TITLE: Energy of a signal simplification
QUESTION [0 upvotes]: I'm learning about signals from the textbook Signals and Systems Second Edition, by Oppenheim and Willsky in order to learn more about Fourier Analysis and I'm having difficulty with the Energy equation that I found for a complex periodic signal. For this type of signal, we have the following basic form for the energy of the signal over 1 period:
$$E_{period} = \int_{0}^{T_0} |e^{j\omega_0t}|^2dt$$
which then simplifies to just:
$$E_{period} = \int_{0}^{T_0} 1 \cdot dt$$
I'm assuming that the text is saying $e^{j\omega_0t} = e^{j\omega_0T} = 1$ but I don't understand how it arrives to this conclusion. Based on earlier information in the text, I know that for any complex periodic signal:
$$x(t) = e^{i\omega_0t} = e^{i\omega_0(t+T)}$$
and this equation is used to obtain :
$$x(t) = e^{i\omega_0t} = e^{i\omega_0t}e^{i\omega_0T} \Rightarrow e^{i\omega_0T} = 1$$
However, I do not see how this can also be used to deduce that $e^{j\omega_0t}=1$ as well. So, could someone please help clear up why the signal inside the integral simplifies to 1?
Note: at this point in the book, it is just going over the different types of signals and has not yet touched Fourier Analysis.

REPLY [1 votes]: Parseval's theorem for the Fourier Transform is defined as follows:
$$E_x = \int_{-\infty}^{\infty}|x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty}|X(\omega)|^2 d\omega $$
To answer your equation, please take note of your notation.
Given the complex signal:
$$ x(t) = e^{j\omega_0 t} $$
This implies:
$$ |x(t)| = |e^{j\omega_0 t}| = 1 $$
Comments: Recall the $|\cdot|$ function means the length of the complex number. $e^{j\omega_0 t}$ is a unit circle in the complex plane, with the angular frequency of $\omega_0$. Even if the complex signal is periodic over $T$, i.e. $e^{j\omega_0 (t+T)}$, the length remains the same, i.e. $|e^{j\omega_0 t}|=|e^{j\omega_0 (t+T)}|=1$. Hope it clears your doubts.


text:
\begin{document}

\begin{abstract}
Fix a positive integer $N\geq2$. For a real number $x\in[0,1]$ and a digit $i\in\{0, 1,...,N-1\}$, let $\Pi_i(x, n)$ denote the frequency of the digit $i$ among the first $n$ $N$-adic digits of $x$. It is well-known that for a typical (in the sense of Baire) $x\in[0, 1]$, the sequence of digit frequencies  diverges as $n\ra\infty$. In this paper we show that for any regular linear transformation $T$ there exists a residual set of points $x\in[0, 1]$ such that the $T$-averaged version of the sequence $(\Pi_i(x, n))_n$ also diverges significantly.
\end{abstract}

\maketitle

\section{Statement of results}

Fix a positive integer $N\geq2$. Throughout this paper, we will consider the unique, non-terminating, base $N$ expansion of a number $x\in[0, 1]$, written as $x = \frac{d_1(x)}{N}+ \frac{d_2(x)}{N^2} + ... + \frac{d_n(x)}{N^n}+ ...$ with $ d_i(x)\in\{0,1,...N-1\}$. For each digit $i\in\{0,1,...,N-1\}$, we will write \[
\Pi_i(x, n) :=\frac{1}{n}\left|\left\{   1\leq j\leq n\; :\; d_j(x)=i    \right\}\right|
\]
for the frequency of the digit $i$ among the first $n$ digits of $x$. Let $\Delta_N$ denote the family of $N$-dimensional probability vectors, that is
\[
\Delta_N :=\left\{\left (p_0, p_1,...,p_{N-1}\right)\; :\; p_i\geq 0,\;\sum_{i=0}^{N-1}p_i = 1  \right\}
\]
and write
$$
 \mathbf{\Pi}(x, n) = \left(\Pi_0(x, n),\Pi_1(x, n),...,\Pi_{N-1}(x, n)\right)
$$
for the vector of frequencies of digits. Then clearly for every natural number $n$ the vector of frequencies $\mathbf \Pi(x, n)$ lies in the simplex $\Delta_N$, and consequently, the set of accumulation points of the sequence $\left(\mathbf\Pi(x, n)\right)_n$ is a subset of $\Delta_N$.

Borel's famous normal number theorem states that for Lebesgue almost every point in $[0,1]$ its vector of frequencies converges towards the uniform probability vector $\left(\frac{1}{N},...,\frac{1}{N} \right)$ with respect to $\norm{.}_1$ as $n\ra\infty$. More generally Birkoff's Ergodic Theorem tells us that for each probability measure $\mu$ of $[0,1]$ which is invariant under the transformation $x\mapsto Nx \mod1$, the sequence $\mathbf \Pi(x, n)$ converges to a specific probability vector $p_\mu\in\Delta_N$ for $\mu$ almost every $x\in[0,1]$ as $n\ra\infty$. Yet still from a topological viewpoint there is a completely different picture.

We recall that in a metric space $X$, a set $R$ is called residual if its complement is of the first category. Also, recall that we say that a typical element $x\in X$ has property $P$ if the set $ R :=\{ x \in X \; :\;  x \text{ has property }P\}$ is residual. We refer the reader to Oxtoby \cite{Ox} for more details. Recently, the study of the limiting behaviour of the sequence of frequencies $\left(\Pi_i(x,n)\right)_n$ from a topological viewpoint has enjoyed a great interest from many authors \cite{Vo, Sa1, APT, CZ, Ol, Sa2, Si, SS, SS2}. In particular, perhaps unexpectedly, Olsen showed that for a typical $x\in[0, 1]$, the set of accumulation points of $(\mathbf\Pi(x, n))_n$ is all of $\Delta_N$.

\begin{theorem}[A, \cite{Ol}]\label{olsen}
The set \[
\left\{x \in [0, 1]\;:\; \text{The set of the accumulation points of $(\mathbf\Pi(x, n))_n$ equals $\Delta_N$}\right\}
\]
is residual.
\end{theorem}

The result above has inspired a series of papers (\cite{Ol2, Ol3, Ol4, MAN, MAN2}) that appeared later in the literature. These later results were sometimes considering different settings such us continued fraction expansions, iterated function systems, finite or countable Markov partitions and other times enhancing the way in which the vectors of frequencies were diverging.
In this paper we offer a generalisation of the second form. Namely, Theorem \ref{main} states that for any regular linear transformation $T$, there exists a residual set of points $x$ whose set of accumulation points of their $T$-averaged vector of frequencies $\left(\Pi^T(x, m)\right)_m$ includes the simplex of all $N$-dimensional probability vectors. The precise definitions are given below.

