Adobe, Acrobat, and Flash are trademarks of Adobe Systems Inc. Macintosh, Mac, and TrueType are trademarks of Apple Computer, Inc. Windows and OpenType are trademarks of Microsoft Corporation. Sabon is a trademark of Linotype Corp. Bembo is a trademark of the Monotype Corporation. Burin Sans and Sackers Gothic are trademarks of Monotype Imaging. Bank Gothic is a trademark of Grosse Pointe Group, LLC. Other trademarks mentioned are the property of their respective owners.


text:
Q: jQuery infinite loop animation problem I'm trying to do a simple loop on a jQuery animation but it won't work, where I'm wrong?
var time = 750;
var go_up = true;

$('#vittorio a').hover(function () {
    checkMovement ();
}, function () {
    $('#heart').stop().animate({
        top:'0px'
    }, 1500);
});

function moveHeart (pos) {
    $('#heart').stop().animate({
        top:pos
    }, time, function () {
        checkMovement();
    });
}

function checkMovement () {
    if (!go_up) {
        go_up = true;
        pos = '3px';
    } else {
        go_up = false;
        pos = '-3px';
    }

     moveHeart (pos);
}

The loop works only once.

A: seems you forgot to use semicolon on checkMovement() call

redpajama_set_name: RedPajamaStackExchange


text:
The 2018 Antiques & Garden Show Preview Party on Feb. 1 kicked off a weekend of events, showcasing excellence in horticultural and interior design. The Thursday evening cocktail party offered patrons a sneak peek of featured garden designs and more than 150 premier antique and horticulture dealers before the doors officially opened the following morning.
The entrance to the Music City Center exhibit hall opened to the first of many spectacular garden creations. The exhibition display from the Cheekwood team led by Peter Grimaldi, Vice President of Gardens and Horticulture, included a garden of white tulips, hyacinth and lush evergreens, encircling a grand arch flanked by cascading waterfall staircases. Prior to the preview party, judges awarded Kathi and Brian Gilleland of Poise & Ivy the show's top prize, the Bryant Fleming Award for Best in Show, for their design of the center garden.
Amid the craft cocktails, hors d'oeuvres and shopping, the preview party also served as a celebration for author, architect and Antiques & Garden Show honorary chairman Gil Schafer. Among his many accolades, Schafer has received VERANDA's "Art of Design Award" and was recently named to the 2017 Architectural Digest AD 100. VERANDA Editor in Chief Clinton Smith was also in attendance, signing copies of his book, "Inspired by Color."
The Antiques & Garden Show's 2018 general chairmen, Kim Holbrook and Robin Puryear, orchestrated a weekend of special events, book signings and panel discussions, highlighted by keynote speaker Charles, 9th Earl Spencer, brother to the late Diana, Princess of Wales, and Viscount of Althorp, his family's 500-year-old estate. Earl Spencer's presentation of "Althorp: The Story of an English House" was one of the most eagerly anticipated events among preview party patrons.
More:Princess Diana's brother on quest to restore Althorp Estate, her final resting place
The 28th annual Antiques and Garden Show of Nashville benefited Cheekwood and ECON Charities.

redpajama_set_name: RedPajamaCommonCrawl


text:
Justin Furstenfeld Tells His Story of Sobriety, Love and Life Through Song Blue October's Frontman Opens Up His Heart To An Adoring Crowd At Hotel Cafe
Derrick Lee, 12 months ago 0 5 min read
LOS ANGELES, CA- On July 12th, 2018, Justin Furstenfeld, the frontman for the chart topping alternative rock band Blue October, started a 5 week "Open Book" acoustic residency at the Hotel Cafe. With a new Blue October album, I Hope You're Happy, slated for release on August 17th, you wouldn't expect Justin to be "working" during his time away from the band prior to preparing for a 30 date domestic tour in support of the new album, but he's got a story to tell … and he tells it well.
I was able to spend a few minutes with Justin to take some portraits prior to his taking the stage in the intimate, 165 capacity room. He greeted me warmly like I was a friend he'd known forever, and went out of his way to accommodate my requests. He was so personable, and kind – not to mention hilarious- during my brief time with him, I just had a sense that the evening's patrons would be in for a real special treat.
Justin Furstenfeld [Portrait] @ Hotel Cafe 7/19/18. Photo by Derrick K. Lee, Esq. (@Methodman13) for www.BlurredCulture.com.
He gave some of his die-hard fans an opportunity to hang with him during soundcheck, giving them a chance to ask him some personal questions. What I took away from his intimate interactions with the soundcheck audience was that he truly cares about his fans. When he was asked about what irks him most when he's on stage in this type of setting, he bluntly stated that at these shows, his fans open up to him with their own troubles and that they should be accorded as much respect and attention that he gets. He can't stand when people in the audience aren't connecting with all of the stories, his and his fans, that are being told.
Thankfully, the audience on this night didn't need a scolding from Justin.
After the general admission ticket holders had entered the performance room, Justin took the stage and regaled us with a chronological musical history of his life story; a story that's still being written but has had so many ups and downs. Justin doesn't shy away from any topic. He's open and vulnerable, really allowing the audience to get into his head and learn about the man behind the music.
When a fan in the audience asked him if he relives the music when he performs it, he admitted:
"It's good to cry […] When I went away to get help, there was something they taught me. They studied my music, and they said that I regurgitated my pain. When I walk off the stage, I need to turn it off, I need to embrace the old me before the drugs [start calling for me]. Now, when I sing the music, I sing it with gratitude."
He sang with gratitude all night long.
He started from the beginning, telling the story about the first song he wrote and the stories kept rolling from there. He weaved into his narrative the backgrounds of how songs came to be, then performing those songs for the audience so they could hear exactly what he was talking about.
There were stories that induced laughter (the story about his "A" tattoo), inspiration (the story about his first band getting "called in" after one of their first performances), love (his wife … man, does he love his wife… and PS, he let's us know that getting backstage passes to a Lil Wayne concert is not a good date) and his sobriety. In a particularly telling moment, he admits the reasons why there are certain songs he just can't sing anymore.
I've got plenty of notes on the stories he told on this night, but there's no adequate way for me to write about them here. Justin's good humor and honesty is what gives all his stories, and his songs, life. You need to hear the regret in his voice when he talks about things that he's ashamed of. You need to hear the way his tone perks up every time he talks about his wife and kids.You need to hear him laugh at his own jokes. You need to hear the awe in his voice when speaks about that moment during his rehab when he had his personal revelation. You just need to see and hear him tell and sing his story.
If I didn't know any better, his performance at Hotel Cafe was more than just a concert. It was a conversation. It was therapy. It was a performance piece. It was insight into a man's soul through his words.
Fans of Blue October will be happy to hear that he performed many of the band's hits including "Hate Me", "Home", "Into The Ocean" and "Calling You", but what made those songs even more powerful were the stories behind them and the way they were told, and to know exactly what I'm talking about, you're just going to have to buy a ticket find out.
CLICK HERE to purchase tickets for the remained two performances of Justin Furstenfeld's residency at Hotel Cafe, and also follow Justin's band Blue October on Instagram, Facebook and Twitter.
Justin Furstenfeld @ Hotel Cafe 7/19/18. Photo by Derrick K. Lee, Esq. (@Methodman13) for www.BlurredCulture.com.
#Acoustic	#Alternative Pop	#Alternative Rock	#Art Rock	#Blue October	#Concert Photography	#Concert Review	#DKL PHOTOS	#Hotel Cafe	#Justin Furstenfeld	#New Wave	#Post-Grunge
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UK hip-hop heavyweight Dabbla drops video for huge new track 'FUTD' Check out his new alien friend...


text:
Wisdom// February 1, 2017
You Were Sick
How Kurt Vonnegut's final novel gives the perfect advice for resistance in Trump's America.
Topher South, Writer, teacher, dad, art media fanboy, fitness and nutrition obsessive, political junkie, generally kind of a doof.
It's been a few years since graduation, but I'll always be an English major at heart. And like many English majors before and after me, Kurt Vonnegut is a big deal to me.
At times I've wanted his quotes tattooed on me, but I'd end up looking like Guy Pearce in Memento. His stuff has just meant too much to me. From sixteen years old, when I read Hocus Pocus and first heard about humanism, then became a pacifist when I devoured Slaughterhouse-Five.
When my wife was pregnant with our son, I prepared myself for fatherhood by re-reading the scene in God Bless You, Mr. Rosewater when the eponymous Eliot Rosewater baptises two little twin babies with one of the most powerful speeches ever committed to ink and paper:
"Hello, babies. Welcome to Earth. It's hot in the summer and cold in the winter. It's round and wet and crowded. At the outside, babies, you've got about a hundred years here. There's only one rule that I know of, babies — 'God damn it, you've got to be kind.'"
I still get choked up when I think about that.
And now, here we are. A historical precipice that would fit right in with the more dystopian and apocalyptic Vonnegut, right between Harrison Bergeron and Galápagos. A billionaire madman has used cutting-edge social technology to win an election by losing the popular vote, and is marching us toward physical and philosophical destruction. It has all the hallmarks of a Vonnegut story. It's desperately grim, with some truly loathsome characters, is compulsively readable (I'm looking at you, my eternally-refreshing Twitter feed), and is shockingly, brutally funny.
And, of course, Ol' Kurt would have loved the Russian golden shower bit. It's the perfect mix of impossibly funny, and impossible to laugh at.
As in many times in my life, I find myself looking to cultural bits and pieces I've crammed into my brain over the years for wisdom, hope, solace, anything. Enter, again, Kurt Vonnegut, with what we need most right now: A call to action.
Plot purists be warned, there be spoilers in these parts.
Released in 1997, Timequake was the last novel he published. It's often (completely unjustly) overlooked in his overall canon, and even the author himself referred to it as the book "that didn't want to be written". In true Vonnegut fashion, it's kind of a mish-mash of autobiography and science fiction. The basic idea is that time rewinds ten years, from 2001 to 1991, and then starts again. Everyone on earth has to relive those ten years, and do it exactly the same. They are on, as he says, automatic pilot, just riding through those years as passengers, watching themselves and everyone around them commit the same mistakes and fall victim to the same traumas. And none of them knows when or if it will ever end.
This is obviously pretty horrifying. And Vonnegut milks it for every last drop of existential dread. An architect who killed himself during that time over having his job replaced by a computer program is brought back to life, and dragged back through the series of events that led to his suicide. A paraplegic must regain the ability to walk, only to jump headfirst into a shallow pool again, knowing what will happen, while everyone around looks on, internally in horror, externally having fun in the sun. A black man must be wrongfully accused for a rape again, the judge knowing he'll be eventually exonerated, and serve an unjust prison sentence, again.
It's pretty heavy stuff.
But the real problem comes when time catches up to itself, the timequake ends (or begins, depending on how you look at it), and the world suddenly has free will again. People haven't been in control for ten years, and now they are, and they don't know what to do. Planes fall from the sky, car accidents litter every city, chaos abounds.
Except for one guy, Vonnegut's career-spanning alter ego character, Kilgore Trout, who led such a determinedly uneventful life that he barely notices when the ten years are up. Apathetic, unhappy, unfazed, he has to step up.
See where I'm going with this?
He goes to that wrongfully convicted ex-convict, now a security guard at a museum, who is terrified to move, lest he be plunged back into his prison sentence, and says the words that were written twenty years too early. "You were sick, but now you're well again, and there's work to do." They then set about waking people from their stupors with those words, essentially saving the world from its disaster and malaise.
It's not his best novel. It's not my favorite of his novels. But that mantra stuck in my craw the moment I read it, eighteen or nineteen years old, and has always floated around in my head, waiting for moments to pop up and be of use. It exploded into mind recently, as the unthinkable happened and we got our very own, all-American, reality show authoritarian.
Many Americans have felt trapped and abused by the political cycle, ignored and forgotten by our leaders. We've been mad as hell for a while now, and that was quite clear in the 2016 presidential election. On the left with Sanders supporters' demands for social justice and economic equity, and on the right, where the Tea Party movement reached its inevitable culmination in Trump, populism made that anger very real, and very loud. And in the end, the angriest, loudest side won out, and on January 20th, we inaugurated…that.
Enough thinkpieces have been published about the causes of this catastrophe, so I won't go further into that. But, it matches up to Timequake strikingly. A long period of frustrating cycling was followed by a descent into chaos when the cycle changed. And now we're in what could be the worst moment in American history.
Except.
A culture of activism has taken over. The Women's March on Washington on January 21st did something, jolting people from their cringing fear and giving them a way to stand up. Since then, there's been a nearly nonstop protest culture in DC, and days of airport protests against the Muslim banning executive order signed on the 27th. Even as I write this, a scientists' march in defense against the wave of anti-science decisions being passed down from the new White House is being organized. These mighty movements are birthing a network, a conscious, organized force to be reckoned with that isn't going away anytime soon.
Countless reports, and many people I know, have commented that they've never done anything overtly political. In their entire lives, they've never been moved to participate in the system, sometimes including voting at all. But on the morning after the inauguration, several million brave, powerful men and mostly women rose up and decided to strike down apathy and inaction.
They said that progress can't simply be reversed in one election.
They said that attacking the weakest among us awakens the strongest among us.
They said that this can be overcome, but overcoming takes work.
And so again, I come to my literary spirit animal, Kurt Vonnegut. When I feel catatonic, broken by the onslaught of this-can't-really-be-happening-here horror, shocked into paralysis, I say, "You were sick, but now you're well again, and there's work to do."
Because there is so much work to do.
— Published on February 1, 2017
Community// June 1, 2019
Do it. You'll never regret it.
by Mita Bhan
Rising Star Fiona Katauskas: "I want to start with a simple premise: Common Ground"
Community// March 6, 2017
Life In The Middle: Learning To Live In The In-Between
by Simon Menelaws


