}-\frac{1}{2}\right)$ where
$\nu_{\alpha\beta}=\frac{\pi}{ 2\alpha\ln\beta}$ and $s=m-t$ as above. Then
there exists $f\in C^{\infty}_{0}(D)$ such that $|r_{\infty} |^{t}
\left(\frac{\partial}{\partial z_1}\right)^m P_{\infty}f$ is not in $L^p(D)$.
\end{lemma}
\begin{proof} Since $P_{Jk}$ maps $W^{p,\delta}(D)\cap A^p(D)$ onto
$W^{p,\delta}(D)\cap A^p_{Jk}(D)$ for all $\delta\geq 0$ it is sufficient to
prove that there exists $f\in C^{\infty}_0(D)$ such that
$P_{Jk}P_{\infty}f\not\in W^{p,s}(D).$ Fix $w\in D, J=0,$ and $k=-2$. Let $f$ be
a nonnegative smooth function with compact support in $D$ such that it depends
on $|z-w|$ and $\int_{D}f=1.$ Then $K_{0,-2}(\cdot,w)=P_{0,-2}P_{\infty}f.$ We
can write $s=m-t$ where $m$  is a nonnegative integer and $0\leq t<1.$ In view
of \eqref{HoloSob} above (adapted to $D$) it suffices to show that
$|r_{\infty}(z)|^{t}\frac{\partial^m}{\partial z_1^m}K_{0,-2}(z,w)\not\in
L^p(D)$ for fixed $w.$  Proposition \ref{Kernel} implies that
\[\frac{\partial^m}{\partial z_1^m}K_{0,-2}(z,w)=
C z_{1}^{\nu_{\alpha\beta}-n/2-i\mu_{\alpha} -m}
+O\left(z_1^{\nu_{\alpha\beta}-n/2+\varepsilon-m}
\right).\]
Let
\begin{multline*}
D'= \Big\{(z_{1},z',z_{n})\in \mathbb{C}^{n}:\re\left( z_{1}e^{-2\alpha
i\ln|z_{n}|}\right) >|z'|^{2},1+\delta<|z_{n}|<\beta-\delta,\\
|z_1|<\delta, \Big|\theta_1-2\alpha \ln|z_n|\Big|<\frac{\pi}{4}\Big\}
\end{multline*}
for suitably small $\delta>0$.
Then $|r_{\infty}|$ is comparable to $|z_1|$ on $D'$ and
\begin{align*}
\int\limits_{D}|r_{\infty}(z)|^{pt}\left|\frac{\partial^m}{\partial
z_1^m}K_{0,-2}(z,w)\right|^{p}dV(z)
&\ge  \int\limits_{D'}|r_{\infty}(z)|^{pt}\left|\frac{\partial^m}{\partial
z_1^m}K_{0,-2}(z,w)\right|^{p}dV(z)\\
 &\ge c \int_{0}^{\delta} r_1^{p\nu_{\alpha\beta}+pt-pm+n-1-pn/2} dr_1
\end{align*}
where $c$ is a positive constant. The last integral above is divergent if $s\geq
\nu_{\alpha\beta}+n\left(\frac{1}{p}-\frac{1}{2}\right).$ Therefore
\[|r_{\infty}(z)|^t\frac{\partial^m}{\partial z_1^m}P_{0,-2}P_\infty f
=|r_{\infty}(z)|^t\frac{\partial^m}{\partial z_1^m}K_{0,-2}(z,w)\not\in
L^p(D)\]
for $s\geq \nu_{\alpha\beta}+n\left(\frac{1}{p}-\frac{1}{2}\right).$
\end{proof}

\section{Acknowledgment}
We would like to  thank the referee for pointing out a mistake  in an
earlier version of this manuscript.