Given a divergent sequence, by considering an averaged version of this sequence one may succeed in producing a convergent sequence. In this paper we prove a somewhat unexpected result that strengthens Theorem \ref{olsen} considerably. To do that we need the following definition.

\begin{definition}[\cite{hard}]
An $\mathbb{N}\times\mathbb{N}$ real valued matrix $T=\left[c_{m,n}\right]$, is called a \emph{linear transformation} if for any real sequence $(s_{n})_{n}$ and any natural number $m$, we have that $t_{m}=\sum_{n\geq1} c_{m,n} s_{n}$ converges to a real number. The sequence $(t_{m})_{m}$ is called the $T$-averaged version of $(s_{n})_{n}$ and if $\lim_{m\to\infty}t_{m}$ exists we call it the $T$-value of $(s_{n})_{n}$.
\end{definition}

\begin{example}
The linear transformation $T=\left[c_{m,n}\right]_{\mathbb{N}\times\mathbb{N}}$ with $c_{m,n}=$ $\begin{cases}
\frac{1}{m}\;\;\;\text{if $1\leq n\leq m$}\\
0\;\;\;\text{otherwise}
\end{cases}$
corresponds to considering the sequence of the natural averages of a sequence.
\end{example}

Namely, we show that for any linear transformation $T$, the $T$-averaged version of the sequence $(\mathbf\Pi(x, n))_n$ accumulates at every point in $\Delta_N$ for a residudal set of points in $[0,1]$. The special case of Cesàro averages was considered in \cite{Ol3}. To state this result a bit more precisely, we define a regular linear transformation below.

Linear transformations are used to help generalise the sense of a limit of sequence, in order to provide a \emph{sensible} value of a limit for a divergent sequence. However, if a sequence is already convergent we would like the averaged version of this sequence to also converge to the same limit. We hence call a linear transformation $T=\left[c_{m,n}\right]$ \emph{regular}, if $T$ maps all convergent sequences to their original limit, that is $T$ is regular if
$$
t_{m}\rightarrow l\,\,\,\text{as}\,\,\,m\rightarrow\infty\;\;\text{ whenever }\;\;s_{k}\rightarrow l\,\,\,\text{as}\,\,\,k\rightarrow\infty.
$$

Fix a regular linear transformation $T=[c_{m,n}]_{\N\times\N}$ and consider the $T$-averaged version of the sequences of digit frequencies
$$
\Pi_i^T(x,m):=\sum_{n=1}^\infty c_{m,n}\,\Pi_i(x,n)\;\;\;\;\;\text{for } 0\leq i\leq N-1.
$$
 Then, the $T$-averaged version of the vector of digit frequencies is given by the sequence with $m$th term  \[
\mathbf\Pi^T(x,m):=\left(\Pi^T_0(x, m),\Pi^T_1(x, m),...,\Pi^T_{N-1}(x, m)\right)
\]

We now state our main theorem.
\begin{theorem}\label{main}
For any regular linear transformation $T$ the set
\[
R_T:=\left \{ x\in[0, 1]\; :\; \text{The sequence $\left(\mathbf\Pi^T(x, m)\right)_m$ accumulates at every point of $\Delta_N$} \right\},
\]
is residual.
\end{theorem}

\begin{remark}
Results similar in nature but in somewhat more general settings were obtained independently by Olsen and West in \cite{Ol4} and Aveni and Leonetti in \cite{ita}.
\end{remark}

The proof of Theorem \ref{main} is given in Section \ref{proof} below. Before proceeding we make two final remarks concerning our result .
Firstly, fix a countable family of regular linear transformations $\mathcal{T}=\{T_1, T_2,...\}$ (Examples include H\"older means, Ces\`aro averages,...). Notice that it clearly follows from Theorem $\ref{main}$ that the countable intersection $\mathcal{R}_\mathcal{T}:=\bigcap_{T\in\mathcal{T}}  R_{T}$ is also residual.

Additionally, also in terms of dimensions, the size of $R_T$ varies between \emph{very big} and \emph{very small} depending on the exact viewpoint. Below we write $\dim_H$ and $\dim_P$ for the Hausdorff dimension and the packing dimension, respectively. The reader is referred to \cite{Fa} for the definitions of the Hausdorff and packing dimensions. It follows from \cite{OW} that $ \dim_H R_T = 0$. On the other hand, using Theorem \ref{main} we can find the packing dimension of the set $R_T$. A result analogue to the following corollary appeared in \cite{Ol} with a slightly inaccurate proof. We postpone our argument until the next section.
\begin{corollary}\label{dimen}
For any regular linear transformation $T$ the set $R_T$ has full packing dimension, that is
\[\dim_P\, R_T = 1\]
\end{corollary}

\section{Proof of Theorem \ref{main}} \label{proof}

Throughout the proof, we will work with a subset of $[0,1]$, namely
\[
\mathbb{I}:= [0, 1]\setminus \left\{x\in [0, 1]\; :\; x \text{ has a terminating $N$-adic expansion}\right\}.
\]
For brevity, write \[
\mathbb{D}:=\mathbb{Q}^N\cap\Delta_N\setminus\{(1,0,...,0),(0,...,0,1)\}
\]
where we excluded two particular vectors for technical reasons, which become apparent in the proof of Claim \ref{dense}.

