text:
\begin{document}

\title{On Weil Sums, Conjectures of Helleseth, and Niho Exponents}

\author{Liem Nguyen}
\address{Department of Mathematics\\
Louisiana State University\\
Baton Rouge, Louisiana, 70808}
\email{lngu145@lsu.edu}

\subjclass[2020]{11T23, 11T24}
\date{\today}
\keywords{Weil sum, character sum, Helleseth conjecture, finite field, Niho exponent}
\thanks{The author is partially supported by NSF DMS \#$1602047$.}

\maketitle

\begin{abstract}
Let $F$ be a  finite field,  $\mu$ be a fixed additive character and $s$ be an integer coprime to $|F^\times|$. For any $a\in F$, the corresponding Weil sum is defined to be $W_{F,s}(a)=\displaystyle\sum_{x \in F} \mu(x^s-ax)$. The Weil spectrum counts distinct values of the Weil sum as $a$ runs through the invertible elements in the finite field. The value of these sums and the size of the Weil spectrum are of particular interest, as they link problems in coding and information theory to other areas  of  math  such  as  number  theory  and  arithmetic  geometry. We prove the Vanishing Conjecture of Helleseth ($1971$) on the presence of zero in the Weil spectrum in the case of Niho exponents. We also state a conjecture for when the Weil spectrum contains at least five elements, and prove it for a certain class of Weil sum.
\end{abstract}

\section{Introduction}
\label{introduction}
\subsection{\bl The Weil sum} \bk

Let $F$ be a finite field of characteristic $p$ and size $q=p^n$. Let $\mu: F \to \mathbb{C}$ be the canonical additive character. Recall that the canonical additive character $\mu (x)= \zeta_p^{\Tr_{F/\mathbb{F}_p}(x)}$, where $\zeta_p=e^{2\pi i/p}$ is a $p$th root of unity and $\Tr_{F/\mathbb{F}_p}(x)$ is the absolute trace function from $F \to \mathbb{F}_p$. If $L$ is an extension of $F$, i.e $|L|=q^m$ for some nonnegative integer $m$, then \bl we take \bk by $\mu(x)=\zeta_p^{\Tr_{F/\mathbb{F}_p}(\Tr_{L/F}(x))}$ where $\Tr_{L/F}(x)$ is the trace function from $L \to F$.

Let $x \mapsto x^s$ be a power mapping from $F$ to $F$ where $s$ is a fixed positive integer. For a point $a \in F$, we define the Weil sum as
\[
W_{F,s}(a)=\sum_{x \in F}\mu(x^s-ax).
\]
If $\gcd(s,q-1)=1$, then $s$ is said to be an \emph{invertible exponent} \bl over $F$. \bk Furthermore,
\begin{align*}\label{W(0)}
W_{F,s}(0)=\sum_{x \in F} \mu(x^s)= \sum_{x \in F} \mu(x)=0,
\end{align*}
since the map $x \mapsto x^s$ permutes the elements of $F$.

Since $W_{F,s}(a)$ is a sum of roots of unity, $W_{F,s}(a)$ is an algebraic integer. The following theorem by Tor Helleseth states the necessary and sufficient conditions for $W_{F,s}(a)$ to be a rational integer for every $a\in F$ \cite{tor}.

\begin{theorem}[Helleseth]
$W_{F,s}(a) \in \mathbb{Z}$ for all $a \in F^\times$ if and only if $s \equiv 1 \pmod{p-1}$.
\end{theorem}
Now, $s$ is said to be \emph{singular} if there is an $a \in F^\times$ such that $W_{F,s}(a)=0$.

\bl

The Weil sum is of interest to us as it relates many problems in coding and information theory to other areas such as number theory and arithmetic geometry. Properties of the Weil sum including its values, number of values over the finite field, and its bounds are still not well understood. These aspects are certainly of interest from a purely number-theoretic standpoint. On the other hand, in sequence design in information theory, the cross-correlation function between two $p$-ary maximal linear recursive sequences measures how similar they are, and can be realized as a character sum, specifically as the Weil sum plus $(-1)$. One important criterion that makes such sequences useful in remote sensing and communications is that they should have low cross-correlation (see \cite{tor, Niho1972MultiValuedCF, Tormaster, sarwate, games, sawarteconjecture, Calderbank3value, WelchConjecture, WelchNihoConjecture}).

In coding theory, a $p$-ary m-sequence (by restricting to a finite subsequence) can be regarded as a codeword in the $p$-ary cyclic code. We can define the Hamming weight of a codeword to measure the number of substitutions to change one string to another. We can relate this weight to the cross-correlation function, which in turn relates to the Weil sum. We are interested in the distinct values of these weights, and the number of the weights, and hence the corresponding properties of the Weil sum (see \cite{mcguire, charpin, feng} for more details).

For a good summary of various research aspects of the Weil sum, as well as their applications, see \cite[Appendix]{katz} and \cite{Katz2018WeilSO} by Daniel Katz.

\bk

\subsection{\bl Properties and Helleseth Conjectures} \bk

It is natural to wonder what kind of value one would get from the Weil sum. We have seen that the $W_{F,s}(a)$ is always $0$ at $a=0$, and interestingly, this presence of zero value is not known for nonzero elements $a$. This prompted Tor Helleseth to propose the following conjecture \cite{Tormaster,tor} in $1971$.

\begin{conjecture}[Helleseth Vanishing Conjecture]
If $q=|F| > 2$ and $s$ is an invertible exponent over $F$ such that
$s \equiv 1 \pmod{p-1}$, then $s$ is singular.
\end{conjecture}

Now, if we put some restrictions on the exponent $s$, some partial results on the Vanishing Conjecture can be obtained. For the finite field $L$ of order $q=p^{2n}$, an exponent $s$ is called a \emph{Niho exponent} if $s$ is not a power of $p \bl \pmod{p^{2n}-1}$ \bk and $s \equiv p^j \pmod{p^n-1}$. If $j=0$, then such exponent is called a \emph{normalized Niho exponent}. Niho exponents \bl were \bk first introduced by Yoji Niho in $1972$ in his PhD thesis on the cross-correlation function between an $m$-sequence and its $d$-decimation \cite{Niho1972MultiValuedCF}. Since then further research has been done using Niho exponents, and it has resulted in various applications in coding theory, sequence design and cryptography \cite{Nihoexponent}. Moreover, the Helleseth Vanishing Conjecture was proved for Niho exponents for a field of characteristic $2$ \cite{charpin}.

One useful fact about Weil sums with Niho exponents is that we can replace them with normalized Niho exponents due to the following result discussed in Aubry, Katz and Langevin paper \cite{cyclotomy}.

\begin{lemma}\cite[Lemma $3.2$]{cyclotomy}\label{powerlemma}
Let $F$ be a finite field of characteristic $p$ and $s$ be an invertible exponent over $F$.
Then $W_{F,s}(a)=W_{F,p^j s}(a)$ for any $a \in F$ and $j \in \mathbb{Z}$.
\end{lemma}

The proof of this lemma relies on the fact that $x^{p^js}$ and $x^s$ are Galois conjugates and thus have the same trace.

The next questions of interest would be how many distinct values $W_{F,s}(a)$ takes as $a$ ranges over $F$, and what they are. We define the \emph{Weil spectrum} for some fixed $s$ to be the set $\{W_{F,s}(a)\mid a\in F^\times\}$, and say that it is \emph{r-valued} if $|\{W_{F,s}(a)\mid a\in F^\times\}|=r$.

If $s$ is a power of $p$ modulo $(q-1)$, we call $s$ to be \emph{degenerate}. For degenerate power $s$ we know that $W_{F,s}(a)$ takes two values via a theorem by Helleseth \cite{tor}.

\begin{theorem}[Helleseth \cite{tor}] \label{2values}
If $s$ is degenerate, $W_{F,s}(a)$ is two-valued over $F$ where
\[
W_{F,s}(a)=
\begin{cases}
q & \text{if $a=1$},\\
0 & \text{otherwise}
\end{cases}
\]
If $s$ is nondegenerate, then $W_{F,s}(a)$ takes at least three values over $F^\times$.
\end{theorem}
So, when exactly is the Weil spectrum three-valued? In the same paper that Helleseth proposed the Vanishing Conjecture in $1971$, he also gave a criteria for when this three-valued property is never met \cite{Tormaster,tor}.
\begin{conjecture}[Helleseth Three-Valued Conjecture, 1971]
Let $F$ be a finite field of characteristic $p$. If $[F:\mathbb{F}_p]$ is a power of $2$, then for any invertible exponent $s$, the spectrum of the Weil sum $W_{F,s}(a)$ is not three-valued.
\end{conjecture}
More progress has been made towards this conjecture in comparison to the Vanishing Conjecture, using various approaches from coding theory, cryptography and number theory \cite{Calderbank3value, WelchConjecture, WelchNihoConjecture, mcguire,charpin,feng,katz,cyclotomy,divisibility,katz_langevin}. Currently, only ten families of three-valued Weil sum are known \cite[Table 1]{cyclotomy}, and these are conjectured to be the only ones that occur. \bl The cases for characteristic $p=2$ and $p=3$ in the Three-Valued Conjecture were proven by Daniel Katz in \cite{katz} and in \cite{divisibility}, respectively \bk. Special families of the three-valued Weil sum for all characteristics $p$ are also addressed via the Welch Conjecture and the Niho Conjecture. Canteaut, Charpin, and Dobbertin gave a proof to the Welch Conjecture in \cite{WelchConjecture} and Hollmann and Xiang proved both the Welch and Niho Conjectures in \cite{WelchNihoConjecture}. More recently, a family of three-valued Weil \bl spectra \bk has been proved \cite{threevaluefamily}.

The organization of this paper is as follows. The first part discusses how a Weil sum can be viewed as an inner product of characters over a finite field. This observation leads to a relation which is part of a method called power moments. In general, the method of power moments studies the summation $\displaystyle\sum W^k_{F,s}(a)$ for any positive integer $k$. The summation can be taken over a finite field, or over different orbits in the field via multiplication action by a subfield. This method is useful in the studies of distribution/averaging behavior and the divisibility of Weil sum \cite{tor,divisibility}.

The second part of the paper gives a proof of the Vanishing Conjecture for the case of a Niho exponent $s$. The discussion then continues with obtaining bounds for the Weil sum in this setting.

\begin{theorem}\label{vanishingp^2n}
Let $L$  be a finite field where $q=p^{2n}$ for some odd prime $p$ and positive integer $n$. Suppose that $s$ is an invertible Niho exponent \rd over $L$. \bk Then $s$ is singular.
\end{theorem}

The last part of the paper focuses on computing the Weil sum for special values $a$ in the finite field and the Weil spectrum for the case of Niho exponent. Based on numerical evidence, we propose a conjecture for the five-valued behavior.

\bl
\begin{conjecture}\label{5valuesconjecture}
Let $L$ be a quadratic extension of a finite field $F$ of order $p^n$, where $p$ is an odd prime. Let $s=1+k(p^n-1)$ be an invertible Niho exponent over $L$, $d_1=\gcd(k, p^n+1)$, and $d_2=\gcd(k-1,p^n+1)$.
If either
\begin{itemize}
    \item[(i)] $d_1+d_2 \geq 5$ or
    \item[(ii)] $d_1+d_2=3$ and $p^n \equiv 11 \pmod{12}$
\end{itemize}
satisfies, then the Weil spectrum over $L$ is at least five-valued.