redpajama_set_name: RedPajamaArXiv


Are you looking for a new stock pick? Check out this inexpensive, income-producing life sciences pick.
What is BioMed Realty Trust? A mid-cap real estate investment trust (REIT) that leases, develops, constructs, redevelops, acquires, finances and owns space utilized by the life science community (with over 18 million square feet of rentable space.
Shares of BioMed have dropped fairly sharply this year as selling in the REIT space overall, seemingly driven by rising U.S. interest rates, has taken its toll. Additionally, BMR shares were pressured earlier this year on news that the company was undergoing a management shakeup, as well as news that it would lose tenant Genzyme, currently occupying 343,000 square feet, to rival Alexandria Real Estate in 2018.
While we do not see interest rate risk abating anytime soon, and acknowledge that if the biotech space falls on tougher times, as far as ease of raising capital, BMR's operating environment will stiffen. With that said, it's definitely not all bad news for BioMed, and we think the 2015 sell off is overdone, which has created an attractive entry point or additional share purchase opportunity.
Results from the firm's first quarter indicate that BMR continues to make headway. While analysts were looking for BioMed to report funds from operations (FFO) of $0.37 per share for Q1, the company delivered $0.41. Additionally, the firm reported revenue of $178 million for the period, outpacing expectations of $169 million.
While the Genzyme news is definitely a loss, we believe that the remaining three years on the lease provides more than adequate time for BioMed to re-tenant the space (potentially with even better terms). Additionally, in June, BMR announced that biotech company Illumina had signed a 20-year lease for a new 155,000 square foot space in the Cambridge area of the U.K.
While BMR could face operating headwinds, including competition, potential consolidation within the life science space, potential increases in construction/redevelopment costs and weakening in capital formation by its clients, we believe that the stock is an attractive investment for those looking to add REIT exposure to their financial holdings as well as those looking for a unique way to access the healthcare space, especially the often more volatile and pricey biotech segment.
Our positive outlook is supported by the belief that BMR will benefit from rent growth in key markets, realize value creation from new and redeveloped space, and will benefit from continued ease of capital access for its prospective and current clients. We also like the firm's continued efforts to increase its all-important academic research space as well as the internal realignment that included shifts within management personnel to help sharpen its core business focus.
Finally, we believe the leasing side of the business will continue to benefit from favorable demographic trends as well as innovation and growth in precision medicine and digital health initiatives within the life sciences industry. We are quite comfortable with BioMed's financial footing and like the company's attractive 5.3% dividend yield.
Look for our next blog post to learn why we like BMR.
For more information on AFAM Capital and all of our investment strategies, click here.
We love a broad and diverse portfolio of stocks. No, we REALLY love diversification.Whether you are looking to build a portfolio or add to existing nest egg, we've got 10 stock picks for you. Download our guide of stocks to get started.
Nothing presented herein is, or is intended to constitute, investment advice, nor sales material, and no investment decision should be made based on any information provided herein. Information provided reflects AFAM Capital, Inc.'s (AFAM) views as of a particular time. Such views are subject to change at any point and AFAM shall not be obligated to provide notice of any change. Any securities information regarding holdings, allocations and other characteristics are presented to illustrate examples of the types of investments or allocations that that AFAM may have bought or pursued as of a particular date. It may not be representative of any current or future investments or allocations and nothing should be construed as a recommendation to purchase or sell a particular security or follow any investment strategy or allocation. Any forward looking statements or forecasts are based on assumptions and actual results are expected to vary from any such statements or forecasts. No reliance should be placed on any such statements or forecasts when making any investment decision. While AFAM has used reasonable efforts to obtain information from reliable sources, we make no representations or warranties as to the accuracy, reliability or completeness of third party information presented herein. No guarantee of investment performance is being provided and no inference to the contrary should be made. There is a risk of loss from an investment in securities. Past performance is not a guarantee of future performance.
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Happy Saturday, friends! It's been a busy few days with the Nordstrom Anniversary Sale madness and keeping the blog updated. I placed a large order on Thursday morning when the sale went live and then we back for more. Twice! I stocked up on closet staples to ensure I'm set for the Fall and Winter and even though it's still summer for the next couple months you can't beat the prices so I'm not about to let this sale pass me by! If you're interested in shopping my pick click on the Nordstrom Anniversary Sale tab on my blog menu to see everything I bought, tried on, and is on my "wishlist"!
This adorable top is not part of the Nordstrom Anniversary Sale and is not even from Nordstrom! I've been eyeing it for the last few weeks so as soon as it went on sale I snatched it up! I would say it runs a size large. I sized down to an x-small. Its just the perfect little throw on tank with shorts or jeans for a put together look. It's currently on sale for $25 so grab it while you can!
I hope everyone is enjoying their weekend! We are relaxing on the beach and after the last few days a couple of glasses of wine and an early night is just what this mama needs!


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Unlike the fraudulent December 2011 Duma election that brought tens of thousands of protesters to the streets chanting "Russia without Putin," the September 18 election had the appearance of propriety. The head of the Central Electoral Commission, a respected human rights activist, Ella Pamfilova, conceded that the election was conducted largely without incident with only three regions in which electoral violations were reported. She does not expect any major challenge to the preliminary results as a consequence of electoral fraud.
A smirking Vladimir Putin interpreted his party's victory as a rebuke to foreign pressure and sanctions. My interpretation is quite different: The electoral victory came through Putin's control of electoral resources and his repressive threats against opposition candidates and their voters. "Administrative measures" overcame declining public support for a government that has given the Russian people three years of recession, sharp drops in living standards, and neglect of its key constituency of retired persons and government employees.
First, Putin used electoral rules to disqualify opposition candidates and parties, among them Alexei Navalny and his Progress Party. (Navalny could not run because he, as Russia's premier corruption fighter, was himself convicted of corruption by a kangaroo court). Russia's Freedom Party was weakened by the unsolved murder of major Putin rival Boris Nemtsov. Putin also prevented coalitions of opposition parties from being recognized.
Second, under Putin's new election rules, half of the members of parliament are now chosen from individual-mandate districts. The intent was to lessen the effect of national economic woes and to rely on the region's control of administrative resources to elect United Russia's candidates. In the voting districts, opposition candidates were denied meeting space, billboards were torn down, and factory workers were told for whom to vote, or risk their jobs. In some cases, voters did not know that opposition candidates were running until they entered the voting booth.
Third, Putin's newly-instituted extremism laws prevented demonstrations, shut down internet campaigns, and otherwise threatened supporters of opposition candidates. Supporters of opposition candidates and candidates themselves had to ask themselves whether it was worth the risk of fines, jail, or lost jobs, all for the sake of a lost cause.
Fourth, disgruntled Russian voters made the purely rational decision to stay home. The difference between Levada's 30 percent approval and United Russia 54 percent vote tally suggests that voters disenchanted with United Russia shunned the polls. The Moscow and Petersburg turnouts were at an all-time low of 28 percent, versus earlier turnout rates of over fifty percent. Voters in Siberia and the Far East also had historically low voter turnout, suggesting opposition to Kremlin policies in the crucial resource-rich areas of Russia.
Fifth, the low turnout of disgruntled voters was by Kremlin design. The election day was moved to late summer from its usual December. Eligible Russian voters decided not to waste one of their last summer weekends at their dachas on a meaningless election. Only in the Chechnya of regional strongman Ramzin Kadyrov was the turnout a Soviet-style 97.99 percent.
Sixth, little attention has been paid to the first election in annexed Crimea. Crimean leaders were under extreme pressure to show high turnout and enthusiastic support for pro-Kremlin parties. High turnout was supposed to show the full integration of Crimea into its Russian motherland. A small scandal broke out when a "falsified" letter was disclosed from Moscow supposedly calling for a turnout of 70 percent. Crimean election results, however, do look peculiar, such as TASS's claim that the Kremlin-hostile Crimean Tatars had among the highest turnout. By the end of voting, Crimean authorities proclaimed a "good" turn out, not far from the Russian average.
A cocky Vladimir Putin has shown that a repressive regime with firm control of mass media and election rules can win elections without ballot-box stuffing, no matter how dire the economic condition. All that is needed is control of the candidate list, access to television, and serious sanctions for those who do not cooperate. The September 2016 Duma elections are a preview of the 2018 presidential elections. The Russian people have now learned that they cannot bring about change in their leadership through the ballot box. Russians have an expression for this: Putin na vsegda – Putin forever. Do they endure or find other means?