Let $\psi:\N\mapsto\N$ be any increasing function such that  $\psi(n)>n^2$ for all $n\in\N$, and write $\psi^{m}$ for the composition of $\psi$ with itself $m$ times. For $h\in\N$ we define the property $P_{h,\psi}$ as follows.
\begin{definition}
We say that a sequence $(\mathbf x_n)_n$ in $\R^N$ has property $P_{h,\psi}$ if for all $i\in\N,\;m\in\N$ and all $\mathbf q\in\mathbb{D}$, there exists $j\in\N$ satisfying the following:
\begin{itemize}
  \item $j\geq i$
  \item $\frac{j}{\psi(j)}\le \frac{1}{h}$
  \item if $j\le n\le \psi^m(j)$ then $\|\mathbf x_n-\mathbf q \|_1<\frac{1}{h}$.
\end{itemize}
\end{definition}

\noindent Our proof of Theorem \ref{main} will consist of three lemmas:
\begin{enumerate}
  \item First we will prove that for any $h\in\N$ and any increasing $\psi:\N\mapsto\N$ such that  $\psi(n)>n^2$ for all $n\in\N$, the following set is residual:
  \begin{align}\label{a}
    A_{h,\psi}:=\left\{ x \in\mathbb{I}\; :\;(\mathbf\Pi(x, n))_n \text{ has property $P_{h,\psi}$}\right\}.
  \end{align}
\item Then we will show that if $T$ is any regular linear transformation and $h\in\N$ then there exists $\psi_{h,T}:\N\mapsto\N$ such that if $x\in A_{6Mh,\psi_{h,T}}$, with $M$ depending on $T$, then  $(\mathbf\Pi^T(x,m))_{m}$ has property $P_{h,\psi_{h,T}}$.

\item Finally, setting $A_T:=\bigcap_{h\in\N} A_{6Mh,\psi_{h,T}}$ we will show that \[ A_{T}\subset R_T\] (recall that $R_T$ is defined in Theorem \ref{main}).
\end{enumerate}

\begin{lemma}\label{family}
For all  $h\in\N$ and all increasing $\psi:\N\mapsto\N$ such that  $\psi(n)>n^2$ for all $n\in\N$, the set $A_{h,\psi}$ is residual (recall that $A_{h,\psi}$ is defined in (\ref{a})).
\end{lemma}

\begin{proof}
For fixed $i\in\N,\; m\in\N$ and $\mathbf q\in\mathbb D$, we define property $P_{h,\psi,i,m,\mathbf q}$, as follows. We say that a sequence $(\mathbf x_n)_n$ has property $P_{h,\psi,i,m,\mathbf q}$ if there exists $j\in\N$ satisfying
\begin{enumerate}
  \item $j\geq i$
  \item $\frac{j}{\psi(j)}<\frac{1}{h}$
  \item if $j<n<\psi^m(j)$ then $\|\mathbf x_n-\mathbf q\|_1<\frac{1}{h}$.
\end{enumerate}

Let $G_{h,\psi,i,m,\mathbf q}:=\left\{ x\in\mathbb I\; :\; (\mathbf\Pi(x, n))_n \text{ has property $P_{h,\psi,i,m,\mathbf q}$}\right\}$. Clearly,
\[
\bigcap_{i\in\N}\bigcap_{m\in\N}\bigcap_{\mathbf q\in\mathbb D}\,G_{h,\psi,i,m,\mathbf q}=A_{h,\psi}
\]

\begin{claim}\label{open}
$G_{h,\psi,i,m,\mathbf q}$ is open.
\end{claim}
\begin{proof}
Let $x\in G_{h,\psi,i,m,\mathbf q}$. Then, there exists a positive integer $j$ such that $j\geq i$, $\frac{j}{\psi(j)}< \frac{1}{h}$, and if $j < n < \psi^m(j )$, then $\left\|\mathbf\Pi(x,n) -\mathbf q \right\|_1<\frac{1}{h}$.
We now choose $\delta$ to equal $\frac{1}{N^{a}}$ where $a>\psi^{m+1}(j)+1$ and since $x\in\mathbb I$ we can also insist that $a\in\N$ is such that the $(a-1)$ and $(a-2)$-nd digits of $x$ are not both either $0$ or $N-1$. . Then all $y\in B(x,\delta)$ have their first $\psi^m(j)$ digits the same as $x$, and so $B(x,\delta)\subset G_{\psi,h,i,m,\mathbf q}$. This completes the proof of Claim \ref{open}.
\end{proof}

\begin{claim}\label{dense}
$G_{h,\psi,i,m,\mathbf q}$ is dense.
\end{claim}

\begin{proof}
Let $x\in\mathbb{I}$ and $\delta>0$. We must now find $y\in B(x,\delta)\cap G_{h,\psi,i,m,\mathbf q}$. Let $t\in\N$ be such that $\frac{1}{N^t}<\delta$. We can clearly choose a positive integer $s\in\N$ and $z_1,...,z_s\,\in\{0,1,...,N-1\}$ such that if $z =\frac{z_1}{N}+\frac{z_2}{N^2}+...+\frac{z_s}{N^s}$ then $\mathbf \Pi(z, s) = \mathbf q$. Let
\[
y = \frac{d_1(x)}{N}+...+\frac{d_t(x)}{N^t}+\sum_{d=0}^\infty\left( \frac{z_1}{N^{t+d s+1}}+\frac{z_2}{N^{t+d s+2}}+...+\frac{z_s}{N^{t+d s+s}} \right)
.\]
Then $y\in B(x,\delta)$ (as $y$ has its first $t$ digits the same as $x$).