Moreover, in case (i), the five values are $\{0, -p^n, p^n, 2\alpha p^n, (2\beta+1)p^n\}$ where $\alpha, \beta \geq 1$ are integers. In case (ii), at least four values are $\{0,-p^n,p^n,2p^n\}$.
\end{conjecture}

\bk

A special case of the condition $d_1+d_2 \geq 5$ in \cref{5valuesconjecture} is $p^n \equiv 2 \pmod{3}$. Hence, we can restate the conjecture with simpler assumptions as follows.

\begin{conjecture}\label{5valuesconjecturep^n2mod3}
Let \bl p be an odd prime, \bk $L$ be a quadratic extension of a finite field $F$ of order $p^n$, and $s=1+k(p^n-1)$ be an invertible Niho exponent over $L$. If $p^n \equiv 2 \pmod{3}$, then the Weil spectrum has at least five values of the form $\{0, -p^n, p^n, 2\alpha p^n, (2\beta+1)p^n\}$ for integers $\alpha,\beta \geq 1$.
\end{conjecture}

\begin{remark}\label{remarkp^n2mod3}
Since s is an invertible exponent over $L$, $\gcd(s,p^{2n}-1)=1$. Hence, if $p^n \equiv 2 \pmod{3}$, then $s \equiv 1$ or $2 \pmod{3}$. Thus, $k \equiv 0 \pmod{3}$ and $(k-1) \equiv 2 \pmod{3}$, or $k \equiv 1 \pmod{3}$ and $(k-1) \equiv 0 \pmod{3}$. Moreover, $p^n+1$ is divisible by $2$ and $3$. Therefore either $d_1$ or $d_2$ in \cref{W1} is divisible by $3$. The same conclusion can be made for the divisibility of either $d_1$ or $d_2$ by $2$. Hence, $d_1+d_2 \geq 5$.
\end{remark}

\bl

We end the paper by showing case (i) of \cref{5valuesconjecture} holds true for sufficiently large primes. Finally, We then proved case (ii).

\begin{theorem}\label{theoremM_i}
Let $L$ be a quadratic extension of a finite field $F$ of order $p^n$, where $p$ is an odd prime and $n \geq 2$ is an integer. Let $k \geq 2$ be an integer such that $k <\displaystyle\frac{p}{2}+1$, and $s=1+k(p^n-1)$ \rd be \bk an invertible Niho exponent over $L$.
Let $d_1=\gcd(k, p^n+1)$, and $d_2=\gcd(k-1,p^n+1)$. If $d_1+d_2 \geq 5$, then the Weil spectrum over $L$ is at least five-valued.
Moreover, four of those five values are $\{0, -p^n, 2\alpha p^n, (2\beta+1)p^n\}$ where $\alpha, \beta \geq 1$.
\end{theorem}

\begin{remark}\label{remarkk}
If $k=0$ or $1$ then $s$ is degenerate. So in general, we can take $2 \leq k \leq p^n$, since $k+p^n+1$ gives the same exponent $s$ $\pmod{p^{2n}-1}$ as $k$ over $L$.

For the case of $n=1$ in \cref{theoremM_i}, taking integer $k$ such that $p^{1/2}>2(k-1)$ would yield the same conclusion.
\end{remark}

\begin{theorem}\label{5valuesconjecturep^n3mod4}
Let $L$ be a quadratic extension of a finite field $F$ of order $p^n$, where $p$ is an odd prime. Let $s=1+k(p^n-1)$ be an invertible Niho exponent over $L$, $d_1=\gcd(k, p^n+1)$, and $d_2=\gcd(k-1,p^n+1)$. If $d_1+d_2=3$ and $p^n \equiv 11 \pmod{12}$, then the Weil spectrum over $L$ is at least five-valued. Moreover, four of those five values are $\{0, -p^n, p^n, 2 p^n\}$.
\end{theorem}
\bk

\section{Preliminaries}
We review some techniques with characters over a general finite field $F$ of order $q=p^n$.

For $a\in F$, let $\mu_a(x)=\mu(ax)$. Then the set of additive characters $\{ \mu_a : a \in F\}$ form an orthonormal  basis, with respect to the following inner product, for the space of functions from $F$ to $\mathbb{C}^\times$. One observes that the additive character $\mu(x)$ in our introduction is $\mu_1(x)$.

\begin{definition}
For all functions $f,g: F \to \mathbb{C}^\times$, we define the inner product
\[
\langle f, g \rangle = \frac{1}{q}\sum_{x \in F}f(x) \overline{g(x)},
\] where $\overline{\,  \cdot \, }$ stands for complex conjugation.
\end{definition}

If we let $f_s$ be the function
$f_s(x):=\mu(x^s)$, our Weil sum is the coordinates (or the Fourier coefficients) up to a factor of $1/q$ of $f_s$ with respect to the orthonormal basis  $\{ \mu_a : a \in F\}$. More precisely, the Weil sum becomes
\[
W_{F,s}(a)=\sum_{x \in F}\mu(x^s-ax)=\sum_{x \in F}\mu(x^s)\overline{\mu_a(x)}=q\cdot \langle f_s, \mu_a \rangle,
\]
and
$$f_s=\frac{1}{q}\sum_{a \in F} W_{F,s}(a) \cdot \mu_a.$$
On the other hand,
$$
 \langle f_s, f_s \rangle= 1,
$$
and hence,
\begin{align*}
    1=\left\langle \sum_{a \in F} \frac{1}{q}W_{F,s}(a)\cdot \mu_a,\sum_{b \in F}\frac{1}{q}W_{F,s}(b)\cdot \mu_b \right\rangle =\frac{1}{q^2}\sum_{a \in F}\bl |W_{F,s}(a)|^2. \bk
\end{align*}

\bl The Weil sum is shown to only take real values \cite[Theorem $2.1$(c)]{katz}, so the relation above becomes \bk
\begin{align}\label{secondpowermoment}
    1=\frac{1}{q^2}\sum_{a \in F} W_{F,s}(a)^2.
\end{align}
Relation \eqref{secondpowermoment} can also be proved using the cross-correlation function in  \cite{katz}. In fact, it is called the second power moment of the Weil sum. In general we can consider the summation of all Weil \bl sums \bk in the finite field raised to a positive integer $m$. This is called the $m$th power moment. For the first few moments we have the following result which was proved in \cite{katz}.

\begin{lemma}\label{powermoment}
Let $F$ be a finite field of order $p^n$ and $s$ be a fixed invertible exponent. Then
\begin{itemize}
    \item[(i)] $\displaystyle\sum_{a \in F} W_{F,s}(a)=p^n$,
    \item[(ii)] $\displaystyle\sum_{a \in F} W_{F,s}(a)^2=p^{2n}$, and
    \item[(iii)] $\displaystyle\sum_{a \in F} W_{F,s}(a)^3=p^{2n} \cdot |R|$, where $R=\{x \in L \mid  (1-x)^s+x^s-1=0\}$.

\end{itemize}

\end{lemma}

As for the settings of a quadratic extension $L$ over $F$, we have the following moment property of the Weil sum in different orbits under the multiplication action of $F^{\times}$ on $L^{\times}$.

\begin{lemma}\label{firstmoment}
Let $F$ be a finite field of order $p^n$ and $L$ be a quadratic extension of $F$. Suppose that $s$ is an invertible exponent \bl over $L$ \bk and $s\equiv 1\pmod{\bl p^n-1 \bk}$.
Then for a fixed $b \in L^{\times}$,
\[
\sum_{a\in F} W_{L,s}(ab)=
\begin{cases}
\bl p^{2n}-1 & \text{if $b \in F$,}\\
0 & \text{otherwise.}
\end{cases}
\]
\end{lemma}

\begin{proof}
The first case for $b \in F$ was proved in \cite[lemma 2.5]{cyclotomy}. So we will show the second equality here.
Observe that
\begin{align*}
    \sum_{a \in F}W_{L,s}(ab) &= \sum_{a \in F}\sum_{x \in L} \mu(x^s-abx)\\
    &=\sum_{x \in L} \mu(x^s)\sum_{a \in F}\zeta_p^{\Tr_{F/\mathbb{F}_p}(-a(Tr_{L/F}(bx)))}.
\end{align*}

If $\Tr_{L/F}(bx) \neq 0$, then the inner sum is \bl the sum of all $p$th roots of unity precisely $p^{n-1}$ times, so it is $0$. \bk

Hence,
\begin{align*}
    \sum_{a\in F}W_{L,s}(ab) &=p^n \cdot \sum_{\substack{x \in L \\ \Tr_{L/F}(bx)=0}}\mu(x^s).
\end{align*}

\bl
Now we consider the equation   $0=\Tr_{L/F}(y)=y^{p^n}+y=y(y^{p^n-1}+1)$ over $L$. Note that the polynomial $y^{p^n}+y$ has formal derivative of $1$ so it is separable over $L[y]$ with $p^n$ distinct roots.
\bk
Let $x_0$ be a non-zero element such that $bx_0$ is a nonzero solution to $\Tr_{L/F}(y)=0$. Then all the roots of the polynomial  are of the form $cbx_0$, where $c \in F$. Note that $\Tr_{L/F}(x_0)\neq 0$ because of $b\not\in F$.  Hence,
\begin{align*}
       \sum_{a\in F}W_{L,s}(ab) &= p^n \cdot \sum_{c \in F} \mu((cx_0)^s) \\
       &=p^n \cdot \sum_{c \in F} \zeta_p^{\Tr_{F/\mathbb{F}_p}(c\Tr_{L/F}(x_0^s))}=0.
\end{align*}

\bl

Note that the second-to-last equality follows from $s \equiv 1 \pmod{p^n-1}$ and $c^s=c$ in $F$.

\bk
\end{proof}

The conclusion of \cref{firstmoment} also implies the first moment property of the Weil sum.

\section{The Vanishing Conjecture and Bounds on $W_{L,s}(a)$}
For the rest of this paper (sections 3 and 4) we turn our focus to Niho exponents. \bl Let $p$ be an odd prime \bk. Our setting is a finite field $F$ of order $q=p^n$, together with a quadratic extension $L$ over $F$. Let $s$ be a nondegenerate invertible exponent. By \cref{powerlemma}, we can take $s \equiv 1 \pmod{p^n-1}$, then $s=k(p^n-1)+1$ for some nonnegative integer $k$.

Our goal in this section is to prove the Vanishing Conjecture for the case of Niho exponent $s$ (i.e \cref{vanishingp^2n}), and study the bounds on the Weil sum for various values of $a$ and $s$. We first start with a lemma that gives a formula for Weil sum $W_{L,s}(a)$ based on the cardinality of a relevant set.

\begin{lemma}\label{divisibilityp^nlemma}
Let $L$ be the quadratic extension of the finite field $F$. Assume that $s$ is an invertible Niho exponent over $L$. Let $K_{a,s}=\{x \in L^\times\mid \Tr_{L/F}(x^s- a x)=0\}$.

Then $|K_{a,s}|$ is a multiple of $(p^n-1)$ and
$$
W_{L,s}(a)=\displaystyle p^n\cdot \frac{|K_{a,s}|}{p^n-1}-p^n.
$$
Furthermore, $W_{L,s}(a)$ is divisible by $p^n$.
\end{lemma}

\begin{remark}
The first statement of the theorem was also proved in \cite{cyclotomy}.
\end{remark}

\begin{proof}

By \cref{powerlemma}, we can replace the condition $s \equiv p^j \pmod{p^n-1}$ by $s \equiv 1 \pmod{p^n-1}$.
Noting that $W(0)=0$ and $$
|K_{0,s}|=|\{x \in L^\times\mid (x^s)^{p^n}- x^s=0\}| =p^n-1,
$$
the identity $W_{L,s}(0)=\displaystyle p^n\cdot \frac{|K_{0,s}|}{p^n-1}-p^n$ holds.  We now assume $a\neq 0$.
We have
\begin{align*}
    W_{L,s}(a)=\sum_{x \in L^\times}\mu(x^s-ax)+\mu(0)=\sum_{x \in L^\times}\mu(x^s-ax)+1.
\end{align*}

For any $y \in L^\times$, we can write $y=bx$ for some $b\in F^{\times}$, and
$$
  \Tr_{L/F}((bx)^s-a(bx))=\Tr_{L/F}(bx^s-abx)=b\Tr_{L/F}(x^s-ax).
$$ Therefore, each element $y$ in the coset $\bar x:=xF^\times$ either lies in $K_{a,s}$ or not depending on whether $x$ lies in $K_{a,s}$ or not.  This implies that $|K_{a,s}|$ is a multiple of $|F^\times|=p^n-1$.

We then rewrite $\displaystyle\sum_{ x \in L^\times}\mu(x^s-ax) $ as follows.

\begin{align*}
    \sum_{ x \in L^\times}\mu(x^s-ax) &=\sum_{ \bl \{\bar x\} \in L^{\times}/F^{\times} \bk }\sum_{b \in F^\times}\zeta_p^{\Tr_{F/\mathbb{F}_p}(\Tr_{L/F}((bx)^s-a(bx)))}\\
    &=\sum_{\bl \{\bar x\} \in L^{\times}/F^{\times} \bk}\sum_{b \in F^\times}\zeta_p^{\Tr_{F/\mathbb{F}_p}(b(\Tr_{L/F}(x^s-ax)))}\\
    &=\sum_{\bl \{\bar x\} \in L^{\times}/F^{\times} \bk}\sum_{b \in F} \zeta_p^{\Tr_{F/\mathbb{F}_p}(b(\Tr_{L/F}(x^s-ax)))}-(p^n+1).\\
\end{align*}

If $x \notin K_{a,s}$, then for a fixed equivalence class {$\bar x$}
the inner sum   $$\displaystyle\sum_{b \in F}\zeta_p^{\Tr_{F/\mathbb{F}_p}(b(\Tr_{L/F}(x^s-ax)))}=\displaystyle\sum_{u \in F}\zeta_p^{\Tr_{F/\mathbb{F}_p}(u)} = 0.$$ Otherwise it is $p^n$.