We got it from Here… Thank You 4 Your service is the sixth and final studio album by American hip-hop group A Tribe Called Quest. The album features contributions from the deceased member Phife Dawg, who died several months before the release.
The album, released on November 11, 2016, via Epic Records, features guest appearances of André 3000, Kendrick Lamar, Jack White, Elton John, Anderson Paak, Talib Kweli from Black Star, including the group's most frequent collaborators Consequence and Busta Rhymes.
The title is an homage to Phife Dawg, who came up with the title before his death.
What have the artists said about the album title?
I don't know … We're just going with it because he [Phife] liked it.
What have the artists said about the death of Phife Dawg?
It's so hard for me to sit in there and hear his voice. Sometimes I just have to like take a break and walk away. It gets heavy. It doesn't necessarily get sad, it just gets heavy. I literally feel the energy from him when I hear his voice.
I'm gonna be missing him for a while, God is in control, And I feel at peace. I feel hopeful. I feel Phife with me.
What have collaborators said about the album?
If you wrote your rhyme somewhere else, you still had to come back and lay your verse in Q-Tip's house. So we pretty much did every song together. Everybody wrote his stuff in front of everybody. Everybody spat their rhymes in front of each other. We were throwing ideas around together.
We recorded so many tracks and ideas. It's one of those scenarios where we're so excited to finally get to work together that it was exploding in a whole different direction. We really didn't know what we were doing, it was just a 'hurry up and press record' kind of moment.
How long back did they start recording this album?
Apparently the ball got rolling on recording the new album just after a late night performance in November 2015, the same night of the Paris attacks.
Is there any symbolism of having 2 "sides" or discs?
I wonder the same because the physical CD is only one disc. I'm sure the vinyl edition will be 2 discs with 4 sides when it comes out. But for now, it's up to the listener to decide.


text:
#ifndef _KERNEL_RENAMES_H
#define _KERNEL_RENAMES_H

/* Move built-in declarations and libm functions out of the way.
  (There should be a better way of doing so. These functions are
  built-in math functions for OpenCL (see Clang's "Builtins.def").
  Functions defined in libc or libm may also
  interfere with OpenCL's functions, since their prototypes will be
  wrong. */
#define abs            _cl_abs
#define abs_diff       _cl_abs_diff
#define acos           _cl_acos
#define acosh          _cl_acosh
#define acospi         _cl_acospi
#define add_sat        _cl_add_sat
#define all            _cl_all
#define any            _cl_any
#define asin           _cl_asin
#define asinh          _cl_asinh
#define asinpi         _cl_asinpi
#define atan           _cl_atan
#define atan2          _cl_atan2
#define atan2pi        _cl_atan2pi
#define atanh          _cl_atanh
#define atanpi         _cl_atanpi
#define bitselect      _cl_bitselect
#define cbrt           _cl_cbrt
#define ceil           _cl_ceil
#define clamp          _cl_clamp
#define clz            _cl_clz
#define copysign       _cl_copysign
#define cos            _cl_cos
#define cosh           _cl_cosh
#define cospi          _cl_cospi
#define cross          _cl_cross
#define degrees        _cl_degrees
#define distance       _cl_distance
#define dot            _cl_dot
#define erf            _cl_erf
#define erfc           _cl_erfc
#define exp            _cl_exp
#define exp10          _cl_exp10
#define exp2           _cl_exp2
#define expm1          _cl_expm1
#define fabs           _cl_fabs
#define fast_distance  _cl_fast_distance
#define fast_length    _cl_fast_length
#define fast_normalize _cl_fast_normalize
#define fdim           _cl_fdim
#define floor          _cl_floor
#define fma            _cl_fma
#define fmax           _cl_fmax
#define fmin           _cl_fmin
#define fmod           _cl_fmod
#define fract          _cl_fract
#define frexp          _cl_frexp
#define hadd           _cl_hadd
#define half_cos       _cl_half_cos
#define half_divide    _cl_half_divide
#define half_exp       _cl_half_exp
#define half_exp10     _cl_half_exp10
#define half_exp2      _cl_half_exp2
#define half_log       _cl_half_log
#define half_log10     _cl_half_log10
#define half_log2      _cl_half_log2
#define half_powr      _cl_half_powr
#define half_recip     _cl_half_recip
#define half_rsqrt     _cl_half_rsqrt
#define half_sin       _cl_half_sin
#define half_sqrt      _cl_half_sqrt
#define half_tan       _cl_half_tan
#define hypot          _cl_hypot
#define ilogb          _cl_ilogb
#define isequal        _cl_isequal
#define isfinite       _cl_isfinite
#define isgreater      _cl_isgreater
#define isgreaterequal _cl_isgreaterequal
#define isinf          _cl_isinf
#define isless         _cl_isless
#define islessequal    _cl_islessequal
#define islessgreater  _cl_islessgreater
#define isnan          _cl_isnan
#define isnormal       _cl_isnormal
#define isnotequal     _cl_isnotequal
#define isordered      _cl_isordered
#define isunordered    _cl_isunordered
#define ldexp          _cl_ldexp
#define length         _cl_length
#define lgamma         _cl_lgamma
#define lgamma_r       _cl_lgamma_r
#define log            _cl_log
#define log10          _cl_log10
#define log1p          _cl_log1p
#define log2           _cl_log2
#define logb           _cl_logb
#define mad            _cl_mad
#define mad24          _cl_mad24
#define mad_hi         _cl_mad_hi
#define mad_sat        _cl_mad_sat
#define max            _cl_max
#define maxmag         _cl_maxmag
#define min            _cl_min
#define minmag         _cl_minmag
#define mix            _cl_mix
#define modf           _cl_modf
#define mul24          _cl_mul24
#define mul_hi         _cl_mul_hi
#define nan            _cl_nan
#define native_cos     _cl_native_cos
#define native_divide  _cl_native_divide
#define native_exp     _cl_native_exp
#define native_exp10   _cl_native_exp10
#define native_exp2    _cl_native_exp2
#define native_log     _cl_native_log
#define native_log10   _cl_native_log10
#define native_log2    _cl_native_log2
#define native_powr    _cl_native_powr
#define native_recip   _cl_native_recip
#define native_rsqrt   _cl_native_rsqrt
#define native_sin     _cl_native_sin
#define native_sqrt    _cl_native_sqrt
#define native_tan     _cl_native_tan
#define nextafter      _cl_nextafter
#define normalize      _cl_normalize
#define popcount       _cl_popcount
#define pow            _cl_pow
#define pown           _cl_pown
#define powr           _cl_powr
#define radians        _cl_radians
#define remainder      _cl_remainder
#define remquo         _cl_remquo
#define rhadd          _cl_rhadd
#define rint           _cl_rint
#define rootn          _cl_rootn
#define rotate         _cl_rotate
#define round          _cl_round
#define rsqrt          _cl_rsqrt
#define select         _cl_select
#define sign           _cl_sign
#define signbit        _cl_signbit
#define sin            _cl_sin
#define sincos         _cl_sincos
#define sinh           _cl_sinh
#define sinpi          _cl_sinpi
#define smoothstep     _cl_smoothstep
#define sqrt           _cl_sqrt
#define step           _cl_step
#define sub_sat        _cl_sub_sat
#define tan            _cl_tan
#define tanh           _cl_tanh
#define tanpi          _cl_tanpi
#define tgamma         _cl_tgamma
#define trunc          _cl_trunc
#define upsample       _cl_upsample