Next we show that $y\in G_{h,\psi,i,m,\mathbf q}$. All $z_i$'s cannot be $0$ or $N-1$, because we excluded the vectors $(1, 0,...,0),\,(0,...,0,1)$. Therefore, $y$ has a non-terminating $N$-adic expansion. Choose $j$ big enough such that
\[
\frac{j}{\psi(j)}<\frac{1}{h}\;\;\;\text{  and  }\;\;\; j\geq\max\left\{ i,t,hN\max_{0\leq\ell\leq N-1}\left\{ {N_\ell (z,s)\left(2+\frac{t}{s}\right)+N_\ell (y,t)} \right\} \right\}
\]
where $N_\ell (x, n) = |\{1\leq k\leq n \; : \; d_k(x) = \ell \}|$. Fix a positive integer $n$ with $j < n < \psi^{m}(j)$ and observe that we can find integers $r$ and $b_\ell$ with $0\leq r < s$ and $0\leq b_\ell < N_\ell(z, s)$, such that $n = t + \left[\frac{n-t}{s}\right] s+ r$, and $N_\ell(y,n) = N_\ell(y, t) +\left[\frac{n-t}{s}\right]N_\ell(z,s)+b_\ell$. We now have,
\begin{align*}
\left\| \mathbf \Pi(y,n)-\mathbf q \right\|_1&=\sum_{\ell=0}^{N-1}  \left|  \frac{N_\ell(y,n)}{n}-\frac{n\frac{N_\ell(z,s)}{s}}{n}  \right|    \\

&= \sum_{\ell=0}^{N-1} \left|  \frac{N_\ell(y,t)+[\frac{n-t}{s}]N_\ell(z,s)+b_\ell}{n}-\frac{(t+[\frac{n-t}{s}]s+r)\frac{N_\ell(z,s)}{s}}{n}  \right|      \\

&\leq \sum_{\ell=0}^{N-1}  \left(  \left|  \frac{[\frac{n-t}{s}]N_\ell(z,s)-\frac{N_\ell(z,s)}{s}(t+r)-N_\ell(z,s)[
\frac{n-t}{s}]}{n}  \right| +\frac{N_\ell(y,t)+b_\ell}{n}  \right)
       \\

&<  \sum_{\ell=0}^{N-1} \left(  \frac{\frac{N_\ell(z,s)}{s}(t+r)+N_\ell(y,t)+N_\ell(z,s)}{n} \right)     \\

&< \sum_{\ell=0}^{N-1}  \left( \frac{N_\ell(z,s)(2+\frac{t}{s})+N_\ell (y,t)}{j}  \right) \leq \frac{1}{h}.

\end{align*}
This shows that $y \in G_{h,\psi,i,m,\mathbf q}$ and completes the proof of Claim \ref{dense}.
\end{proof}

It follows from Claim \ref{open} and Claim \ref{dense}, that $A_{h,\psi}$ is the countable intersection of open and dense sets, and hence residual. This completes the proof of Lemma \ref{family}.
\end{proof}

\begin{lemma}\label{extreme}
For any regular linear transformation $T$ and any $h\in\N$ we can:
\begin{enumerate}
    \item construct an increasing function $\psi_{h,T}:\N\mapsto\N$ such that $\psi(n)>n^2$ for all $n\in\N$ and,
    \item show that for some $M$ depending on $T$, we have that for each\[
    x\in A_{6Mh,\psi_{h,T}}\text{ the sequence $\left(\mathbf\Pi^T(x,m)\right)_m$ has property $P_{h,\psi_{h,T}}$.}
    \]
\end{enumerate}

\end{lemma}

Before proceeding to prove Lemma \ref{extreme}, let us first recall a very useful characterisation result for \emph{regular} linear transformations. Given a linear transformation $T$ corresponding to an $\N\times\N$ real valued matrix $\left[c_{m,n}\right]$ and a sequence $(s_n)_n$ we are interested in the behaviour of its averaged version, that is the sequence whose $m$-th term is given by $t_m=\sum_{n\geq1} c_{m,n} s_n$.

\begin{proposition}[\cite{hard}]\label{regular}
A linear transformation $T=\left[c_{m,n}\right]_{\mathbb{N}\times\mathbb{N}}$ is regular if and only if the following conditions are satisfied:
\begin{enumerate}
\item there exists $M\in\N$ such that for all $m\in\mathbb{N}$ we have that
$\sum_{n=1}^{\infty}|c_{m,n}|<M$,
\item for all $n\in\mathbb{N}$, $\lim_{m\to\infty}c_{m,n}=0$ and
\item we have that $\lim_{m\rightarrow\infty}\sum_{n=1}^{\infty}c_{m,n}= 1$.
\end{enumerate}
\end{proposition}

\begin{proof}[Proof of Lemma \ref{extreme}]
Fix $h\in\N$ and $T=[c_{m,n}]$ a regular linear transformation. Firstly, we construct $\psi_{h,T}:\N\mapsto\N$. To do that recall that from Proposition \ref{regular} we can associate to $T$ a number $K\in\N$ and sequences of natural numbers $(M_k)_k$ and $(N_k)_k$ such that:
 \begin{enumerate}
     \item for each $m\in\N$ we have $\sum_{n\geq M_m} |c_{m,n}|\leq\frac{1}{12h}$,
     \item for each $n\in\N$ we have $|c_{m,n}|<\frac{1}{2^{n+10}h}$ whenever $m>N_n$,
     \item for $m>K$ we have $\left|1-\sum_{n\geq1}c_{m,n}    \right|<\frac{1}{2h}.$
 \end{enumerate}
Recall also $M\in\N$ the uniform bound of the absolute row sums of $T$. Having fixed these constants depending on the linear transformation $T$ and the natural number $h$, we proceed by defining our function $\psi$ by $\psi(0)=K+2$ and recursively for $n\geq1$:
$$
\psi_{h,T}(n):=M_n+N_n+n^2+\psi_{h,T}(n-1).
$$
This function $\psi=\psi_{h,T}$ is increasing and crucially $\psi(n)>\max\{n^2,M_n,N_n,K\}+1$ for all $n\geq0$.

Then, fix $i\in\N$, $m\in\N$ and $\mathbf q\in\mathbb D$. For $x\in A_{6Mh,\psi}$, the sequence $\left(\mathbf\Pi(x,n)\right)_n$ satisfies property $P_{6Mh,\psi}$.