Thus,

$$
W_{L,s}(a)=\frac{p^n|K_{a,s}|}{p^n-1}-(p^n+1)+1= p^n\cdot \frac{|K_{a,s}|}{p^n-1}-p^n.
$$

This completes the proof.

\end{proof}

Now we are ready to give a proof of \cref{vanishingp^2n}.

\begin{proof}[Proof of Theorem \ref{vanishingp^2n}]

By \cref{divisibilityp^nlemma}, $W_{L,s}(a)=p^n \cdot h_a$ for some $h_a \in \mathbb{Z}$. Specifically, $h_0=0$ since $W_{L,s}(0)=0$. Applying this and relation \eqref{secondpowermoment} to the setting of a field $L$ of order $q=p^{2n}$, we have
\begin{align}\label{secondpowermomentL}
    q=p^{2n}=\displaystyle\sum_{a \in L^{\times}}h^2_a.
\end{align}

If $h_a=0$ for some $a \in L^\times$, then the Vanishing conjecture holds. To prove this, we use proof by contradiction and assume that $h_a\neq 0$ for all  $a \in L^\times$.
If $|h_a|=1$ for all $a \in L^\times$, then from $\eqref{secondpowermomentL}$, we have that $q-1=q$, which is not possible.
So $|h_{a'}| \geq 2$ for some $a' \in L$,
then
\[
\sum_{a \in L^{\times}}h_a^2
\geq \sum_{\substack{a \in L^\times \\ a \neq a'}}h_a^2+2^2=(q-2)+4=q+2>q,
\]
which also contradicts $\eqref{secondpowermomentL}$.

So at least $W_{L,s}(a)=0$ for some $a \in L^\times$.
\end{proof}

As a consequence to \cref{vanishingp^2n}, the Vanishing Conjecture holds true for $\mathbb{F}_{p^2}$.

\begin{corollary}
Suppose $s$ is an invertible exponent \bl over $\mathbb{F}_{p^2}$ \bk and $s \equiv 1 \pmod{(p-1)}$, then the Vanishing Conjecture holds for the field $\mathbb{F}_{p^2}$.
\end{corollary}

Lemma \ref{divisibilityp^nlemma} gives a formula for $W_{L,s}(a)$ based on the cardinality of the set $K_{a,s}$. By identifying field elements in $K_{a,s}$, we can bound $|K_{a,s}|$ in order to deduce bounds on $W_{L,s}(a)$.

\begin{proposition} \label{prop1}
Let $a \in F$ and \bl $p$ be an odd prime. \bk Suppose $x^{2(p^n-1)}=1$ and $x \notin F$, then $\Tr_{L/F}(x^s-ax)=0$.
\end{proposition}

\begin{proof}

Since $x^{2(p^n-1)}=1$ and $x \notin F$, $x^{p^n-1}=-1$.

We have that $x^{(p^n-1)^2}=x^{p^{2n}-1-2(p^n-1)}=(x^{2(p^n-1)})^{-1}=1$.

Now,
\begin{align*}
    \Tr_{L/F}(x^s-ax) &= \Tr_{L/F}(x^s)-a\Tr_{L/F}(x)\\
                &= x^s+x^{sp^n}-a(x+x^{p^n})\\
                &=x^s(1+x^{(k(p^n-1)+1)(p^n-1)})-ax(1+x^{p^n-1})\\
                &=x^s(1+x^{p^n-1})-ax(1+x^{p^n-1})\\
                &=0.
\end{align*}
\end{proof}
Note that there are $2(p^n-1)$ solutions for the equation $x^{2(p^n-1)}=1$ in $L$, since $\gcd(2(p^n-1),p^{2n}-1)=2(p^n-1)$. This gives a bound on the size of $K_{a,s}$, hence a bound on the Weil sum.

\begin{proposition} \label{prop2}
Let $p^n \equiv 2 \pmod{3}$. If $x^{3(p^n-1)}=1$, then $\Tr_{L/F}(x^s-x)=0$.
\end{proposition}

\begin{proof}
We have that
\begin{align}
\Tr_{L/F}(x^s-x) &= \Tr_{L/F}(x^s)-\Tr_{L/F}(x) \nonumber\\
                &= x^s+x^{sp^n}-(x+x^{p^n}) \nonumber\\
                &=x^{k(p^n-1)+1}(1+x^{(k(p^n-1)+1)(p^n-1)})-x(1+x^{p^n-1}) \nonumber\\
                &=x(x^{k(p^n-1)}+x^{(p^n-1)(2k+1)}-1-x^{(p^n-1)}), \label{trace1}\\ \nonumber
\end{align}
using the relation $x^{(p^n-1)^2}=x^{p^{2n}-1-2(p^n-1)}=x^{-2(p^n-1)}=x^{(p^n-1)}$.

If $k \equiv 0  \pmod{3}$ or $k \equiv 1 \pmod{3}$, then the expression \eqref{trace1} becomes $0$.

If $k \equiv 2 \pmod{3}$, then $s \equiv 0 \pmod{3}$, but $q-1=p^{2n}-1 \equiv 0 \pmod{3}$. So $\gcd(s,q-1) \geq 3$, which is a contradiction.
\end{proof}

\begin{theorem}\label{generalboundtheorem}
For an \bl odd prime $p$\bk, we have the following bounds on $W_{L,s}(a)$:
\begin{itemize}
    \item[(1)] If $a \in L$, then $W_{L,s}(a)\geq-p^n$.
    \item[(2)] If $a \in F$, then $W_{L,s}(a) \geq 0$.
    \item[(3)] In particular, $W_{L,s}(1) \geq p^n$.
                If $p^n \equiv 2 \pmod{3}$, then $W_{L,s}(1) \geq 3p^n$.
\end{itemize}
\end{theorem}

\begin{remark}
Since $W_{L,s}(a)$ is a sum of roots of unity, $|W_{L,s}(a)| < q$ for nondegenerate $s$.
\end{remark}

\begin{proof}
Since $|K_{a,s}|\geq 0$, $|W_{L,s}(a)| \geq-q$ for $a \in L$.

If $a \in F$, then by \cref{prop1} there are at least $2(p^n-1)-(p^n-1)=p^n-1$ points in $K_{a,s}$. So $W_{L,s}(a) \geq 0$ by \cref{divisibilityp^nlemma}.

For part $(3)$, if $x \in F$, then $x^s=x$ and $\Tr_{L/F}(x^s-x)=0$. So such $x$ lies in $K_{1,s}$. Combining this fact and \cref{prop1}, there are at least $2(p^n-1)$ points in $K_{1,s}$. Therefore, $W_{L,s}(1) \geq p^n$. Moreover, if $p^n \equiv 2 \pmod{3}$, then there are $3(p^n-1)$ solutions to the equation $x^{3(p^n-1)}=1$, and by \cref{prop2} and \cref{divisibilityp^nlemma}, $W_{L,s}(1) \geq 3p^n$.
\end{proof}

\section{The Weil Spectrum}
\bl Let $L$ be quadratic extension of a finite field $F$ of order $p^n$, where $p$ is an odd prime. \bk In this section we give formulas for the Weil sum at specific elements \bl of $L$. Let $s$ be an invertible Niho exponent over $L$. \bk We recall that $s=1+k(p^n-1)$ and
as noted in \cref{remarkk}, we can take $2 \leq k \leq p^n$.
Our discussion begins by considering the value of the Weil sum at a root of unity in the field for certain primes $p$. The formula is obtained by realizing the relation between the elements in the set $K_{a,s}=\{x \in L^\times\mid\Tr_{L/F}(x^s-ax)=0\}$ in  \cref{divisibilityp^nlemma} and the root of unity.

\begin{proposition}\label{W1}
\bl Let $s=1+k(p^n-1)$ be an invertible Niho exponent over $L$, where $2 \leq k \leq p^n$. \bk Let $d_1=\gcd(k,p^n+1)$, $d_2=\gcd(k-1,p^n+1)$, and $t$ be a positive integer with  \bl $t\mid p^{n}+1$. \bk Let $\zeta_t$ be a primitive $t$-th root of unity in $L$.
For $i=1$ or $2$, let
\[
\delta_{i,t}=\begin{cases}
      1 & \text{if $t \mid \frac{p^n+1}{d_i}$,} \\
      0 & \text{otherwise.}
   \end{cases}
   \]
Then \[
W_{L,s}(\zeta_t)=\begin{cases}
      p^n(d_1+d_2-2) & \text{if $t=1$,} \\
      p^n(d_1\delta_{1,t}+d_2\delta_{2,t}-1) & \text{otherwise.}
   \end{cases}
\]
\end{proposition}

\begin{proof}
We compute $|K_{\zeta_t,s}|$ in \cref{divisibilityp^nlemma}. Let $x \in K_{\zeta_t,s}$ then $\Tr_{L/F}(x^s)=\Tr_{L/F}(\zeta_tx)$.
We also have that $N_{L/F}(x^s)=N_{L/F}(\zeta_tx)$, since $\zeta^{p^n+1}_t=1$. Hence, $\zeta_tx$ and $x^s$ satisfy the same degree two minimal polynomial over $F$. So we can consider two cases $x^s=\zeta_tx$ or $x^s=(\zeta_tx)^{p^n}$.
Let $$L^{\times}=\langle g \rangle$$ for some generator $g$ in the field. Then $x=g^i$ for some $i \in \mathbb{Z}_{p^{2n}-1}$. We can pick $\zeta_t=g^{(p^{2n}-1)j/t}$ where $\gcd(j,t)=\gcd(j,p^{2n}-1)=1$.
For the case $x^s=\zeta_tx$, we have that $x^{k(p^n-1)}=\zeta_t$. Then $g^{ik(p^n-1)}=g^{(p^{2n}-1)j/t}$, so \begin{align}\label{w1_1}
    (p^n-1)ik \equiv \frac{(p^{2n}-1)j}{t} \pmod{p^{2n}-1},
\end{align}
which implies that
\begin{align}\label{w1_2}
    ik \equiv \frac{(p^n+1)j}{t} \pmod{p^n+1}.
\end{align}
Let $d_1=\gcd(k, p^n+1)$. Then \eqref{w1_2} is solvable if $\frac{p^n+1}{t} \equiv 0 \pmod{d_1}$. If it is solvable then there are $d_1$ solutions. When $t=1$, \eqref{w1_2} is always solvable. Hence \eqref{w1_1} has $d_1(p^n-1)$ solutions if $t=1$, and $d_1\delta_{1,t}(p^n-1)$ solutions otherwise.
Similarly, for the case $x^s=(\zeta_tx)^{p^n}=\zeta^{-1}_tx^{p^n}$, we have that $x^{(k-1)(p^n-1)}=\zeta^{-1}_t=g^{-(p^{2n}-1)j/t}$. Let $d_2=\gcd(k-1,p^n+1)$, then there are $d_2(p^n-1)$ solutions to this case if $t=1$ and $d_2\delta_{2,t}(p^n-1)$ for other values of $t$.

\bl
For both cases to have simultaneous solutions, we have that $\zeta_t x=x^s=(\zeta_tx)^{p^n}$. This means $x^s=\zeta_t x \in F^{\times}$ and thus, $x^{s(p^n-1)}=1$. On the other hand, $x^{p^{2n}-1}=1$. Since $s$ is invertible over $L$, $\gcd(s,p^{2n}-1)=1$. So there exists integers $y_1$ and $y_2$ sucht that $y_1s+y_2(p^{2n}-1)=1$. Now, $x^{p^n-1}=x^{(p^n-1)(y_1s+y_2(p^{2n}-1))}=1$, so $x \in F^{\times}$. We also note that $\zeta_t=x^{s-1}=x^{k(p^n-1)}=1$.

