New Book on the History of Pavement Art!
The FIRST book of its kind to tell the history of pavement art.
From its origins in pre-history to the Victorians, Edwardians and the present day, this new book will cover it all. Author Philip Battle decided to crowd fund this publication through Sponsume (See more at: http://www.sponsume.com/project/all-my-own-work-history-pavement-art#sthash.ihZDxgKT.dpuf). His goal is to raise £8,000 by October 6, 2014.
Authored by professional pavement artist & writer; Philip Battle, forwarded by the world acknowledged master of 3D pavement art KURT WENNER, and edited by professional publisher and copywriter Fiona Shaw.
It includes the personal stories of individual artists, plucked from historical obscurity and told here for the very first time; like Alice G Colman (1874-1934), Britain's first lady pavement artist. Liverpool's own child pavement artist James William Carling (1857-1887) with the personal stories of many others, and the social context in which they lived. It brings the history right up to date with a description of modern day pavement art; festivals, events and the popularization of the art-form across the globe, with examples of 3D anamorphic art, and other modern trends. It also includes lots of rare photos of the artists, that have been painstakingly collected by Philip over the years.
This is your chance to be a part of street painting history and help a fellow skreever and street painter to publish this important document. Every street artist should own a copy!
We wish Philip, an ISPS Board Member, all the luck and hope you will support this important work.
What Should I Charge For My Street Paintings?
Many artists struggle with this question. Someone wants to hire me? Ack! What should I charge? The biggest mistake is to throw out a quick number, without thinking it through.
First - ask for more details. Try to get them to email the details to you in written format. This gives you time to think and plan. And if they don't give you enough information, ask more questions.
How large do you want the art?
Do you want a reproduction or will I need to design custom art?
How many people will need to approve the art?
What dates do you want the art done?
Third - Submit a written quote, either in an email format or fax. Spell out as much of the details as you can, so if there is a problem, it will be caught early, before you are committed. If you need to purchase tickets for travel, request at least half up front, so you don't get stuck with the cost.
Fourth - have the client sign and date the quote and return to you, as their approval of the costs. Keep all you emails and faxes until you are paid the full amount.
Fifth - The remaining amount should be paid once the art has been completed (rain or shine). Usually clients will give you a check at the event, or you can have money transferred electronically.
Do not undervalue your art or your time! You need to live and make a living just like everyone else. You may encounter some clients who think you will donate your time for free, but there are plenty of others out there that have a budget and will pay you. Who do you want to work for? If all of the artists charge fair prices for their time and art, we all win in the end.
How to treat a featured artist.
Many new chalk festivals want to have great street art at their events. So they set out to attract great artists, but go about it the wrong way. Event planners need to think of chalk artists/street painters in the same way that they think about any entertainment that they hire to enhance their event. Most events have a budget set aside for entertainment (bands, kids activities, etc.). Very few events can happen successfully without "entertainment." It's what attracts people to your event.
Once you see street painting as entertainment, it then makes sense to treat the chalk artists just like you would a band. Some events want a lot of featured artists, and others only want 2 or 3. It can work either way, just make sure that the artists you bring in are really good at chalk art, and will be good ambassadors of the art form and actually "entertain" while they are on site chalking. They should provide photos to be used for advertising the upcoming event, and be available for interviews with local press. They can also be brought in a day or two early to lead a street painting workshop for local artists.
Artists need to eat, too.
There are a couple of ways you can compensate chalk artists. One is to approach the artists, explain what you want and request a quote for their services. The quote should include travel costs, food and art materials. You also need to remember to include any street cleaning, barricades, security, tents, and other items in your costs when figuring out the total costs. Some events will arrange for lodging, since they get special discounts or donated rooms from local hotels in exchange for sponsorship acknowledgment.
Another option is to offer a stipend. A set amount would be given to every artist that is invited, and they can use the money for their travel, lodging, materials, food and whatever else is not covered by the event. This seems like the most "fair" way to do it, but it makes it harder to attract artists from father away, since their travel costs might be too high.
The event should include a light breakfast, coffee and juices, cold water, lunches and snacks each day. Most featured artists will bring their own supplies and chalk.
It's also nice when the event has a party or get together on the evening before the event for the artists and sponsors to get together and mingle. When the artists are working during the day, it is tough to have time to talk to everyone, and a pre-party is a great way for everyone to talk in a more relaxed atmosphere.
The event should either arrange the artist's flights or travel and pay for it, or advance money to the artist ahead (half up front is normal), so the artist can purchase their flights. The rest of the amount due should be paid at the end of the event directly to the artist.
Don't charge the artists fees to participate or for anything (t-shirts, programs, etc.). Street painting is a physically demanding art form, and the artists take it very seriously. Many started by donating many weekends to the art form. Asking them to "chip in" is like asking a volunteer at an event to pay for parking and a ticket to the event.
The important thing to remember is to treat your artists well if you want to grow your event into a premier art happening. An event that respects and treats the artists as special and talented guests will become known as a special event, and you will have your pick of great artists.
Director of the International Street Painting Society, Jennifer is a chalk artist with 9 years of street painting experience, over 25 years of advertising experience and a lifetime of experience as an artist.


From here on in I will be listing the games that I am currently working on and the games I am looking to boost now. My back log is long so having them all here is a bit too much!
Mortal Kombat - Just solo left.
Posted by MadCarbon on 24 February 15 at 18:27 | There are 5 comments on this blog post - Please log in to comment on this blog.
Glad you enjoyed this! I might get to play it by the time TLoU 3 comes out.
Definitely worth playing for sure! The controls did my head in though and I died a lot because of them. But still looks amazing considering its age!
Congratulations! That's such a good effort - Any tips for me?
Grats!! Thats an excellent completion and game!
Great job. Was this the PS3 game?
Grats! Oh god, have fun with Brutal dying in 2 shots. I'm not even kidding, it's so dumb.
Very well done. Are you getting the other ones?
will look to start on 2 hopefully later this year.
LiquidCode status: Happy Easter, where's my f**cking Easter Egg??!!!!!!
Wish I could've found a better quality vid that had that soundbyte but oh well.
Marble Blast Ultra was a fun little game Though my fave game of that style is Mercury Hg!
Yeah, was fun, takes me back. I need to play Mercury Hg, had to buy it from NA store due to it not being delisted in EU.
I mean, it being delisted in EU ofc.
TrueTrophy shared the news item The TrueTrophies News & Editorial Team Is Recruiting!


text:
\section{Introduction}
The question of whether the Schr\"odinger operator
\begin{align}\label{Schrodinger operator}
-\Delta+V
\end{align}
with a decaying potential $V$ can have eigenvalues embedded in the continuous spectrum is notoriously difficult. On physical grounds, one has good reasons to expect that the potential cannot prevent a particle from escaping to spatial infinity, due to quantum mechanical tunneling. This argument would lead one to believe that embedded eigenvalues cannot occur. It came as a big surprise in the early days of quantum mechanics when Wigner and von Neumann \cite{wigner1929merkwurdige} (see also \cite[Section~XII.13]{MR0493421} for a corrected proof) found an example of a bounded potential that decays like $1/|x|$ at infinity such that the operator \eqref{Schrodinger operator} has an embedded eigenvalue $\lambda=1$. The crucial feature of the potential is its delicate oscillatory pattern, which causes waves to be reflected coherently. On the other hand, Kato~\cite{MR0108633} proved that if $V$ is bounded and $V(x)=o(1/|x|)$ as $|x|\to\infty$, then there are no embedded eigenvalues (see  \cite{MR0276624} and \cite{MR0247300} for the case of long-range perturbations, which we will not discuss here.)

A different example leading to an embedded eigenvalue for \eqref{Schrodinger operator} was constructed by Ionescu and Jerison \cite{MR2024415}. The Ionescu-Jerison potential is bounded, decays like $1/|x|$ in one coordinate direction and like $1/|x|^2$ in the others. While the Wigner-von Neumann construction is reduced to a one-dimensional problem by spherical symmetry, the Ionescu-Jerison construction is inherently multi-dimensional. Note that the Wigner-von Neumann potential is in $L^q({\mathbb R}^d)$ for any $q>d$, while the Ionescu-Jerison potential is in $L^q({\mathbb R}^d)$ for any $q>(d+1)/2$. Koch and Tataru \cite{KochTataru2006} proved the absence of embedded eigenvalues for \eqref{Schrodinger operator} for a large class of potentials including $L^{(d+1)/2}({\mathbb R}^d)$. This is sharp in view of the Ionescu-Jerison example.

Recently, Frank and Simon \cite{2015arXiv150401144F} adapted and simplified the two examples to show that embedded eigenvalues may occur for potentials with arbitrary small $L^q$ norms, More precisely, they proved the following.
\begin{itemize}
\item[(A)] There exists a sequence of potentials $V_n\in L^q({\mathbb R}^d)$ for any $q>(d+1)/2$ such that $-\Delta+V_n$ has eigenvalue $\lambda=1$ and $\lim_{n\to\infty}\|V_n\|_{L^q({\mathbb R}^d)}= 0$.
\item[(B)] There exists a sequence of \emph{radial} potentials $V_n\in L^q({\mathbb R}^d)$  for any $q>d$ such that $-\Delta+V_n$ has eigenvalue $\lambda=1$ and $\lim_{n\to\infty}\|V_n\|_{L^q({\mathbb R}^d)}= 0$.
\end{itemize}
The potentials can be chosen real-valued, so that \eqref{Schrodinger operator} is self-adjoint. Conversely, they prove that even for complex-valued potentials $V\in L^q({\mathbb R}^d)$, with $d/2\leq q\leq (d+1)/2$ in the general case and $d/2\leq q\leq d$ in the radial case (in both cases $q\neq d/2$ if $d=2$), every eigenvalue $\lambda\in{\mathbb C}$ of \eqref{Schrodinger operator} satisfies
\begin{align}\label{Lieb-Thirring}
|\lambda|^{1-d/(2q)}\leq C_{d,q}\|V\|_{L^q({\mathbb R}^d)}.
\end{align}
Since the exponent of $|\lambda|$ is nonnegative, \eqref{Lieb-Thirring} implies the absence of eigenvalues in $(0,\infty)$ for potentials with small $L^q$ norm.
The result in the general case was proved before by Frank \cite{MR2820160} for non-embedded eigenvalues, i.e.\ $\lambda\in{\mathbb C}\setminus[0,\infty)$. In view of (A)--(B), the assumption on the upper bound for $q$ are optimal. Counterexamples for $q<d/2$ are of a very different nature and we will not consider them here. We just mention \cite{MR1809288} and \cite{MR1880829} where examples of unbounded, compactly supported potentials $V\in L^q({\mathbb R}^d)$ (with $q<d/2$) were constructed such that \eqref{Schrodinger operator} has eigenvalue $\lambda=1$.