// We provide our own printf
// Note: We declare our printf as taking a constant format string, but
// we implement it in C using a const format string (i.e. a format
// string living in a different address space). This works only if all
// address spaces are actually the same, e.g. on CPUs.
int __cl_printf(__constant const char* format, ...);
#define printf __cl_printf

#define atom_add     atomic_add
#define atom_sub     atomic_sub
#define atom_xchg    atomic_xchg
#define atom_inc     atomic_inc
#define atom_dec     atomic_dec
#define atom_cmpxchg atomic_cmpxchg
#define atom_min     atomic_min
#define atom_max     atomic_max
#define atom_and     atomic_and
#define atom_or      atomic_or
#define atom_xor     atomic_xor

#endif

redpajama_set_name: RedPajamaGithub


text:
A time for working together
Like most areas of business, the seafood sector in Scotland is in the midst of a very testing and trying time. Lost markets for a number of high-end species such as Scallop and prawns and significantly reduced demand for whitefish has led to a high number of vessels tying to the quayside. A limited number of vessels continue to fish for haddock for the domestic market with species such as monkfish, hake, ling and megrim shipped to foreign markets mainly France and Spain. These exports are traditional and form a significant element of the trade from fish processors in the north east of Scotland who send regular shipments on a daily basis.
That trade is now threatened after a section of the French industry applied pressure through political routes to the large retailers to ensure that only fish of French origin is purchased at auction. The genesis of their actions is unclear but understood to be linked to poor returns for a quantity of French fish. What is clear is the impact their actions will have for Scottish fishermen with exports of fish to France now reduced to very low levels. Their actions have a direct impact on Scottish markets with prices for a number of key species dropping sharply.
Scottish waters are fished quite heavily by around thirty large French vessels. These vessels land their catch into Scottish ports and truck the produce to France. Understandably, and in response to the protectionist measures applied in France, a number of Scottish fishermen are saying that this practice must now be brought to a halt.
In the hope of deescalating this whole affair we are calling on our French colleagues to step back from this agenda. Far more will be gained from working together than operating in isolation.