how to screen sand gravel Welcome to Shanghai CNC Mining and Construction Machinery Co., Ltd.
The screens or plates measure up to 10 ft ( m) ... The production of sand and gravel in many areas has come under increasingly stringent restrictions.
Quality Sand Gravel Screen for sale, Sand Gravel Screen manufacturers of Zhenyuan Coal Crusher and Sand Gravel Screen. The Sand Gravel Screens are usually .
Industry News. Pembroke Sand & Gravel - of New Hampshire. Pembroke Sand and Gravel of NH offering a wide variety of aggregrate products and materials.
Feb 15, 2011· Author Topic: Question about screening gravel (Read 6815 times) 0 ... As for size I want to screen out the sand from the bigger gravel.
Sep 12, 2011· Want To Build Small Gravel Screen Have Questions planning on building a small gravel screen plant been thinking about ... What are you going to screen? topsoil? sand?
How To Screen Gravel And Sand - bdcain. how to make a screen for sand and gravel? Yahoo Answers Mar 06, 2006 How to make a screen for sand and gravel?
The filter pack of uniform, relatively coarse, sand or gravel surrounding the well screen may be provided in two ways (E. E. Johnson, 1955).


The JD20XT from Penton Audio is a full-range weather-resistant mini speaker measuring just 6" x 9" x 6". Available in black or white, models include a constant voltage transformer, full-range 20-watt, 5" coaxial loudspeaker and a mounting bracket. As with most Penton products, it is strikingly handsome. Visually identical, the JD20X16 is 16 ohms.
This model's shaped frequency response reproduces music with high accuracy, featuring polypropylene cone with rubber surround, high-temperature voice coil and a tweeter with Mylar diaphragm and neodymium magnet.
Concealing a 70.7-volt transformer with screwdriver-adjustable taps, its ABS housing has UV inhibitors for long life without noticeable deterioration. For added strength, its grille screen is CRS and has an epoxy coating.
Secured with a stainless steel set screw, its patented aluminum mounting bracket allows attachment to single gang or double-gang box (patent applied for). The entire model is finished in your choice of white (-W) or black (-B). Connection is via chrome-plated push terminals.
Mini loudspeakers are very practical for providing full-range audio with an extremely low profile. If desired the JD20XT and JD20X16 can be well hidden on walls and painted with epoxy paints. Mainly utilized for background and foreground music, it still allows vocals and important announcements to cut through.
Typical applications include boutiques, restaurants, taverns, atriums, shopping centers, boardrooms and corporate offices.


International Partner Buro S.R.L. can make 3D laser scanning s, models and presentations of accurate data from the scanned locations.
All the data collected in one location can be used for delivery of products and services with the purpose of recreating and viewing of interest areas.
These products differ from 2D plans to models generated using graphic stations and complex animations.
Views and reconstructions often help in solving cases and are a great help in various problems and in minimizing costs.


Paul Buchheit is an advocate for social and economic justice. His essays, videos and poems can be found at You Deserve Facts. He has contributed for several years to Common Dreams, Truthout and AlterNet.
Paul Buchheit examines the factors that have led to lower wages and less job security for most workers in the US.


text:
Young Dorothy finds herself in a magical world where she makes friends with a lion, a scarecrow and a tin man as they make their way along the yellow brick…
Genre: Adventure, Family, Fantasy
Naive and idealistic Jefferson Smith, leader of the Boy Rangers, is appointed to the United States Senate by the puppet governor of his state. He soon discovers, upon going to…
The spoiled daughter of a well-to-do plantation owner is forced to use every means at her disposal to claw her way out of poverty, following Maj. Gen. William Sherman's "March…
The Story of Vernon and Irene Castle (1939)
In 1911, minor stage comic, Vernon Castle meets the stage-struck Irene Foote. A few misadventures later, they marry and then abandon comedy to attempt a dancing career together. While they're…
Genre: Biography, Drama, Music, Romance, War
The Outsider (1939)
A controversial osteopath sets out to cure the daughter of a famous surgeon – and falls in love with her in the process.
Stagecoach (1939)
A group of people traveling on a stagecoach find their journey complicated by the threat of Geronimo, and learn something about each other in the process.
Genre: Adventure, Drama, Western
Made for Each Other (1939)
A couple struggle to find happiness after a whirlwind courtship.
Jesse James (1939)
After railroad agents forcibly evict the James family from their family farm, Jesse and Frank turn to banditry for revenge.
Genre: Biography, Crime, Drama
The Dark Eyes of London (1939)
Insurance agent-physician collects on policies of men murdered by a disfigured resident of the home for the blind where he acts as doctor-on-call.
Poison Pen (1939)
The inhabitants of a peaceful village begin receiving mysterious hate mail penned by someone with malicious thoughts.
The Four Feathers (1939)
A disgraced officer risks his life to help his childhood friends in battle.
Genre: Adventure, Drama, Romance, War
Three Smart Girls Grow Up (1939)
A businessman's (Charles Winninger) youngest daughter (Deanna Durbin) helps her innocent sisters in love.


text:
//========= Copyright © 1996-2005, Valve Corporation, All rights reserved. ============//
//
// Purpose:
//
// $NoKeywords: $
//=============================================================================//

#ifndef PANELLISTPANEL_H
#define PANELLISTPANEL_H

#ifdef _WIN32
#pragma once
#endif

#include <utllinkedlist.h>
#include <utlvector.h>
#include <vgui/VGUI.h>
#include <vgui_controls/Panel.h>

class KeyValues;

namespace vgui
{

//-----------------------------------------------------------------------------
// Purpose: A list of variable height child panels
//  each list item consists of a label-panel pair. Height of the item is
// determined from the label.
//-----------------------------------------------------------------------------
class PanelListPanel : public Panel
{
	DECLARE_CLASS_SIMPLE( PanelListPanel, Panel );

public:
	PanelListPanel( vgui::Panel *parent, char const *panelName );
	~PanelListPanel();

	// DATA & ROW HANDLING
	// The list now owns the panel
	virtual int AddItem( Panel *labelPanel, Panel *panel );
	int	GetItemCount() const;
	int GetItemIDFromRow( int nRow ) const;
	int GetVisibleItemCount();

	// Show / hide an item
	void SetItemVisible( int nItemID, bool bVisible );
	bool IsItemVisible( int nItemID ) const;

	// Iteration. Use these until they return InvalidItemID to iterate all the items.
	int FirstItem() const;
	int NextItem( int nItemID ) const;
	int InvalidItemID() const;

	virtual Panel *GetItemLabel(int itemID);
	virtual Panel *GetItemPanel(int itemID);

	virtual void RemoveItem(int itemID); // removes an item from the table (changing the indices of all following items)
	virtual void DeleteAllItems(); // clears and deletes all the memory used by the data items
	void RemoveAll();
	void HideAllItems();

	// painting
	virtual vgui::Panel *GetCellRenderer( int row );

	// layout
	void SetFirstColumnWidth( int width );
	int GetFirstColumnWidth();
	void SetNumColumns( int iNumColumns );
	int GetNumColumns( void );
	void MoveScrollBarToTop();
	void OverrideChildPanelWidth( bool bOverride );			// if true, width of child panels is set to fill this panel's width

	// selection
	void SetSelectedPanel( Panel *panel );
	Panel *GetSelectedPanel();
	/*
		On a panel being selected, a message gets sent to it
			"PanelSelected"		int "state"
		where state is 1 on selection, 0 on deselection
	*/

	void		SetVerticalBufferPixels( int buffer );
	int			GetVerticalBufferPixels() { return m_iPanelBuffer; }

	void		ScrollToItem( int itemNumber );

	// scrollbar
	void		SetShowScrollBar( bool bShow );
	bool		GetShowScrollbar() { return m_bShowScrollBar; }
	ScrollBar	*GetScrollBar() { return m_vbar; }

	// mouse wheel
	void		AllowMouseWheel( bool bAllow );

	CUtlVector< int > *GetSortedVector( void )
	{
		return &m_SortedItems;
	}

protected:
	// overrides
	virtual void OnSizeChanged(int wide, int tall);
	MESSAGE_FUNC_INT( OnSliderMoved, "ScrollBarSliderMoved", position );
	virtual void PerformLayout();
	virtual void ApplySchemeSettings(vgui::IScheme *pScheme);
	virtual void OnMouseWheeled(int delta);

	virtual void ApplySettings( KeyValues *inResourceData );
	virtual void GetSettings( KeyValues *outResourceData );
	virtual const char *GetDescription( void );

private:
	int	ComputeVPixelsNeeded();

	enum { DEFAULT_HEIGHT = 24, PANELBUFFER = 5 };

	typedef struct dataitem_s
	{
		// Always store a panel pointer
		Panel *panel;
		Panel *labelPanel;
	} DATAITEM;

	// list of the column headers

	CUtlLinkedList<DATAITEM, int>		m_DataItems;
	CUtlVector<int>						m_SortedItems;

	ScrollBar				*m_vbar;
	Panel					*m_pPanelEmbedded;

	PHandle					m_hSelectedItem;
	int						m_iFirstColumnWidth;
	int						m_iNumColumns;
	int						m_iDefaultHeight;
	int						m_iPanelBuffer;
	bool					m_bShowScrollBar;
	bool					m_bAllowMouseWheel;
	bool					m_bOverrideChildPanelWidth;
};

}
#endif // PANELLISTPANEL_H

redpajama_set_name: RedPajamaGithub


text:
Q: Dynamically loaded jQuery not defined when using inline script Situation:

*

*jQuery is dynamically loaded together with other scripts by one file javascripts.js in the <head> section of the html file

*Each html file has it's own javascript code executed on jQuery(document).ready() in the <body> section of the html file

Problem:

*

*Error: jQuery is not defined for javascript in the <body> section

*Modifying the html file is not an option (+1000 files with same problem)

Example html file:
<html>
  <head>
    <title>JS test</title>
    <script src="javascripts.js" type="text/javascript"></script>
  </head>
<body>
  <input type="text" class="date">
  <script>
    jQuery(document).ready(function() {  // Error: jQuery not defined
      jQuery('.date').datepicker();
    });
  </script>
</body>
</html>

javascripts.js:
// Load jQuery before any other javascript file
function loadJS(src, callback) {
    var s = document.createElement('script');
    s.src = src;
    s.async = true;
    s.onreadystatechange = s.onload = function() {
        var state = s.readyState;
        console.log("state: "+state);
        if (!callback.done && (!state || /loaded|complete/.test(state))) {
            callback.done = true;
            callback();
        }
    };
    document.getElementsByTagName('head')[0].appendChild(s);
}

loadJS('javascripts/jquery-1.8.3.min.js', function() {
    var files = Array(
      'javascripts/functions.js',
      'javascripts/settings.js'
    );

    if (document.getElementsByTagName && document.createElement) {
        var head = document.getElementsByTagName('head')[0];

        for (i = 0; i < files.length; i++) {
            var script = document.createElement('script');
            script.setAttribute('type', 'text/javascript');
            script.setAttribute('src', files[i]);
            script.async = true;
            head.appendChild(script);
        }
    }
});