Hence, when $t=1$ the solutions for both cases were counted twice for $x \in F^{\times}$. Therefore,

\bk

\[
|K_{\zeta_t,s}|=\begin{cases}
    (p^n-1)(d_1+d_2-1) & \text{if $t=1$,}\\
    (p^n-1)(d_1\delta_{1,t}+d_2\delta_{2,t}) & \text{otherwise.}
\end{cases}
\]
Apply this to the formula for $W_{L,s}(\zeta_t)$ in \cref{divisibilityp^nlemma}, we have
\[
W_{L,s}(\zeta_t)=\begin{cases}
      p^n(d_1+d_2-2) & \text{if $t=1$,} \\
      p^n(d_1\delta_{1,t}+d_2\delta_{2,t}-1) & \text{otherwise.}
   \end{cases}
\]
\end{proof}

\begin{remark}
Theorem \ref{generalboundtheorem}(3), can be obtained by \cref{W1}. As noted in \cref{remarkp^n2mod3}, \bl for an odd prime $p$\bk, either $d_1$ or $d_2$ must be divisible by $2$, so $d_1+d_2 \geq 3$.
Moreover, if $p^n \equiv 2 \pmod{3}$, then $d_1+d_2 \geq 5$, and thus, $W_{L,s}(1)=p^n(d_1+d_2-2) \geq 3p^n$.
\end{remark}

From \cref{W1}, we deduce the following corollary for the Weil sum at $a=-1$.

\begin{corollary}\label{W-1}
    Let \bl $p$ be an odd prime, \bk $L$ be a quadratic extension \bl of order $p^{2n}$ \bk over \bl the finite field \bk $F$, $s$ be an Niho exponent \bl over $L$, \bk and $d_1,d_2$ be defined as in \cref{W1}. Then
\[
W_{L,s}(-1)=p^n\left(d_1 \cdot \frac{1+(-1)^{(p^n+1)/d_1}}{2}+d_2 \cdot \frac{1+(-1)^{(p^n+1)/d_2}}{2}-1\right).
\]
If $p^n \equiv 3 \pmod{4}$ \bl and \bk $d_1+d_2=3$, then $W_{L,s}(-1)=2p^n$.
\end{corollary}

We state a result from Katz in \cite{cyclotomy} that is useful in the next step of our discussion.

\begin{lemma}\cite[Corollary 3.4]{cyclotomy}\label{cyclotomykatz}
Let $F$ be a finite field of characteristic $p$, and let $L$ be an extension of $F$ with $[L:F]$ a power of a prime $\ell$ distinct from $p$. Let $s$ be degenerate over $F$ but not over $L$. Then $W_{L,s}(1) \equiv |F| \pmod{\ell}$ and $W_{L,s}(a) \equiv 0 \pmod{\ell}$ for every $a \in F\setminus \{1
\}$.
\end{lemma}

Combining this lemma with our previous results, we can deduce the following four values in the Weil spectrum in the cases of \cref{5valuesconjecture}.

\begin{corollary}\label{4valued}
\bl Let $L$ be a quadratic extension of a finite field $F$ of order $p^n$, where $p$ is an odd prime. Let $s=1+k(p^n-1)$ be an invertible Niho exponent over $L$, $d_1=\gcd(k, p^n+1)$, and $d_2=\gcd(k-1,p^n+1)$.
\begin{itemize}
    \item[(i)] If $d_1+d_2 \geq 5$, then the spectrum of $W_{L,s}(a)$ contains at least $4$ values of the form $\{0, -p^n, 2\alpha p^n, (2\beta+1)p^n\}$ for some integers $\alpha,\beta \geq 1$.
    \item[(ii)] If $d_1+d_2=3$ and $p^n \equiv 11 \pmod{12}$, then the spectrum of $W_{L,s}(a)$ contains $0, -p^n, 2p^n, p^n$.
\end{itemize}
\bk
\end{corollary}

\begin{proof}
\bl
In both cases: By \cref{2values}, the Weil spectrum contains at least three values, and one of which is $0$ by \cref{vanishingp^2n}. If the other two values are both positive we would have $\left(\displaystyle\sum_{a \in L}W_{L,s}(a)\right)^2 > \displaystyle\sum_{a \in L}W^2_{L,s}(a)$, and this is contradiction to the first and second moments in \cref{powermoment}. Hence, the spectrum must contain at least a negative value. From \cref{divisibilityp^nlemma} and \cref{generalboundtheorem} (part $1$), this negative value must be $-p^n$.

For case (i): Apply \cref{cyclotomykatz} to our setting of the quadratic extension $L$ over $F$, \bl the prime $\ell=[L:F]=2$. Then the Weil sum $W_{L,s}(a)$ admits an odd value for $a=1$ and even values for all $a \in F \setminus \{1\}$. By \cref{W1},
 $W_{L,s}(1)=(d_1+d_2-2)p^n \geq 3p^n$. If $W_{L,s}(a)=0$ for all $a \in F \setminus \{1 \}$, then taking $b=1$ in \cref{firstmoment} yields $p^{2n}-1=\displaystyle\sum_{a \in F}W_{L,s}(a)=W_{L,s}(1)$. This is a contradiction since $p^n \mid W_{L,s}(1)$ but $p^n \nmid p^{2n}-1$. Hence, there is a nonzero even value for some $a \in F \setminus \{1\}$. Therefore, the Weil spectrum consists of $-p^n,0,2\a p^n$, and $(2\beta+1)p^n$, where $\a,\b \geq 1$.

 For case (ii): By \cref{W1},
 $W_{L,s}(1)=(d_1+d_2-2)p^n = p^n$. By \cref{W-1}, $W_{L,s}(-1)=2p^n$. Hence the Weil spectrum in this case consists of $-p^n, 0, p^n$, and $2p^n$.
\end{proof}

\bk

Our numerical evidence suggests a stronger conclusion than \cref{4valued} implies. This leads to \cref{5valuesconjecture} and \cref{5valuesconjecturep^n2mod3} in the introduction.

For the rest of the paper we discuss partial results towards \cref{5valuesconjecture} for certain families of Niho exponents $s$. For such an $s$, we give a count on the solution set $R=\{x \in L\mid (1-x)^s+x^s-1=0\}$ in \cref{powermoment}.

\begin{lemma}\label{xx^slemma}
Let $L$ be a quadratic extension of $F$, and $|F|=p^n$ and $k \geq 2$. Let $d_1=\gcd(k,p^n+1)$ and $d_2=\gcd(k-1,p^n+1)$. Then $|R|=p^n+(d_1-1)(d_1-2)+(d_2-1)(d_2-2)$.
\end{lemma}

\begin{proof}
Clearly, all elements in $F$ are in $R$. So $|R| \geq p^n$. Now suppose $x \in R \setminus F$.

We have that $(1-x)^s=1-x^s$. Computing the norm $N_{L/F}$ of both sides, we get
\[
N_{L/F}(1-x^s)=1-x^{sp^n}-x^s+x^{s(p^n+1)}=1-\Tr_{L/F}(x^s)+N_{L/F}(x^s)
\]
and
\[
N_{L/F}((1-x)^s)=1-x^{p^n}-x+x^{p^n+1}=1-\Tr_{L/F}(x)+N_{L/F}(x).
\]

As $s=1+k(p^n-1)$, we know $N_{L/F}(x)=N_{L/F}(x^s)$. Equating the norm of $1-x^s$  gives us $\Tr_{L/F}(x)=\Tr_{L/F}(x^s)$.

As $s=1+k(p^n-1)$, $N_{L/F}(x)=N_{L/F}(x^s)$.
Since the norm and trace of $x$ and $x^s$ are the same, they must satisfy the same degree-two minimal polynomial over $F$. Hence $x^s=x$ or $x^s=x^{p^n}$.

\underline{Case 1}: $x^s=x$.

This implies $x^{k(p^n-1)}=1$.

Since $x \notin F$, $x^{p^n-1} \neq 1$. Now, $1=x^{p^{2n}-1}=x^{(p^n-1)(p^n+1)}$. So a solution in this case must satisfy $x^{d_1({p^n-1})}=1$, where $d_1=\gcd(k,p^n+1)$. Let $L^{\times}= \langle g \rangle$ and $h=g^{(p^{2n}-1)/d_1}$ be an element of order $d_1$ in $L^\times$. Then $x^{p^n-1}$ must be in $\langle h \rangle$. Without loss of generality, let $x^{p^n-1}=h^{t_1}$, where $ 1\leq t_1 \leq d_1-1$. Since $x \notin F$, $t_1 \neq t_2$.
On the other hand, $1=x^s+(1-x)^s=x+(1-x)^s$. This implies $(1-x)^{s-1}=1$ or $(1-x)^{k(p^n-1)}=1$. Using the similar argument from above, we can say $(1-x)^{p^n-1}=h^{t_2}$, where $1 \leq t_2 \leq d_1-1$.

Now,
\begin{align*}
     (1-x)^{p^n-1} = h^{t_2}
    &\implies 1-x^{p^n} = h^{t_2}(1-x)\\
    &\implies 1-h^{t_1}x =h^{t_2}-h^{t_2}x\\
    &\implies  x=\frac{1-h^{t_2}}{h^{t_1}-h^{t_2}}.
\end{align*}
With $1 \leq t_1,t_2 \leq d_1-1$ and $t_1 \neq t_2$, there are $(d_1-1)(d_1-2)$ choices for solution $x$.

\bl
Now, to show such $(d_1-1)(d_1-2)$ choices are in $R$, consider $x=\displaystyle\frac{1-h^{t_2}}{h^{t_1}-h^{t_2}}$, where $h$ is defined as above, and $1 \leq t_1, t_2 \leq d_1-1$.

Since $h$ has order $d_1=\gcd(k,p^n+1)$, $h^k=h^{p^n+1}=1$. Thus, $h^{p^n}=h^{-1}$.

We have that
\begin{align}\label{denom}
  (h^{t_1}-h^{t_2})^s=(h^{t_1}-h^{t_2})^{(1+k(p^n-1))} &= (h^{t_1}-h^{t_2})^{(1-k)}(h^{t_1p^n}-h^{t_2p^n})^k \\ \nonumber
  &= (h^{t_1}-h^{t_2})^{(1-k)}(h^{-t_1}-h^{-t_2})^k \\ \nonumber
  &=h^{-t_1k}h^{-t_2k}(h^{t_1}-h^{t_2})^{(1-k)}(h^{t_2}-h^{t_1})^k \\ \nonumber
  &=(h^{k})^{-t_1}(h^{k})^{-t_2}(-1)^k(h^{t_1}-h^{t_2}) \\ \nonumber
  &=(-1)^k(h^{t_1}-h^{t_2}) \\
\end{align}

Similarly, we can show that
\begin{align}\label{num1}
    (h^{t_1}-1)^s=(-1)^k(h^{t_1}-1),
\end{align}
and
\begin{align}\label{num2}
(1-h^{t_2})^s=(-1)^k(1-h^{t_2}).
\end{align}

Now,
\begin{align*}
    (1-x)^s+x^s-1 &= \left(1-\frac{1-h^{t_2}}{h^{t_1}-h^{t_2}}\right)^s+\left(\frac{1-h^{t_2}}{h^{t_1}-h^{t_2}}\right)^s-1 \\
    &=\left(\frac{(h^{t_1}-1)^s+(1-h^{t_2})^s}{(h^{t_1}-h^{t_2})^s}\right)-1=0,
\end{align*}
with the last equality following from relations \eqref{denom}, \eqref{num1}, and \eqref{num2}.
Hence, such choice $x$ is in $R$.

\bk

\underline{Case 2}: $x^{p^n}=x^s=x^{1+k(p^n-1)}$.

This implies $x^{(k-1)(p^n-1)}=1$. Suppose $x \notin F$. Similar to the argument in case $1$ we let $d_2=\gcd(k-1,p^n+1)$ and find solutions to the equation $x^{d_2(p^n-1)}=1$. \bl Let $\ell=g^{(p^{2n}-1)/d_2}$ be an element of order $d_2$, where $g$ is the generator of $L^{\times}$ as in case $1$.
On the other hand, A quick check also yields us the relation $(1-x)^{(k-1)(p^n-1)}=1$.Then $x^{p^n-1}$ and $(1-x)^{p^n-1}$ must be in $\langle \ell \rangle$. Using the similar argument from above, we obtain $x=\displaystyle\frac{1-\ell^{r_2}}{\ell^{r_1}-\ell^{r_2}}$, where $1 \leq r_1,r_2 \leq d_2-1$.
There are $(d_2-1)(d_2-2)$ choices for such $x$.

In the reverse direction, we first note that $\ell^{k}=\ell$ and $\ell^{p^n}=\ell^{-1}$, since $\ell$ has order $d_2=\gcd(k-1,p^n+1)$.