We can therefore find $j'\geq i$, with $\frac{j'}{\psi(j')}<\frac{1}{6Mh}$ and such that if $j'<n<\psi^{m+2}(j')$, then \begin{align}\label{step}
  \left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1<\frac{1}{6Mh}.
\end{align}
Set $j=\psi(j')$. Then, clearly $j\geq i$ and $\frac{j}{\psi(j)}=\frac{\psi(j')}{\psi(\psi(j'))}<\frac{1}{\psi(j')}<\frac{j'}{\psi(j')}<\frac{1}{h}$. For all $j<r<\psi^m(j)$, that is $\psi(j')<r<\psi^{m+1}(j')$, we have

\[
\left\| \mathbf \Pi^T(x,r)-\mathbf q \right\|_1\leq\left\| \sum_{n\geq1} c_{r,n} \left(\mathbf\Pi(x,n)-\mathbf q\right)  \right\|_1+\left\| \mathbf q (1-\sum_{n\geq1}c_{r,n}) \right\|_1\leq \sum_{n\geq1} |c_{r,n}| \left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1+\frac{1}{2h}
\]
since $r>\psi(j')>K$ and $\norm{\mathbf q}_1=1$. Now we proceed by splitting our summation into three parts which we bound separately.
\begin{align*}

 \sum_{n\geq 1} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1&=\sum_{n=1}^{j'} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q  \right\|_1\\

 &+ \sum_{n\geq j'+1}^{\psi^{m+2}(j')-1} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1\\

&+ \sum_{n\geq \psi^{m+2}(j')} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1
\end{align*}
Firstly, using the fact that $r>\psi(j')$ we get that $r>\max\{N_1,...,N_{j'}\}$ and hence
\begin{align*}
  \sum_{n\leq j'} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q   \right\|_1\leq 2\sum_{n\leq j'} |c_{r,n}|\leq \sum_{n\leq j'} \frac{1}{2^{n+9}h}\leq \frac{1}{6h}.
\end{align*}
Then using (\ref{step}) we can bound the second summand by
\[
\sum_{n\geq j'+1}^{\psi^{m+2}(j')-1} |c_{r,n}|\left\|  \mathbf\Pi(x,n)-\mathbf q   \right\|_1\leq \frac{1}{6 Mh} \sum_{n\geq 1} |c_{r,n}|\leq \frac{1}{6h}.
\]
Finally, for $n\geq\psi^{m+2}(j')=\psi(\psi^{m+1}(j'))>\psi(r)$ we have that $n>M_r$ hence
\[
\sum_{n\geq \psi^{m+2}(j')} |c_{r,n}|\left\| \mathbf\Pi(x,n)-\mathbf q \right\|_1\leq 2\sum_{n\geq M_r} |c_{r,n}|\leq \frac{1}{6h}.
\]
Putting all these together we get that for all $j<r<\psi^m(j)$  \[
\left\| \mathbf \Pi^T(x,r)-\mathbf q \right\|_1\leq \frac{1}{h}
\]
The choices of $i,m\in\N$ and $\mathbf q\in\mathbb D$ were arbitrary hence the sequence $\left(\mathbf\Pi^T(x,m)\right)_m$ has property $P_{h,\psi_{h,T}}$. This completes the proof of Lemma \ref{extreme}.

\end{proof}

\begin{lemma}\label{subset} Let $T=[c_{m,n}]$ be a regular linear transformation and recall the set $A_T:=\bigcap_{h\in\N} A_{6Mh,\psi_{h,T}}$. Then
$$A_T\subset R_T:=\left \{ x\in[0, 1]\; :\; \text{The sequence $\left(\mathbf\Pi^T(x, m)\right)_m$ accumulates at every point of $\Delta_N$} \right\}.$$
\end{lemma}
\begin{proof}
Let $x\in A_T$. To prove that $x\in R_T$, it suffices to show that each $\mathbf p\in \Delta_N$ is an accumulation point of $\left(\mathbf \Pi^T(x,m)\right)_m$. Therefore, we start by fixing $\mathbf p \in\Delta_N$ and $k\in \N$ and choose a vector $\mathbf q \in\mathbb D$ such that $\|\mathbf p-\mathbf q\|_1<\frac{1}{k}$. We claim that we can find $n_k>k$ such that
\begin{align}
    \left\|\mathbf q-\mathbf \Pi^T(x, n_k)\right\|_1\leq\frac{1}{k}.
\end{align}

We now prove the claim above. Indeed, for any for $x \in A_T$, by using Lemma \ref{extreme} we have that $(\mathbf \Pi^T(x,m))_m$ has property $P_{k,\psi_{k,T}}$. In particular, we can find $j \in N$ with $j\geq k$ and such that if $j < n < \psi_{k,T}(j)$ then
 $ \|\mathbf q-\mathbf \Pi^T(x, n)\|_1\leq\frac{1}{k}$. Hence if $n_k$ is any integer with $j < n_k < \psi_{k,T}(j)$ then \[  \|\mathbf q-\mathbf \Pi^T((x, n_k)\|_1\leq\frac{1}{k},\]
and in particular $n_k>j>k$ as required. Then we get that,
\begin{align}\label{convergence}
    \left\| \mathbf p -\mathbf \Pi^T(x,n_k)    \right\|_1\leq \left\|  \mathbf p- \mathbf q   \right\|_1 + \left\|  \mathbf q -\mathbf \Pi^T(x,n_k)    \right\|_1\leq \frac{2}{k}.
\end{align}

Since $n_k > k$, we can extract an increasing subsequence $(n_{k_u})_u$ of $(n_k)_k$. It now follows from (\ref{convergence})
that $\mathbf \Pi^T(x,n_{k_u})  \ra\mathbf p$ as $u\ra\infty$ with respect to $\|.\|_1$. Hence $\mathbf p$ is an accumulation point of
$(\mathbf \Pi^T(x,m))_m$. This completes the proof of Lemma \ref{subset}.
\end{proof}