A: This is happening, as many in the comments have pointed out, because you are loading jQuery asynchronously. Asynchronous means the rest of the code is executed, and so your document-ready handler (DRH) line is running before jQuery is present in the environment.
Here's a really hacky way of resolving this. It involves making a temporary substitute of jQuery whose job is just to log the DRH callbacks until jQuery has arrived. When it does, we pass them in turn to jQuery.
JS:
//temporary jQuery substitute - just log incoming DRH callbacks
function jQuery(func) {
    if (func) drh_callbacks.push(func);
    return {ready: function(func) { drh_callbacks.push(func); }};
};
var $ = jQuery, drh_callbacks = [];

//asynchronously load jQuery
setTimeout(function() {
    var scr = document.createElement('script');
    scr.src = '//ajax.googleapis.com/ajax/libs/jquery/1.11.0/jquery.min.js';
    document.head.appendChild(scr);
    scr.onload = function() {
        $.each(drh_callbacks, function(i, func) { $(func); });
    };
}, 2000);

HTML:
jQuery(document).ready(function() { alert('jQuery has loaded!'); });

Fiddle: http://jsfiddle.net/y7aE3/
Note in this example drh_callbacks is global, which is obviously bad. Ideally hook it onto a namespace or something, e.g. mynamespace.drh_callbacks.

A: I believe this simple solution should do the trick.  The changed line in the html changes the jquery onload function to a regular function.  The jquery onload function will sometimes happen before the jquery is loaded and we can't have that.  It's unreliable.  We need that function not to execute on page load, but AFTER the jquery has loaded.
To that end, the three lines I've added in the javascript.js are inside the code that is executed immediately after jQuery has finished loading.  They test to see if the pageLoaded function has been defined (so you don't have to put one on every page, only the ones that need it) and then execute it if it's there.
Now, because the change to the HTML is simple, you can just do a regex search and replace on those 1000 files to fix them.  Tools like Sublime, Eclipse or TextPad are suited for that task.
Cheers!
Example html file:
<html>
  <head>
    <title>JS test</title>
    <script src="javascripts.js" type="text/javascript"></script>
  </head>
<body>
  <input type="text" class="date">
  <script>
    function pageLoaded() { // changed
      jQuery('.date').datepicker();
    } // changed
  </script>
</body>
</html>

javascripts.js:
// Load jQuery before any other javascript file
function loadJS(src, callback) {
    var s = document.createElement('script');
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text:
\section{Introduction}
The Dirichlet-to-Neumann operator, which maps a function on the boundary of a domain to the normal derivative of its harmonic extension in the domain, is an object with a revered history in spectral theory, its eigenvalues and eigenfunctions still being the subject of active research. In this note we study the harmonic extensions of eigenfunctions of the Dirichlet-to-Neumann problem, usually called Steklov eigenfunctions; these have attracted a good amount of attention in the past years in their own right, as an interior counterpart to the eigenfunctions of the Dirichlet-to-Neumann operator.

For $\Omega$ a bounded domain in ${\mathbf R}^d$, where $d\geq 2$, the Steklov eigenvalue problem on $\Omega$ is informally given by
\begin{align}
\label{problem}
    \begin{cases}
    \Delta u=0 \qquad \: \text{in} \ \Omega,\\
    \partial_{\nu}u={\lambda} u \quad \ \text{on} \ \partial \Omega. \end{cases}
\end{align}
Note that the spectral parameter $\lambda$ is in the boundary conditions; a number ${\lambda}$ for which there is a solution to \eqref{problem} is called a Steklov eigenvalue, and a corresponding solution is called a Steklov eigenfunction. Of course, for problem \eqref{problem} to make sense some regularity on $\partial\Omega$ must be imposed. In most of this work we will require that $\Omega$ be a Lipschitz domain; we defer the precise weak definition of problem \eqref{problem} in this setting to Section 2.\\

We refer to the survey \cite{GP} for many results and open questions about the Steklov problem, with an angle towards spectral geometry. In particular, a question was raised there about density of nodal (i.e. zero) sets of solutions of \eqref{problem}. It is a well known property that eigenfunctions of the Laplace-Beltrami operator on a compact Riemannian manifold have wavelength dense nodal sets, meaning that every ball of wavelength radius in the manifold contains a zero of the eigenfunction; the authors of \cite{GP} ask whether a similar property holds for the interior nodal sets of solutions of \eqref{problem}, or for nodal sets on the boundary, which are the nodal sets of eigenfunctions of the Dirichlet-to-Neumann operator. Note that the right density scale should be $C/{\lambda}$; it is easy to convince oneself of this either by the scaling of \eqref{problem} or by looking at the explicit example of a ball.

A negative answer to the first question about interior nodal sets was given recently in \cite{BG}: the authors construct a dense family of simply connected domains in ${\mathbf R}^2$ with real analytic boundary such that for any such domain each Steklov eigenfunction $u_{{\lambda}}$ does not change sign in a ball inside the domain, the location of the ball depending on ${\lambda}$ but the radius being ${\lambda}$-independent. They actually show more, namely that $u_{{\lambda}}$ is well approximated in a fixed ball inside the domain by a function with only a finite number, independent of ${\lambda}$, of non-zero Fourier modes. \\

The density question for boundary nodal sets is, to the best of our knowledge, in general still open; it follows from the main result of \cite{S}, Theorem 2.1, that it is true for smooth simply-connected planar domains. A closer inspection of the proof of Theorem 2.1 in \cite{S} reveals that $\mathcal{C}^{\infty}$-smoothness is not needed, and $\mathcal{C}^4$ should be sufficient. Note also that the boundary density property cannot possibly hold for every Lipschitz domain, or even for every piecewise smooth one, as the explicit example of a rectangle shows: eigenfunctions with zeros only on two sides of the rectangle exist (it is easy to compute the eigenfunctions explicitly as products of trigonometric and hyperbolic functions, see \cite{GP}). Our first contribution in the present note lies in between the interior and boundary cases: we show that balls of radius $C/{\lambda}$ centered at the boundary do contain zeros of $u_{{\lambda}}$. The precise statement follows.

\begin{theorem}
\label{upper}
Let $\Omega$ be a Lipschitz domain in ${\mathbf R}^d$, $d\geq 2$, and let $u=u_{{\lambda}}$ be a solution of \eqref{problem}, where we assume ${\lambda}\neq 0$. Let $\mathcal{Z}_{u_{{\lambda}}}=\{x\in \Omega : u_{{\lambda}}(x)=0\}$. Then there exists a constant $C=C(\Omega)$ such that
\begin{align}
\label{density}
    \mathcal{Z}_{u_{{\lambda}}}\cap B\neq \emptyset
\end{align}
for any ball $B$ in ${\mathbf R}^d$ of radius $C/{\lambda}$ centered at a point in $\partial\Omega$.
\end{theorem}
Some remarks are in order.
\begin{remark}
     \textit{(i)} The result is only interesting when ${\lambda}$ is large enough: by orthogonality to constants on the boundary (that is, eigenfunctions with ${\lambda}=0$) any solution of \eqref{problem} with ${\lambda}\neq 0$ has to change sign on the boundary, and hence satisfy \eqref{density} for $C$ large enough if ${\lambda}$ is bounded.\\
     \textit{(ii)} Our theorem does not exclude the chance that the zero set is essentially contained in a corona around the boundary of size shrinking with ${\lambda}$; the results of \cite{BG} leave open the possibility that zeros are contained in any ball of radius $C/{\lambda}$ centered at a point in a collar neighbourhood of the boundary of size independent of ${\lambda}$, albeit small. We are inclined to think that the latter statement should be true.\\
     \textit{(iv)} The result of \cite{BG} says essentially that Steklov eigenfunctions need not oscillate deep in the interior, where they are small; there are also known instances where eigenfunctions concentrate on parts of the boundary, see for example \cite{LPPS}. Our result says that Steklov eigenfunctions need to oscillate near the boundary just because they satisfy the Steklov condition, regardless of whether they are small or large there. We thank Iosif Polterovich for bringing up this point.

\end{remark}

\subsection*{Inner radius}
We denote by $\mathcal{D}_{{\lambda}}$ a nodal domain for the Steklov eigenfunction $u_{{\lambda}}$, that is a connected component of the set $\Omega\setminus \mathcal{Z}_{u_{{\lambda}}}$. Denote by $\rho(\mathcal{D}_{{\lambda}})$ the radius of the largest ball $B$ centered at a point $y\in\partial \Omega\cap\overline{\mathcal{D}_{{\lambda}}}$ such that $B\cap\Omega\subset \mathcal{D}_{{\lambda}}$; we call it the modified inner radius of $\mathcal{D}_{{\lambda}}$. Theorem \ref{upper} can then be restated as
\begin{align*}
    \rho(\mathcal{D}_{{\lambda}})\leq \frac{C(\Omega)}{{\lambda}}.
\end{align*}
Note that we do not say anything about balls in $\mathcal{D}_{{\lambda}}$ centered at points not in $\partial\Omega$; as shown in \cite{BG}, it is possible that deep inside $\mathcal{D}_{{\lambda}}$ one could inscribe balls of ${\lambda}$-independent radius. We think that centering the balls at $\partial\Omega$ provides the correct analogy with the Laplace eigenfunctions case. In this setting, the second main contribution of this note is a sharp lower bound for the modified inner radius in dimension two.

\begin{theorem}
\label{lower}
Let $\Omega\subset {\mathbf R}^2$ be a Lipschitz domain, and let $\mathcal{D}_{{\lambda}}$ be a nodal domain of $u_{{\lambda}}$, where ${\lambda}\neq 0$. There exists a constant $c(\Omega)>0$ such that
\begin{align}
\label{inrad}
    \rho(\mathcal{D}_{{\lambda}})\geq \frac{c(\Omega)}{{\lambda}}.
\end{align}
\end{theorem}

In words, one can inscribe inside $\mathcal{D}_{{\lambda}}$ the intersection of $\Omega$ and a ball of radius $c(\Omega)/{\lambda}$ centered somewhere on $\partial \Omega\cap\overline{\mathcal{D}_{{\lambda}}}$. Note that in this way we also get a lower bound for the ``genuine'' inner radius, by inscribing a full ball of radius $c_1/{\lambda}$ inside the intersection of $\Omega$ and the ball above.

\begin{remark}
     \textit{(i)} It is possible to obtain a non-sharp estimate (polynomial in ${\lambda}^{-1}$) of the same type in higher dimensions as well, at least under some higher regularity assumption on the boundary ($\mathcal{C}^2$ suffices), by following the method of \cite{M}: this entails proving a local asymmetry estimate for Steklov eigenfunctions and combining it with a Poincaré inequality. As the flavour of the asymmetry estimate is somewhat different from the rest of the results in this paper, we do not pursue this here and defer it to later work. We will comment further on the issue in Section 4.
\end{remark}

\subsection*{Hausdorff measure bounds}
In the Laplace eigenfunction case, a density result for zeros is the starting point in the proof of a sharp lower bound on the Hausdorff measure of the nodal set, conjectured by Yau. The result is due to Logunov in \cite{L}: the proof is not easy and uses some clever insights about harmonic functions. In upcoming work, we combine Theorem \ref{upper} with the ideas of \cite{L} to improve on the existing lower bound for the Hausdorff measure of the nodal set of a Steklov eigenfunction: namely, we prove that if $\Omega$ is a domain with $\mathcal{C}^2$-smooth boundary, for a Steklov eigenfunction $u_{{\lambda}}$ the bound
\begin{align*}
    \mathcal{H}^{d-1}(\mathcal{Z}_{u_{{\lambda}}})\geq c(\Omega)
\end{align*}
holds with a constant $c(\Omega)$ independent of ${\lambda}$. The previous best result, established in \cite{SWZ}, yielded a lower bound polynomially decaying in ${\lambda}$ (for $d>2$), of order ${\lambda}^\frac{2-d}{2}$. In the same article we also obtain an upper bound on the Hausdorff measure of the nodal set.