In today's fast-paced market, business needs change quickly and there may be times when you find yourself urgently requiring new, additional, or specific project office space.
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The Dawnguard will NOT be happy if you become a Vampire Lord. They will not help you, or give you side quests (though you can still finish the main story quest). They will not help you, or give you side quests (though you can still finish the main story quest). how to give yogurt for school lunch Dawnguard Side: The Dawnguard side of this quest is much like the Vampire side, except that after acquiring the bow, Serana will tell you to head back to Isran. Once at Fort Dawnguard , talk to Isran, who will gather the rest of the Dawnguard.
If you sided with the Dawnguard, while a Vampire Lord they will refuse any interactions with you, stating that you need to get cured immediately. This starts Rising at Dawn . She can also transform any elven arrows in your possession into Bloodcursed elven arrows after obtaining Auriel's Bow . how to get my ex back if he blocked me If you get infected again, you can repeat this quest. Once you are cured, Dawnguard will allow you to join them again. In fact, their members will act like nothing even happened, and you can continue the questline as normal.
10/08/2013�� Alternately, what if the Dawnguard discovers a nest of vampires that are behaving themselves -- restricting their feeding to spouses, followers, and other non-thrall volunteers, for example.
This is the Vampire equivilant of the Dawnguard mission 'Hunting the Monster'. You need to kill the Dawnguard without being seen. One other tactic is to transform into a Vampire Lord first, kill your target, flee and then revert back into human form without being seen.
If you joined the Dawnguard and then became a Vampire, the other Vampires will still attack you, the Dawnguard won't because you helped them out, but they will refuse to help you in any way, calling you a traitor or a monster and telling you to talk to Isran which will redirect you to Falion, allowing you to redeem yourself amongst the Dawnguard.


Please submit papers via CyberChairPro.
All papers must conform to the two-column ACM conference format.
they use any condensed formatting such reducing line-spacing, extending paper width and height, making font size smaller etc.


text:
COUPLE CHARGED WITH SEVERE ABUSE OF THEIR 8 CHILDREN. Officials say a Williamsport couple have been charged with mistreating their eight children with abuse that included severe beatings, starvation and molestation.
Home / General / ABOUT SEX OFFENDERS AND OTHERS / COUPLE CHARGED WITH SEVERE ABUSE OF THEIR 8 CHILDREN. Officials say a Williamsport couple have been charged with mistreating their eight children with abuse that included severe beatings, starvation and molestation.
Josephine Rodriguez December 21, 2020 ABOUT SEX OFFENDERS AND OTHERS, EMOTIONAL AND PHYSICAL EFFECT, FAMILY MEMBERS, FEMALE SEX OFFENDERS AND OTHERS, GOOD POLICING CATCHES SEX OFFENDERS AND OTHERS, SEX OFFENDERS AND OTHERS APPREHENDED, SEX OFFENDERS AND OTHERS COMMITTING ADDITIONAL CRIMES, SEX OFFENDERS AND OTHERS IN POSITIONS OF TRUST, SEX OFFENDERS AND OTHERS WHO REOFFEND, UNBELIEVABLE UNFORGETTABLE, WORST OF THE WORST
WILLIAMSPORT, PA. (AP) — A Williamsport couple who mistreated their eight children with abuse that included severe beatings, starvation and molestation for over a decade have been charged with endangerment and other counts, authorities said Friday.
Their arrests capped a seven-month investigation that began when the father was arrested for firing a gun twice out the window of their home during an argument with his wife, state Attorney General Josh Shapiro said.
The children, who now range in age from 14 to 26, were interviewed after their father's arrest and disclosed emotional and physical abuse inflicted by both parents, including punches and beatings with a leather belt. They also said they were forced to drink noxious liquids and were locked in a closet for as long as a week without food or water.
The father also inappropriately touched the daughters on multiple occasions, particularly when he was drunk, Shapiro said.
The children told investigators that while their father typically inflicted the punishments, their mother would watch and at times assisted.
The Associated Press typically does not name victims of sexual abuse and is not naming the parents to protect their children's identities.
"These helpless children experienced brutal abuse over a sustained period of time. The conduct of these parents was reprehensible and we will prosecute them to the fullest extent of the law," Shapiro said, adding that the abuse occurred from 1994 to 2020.
The three minor children — 14- and 17-year-old girls and a 15-year-old boy who has cerebral palsy — have been placed in temporary foster care, authorities said.
The mother and father are each charged with eight counts of child endangerment, one for each child, Shapiro said. The father is also charged with intimidating a witnesses, obstructing government function or administration of law, and indecent assault.
It wasn't known if the couple have retained attorneys.
https://www.usnews.com/news/best-states/pennsylvania/articles/2020-12-04/couple-charged-with-severe-abuse-of-their-8-children
UCDA: HUDSON VALLEY MAN COMMITTED "HEINOUS" SEX CRIMES WITH KIDS
ALTOONA MAN GETS FOUR YEARS FOR SEXUALLY ABUSING TWO GIRLS