There is an interesting connection between the above two examples (Wigner-von Neumann and Ionescu-Jerison) and counterexamples in the Fourier restriction problem. In its dual form, the latter asks for which exponents $p,q\in [1,\infty]$ the inequality
\begin{align}\label{restriction problem}
\|\widehat{g{\rm d}\sigma}\|_{L^{p'}({\mathbb R}^d)}\leq C\|g\|_{L^{q'}(S^{d-1})}
\end{align}
holds for any $g\in L^{q'}(S^{d-1})$. Here, ${\rm d}\sigma$ is the canonically induced surface measure on $S^{d-1}$. Necessary conditions for \eqref{restriction problem} to hold are
\begin{align}\label{necessary conditions restriction}
1\leq p< \frac{2d}{d+1},\quad \frac{1}{q}\geq \frac{d+1}{d-1}\frac{1}{p'},
\end{align}
see e.g. \cite{MR1232192}. Whether these conditions are also sufficient is one of major open problems in harmonic analysis, known as the restriction conjecture. For $q=2$ it is known to be true, i.e.\ \eqref{restriction problem} holds for
\begin{align}\label{Stein Tomas exponents}
1\leq p\leq \frac{2(d+1)}{d+3}.
\end{align}
This is the content of the Stein-Tomas theorem (see \cite[Proposition IX.2.1]{MR1232192}). If one restricts the class of $g$ in \eqref{restriction problem} to radial functions, then the first condition
in \eqref{necessary conditions restriction} is in fact sufficient (see \cite[Chapter 8, Proposition 5.1]{MR2827930}). The necessity follows from the fact that
\begin{equation}\label{asymptotic surface measure}
\begin{split}
\widehat{{\rm d}\sigma}(x)&=2\pi|x|^{-\frac{d-2}{2}}J_{\frac{d-2}{2}}(|x|)\\
&=2\sqrt{2\pi}|x|^{-\frac{d-1}{2}}\cos\left(|x|-\frac{\pi(d-1)}{4}\right)+\mathcal{O}(|x|^{-\frac{d+1}{2}})
\end{split}
\end{equation}
as $|x|\to\infty$ and by taking $g=1$ in \eqref{restriction problem}. In the nonradial case, the necessity of the second condition in \eqref{necessary conditions restriction} follows from the so-called Knapp example (see e.g.\ \cite{MR3243741} or \cite{MR3052498} for a textbook presentation). The connection to the problem of embedded eigenvalues is twofold. First, the numerology is such that
\begin{align*}
\mathcal{B}(L^p({\mathbb R}^d),L^{p'}({\mathbb R}^d))&=L^d({\mathbb R}^d)\quad\mbox{for}\quad p=\frac{2d}{d+1},\\
\mathcal{B}(L^p({\mathbb R}^d),L^{p'}({\mathbb R}^d))&=L^{\frac{d+1}{2}}({\mathbb R}^d)\quad\mbox{for}\quad p=\frac{2(d+1)}{d+3}.
\end{align*}
Here $\mathcal{B}(L^p({\mathbb R}^d),L^{p'}({\mathbb R}^d))$ denotes the space of bounded operators from $L^p$ to $L^{p'}$.
On the left, we have precisely the endpoints in \eqref{necessary conditions restriction}, \eqref{Stein Tomas exponents}. On the right, we have the endpoints for $q$ in (A), (B). Second, the eigenfunction used in \cite{2015arXiv150401144F} for the Wigner-von Neumann example is $u(x)=\widehat{{\rm d}\sigma}(x)w(x)$, where $w$ is a positive radial function. This is the first (and quite obvious) connection. The eigenfunction in the simplest version (the resulting potential is complex-valued in this case) of the Ionescu-Jerison example \cite{MR2024415} has the form
\begin{align}\label{Ionescu-Jerison eigenfunction}
u(x)={\rm e}^{\I x_d}(1+|x'|^4+|x_d|^2)^{-N/2},
\end{align}
where $x=(x',x_d)\in {\mathbb R}^{d-1}\times{\mathbb R}$ and $N>(d+1)/4$ (this ensures that $u\in L^2({\mathbb R}^d)$). The parabolic scaling is characteristic for Knapp type examples. In fact, it is not difficult to show that $g=\widehat{u}$ is a
superposition of infinitely many Knapp examples (see Section \ref{Section Knapp}).

The significance of the unit sphere in the above discussion is that it coincides with the Fermi surface for the dispersion relation $T(\xi)=|\xi|^2$ at energy $\lambda=1$.
For arbitrary smooth kinetic energies (dispersion relations) $T$, the Fermi surface at energy $\lambda$ (assumed to be a regular value of $T$) is given by
\begin{align}\label{Fermi surface}
M_{\lambda}=\set{\xi\in{\mathbb R}^d}{T(\xi)=\lambda}.
\end{align}
If $M_{\lambda}$ is compact and has everywhere non-vanishing Gussian curvature, the conclusion of the Stein-Tomas theorem still holds, i.e.\ \eqref{restriction problem} with $q=2$ and $p$ as in \eqref{Stein Tomas exponents} is true. Moreover, analogues of the bound \eqref{Lieb-Thirring} were proved by the author in \cite{MR3608659} for some special choices of the kinetic energy.

The aim of this note is to construct examples of embedded eigenvalues for the generalized Schr\"odinger operators, i.e.\ operators of the form
\begin{align}\label{generalized Schrodinger operator}
T(D)+V.
\end{align}
Here $V$ is a decaying potential and the translation-invariant kinetic energy operator $T(D)$ is a Fourier multiplier acting on Schwartz functions $f$ as
\begin{align}\label{def. T(D)}
(T(D)f)(x):=(2\pi)^{-d}\int_{{\mathbb R}^d}{\rm e}^{\I x\cdot\xi}T(\xi)\widehat{f}(\xi){\rm d} \xi.
\end{align}
The convention for the Fourier transformation used here is
\begin{align*}
\widehat{f}(\xi):=\int_{{\mathbb R}^d}{\rm e}^{-\I x\cdot\xi}f(x){\rm d} x,\quad f\in L^1({\mathbb R}^d).
\end{align*}
All integrals will be over ${\mathbb R}^d$ unless otherwise indicated and $D_j=-\I\partial_j$ for all $j=1,\ldots,d$.

The following are the main results of this note.
\begin{theorem}[Complex-valued potentials]\label{thm. Complex-valued potentials
Let $T:{\mathbb R}^d\to{\mathbb C}$ be smooth and polynomially bounded, and let $\lambda\in{\mathbb C}$ be a regular value. Then there exists a sequence of smooth potentials $V_n:{\mathbb R}^d\to{\mathbb C}$, $n\in{\mathbb N}$, satisfying (possibly after rotating coordinates)
\begin{align}\label{decay V nonradial}
|V_n(x)|\lesssim(n+|x_1|^2+\ldots +|x_{d-1}|^2+|x_d|)^{-1}
\end{align}
and such that $\lambda$ is an eigenvalue of $T(D)+V_n(x)$ in $L^2({\mathbb R}^d)$for every $n\in{\mathbb N}$. In particular, for any $q>(d+1)/2$, we have that $\lim_{n\to\infty}\|V_n\|_{L^q}= 0$.
\end{theorem}
\begin{theorem}[Complex-valued potentials II]\label{thm. Complex-valued potentials II}
Let $T:{\mathbb R}^d\to{\mathbb R}$ be smooth and polynomially bounded, and let $\lambda\in{\mathbb R}$ be a regular value (hence $M_{\lambda}$ is a smooth hypersurface). Assume that $M_{\lambda}$ has $k<d-1$ non-vanishing principal curvatures at some point. Then there exists a sequence of smooth potentials $V_n:{\mathbb R}^d\to{\mathbb C}$, $n\in{\mathbb N}$, satisfying (possibly after rotating coordinates)
\begin{align}\label{decay V nonradial k nonzero curvatures}
|V_n(x)|\lesssim(n+|x_1|^2+\ldots +|x_{k}|^2+|x_{k+1}|^3+\ldots+|x_{d-1}|^3+|x_d|)^{-1}
\end{align}
and such that $\lambda$ is an eigenvalue of $T(D)+V_n(x)$ in $L^2({\mathbb R}^d)$, for all $n\in{\mathbb N}$. In particular, for any $q>(k+2)/2+(d-1-k)/3$, we have that $\lim_{n\to\infty}\|V_n\|_{L^q}= 0$.
\end{theorem}
\begin{theorem}[Real-valued, radial potentials]\label{thm. Radial potentials}
Let $T:{\mathbb R}^d\to{\mathbb R}$ be a radially symmetric polynomial, and let $\lambda\in{\mathbb R}$ be a regular value. Assume in addition that $M_{\lambda}$ is compact. Then there exists a sequence of smooth radial potentials $V_n:{\mathbb R}^d\to{\mathbb C}$, $n\in{\mathbb N}$, satisfying
\begin{align}\label{V decay Radial potentials}
|V_n(x)|\lesssim(n+|x|)^{-1}.
\end{align}
and such that $\lambda$ is an eigenvalue of $T(D)+V_n(x)$ in $L^2({\mathbb R}^d)$, for all $n\in{\mathbb N}$. In particular, $\lim_{n\to\infty}\|V_n\|_{L^q}= 0$ for any $q>d$.
\end{theorem}

\begin{theorem}[Chandrasekhar-Herbst operator]\label{thm Chandrasekhar-Herbst}
Let $H=\sqrt{-\Delta+1}-1$ be the Chandrasekhar-Herbst operator with mass $1$, and let $\lambda>0$. Then there exists a sequence of smooth potentials $V_n:{\mathbb R}^d\to{\mathbb R}$, $n\in{\mathbb N}$, satisfying
\begin{align}\label{V decay Chandrasekhar-Herbst}
|V_n(x)|\lesssim(n+|x'|^2+|x_d|)^{-1}
\end{align}
holds and such that $\lambda$ is an eigenvalue of $H+V_n$, for every $n\in{\mathbb N}$.
\end{theorem}
The last theorem generalizes a recent result of \cite{MR3661408} to nonradial potentials. Finally, as an example of a matrix-valued differential operator, we consider the Dirac operator (see e.g.\ \cite{MR1219537} for the definition of the Dirac matrices $\alpha_j$).
\begin{theorem}[Dirac operator]\label{thm. Dirac}
Let $\mathcal{D}=\alpha\cdot D+\beta$ be the Dirac operator with mass~$1$ and let $\lambda\in (-\infty-1)\cup (1,\infty)$. There exists a sequence of smooth (anti-hermitian) potentials $V_n$, $n\in{\mathbb N}$, satisfying
\begin{align}\label{V decay Dirac}
|V_n(x)|\lesssim(n+|x'|^2+|x_d|)^{-1}
\end{align}
and such that $\lambda$ is an eigenvalue of $\mathcal{D}+V_n$ for every $n\in{\mathbb N}$.
\end{theorem}

\begin{remark}
All of the above theorems also hold when $\lambda$ is a critical value of~$T$. In fact, this case is easier and somewhat less interesting since the distiction between the radial and the nonradial case disappears. An inspection of the proof of Theorem \ref{thm. Complex-valued potentials} will show that if $\lambda$ is a critical value, then one can even achieve the decay $$|V_n(x)|\lesssim (n+|x|)^{-2}.$$
\end{remark}

\begin{remark}
Using \cite[Lemma 3.3]{MR3608659} together with a Birman-Schwinger argument one can show that if $T:{\mathbb R}^d\to{\mathbb R}$ is smooth and proper (i.e.\ preimages of compact sets are compact), and if $M_{\lambda}$ has everywhere nonvanishing Gaussian curvature, then $\lambda$ cannot be an eigenvalue of $T(D)+V$, provided $\|V\|_{L^{(d+1)/2}}+\|V\|_{L^{\infty}}$ is sufficiently small.
\end{remark}

\begin{remark}
Similarly, one can show that if $M_{\lambda}$ has at least $k$ nonvanishing principal curvatures at every point, then $\lambda$ cannot be an eigenvalue, provided $\|V\|_{L^{(k+2)/2}}+\|V\|_{L^{\infty}}$ is sufficiently small. Note that there is a gap between the $q$ in Theorem \ref{thm. Complex-valued potentials II} and $q=~(k+2)/2$ here. The reason is that the latter does not consider worst-case scenarios. For instance, if $T(\xi)=\xi_1^2+\ldots+\xi_{k+1}^2$ where $k<d-1$ (i.e.\ $T$ depends on less than $d$ variables), then, e.g.\ for $\lambda=1$, the Fermi surface $M_{\lambda}$ is a cylinder, and the proof of Thereom \ref{thm. Complex-valued potentials II} shows that one can get
\begin{align*}
|V_n(x)|\lesssim(n+|x_1|^2+\ldots +|x_{k}|^2+|x_{k+1}|^m+\ldots+|x_{d-1}|^m+|x_d|)^{-1}
\end{align*}
for arbitrary positive $m$. Then $V_n\in L^q({\mathbb R}^n)$ for all $q>(k+2)/2+(d-1-k)/m$. This shows that the exponent $q=(k+2)/2$ is in general optimal.
\end{remark}