\subsection*{Plan of the paper}
We prove Theorem \ref{upper} in Sections 2 and 3; in Section 2 we discuss some regularity for Steklov eigenfunctions and we describe an auxiliary boundary value problem that we need in the argument, and Section 3 gives the actual proof of the density statement (two of them, as a matter of fact). The proof of Theorem \ref{lower} is given in Section 4.

\section{On boundary value problems}
\subsection*{The Steklov problem}
In the statement of Theorem \ref{upper} we assume that $\Omega$ is a Lipschitz domain: by this we mean that for every point $p\in \partial \Omega$ there is a hyperplane $H$ through $p$ such that locally $\partial \Omega$ is the graph of a Lipschitz function on $H$ and $\Omega$ is the set of points below that graph. In this setting, a weak solution of \eqref{problem} is a function $u\in H^1(\Omega)$ such that
\begin{align}
\label{weak}
    \int_{\Omega}\nabla u \nabla v = {\lambda} \int_{\partial\Omega} uv
\end{align}
for all $v\in H^1(\Omega)$ (we denote by $H^1(\Omega)$ the Sobolev space $W^{1,2}(\Omega)$); a number ${\lambda}$ for which the above equation has a nontrivial solution is called an eigenvalue of the Steklov problem, and a solution $u=u_{{\lambda}}$ is a Steklov eigenfunction. Since $\Omega$ is Lipschitz, the trace operator $H^1(\Omega)\rightarrow L^2(\partial\Omega)$ is compact and hence the spectrum of \eqref{problem} is discrete and accumulating to infinity; the eigenvalues can be arranged in a sequence $0={\lambda}_0<{\lambda}_1\leq {\lambda}_2\leq\dots\rightarrow \infty$, with corresponding eigenfunctions $u_{{\lambda}_j}$ forming an orthonormal basis of $L^2(\partial\Omega)$. This is a folklore result, but see \cite{AM} for a proof.

It is possible to prove, using a Moser iteration type argument, that a solution of \eqref{weak} is actually in $L^{\infty}(\Omega)$, with bounds for the $\sup$ norm depending on ${\lambda}$ and on $\Omega$ (see for instance \cite{BGT}). One thus sees that a Steklov eigenfunction is the solution of a Neumann boundary value problem with Neumann data in $L^{\infty}(\partial\Omega)$; in \cite{BH}, a solution to the Neumann problem with bounded Neumann data is constructed, and this solution is continuous up to the boundary. By the uniqueness up to constants of weak solutions of Neumann boundary value problems in Lipschitz domains (see \cite{JK}), one obtains then that solutions to \eqref{weak} actually are in $\mathcal{C}(\overline{\Omega})$. This fact is not really needed in the next section, but in our opinion it makes it easier to think about Steklov eigenfunctions in Lipschitz domains.

\subsection*{A Steklov-Dirichlet problem}
Our strategy to prove density for the zero sets of Steklov eigenfunctions near the boundary relies on an auxiliary boundary value problem, for which we need a description of the spectrum analogous to that of the Steklov problem.

Let $W$ be a bounded Lipschitz domain in ${\mathbf R}^d$; we assume the boundary consists of two non-empty relatively open sets, $F$ and $S$, with common Lipschitz boundary. We call a Steklov-Dirichlet (sometimes called mixed Poincar{\'e}-Steklov as in \cite{A}) problem the following mixed spectral problem:
\begin{align}
\label{mixed}
    \begin{cases}
    \Delta u=0 \qquad \: \text{in} \ W,\\
    u=0 \qquad \ \ \; \text{on} \ S,\\
    \partial_{\nu}u=\sigma u \quad \ \text{on} \ F.\\
    \end{cases}
\end{align}
Under our hypotheses, it is well known that the spectrum is discrete and consists of eigenvalues $0<\sigma_1\leq \sigma_2\leq\dots \rightarrow \infty$, each having finite multiplicity (see \cite{A}, \cite{BKPS}); moreover, the corresponding eigenfunctions are orthogonal in $L^2(F)$. For completeness of the exposition we provide an outline of a proof, following the arguments of \cite{AM} and \cite{AM2} for the Steklov case. As in \cite{A}, \cite{BKPS}, we denote
$$H_0^1(W;S)=\{u\in H^1(W) : u=0 \ \text{on} \ S\};$$
here and in the sequel, we consider elements in $H^1$ restricted to the boundary in the sense of Sobolev traces. We introduce now an operator of the Dirichlet-to-Neumann type: let $\mathcal{D}_{M}:L^2(F)\to L^2(F)$ be the unbounded operator defined by
\begin{align*}
    &dom(\mathcal{D}_{M})=\{\phi\in L^2(F) \ \text{such that} \ \exists u \in H_0^1(W;S) : \Delta u=0 \ \text{in} \ W ,\\ & \qquad \qquad \qquad u_{|_F}=\phi \ \text{and} \ \partial_{\nu}u\in L^2(F) \};\\
    &\mathcal{D}_{M}\phi=\partial_{\nu}u_{|_F}.
\end{align*}
Note that the spectrum of $\mathcal{D}_M$ coincides with that of problem \eqref{mixed}, so by general spectral theory one needs only to prove that $\mathcal{D}_M$ is selfadjoint, bounded below, and with compact resolvent. If $S=\emptyset$, in which case $\mathcal{D}_M$ is the standard Dirichlet-to-Neumann operator, this was proved in \cite{AM2},\cite{AM}. We follow closely their outline to check that the same holds when $S$ is non-empty. Let $V=\{u_{|_F} : u\in H_0^1(W;S)\}$ be the trace space, with $\|u_{|_F}\|_{V}=\|u\|_{H_0^1(W;S)}$; since $W$ is a Lipschitz domain the embedding $V\to L^2(F)$ is compact. We denote also
\begin{align*}
    \mathcal{H}=\{u \in H_0^1(W;S) : \Delta u=0\}.
\end{align*}
Since $0$ is not a Dirichlet eigenvalue, from the solvability of the Dirichlet problem in Lipschitz domains (see \cite{D},\cite{JK}) it follows that we have the (non-orthogonal) direct sum
$$H_0^1(W;S)=H_0^1(W)\oplus\mathcal{H}.$$
Using this, one sees that the linear map $\mathcal{H}\ni u\to u_{|_F}\in V$ is bijective. Define now the bilinear form $a : V\times V\to {\mathbf R}$ by
$$a(\phi,\psi)=\int_{W}\nabla u \nabla v,$$
where $u,v \in \mathcal{H}$ with $u_{|_F}=\phi$ , $v_{|_F}=\psi$. We recall now an abstract lemma, which can be found as Lemma 2.3 in \cite{AM2}:
\begin{lemma}
    Let $X_1,X_2,X_3$ be Banach spaces, $X_1$ reflexive. Let $T: X_1\to X_3$ be a compact linear operator and $S: X_1\to X_2$ a bounded injective linear operator. Then for any $\varepsilon>0$ there exists $c>0$ such that for any $x\in X_1$, $\|Tx\|^2_{X_3}\leq \varepsilon \|x\|^2_{X_1}+c\|Sx\|^2_{X_2}$.
\end{lemma}
Using the lemma for $T$ being the embedding $\mathcal{H}\to L^2(W)$ and $S$ being the trace operator $\mathcal{H}\to L^2(F)$, it is immediate to show that $a$ is coercive (elliptic) on $V$. There is a linear operator on $L^2(F)$ associated to $a$, call it $B$; let $u\in \mathcal{H}$, $b\in L^2(F)$. Then $u_{|_F}\in dom(B)$ and $Bu_{|_F}=b$ if and only if
\begin{equation}
\label{form}
    \int_{W}\nabla u \nabla v=\int_{F} bv
\end{equation}
for all $v\in\mathcal{H}$. Using that $\Delta u=0$, one gets that \eqref{form} holds also for $v\in H_0^1(W)$; remembering that $H_0^1(W;S)=H_0^1(W)\oplus\mathcal{H}$, \eqref{form} holds for all $v\in H_0^1(W;S)$, and consequently $b=\partial_{\nu} u$; this implies that $u_{|_F}\in dom(\mathcal{D}_M)$ and $\mathcal{D}_M u_{|_F}=Bu_{|_F}$. It is also easy to see that $dom(\mathcal{D}_M)\subset dom(B)$, so that $\mathcal{D}_M=B$ is the operator associated with the form $a$. The fact that $\mathcal{D}_M$ is bounded below and with compact resolvent now follows immediately by coercivity of $a$ and the embedding $V\to L^2(F)$ being compact. \\

The description of the spectrum of $\mathcal{D}_M$ is now given by the usual max-min procedure with the form $a$; in particular, the first eigenvalue of $\mathcal{D}_M$, and thus of problem \eqref{mixed}, is given by
\begin{align}
\label{eigenvalue}
    \sigma_1=\min_{{u\in H_0^1(W ; S)}}\frac{\int_W |\nabla u|^2}{\int_F |u|^2}.
\end{align}
Let now $u_j$, $u_k$ correspond to eigenvalues $\sigma_j\neq \sigma_k$; one has the string of equalities
\begin{align*}
    \sigma_j\int_F u_j u_k = \int_F \partial_{\nu}u_j u_k = \int_W \nabla u_j \nabla u_k = \int_{F} u_j \partial_{\nu}u_k = \sigma_k\int_F u_j u_k,
\end{align*}
which implies that $u_j$ and $u_k$ are orthogonal in $L^2(F)$.
We record the following fact, which is proved in the same way as in the Dirichlet case:
\begin{proposition}
    The first eigenvalue $\sigma_1$ of problem \eqref{mixed} is simple and the corresponding eigenfunction is positive in $W$.
    \begin{proof}
    Let $u$ be an eigenfunction corresponding to $\sigma_1$. Since $u$ is in $H_0^1(W;S)$, also $|u|\in H_0^1(W;S)$. Moreover, $|\nabla|u||=|\nabla u|$, so that the variational quotient in \eqref{eigenvalue} is minimized by $|u|$ as well; as a consequence, $|u|$ is harmonic in $W$, and therefore $u$ does not change sign in $W$. Without loss of generality, we can assume $u$ is positive. If there were two linearly independent eigenfunctions $u_1$ and $u_2$ corresponding to $\sigma_1$, then we could find a linear combination of them which changes sign in $W$. By the above, this implies that $\sigma_1$ is simple.
    \end{proof}
\end{proposition}