Using a similar argument as in case $1$, we have that
\begin{align}\label{denomcase2}
  (\ell^{r_1}-\ell^{r_2})^s &= (\ell^{r_1}-\ell^{r_2})^{(1-k)}(\ell^{r_1p^n}-\ell^{r_2p^n})^k \\ \nonumber
  &= (\ell^{r_1}-\ell^{r_2})^{(1-k)}(\ell^{-r_1}-\ell^{-r_2})^k \\ \nonumber
  &=\ell^{-r_1k}\ell^{-r_2k}(\ell^{r_1}-\ell^{r_2})^{(1-k)}(\ell^{r_2}-\ell^{r_1})^k \\ \nonumber
  &=(-1)^k\ell^{-r_1}\ell^{-r_2}(\ell^{r_1}-\ell^{r_2}) \\ \nonumber
  &=(-1)^k(\ell^{-r_2}-\ell^{-r_1}).
\end{align}

Similarly, we can show that
\begin{align}\label{num1case2}
    (\ell^{r_1}-1)^s=(-1)^k(1-\ell^{-r_1}),
\end{align}
and
\begin{align}\label{num2case2}
(1-\ell^{r_2})^s=(-1)^k(\ell^{-r_2}-1).
\end{align}

By relations \eqref{denomcase2}, \eqref{num1case2}, and \eqref{num2case2}, we can show that $(1-x)^s+x^s-1=0$. Thus, such choice $x$ is in $R$.

\bk

Hence there are $(d_2-1)(d_2-2)$ solutions of $x$ in this case.

Note that since $k$ and $k-1$ are coprime, $d_1$ and $d_2$ are coprime as well. Therefore, the solutions in case $1$ and case $2$ for $x \notin F$ are distinct. Accounting for solutions $x \in F$ we have $|R|=p^n+(d_1-1)(d_1-2)+(d_2-1)(d_2-2)$.
\end{proof}

\begin{corollary}
Let $s=1+k(p^n-1)$ for some integer $k$, $0 \leq k \leq p^n$. Then $k$ and $2-k+p^n$ gives the same number of solutions to the equation $(1-x)^s+x^s-1=0$ for $x \in L$.
\end{corollary}

\begin{proof}
From the proof of \cref{xx^slemma}, the exponents  $s$ and $sp^n$ give the same number of solutions to the equation $(1-x)^s+x^s-1=0$ for $x \in L$. Now, $sp^n=1+(1-k)p^n$, and $0 \leq 2-k+p^n \leq p^n$ gives the same exponent as $1-k$.
\end{proof}

\bl

Finally, we end our discussion by giving some partial results towards \cref{5valuesconjecture}.
The idea behind the proofs of \cref{5valuesconjecturep^n3mod4} and \cref{5valuesconjecturep^n2mod3} is to apply the power moments in \cref{powermoment} to the four Weil sum values and the count for the set $R$ in \cref{xx^slemma} to derive a contradiction.

\bk

\begin{proof}[Proof of Theorem \ref{theoremM_i}]
\bl According to \cref{4valued}, there are four values in the Weil spectrum. Suppose that these are the only ones in the spectrum. \bk Let $m_1,m_2,m_3$ and $m_4$ be the number of elements whose Weil sum value is $-p^n, 0, 2\alpha p^n$ and $(2 \beta +1)p^n$, respectively, for integers $\alpha, \beta \geq 1$. In here $2 \beta+1=d_1+d_2-2$ from \cref{W1}.
By \cref{powermoment} we have the following system of equations:
\begin{numcases}

 \ m_1+m_2+m_3+m_4=p^{2n} \\ \label{1}
 -m_1+2\alpha m_3 + (d_1+d_2-2)m_4=p^n \\ \label{2}
 m_1+4{\alpha}^2 m_3+(d_1+d_2-2)^2 m_4 =p^{2n}\\ \label{3}
 -m_1+8{\alpha}^3 m_3 +(d_1+d_2-2)^3 m_4=p^n |R|, \label{4}

\end{numcases}
where $|R|=p^n+(d_1-1)(d_1-2)+(d_2-1)(d_2-2)$.

The above system has a unique solution over $\mathbb{Q}$, which is
\begin{align*}
m_1 &= \bl -\displaystyle\frac{p^n(d_1^2 + d_2^2+(2\a-p^n-3)(d_1+d_2) - p^n(2\a-3) + 4(1-\a))}{(2\a + 1)(d_1 + d_2 - 1)} \bk \nonumber \\
m_2 &=\displaystyle\frac{1}{2} \cdot\displaystyle\frac{p^n((d_1+d_2)(2\a(p^n+1)-p^n-4)+d_1^2+d_2^2-6\a(p^n+1)+2(2p^n+3))}{\a(d_1 + d_2 - 2)} \nonumber \\
m_3 &=\displaystyle\frac{1}{2} \cdot\displaystyle\frac{p^n(d_1^2 + d_2^2 - (p^n+4)(d_1 + d_2) + 2(2p^n + 3))}{(2\a - d_1 - d_2 + 2)(2\a + 1)\a} \nonumber \\
m_4 &= -\displaystyle\frac{p^n(d_1^2 + d_2^2 - 2p^n(\a-1) - 3(d_1 + d_2) + 4-2\a)}{(2\a - d_1 - d_2 + 2)(d_1 + d_2 - 1)(d_1 + d_2 - 2)}.
\end{align*}
\bl
From the numerator of $m_3$, we have that
\begin{align}
d_1^2 + d_2^2 - (p^n+4)&(d_1 + d_2) + 2(2p^n + 3) \nonumber \\ &=(d_1-2)^2+(d_2-2)^2-(d_1+d_2-4)p^n \bl - 2 \nonumber \\
&\leq {(d_1-2)^2+(d_2-2)^2-p^n \bl - 2}, \label{m1}
\end{align}
since $d_1+d_2-4\geq 1$.
Note that since $k < \displaystyle\frac{p}{2}+1$, $d_1<\displaystyle\frac{p}{2}+1$ and $d_2 <\displaystyle\frac{p}{2}$. Hence, we can bound \eqref{m1} by  $\left(\displaystyle\frac{p}{2}-1\right)^2+\left(\displaystyle\frac{p}{2}-2\right)^2-p^n-2 = \displaystyle\frac{1}{2}p^2-3p-p^n+3 \leq -\displaystyle\frac{1}{2}p^2-3p+3<0$, since $n \geq 2$. So the numerator of $m_3$ is negative. Thus, the denominator of $m_3$, i.e $2(2\a - d_1 - d_2 + 2)(2\a + 1)\a$ is negative, which implies the factor $(2\a - d_1 - d_2 + 2)<0$.

Now, this forces the denominator of $m_4$ to be negative, which implies that the expression $(d_1^2 + d_2^2 - 2p^n(\a-1) - 3(d_1 + d_2) + 4-2\a)$ in the numerator of $m_4$ must be positive. If $\a \geq 2$, using the bounds for $d_1$ and $d_2$, we can bound the numerator of $m_4$ by
\begin{align*}
    (d_1^2 + d_2^2 - 2p^n(\a-1) &- 3(d_1 + d_2) + 4-2\a) \\ &\leq d_1^2+d_2^2-2p^n-3(d_1+d_2) \\
    &=\left(d_1-\frac{3}{2}\right)^2+\left(d_2-\frac{3}{2}\right)^2-\frac{9}{2}-2p^n\\
    &< \left(\frac{p+2}{2}-\frac{3}{2}\right)^2+\left(\frac{p}{2}-\frac{3}{2}\right)^2-\frac{9}{2}-2p^n\\
    &=\frac{1}{2}p^2-2p-2p^n-2 \leq -\frac{1}{2}p^2-2p-2<0,
\end{align*}
which is a contradiction.

Hence, $\a$ must be 1.
Replacing this for $m_4$, we have $$m_4=\displaystyle\frac{1}{2}\displaystyle\frac{p^n(d_1^2+d_2^2-3(d_1+d_2)+2)}{(d_1+d_2-4)(d_1+d_2-2)(d_1+d_2-1)}.$$
Observe that the factors in the denominator $(d_1+d_2-4)<(d_1+d_2-2)<(d_1+d_2-1)<\displaystyle\frac{p+2}{2}+\frac{p}{2}-1=p$. Moreover, $(d_1+d_2-2), (d_1+d_2-1) \geq 3$, so these two factors do not divide $p$. Hence, for $m_4$ to be an integer, they must divide $d_1^2+d_2^2-3(d_1+d_2)+2$.

However,
\begin{align*}
    d_1^2+d_2^2-3(d_1+d_2)+2&=(d_1+d_2-4)(d_1+d_2-1)-2(d_1-1)(d_2-1)\\
    &<(d_1+d_2-2)(d_1+d_2-1).\\
\end{align*}
This is a contradiction.

Therefore, there must be a fifth value in this Weil spectrum.
\end{proof}

\bl

\begin{remark}
For the case $n=1$ in \cref{theoremM_i}, taking odd prime p such that $p^{1/2}>2(k-1)$, and following the argument in the proof of \cref{theoremM_i} with this bound would yield the same conclusion (i.e the Weil spectrum has at least five values).
\end{remark}

\bl

\begin{proof}[Proof of Theorem \ref{5valuesconjecturep^n3mod4}]
This proof is in a similar flavor to the proof of \cref{theoremM_i}. Since $p^n \equiv 11 \pmod{12}$, $p^n \equiv 3 \pmod{4}$. By \cref{W-1}, $W_{L,s}(-1)=2p^n$. Since $d_1+d_2=3$, $W_{L,s}(1)=p^n$ by \cref{W1}. As in the last proof, let $m_1,m_2,m_3$ and $m_4$ be the number of elements whose Weil sum value is $-p^n, 0, 2\alpha p^n$ and $(2 \beta +1)p^n$, but here we take $\a=1,\b=0$, specifically.
Now, $d_1+d_2=3$, where $d_1, d_2 \geq 1$ (since $k \geq 2$) implies that one of them is $1$ and the other one is $2$. Hence, $d_1^2+d_2^2=5$.
Replacing these values of $\a,\b, d_1+d_2=3$, and $d_1^2+d_2^2=5$ in the solutions of $m_1,m_2,m_3$, and $m_4$ in the proof of \cref{theoremM_i}, we obtain $
m_1 = \displaystyle\frac{p^n(p^n-1)}{3},
m_2 =\displaystyle\frac{p^n(p^n-1)}{2},
m_3 =\displaystyle\frac{p^n(p^n-1)}{6},$ and $m_4 = p^n.$

Since $p^n \equiv 11 \pmod{12}$, $p \neq 3$. Since $m_1$ must be an integer, $p^n \equiv 1 \pmod{3}$, but this is a contradiction. Therefore, there exists a fifth value in this Weil spectrum.

\end{proof}

\bk

\section*{Acknowledgements}
The author would like to thank  Qing Xiang for suggesting the problem, Daniel Katz, \bl Wen-Ching Winnie Li, Bao Pham, and Fang-Ting Tu for helpful comments on an earlier draft to this paper, and Ling Long for fruitful discussions and feedback to this project. \bl The author is also grateful for the elaborate comments from the anonymous referee. All suggestions have improved this paper greatly. \bk

\bibliographystyle{amsplain}
\bibliography{ReferenceFull}

\end{document}


TITLE: Prove that $(x+1)^x-x^x(x-1)$ only has one (real) root
QUESTION [6 upvotes]: For $x>0$, I want to prove that $(x+1)^x-x^x(x-1)$ only has one root

REPLY [1 votes]: My answer is definitely going to pull from some ideas of Robert Israel.
So first we re-write the equation as
$$(1+1/x)^x = x-1$$ as previously suggested.  Then we consider these as two separate functions and ask when they intersect.
Let $f(x) = (1+1/x)^x$ and let $g(x)=x-1$.
Motivated from the following answer How to prove $(1+1/x)^x$ is increasing when $x>0$?
we see that $\log(f(x))' = \log(1+\frac{1}{x})-\frac{1}{x+1}$ is increasing and hence $f(x)$ is too. Likewise one can easily show that $\log(f(x))'' = -\frac{1}{x(1+x)^2}<0$ this shows that $f$ is log-convex and hence convex (this follows from the following The composition of two convex functions is convex since $Exp$ is convex and $\log(f)$ is convex hence $f=Exp[\log(f)]$ is too).
$g(1)=0$ and $f(1)=2$ so $g<f$ at $x=1$. Now that we have established $f$ is convex it follows that once $g$ (a straight line) is greater than $f$ it is always greater. It follows there is one and only one solution for the equation equivalently $g$ intersects $f$ exactly once.


TITLE: Is total energy density in energy-momentum tensor equal to total energy density with relativistic mass density?
QUESTION [0 upvotes]: I am studying the Tolman-Oppenheimer-Volkhoff (TOV) equations of hydrostatic equilibirium:
     $$\frac{dP}{dr}= −(G/c^2)(\epsilon + P)\frac{M + \frac{4\pi(r^3)P}{c^2}}{r^2− \frac{2GMr}{c^2}} \tag{1}$$
I want to choose a total central energy density $\epsilon$ and I am confused about two possible formulations
$$\epsilon=\rho c^2 \tag{2}$$
and
$$\epsilon= (1+a)\rho + k*{\rho^{\Gamma}/\Gamma} − 1 \tag{3}$$
Can anyone tell me in what case which equation is best for chosen energy density in solving equation (1) and also clarify, me whether the $\rho$ in equations (2) and (3) are relativistic energy densities or rest mass density?
I choose $\rho({\rm central})=5.3452·10^{14}$ g/cc but not deciding how to take $\epsilon$.