\begin{remark}
By a tranference-type argument one can prove analogues of Theorems \ref{thm. Complex-valued potentials}--\ref{thm. Radial potentials} to discrete Schr\"odinger operators on $\ell^2({\mathbb Z}^d)$. In this case $T$ is a function on the torus $[0,2\pi]^d$ and acts as (by slight abuse of notation)
\begin{align}\label{def. T(D) discrete case}
(T(D)f)(n):=(2\pi)^{-d}\int_{[0,2\pi]^d}{\rm e}^{\I x\cdot\xi}T(\xi)\widehat{f}(\xi){\rm d} \xi,\quad n\in{\mathbb Z}^d,
\end{align}
on functions $f:{\mathbb Z}^d\to{\mathbb C}$ with rapid decay. Here,
\begin{align*}
\widehat{f}(\xi):=\sum_{n\in{\mathbb Z}^d}{\rm e}^{-\I n\cdot\xi}f(n)
\end{align*}
is the discrete Fourier transform of $f$. An argument involving the Poisson summation formula shows that if $f|_{{\mathbb Z}^d}$ is the restriction of a smooth function $f:{\mathbb R}^d\to{\mathbb C}$ with sufficient decay, then
\begin{align}\label{Poisson summation}
(T(D)f|_{{\mathbb Z}^d})(n)=(T_{\rm per}(D)f)(n),\quad n\in{\mathbb Z}^d.
\end{align}
Here $T_{\rm per}:{\mathbb R}^d\to{\mathbb C}$ is the periodization of $T:[0,2\pi]^d\to{\mathbb C}$. On the left hand side $T(D)$ is understood as a multiplier on ${\mathbb Z}^d$ as in \eqref{def. T(D) discrete case}, while on the right hand side $T_{\rm per}(D)$ is a multiplier on ${\mathbb R}^d$ in the sense of \eqref{def. T(D)}. Since Theorems \ref{thm. Complex-valued potentials}--\ref{thm. Radial potentials} can be applied to $T_{\rm per}$ and the eigenfunctions in the proofs may be chosen smooth and with arbitrary decay, the discrete analogues follow directly from \eqref{Poisson summation}.
\end{remark}

\begin{remark}
An interesting example where the discrete analogue of Theorem \ref{thm. Complex-valued potentials II} applies is the (standard) discrete Schr\"odinger operator
\begin{align*}
Hf(n):=-\frac{1}{4}\sum_{j=1}^d\left(f(n+e_j)-f(n-e_j)\right)+\frac{d}{2}f(n)+V(n)f(n)
\end{align*}
on $\ell^2({\mathbb Z}^d)$. The symbol of the kinetic energy is
\begin{align*}
T(\xi)=d-\sum_{j=1}^d\cos(\xi_j),
\end{align*}
and is well known (see e.g.\ \cite[Section 4.2]{MR3449177}) that in $d\geq 3$ dimensions its Fermi surface $M_{\lambda}$ has points of vanishing Gaussian curvature, i.e.\ $k<d-1$ in Theorem~\ref{thm. Complex-valued potentials II}.
\end{remark}

We briefly sketch the outline of this note. In Section 2 we prove Theorems~\ref{thm. Complex-valued potentials} and Theorem \ref{thm. Complex-valued potentials II} as well as some basic, but useful auxiliary results on mapping properties of pseudodifferential operators between spaces of functions with (anisotropic) symbol-type behavior. In Section 3 we connect the construction of the eigenfunctions used in the previous section to the Knapp example in harmonic analysis. In Section 4 we show that the potentials in Theorems \ref{thm. Complex-valued potentials}--\ref{thm. Complex-valued potentials II} may be chosen real-valued under additional assumptions on the kinetic energy (see Proposition~\ref{proposition real potential} and Corollary~\ref{corollary real valued V}). In Section 5 we prove Theorems \ref{thm. Radial potentials}--\ref{thm. Dirac}.
\section{Proofs of Theorems \ref{thm. Complex-valued potentials}--\ref{thm. Complex-valued potentials II}}

\subsection{Anisotropic symbols}
Let $\rho:{\mathbb R}^d\to[1,\infty)$ be a temperate weight, i.e.\ a continuous function with the property that there exists $s>0$ such that for every $x,y\in{\mathbb R}^d$
\begin{align}\label{def. temperate weights}
\rho(x)\lesssim \rho(y)\langle x-y\rangle^{s}.
\end{align}
Note that this implies in particular that
\begin{align}
\langle x-y\rangle^{-s}\lesssim \rho(x)\lesssim \langle x-y\rangle^{s}.
\end{align}
We fix $\gamma=(\gamma_1,\ldots,\gamma_d)\in(0,1]^d$ such that
\begin{align}\label{special case wj powers of a fixed weight bis}
\gamma_d\geq \gamma_{d-1}\geq \ldots\geq \gamma_1.
\end{align}
\begin{definition}
Let $\ell\in{\mathbb R}$ and assume that $\rho,\gamma$ satisfy \eqref{def. temperate weights}, \eqref{special case wj powers of a fixed weight bis}. We say that $f\in S^{\ell}_{\gamma}({\mathbb R}^d)$ if $f\in C^{\infty}({\mathbb R}^d)$ and for any $\alpha\in{\mathbb N}_0^d$ there exist constants $C_{\alpha}>0$ such that for all $x\in{\mathbb R}^d$
\begin{align}\label{def. Sw}
|\partial^{\alpha}f(x)|\leq C_{\alpha}\rho(x)^{\ell}(x)\rho(x)^{-\gamma_1\alpha_1}\ldots \rho(x)^{-\gamma_d\alpha_d}.
\end{align}
\end{definition}
We recall that $a\in C_{\rm pol}^{\infty}({\mathbb R}^d)$, the space of smooth,  polynomially bounded functions, if $a\in C^{\infty}({\mathbb R}^d)$ and for any $\alpha\in{\mathbb N}_0^d$, there exist constants $C_{\alpha}>0$, $m_{\alpha}\in{\mathbb R}$ such that for all $x\in{\mathbb R}^d$
\begin{align}\label{def. smooth, polynomially bounded}
|\partial^{\alpha}a(\xi)|\leq C_{\alpha} \langle\xi\rangle^{m_{\alpha}}.
\end{align}
Both $S^{\ell}_{\gamma}({\mathbb R}^d)$ and $C_{\rm pol}^{\infty}({\mathbb R}^d)$ are locally convex spaces when equipped with the seminorms
\begin{align*}
\|f\|_{n}&:=\sup_{|\alpha|\leq n}\sup_{x\in{\mathbb R}^d}|\rho(x)^{-\ell}\rho(x)^{\gamma_1\alpha_1}\ldots \rho(x)^{\gamma_d\alpha_d}f(x)|,\quad f\in S^{\ell}_{\gamma}({\mathbb R}^d),\\
\|a\|_{n,m}&:=\sup_{|\alpha|\leq n}\sup_{x\in{\mathbb R}^d}|\langle \xi \rangle^{-m}\partial^{\alpha}a(\xi)|,\quad a\in C_{\rm pol}^{\infty}({\mathbb R}^d),
\end{align*}
respectively.
Let $h\in (0,1]$ and $\xi\in{\mathbb R}^d$. By abuse of notation we write
\begin{align}\label{def. h times xi}
h\xi:=(h^{\gamma_1}\xi_1,\ldots,h^{\gamma_d}\xi_d)\in{\mathbb R}^d.
\end{align}
Here $h$ plays the role of a semiclassical parameter.
Given $a\in C_{\rm pol}^{\infty}({\mathbb R}^d)$ and $f\in S^{\ell}_{\gamma}({\mathbb R}^d)$,
we define
\begin{align}\label{def. semiclassical psdo}
(a(hD)f)(x):=(2\pi)^{-d}\lim_{\epsilon\to 0}\int\int{\rm e}^{\I (x-y)\cdot\xi}\varphi(\epsilon\xi)\varphi(\epsilon y) a(h\xi)f(y){\rm d} y{\rm d}\xi
\end{align}
where $\varphi\in\mathcal{S}({\mathbb R}^d)$ is fixed and satisfies $\varphi(0)=1$.
Note that the integral in \eqref{def. semiclassical psdo} is absolutely convergent, and an integration by parts argument (as in the proof of Proposition \ref{proposition pseudodifferential mapping properties} below) shows that the definition is independent of the choice of $\varphi$. For $h=1$ the definition \eqref{def. semiclassical psdo} coincides with the one of \eqref{def. T(D)}. We write it in this way to avoid the semiclassical Fourier transformation. We will set $h=1/n$ later on. Alternatively, we could stick to $h=1$, but then we will have to consider functions
depending on $n$ in the subsequent proofs. Both points of views are equivalent, and we freely swith between one and the other in later sections.

\begin{proposition}\label{proposition pseudodifferential mapping properties}
Let $a\in C_{\rm pol}^{\infty}({\mathbb R}^d)$
Then
\begin{align}\label{eq. mapping properties a(x,D)}
a(hD):S^{\ell}_{\gamma}({\mathbb R}^d)\to S^{\ell}_{\gamma}({\mathbb R}^d)
\end{align}
as a continuous map, with seminorm bounds independent of $h\in (0,1]$.
\end{proposition}
\begin{proof}
We prove that $g=a(hD)f$ satisfies \eqref{def. Sw} for $\alpha=0$. The proof of the general case is similar. For $f\in S^{\ell}_{\gamma}({\mathbb R}^d)$ and $\epsilon\in (0,1]$ let
\begin{align*}
g_{\epsilon}(x):=
(2\pi)^{-d}\int\int{\rm e}^{\I (x-y)\cdot\xi}\varphi(\epsilon y)\varphi(\epsilon\xi) a(h\xi)f(y){\rm d} y{\rm d} \xi.
\end{align*}
Integration by parts yields
\begin{align*}
g_{\epsilon}(x)=(2\pi)^{-d}\int\int{\rm e}^{\I(x-y)\cdot\xi}L_1^{l_1}L_2^{l_2}\varphi(\epsilon\xi)\varphi(\epsilon y)a(h\xi) f(y){\rm d} y{\rm d} \xi
\end{align*}
where $l_1,l_2\in{\mathbb N}$ and
\begin{align*}
L_1=\frac{1+(x-y)\cdot D_{\xi}}{1+|x-y|^2},\quad L_2=\frac{1-\xi\cdot  D_y}{1+|\xi|^2}.
\end{align*}
Hence, there exists $m(l_1)>0$ such that
\begin{align*}
|g_{\epsilon}(x)|\lesssim\|f\|_{l_2}\|a\|_{l_1,m(l_1)} \int\int\langle x-y\rangle^{-l_1}\langle \xi\rangle^{-l_2+m(l_1)} \rho(y)^{\ell}{\rm d} y{\rm d}\xi.
\end{align*}
By changing the roles of $x$ and $y$ in \eqref{def. temperate weights} we see that $\rho(y)\lesssim \rho(x)\langle x-y\rangle^s$. Choosing first $l_1$ so large that $-l_1+ls<-d$ and then $l_2$ so large that $-l_2+m(l_1)<-d$, we get
\begin{align*}
|g_{\epsilon}(x)|\lesssim \rho(x)^{\ell}
\end{align*}
where the implicit constant is independent of $\epsilon$ and $h$. The claim follows by letting $\epsilon\to 0$.
\end{proof}
\begin{definition}
For a smooth function $a:{\mathbb R}^d\to{\mathbb C}$ the \emph{Taylor support} of $a$ at the origin is the set
\begin{align*}
\mathcal{T}(a):=\set{\alpha\in{\mathbb N}_0^d}{\partial^{\alpha}a(0)\neq 0}.
\end{align*}
The Newton polyhedron of $a$, denoted by $\mathcal{N}(a)$, is the convex hull of the set $\set{\alpha+{\mathbb R}_+^d}{\alpha\in \mathcal{T}(a)}$. The set of vertices of $\mathcal{N}(a)$ is denoted by $\mathcal{V}(a)$ and is a subset of $\tau(a)$.
\end{definition}
\begin{proposition}\label{proposition symbols}
Let $a\in C_{\rm pol}^{\infty}({\mathbb R}^d)$, $f\in S^{\ell}_{\gamma}({\mathbb R}^d)$ and assume that \eqref{special case wj powers of a fixed weight bis} holds. Moreover, assume that
\begin{align}\label{assumption Taylor support of a}
\emptyset\neq \tau(a)\subset\Set{\alpha\in{\mathbb N}_0^d}{\sum_{j=1}^d \gamma_j\alpha_j\geq 1}.
\end{align}
Then $a(hD)f\in h  S^{\ell-1}_{\gamma}({\mathbb R}^d)$.
\end{proposition}