\section{Density of nodal sets near the boundary}
We give here two different proofs of Theorem \ref{upper}. Throughout the section we denote by $u=u_{{\lambda}}$ a solution of problem \eqref{problem}, corresponding to the eigenvalue ${\lambda}$. We can assume that ${\lambda}$ is greater than some large enough constant ${\lambda}_{\Omega}$ (see Remark \textit{(i)} following Theorem \ref{upper}). The two proofs are quite similar in spirit, and both involve the mixed Steklov-Dirichlet problem introduced in Section 2; while the second proof may look more natural to those familiar with Laplace eigenfunctions, it appears to require a higher degree of regularity on the domain. The first proof works under the Lipschitz assumption only and, moreover, one does not really need to think about nodal domains for the argument to carry on.
\subsection*{First proof} Let $x_0$ be a point on $\partial\Omega$, and let $W=B(x_0,s)\cap \Omega$. We take $s$ small enough so that $W$ can be written as the set of points below the graph of a Lipschitz function. Assume that $u$ does not change sign in $W$, say $u$ is positive. To keep the notation of Section 2, we call $S=\partial B(x_0,s)\cap \Omega$ and $F=\partial\Omega\cap B(x_0,s)$. We let $\sigma_1$ and $h$ denote respectively the first eigenvalue and the first eigenfunction of the Steklov-Dirichlet problem in $W$ with boundaries $S$ and $F$; that is, $h$ solves the problem
\begin{align}
\label{auxiliary}
    \begin{cases}
    \Delta h=0 \qquad \ \: \text{in} \ W,\\
    h=0 \qquad \ \ \ \  \text{on} \ S,\\
    \partial_{\nu}h=\sigma_1 h \quad \ \text{on} \ F.\\
    \end{cases}
\end{align}
The key to the proof is the following lemma, which is a consequence of the variational principle \eqref{eigenvalue}. We warn the reader that constants denoted by $C$ may change value from line to line.
\begin{lemma}
\label{key}
There exists a small enough constant $c_{\Omega}$ and a dimensional constant $C>0$ such that if $s<c_{\Omega}$, then $\sigma_1<C/s$.
\begin{proof}
We select $c_{\Omega}$ such that for $s<c_{\Omega}$,
\begin{align}
\label{measure}
    \mathcal{H}^{d-1}(F\cap B(x_0,s/2))\gtrsim s^{d-1},
\end{align}
where $\mathcal{H}^{d-1}$ denotes $d-1$-dimensional Hausdorff measure and the symbol $\gtrsim$ hides a dimensional constant. Here we used the hypothesis that the domain is Lipschitz. Consider now the function
\begin{align*}
    \psi(x)=1-\frac{|x-x_0|^2}{s^2},
\end{align*}
and note that $\psi\in H_0^1(W ; S)$ so that $\psi$ is allowed as a test function in the variational quotient in \eqref{eigenvalue}. We have that $$\int_W |\nabla \psi|^2\lesssim \frac{1}{s^4}\int_{B(x_0,s)}|x-x_0|^2\lesssim s^{d-2}.$$
Note that for $x\in W\cap B(x_0,s/2)$ we have $\psi(x)\geq 3/4$, so that by \eqref{measure}
$$\int_{F}|\psi|^2\geq \int_{F\cap B(x_0,s/2)}|\psi|^2\gtrsim s^{d-1}.$$
Plugging the above estimates in \eqref{eigenvalue}, we obtain $\sigma_1\lesssim 1/s$.
\end{proof}
\end{lemma}

We take now the function $v=h^2$, where $h$ is as in \eqref{auxiliary}. Note that $\Delta v=2|\nabla h|^2\geq 0$, $v=\partial_{\nu}v=0$ on $S$ and $\partial_{\nu}v=2\sigma_1 v$ on $F$. By Green's formula (see for instance \cite{MZ}, Chapter 6, for a proof of Green's formula for Lipschitz domains) we thus have
\begin{align*}
    0=\int_{W}v\Delta u = \int_{W}u\Delta v + \int_F v\partial_{\nu}u - u\partial_{\nu} v.
\end{align*}
Since $u$ is positive in $W$, the first term on the right hand side is positive, so that by Lemma \ref{key} we obtain
\begin{align*}
    0\geq \int_F ({\lambda}-2\sigma_1)uv \geq \int_F ({\lambda}-C/s)uv,
\end{align*}
whence the condition $s<C/{\lambda}$. The proof of Theorem \ref{upper} is complete.

\subsection*{Second proof}
The main purpose in giving another proof is to record the observation contained in Proposition \ref{nodal}, which is certainly part of the folklore and has been used implicitly for instance in \cite{BKPS}. We assume here that $\partial\Omega$ has some higher regularity than Lipschitz, say $\partial\Omega$ is $\mathcal{C}^2$; under this hypothesis, a Steklov eigenfunction is in $\mathcal{C}^1(\overline{\Omega})$. Since we were unable to locate this fact in the literature, we state it as a proposition and outline a proof.
\begin{proposition}
Let $\Omega$ be a $\mathcal{C}^{1,1}$ domain, and let $u=u_{{\lambda}}$ be a Steklov eigenfunction. Then $u\in\mathcal{C}^1(\overline{\Omega})$.
\begin{proof}
    The argument relies on elliptic regularity results contained in Chapter 2 of \cite{Gri}; these results are used in \cite{LE} to prove analogous regularity statements for Neumann and Robin eigenfunctions. In particular, we will use the following statement, a special case of Proposition 2.5.2.3 in \cite{Gri}:\\
    \textit{Let $\Omega$ be a $\mathcal{C}^{1,1}$ domain, and let $u\in W^{1,p}(\Omega)$ be a weak solution of
    \begin{align*}
    \begin{cases}
    -\Delta u+u=f \quad  \text{in} \ \Omega,\\
    \partial_{\nu}u=g \qquad \ \text{on} \ \partial \Omega, \end{cases}
    \end{align*}
    with $f\in L^p(\Omega)$ and $g\in W^{1-1/p,p}(\partial\Omega)$; then $u\in W^{2,p}(\Omega)$.
    }\\
    Let now $u$ be a Steklov eigenfunction; then $u$ is a solution of the problem above with $f=u$ and $g={\lambda} u$. We first apply the result for $p=2$ (note that since a Steklov eigenfunction is in $W^{1,2}(\Omega)$, its trace on $\partial\Omega$ is indeed in $W^{1/2,2}(\partial\Omega)$), obtaining that $u\in W^{2,2}(\Omega)$. Then, by Sobolev embedding, $u\in W^{1,p}(\Omega)$ for some $p>2$; it follows that $f\in L^p(\Omega)$ and $g\in W^{1-1/p,p}(\partial\Omega)$ for this $p$, and we apply again the theorem to get $u\in W^{2,p}(\Omega)$. It is now clear how to bootstrap the argument to obtain $u\in W^{2,p}(\Omega)$ for any $p<\infty$, and Sobolev embedding theorems then imply that $u\in\mathcal{C}^1(\overline{\Omega})$.
\end{proof}
\end{proposition}
While Proposition \ref{nodal} below is quite intuitive and mirrors what happens for Laplace eigenfunctions, the proof needs some care due to the fact that nodal domains may not be Lipschitz domains; a workaround to this was suggested in \cite{BM}, Appendix D, for the Dirichlet eigenfunction case, and then in \cite{LE} for Neumann and Robin eigenfunctions. We follow these two references in the proof below.

\begin{remark}
     A word on notation: since a Steklov eigenfunction $u$ is $\mathcal{C}^1$ up to the boundary, in the following we implicitly consider an extension of it in a neighbourhood of $\overline{\Omega}$; when we talk about nodal domains of $u$, we are thinking about nodal domains of this extension. In this way the intersection of a nodal domain with $\partial\Omega$ is a relatively open set in $\partial\Omega$.
\end{remark}
\begin{proposition}
\label{nodal}
    Assume that $\partial\Omega$ is $\mathcal{C}^2$, Let $u=u_{\lambda}$ be a solution of \eqref{problem}, and denote by $\mathcal{D}$ a nodal domain of $u$. Then $u$ is the first eigenfunction of a Steklov-Dirichlet problem in $\mathcal{D}$, with $S=\partial{\mathcal{D}}\cap\Omega$ and $F=\mathcal{D}\cap\partial\Omega$ in the notation of \eqref{mixed}; that is to say that
    \begin{align}
    \label{nodalsteklov}
        {\lambda}=\frac{\int_{\mathcal{D}}|\nabla u|^2}{\int_{F}|u|^2}=\inf \frac{\int_{\mathcal{D}}|\nabla \phi|^2}{\int_{F}|\phi|^2},
    \end{align}
    where the $\inf$ is taken over all $\phi\in\mathcal{C}^{\infty}(\overline{\mathcal{D}})$ which are compactly supported in $\mathcal{D}\cup F$.
    \begin{proof}
    As remarked earlier, every nodal domain of $u$ intersects $\partial \Omega$. Suppose that $u$ is positive in $\mathcal{D}$. If $\mathcal{D}$ were a Lipschitz domain, the proposition would immediately follow from the considerations in Section 2. Following \cite{BM} and \cite{LE} we consider instead the set $\mathcal{D}_{\alpha}=\{x\in\mathcal{D}: u(x)>\alpha\}$, for $\alpha$ small enough. If $\alpha$ is a regular value both of $u$ and of $u$ restricted to $\partial\Omega$ (recall that $u\in \mathcal{C}^1(\overline\Omega)$), then $\mathcal{D}_{\alpha}$ is a Lipschitz domain. Set $S_{\alpha}=\partial\mathcal{D}_{\alpha}\cap\Omega$, $F_{\alpha}=\mathcal{D}_{\alpha}\cap\partial\Omega$, and $u_{\alpha}=u-\alpha$. By Green's formula in $\mathcal{D_{\alpha}}$ and the Steklov condition we get that
    \begin{align*}
        \int_{\mathcal{D}_{\alpha}}|\nabla u|^2=\int_{F_{\alpha}}u_{\alpha}\partial_{\nu}u_{\alpha}={\lambda} \int_{F_{\alpha}}|u|^2-\alpha{\lambda}\int_{F_{\alpha}}u.
    \end{align*}
    Sard's Lemma gives a sequence of regular values for both $u$ and its restriction to the boundary approaching $0$, so that passing to the limit in the equality above we obtain
    \begin{align}
    \label{green}
        \int_{\mathcal{D}}|\nabla u|^2={\lambda} \int_{F}|u|^2,
    \end{align}
    which gives the first equality in \eqref{nodalsteklov}. We consider now another auxiliary function in $\mathcal{D_{\alpha}}$, where $\alpha$ is a regular value as above; namely, we let $v_{\alpha}$ be the first eigenfunction of the mixed Steklov-Dirichlet problem in $\mathcal{D_{\alpha}}$, with Steklov condition on $F_{\alpha}$ and Dirichlet condition on $S_{\alpha}$. As in Section 2, we denote by $\sigma_1(\mathcal{D_{\alpha}})$ the corresponding first Steklov-Dirichlet eigenvalue. By Green's formula, we have that
    \begin{align*}
        \int_{\mathcal{D_{\alpha}}}\nabla u \nabla v_{\alpha}=\int_{S_{\alpha}}u\partial_{\nu}v_{\alpha}+\int_{F_{\alpha}}u\partial_{\nu}v_{\alpha}=\int_{S_{\alpha}}u\partial_{\nu}v_{\alpha}+\sigma_1(\mathcal{D_{\alpha}})\int_{F_{\alpha}}uv_{\alpha}
    \end{align*}
    and also that
    \begin{align*}
        \int_{\mathcal{D_{\alpha}}}\nabla u \nabla v_{\alpha}=\int_{S_{\alpha}}v_{\alpha}\partial_{\nu}u+\int_{F_{\alpha}}v_{\alpha}\partial_{\nu}u={\lambda}\int_{F_{\alpha}}uv_{\alpha},
    \end{align*}
    and therefore
    \begin{align*}
        {\lambda}\int_{F_{\alpha}}uv_{\alpha}=\sigma_1(\mathcal{D_{\alpha}})\int_{F_{\alpha}}uv_{\alpha}+\int_{S_{\alpha}}u\partial_{\nu}v_{\alpha}.
    \end{align*}
    Without loss of generality, let $v_{\alpha}$ be positive in $\mathcal{D_{\alpha}}$; then on $S_{\alpha}$ we have $\partial_{\nu}v_{\alpha}\leq 0$, from which we obtain by the equation above that
    \begin{align}
    \label{above}
        \sigma_1(\mathcal{D_{\alpha}})\geq {\lambda}.
    \end{align}

Let us call $\sigma_1(\mathcal{D})$ the $\inf$ on the right hand side of \eqref{nodalsteklov}; since $\mathcal{D}$ may not be a Lipschitz domain, this is a slight abuse of notation. Using \eqref{green}, we have the string of inequalities
\begin{align*}
    {\lambda}\int_{F}|u|^2=\int_{\mathcal{D}}|\nabla u|^2\geq \int_{\mathcal{D_{\alpha}}}|\nabla u_{\alpha}|^2\geq \sigma_1(\mathcal{D_{\alpha}})\int_{F_{\alpha}}|u_{\alpha}|^2\geq \sigma_1(\mathcal{D}) \int_{F_{\alpha}}|u_{\alpha}|^2,
\end{align*}
from which, letting $\alpha\to 0$, we obtain ${\lambda}\geq \sigma_1(\mathcal{D})$.