REPLY [1 votes]: Eq. (3) of OP seems to be polytropic EoS in which $\rho$ seems to be a rest mass density. Eq. (2) in this context wrong when $\epsilon$ is supposed to be the (internal) energy density or the $T_0{}^0$ component of the energy-momentum tensor of an ideal fluid.
Let me stress that the entire equation of state (EoS) and (internal) energy density vs. rest mass density "business" has nothing to do with "$E=m c^2$". Not saying anything about relativity: the internal energy density is in general not equal to the rest mass density; that is just thermodynamics. Maybe a personal remark: Einstein's "$E=m c^2$" is way to popular and a brutal understatement of the man's genius. When looking at it in geometrical units this relation is a triviality. $E^2=m^2 c^4 +p^2 c^2$ tells a story and has physical meaning but I am getting carried away.
The TOV eq. have to thermodynamic quantities in them: Pressure and (internal) energy density: no rest mast density or number density or anything. The eq. (1) of OP, my eq. (4) and an EoS in the form $\epsilon(P)$ form a complete and solvable system.
\begin{equation}
\frac{d M}{d r}=4\pi\, \frac{\epsilon(P(r))}{c^2}\, r^2\tag{4}
\end{equation}
I dislike the form of eq. (3) because it lacks proper definitions of the involved quantities and the pressure coefficient $k$ is not defined dimensionless apart from this it looks to miss some factors of $c$. And I have no idea what $a$ is supposed to be. If one assumes a polytropic EoS
\begin{equation}
P\equiv \kappa \epsilon_0 \left(\frac{n}{n_0}\right)^\gamma,\tag{5}
\end{equation}
then the first law of thermodynamic (at zero temperature)
\begin{equation}
\epsilon=\frac{d U}{d V}=-P+\mu\frac{dN}{dV}=-P+\mu n = -P + \left(\frac{d\epsilon}{d n}\right)n\tag{6}
\end{equation} dictates
\begin{equation}
\epsilon=m_B c^2 n + \frac{\kappa \epsilon_0}{\gamma-1}\left(\frac{n}{n_0}\right)^\gamma \tag{7}
\end{equation}
as a solution of the differential equation (6).
Where $P$ is pressure, $\epsilon$ is energy density $\mu$ is chemical potential $\mu=d\epsilon / d n$ and $n$ is number density. The constants related to the EoS are particle rest mass $m_B$, an arbitrary number density $n_0$ an arbitrary energy density $\epsilon_0$ a dimensionless pressure coefficient $\kappa$ and a relativistic polytropic index $\gamma$. I only introduced $n_0$ and $\epsilon_0$ so that $\kappa$ is dimensionless and independent of $\gamma$; if you do not do it this way, then $\kappa$ becomes dependent of $\gamma$, like $k$ of OP is dependent of $\Gamma$ just to ensure dimensional homogeneity.
$\epsilon$ has a rest mas contribution $m_B c^2 n=\rho c^2$ but also a pressure term. This is a consequence of the ansatz (5) and the classical first law of thermodynamics.
One can combine eq. (5) and (7) to
\begin{equation}
\epsilon=\frac{m_B c^2}{n_0}\left(\frac{P}{\kappa \epsilon_0}\right)^{1/\gamma}+\frac{P}{\gamma-1}, \tag{8}
\end{equation}
which is an EoS of the form $\epsilon(P)$  which can be used in the TOV eqs. I gave some working numbers for this kind of polytrope in my answer https://physics.stackexchange.com/q/296297
Eq. (3) of OP without a relation to the pressure is not enough to solve the TOV eqs. If one wants to integrate the TOV eqs. one has to choose the EoS, for the polytrope I presented meaning choosing $\kappa$, $\gamma$, $n_0$ and $\epsilon_0$, and a central pressure. IF you want some specific central energy density or number density (which can be related to rest mass density $n=\rho/m_B$) you could use eq. (4) to find the corresponding pressure, granted you choose some polytropic parameters.


text:
\section{Preliminaries}

We will mainly work with compact matrix quantum groups as defined by Worono-wicz in \cite{woronowicz91}. If $A$ is a *-algebra and $U \in \mat{n}{A}$ we denote by $\overline{U}$ the matrix whose entries are conjugated, i.e. $\overline{U}_{ij} = (U_{ij})^*$. \\
A pair $(A,U)$ of a C*-algebra $A$ and a unitary $U \in \mat{n}{A}$ is called a compact matrix quantum group if
\begin{itemize}
 \item $A$ is generated by the entries of $U$,
 \item there is a *-homomorphism $\Delta:A \rightarrow A \otimes_\textup{min} A$ mapping $u_{ij}$ to $\sum_k u_{ik} \otimes u_{kj}$,
 \item the matrix $\overline{U}$ is invertible.
\end{itemize}

A morphism of compact matrix quantum groups $(A,U) \stackrel{\phi}{\rightarrow} (B,V)$ is a *-homo-morphism $A \rightarrow B$ such that $\phi(u_{ij}) = v_{ij}$ where $U$ and $V$ must have the same size. There is at most one morphism from one quantum group to another. If there is a morphism $(A,U) \rightarrow (B,V)$ then we say that $(B,V)$ is a quantum subgroup of $(A,U)$. \\
Every compact matrix quantum group is also a compact quantum group, i.e. a C*-algebra $A$ with a *-homomorphism $\Delta:A \rightarrow A \otimes_\textup{min} A$ such that
\begin{itemize}
 \item $(\Delta \otimes \id) \circ \Delta = (\id \otimes \Delta) \circ \Delta$,
 \item $\cspan(A \otimes 1)\Delta(A) = \cspan(1 \otimes A)\Delta(A) = A \otimes A$.
\end{itemize}
A morphism of compact quantum groups $(A,\Delta_A) \stackrel{\phi}{\rightarrow} (B,\Delta_B)$ is a *-homomorphism from $A$ to $B$ such that $\Delta _B\circ \phi = (\phi \otimes \phi) \circ \Delta_A$.
Every morphism of compact matrix quantum groups is also a morphism of compact quantum groups.\\
We will also refer to a quantum group $(A,U)$ or $(A,\Delta)$ as $A$. \\
If $(A,\Delta_A)$ and $(B,\Delta_B)$ are quantum groups, then we denote by $(A,\Delta_A) \otimes (B,\Delta_B)$ the direct sum of quantum groups and by $(A,\Delta_A) * (B,\Delta_B)$ their free product. We will also write $A \otimes B$ and $A * B$. \\
A unitary corepresentation matrix of $(A,\Delta)$ is a unitary matrix $V \in \mat{m}{A}$ such that $\Delta(v_{ij}) = \sum_k v_{ik} \otimes v_{kj}$. In particular a one dimensional corepresentation matrix is just a unitary group-like element of $A$.

config: arxiv


TITLE: Electric field diagrams
QUESTION [0 upvotes]: I have these 2 diagrams of an electric field, for the second one i’m not too sure how to draw it because how do i respect the 2 conditions:

the field lines need to be perpendicular to the equipotential lines.
the equipotential lines are parallel to the surface.
is there something wrong im doing or misunderstanding?
Any suggestions as a to how the sketches should look?

REPLY [0 votes]: Do you know of the principle of mirror charges, it is the easiest way of drawing the pictures.the potential lines are parallel only very close to the boundary. Your picture has alamos no lines perpendicular to the metal?


TITLE: Understanding algorithm to show $(x+\partial_x)e^{-x^2/2}=0$ using python?
QUESTION [4 upvotes]: I am given the following Python code which is supposed to verify numerically that
$$(\partial_x + x)e^{-x^2/2}=0.$$
The algorithm does this, by transforming everything into a Fourier basis and then verifying it using the Fourier transform. However, I have some difficulties understanding the mathematical basis of this algorithm. In short, my question is:
When the final result is computed as np.dot( D, yl ), what is that actually in mathematical terms that we are computing here?
 import numpy as np

 ## non-normalized gaussian with sigma=1
 def gauss( x ):
return np.exp( -x**2 / 2 )

  ## interval on which the gaussian is evaluated
 L = 10
 ## number of sampling points
 N = 21
 ## sample rate
 dl = L / N
 ## highest frequency detectable
 kmax= 1 / ( 2 * dl )

 ## array of x values
 xl = np.linspace( -L/2, L/2, N )
 ## array of k values
 kl = np.linspace( -kmax, kmax, N )

 ## matrix of exponents
## the Fourier transform is defined via sum f * exp( -2 pi j k x)
  ## i.e. the 2 pi is in the exponent
 ## normalization is sqrt(N) where n is the number of sampling points
 ## this definition makes it forward-backward symmetric
 ## "outer" also exists in Matlab and basically does the same
exponent = np.outer( -1j * 2 * np.pi * kl, xl )
## linear operator for the standard Fourier transformation
 A = np.exp( exponent ) / np.sqrt( N )

 ## nth derivative is given via partial integration as  ( 2 pi j k)^n f(k)
 ## every row needs to be multiplied by the according k
 B = np.array( [ 1j * 2 * np.pi * kk * An for kk, An in zip( kl, A ) ] )

 ## for the part with the linear term, every column needs to be multiplied
## by the according x or--as here---every row is multiplied element
 ## wise with the x-vector
C = np.array( [ xl * An for An in  A ] )

 ## thats the according linear operator
 D = B + C

 ## the gaussian
 yl = gauss( xl )

 ## the transformation with the linear operator
 print(  np.dot( D, yl ).round( decimals=9 ) )
## ...results in a zero-vector, as expected

REPLY [3 votes]: The line

print(  np.dot( D, yl ).round( decimals=9 ) )

calculates the integrals of the Fourier transform at set of frequencies using the rectangle integration method:
\begin{equation}
F(k) = \int\limits^{\infty}_{-\infty}e^{-ikx}\left(\partial_x + x\right)e^{-\frac{x^2}{2}}dx=\left.e^{-ikx}e^{-\frac{x^2}{2}}\right\vert^{\infty}_{-\infty} +\int^{\infty}_{-\infty}(ik + x)e^{-ikx}e^{-\frac{x^2}{2}}dx=\int^{\infty}_{-\infty}(ik + x)e^{-ikx}e^{-\frac{x^2}{2}}dx\approx\int\limits^{5}_{-5}\left(ik + x\right)e^{-ikx}e^{-\frac{x^2}{2}}dx\approx\sum\limits^{N}_{j = 0}(ik + x_j)e^{-ikx_j}\Delta{x}e^{-\frac{x^2_j}{2}}=\sum\limits^{N}_{j = 0}v^{(k)}_jf_j=(\vec{v}^{(k)}, \vec{f}),
\end{equation}
where $\Delta{x}$ is the coordinate step, $v^{(k)}_j = (ik + x_j)e^{-ikx_j}\Delta{x}$, $f_j = e^{-\frac{x^2_j}{2}}$. Since we want to calculate the following ressult vector $\vec{r}$: $r_p = F(k_p)$, we can to introduce the matrix $D$:
\begin{equation}
D =
\begin{pmatrix}
(\vec{v}^{(k_0)})^{\intercal}\\
(\vec{v}^{(k_1)})^{\intercal}\\
\vdots\\
(\vec{v}^{(k_n)})^{\intercal}
\end{pmatrix}
\end{equation}
or
\begin{equation}
D =
\begin{pmatrix}
v^{(k_0)}_0 & v^{(k_0)}_1 &\ldots & v^{(k_0)}_N\\
v^{(k_1)}_0 & v^{(k_1)}_1 & \ldots & v^{(k_1)}_N\\
& &\ldots &\\
v^{(k_n)}_0 & v^{(k_n)}_1 & \ldots & v^{(k_n)}_N
\end{pmatrix},
\end{equation}
so the vector $\vec{r}$ can be calculated by the following way
\begin{equation}
\vec{r} = D\vec{f}.
\end{equation}
The last equation corresponds to np.dot( D, yl ) in the code.
It is also important to note that according to the Kotelnikov-Nyquist–Shannon theorem, there will be no loss of information only if (spectral density) the highest frequency of the useful signal is equal to half or less than the sampling frequency. Half of the sampling frequency is called the Nyquist frequency:

kmax= 1 / ( 2 * dl )


TITLE: Find the sum of all possible values of $\cos 2x + \cos 2y + \cos 2z.$
QUESTION [1 upvotes]: Let $x$, $y$, and $z$ be real numbers such that $ \cos x + \cos y + \cos z = \sin x + \sin y + \sin z = 0 $. Find the sum of all possible values of $ \cos 2x + \cos 2y + \cos 2z $.
Here is what I have done so far
$$ \cos x + \cos y = -\cos z $$
$$ (\cos x + \cos y)^{2} = (-\cos z)^{2} $$
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y = \cos^{2}z $$
likewise,
$$ \sin x + \sin y = -\sin z $$
$$ (\sin x + \sin y)^{2} = (-\sin z)^{2} $$
$$ \sin^{2} x + \sin^{2} y + 2 \sin x \sin y = \sin^{2}z $$
from this, you get
$$ \cos^{2} x + \cos^{2} y + 2\cos x \cos y + \sin^{2} x + \sin^{2} y + 2\sin x \sin y = \sin^{2}z + cos^{2}z $$
$$ 1 + 1 + 2(\cos x \cos y + \sin x \sin y) = 1 $$
$$ \cos x \cos y + \sin x \sin y = -\frac{1}{2} $$
$$ \cos(x-y) = -\frac{1}{2} $$
$$ x-y = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
likewise,
$$ \cos (x-z) = -\frac{1}{2} $$
$$ x-z = \frac{2 \pi}{3}, \frac{4 \pi}{3} $$
where do I go from here?