\begin{proof}
Choose $k\in{\mathbb N}$ such that $k\gamma_1\geq 1$. By Taylor's theorem, we have
\begin{align*}
a(\xi)=\sum_{|\alpha|\leq k}\frac{1}{\alpha!}\partial^{\alpha}a(0)\xi^{\alpha}+\sum_{|\alpha|=k}\xi^{\alpha}b_{\alpha}(\xi)
\end{align*}
where
\begin{align*}
b_{\alpha}(\xi)=k\sum_{|\gamma|=k}\frac{\xi^{\gamma}}{\gamma!}\int_0^1(1-\theta)^{k-1}\partial^{\gamma}a(\theta\xi){\rm d}\theta.
\end{align*}
Clearly $b_{\alpha}\in C_{\rm}^{\infty}({\mathbb R}^d)$. By Proposition \ref{proposition pseudodifferential mapping properties}, we thus have that $b_{\alpha}(hD)=\mathcal{O}_{S^{\ell}_{\gamma}\to S^{\ell}_{\gamma}}(1)$. By \eqref{special case wj powers of a fixed weight bis} and the choice of $k$, it follows that
\begin{align*}
\big|\sum_{|\alpha|=k}(hD)^{\alpha}b_{\alpha}(hD)f(x)\big|\lesssim h \rho(x)^{\ell-1}.
\end{align*}
Define
\begin{align*}
\mathcal{V}_k(a):=\mathcal{V}(a)\cap \set{\alpha\in {\mathbb N}_0^d}{|\alpha|\leq k}\subset\tau(a).
\end{align*}
Then we can write
\begin{align*}
\sum_{|\alpha|\leq k}\frac{1}{\alpha!}\partial^{\alpha}a(0)\xi^{\alpha}=\sum_{\alpha\in \mathcal{V}_k(a)}\xi^{\alpha}p_{\alpha}(\xi)
\end{align*}
where $p_{\alpha}$ are polynomials. In particular, $p_{\alpha}(hD)=\mathcal{O}_{S^{\ell}_{\gamma}\to S^{\ell}_{\gamma}}(1)$, so that \eqref{assumption Taylor support of a} implies
\begin{align*}
\big|\sum_{\alpha\in \mathcal{V}_k(a)}(hD)^{\alpha}p_{\alpha}(hD)\big|\lesssim h \rho(x)^{\ell-1}.
\end{align*}
The claim is proved.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm. Complex-valued potentials}]
Let $\eta\in M_{\lambda}$. Since $\lambda$ is a regular value, we have $\nabla T(\eta)\neq 0$. By rotating coordinates if necessary, we may assume that $\nabla T(\eta)=|\nabla T(\eta)|e_d$. We fix
\begin{align}\label{gamma, rho thm 1}
\rho(x):=(1+|x'|^4+|x_d|^2)^{1/2},\quad \gamma=(1/2,\ldots,1/2,1)
\end{align}
and define
\begin{align}\label{Ionescu-Jerison type example}
\psi(x):=\rho(x)^{-N},\quad u(x):={\rm e}^{\I \eta\cdot x}\psi(x),
\end{align}
where $N>(d+1)/4$. A change of variables shows that
\begin{align*}
\int_{{\mathbb R}^d}|u(x)|^2{\rm d} x=\left(\int_{{\mathbb R}^{d-1}}(1+|x'|^4)^{-N}{\rm d} x'\right)\left(\int_{{\mathbb R}}(1+|x_d|^2)^{-N+\frac{d-1}{4}}{\rm d} x_d\right)<\infty.
\end{align*}
Proposition \ref{proposition pseudodifferential mapping properties} then implies that $T(D)u\in L^2({\mathbb R}^d)$ as well. Hence $u$ is in the domain of $T(D)$. Note that since
\begin{align}\label{commutation T(D) with plane wave}
T(D){\rm e}^{\I \eta\cdot x}={\rm e}^{\I \eta\cdot x}T(D+\eta)
\end{align}
as operators on smooth functions with bounded derivatives, we may assume without loss of generality that $\eta=0$. For $n\geq 1$ let $h=1/n$ and $u_n(x):=h^Nu(hx)$, where $hx$ is defined as in \ref{def. h times xi}. If $V_n(x)=W_h(hx)$, then
\begin{align*}
(T(D)-\lambda+V_n)u_n=0\iff (T(hD)-\lambda+W_h)u=0.
\end{align*}
Since $u$ has no zeros, we may write the second equation as $W_h=-(T(hD)-\lambda)u/u$.
We will prove that
\begin{align}\label{estimate to prove in thm. Complex-valued potentials}
W_h(x)|\lesssim h(1+|x'|^2+|x_d|)^{-1}.
\end{align}
This is equivalent to \eqref{decay V nonradial}. Since \eqref{estimate to prove in thm. Complex-valued potentials} implies that $W_h\in L^q({\mathbb R}^d)$ for $q>(d+1)/2$, it also follows that
\begin{align*}
\|V_n\|_{L^q({\mathbb R}^d)}=h^{-\frac{d+1}{2q}}\|W_h\|_{L^q({\mathbb R}^d)}=\mathcal{O}(h^{1-\frac{d+1}{2q}})\to 0
\end{align*}
as $h\to 0$.
To prove \eqref{estimate to prove in thm. Complex-valued potentials}
we will apply Proposition \ref{proposition symbols} to $a(\xi)=T(\xi)-\lambda$ and $f=\psi$. It easy to check that $\psi\in S^{-N}_{\gamma}({\mathbb R}^d)$ if $\gamma,\rho$ are defined as in \eqref{gamma, rho thm 1}. Moreover,
condition \eqref{assumption Taylor support of a} is clearly satisfied. Proposition \ref{proposition symbols} now yields that $W_h\in  h  S^{-1}_{\gamma}({\mathbb R}^d)$; in particular, \eqref{estimate to prove in thm. Complex-valued potentials} holds.
\end{proof}
\begin{proof}[Proof of theorem \ref{thm. Complex-valued potentials II}]
Let $\eta\in M_{\lambda}$ be such that only $k<d-1$ principal curvatures are nonzero at this point. We can again assume that $\eta$ is the origin and that $\nabla T(\eta)=|\nabla T(\eta)|e_d$. Write $x=(x',x'',x_d)\in{\mathbb R}^k\times{\mathbb R}^{d-1-k}\times{\mathbb R}$ and fix
\begin{align}\label{gamma, rho thm 2}
\rho(x):=(1+|x'|^4+|x''|^6+|x_d|^2)^{1/2},\quad \gamma=(\underbrace{1/2,\ldots,1/2}_{k \mbox{ times }},\underbrace{1/3,\ldots,1/3}_{d-1-k \mbox{ times }},1).
\end{align}
We then define $u$ as in \eqref{Ionescu-Jerison type example}, but with $\rho$ as in \eqref{gamma, rho thm 2}.
Near the origin, $M_{\lambda}$ is the graph of a smooth function $\Phi:{\mathbb R}^{k}\times{\mathbb R}^{d-1-k}\to{\mathbb R}$ over the last coordinate axis, and $\Phi(0)=0$, $\nabla\Phi(0)=0$. By the curvature assumption, we can write (after perhaps a linear change of the $(\xi',\xi'')$ coordinates)
\begin{align*}
\Phi(\xi',\xi'')=\frac{1}{2}\sum_{j=1}^{k}s_j\xi_j^2+\mathcal{O}(|(\xi',\xi'')|^3),\quad s_j=\pm 1,
\end{align*}
as $(\xi',\xi'')\to 0$, with similar estimates for all derivatives.
By differentiating the identity $T(\xi',\xi'',\Phi(\xi',\xi''))=\lambda$ twice, we find that
\begin{align*}
\partial_{i}\partial_j T(0)=-(\partial_d T(0))^{-1}\partial_{i}\partial_j\Phi(0),\quad i,j\leq d-1,
\end{align*}
and hence the Taylor support of $T(\xi)-\lambda$ at $\xi=0$ is contained in the set
\begin{align*}
\Set{\alpha\in{\mathbb N}_0^d}{\frac{1}{2}(\alpha_1+\ldots+\alpha_{k})+\frac{1}{3}(\alpha_{k+1}+\ldots+\alpha_{d-1})+\alpha_d\geq 1}.
\end{align*}
We define $W_h:=-(T(hD)-\lambda)u/u$ and $V_n(x):=W_h(hx)$ for $h=1/n$.
As in the proof of Theorem~\ref{thm. Complex-valued potentials} it is sufficient to prove that
\begin{align}\label{estimate to prove in thm. Complex-valued potentials II}
|W_h(x)|\lesssim h(1+|x'|^2+|x''|^3+|x_d|)^{-1}.
\end{align}
Indeed, \eqref{estimate to prove in thm. Complex-valued potentials II} is equivalent to \eqref{decay V nonradial k nonzero curvatures}, and for any $q>(k+2)/2+(d-1-k)/3$, we have that
\begin{align*}
\|V_n\|_{L^q({\mathbb R}^d)}=h^{-\frac{1}{q}-\frac{k}{2q}-\frac{d-k-1}{3q}}\|W_h\|_{L^q({\mathbb R}^d)}=\mathcal{O}(h^{1-\frac{1}{q}-\frac{k}{2q}-\frac{d-k-1}{3q}})\to 0
\end{align*}
as $h\to 0$. The bound \eqref{estimate to prove in thm. Complex-valued potentials II} follows from Proposition \ref{proposition symbols}.
\end{proof}

\section{Connection to Knapp's example}\label{Section Knapp}
Let us now quickly recall Knapp's homogeneity argument (see e.g.\ \cite[Example~ 10.4.4]{MR3243741}) for the sphere $S^{d-1}$. The same argument yields a necessary condition for the restriction problem of $M_{\lambda}$ if the latter has everywhere nonzero Gaussian curvature. For the present purpose it is more convient to consider the restriction problem in its original (not adjoint) form
\begin{align}\label{original restriction problem}
\|\widehat{f}\|_{L^q(S^{d-1})}\leq C\|f\|_{L^p({\mathbb R}^d)},\quad \mbox{for all  }f\in\mathcal{S}({\mathbb R}^d).
\end{align}
This is equivalent to \eqref{restriction problem}. Since the problem is translation-invariant, we may assume that the north pole of the unit sphere is the origin of our coordinate system. Let $f=\chi_R$ be a smoothed version of the characteristic function for a rectangle $R$. Then for $0<\delta<1$ and $f_{\delta}(x):=f(\delta x',\delta^2 x_d)$ one easiy checks that
\begin{align*}
\|f_{\delta}\|_{L^p(R^{d})}\sim \delta^{-\frac{d+1}{p}},\quad
\|\widehat{f}_{\delta}\|_{L^q(S^{d-1})}\sim \delta^{\frac{d-1}{q}-d-1}.
\end{align*}
Since $\delta<1$, this shows that \eqref{original restriction problem} can only hold if the second condition in \eqref{necessary conditions restriction} is satisfied.

It is customary to call $f_{\delta}$ a Knapp example.
If we look at its phase-space portrait we see that it is supported on a rectangle $R_{\delta}$ of size $\delta\times \delta^{-2}$ and that its Fourier transform decays rapidly off the dual rectangle $$R_{\delta}^*:=\set{\xi\in{\mathbb R}^d}{|\xi|'\lesssim\delta,\, |\xi_d|\lesssim \delta^{2}}.$$ Since
\begin{align*}
T(\xi)=\lambda+(\partial_1T(0))\xi_d+\mathcal{O}(|\xi|^2)
\end{align*}
as $|\xi|\to 0$, we thus have
\begin{align*}
T(D)f_{\delta}(x)\sim T|_{R_{\delta}^*}f_{\delta}(x)\sim \lambda f_{\delta}(x)+\mathcal{O}(\delta^2).
\end{align*}
Since pseudodifferential operators do not move the support too much, this function is essentially supported (up to rapidly decaying tails) on the double of $R_{\delta}$. Cover ${\mathbb R}^d\setminus B(0,1)$ by finitely overlapping rectangles $R_j$ of size $2^j\times 4^j$, such that for $j\in {\mathbb N}$,
\begin{align*}
x\in R_j\implies |x'|^2+|x_d|\sim 4^j.
\end{align*}
This can be done by a slight modification of the standard Littlewood-Paley decomposition. Let $\{\chi_j\}_{j=1}^{\infty}$ be a partition of unity subordinate to this cover and define
\begin{align}\label{superposition of Knapp examples}
u(x):=\sum_{j=1}^{\infty} 4^{-Nj} \chi_{j}(x),\quad x\in {\mathbb R}^d\setminus B(0,1).
\end{align}
The function $u$ may be viewed as a superposition of Knapp examples $\{\chi_{j}\}_{j=1}^{\infty}$ corresponding to rectangles $\{R_{j}\}_{j=1}^{\infty}$.
It is not difficult to show that, for $x$ in the exterior of the unit ball, we have that
\begin{align}\label{connection Ionescu-Jerison to Knapp}
u(x)\sim (|x'|^2+|x_d|)^{-N}
\end{align}
as well as
\begin{align*}
T(D)u(x)\sim \sum_{j=1}^{\infty}4^{-Nj} T|_{R_{j}^*}\chi_{R_j}(x)\sim \lambda u(x)+\mathcal{O}((|x'|^2+|x_d|)^{-N-1}).
\end{align*}
Modifying $u$ in \eqref{superposition of Knapp examples} by a compactly supported function, we easily arrange that $u>0$ and that $V=-(T(D)-\lambda)u/u$ is smooth and satisfies \eqref{decay V nonradial}. The relation~\eqref{connection Ionescu-Jerison to Knapp} connects the Knapp example and the Ionescu-Jerison example (i.e.\ the eigenfunction used in the proof of Theorem \ref{thm. Complex-valued potentials}). There is a similar connection between a Knapp example for surfaces with $k$ non-vanishing principal curvatures and the eigenfunction used in the proof of Theorem~\ref{thm. Complex-valued potentials II}.