For the converse inequality, let $\varepsilon>0$ and pick $\phi$ compactly supported in $\mathcal{D}\cup F$ such that
\begin{align*}
    \frac{\int_{\mathcal{D}}|\nabla \phi|^2}{\int_{F}|\phi|^2}\leq \sigma_1(\mathcal{D})+\varepsilon.
\end{align*}
We take now $\alpha$ a regular value as above so small that the support of $\phi$ is contained in $\mathcal{D}_{\alpha}\cup F_{\alpha}$; note that here we can exclude the case of a point $x_0\in\partial\Omega$ such that $u(x_0)=0$ and $u>0$ in a neighbourhood of $x_0$ by Hopf's lemma combined with the Steklov condition. Using \eqref{above}, we then have
\begin{align*}
    {\lambda}\leq \sigma_1(\mathcal{D_{\alpha}})\leq \frac{\int_{\mathcal{D}}|\nabla \phi|^2}{\int_{F}|\phi|^2}\leq \sigma_1(\mathcal{D})+\varepsilon.
\end{align*}
Since this holds for any $\varepsilon>0$, we finally obtain ${\lambda}=\sigma_1(\mathcal{D})$, which gives the second equality in \eqref{nodalsteklov} and concludes the proof of the proposition.
\end{proof}
\end{proposition}

The proof of Theorem \ref{upper} is now quite easy, and the main ingredient is once again provided by Lemma \ref{key}. Suppose that a nodal domain of $u$, $\mathcal{D}$, contains a set $W=B(x_0,s)\cap \Omega$, for some $x_0\in \partial \Omega$ and some $s>0$. We assume $s$ is small enough to apply Lemma \ref{key}. Consider problem \eqref{mixed} on $W$, with $S=\partial B(x_0,s)\cap \mathcal{D}$ and $F=\partial\Omega\cap B(x_0,s)$; by Lemma \ref{key}, we have $\sigma_1(W)<C/s$. On the other hand, by Proposition \ref{nodal}, ${\lambda}=\sigma_1(\mathcal{D})$, where as in the proof of Proposition \ref{nodal} by $\sigma_1(\mathcal{D})$ we denote the $\inf$ on the right hand side of \eqref{nodalsteklov}. This definition immediately implies weak domain monotonicity, $\sigma_1(W)\geq \sigma_1(\mathcal{D})$. We obtain
\begin{align*}
    C/s>\sigma_1(W)\geq \sigma_1(\mathcal{D})={\lambda},
\end{align*}
so that $s<C/{\lambda}$, and the proof is once again complete.

\section{Lower bound on the inner radius}
We discuss here the proof of Theorem \ref{lower}. The argument is quite elementary, and the ideas follow those for Laplace eigenfunctions in \cite{M} and especially \cite{M2}; the boundary presents some additional difficulties.

From now on, $\Omega$ will be a Lipschitz domain in ${\mathbf R}^2$. We start by covering $\partial\Omega$ with a finite number $K=K(\Omega)$ of (possibly overlapping) rectangles centered at $\partial\Omega$, such that in each such rectangle $\Omega$ is the set of points below the graph of a Lipschitz function; let us denote the Lipschitz constant of this function by $\Gamma$. In the sequel we will refer to such rectangles as "large rectangles". After an orthogonal transformation we have the following description of $\Omega$ in a large rectangle $Q$:
\begin{align}
\label{lip}
    \Omega\cap Q=\{(x,y)\in Q:-1<x<1 , -2\Gamma< y < g(x)\},
\end{align}
where $g$ is a Lipschitz function with $g(0)=0$, $|g(x_1)-g(x_2)|\leq \Gamma |x_1-x_2|$ for $x_1, x_2 \in [-1,1]$. The boundary is described by $\partial\Omega\cap Q=\{(x,y)\in Q: y=g(x)\}$. Note that we choose the rectangles in a way that the graph of $g$ intersects the boundary of each rectangle only on the vertical sides.\\

Let now $u_{{\lambda}}$ be a Steklov eigenfunction and $\mathcal{D}$ any of its nodal domains. As in Section 3, we use the notation $F=\partial\Omega\cap \overline{\mathcal{D}}$ and $S=\mathcal{Z}_{u_{{\lambda}}}\cap\overline{\mathcal{D}}$ (note that by our definition $\mathcal{Z}_{u_{{\lambda}}}\subset\Omega$). We have that
\begin{align}
\label{ray}
    {\lambda}=\frac{\int_{\mathcal{D}}|\nabla u_{{\lambda}}|^2}{\int_F |u_{{\lambda}}|^2}.
\end{align}

As in the Introduction, we denote by $\rho(\mathcal{D})$ the modified inner radius of $\mathcal{D}$; we repeat that by this we mean the radius of the largest ball $B$ centered at $F$ such that $B\cap\Omega\subset \mathcal{D}$. Fix
\begin{align*}
    h=100\rho(\mathcal{D});
\end{align*}
we can assume that $h$ is small enough. For each rectangle $Q$ in the above covering of $\partial\Omega$ we cover $F\cap Q$ by non-overlapping almost rectangles $q$; by almost rectangle we mean a set of the form
\begin{align}
\label{cubes}
    q=\{(x,y)\in Q:x_q-h<x<x_q+h, g(x)-2\Gamma h < y < g(x)+2\Gamma h\},
\end{align}
for some $x_q\in (-1+h,1-h)$ and $g$ as above. Again the choice of the sides of the almost rectangle is made to ensure that the graph of $g$ intersects each $q$ on the vertical sides only.
Let now
\begin{align*}
    u=
    \begin{cases}
      u_{{\lambda}} \quad \text{on } \mathcal{D};\\
      0 \quad \ \; \text{on } \Omega\setminus\mathcal{D}.
    \end{cases}
\end{align*}
Using \eqref{ray} and remembering that almost rectangles do not overlap, it is easy to verify the following fact.
\begin{claim}
\label{special}
There exists at least one almost rectangle $q$ such that
\begin{align*}
    \int_{q\cap\Omega}|\nabla u|^2\leq K{\lambda} \int_{q\cap F}|u|^2,
\end{align*}
where $K$ is as above the number of large rectangles in the covering of $\partial\Omega$.
\end{claim}

If we can show that in the almost rectangle given in Claim \ref{special} we have
\begin{align}
\label{poin}
    \int_{q\cap F}|u|^2\leq Ch \int_{q\cap\Omega}|\nabla u|^2,
\end{align}
with constant $C$ depending only on $\Omega$ (actually on $\Gamma$), Theorem \ref{lower} follows. From now on, the letter $C$ will denote constants that may change value at every occurrence but depend on $\Gamma$ only. We will assign explicit numerical values to constants independent of $\Gamma$.

We now prove \eqref{poin} using some Poincaré-type inequalities, in the spirit of \cite{M2}. Let us first note a trace inequality; the proof could be obtained by general Sobolev trace inequalities and scaling, but since it is simple we give it here for the reader's convenience. We denote
\begin{align}
\label{map}
    \varphi(x,y)=y-g(x).
\end{align}
In the following, for brevity, we consider $\varphi$ as defined in $q$, so that we denote by $\{\varphi=t\}$ the set $\{(x,y)\in q : \varphi(x,y)=t\}$.
\begin{lemma}
In the notation above, we have that
\begin{align}
\label{trace}
    \int_{q\cap F}|u|^2\leq C\left(\frac{1}{h}\int_{q\cap\Omega}|u|^2+h\int_{q\cap\Omega}|\nabla u|^2\right),
\end{align}
where $C$ is independent of $h$.
\begin{proof}
By the co-area formula and the Lipschitz bound of $g$,
\begin{align*}
    \int_{q\cap\Omega}|u|^2\geq \frac{1}{1+\Gamma}\int_{q\cap\Omega}|u|^2|\nabla \varphi|=\frac{1}{1+\Gamma}\int_{-2\Gamma h}^{0}\left(\int_{\{\varphi=t\}}|u|^2\right)dt,
\end{align*}
from which it follows that there is a $t_0\in(-2\Gamma h,0)$ such that
\begin{align*}
    \int_{\{\varphi=t_0\}}|u|^2\leq \frac{C}{h} \int_{q\cap\Omega}|u|^2,
\end{align*}
with $C$ depending on $\Gamma$. Take now a point $(x,g(x))\in q\cap F$; we have that
\begin{align*}
    |u(x,g(x))|^2\leq 2|u(x,g(x)+t_0)|^2+2|u(x,g(x))-u(x,g(x)+t_0)|^2
    \\
    \leq 2|u(x,g(x)+t_0)|^2+Ch\int_{-2\Gamma h}^{0}|\nabla u|^2(x,g(x)+t)dt.
\end{align*}
Integrating along $q\cap F$, we obtain
\begin{align*}
    \int_{q\cap F}|u|^2\leq 2 \int_{\{\varphi=t_0\}}|u|^2+ Ch\int_{q\cap\Omega}|\nabla u|^2 \leq C\left(\frac{1}{h}\int_{q\cap\Omega}|u|^2+h\int_{q\cap\Omega}|\nabla u|^2\right).
\end{align*}
\end{proof}
\end{lemma}

The almost rectangle $q$ identified in Claim \ref{special} will be fixed; we call the point $(x_q,g(x_q))$ the center of $q$. We denote by $(1/4)q$ the almost rectangle with center $(x_q,g(x_q))$ and with side lengths the ones of $q$ divided by $4$. By the choice of $h$, we have that
\begin{align}
\label{deep}
    \frac{1}{4}q\cap (\Omega\setminus\mathcal{D})\neq \emptyset.
\end{align}
(Otherwise, $(1/4)q\cap\Omega \subset \mathcal{D}$ and hence $\mathcal{D}$ would contain the intersection with $\Omega$ of a ball of radius $2\rho(\mathcal{D})$, which contradicts the definition of modified inner radius). Take now any point $p$ in the above intersection; following the notation of \cite{M2}, we denote by $hole(p)$ the nodal domain of $u_{{\lambda}}$ that contains $p$. By the maximum principle, $hole(p)$ must intersect $\partial(q\cap\Omega)$. There are essentially 3 configurations that can arise; we will analyse them separately.

\subsection*{Configuration 1:} $hole(p)\subset q\cap \overline{\Omega}$. This means that $hole(p)$ exits $q\cap\Omega$ through $\partial\Omega$, that is it intersects $\partial(q\cap\Omega)$ only on $\partial\Omega$. As in Section 2 and 3, we denote by $\sigma_1(q\cap\Omega)$ the first eigenvalue of the Steklov-Dirichlet problem with Steklov condition on $q\cap F$, Dirichlet on $\partial(q\cap\Omega)\setminus F$. The following claim is proved with an easy Poincaré inequality.
\begin{claim}
The estimate
\begin{align*}
    \sigma_1(q\cap\Omega)\geq \frac{c}{h}
\end{align*}
holds with a constant $c>0$ independent of $h$.
\begin{proof}
Let $v$ be any function in $H^1(q\cap\Omega)$ which has zero trace on $\partial(q\cap\Omega)\setminus F$. In particular, with $\varphi$ as in \eqref{map}, $v=0$ on $\{\varphi=-2\Gamma h\}$. For $(x,g(x))\in F$, we then have
\begin{align*}
    |v(x,g(x))|^2=|v(x,g(x))-v(x,g(x)-2\Gamma h)|^2=|\int_{-2\Gamma h}^{0} \partial_y v(x,g(x)+t)dt|^2\\
    \leq 2\Gamma h\int_{-2\Gamma h}^{0}|\nabla v|^2 (x,g(x)+t)dt.
\end{align*}
Integrating along $F$, we obtain
\begin{align*}
    \int_{q\cap F}|v|^2\leq Ch \int_{q\cap\Omega}|\nabla v|^2;
\end{align*}
the claim now follows by recalling the variational characterization of $\sigma_1$ in \eqref{eigenvalue}.
\end{proof}
\end{claim}
Since $hole(p)$ is a nodal domain of $u_{{\lambda}}$, we have that
\begin{align*}
    {\lambda}=\frac{\int_{hole(p)}|\nabla u_{{\lambda}}|^2}{\int_{\overline{hole(p)}\cap F}|u_{{\lambda}}|^2}.
\end{align*}
Recall that $hole(p)\subset q\cap \overline{\Omega}$; the function $u_{{\lambda}}\chi_{hole(p)}$ is therefore in competition in the inf defining $\sigma_1(q\cap\Omega)$. By the above claim, we then obtain
\begin{align*}
    h\geq \frac{c}{{\lambda}}.
\end{align*}
Recalling that $h=100\rho(\mathcal{D})$, Theorem \ref{lower} is proved in this configuration.