We can now prove Theorem \ref{main}.
\begin{proof}[Proof of Theorem \ref{main}]
By Lemma \ref{family} it follows that for each $h\in\N$ the set $A_{6Mh,\psi_{h,T}}$ is residual in $\mathbb I$. Since $A_T:=\bigcap_{h\in\N} A_{6Mh,\psi_{h,T}}$ it follows that the set $A_T:=\bigcap{h\in\N}A_{6Mh,\psi_{h,T}}$ is residual in $\mathbb I$ as a countable intersection of residual sets. Then, by Lemma \ref{subset} since for each regular linear transformation $T$, the set $A_T$ is a subset of $R_T$ we get that $R_T$ is residual in $\mathbb I$. Finally, it can be easily seen that $[0, 1]\setminus \mathbb I$ is a countable union of nowhere dense sets and so $R_T$ is residual in $[0, 1]$.
\end{proof}

\begin{proof}[Proof of Corollary \ref{dimen}]
Assume for the sake of contradiction that $\dim_P R_T<1$. Recall that for $F\subseteq\R$ if we let $\overline{\dim}_B(F)$ denote the upper box dimension of $F$ then we have that
$$
\dim_P(F)=\inf\left\{ \sup_{n\in\N} \overline{\dim}_B(E_n)\; :\; E_1,E_2,...\subseteq\R\;\;\text{and}\;\; F\subseteq\bigcup_{n\in\N} E_n    \right\}.
$$
Then by assumption we can find a countable family of subsets $E_1,E_2,...$ of $\R$ with $R_T\subseteq\bigcup_n E_n$ such that $\sup_n \overline{\dim}_B(E_n)<1$. Since $R_T\subseteq[0,1]$ and $R_T\subseteq\bigcup_n E_n$ we get that $$
R_T\subseteq\bigcup_n\big( [0,1]\cap E_n  \big).
$$
To arrive at the desired contradiction we show that for all $n\in\N$ the set $[0,1]\cap E_n$ is nowhere dense in $[0,1]$ contradicting Theorem \ref{main}. Assume this is not the case. Then for some $n\in\N$ and some open set $U\subseteq \R$ we have that$$
[0,1]\cap U\neq \emptyset \;\;\;\;\text{and moreover}\;\;\;\;[0,1]\cap U\cap E_n\text{  is dense in  }[0,1]\cap U.
$$
In particular, we deduce that $[0,1]\cap U\subseteq \overline{[0,1]\cap U\cap E_n}$ and therefore:
\begin{align*}
    \overline{\dim}_B(E_n)&=\overline{\dim}_B(\overline{E_n})\\
    &\geq \overline{\dim}_B(\overline{[0,1]\cap U\cap E_n})\\
    &= \overline{\dim}_B(\overline{[0,1]\cap U})\\
    &\geq  \dim_P([0,1]\cap U)=1,
\end{align*}where we have used that $\dim_P([0,1]\cap U)=1$ for all open sets $U$ that intersect $[0,1]$. This is a contradiction since by assumption $\sup_n \overline{\dim}_B(E_n)<1$. This completes the proof.
\end{proof}

\begin{ack}
The author would like to thank Professor Lars Olsen for presenting him with this problem and Paolo Leonetti for useful discussions.
\end{ack}

config: arxiv


TITLE: Prove that {(-1)^n} is not a Cauchy Sequence by using the definition.
QUESTION [0 upvotes]: I need to use the definition of a Cauchy Sequence to prove that {$(-1)^n$} is not a Cauchy sequence.
I believe I have the main idea down by am lacking a part in my reasoning.
Proof:
For sake of contradiction assume that {$(-1)^n$} is a cauchy sequence. Take $\epsilon =1$. Then there is a $N \in \mathbb{N}$ such that for all m,n $\geq N$ $\Rightarrow |(-1)^m -(-1)^n|$. Let m= N and n = N+1.
Then, $|(-1)^m -(-1)^n| = |(-1)^N-(-1)^{N+1}| = |1-(-1)^{N+1}|$.
This is where I am stuck. I am not sure how to properly show that $(-1)^{N+1}$ reduces down to (-1) which give us |1-(-1)| = 2 >1 thus it is not a cauchy sequence.

REPLY [1 votes]: It isn’t necessarily true that $(-1)^N=1$: that is the case if and only if $N$ is even. You could divide the argument into two cases, one for even $N$ and one for odd $N$, but it’s neater simply to factor out $(-1)^N$:
$$\left|(-1)^N-(-1)^{N+1}\right|=\left|(-1)^N\right|\cdot|1-(-1)|=1\cdot 2=2\,.$$
If you choose to use cases, when $N=2m$ is even you have $$(-1)^N=(-1)^{2m}=\big((-1)^2\big)^m=1^m=1$$ and $$(-1)^{N+1}=(-1)^N(-1)=(1)(-1)=-1\,,$$
and when $N=2m+1$ is odd you have
$$(-1)^N=(-1)^{2m+1}=(-1)^{2m}(-1)=(1)(-1)=-1$$
and
$$(-1)^{N+1}=(-1)^N(-1)=(-1)(-1)=1\,.$$


TITLE: Ramification in local fields
QUESTION [2 upvotes]: Question: I know how to detect ramified or unramified fields in the case of number fields. But I have no feelings about how to do this for local fields. How should I find that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is ramified or unramified? I can answer that $\mathbb{Q}_p(\sqrt[3]{2})=\mathbb{Q}_p[x]/(x^3-2)$ is (totally) ramified, because $x^3-2$ is Eisenstein at $2$ (This is not correct. See the comments). But then how should I realize that $\mathbb{Q}_p(\sqrt[3]{2}, \zeta_3)=\mathbb{Q}_p[x]/(x^3-2, x^2+x+1)$ is ramified or unramified? How should I find that $\mathbb{Q}_p[x]/(x^5-x-2)$ is ramified or unramified? How can I find the uniformizer in the local field $\mathbb{Q}_p[x]/P(x)$, where $P(x)$ is an irreducible monic polynomial, and how can I find the ramification index?
My attempt: This is not actually an attempt. I just want to say what I know in the case of number fields, and I have some questions and solutions (probably unsuccessful solutions) based on it. In the case of number fields I used the discriminant. Also Dedekind-Kronecker theorem was very helpful to find the ramification indices, $e_i$'s, and reside degrees $f_i$'s for almost all primes. Do we have a version of Dedekind-Kronecker theorem for local fields?