\section{Real-valued potentials}\label{Section Real-valued potentials}

\begin{lemma}\label{lemma time reversal symmetry}
Let $T\in C_{\rm pol}^{\infty}({\mathbb R}^d)$ be real-valued and time-reversal symmetric, i.e.\ $T(\xi)=T(-\xi)$. Let $\varphi(x)=\sin(\eta\cdot x)\psi(x)$ where $\psi\in C_b^{\infty}({\mathbb R}^d)$ is any real-valued function and $\eta\in{\mathbb R}^d$. Then the function $T(D)\varphi$ is real-valued.
\end{lemma}

\begin{proof}
A straightforward calculation using \eqref{commutation T(D) with plane wave} yields
\begin{align*}
T(D)\varphi(x)&=\frac{1}{2\I}\left({\rm e}^{\I\eta\cdot x}T(D+\eta)-{\rm e}^{-\I\eta\cdot x}T(D-\eta)\right)\psi(x).
\end{align*}
Changing variables $\xi\to-\xi$ in the resulting integrals and using the fact that $\widehat{\psi}(\xi)=\widehat{\psi}(-\xi)$, one finds that $\overline{T(D)\varphi(x)}=T(D)\varphi(x)$.
\end{proof}

\begin{proposition}\label{proposition real potential}
Assume that the assumptions of Theorem~\ref{thm. Complex-valued potentials} or Theorem~\ref{thm. Complex-valued potentials II} hold and that $T$ is a polynomial such that $T(\xi)=T(-\xi)$ for all $\xi\in{\mathbb R}^d$. In addition, assume that, possibly after a linear bijection $L:{\mathbb R}^d\to{\mathbb R}^d$, we have $e_1\in M_{\lambda}$ and
\begin{align}\label{technical condition real potentials}
\tau(T(e_1+\cdot)\cap\set{\alpha\in {\mathbb N}_0^d}{|\alpha|\in 2{\mathbb N}-1\mbox{ and }\alpha_1=m}=\{m e_1\}
\end{align}
for some odd integer $m$. Then the conclusion of Theorem \ref{thm. Complex-valued potentials} or Theorem \ref{thm. Complex-valued potentials II} holds with real-valued potentials.
\end{proposition}

\begin{proof}
We first note that the assumptions on $T$ are invariant under linear bijections $L:{\mathbb R}^d\to{\mathbb R}^d$. Moreover, if $u$ is a real-valued function, then so is $u_L(x):=u(^tLx)$.
Since
\begin{align*}
\widehat{u_L}(\xi)=|\det L|^{-1}\widehat{u}(L^{-1}\xi),
\end{align*}
we have that $T(D)u_L(x)=T_L(D)u(^tLx)$, where $T_L(\xi)=T(L\xi)$. If $\eta\in M_{\lambda}\setminus\{0\}$ we can arrange that $Le_1=\eta$. Since $T(\xi)=T(-\xi)$, the origin is a critical point, so there is no loss of generality in omitting it.

Assume now that the assumptions of the proposition hold with $L$ being the identity. We will prove the claim under the assumptions of Theorem \ref{thm. Complex-valued potentials}. The case of Theorem \ref{thm. Complex-valued potentials II} is analogous. We also assume that $n=1$ since the general case follows by scaling as in the proof of Theorem \ref{thm. Complex-valued potentials}.

We would like to choose the eigenfunction as in Lemma \ref{lemma time reversal symmetry}, i.e.
\begin{align*}
\widetilde{u}(x):=\sin(x_1)\psi(x)
\end{align*}
where $\psi$ is given by
\begin{align*}
\psi(x):=(1+|P_{\nu}x|^2+|P_{\nu}^{\perp}x|^4)^{-N/2}
\end{align*}
with $N>d/2$. Here $P_{\nu}$ is the orthogonal projection onto the subspace spanned by the unit vector $\nu=\nabla T(e_1)/|\nabla T(e_1)|$ and $P_{\nu}^{\perp}=\rm{id}-P_{\nu}$.
The problem is thatv$u$ now vanishes on the hypersurfaces $x_1=k\pi$, $k\in{\mathbb Z}$, so we cannot solve $(T(D)-\lambda+V)\widetilde{u}=\widetilde{u}$ for $V$ by dividing with $\widetilde{u}$. We follow the basic strategy of Ionescu and Jerison \cite{MR2024415} to remove the singularities. We are looking for a function $u$ with the same zero set (including multiplicities) as $\widetilde{u}$ such that $(T(D)-\lambda) u=0$ on this set.
We first compute
\begin{align}\label{T(D) applied to sin}
T(D)\widetilde{u}(x)&=\frac{1}{2\I}{\rm e}^{\I x_1}(T(D+e_1)-T(D-e_1))\psi(x)+\sin(x_1)T(D-e_1)\psi(x).
\end{align}
Set $f=(T(D)-\lambda)\widetilde{u}$. As in the proof of Theorem \ref{thm. Complex-valued potentials} we use Proposition \ref{proposition symbols} to infer that
\begin{align}\label{eq. derivative bounds f}
|\partial^{\alpha} f(x)|\leq C (1+|P_{\nu}x|+|P_{\nu}^{\perp}x|^2)^{-1}\psi(x),\quad |\alpha|\leq m.
\end{align}
In fact, $a(\xi)=T(\xi+e_1)-T(\xi-e_1)$ as well as $a(\xi)=T(\xi-e_1)-\lambda$ satisfy the assumptions of Proposition \ref{proposition symbols} (after relabelling coordinates). For example, in the first instance, $a(0)=T(e_1)-T(-e_1)=0$ and $\nabla a(0)=\nabla T(e_1)-\nabla T(-e_1)=2\nabla T(e_1)$. Here we used the time-reversal symmetry $T(\xi)=T(-\xi)$ and its consequence $\nabla T(\xi)=-\nabla T(-\xi)$.
Define
\begin{align}\label{def. w}
w(x):=\sum_{k\in{\mathbb Z}}\cos(k\pi)(x_1-k\pi)^m\chi(x_1-k\pi)f(k\pi,x')
\end{align}
where $x'=(x_2,\ldots,x_d)\in {\mathbb R}^{d-1}$ and $\chi\in C_c^{\infty}({\mathbb R},[0,1])$ is supported on $[-c,c]$ and equals $1$ on $[-c/2,c/2]$ for some small $c>0$ to be determined later. We then set
\begin{align}\label{def. regularized u}
u(x):=\sin(x_1)(\psi(x)-\kappa w(x)),\quad \kappa:=\left[(-1)^{\frac{m+1}{2}}\partial_1^mT(e_1)\right]^{-1}.
\end{align}
By \eqref{eq. derivative bounds f} we have
\begin{align}\label{lower bound for u}
|u(x)|\geq |\sin(x_1)|\psi(x)(1-C\kappa c^m)\gtrsim|\sin(x_1)|\psi(x)\gtrsim\mathrm{dist}(x_1,\pi{\mathbb Z})\psi(x)
\end{align}
if $c$ is chosen sufficintly small. We now fix $c<\pi/10$ such that \eqref{lower bound for u} holds. This also guarantees that the functions $\chi(\cdot-k\pi)$, $k\in{\mathbb Z}$ have mutually disjoint supports. A straightforward computation yields
\begin{align*
(T(D)-\lambda) u(x)&=f(x)
-\frac{\kappa}{2\I}\sum_{|\alpha|\geq 1}\frac{1}{\alpha!}(e^{\I x_1}-(-1)^{|\alpha|}e^{-\I x_1})\partial^{\alpha}T(e_1)D^{\alpha} w(x)\\
&=f(x)-\kappa\sin(x_1)\sum_{\alpha\in 2{\mathbb N}}\frac{1}{\alpha!}\partial^{\alpha}T(e_1)(-1)^{\frac{|\alpha|}{2}}\partial^{\alpha} w(x)\\
&-\kappa\cos(x_1)\sum_{\alpha\in 2{\mathbb N}-1}\frac{1}{\alpha!}\partial^{\alpha}T(e_1)(-1)^{\frac{|\alpha|+1}{2}}\partial^{\alpha} w(x).
\end{align*}
This together with \eqref{eq. derivative bounds f}, \eqref{def. w} implies that
\begin{align}\label{bound on (T-lambda) u}
|(T(D)-\lambda) u(x)|\lesssim (1+|P_{\nu}x|+|P_{\nu}^{\perp}x|^2)^{-1}\psi(x).
\end{align}
Using \eqref{technical condition real potentials} we check that, for $k\in{\mathbb Z}$,
\begin{align*}
&(T(D)-\lambda) u(k\pi,x')=f(k\pi,x')\\
&-\kappa
\sum_{|\alpha|\in 2{\mathbb N}-1}\frac{1}{\alpha!}\partial^{\alpha}T(e_1)(-1)^{\frac{|\alpha|+1}{2}}\partial^{\alpha}((x_1-k\pi)^m\chi(x_1-k\pi)f(k\pi,x'))|_{x_1=k\pi}\\
&=f(k\pi,x')(1-\kappa(-1)^{\frac{m+1}{2}}\partial_1^mT(e_1))=0.
\end{align*}
To see this, split the sum into a part where $\alpha=me_1$ and its complement. The first part gives exactly the expression in the last line. The complementary part can be further split into three subparts where $\alpha_1<m$, $\alpha_1>m$ or $\alpha_1=m$. The first subpart is zero because there are too few $x_1$-derivatives that can fall onto the monomial $(x_1-k\pi)^m$. The second subpart is zero because there is at least one $x_1$-derivative that must fall onto the cutoff function $\chi$, and this is constant near $x_1=k\pi$. The third subpart is zero by assumption \eqref{technical condition real potentials}.