\subsection*{Configuration 2:} $hole(p)$ intersects $\partial(q\cap\Omega)$ only in $\{\varphi=-2\Gamma h\}$. In this situation, since $p\in (1/4)q$, every curve of the form $\{\varphi=t\}$ for $t\in(-2\Gamma h,-\Gamma h)$ intersects the zero set of $u$. Consider the set
\begin{align*}
    E_1=\{(x,y)\in q: g(x)-2\Gamma h\leq y \leq g(x)-\Gamma h\}.
\end{align*}
\begin{claim}
Under the above hypothesis,
\begin{align*}
    \int_{E_1}|u|^2\leq Ch^2 \int_{E_1}|\nabla u|^2.
\end{align*}
\begin{proof}
    By the co-area formula,
    \begin{align*}
        \int_{E_1}|u|^2\leq \int_{-2\Gamma h}^{-\Gamma h}\left(\int_{\{\varphi=t\}}|u|^2\right)dt;
    \end{align*}
    since for any $t\in(-2\Gamma h,-\Gamma h)$ there is a point $p_t$ on the curve $\{\varphi=t\}$ such that $u(p_t)=0$, a change of variables and a 1-dimensional Poincaré inequality give
    \begin{align*}
        \int_{\{\varphi=t\}}|u|^2\leq Ch^2 \int_{\{\varphi=t\}}|\nabla u|^2.
    \end{align*}
    Plugging it in the above and using again the co-area formula we obtain
    \begin{align*}
        \int_{E_1}|u|^2\leq Ch^2\int_{-2\Gamma h}^{-\Gamma h}\left(\int_{\{\varphi=t\}}|\nabla u|^2\right)dt\leq Ch^2\int_{E_1}|\nabla u|^2|\nabla \varphi|\leq C'h^2 \int_{E_1}|\nabla u|^2.
    \end{align*}
\end{proof}
\end{claim}

Note that there is at least one $t_0\in(-2\Gamma h,-\Gamma h)$ such that
    \begin{align*}
        \int_{\{\varphi=t_0\}}|u|^2\leq \frac{C}{h}\int_{E_1}|u|^2.
    \end{align*}
We now procede as in the proof of \eqref{trace}: for $(x,g(x))\in q\cap F$,
\begin{align*}
    |u(x,g(x))|^2\leq 2|u(x,g(x)+t_0)|^2+2|u(x,g(x))-u(x,g(x)+t_0)|^2
    \\
    \leq 2|u(x,g(x)+t_0)|^2+Ch\int_{-2\Gamma h}^{0}|\nabla u|^2(x,g(x)+t)dt.
\end{align*}
Integrating along $F$ and using Claim 3 we obtain
\begin{align*}
    \int_{q\cap F}|u|^2\leq 2\int_{\{\varphi=t_0\}}|u|^2+Ch\int_{q\cap\Omega}|\nabla u|^2\leq \frac{C}{h}\int_{E_1}|u|^2+Ch\int_{q\cap\Omega}|\nabla u|^2\\
    \leq Ch\int_{E_1}|\nabla u|^2+Ch\int_{q\cap\Omega}|\nabla u|^2\leq Ch \int_{q\cap\Omega}|\nabla u|^2,
\end{align*}
which is the inequality \eqref{poin} we needed to show to prove Theorem \ref{lower}.

\subsection*{Configuration 3:} $hole(p)$ intersects $\partial(q\cap\Omega)$ in at least one of the vertical sides of $q$; say it crosses the left side. Since $p\in(1/4)q$, we have that every line segment of the form \begin{align*}
    I_s=\{(x,y)\in q\cap \Omega: x=s\}
\end{align*}
for $s\in (x_q-h,x_q-h/2)$ intersects the zero set of $u$. Let
\begin{align*}
    E_2=\bigcup_{s\in (x_q-h,x_q-h/2)}I_s.
\end{align*}
The following claim is proved in the same way as Claim 3.
\begin{claim}
\begin{align*}
    \int_{E_2}|u|^2\leq Ch^2\int_{E_2}|\nabla u|^2.
\end{align*}
\end{claim}
By the co-area formula we have that there exists some $s_0$ such that \begin{align*}
    \int_{I_{s_0}}|u|^2\leq \frac{C}{h}\int_{E_2}|u|^2.
\end{align*}
Consider now any point $p\in q\cap\Omega$, which we write as $p=(s,g(s)-t)$; let $p_0=(s_0,g(s_0)-t)\in I_{s_0}$. Note that both $p$ and $p_0$ lie on the curve $\{\varphi=t\}$. We have that
\begin{align*}
    |u(p)|^2\leq 2|u(p_0)|^2+2|u(p)-u(p_0)|^2\leq 2|u(p_0)|^2+Ch\int_{\{\varphi=t\}}|\nabla u|^2.
\end{align*}
We integrate the above over $q\cap\Omega$ using the co-area formula:
\begin{align*}
    \int_{q\cap\Omega}|u(p)|^2\leq \int_{-2\Gamma h}^0 \left(\int_{\{\varphi=t\}}|u(p)|^2\right)dt\leq Ch\int_{-2\Gamma h}^0|u(p_0)|^2dt \\ + Ch^2\int_{-2\Gamma h}^0 \left(\int_{\{\varphi=t\}}|\nabla u|^2\right)dt.
\end{align*}
By the choice of $s_0$ and Claim 4, we then obtain
\begin{align*}
    \int_{q\cap\Omega}|u|^2\leq Ch^2\int_{q\cap\Omega}|\nabla u|^2.
\end{align*}
Finally, by the above and the trace inequality \eqref{trace}, we get
\begin{align*}
    \int_{q\cap F}|u|^2\leq Ch \int_{q\cap\Omega}|\nabla u|^2.
\end{align*}
This is once again the desired inequality \eqref{poin}, which concludes our analysis. Theorem \ref{lower} is proved.

\subsection*{Comments about higher dimensions} The proof works in dimension two because the length of some projection of $hole(p)$ on a side of the almost rectangle is large enough (independent of ${\lambda}$). In higher dimensions we cannot quite say the same. For Laplace eigenfunctions, a partial workaround is used in \cite{M}: one estimates the local asymmetry of an eigenfunction, that is the volume of the set of positivity in a ball, and then uses a Poincaré inequality of Maz'ya \cite{MZ} involving capacities. This gives a lower bound on the inner radius for Laplace eigenfunctions that is polynomial in the reciprocal of the eigenvalue (see \cite{M} for the dimensional exponent); as far as we know, in dimension greater than two the conjectured optimal lower bound has not been obtained.
A similar approach can be used for Steklov eigenfunctions, if the boundary is regular enough; in this case one can extend a Steklov eigenfunction across the boundary, which facilitates the proof of a local asymmetry estimate for the eigenfunction near the boundary. One can then prove a Poincaré inequality in the spirit of \cite{MZ} and obtain a lower bound on the inner radius that is polynomial in ${\lambda}^{-1}$ (with an explicit dimensional exponent). As mentioned in the introduction, we choose not to give the details in the present work and reserve it to subsequent work.

\section*{Acknowledgements}
This work owes a lot to the patient guidance of Eugenia Malinnikova; discussing the article with her has contributed greatly to both the contents and the presentation, and her reading of various drafts helped spot mistakes and inaccuracies. Many thanks are due to Aleksandr Logunov for his encouragement, as well as for reading a previous draft and asking useful questions. Iosif Polterovich read this work at a previous stage and provided a wealth of helpful comments, for which I am very grateful. I also wish to thank the reviewers for their thorough reading and several suggestions that helped improve the presentation.

This work was started while I was a Visiting Student Researcher at the Department of Mathematics at Stanford University; it is a pleasure to thank the department for the hospitality and the nice working conditions.

The author is supported by Project 275113 of the Research Council of Norway.

redpajama_set_name: RedPajamaArXiv


"'Bharat' welcomes you back home Priyanka. See you soon. By the way humari film Hindi hai (our film is in Hindi)," Salman tweeted on Wednesday with a wink emoji.
Actress Priyanka Chopra, who has been busy with international projects, reminded Salman Khan that she is a "desi girl forever" after he welcomed her on board for the Bollywood project "Bharat" with a quirky message.
Priyanka, who is teaming up with Salman on the big screen after a gap of 10 years, responded: "UP Bareilly ki hun janaab... Desi girl forever. Very happy to be a part of 'Bharat' and see all of you on set."
Priyanka is doing a Bollywood project after a while, and so the excitement about her "homecoming" is at its peak.
"Bharat" will be directed by Ali Abbas Zafar, with whom she has earlier worked in "Gunday".
"Swagat hai aapka Priyanka. 'Bharat' ki mitti aur hawa ko apka intezaar hai. (We welcome you Priyanka. The soil and air of 'Bharat' is waiting for you)," Zafar tweeted.


text:
import {Injectable} from '@angular/core';

/* Datepicker data that requires internationalization. */
@Injectable()
export class MdDatepickerIntl {
  /* A label for the calendar popup (used by screen readers). */
  calendarLabel = 'Calendar';

  /* A label for the button used to open the calendar popup (used by screen readers). */
  openCalendarLabel = 'Open calendar';

  /* A label for the previous month button (used by screen readers). */
  prevMonthLabel = 'Previous month';

  /* A label for the next month button (used by screen readers). */
  nextMonthLabel = 'Next month';

  /* A label for the previous year button (used by screen readers). */
  prevYearLabel = 'Previous year';

  /* A label for the next year button (used by screen readers). */
  nextYearLabel = 'Next year';

  /* A label for the 'switch to month view' button (used by screen readers). */
  switchToMonthViewLabel = 'Change to month view';

  /* A label for the 'switch to year view' button (used by screen readers). */
  switchToYearViewLabel = 'Change to year view';
}


text:
'Arrow' Star Stephen Amell Gets Completely Naked For Hot New Poolside Pic
Celebrities, Entertainment, Fashion and Lifestyle, Hollywood, Hollywood Actors, Hollywood Celebrities
pics from Sonam Kapoor and Anand Ahuja's wedding ceremony that's winning the internet!!
Bollywood, Bollywood Actors, events, Fashion and Lifestyle, Hollywood Celebrities
Emmy Awards 2017 red carpet: Priyanka Chopra, Nicole Kidman, Viola Davis, Issa Rae were the best dressed at the event
events, Fashion and Lifestyle, Red Carpet
EXCLUSIVE! "Main Gay Nahin Hoon", says Sajid Khan
Sajid Khan talks about Ram Gopal Varma, his alleged brawl with his watchman, his sexual inclination and much more…
Not many are aware of it but Sajid Khan made his directorial debut with the short film in Ram Gopal Varma's Darna Mana Hai. Interestingly, the man who is known for his comedy as a performer and a filmmaker chose to make his directing debut with a horror. And for all those who haven't seen the film, it's quite good. Anyways, when I caught up with Sajid Khan last month on the occasion of the 10 anniversary of Heyy Babyy, his first full length feature film, I asked him about his association with Ram Gopal Varma, how he surprised everyone by choosing to direct a horror film and why RGV shouldn't be written off despite a slew of flops. In a BollywoodLife exclusive he said that Ram Gopal Varma shouldn't be written off just because he gave few flops. He asserted that it all depends on a Friday. He even included Hollywood films as well and stated that every five years a big star is written off but once he delivers a successful film people go, "Oh he's back!"
He however mentioned that it's sad if people are saying that he's finished and maintained that people might be saying the same about him. Giving examples of Hollywood stars like Nicholas Cage, Robert Downey Jr and John Travolta, he said that all it takes is one Friday for anyone to bounce back.
Further talking about his horror short in Darna Mana Hai, he revealed that he is a very "Bhoot fearing person". He also revealed that he gets very scared of the paranormal activities. He also claimed that the day he decides to make a scary film, he'll end up surprising many with the shock value in it.
He also spoke about the rumours that have been doing the rounds about his drinking habits, hitting his watchman or even his sexual inclination. He refuted all of it and went on to spell it out for each and everyone when he said, "Main Gay Nahin Hoon."
Bigg Boss 11: Hina Khan, Luv Tyagi, Sapna Choudhary – who should be evicted this week?
Secret Superstar celeb review: Aamir Khan and Zaira Wasim's film leaves Rajkummar Rao, Nawazuddin Siddiqui impressed – read tweets
Rustom Movie Review


text:
Matthew Burke (born 4 February 1997) is an Irish rugby union player who is currently a member of the Connacht academy. He plays as a prop, and represents Corinthians in the All-Ireland League.