text:
Affordable Catholic Education
Xavier Catholic Schools and the Archdiocese of Dubuque are committed to making sure Catholic education is available to families who desire it for their children. Our schools will not become a school available to only those who can afford the full tuition. To that end, tuition at Xavier Catholic Schools is subsidized by our partner parishes (for their respective XCS schools) for families who are active in their parish. Our schools will have supporting parish, non-supporting parish, and non-Catholic tuition rates. Contact your desired school to learn more about specific tuition rates.
The Our Faith, Our Children, Our Future School Tuition Organization (STO), sponsored by the Archdiocese of Dubuque, combined with local assistance programs at each of our schools offer expanded opportunities to help qualifying families afford the Catholic school of their choice. This STO Program is available to both Catholic and non-Catholic families.
Two Types of Tuition Assistance Available...
School Tuition Organization (STO)
The Our Faith, Our Children, Our Future School Tuition Organization (STO) offers financial assistance to qualifying families so they can afford the Catholic school of their choice. The STO Program is available to both Catholic and non-Catholic families. STO applications are submitted online via FACTS Management.
Each Xavier Catholic School also has its own tuition assistance program. It is open to members of our supporting parishes (not just those who qualify for STO). These programs are in addition to the STO funds, and families may qualify for one or both. To be eligible for local assistance, families must first complete an STO application through FACTS Management. We encourage you to contact your specific desired school for additional information regarding their local assistance program.
STO Information for 2020-2021 School Year
The Our Faith, Our Children, Our Future School Tuition Organization (STO) and Xavier are pleased to once again provide tuition assistance to qualifying families. This program, supported by over 1,000 donors, provides expanded opportunities to help qualifying families afford the Catholic School of their choice. Tuition assistance through the STO is available to Iowa residents STO funds are only available for K-12th Grade students. Early childhood or preschool fees/tuition are not eligible. We strongly encourage families to apply for tuition assistance through the STO for the 2020-2021 school year. Families whose taxable income is under 400% of the federal poverty rate may qualify. The priority deadline to apply is April 15, 2020.
Complete the STO Application Process Through FACTS
The STO program will be utilizing FACTS Management once again to assist in determining the level of assistance to be awarded. FACTS Management has an application that all families need to complete to be considered for assistance. That application can be found here: online.factsmgt.com/aid (application is in both English and Spanish). Families must also upload their completed 2019 federal tax returns. Your personal information will be held in strictest confidence at all times. More information about income guidelines and application requirements can be found at ourfaithsto.org/faqs-for-parents. Only one application needs to be completed per family (application fees may vary by school).
The online FACTS application for STO award filing for Round 1 closed on April 15. Round 1 awards for STO or Local Assistance are expected to be assessed and communicated directly to families by the end of May. If families did not complete an application by the Round 1 deadline, there is a July 15 deadline for applications for Round 2 awards. Those awards should be assessed and communicated by the beginning of August. The FACTS application can be accessed at https://online.factsmgt.com/aid. Remember that a completed FACTS application is required for STO and Local Tuition Assistance.
To assist you in completing the application, the FACTS call center is open Monday through Thursday from 7:30am-7:00pm and Fridays from 7:30am -5:00pm at 1-866-315-9262. The FACTS online application also has a chat feature to assist parents available Monday-Friday from 8:00am-5:00 pm. You may also contact Tom Drish in the XHS Business Office for additional assistance at tom.drish@xaviersaints.org or 319-294-6635 ext. 307.
Note Regarding Local Assistance: XCS schools also have local funding to provide tuition assistance over and above the STO program, thus providing supplementary tuition assistance. Local assistance programs are open to all locally qualifying families, not just those eligible for the STO. To qualify for local assistance families must first submit a FACTS application.
Award Notification Letters
Award notification letters will be sent on or before May 22, 2020. A summary timeline with specific information concerning the application deadline is found below.
We strongly encourage interested parents to complete an application. Contact the FACTS call center at 1-866-315-9262 with specific questions about the application. This is a wonderful opportunity for qualifying families.
Parents are urged to complete the FACTS application and submit required tax documentation several weeks before deadlines to allow communication of missing/incomplete materials and processing time
NOW – Complete STO applications available on-line at online.factsmgt.com/aid.
April 15 - Round 1 Application Deadline: Completed applications must be submitted with all required 2019 federal tax returns and other required materials scanned or faxed by this date. Applications with missing information cannot be processed.
May 22 - Award notification letters sent to all families who applied.
June 8 - Families receiving assistance send confirmation they are accepting/declining the assistance back to the school.
July 15 - Round 2 Application Deadline: Completed applications must be submitted with all required 2019 federal tax returns and other required materials scanned or faxed by this date. Applications with missing information cannot be processed.
August 7 - Award notification letters sent to all families who applied.
STO Tuition Assistance Guidelines
Persons in Household - Maximum Qualifying Taxable Income Levels*
Persons in Household - 400% of Poverty Rate
2 - $68,960
4 - $104,800
*Important Notes about Income Guidelines
STO qualifying levels listed are based upon 400% of 2020 Federal Poverty Levels. Final income qualifying levels will be determined by the "Our Faith" STO Board based on identified need and available funds.
Families are eligible for STO assistance if they are Iowa residents AND their taxable income is at or below the above listed income levels. Falling within these guidelines does not necessarily guarantee tuition assistance awards.
Family size includes parents/guardians plus total number of dependents claimed on tax return.
For families/households with more than eight persons, add $17,920 for each additional person.
Any available local tuition assistance beyond STO awards may not be based upon the above income levels - but DOES require completion of a FACTS application.


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'2x more shabu confiscated under Duterte admin'
The government was able to double its confiscation of shabu under the Duterte administration compared to the Aquino administration, according to the Philippine Drug Enforcement Agency.
Interviewed over ANC, PDEA spokesman Derrick Carreon said "in fact, the Philippine National Police also admitted it was [just] a mental exercise on their part."
He made his statement even as Senator Christopher "Bong" Lawrence Go reiterated his stand that the government must use proper force to solve the drug problem.
"You should kill when necessary. You should threaten, kill. If you don't kill the drug lord, nothing happens," Go said.
"It is hard to totally eradicate illegal drugs due to greed. They [drug lords] can buy anybody."
Go and Carreon issued their statements in response to Vice President Leni Robredo's recent statement that President Rodrigo Duterte's war on drugs was a "massive failure."
Robredo said she got the information and data from the PNP and other law enforcement agencies.
"They just wanted to have a picture of what industry we are up against, so we came up with a hypothetical figure," Carreon said.
"This is why we are disputing the grade that was given to us, that was 1 percent."
PDEA director general Aaron Aquino said Robredo's figure was based on an "assumption."
Carreon took a dig at the remarks of the vice president, who claimed the anti-drug campaign was a failure.
"As if they are downplaying the sacrifices of all those people who put their lives and limbs [at risk] every day," he said.
"We have lost 10 agents. The PNP has its own casualties. Scores are wounded. How can you call that [drug war] a massive failure?"
PDEA had seized 5.24 tons of shabu since 2016 against the Aquino administration's 3.29 tons in six years, Carreon said.
He said they were able to dismantle 14 clandestine laboratories that could have produced 96.65 tons of shabu a year.
"If you [Robredo] want to play an assumption game, we also have our own figure," he said. With Macon Ramos-Araneta
Tags: ANCBong GoDerrick CarreonLeni RobredoPhilippine Drug Enforcement AgencyPhilippine National PoliceRodrigo Duterte
Rio N. Araja