Combining the result of the previous computation with \eqref{bound on (T-lambda) u} we get
\begin{align*
|(T(D)-\lambda) u(x)|\lesssim (1+|P_{\nu}x|+|P_{\nu}^{\perp}x|^2)^{-1}\mathrm{dist}(x_1,\pi{\mathbb Z})\psi(x).
\end{align*}
This together with \eqref{lower bound for u} yields that
\begin{align*}
|V(x)|=\frac{|(T(D)-\lambda) u(x)|}{|u(x)|}\lesssim (1+|P_{\nu}x|+|P_{\nu}^{\perp}x|^2)^{-1}.
\end{align*}
That $V$ is real-valued follows from Lemma \ref{lemma time reversal symmetry}.
\end{proof}

\begin{corollary}\label{corollary real valued V}
Assume that the assumptions of Theorem~\ref{thm. Complex-valued potentials} or Theorem~\ref{thm. Complex-valued potentials II} hold and that $T$ is a radial polynomial. Then the conclusion of Theorem \ref{thm. Complex-valued potentials} or Theorem \ref{thm. Complex-valued potentials II} holds with real-valued potentials.
\end{corollary}

\begin{proof}
The assumptions imply that $T$ is of the form $T(\xi)=T_0(|\xi|)$ where
\begin{align*}
T_0(r)=\sum_{j=0}^Kc_lr^{2j}
\end{align*}
for some $K\in{\mathbb N}$ and $(c_l)_{l=1}^K\subset{\mathbb R}$. If $\eta\in M_{\lambda}$, then $|\eta|e_1\in M_{\lambda}$. The linear bijection $L\xi=|\eta|\xi$ preserves spherical symmetry, and $T_L(e_1)=T(Le_1)=\lambda$.
We assume, as we may, that $K\neq 0$. Then
\begin{align*}
|\xi+e_1|^{2K}=(1+2\xi_1+\xi_1^2+\ldots+x_d^2)^K=\xi_1^{2K}+2K\xi_1^{2K-1}+Q_K(\xi)
\end{align*}
where $Q_K$ is a sum of monomials that do not contain the factor $\xi_1^{2K-1}$. Similarly, none of the monomials appearing in $|\xi+e_1|^{2l}$, $l\leq K-1$, contain this factor. Therefore, condition \eqref{technical condition real potentials} holds with $m=2K-1$.
\end{proof}

\section{Proofs of Theorems \ref{thm. Radial potentials}--\ref{thm. Dirac}}

We first state a generalization of Propositions \ref{proposition pseudodifferential mapping properties} and \ref{proposition symbols} to symbols $a(x,\xi)$ depending on~$x$. We restrict ourselves to the case
\begin{align}\label{rho gamma special case}
\rho(x)=(1+|x|^2)^{1/2},\quad\gamma=(1,\ldots,1);
\end{align}
more general cases will not be needed. The proof is a straightforward adaptation of that of Propositions \ref{proposition pseudodifferential mapping properties} and \ref{proposition symbols} and will be omitted.

\begin{lemma}\label{lemma psdos radial weights}
Assume that $a\in C^{\infty}({\mathbb R}^d\times{\mathbb R}^d)$ and that
\begin{align*}
|\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\leq C_{\alpha,\beta}(1+|x|)^{k-|\alpha|}(1+|\xi|)^{m_{\alpha,\beta}}.
\end{align*}
Then $a(x,h D):S_{\gamma}^{\ell}({\mathbb R}^d)\to S_{\gamma}^{\ell+k}({\mathbb R}^d)$ as a continuous map, with seminorms bounded independent of $h\in (0,1]$. Moreover, if $a(x,0)=0$ for all $x\in{\mathbb R}^d$, then $a(x,h D):S_{\gamma}^{\ell}({\mathbb R}^d)\to hS_{\gamma}^{\ell+k-1}({\mathbb R}^d)$.
\end{lemma}

\begin{proof}[Proof of Theorem \ref{thm. Radial potentials}]
We will prove the claim for $n=1$; the general case follows from a scaling argument similar to that in the proofs of Theorems \ref{thm. Complex-valued potentials}--\ref{thm. Complex-valued potentials II}.

Let $\sigma_{\lambda}$ be the uniform surface measure of $M_{\lambda}$. Since $T$ is radial, the Fermi surface $M_{\lambda}$ is a sphere and hence $\sigma^{\vee}_{\lambda}$ is radial. We fix $r_0\gg 1$ and pick a smooth radial function $\varphi$ that is positive for $|x|\leq r_0$ and equal to $\sigma^{\vee}_{\lambda}$ for $|x|>r_0$. We then define the radial function
\begin{align*}
\widetilde{u}(x):=\varphi(x)\psi(x),
\end{align*}
where $\psi(x):=(1+|x|^2)^{-N/2}$, with $N>d/2$.
A straightforward calculation using Fubini and the convolution theorem yields that, for $|x|>r_0$,
\begin{align*}
f(x):=(T(D)-\lambda)\widetilde{u}(x)=a(x,D)\psi(x)
\end{align*}
where
\begin{align*}
a(x,\xi)=(2\pi)^{-d}\int_{M_{\lambda}}{\rm e}^{\I \eta\cdot x}(T(\xi+\eta)-\lambda){\rm d}\sigma_{\lambda}(\eta).
\end{align*}
Since $M_{\lambda}$ is compact, it follows from stationary phase (see e.g.\ \cite[(1.2.8)]{MR1205579}) that
\begin{align}\label{stationary phase estimate for b}
|\partial_x^{\alpha}\partial_{\xi}^{\beta}a(x,\xi)|\leq C_{\alpha,\beta}(1+|x|)^{-\frac{d-1}{2}-|\alpha|}(1+|\xi|)^{m_{\alpha,\beta}}.
\end{align}
Since $a(x,0)=0$, Lemma \ref{lemma psdos radial weights} yields
\begin{align}\label{radial estimate}
|f(x)|\lesssim(1+|x|)^{-\frac{d-1}{2}-1}\psi(x)
\end{align}
After rescaling, we may assume that $M_{\lambda}$ is the unit sphere $S^{d-1}$.
From \eqref{asymptotic surface measure} it follows that
that the zeros of $\widetilde{u}$ are simple and the distance between consecutive zeros is uniformly bounded below by a constant $\delta>0$.
Let us denote these zeros by $Z:=\{r_k\}_{k=1}^{\infty}\subset(r_0,\infty)$.
Then we have that
\begin{align}\label{zeros of Bessel are simple}
|\widetilde{u}(x)|\gtrsim \mathrm{dist}(|x|,Z)(1+|x|)^{-\frac{d-1}{2}}\psi(x).
\end{align}
Pick $\chi\in C_c^{\infty}({\mathbb R},[0,1])$ as in the proof of Proposition \ref{proposition real potential}, i.e.\ $\chi$ is supported on $[-c,c]$ and equals $1$ on $[-c/2,c/2]$ for some sufficiently small $c>0$; in particular, we require $c<\delta/4$, so that the functions $\chi(\cdot-r_k)$ have mutually disjoint support. By assumption, $T(D)$ is a differential operator of the form
\begin{align*}
T(D)=\sum_{j=0}^mc_j(-\Delta)^j
\end{align*}
where $\{c_j\}_{j=0}^m\subset {\mathbb R}$, $c_m\neq 0$. We may clearly assume that $m\geq 1$.
Define the radial functions
\begin{align*}
w(x):=\sum_{k=1}^{\infty}(|x|-r_k)^{2m}\chi(|x|-r_k)f(r_k),
\end{align*}
and
\begin{align*}
u(x):=\widetilde{u}(x)-\kappa w(x),\quad \kappa:=c_m^{-1}(2d)^{-m}.
\end{align*}
It follows from \eqref{radial estimate} and \eqref{zeros of Bessel are simple} that
\begin{align}\label{radial 1}
|u(x)|\gtrsim \mathrm{dist}(|x|,Z)\psi(x)(1+|x|)^{-\frac{d-1}{2}}(1-\mathcal{O}(c^{2m-1})).\end{align}
Moreover, for $|x|=r_k$ we have
\begin{align*}
(T(D)-\lambda)u(x)=f_0(r_k)-\kappa c_m(2d)^mf_0(r_k)=0,
\end{align*}
where we used the notation $f(x)=f_0(|x|)$. This together with \eqref{radial estimate} yields
\begin{align}\label{radial 2}
|(T(D)-\lambda)u(x)|\lesssim \mathrm{dist}(|x|,Z)\psi(x)(1+|x|)^{-\frac{d-1}{2}-1}.
\end{align}
Hence, for $c$ suffiently small, we have by \eqref{radial 1}--\eqref{radial 2}
\begin{align*}
|V(x)|=\frac{|(T(D)-\lambda)u(x)|}{|u(x)|}\lesssim (1+|x|)^{-1}.
\end{align*}
It is clear that $V$ is radial.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm Chandrasekhar-Herbst}]
Here we combine our previous results with ideas from \cite{MR3661408} and \cite{2015arXiv150401144F}.
Observe that if $\kappa:=\sqrt{(\lambda+1)^2-1}$, then $\kappa e_d\in M_{\lambda}$.
As before, we may assume (after a rescaling) that $\kappa=1$ (and hence $\lambda=\sqrt{2}-1$).
First define
\begin{align*}
g(t)&:=\int_0^{t}\sin^2(y){\rm d} y,\\
w_n(x)&:=(n^2+|x'|^4+g(x_d)^2)^{-N/2},
\end{align*}
where $n\in {\mathbb N}$ and $N>(d+1)/4$. We then set
\begin{align*}
\phi_n(x)&:=(\sqrt{(D+e_d)^2+1}+\sqrt{(D-e_d)^2+1})w_n(x),\\
u_n(x)&:=\sin(x_d)\phi_n(x).
\end{align*}
The eigenvalue equation will hold if we set $V_n=-(T(D)-\lambda)u_n/u_n$, as usual. We have to prove that this is a smooth function decaying as in \eqref{decay V nonradial}. An elementary computation (compare \eqref{T(D) applied to sin}) yileds
\begin{align}\label{Chandrasekhar V}
V_n(x)={\rm e}^{\I x_1}\frac{\partial_{x_d}w_n(x)}{\sin(x_d)\phi_n(x)}-\frac{(T(D-e_d)-\lambda)\phi_n(x)}{\phi_n(x)}.
\end{align}
This is fortunate since the term with the sine in the denominator now has a differential (i.e.\ local) operator in the nominator (similar to the situation in Proposition~\ref{proposition real potential}). The zeros of the denominator will be cancelled by the choice of~$w_n$. In fact, using that $1+g(x_d)\gtrsim |x_d|$, we have
\begin{align}\label{Chandrasekhar 1}
\left|\frac{\partial_{x_d}w_n(x)}{\sin(x_d)}\right|\lesssim (n+|x'|^2+|x_d|)^{-N/2-1}
\end{align}
for large enough $n$. Next, we would like to apply Proposition \ref{proposition symbols} to the denominator in the second term of $V_n$ with
\begin{align*}
a(\xi):=(T(\xi-e_d)-\lambda)(\sqrt{(\xi+e_d)^2+1}+\sqrt{(\xi-e_d)^2+1})
\end{align*}
and $f=w_n$. Notice that $a(0)=0$ by the assumption that $e_1\in M_{\lambda}$. Here we cannot use the full strength of Proposition \ref{proposition symbols} since $w_n$ is not a symbol. It only satisfies the weaker estimates
\begin{align*}
|w_n(x)|&\lesssim (n+|x'|^2+|x_d|)^{-N/2},\\
|\partial^{\alpha}w_n(x)|&\lesssim (n+|x'|^2+|x_d|)^{-N/2-1},\quad |\alpha|\geq 1.
\end{align*}
However, as is obvious from the proof of Proposition \ref{proposition symbols}, this still implies that
\begin{align}\label{Chandrasekhar 2}
|a(D)w_n(x)|\lesssim (n+|x'|^2+|x_d|)^{-N/2-1}.
\end{align}
It remains to prove that
\begin{align}\label{Chandrasekhar 3}
|\phi_n(x)|\gtrsim (n+|x'|^2+|x_d|)^{-N/2}
\end{align}
since \eqref{Chandrasekhar V}--\eqref{Chandrasekhar 3} then imply the desired properties of $V_n$. We write
\begin{align*}
\phi_n(x)&=2\sqrt{2}w_n(x)+a(D)w_n(x),\\
a(\xi)&:=\sqrt{(\xi+e_d)^2+1}+\sqrt{(\xi-e_d)^2+1}-2\sqrt{2}.
\end{align*}
Since $a(0)=0$, the same modification of Proposition \ref{proposition symbols} as above yields
\begin{align*}
|\phi_n(x)|&\geq 2\sqrt{2}w_n(x)-\mathcal{O}(1)(n+|x'|^2+|x_d|)^{-N/2-1}\\
&\gtrsim (n+|x'|^2+|x_d|)^{-N/2}(1-\mathcal{O}(1/n))
\end{align*}
Hence \eqref{Chandrasekhar 3} will be satisfied for $n$ sufficiently large.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm. Dirac}]
Similarly as in the proof of Theorem \ref{thm Chandrasekhar-Herbst} we may assume witout loss of generality that $\lambda=\sqrt{2}$. Let $K$ be the dimension of the spinor space, i.e.\ $\alpha_j$ and $\beta$ are $K\times K$ matrices, and we are seeking an eigenfunction $u\in L^2({\mathbb R}^d;{\mathbb C}^K)$. From the properties of the Dirac matrices it follows that the matrix $\alpha_d+\beta$ has eigenvalues $\sqrt{2}$ and $-\sqrt{2}$. We fix an eigenvector $\mathbf{v}\in {\mathbb C}^K$ corresponding to $\lambda=\sqrt{2}$ and define
\begin{align*}
u_n(x)&:={\rm e}^{\I x_d}\psi_n(x)\mathbf{v},\\
\psi_n(x)&:=(n^2+|x'|^4+|x_d|^2)^{-N/2}
\end{align*}
with $n\in{\mathbb N}$ and $N>(d+1)/4$. We then compute
\begin{align*}
\mathcal{D}u(x)&={\rm e}^{\I x_d}\psi_n(x)(\alpha_d+\beta)\mathbf{v}+{\rm e}^{\I x_d}\sum_{j=1}^dD_j\psi_n(x)\alpha_j\mathbf{v}\\
&=\lambda u_n(x)+{\rm e}^{\I x_d}\sum_{j=1}^dD_j\psi_n(x)\alpha_j\mathbf{v}.
\end{align*}
If we set
\begin{align}\label{V Dirac}
V_n(x):=-\frac{1}{\psi_n(x)}\sum_{j=1}^dD_j\psi_n(x)\alpha_j,
\end{align}
then we obtain
\begin{align*}
(\mathcal{D}+V_n(x)-\lambda)u_n(x)={\rm e}^{\I x_d}\left(\sum_{j=1}^dD_j\psi_n(x)\alpha_j+V_n(x)\psi_n(x)\right)\mathbf{v}=0.
\end{align*}
Since $\alpha_j$ are hermitian matrices, $\psi_n$ is real-valued and $D_j=-\I\partial_j$, it follows that $V_n$ is anti-hermitian.
\end{proof}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