Early life
Born in Lewisham, London, Burke moved to Ireland when he was one year old, settling in Ballinrobe, County Mayo, where his fathers parents were from. He first began playing rugby aged 6 with the local club in Ballinrobe, before joining Galwegians in 2015, and represented Connacht at various age grade levels, as well as winning a cap for the Ireland schoolboys team in February 2015.

Connacht
Whilst still in the academy, Burke made his senior competitive debut for Connacht in their 34–14 away defeat against English side Sale Sharks during the 2018–19 Challenge Cup on 20 October 2018. Burke signed a professional contract with the province in April 2019, which will see him join the senior squad ahead of the 2019–20 season.

References

External links
Connacht Academy Profile
Pro14 Profile

1997 births
Living people
People from Lewisham
Irish rugby union players
Galwegians RFC players
Connacht Rugby players
Rugby union props
Rugby union players from Lewisham

redpajama_set_name: RedPajamaWikipedia


If we desire to be successful we require having the willingness to give up, sacrifice, suffer, and accept anything on the root of success. Few among us expect success to be automatic, instant, and excessively effortless.
We usually call these folks 'dreamers'. Instead of being just dreamers try to focus on the fact that, with hard work, you can make your dreams a reality.
Don't allow anything to stand in your way – not even the negative words of others.
Being human, we're all susceptible to not following through on our goals.
We get occupied, lazy, or find ourselves fighting due to health concerns.
However, if we truly want to surmount our fears and succeed in achieving our goals, we have to get up and work hard.
The ultimate success of our lives depends on the actions of ourselves and not others, it is important that we try at all times to carry ourselves with self-esteem and confidence.
It doesn't really matter how many others have or don't have faith in us, what matters is how much faith we have in ourselves.
It's more than you crave it so bad that you want to breathe that's not enough, you got to be willing to face every fear, climb every mountain, defy every odd, on-slot every obstacle.
Take one step at a time, one day at a time, only you are left standing – there's no one else and there's nothing in your way. Until you reach the top.


Not everyone needs an Ecommerce site but they do need an amazing Corporate website or as some may call it a company website. Either way this type of site demands just as much attention as do the product or service selling sites. It is important to build your brand and the ideal way of doing this is will a Corporate website that is going to represent you well. One big mistake that is made with the business website is that once it has been fully built it is often forgotten about. Just take a look around the web and you will see a multitude of beautifully designed Company websites however they have not been updated for years. There is no date but you cannot just assume that all the information now contained here is fresh and up to date. You may after purusing the site for a while soon detect something is not quite right.
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The faithful church listens to the orthodox traditions of the law and the prophets, and especially in the ways these are interpreted by Jesus, and most radically in the Sermon on the Mount (Matthew 5-7). Orthodoxy can be abused, in our temptations toward empire or our captivity to ego, but this is our error—this does not negate the need for orthodoxy, which is the consensus of the church's vision about God and neighbor, and our relationship with each. Orthodoxy is not sufficient in and of itself; our right opinions "are the slenderest part of religion", John Wesley wrote, and yet our thinking, praying and praising form a root system that helps us to bear fruit.
My prayer is that those who gather in Portland share a common orthodox faith—one that envisions the Triune God, incarnate in Jesus Christ, at the center; one that professes our need for salvation through his grace; one that depends on the power of the Holy Spirit which is not a feeling or emotion, but rather the source of spiritual gifts and the voice that guides us into the truth. This common orthodox faith is present for us in the depth and diversity of the scriptures, which lead us in paths of righteousness and reform and renew us as God's people.
+For more on faithfulness, see "Confessing and Reconciling," here, and "Always Being Reformed According to the Word of God," here.
The fruitful church knows that she does not exist for herself, but to nurture and give life to new generations of men and women who will discover paths of faithfulness. For us, this is marked by the Methodist way of being Christian—we believe that there are ordinary channels through which the grace of God flows—reading scripture, receiving communion, offering praise to God, being accountable to other disciples, living with the poor, seeking to transform unjust systems. These practices, repeated faithfully, become habits and disciplines, and in God's time and through God's providence, they bear fruit. A fruitful church is clear about her mission: to make disciples of Jesus Christ, for the transformation of the world (Book of Discipline, 120).
I am praying for a fruitful, generative church, one that grows spiritually, missionally and numerically. The diminishing presence of the disciples of Jesus in many strategic regions of our nation and world is cause for lament and self-examination. The signs of life—new disciples, younger disciples, more diverse disciples—are occasions for rejoicing, stories to be shared and practices to be replicated.
+For more on fruitfulness, see "Making Disciples in a Postmodern World," here and "Generative Christians, Generative Congregations," here.
The unity of the church is God's gift (Acts 2) and Jesus' prayer (John 17). There are many who confess that unity is not a present reality, or admit that it is not desirable, and state that unity comes at the cost of a compromise they cannot accept. Some place unity in opposition to truth, and yet this is artificial—the Lord who prays for the unity of his disciples, in order that the world might believe (John 17) is the One who is the way, the truth and the life (John 14).
The culture is our nation is increasingly fragmented. The climate of our political processes is undeniably toxic. The condition of relationships within families is often fragile. The church at every level, from a local congregation to the general conference, often reflects these realities. The call to unity in the present moment is a response to the challenge of the Apostle Paul: "Do not be conformed to the world, but be transformed by the renewing of your minds" (Romans 12). And the call to unity, for disciples, is inseparable from the ministry of reconciliation (2 Corinthians 5) that is God's dream for the world.
+For more on unity, see "A Catholic Spirit Reconsidered" here, and "Treasure Hidden in a Field" here.
How are we to think and believe? How are we to worship and pray? How might we live with each other? These are at the heart of my prayers for the church, and specifically for her faithfulness, fruitfulness and unity. Because I know—and am known by—a merciful God, I believe that God desires these qualities in us. I have the sense that when I pray, I am joining in a prayer that the Holy Spirit is already breathing. Perhaps this is why these three qualities—faithfulness, fruitfulness, unity—come to me so naturally, almost as natural as breathing when I stop to kneel and pray.
Across the Florida Annual Conference, we have set aside today as a time to kneel, wherever we are, and to pray for the 2016 United Methodist General Conference. I hope you will join me—and I am bold to say, I hope you will join the Holy Spirit—in prayer for the faithfulness, fruitfulness and unity of the United Methodist Church.


n. 1. A measure of length or distance, varying in different countries from about 2.4 to 4.6 English statute miles of 5,280 feet each, and used (as a land measure) chiefly on the continent of Europe, and in the Spanish parts of America. The marine league of England and the United States is equal to three marine, or geographical, miles of 6080 feet each.
1. An alliance or combination of two or more nations, parties, organizations, or persons, for the accomplishment of a purpose which requires a continued course of action, as for mutual defense, or for furtherance of commercial, religious, or political interests, etc.
'Twixt us and them no league, nor amity.
2. (Sports) An association of sports teams that establishes rules of play, decides questions of membership in the league, and organizes matches between the member teams. In some cases a sports league is called a conference, as in the National Football Conference.
an alliance of Roman Catholics formed in 1576 by influence of the Duke of Guise for the exclusion of Protestants from the throne of France.
an association, organized in Dublin in 1879, to promote the interests of the Irish tenantry, its avowed objects being to secure fixity of tenure, fair rent, and free sale of the tenants' interest. It was declared illegal by Parliament, but vigorous prosecutions have failed to suppress it.
v. i. 1. To unite in a league or confederacy; to combine for mutual support; to confederate.
v. t. 1. To join in a league; to cause to combine for a joint purpose; to combine; to unite; as, common interests will league heterogeneous elements.


The first R8 arrived back in 2007, and we weren't expecting much from it. Audi had no history of making supercars, so anything better than just ok would have been hailed as a success.
As it was, the R8 turned out to be a sensationally brilliant car. It was perhaps a little clinical in its execution, lacking some of the flair of its Italian and even other German rivals, but the balance it struck between being fast and practical was judged near perfectly.
Now it's time for the new model, and that means there's currently only a 5.2-litre V10 engine to choose from and just the one gearbox – the manual has gone as there was little demand, so you're stuck with a seven-speed semi-auto box.
That engine is available in two forms, the V10 Plus producing a joyous 602bhp or a plain old V10 built for wimps that manages just 533bhp. Naturally, it's the former that I'm testing. That power translates to a 0-62mph time of just 3.2 seconds and a maximum speed of 205mph.
What that means in reality is that you'll barely ever have to think about an overtake as a simple stab of the throttle makes the scenery turn all blurry and drags the horizon towards you at an alarming rate of knots, accompanied by a dramatic boom from the exhausts, with extra pops, bangs and crackles as you lift off the throttle. With the drive selector in Sport, it's pure drama.
Grip levels match the volume, with tyres as wide as the Mississippi gripping tenaciously. Even on the twisty wet roads I faced on the launch, you've got to be doing something seriously wrong to upset the car. It's unstoppable, like a supersonic freight train with Formula 1 reactions. Act the fool and the four-wheel drive quattro system will help keep physics at bay just a little longer as it sends the power to the front or rear, or to any kind of split between the two.
Switch it to Comfort and things change dramatically. The adaptive suspension softens up, the exhaust quietens down to a subtle rumble and the gearbox shifts up gears earlier. It turns it in to a sensible supercar that you could use everyday, assuming you don't need practicalities like rear seats.
The interior is reminiscent of the latest Audi models, but turned up to 11. A 12.3-inch 'virtual cockpit' dominates the driver's view, the system replacing the usual dials with a digital screen displaying everything from engine modes and driving data to music information and sat-nav instructions. The sat-nav, if you've got a SIM card plugged in, even downloads graphical data from Google Earth to present a real-world map of the road ahead.
It's of typical high quality, with delightful design touches (such as the pistol trigger heating controls) that make the cabin feel every bit as premium as the asking price suggests it should be.
It's even possible to make a reasonable case for the R8 with regard to practicality, but the lack of rear seats will always hamper it. The remaining two seats are spacious enough, with a small amount of storage behind them, while a boot at the front of the car will take a couple of weekend bags.
Car tax is at the highest category, as you might expect, meaning group M at £505 a year. If you're lucky enough to get an R8 as a company car, there's a Benefit in Kind rate of 37%. If you're not, you'll be looking at fuel economy of 21.9mpg, although it's surprisingly frugal when cruising along the motorway. There's even cylinder-on-demand as standard that turns half the engine off when it's trundling along.
Only the options list lets the side down. After you've spent your £132,715, you're faced with a list of choices, some of which you might feel are essential. Laser headlights that light up twice as much tarmac as anything else? £3,000. A reversing camera to combat the dreadful rear visibility? £600. Sports exhaust, dynamic steering, adjustable dampers and the nice Nappa leather interior? That's more than £5,000. Even cruise control is a £275 extra.
Despite those small flaws, make no mistake: The R8 is a masterpiece, a brilliant execution of what a sports car should be. That 5.2-litre engine delivers perfectly linear power, always allowing you to assess exactly what power you have available and how the car will react.
Audi's halo model rivals virtually anything on the road, and that includes such exalted competitors as those from Porsche, Ferrari and even McLaren.
It really is that good, as it should be when you consider that 56% of the car is shared with Audi's utterly dominant Le Mans car. And when you consider that, it's something of a bargain, options list or not.