Do Dresses Made From Wine Get More Fashionable As They Age?
First, the bad news. You can't get drunk wearing this wine dress. So you'll have to find another secret way to get your morning buzz on. But the good news is that researchers from the University of Western Australia have found a way to grow fabric using a similar fermentation process to making booze.
The school's Micro'be' fermented fashion is made from pieces of organic fabric grown using Acetobacter bacteria. It's responsible for turning wine into vinegar, but can also be used to create small fibers of cotton-like cellulose as it does for this unique application. The dress's red hue comes from the fact that red wine was used in the initial fermentation process, but white wine, and even beers like Guinness have been successfully used to produce a wider range of color options.
So maybe one day you'll be able to easily pick up a six-pack of cocktail dresses on the way home from work. And we're assuming the characters on Mad Men would have wholeheartedly embraced this technology had it existed back in the 1960s.


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Home Bitcoin News
Record Unemployment Has Contributed To A 'Striking' Shift Toward Bitcoin, New Research Reveals
in Bitcoin News
Bitcoin interest has surged in the wake of the coronavirus pandemic and the unprecedented measures undertaken by governments to contain it while propping up the global economy.
The bitcoin price has bounced back to almost $10,000 per bitcoin after crashing to under $4,000 during the broad March coronavirus sell-off.
Now, new research has revealed the extent public opinion has shifted toward bitcoin since 2017—with almost half of millennials "likely" to buy bitcoin within the next five years.
MORE FROM FORBESCoronavirus Has Revealed 'Holes Within Traditional Financial Systems'-And Demand For Bitcoin Is Suddenly 'Surging'By Billy Bambrough
Faith in bitcoin has surged over recent years, with almost half of people surveyed now found to … [+] prefer bitcoin over stocks, real estate, and gold.
"With over 20 million Americans currently unemployed, the public narrative towards bitcoin has changed," researchers at financial information platform The Tokenist wrote, summarizing their findings.
"The results are striking. We found increased knowledge of, and growing confidence in, bitcoin among all age and gender groups surveyed. This effect was most pronounced in millennial respondents, 45% of whom would now preferentially invest in bitcoin over stocks, real estate and gold."
As the coronavirus Covid-19 swept across the U.S., the unemployment rate soared to almost 15% in April, the highest level on record, with millenials among the hardest hit. In February, the unemployment rate was at a half-century low of 3.5%.
The survey, which gauged how market fluctuations caused by Covid-19 has affected people's view of bitcoin, revealed 47% of people trust bitcoin over big banks, an increase of 29 percentage points over the past three years.
Meanwhile, 43% of respondents, and 59% of millennials, thought that most people will be using bitcoin within the next decade and 44% of millennials report that they are likely to buy bitcoin in the next five years.
"These data indicate that bitcoin has a bright future, and will likely benefit significantly from the current market crisis," the researchers wrote.
"With confidence in traditional investment instruments decreasing, bitcoin stands poised to offer investors an alternative, long-term store of value."
MORE FROM FORBESThe Bitcoin Market Is Totally Dominated By A Tiny Number Of Pro TradersBy Billy Bambrough
Trust in bitcoin has surged over the last three years while faith in big banks, such as JPMorgan, … [+] Goldman Sachs, and Wells Fargo, has plummetted.
The Tokenist
The research chimes with reports from the bitcoin and cryptocurrency community that coronavirus and its ramifications has boosted interest in bitcoin.
"The surging activity we've seen since the beginning of 2020 has been in part inspired by the Federal Reserve's unprecedented monetary intervention," Alex Leishman, the founder of River Financial, a San Francisco-based startup bitcoin brokerage and financial services firm that has seen its client base double every month this year, said last week.
Growing interest in bitcoin from "seasoned macro investors" like Paul Tudor Jones, Dan Tapiero and Raoul Pal is legitimizing bitcoin "within the mainstream finance community," according to Leishman.
"The evolution of finance is only happening faster in the wake of the current global economic crisis, which has illuminated holes within traditional financial systems that can potentially be filled by bitcoin," said Olaf Carlson-Wee, founder of hedge fund Polychain Capital, who recently led a River funding round.
The Tokenist surveyed around 5,000 people through April 2020, comparing the findings to surveys carried out by brokerage eToro, investor Blockchain Capital, and personal finance website BankRate in 2017.


These resource are to help you better know St. Thomas as well as grow spiritually.
God, in His grace gives each member of the church community gifts to serve in ministry. This inventory will assist you in determining your gifting so that you might be the best servant you can be with your brothers and sisters at St. Thomas. Click here to take the spiritual test. Choose the test that best fits you: adult or youth. Then go to the bottom and enter as "guest". This should take 5-10 minutes and results are immediate.
Have you ever heard The Gospel? Are you wondering what it really is? What does it even mean? Click on the "What is the Gospel?" picture for answers to your questions. After watching the video, please contact us if you have any further questions.