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text:
As I Lay Dying — американская металкор-группа из Сан-Диего (Калифорния).

История

Формирование и первые релизы (2000—2004)
Коллектив был образован в начале 2000-го в Сан-Диего экс-гитаристом Society Finest Тимом Ламбезисом, поскольку он начал самостоятельно писать песни и хотел стать барабанщиком в новоиспеченном коллективе. Но Тим был никудышным барабанщиком, поэтому ему пришлось взять в руки микрофон.

Изначально As I Lay Dying были трио, позже в состав пришёл гитарист Эван Уайт и барабанщик Джордан Манчино, уже через месяц группа засела в студию Pluto records для записи дебютного релиза. Несмотря на приход новых лиц, Тим так и остался фронтменом, поджигая остальных своей творческой харизмой.

Он рос на металле и депрессивном панке, поэтому играть именно в этих стилях для него было нормальным, даже, несмотря на то что с годами больше начинал испытывать интерес к творчеству Guns N' Roses и Darkness.

Дебютный альбом Beneath The Encasing Of Ashes, который группа выпустила в июне 2001-го, хорошо продавался, став, в конце концов, самым продаваемым среди всех релизов выпущенных на Pluto records.

В августе 2002-го As I Lay Dying выпустили совместный сплит с сан-диеговским коллективом American Tragedy.

В марте 2003-го As I Lay Dying подписали контракт с лейблом Metal Blade, а в июле выпустили свой третий релиз Frail Words Collapse. В то время группе были предложены контракты с рядом более весомых лейблов, но, изучив историю каждого, As I Lay Dying пришли к выводу, что лучшим выбором будет Metal Blade. И именно после выхода Frail Words Collapse коллектив был замечен всеми. Большого успеха на телевидении добились видео на песни «94 Hours» и «Forever», которые активно ротировали на MTV2 и Fuse.

За год As I Lay Dying отыграли совместные туры с Killswitch Engage, In Flames, Shadows Fall, Lamb Of God, Hatebreed, Everytime I Die, The Black Dahlia Murder, Scarlet, Chimaira, Soilwork, Papa Roach, Soul Embrace, Bleeding Through, Sworn Enemy, Danzig, 3 Inches Of Blood, Haste, Zao, Six Feet Under, The Agony Scene и другими группами.

Многие американские металкор-коллективы отталкивались от шведского дэта, но на звучании As I Lay Dying это отразилось в значительной степени. Зимой 2004-го группа отправилась в качестве хедлайнеров в Германию, отыграв по дороге несколько концертов в Голландии. Это был первый европейский выезд, а за полугодовой период постоянных туров As I Lay Dying удалось провести дома всего лишь три недели.

После сильной загруженности и напряжённых отношений внутри коллектива Крис и Эван покинули состав группы. На их место пришли гитаристы Фил Сгроссо и Ник Хип, а также басист Клинт Норисс. Ушедший Эван Уайт присоединился к Noise Ratchet, где уже играл бывший басист As I Lay Dying Джон Джеймсон.

Коммерческий успех и последующие релизы (2005—2012)
В 2005-м вышел новый альбом — Shadows are Security на лейбле Metal Blade, заняв по выходу 35-е место в чарте Биллборда. Первым хитом с нового альбома стал «Through Struggle», получивший широкую ротацию. В поддержку нового релиза As I Lay Dying совершили тур по штатам вместе с Norma Jean, Madball и Life Once Lost. As I Lay Dying выступили в Японии вместе с Shadows Fall и Every Time I Die, а летом их пригласили хедлайнерами на христианский фест Cornerstone 2005-го. Но главным событием для группы стало официальное приглашение на Ozzfest 2005, где выступали Megadeth, Iron Maiden, Rob Zombie и множество других, включая и самого виновника торжества Оззи Осборна. Британский журнал Kerrang назвал As I Lay Dying лучшей группой из всех присутствовавших на том фестивале. Тем временем Тим в содружестве с Pluto records образовал свой собственный лейбл Clockwork Recordings.

Осенью 2005-го группа отправилась в тур Hell on Earth вместе с Heaven Shall Burn, Evergreen Terrace, Agents of Man, End of Days и Neaera. Примерно в это же время они получили премию San Diego Music Awards в категории «Группа года».

К зиме 2005-го As I Lay Dying отправились во второй раз в Германию, где часть тура прошла со Slipknot, а позже группа делила сцену с Deftones, Thrice, Atreyu, Story of the Year, Dredg и другими.

В 2006-м группу пригласили на правах хедлайнеров участвовать в туре Sound and Underground, который прошёл по Штатам и Канаде. В туре, помимо As I Lay Dying, принимали участие In Flames, Cannibal Corpse, Trivium, Terror, Gwar, The Black Dahlia Murder, Behemoth, The Chariot и Through the Eyes of the Dead.

Перебравшись на лейбл Sworn Enemy Records, As I Lay Dying записали альбом An Ocean Between Us, выход которого намечен на август 2007-го.
20 августа 2008 года музыканты дали концерт в Москве в клубе «Точка», который с трудом смог вместить всех зрителей.

В марте группа анонсировала новый альбом под названием The Powerless Rise. Дата североамериканского релиза была назначена на 11 мая 2010 года. В соответствии с ранее объявленными планами альбом в США вышел 11 мая 2010 года. В Европе альбом стартовал 7 мая на лейбле Metal Blade.

25 сентября 2012 года группа выпустила шестой альбом, получивший название Awakened.

Судебное дело Ламбезиса и приостановка деятельности группы (2013—2016)
В мае 2013 года Тим Ламбезис был задержан по обвинению в подготовке убийства своей жены. О преступных намерениях Ламбезис сказал своему тренеру, а тот в свою очередь передал эту информацию полиции.

Из-за ареста фронтмена группы, все концертные туры As I Lay Dying были отменены, в том числе и концерты в России.

17 сентября 2013 года Тим Ламбезис впервые предстал перед судом по подозрению в попытке нанять киллера для убийства своей жены. На предварительном слушании было принято решение передать дело на дальнейшее рассмотрение в суд и отпустить Т.Ламбезиса под залог в 3 млн долларов Адвокат Ламбезиса утверждает, что его подопечный пошел на преступление из-за употребления стероидов. В феврале 2014 Ламбезис сознался в намерении убить свою супругу. Музыкант заявил, что мотивом к преступлению было обязательство выплачивать 60 % заработка жене в качестве алиментов. 16 мая 2014 года Ламбезис был приговорен к 6 годам лишения свободы, с учётом 48 дней, проведенных в заключении до выплаты залога и изменения меры пресечения на домашний арест. К этому моменту группу покинули все участники, за исключением барабанщика Джордана Манчино. Ламбезис и Манчино, также владеющий частью прав на группу, решили заморозить проект на неопределенный срок. Ушедшие участники вместе с Манчино образовали группу Wovenwar.

Возобновление активности (2016—2019)
Ламбезис был досрочно освобождён в декабре 2016 года. В 2017 году 19 июня на Facebook группы был выложен пост с единственным содержащимся словом в нём, «Activity», намекающий на возможное возвращение группы. 2 сентября Metal Injection сообщили, что Ламбезис работает над новой музыкой и планирует выпустить её под брендом As I Lay Dying. Первоначальная информация сообщала, что ни один из прежних членов группы, включая Манчино, не вернётся, делая таким образом Ламбезиса единственным оригинальным участником группы с 2000 года.

1 июня 2018 года на официальной странице группы в Facebook появился тизер клипа с отрывком новой песни. As I Lay Dying собрались в том же составе, в котором они приостановили деятельность в 2014 году.

8 июня 2018 года вышла песня «My Own Grave» и видеоклип на неё.

5 апреля 2019 года группа анонсировала большой осенний европейский тур, совместно с Chelsea Grin, Unearth.

11 апреля вышла песня «Redefined» и видеоклип на неё.

20 сентября 2019 года вышел 7-й полноформатный альбом Shaped By Fire.

Европейский лейбл и возвращение в хэдлайнеры метал-сцены (2019-2021)

Сразу после выхода Sharped by Fire на немецком Nuclear Blast Records стартовал большой тур по странам Европы в поддержку релиза, с sold out концертами в Москве, Санкт-Петербурге, Стокгольме, Будапеште. Впоследствии также были проданы все билеты на шоу в Германии.

Европейские шоу оказались очень коммерчески успешными для группы, во время их путешествия по Европейскому союзу альбом поднялся в топ-10 Metal и Rock чартов Германии, Великобритании, США и держался в нем всю осень 19-го года. Также альбом попал в US Billboard 200 на 50-ое место. Далее группа отправилась в тур по азиатским странам, Индии. Презентация пластинки закончилась туром по США совместно с After the Burial и Emmure.

Во время гастролей был отснят мини-фильм в формате влога о концертах, подготовке к ним и комментариями участников группы, в частности, о концерте в Москве. Все серии доступны на канале группы в Youtube.

Эпидемия COVID-19
Запланированные на весну, лето и осень 2020-го года концерты в Европе, США были перенесены на 2021-ый год из-за эпидемии COVID-19 в мире.

27-го марта 2020г., в разгар пандемии коронавируса, вышел сингл "Destruction Or Strength". К синглу группа прикрепила следующее сообщение:

"Это беспрецедентные времена. В условиях неопределенности относительно того, как долго продлится пандемия COVID-19 и насколько серьезными будут ее последствия, бесчисленное количество людей отчаянно борются с этим "штормом". Стремясь внести позитивный вклад в развитие наших сообществ и самых близких нам людей, мы выпускаем неизданный трек из наших сессий альбома Shaped By Fire под названием "Destruction or Strength". Он будет доступен на всех стриминговых сервисах и в цифровых магазинах, в том числе Bandcamp в качестве загрузки по схеме «Заплати сколько хочешь».

В 2021 году запланирован масштабный хэдлайн-тур по штатам, а также участие в самых крупных метал-фестивалях Евросоюза, таких как Wacken, Summerblast, Impericon и др.

15 августа появилась информация, что гитарист Ник Хипа покинул коллектив. Ламбезис не подтвердил, но и не опроверг это.

16 мая 2022 года стало известно, что басист и вокалист Джош Гилберт покинул коллектив.
9 июня барабанщик Джордан Манчино сообщил, что не примет участие в ближайших концертных выступлениях группы до разрешения некоторых разногласий с группой. В состоявшемся туре ушедших музыкантов заменили басист и вокалист Miss May I Райан Нефф и бывший барабанщик Unearth Ник Пирс.