REPLY [4 votes]: You said you know about using the discriminant of a $\mathbf Z$-basis of the integers of a number field to detect ramification.  You can do the same thing in a local field: if $K$ is a finite extension of $\mathbf Q_p$ then the discriminants of all $\mathbf Z_p$-bases of $\mathcal O_K$ are equal to multiplication by the square of a unit in $\mathbf Z_p$, and $K/\mathbf Q_p$ is ramified if and only if the discriminant of a $\mathbf Z_p$-basis of $\mathcal O_K$ is divisible by $p$ (equivalently, the discriminant is not a unit in $\mathbf Z_p$).


TITLE: Sum of the powers of roots of unity
QUESTION [2 upvotes]: I was reading this proof of the roots of unity filter and I point can't understand is why:
$$ S = \sum_{j=0}^{n-1} w^{kj} =0$$
If $ \omega $ is the nth root of unity and $k$ is some positive integer not equal to zero and n doesn't divide k
I understand that the sum of nth roots of unity are zero as in:
$$ S = \sum_{j=0}^{n-1} w^j = 0$$
But I can't understand the powers of them should be as well. The reason I find it strange is that the nth roots of unity solves the following polynomial equation
$$ P(x) =(x-1)( \sum_{i=0}^{n-1} x^i)$$
So why should it be that any of their powers solve the above equation as well?

REPLY [1 votes]: Here's the polynomial which solves it:
$$  1 + x^k + x^{2k}..+x^{(n-1)k} = \frac{x^{nk} -1}{x-1}$$
Or,
$$ (x-1)\left[ 1 + x^k + x^{2k}..+x^{(n-1)k}  \right]= x^{nk}-1 = P(x)$$
Clear to see power of nth roots of unity solves this equation

REPLY [0 votes]: $w=e^{2i j\pi/n}, j=0,1,2,....n-1,$
$$S_n=\sum_{j=0}^{n-1} w^{k}= \sum_{j=0}^{n-1} e^{2\pi  k j/n}~~~~(1) $$
Use sun of a finite GP, to get
$$\implies S_n=\frac{e^{2i\pi k }-1}{e^{2i\pi k/n}-1}$$
If $k\ne pn$, $S_n=0$,
If $k=p n$ then $S_n=0/0$. So go (1) and put $k=pn$, where $p$ is an integer.
We get $S_n=n$ when $k$ is an integer multiple of $n$.


TITLE: For which values of $a$ does $\lim\limits _{x\to a}\frac{|x^2-(2+a)x+2a|}{x-a}$ exist?
QUESTION [0 upvotes]: For which values of $a\in{\mathbb{R}}$ the limit $$\lim_{x\to{a}}{}{\frac{|x^2-(2+a)x+2a|}{x-a}}$$ exists?

I suspect that I need to remove the absolute value and then factor the numerator and last divide with $x-a$ in order to remove the undefined form of $\frac{0}{0}$, but am I correct?

REPLY [5 votes]: Hint. Note that
$$x^2-(2+a)x+2a=(x-2)(x-a).$$
Moreover, pay attention to the fact that $|x-a|/(x-a)=1$ if $x>a$ and it is equal to $-1$ for $x<a$.


TITLE: probability distribution with discrete random variable
QUESTION [2 upvotes]: I have tried it out, but finding it difficult to post them in proper format.
Please help me with the solution.

REPLY [1 votes]: (a) Here $X$ is the number of trials until the first "success," where the probability of success on any trial is $p$, and the trials are independent.
We have $X\gt n$ if and only if the first $n$ trials result in $n$ failures. The probability that on any trial we have failure is $1-p$. So the probability of $n$ failures in a row is $(1-p)^n$.
(b) Let $B$ be the event $X\gt n+k$, and let $A$ be the event $X\gt n$. By the definition of conditional probability, we have
$$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.\tag{1}$$
In our case, $A\cap B$ is just the event $X\gt n+k$.
By the calculation in (a), we have $\Pr(X\gt n+k)=(1-p)^{n+k}$, and $\Pr(X\gt n)=(1-p)^n$. Thus, substituting in (1), we get
$$\Pr(B|A)=\frac{(1-p)^{n+k}}{(1-p)^n}=(1-p)^k.$$
But by (a), $(1-p)^k=\Pr(X\gt k)$. This completes the proof.
(c) You will need to explain this yourself, in language that is natural to you. We give a brief guide.
We are flipping a coin that has probability $p$ of landing heads, where $0\lt p\le 1$. The random variable $X$ is the number of flips until the first head.
Suppose we have flipped $n$ times without success. The probability that we will need a total of more than $n+k$ flips is the same as the probability that we will need at least $k$ flips if we start a fresh sequence of flips. That's because the coin does not remember that it let you down $n$ times in a row.


TITLE: What is the exact and precise definition of an ANGLE?
QUESTION [24 upvotes]: On wikipedea I found a definition of an Angle as  such:

"In order to measure an angle θ, a circular arc centered at the vertex of the angle is drawn, e.g. with a pair of compasses. The length of the arc s is then divided by the radius of the arc r, and possibly multiplied by a scaling constant $k$ (which depends on the units of measurement that are chosen):
   $$\theta = k \frac{s}{r}.$$
  The value of θ thus defined is independent of the size of the circle: if the length of the radius is changed then the arc length changes in the same proportion, so the ratio $s/r$ is unaltered.'

And the definition of circular arc is given as:

"a circular arc is a segment of a circle, or of its circumference (boundary) if the circle is considered to be a disc."