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Q: Doing a input type="text" with CSS3 and fixed image on the background I am developing my first HTML5 / CSS3 website and I want to do all I can correctly and with well methods and being usable too.
I have to reproduce the next search input in this website:

What I don't know how to do it is to implement the background of the image, or if the image needs to be a <input type="image" />.
At the moment, what I've is the next:
#header-search input {
    -webkit-border-radius: 5px;
    -moz-border-radius: 5px;
    border-radius: 5px;
    border:1px solid #8e8e8e;
    background-color:#f5f5f5;
    height:16px;
    padding:4px;
    color:#4a4a4a
}

This creates an input with the same radius and all the area is input text. There is no space for the magnifying glass and if it was a background the image can not be clicked. In this point, what is best? Let the input only triggered by  or... how will you do it?
Anyway you will do it, please tell me.
Thank you in advance!

A: You can use a submit button with a background-image and position it over the searchbox.
Do not forget to add padding to the right of your input box to allow for the button without overlap. (i have also added a class to your search box so the rules do not target the same elements)
Html
<form id="header-search">
    <input type="text" class="searchbox" /><input type="submit" class="button" value="" />
</form>

Css
#header-search{overflow:auto;}

#header-search input.searchbox {
     -webkit-border-radius: 5px;
     -moz-border-radius: 5px;
     border-radius: 5px;
     border:1px solid #8e8e8e;
     background-color:#f5f5f5;
     height:16px;
     padding:4px;
     padding-right:28px;
     color:#4a4a4a;
     float:left;
 }

#header-search input.button{
    border:0;
    padding:0;
    margin:0 0 0 -24px;
    width:24px;
    height:24px;
    background:transparent url('your-search-icon-path-here') center center  no-repeat;
    float:left;
}

the end result is

demo http://jsfiddle.net/gaby/F5Nmp/1/

A: You should try :
background-image : url(path/to/your/image/relative/to/css/file) no-repeat;
background-position : right;

You don't need
<input type="image"/> <!-- this can be used to replace a submit control
                           for example, but not what you need I think -->

You should also consider using a more specific CSS rule (to prevent styling other inputs)
#header-search input[type="text"]

A: The best I could come up with is the following, which is a tad more complex than I'd like:
#header-search, button {
    display: inline-block;
    vertical-align: baseline;
    border:1px solid #8e8e8e;
    background-color:#f5f5f5;
    height:30px;
    padding:4px;
    color:#4a4a4a;
    margin: 0;
    float: left;
}
#header-search {
    -webkit-border-radius: 5px 0 0 5px;
    -moz-border-radius: 5px 0 0 5px;
    border-radius: 5px 0 0 5px;
    border-right-width: 0;
}
button {
    -webkit-border-radius: 0 5px 5px 0;
    -moz-border-radius: 0 5px 5px 0;
    border-radius: 0 5px 5px 0;
    border-left-width: 0;
}

And involves amending the mark-up (and adding a button type="submit" to contain the magnifying glass image, sigh...):
<input type="search" id="header-search" />
<button type="submit">
    <img src="http://davidrhysthomas.co.uk/linked/search.png" />
</button>

JS Fiddle demo.
Note: please be aware that I, for some reason currently unknown to me, used the id of #header-search on the input itself, rather than the parent element. So your selector will need to be amended from the one I used in the above example.

Edited because it looked wrong, with the outline, on focus, following the rectangle of the search input (check the original JS Fiddle demo). So, I replaced the outline with a border colour-change, to make it look a little more 'natural':

#header-search:focus {
    border-color: #f90;
    border-right-width: 0;
}

#header-search:focus + button[type=submit] {
    border-color: #f90;
    border-left-width: 0;
}

Revised JS Fiddle demo.


If you are a Desi girl absolutely in love with your ethnic wear then you must be having at least five different Kurti pieces in your wardrobe. Indian designs and prints are not only beautiful but also mysterious. The colour and silhouettes are unlike any other western design. They are not only unique to Indian body shapes and sizes but also apt with our traditions which again separates them from any other Asian design. And ethnic kurtis epitomize that desi look! In time they have also blended well with our western wears to create a perfect indo-western fusion while maintaining their perfect traditional vibes. Pairing up a Kurta with just a skirt or jeans or leggings is so cliche and so is the time to experiment a little a create that perfect millennial look for girls of today. Gear up for some trendy ethnic looks.
The cape kurtis or a long shrug kurti can perfectly be worn along with wrap around pants. A silky silhouette is what we need exactly here along with statement jewelry and trinkets to add that bohemian appeal.These carefree pants perfectly compliment to laid back vibes during a brunch or a pool party.
Who said the very traditional ethnic coverall kurti can not be worn with those super hot shorts or hot pants? When it comes to our millennial generation,nothing is impossible!You can team up your slit kurtis with your most worn out and ripped hot pants and be the head turner instantly.
Oh! Dhoti pants are such a rage! The peplum kurtis or flared short tunics are perfect to be paired up with these stylish Dhoti-pants. Just choose most pop coloured Dhoti Pants and don't forget to add four inches of high heels with this top of the trend look. This Indian wear is surely one of the most trendy ethnic look of 2018 when we talk Indian designs completely.
If you are just not a regular girl then this one of the most outrageous desi look is just for you. The denim style kurtis are perfect pick for your drawer of stockings. Just make sure to stick to minimal makeup and minimal accessories. Just stick to bare minimal look to rock a trendy ethnic look.
Remember all those house wives working on their fitness regime in a salwar suit and a pair of sports shoes. Well it seems like, it has caught attention of our fashion pioneers and Instagram stars and now is the perfect time to sport that grunge boot with your ethnic kurta. Many Bollywood celebrity has been spotted flaunting this look effortlessely. Though checkered boots are complete no no with heavily embroidered kurtas and also you should always go for pop coloured kurtis with those grunge leather boots of yours. Pendants or cool lockets just seal this all 'Oh! So cool look' completely.
It's too boring to stick to lame ethnic style,Challenge your limits and rise a DIVA!