Состав группы
Текущий состав
 Тим Ламбезис — экстрим-вокал, иногда чистый вокал (2000—2014, 2017—настоящее время), клавишные (2000—2003)
 Фил Сгроссо — ритм-гитара, бэк-вокал, клавишные (2003—2014, 2018—настоящее время)

Бывшие участники
 Ноа Чейз — бас-гитара (2001)
 Эван Уайт — соло-гитара (2001—2003), бас-гитара (2001)
 Томми Гарсия — бас-гитара, ритм-гитара, бэк-вокал (2001—2003; сессионный участник 2003—2010)
 Джейсон Кребс — ритм-гитара (2002—2003)
 Аарон Кеннеди — бас-гитара (2003)
 Клинт Норрис — бас-гитара, чистый вокал (2003—2006)
 Ник Хипа — соло-гитара, бэк-вокал (2004—2014, 2018—2020)
 Джош Гилберт — бас-гитара, чистый вокал (2007—2014, 2018—2022)
 Джордан Манчино — ударные (2000—2014, 2018—2022)

Гастрольные участники
 Чед Акерман — ритм-гитара (2001—2002), бэк-вокал (сессионно, 2007)
 Брэндон Хэйс — бас-гитара (сессионный участник 2002—2003)
 Дэвид Артур — чистый вокал (2005)
 Дуэйн Рид — бэк-вокал (2007)
 Джастин Фоули — ударные (2009)
 Джоуи Брэдфорд — бэк-вокал (2012)

Временная шкала

Дискография

 «Beneath the Encasing of Ashes» (2001)
 «Frail Words Collapse» (2003)
 «Shadows Are Security» (2005)
 «An Ocean Between Us» (2007)
 «The Powerless Rise» (2010)
 «Awakened» (2012)
 «Shaped by Fire» (2019)

Видеография
 «94 Hours» (Frail Words Collapse) − 2003
 «Forever (Live)» (Frail Words Collapse) − 2003
 «Confined» (Shadows Are Security) − 2005
 «The Darkest Nights» (Shadows Are Security) − 2005
 «Through Struggle» (Shadows Are Security) − 2005
 «Nothing Left» (An Ocean Between Us) − 2007
 «The Sound of Truth» (An Ocean Between Us) − 2007
 «Within Destruction» (An Ocean Between Us) − 2007
 «I Never Wanted (Live)» (An Ocean Between Us) − 2007
 «Parallels» (The Powerless Rise) − 2010
 «Anodyne Sea» (The Powerless Rise) − 2010
 «Electric Eye» (Decas) — 2011
 «A Greater Foundation» (Awakened) — 2012
 «My Own Grave» (Shaped by Fire) — 2018
 «Redefined» (Shaped by Fire) — 2019
 «Shaped by Fire» (Shaped by Fire) — 2019
 «Blinded» (Shaped by Fire) — 2019
 «Torn Between» (Shaped by Fire) - 2020

Примечания

Ссылки

 Профиль группы на MySpace.com

Металкор-группы США
Христианские рок-группы США
Музыкальные коллективы из Сан-Диего
Музыкальные коллективы, появившиеся в 2000 году
Исполнители Metal Blade Records
Хеви-метал-группы из Калифорнии
Исполнители Nuclear Blast

redpajama_set_name: RedPajamaWikipedia


text:
Trans-Pennine rail link proposal is potential boon for Yorkshire construction
by Integrity Software Webmaster
Chancellor George Osborne has proposed a third high-speed rail link to connect Manchester and Leeds, a move he says will help to create a 'Northern powerhouse' city economy to rival that of London.
A move many will see as long overdue, the Chancellor's trans-Pennine high speed network will theoretically help these Northern cities to compete with London and the Southeast.
'We need an ambitious plan to make the cities and towns here in this Northern belt radically more connected from east to west – to create the equivalent of travelling around a single global city,' said Osborne.
'I want us to start thinking about whether to build a new high-speed rail connection east-west from Manchester to Leeds.'
Should the proposal go ahead, a high-speed rail link between Leeds and Manchester would certainly be a boon for businesses and commuters in the North. However, the significant infrastructure project such a move would necessitate is also good news for the region's construction firms.
The Chancellor suggested that the new high-speed connection would be 'based on the existing rail route, but speeded up with new tunnels and infrastructure.' Naturally, this would require significant support from infrastructure construction firms in Greater Manchester and West Yorkshire.
It remains to be seen whether Osborne's proposal is a genuine possibility or merely a tactic to win over floating voters in two cities that are traditionally known as Labour strongholds. However, West Yorkshire construction firms should be aware that such infrastructure projects could be on the horizon. Our construction industry job costing software is just what you need to determine whether or not a contact is worth pursuing.
The construction industry in 2014: a mid-year, sector-by-sector review
As we near the halfway mark for 2014, it's time to review the state of the construction industry.
Construction industry growth continues but concerns emerge over costs and skills
The revival of the UK's construction industry is causing new problems for construction companies.
Help to Buy heat map: the future of construction industry growth
At first glance, new figures suggest that the Help to Buy scheme isn't responsible for house price inflation. We take a closer look.
Next up: Planned new homes at highest number since 2008
Snashall Steel Fabrications have chosen to upgrade to Evolution M

redpajama_set_name: RedPajamaCommonCrawl


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Alex Rodriguez received the Latinobaseball.com 2015 Player of the Year Award at Yankee Stadium prior to the Yankees game with the Chicago White Sox.
Thais Schenk of Dominican Republic Tourism, Manny Gonzalez of Hennessy and Lou Melendez of Sports Media Partners joined Ralph Paniagua, the Publisher of Latinobaseball.com to present the award to Alex at home plate at Yankee Stadium before the game.
"The fans at the stadium applauded for Alex, that was great to hear," said Thais Schenk, representing the Dominican Republic, which has more players in professional baseball than any country other than the U.S.
"What a change from Opening Day, when he was received with jeers rather than cheers," said Manny Gonzalez of Hennessy, whose tagline "Never Stop, Never Settle" fit perfectly with Mr. Rodriguez's return year to the diamond.
"Mr. Rodriguez received this year's award for his remarkable performance both on and off the field," stated Ralph Paniagua, Publisher of Latinobaseball.com. "After his one year suspension and advanced age, Alex Rodriguez won back the fans and was an important cog in getting the Yankees into the Playoffs," he concluded.
As the Yankees Designated Hitter during 2015, Alex's numbers exceeded everyone's expectations with 32 Homeruns, 83 RBIs and 81 runs scored to date with one week left in the season.
In 2015, Alex celebrated his 3,000th hit, surpassed Willie Mays Homerun total, broke Lou Gehrig's Yankee RBI record and hit more Grand Slams than anyone in baseball history.
Alex won over his teammates, impressing them with his clutch play and his positive attitude in the Clubhouse. He also won over the most difficult and demanding fans, the New York Baseball fans.
Alex humbly and without much fanfare was a leader in the community, by supporting youth sports and health programs by donating millions of dollars to the Boys and Club Clubs and local health programs.
Latinobaseball.com is proud to present Alex A-Rod Rodriguez with the 2015 Latinobaseball.com Player of the Year.


Friday The 13th...The NASCAR Camping World Truck Series heads to the Windy City for the EnjoyIllinois.com 225 at Chicagoland Speedway on Friday, September 13. Paludo has two previous starts at the 1.5-mile D-shaped oval, with a best finish of eighth in 2011. Paludo has also started in the top 10 in both of his previous visits to Chicago.
This Week's Silverado...Paludo and the No. 32 BRANDT do Brasil team will unload chassis TSM-600 this weekend at Chicagoland Speedway. TSM-600 has never started or finished outside the top 10 in the four races in which it has been utilized, earning the pole position and a best finish of second at Pocono Raceway. The truck has both an average start and average finish of sixth, and most recently hit the track at Michigan International Speedway where Paludo finished fifth.
Watch And Listen...The EnjoyIllinois.com 225 can be seen live Friday, September 13 on FOX Sports 1, with coverage beginning at 8:00 p.m. EDT. The race can also be heard live on MRN Radio and SiriusXM Satellite Radio, channel 90.
Previous Race Recap...Paludo entered the weekend at Iowa Speedway with momentum on his side and the "perfect" truck before a stuck throttle resulted in a mid-practice wreck, forcing the No. 32 team to utilize a backup for the Fan Appreciation 200. Paludo and crew persevered in a truck that did not make a lap of practice, fighting back from a lap down to finish 14th.
"I can't wait to get to Chicagoland. I'm very proud to be representing BRANDT do Brasil. A lot of people think of Brazil as beaches and big cities, but agriculture and companies like BRANDT play a huge role in many parts of the country, including in the south where I grew up. Chicagoland is BRANDT's home track and they always have a huge number of guests supporting our team at the racetrack. I don't think there would be anything more special than to go to victory lane in front of BRANDT while representing my home country. I've run well in Chicago in the past and we're bringing a chassis that we've been very successful with so far, so I think this is going to be a great weekend."


text:
Daniel Barbiero / Cristiano Bocci — Nostos
(Acustronica, 2015, CD)
For this collaborative outing, Cristiano Bocci, who is a bass player, left the bass playing almost entirely to Daniel Barbiero. Bocci provides all manner of electronic processing to the sounds that Barbiero made, and much of the resulting sound is highly abstract, with bowed tones often recognizable, but often twisted and manipulated into obscure clouds of sound. Barbiero is an experimentalist himself, so he freely elicits all manner of noises from his instrument. Thumps, cracks, bow bouncing on strings, all become fodder for Bocci's treatments. Sometimes, as on "Contrabbassi in Marcia," the sounds are edited and looped into a rhythmic backing over which bowed notes float freely and are treated to buzzing distortion. At times, the electronics dominate, presenting misty washes of indistinct origin. Most of the eight tracks tend towards somber atmospheres and low tones, with the rich overtones from the bow providing the full sound we expect from the double bass. As is usually the case with recordings that rely heavily on processing, it's hard to tell what started where, but we can leave that to the technical notes (if there were such things for Nostos). The set ends with "L'Ultimo Treno per Follonica," where Bocci gets out his electric bass for a virtual duet with Barbiero's recorded part, with relatively little electronic processing. It's a meditative, restful way to end the proceedings after some of the strange noises that came before. Nostos should be of interest both to bass lovers and to listeners curious about the possibilities of processing, provided there's a tolerance for the abstract.
Filed under: New releases, 2015 releases
Related artist(s): Daniel Barbiero, Cristiano Bocci
Djam Karet - The Heavy Soul Sessions – Over the last four or five years, Djam Karet has gone through some evolutionary changes. Gayle Ellett has switched from guitars to primarily analog keyboards, and a fifth member Mike Murray has been... (2011) » Read more
Search Artist Title Release Label
Mono and interpretive dance
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My names Liam! How are you all doing?
I'm doing great myself thanks.
A little back story about me, my computer broke just over 2 years ago now and I've not played any computer games since. I've recently purchased a computer and am looking forward upon the release of SkyBlock. A few of my friends have been raving on non stop about this server and since me following along with the journey of @CosmicSkyblock on twitter I've found myself hooked.
I know very little about SkyBlock but am extremely excited to dive into it and learn!
I've played CosmicPvP before and even used to be helper for just over a year but as I said above my computer broke and things just fell apart from there.
Hopefully I can catch some of you guys tuning in and watching me on my journey of trying to grow the best alliance on Cosmic SkyBlock over on my twitch which you can find a link to in my signature below!
I wish everyone the best of luck on the server although lets not all get to friendly, as out there on the battlefield things will get very competitive for sure!
@Harrypotterfan07 Would you recommend checking out CosmicPrisons or would you say it's to late to get into it?
I wish I came back earlier than I have, missed it alot to be honest! Hopefully I can express the same here on CosmicSky even though I'm not staff ahha! I'll still be trying to help out so if you need anything just let me know!
Lol how many people are in Hive?
Currently around 10 of us! Not sure if the leader has more people up his sleeve but there is 10 of us that are currently actively speaking and planning out anyways!
I used to be on there all the time! Around three years ago or so!
I know right, I was way easier back then when the custom enchants was few and easy to understand. Now there is so many I just loose track and it's hard to learn so many. Looking forward to release and will be on like 24/7 lol. Thanks and hope to see you having fun on there too. I'll be streaming release so feel free to stop by, would be great to see you there!