REPLY [0 votes]: Continuing from your work, we have $y-z\equiv z-x\equiv x-y\equiv\frac23 r\pi\;\,(\!\!\!\!\mod2\pi)$, with $r\in\{1,2\}$ ($r$ is constant, since the three differences sum to zero, modulo $2\pi$). So $$\cos2x+\cos2y+\cos2z=\cos2x+\cos(2x-\tfrac43r\pi)+\cos(2x+\tfrac43r\pi)\qquad\qquad$$
$$=\cos2x+2\cos2x\cos\tfrac43r\pi$$
$$=0.\qquad\qquad\qquad\qquad\quad$$


text:
\section{Introduction}\label{Introduction}
Environmental awareness, supportive policy mechanisms and falling costs of solar photovoltaic (PV) components are driving an accelerating uptake of distributed energy resources in low-voltage (LV) residential areas. It has become a challenge for many distribution network service providers (DNSPs) to host new solar PV installations~\cite{Chaudhary2018}. Additional PV output generated in the same area results in voltage surges, thus increasing the risk of appliance damage. The consequent overvoltage protection settings will disconnect PV inverters and can also initiate voltage fluctuations across the network~\cite{Farivar2011}.The voltages issues are exacerbated in the networks with a high build-up of, so-called, \emph{legacy} inverters without grid support functions~\cite{Wyndham2016}.

Multiple studies have shown improved voltage control using autonomous inverters with reactive power (Volt/VAr) droop curve~\cite{Seuss2016, Anonas2018, Ali2018}, active power (Volt/Watt) droop curve~\cite{GhapandarKashani2017, Kraiczy2018, Ali2015}, and a combination of both~\cite{Giraldez2017,Giraldez2018}. All new solar PV installations must include autonomous voltage regulation based on a droop control strategy as required under the IEEE 1547 Standard on Smart Inverters~\cite{Stice2019}. Autonomous inverters are network-agnostic, which makes them simple and easy to deploy, but it also means that they provide only suboptimal operations of the network~\cite{Baker2018}, particularly in the areas with a large number of customers per distribution transformer~\cite{Parajeles2017}. While passive inverter control can reduce the number of customers experiencing overvoltages~\cite{Hoke2018}, the limitations of such control approach become particularly obvious when operating in a network with high PV penetration~\cite{Lusis2019}.

Improved voltage regulation in a PV-rich distribution network was achieved in~\cite{Dallranese2015} by implementing bidirectional communication between  inverters and a central node. It utilises smart meter data of customers' voltages and net loads across the network to determine and deploy the optimal active and reactive power setpoints to each of the controllable inverters. Existing studies have demonstrated the operation of coordinated inverters alone~\cite{Guggilam2016}, in combination with battery storage~\cite{DallAnese2018} and electric vehicles~\cite{Chapman2018}. To the authors' knowledge, all previous studies assume that all distributed energy resources (DER) in the network can be controlled. This is optimistic, given the wide deployment of passive inverters. Moreover, some customers may be unwilling to share inverter data or give up the control over their assets~\cite{Zeraati2018}. Thus, in this paper, we simulate the deployment of coordinated inverters in a LV network with a significant share of legacy and autonomous inverters and demonstrate the aggregated ability of coordinated inverters to mitigate voltage disturbances.

Centralised PV inverter control can also enable grid support services (GSS) and establish new relationships between the system operator and the entity in charge of DER~\cite{Rasmussen2018}. The aggregated output capacity of coordinated inverters can be seen as a \emph{virtual power plant} (VPP) and the entity in charge of coordinating inverters as a \emph{VPP aggregator}~\cite{Pasetti2018}. PV inverters have the flexibility and fast response needed to provide up-regulation ($\mathit{UR}$) and down-regulation ($\mathit{DR}$) services. That means that it can meet the need to change the frequency or respond to an event by increasing/decreasing the power output. The calculation of feasible and flexible operating regions and maximum grid support capacity for a VPP was demonstrated in~\cite{Riaz2019}. However, the network operator may not see the expected change in the frequency following a response from the VPP: since ($\mathit{UR}$) and ($\mathit{DR}$) delivery by coordinated inverters alter local voltages, it will trigger inverters with passive control to change their output (as per local voltage control settings). Autonomous inverters' sliding along the droop curve or potential disconnection due to overvoltages induced by the VPP aggregator should be quantified and accounted for before making a GSS capacity offer to the market operator.

By simulating increasing PV penetration levels, we demonstrate that coordinated inverters are significantly more effective in voltage regulation than autonomous inverters while avoiding unnecessary curtailment, which occurs in the scenario with non-exporting inverters. We have also demonstrated that offering grid support services can serve as a financial incentive for the CIC deployment. This paper can be distinguished from other studies by the fact that we have considered the existing inverter infrastructure in LV distribution networks, with 30\% PV penetration level consisting of existing autonomous and legacy inverters. The cumulative effects of inverter interaction are quantified in terms of voltage disturbances and power curtailment. For the given networks, it is shown that, by deploying as many coordinated inverters as the existing passive inverters, voltage issues can be prevented in 85\% of PV locations.

This paper is organised as follows: Inverter models are introduced in Section 2, and the network operation and simulation setup are explained in Section 3. The provision of grid support services by coordinated inverters is formulated in Section 4. The results are discussed in Section 5 and a summary of the results and conclusions is given in Section 6.


TITLE: Showing if a function is injective or surjective problem
QUESTION [1 upvotes]: $F : \Bbb{P}(X) \rightarrow \Bbb{P}(X) ; U \rightarrow (U-A) \cup (A-U)$
My intuition has been telling me that this function is bijective but I having the most difficult time trying to show this. Any help would be appreciated, thank you!!
edit: So far, I introduced sets $U_1$ and $U_2$, I am assuming $f(U_1)=f(U_2)$ and trying show $U_1=U_2$ in order to show injection. So I have $(U_1-A)\cup(A-U_1)=(U_2-A)\cup(U_2-A)$ and I'm unfamiliar with how to manipulate this into $U_1=U_2$. I haven't worked much with set notation before.
As for showing that f is surjective, I've tried drawing some Venn diagrams to help me process the question but I'm not sure where to go. Thank you for the quick replies!

REPLY [0 votes]: Hint $\Delta$ is associative, hence
$$A \Delta  (A \Delta W)= (A \Delta A) \Delta W=W$$


TITLE: How to solve this arithmetic sequence with distance $a$?
QUESTION [0 upvotes]: Let $\{a_{n}\}$ be an arithmetic sequence with distance $a$.  Find
$$\lim_{n\to\infty}\frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\dotsb +\frac{1}{\sqrt{a_{n}}+\sqrt{a_{n+1}}}\right).$$
I don't know how to solve it. Hope you can help me.

REPLY [0 votes]: \begin{align}\lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}+\sqrt{a_{i+1}}} &= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}+\sqrt{a_{i+1}}} \frac{\sqrt{a_{i+1}}-\sqrt{a_i}}{\sqrt{a_{i+1}}-\sqrt{a_i}} \\
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}}\sum_{i=1}^{n} \frac{\sqrt{a_{i+1}}-\sqrt{a_i}}{a} \\
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} \frac{\sqrt{a_{n+1}}-\sqrt{a_1}}{a} \\
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} \frac{\sqrt{a_1+na}-\sqrt{a_1}}{a} \\
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{n}} \frac{na}{a}\frac{1}{\sqrt{a_1+na}+\sqrt{a_1}} \\
&= \lim_{n \rightarrow \infty} \frac{\sqrt{n}}{\sqrt{a_1+na}+\sqrt{a_1}} \\
&= \lim_{n \rightarrow \infty} \frac{1}{\sqrt{\frac{a_1}{n}+a}+\sqrt{\frac{a_1}{n}}}
\\&=\frac{1}{\sqrt{a}}\\
\end{align}


TITLE: Fourier Transform of Dirac Delta Function
QUESTION [3 upvotes]: Dirac's delta function represents a wave whose amplitude goes to infinity as its duration in time goes to zero.  It is a pulse of infinite intensity but infinitesmal duration.
Please provide an exact description of what the Fourier transform of the delta function looks like.

REPLY [1 votes]: To complete Marvis' answer, the Dirac comb $\delta_0$ is a measure, and there is a well developed theory of Fourier Transform for positive definite measures and also for finite measures ($\delta_0$ is both). In this setting the Fourier Transform of $\delta_0$ is actually the Lesbegue measure on $\hat{\mathbb R}$.
Also, in this context, the meaning of the relation
$$\widehat{\delta_0} = 1_{\mathbb R}$$
is just the Bochner's Theorem for positive definite functions: The Fourier Transform of a finite positive measure is just the Positive definite function given by Bochner's Theorem.


TITLE: Relation between averages in the sphere and the ball
QUESTION [2 upvotes]: Let $\Omega$ be a domain in $\mathbb R^N$ and $u \in C(\Omega)$. If $\omega_N$ denotes the volume of $B_1(0) \subset \mathbb R^N$ and $0 < r < d(x, \partial \Omega)$, the averages are
$$
A_b(x) = \frac{1}{\omega_N r^N} \int_{B_r(x)} u(y) \ dy
$$
in the ball, and
$$
A_s(x) = \frac{1}{N \omega_N r^{N - 1}} \int_{\partial B_r(x)} u(y) \ d\sigma(y)
$$
in the sphere.
We know that if $u$ is harmonic, then $A_b(x) = A_s(x) = u(x)$.

In general, can one say something like $A_s \leq (\geq) A_b$?

I kindly ask for a hint, in case the answer is positive, or for examples, in case the answer is negative.
EDIT:
Ted's hint made me think of the following: take $u$ to be negative inside the ball and $0$ on the boundary. Then of course $A_b < 0 = A_s$. On the other hand, if $u$ is positive, it is clear that $A_b > 0 = A_s$. We conclude that the answer to the question is negative in general.
This seems to be very simple, am I missing something?

REPLY [2 votes]: The answer seems to be no in the case $N=3$. Fix $\epsilon > 0$ small, and
take $u$ to be a
continuous function with $u(x,y,z) = 1$ when $x > \epsilon$
and $u(x,y,z) = 0$ when
$x < 0$. Consider averages over $B = B_2(1,0,0)$ and $B' = B_2(-1,0,0)$ and their boundaries.
The sphere $\partial B$ has, modulo $\epsilon$,  $3/4$ of a diameter
in the region where $u= 1$, and by a result of Archimedes also has
$3/4$ of its area there, so that average is $3/4$. But $B$ itself has more
than $3/4$ of its volume in the region where $u=1$, so
$$ A_b > A_s $$
for this case.
The sphere $\partial B'$ has, modulo $\epsilon$, $1/4$ of its area in the
region where $u = 1$, so that average is $1/4$. But $B'$ has less than
$1/4$ of its volume in that region, so
$$
 A_b < A_s
$$
for this case.
I don't know what happens in other dimensions.


TITLE: Symmetries: from a classical field theory to quantum
QUESTION [4 upvotes]: My background about this argument:
a) Let's consider a classical field theory, where  $\mathcal{L}(\phi(x),\partial_{\mu}\phi_(x))$ is the lagrangian density.
A symmetry is a transformation $\phi(x) \rightarrow\phi'(x')=f(\phi(x))$ of the fields that leaves the lagrangian density $\mathcal{L}$ invariant (more general definitions are possible, but I hope this one is sufficient). Noether's theorem establishes that continous symmetry groups correspond to conserved currents and charges.
b) In the corresponding quantum field theory, fields are promoted to fields operators acting on the Hilbert space of states.
In this case, symmetries are represented by probability conserving operators. According to Wigner's theorem, these are unitary operators (or antiunitary) on the Hilbert space of states.
Conserved quantities should instead arise as operators that commute with the Hamiltonian.
The latter is "imported" in the quantum field theory just by considering directly the classical expression, where field operators appear instead of classical fields.
Example:
Dirac free lagrangian density is:
$$\mathcal{L}=\overline{\psi}(i\gamma^\mu\partial_\mu-m)\psi$$
Parity is a (discrete) symmetry of this lagrangian at a classical level, since ($'$ denotes parity-transformed quantities):
$$\mathcal{L'}=\overline{\psi'}(i\gamma^\mu\partial_\mu'-m)\psi'=\overline{\gamma^0\psi}(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+(i\gamma^\mu\partial_\mu'-m)\gamma^0\psi=\psi^+\gamma^0(i\gamma_\mu\partial^\mu-m)\psi=\mathcal{L}.$$
Questions:
a) Does the classical symmetry imply the existence of an operator that in some sense "represents" parity symmetry at a quantum level?
b) If yes, what is the connection between the operator and the classical parity transformation?
c) Does a conserved quantity at a quantum level (operator that commute with hamiltonian) arise in this case as a consequence of the existence of the parity symmetry?
d) Is it possible to generalize the considerations about the special simple case of parity?

REPLY [4 votes]: Often, a classical symmetry will become a symmetry of the quantum theory, but not always. A symmetry of the classical theory which fails to be a symmetry of the quantum theory is called an anomaly.
One way to express the idea of a conserved current for a symmetry of a quantum level is the Ward identity (see, eg, Eq 6.45 of the chapter "Symmetries in QFT" in David Skinner's Quantum Field Theory II course)
\begin{equation}
\partial_\mu \big\langle j^\mu(x) \prod_{i=1}^N O(x_i)\big\rangle = -\sum_{n=1}^N \delta(x-x_i)\big\langle \delta O(x_i) \prod_{j\neq i} O(x_j)\big\rangle
\end{equation}
where $j^\mu(x)$ is the Noether current for the symmetry (which is an operator in QFT), $O(x_i)$ is a local operator defined at the point $x_i$, and $\delta O(x_i)$ is the transformation of $O(x_i)$ under the symmetry. The terms on the right hand side are called contact terms, and are  quantum corrections to the classical Noether's theorem.
When the symmetry is anomalous, there is an additional term in the Ward identities due to the anomaly
\begin{equation}
\partial_\mu \big\langle j^\mu(x) \prod_{i=1}^N O(x_i)\big\rangle = -\sum_{n=1}^N \delta(x-x_i)\big\langle \delta O(x_i) \prod_{j\neq i} O(x_j)\big\rangle + \big\langle F(x) \prod_{i=1}^N O(x_i) \big\rangle
\end{equation}
where the last term represents the failure of the current to be conserved due to the anomaly.
Note that when we take the $O(x_i)$ to simply be $1$, we get the special case
\begin{equation}
\partial_\mu \big\langle j^\mu(x) \big\rangle = F(x)
\end{equation}
where $F=0$ if the symmetry is not anomalous.


TITLE: Graph automorphism group conjugacy classes
QUESTION [0 upvotes]: Let $G$ be the automorphism group on a graph $\Gamma(V,E)$. Why is it that group elements belonging to the same conjugacy class have the same cycle type? For example, if $a,b\in G$ are two automorphisms on $\Gamma$, such that $g^{-1}ag=b$ for some $g\in G$, then $a,b$ have the same cycle type?

REPLY [1 votes]: This is because $$\sigma (a_1\ \ldots \ a_n)(b_1\ \ldots \ b_k)\sigma^{-1}=\sigma (a_1\ \ldots \ a_n)\color{red}{\sigma^{-1}\sigma}(b_1\ \ldots \ b_k)\sigma^{-1}$$
and $$\sigma (a_1\ \ldots \ a_n)\sigma^{-1}=(\sigma(a_1)\ \ldots \ \sigma(a_n))$$


TITLE: How to find the position vector at time t, when given plane curve equation?
QUESTION [0 upvotes]: I have a question about the following:
An object moves to the right along the plane curve $y=x^2$ with constant speed $v=5$. find the velocity and acceleration of the object when it is at the point (1,1).
It is concluded from this that the position of the object at time t is: $$\textbf{r}=x\textbf{i}+x^2\textbf{j}$$
But why is this exactly, how can we conclude this from the given info of this question?
If you know the answer and would like to share it with me, it would be very much appreciated :)
thanks in advance

REPLY [1 votes]: Let's say we denote the time as $t$ and the position of the object is
$$
r(t) = \begin{pmatrix} x(t) \\ y(t) \end{pmatrix}
$$
or in your notation
$$
r(t) = x(t) \mathbf{i} + y(t) \mathbf{j} = x(t) \begin{pmatrix}1 \\ 0 \end{pmatrix} + y(t) \begin{pmatrix}0 \\ 1 \end{pmatrix}.
$$

Since it has to be on the curve $y = x^2$, we know that $y(t) = x(t)^2$, hence
$$
r(t) = \begin{pmatrix} x(t) \\ x(t)^2 \end{pmatrix}
$$

It is important to notice that $r(t) \neq \begin{pmatrix}t \\ t^2 \end{pmatrix}$! Since then it would not have the correct speed! So, in your question, the variable $x$ depends on time!

It's velocity should be $v = 5$, so let's compute the velocity:
$$
v = \Vert \dot r(t) \Vert = \left\Vert \begin{pmatrix} \dot x(t) \\ 2 x(t) \dot x(t) \end{pmatrix} \right\Vert = \sqrt{ (1+4 x(t)^2) } \Vert \dot x(t) \Vert.
$$

From this equation, you can see that
$$
\dot x(t) = \pm \frac{5}{\sqrt{1+4x(t)^2}}.
$$
I leave the rest to you. You can use the above equation to determine $\dot x(t)$ and with that, you can find $\dot y(t)$ as well.


TITLE: The Praying Eyes theorem generalized
QUESTION [2 upvotes]: Is there an obvious way to slice two spheres (no necessarily equal)
simultaneously such that the sections share common areas? I mean, I cannot see another way (easier) different from the way provided by the Praying Eyes theorem and related theorems.
You may want to take a look of this first:
http://geometriadominicana.blogspot.com/2014/03/praying-eyes-theorem.html
Edit: just wanted to say that the construction is possible for three spheres too. For the construction in 2D take a look of this page http://www.cut-the-knot.org/pythagoras/LightingTheBall.shtml

REPLY [2 votes]: EDIT: the construction far below suffices when the center P is outside the circle with center Q, including when the circles do not intersect at all. For the missing case, I'm thinking a segment through an intersection point of the two circles, the only requirement being same chord length. I'll think about it.
EDIT TOO: if the circles intersect, we can simply take their intersection as the collinear chords of equal length:

ORIGINAL:This is actually a two dimensional construction, no continuity argument required; also, it is sometimes impossible.
Start with a circle with center P of small diameter (red) and a circle of center Q of larger diameter. Draw the red segment as a chord of the second circle, and find the segment midpoint W. Draw another circle around center Q, this time with radius QW.
Next, draw the midpoint of the segment PQ, at M. Draw a semicircle around M with radius MQ. This intersects the QW circle in a point, call this T.
By construction, QTP is a right angle. Since length QT is the same as QW, the intersection of the line PT with the larger circle is the same length as the diameter of the smaller circle, always in red.
In higher dimension, instead of just the line PT, take the codimension one hyperplane... that is, take any 2-plane through both centers and perform this construction, then fill in


TITLE: Support of a probability measure on a disconnected space
QUESTION [1 upvotes]: Let $\mu$ be a Borel probability measure with compact support on a disconnected locally compact Hausdorff space $X$. Let $Y$ be a compact open subset of $X$ such that the intersection of $Y$ with the support of $\mu$ is non-empty. Then when can we conclude that $\mu(Y)=1$ ?
Thank you very much, in advance.

REPLY [0 votes]: There maybe missing hypothesis (or possibly I misunderstood the OP)  but the answer to you question is: no, we can't.

The probability measure
$$
\mu(dx):=\frac{1}{2}\delta_0(dx) +\frac{1}{2}\sum_{n\in\mathbb{N}}2^{-n}\delta_{1/n}(dx)
$$
is supported in a totally disconnected, Hausdortt and compact set $X=\{1/n:n\in\mathbb{N}\}\cup\{0\}$.

The set $Y=\{1\}$ is clopen and $Y\cap X=Y\neq\emptyset$

$\mu(Y)=\frac{1}{4}$.


TITLE: Inequality with number of digits
QUESTION [0 upvotes]: Let $a$ and $b$ be two natural numbers that have an equal number $n$ of digits in their decimal expansions. The first $m$ digits (from left to right) of the numbers $a$ and $b$ are equal. Prove that if $m > \dfrac{n}{2}$, then $$a^{\frac{1}{n}}-b^{\frac{1}{n}} < \dfrac{1}{n}.$$

Let $a > b$. We notice that $a-b < 10^{n-m}$ and so $a-b < 10^{\frac{n}{2}}$. How do we continue from here?

REPLY [1 votes]: From the hypothesis, we can write
$$a=10^{n-m}x+y,\qquad\text{and}\qquad b=10^{n-m}x+y',$$
where $x,y$ and $y'$ are positive integers such that $y,y'<10^{n-m}$. Now,
$$a^{1/n}=(10^{n-m}x+y)^{1/n}\leq(10^{n-m}(x+1))^{1/n}=10^{1-m/n}(x+1)^{1/n}$$
and
$$b^{1/n}=(10^{n-m}x+y')^{1/n}\geq(10^{n-m}x)^{1/n}=10^{1-m/n}x^{1/n}.$$
Therefore
$$a^{1/n}-b^{1/n}\leq 10^{1-m/n}(x+1)^{1/n}-10^{1-m/n}x^{1/n}=
10^{1-m/n}\left((x+1)^{1/n}-x^{1/n}\right).$$
But $m/n\geq1/2$, so $1-m/n\leq 1/2$, and hence
$$a^{1/n}-b^{1/n}\leq \sqrt{10}\left((x+1)^{1/n}-x^{1/n}\right).$$
Now, since $(x+1)^{1/n}-x^{1/n}\leq \frac1nx^{1/n-1}$ (see below),
$$a^{1/n}-b^{1/n}\leq \sqrt{10}\frac1nx^{1/n-1}\leq \frac{\sqrt{10}}{9^{1-1/n}}\frac1n.$$
For $n\geq3$,
$$a^{1/n}-b^{1/n}\leq \frac{\sqrt{10}}{9^{2/3}}\frac1n<\frac1n,$$
and for $n<3$ there is nothing to prove.
Note (proof of the inequality $(x+1)^{1/n}-x^{1/n}\leq \frac1nx^{1/n-1}$): we write
$$(x+1)^{1/n}-x^{1/n}=x^{1/n}((1+1/x)^{1/n}-1)$$
and apply Bernoulli's inequality (note that $1/n<1$):
$$x^{1/n}((1+1/x)^{1/n}-1)\leq x^{1/n}\frac1{nx}=\frac1nx^{1/n-1}.$$


TITLE: Is the function $G$ right continuous?
QUESTION [1 upvotes]: Suppose $S\subset\mathbb R$ such that $S\cap(x,\infty)\neq\phi$ for every $x\in\mathbb R$ and suppose that $g:S\to\mathbb R$ is a non-decreasing bounded function. Define $G:\mathbb R\to\mathbb R$ as: $$G(x)=\inf_{y\in S\cap(x,\infty)}g(y)$$for any $x\in\mathbb R$.
Is the function $G$ right continuous?

I observe that $G$ is non-decreasing. Somehow I believe that it is not true that $G$ is right continuous. But I cannot really prove it. I suppose the reason why I believe this is that to say that $G$ is right continuous, I would need $g$ to be right continuous, which is not given. Again, I may be wrong.

REPLY [2 votes]: Well, note that for every decreasing sequence $x_n \to x$ you have $\bigcup_n (x_n,\infty) = (x,\infty)$. This implies $\lim_{n \to \infty} \inf_{S\cap(x_n,\infty)} g =\inf_n \inf_{S\cap(x_n,\infty)} g = \inf_{S\cap(x,\infty)}g$. Since you have in general $\inf \bigcup_{i\in I} A_i = \inf_{i\in I} \inf A_i$