Now since circular arc is a segment of a circle we are not supposed to have circular arc of length more than circumference of the circle. So we cannot define angles more than $2\pi$ in this manner. But I found a topic in the book, "Unified algebra and trigonometry" (Addison-Wesley mathematics series) art 3-5. I've posted it as an answer to the question
https://physics.stackexchange.com/questions/87057/angular-displacement/87066#87066.
There the author of the book explains that we can have circular arc more than the circumference of a circle as:
 "....where the circular arc $\stackrel \frown {PP}^{′}$ denoted by $s$,.... since the number of revolutions of any angle is determined by the ratio of the intercepted circular arc length $s$ to circumference of the circle we define, magnitude of an angle in revolutions as
Angle in revolutions$ = s/2πr$..."
means there can be angles of more than $2\pi$ .
Question1:Does the definition of an angle depends upon its use in physics?
Question2: if we assign a point on a solid body named $P$ having an initial angle $\theta$ after this solid body rotates a full circle can we assume the angle of point $P$ after the revolution to be again $\theta$ or $\theta +2\pi$. Please state a precise definition to support the validity of this argument.

As I understands if we assign polar-coordinates to point $P$ these polar coordinates  will be $(\pm r,\theta\pm 2n\pi)$ and but if rotation is involved should we consider the angle used as "oriented" or "non-oriented".

REPLY [29 votes]: "Angle" is a touchy subject. There are (at least) the following three interpretations, applicable depending on circumstances.
(a) The simplest is the following: Given two nonzero vectors ${\bf x}$, ${\bf y}\in{\mathbb R}^n$, $\>n\geq2$,  the (nonoriented) angle $\alpha$ between them is the nonnegative number
$$\alpha:=\arccos{{\bf x}\cdot{\bf y}\over|{\bf x}|\>|{\bf y}|}\in[0,\pi]\ .$$
In particular: Given two rays emanating from the same point $P$ in space the "enclosed angle $\alpha$" is a number between $0$ and $\pi$ inclusive, and is equal to the length of the shorter arc cut out on the unit circle in the plane of  these two rays.
This idea of angle is also at the basis of spherical trigonometry, where the angle between two points ${\bf u}$, ${\bf v}\in S^2$ is considered as distance between these two points. As such it satisfies the triangle inequality.
(b) When some rotation about the origin in ${\mathbb R}^2$ is involved then  the group $SO(2)$ steps into action, and it makes sense to talk about oriented angles. An oriented angle is an equivalence class of real numbers modulo $2\pi$. Each such class has a unique representant in the interval $[0,2\pi[\ $, or in the interval $\ ]{-\pi},\pi]$, and the set of these classes is bijectively related to $SO(2)$.
An example: The map
$$T:\quad (x,y)\mapsto(x\cos\alpha-y\sin\alpha,\  x\sin\alpha+y\cos\alpha)$$
is a rotation of the euclidean plane about the angle $\alpha\in{\mathbb R}/(2\pi{\mathbb Z})$. When the rotation is physically performed in time and all points are in fact rotated carousel-like $n$ full turns before stopping at the right place the information about $n$ is not present in $T$.
(c) The circle group $SO(2)$ has the full real line ${\mathbb R}$ as its "universal cover". Sometimes it is desirable to work in ${\mathbb R}$ when talking about angles, e.g., when studying the logarithmic spiral
$$\sigma:\quad t\mapsto (e^t\cos t,e^t\sin t)\qquad(-\infty<t<\infty)\ .$$
Another important example is the following: You have a closed curve
$$\gamma:\quad t\mapsto {\bf z}(t)=\bigl(x(t),y(t)\bigr)\in\dot{\mathbb R}^2\qquad(0\leq t\leq L)$$
which encircles the origin a certain number $n$ of times. In order to compute this $n$ we cumulate the infinitesimal changes of $\phi(t):=\arg\>{\bf z}(t)$ and in the end divide by $2\pi$. Here $\phi(t)$ is only "defined up to $2\pi$", as in (b), but $\phi'(t)$ is well defined, and is given by
$$\phi'(t)={x(t)y'(t)-x'(t)y(t)\over x^2(t)+y^2(t)}\qquad(0\leq t\leq L)\ .$$
It follows that the total argument increase $\Delta\phi\in{\mathbb R}$ along $\gamma$ is given by
$$\Delta\phi=\int_0^L\phi'(t)\ dt=\int_0^L {x(t)y'(t)-x'(t)y(t)\over x^2(t)+y^2(t)}\ dt\ ,$$
from which we then obtain $n={\Delta\phi\over 2\pi}$. In complex analysis this number appears as
$$n(\gamma,0)={1\over2\pi i}\int\nolimits_\gamma{dz\over z}\ .$$


text:
\begin{document}

\begin{abstract}
Affinely closed homogeneous spaces $G/H$, i.e.,
affine homogeneous spaces that admit only the trivial affine
embedding, are characterized for any affine algebraic
group~$G$. As a corollary, a description of affine $G$-algebras with finitely
generated invariant subalgebras is obtained.
\end{abstract}

\maketitle

\section{Introduction}
Let $G$ be an affine algebraic group over
an algebraically closed field $\kk$ of characteristic zero and $H$
an algebraic subgroup of $G$.  By Chevalley's Theorem, the
homogeneous space $G/H$ admits the canonical structure of a
quasiprojective variety. {\it An embedding} of the homogeneous
space $G/H$ is an algebraic $G$-variety $X$ with a base point
$x\in X$ such that the orbit $Gx$ is dense (and open) in $X$ and
the stabilizer $G_x$ equals $H$. We denote an embedding as
$G/H\hookrightarrow X$. We say that an embedding is {\it trivial}
if $Gx=X$. An embedding $G/H\hookrightarrow X$ is said to be {\it
affine} if the variety $X$ is affine. It is easy to show (for
example, see~\cite[Th.1.6]{pv}) that a space $G/H$ admits an
affine embedding if and only if $G/H$ is a quasi-affine variety,
or, equivalently, $H$ may be realized as the stabilizer of a
vector in a finite-dimensional $G$-module. In this case the
subgroup $H$ is said to be {\it observable} in $G$. An effective
description of observable subgroups in an affine algebraic group $G$ was
obtained by A.~Sukhanov~\cite{suh}.

\smallskip

The following definition was introduced in~\cite{at}.

\begin{definition}
A homogeneous space $G/H$ is called {\it affinely closed} if it admits
only the trivial affine embedding.
\end{definition}

An affinely closed homogeneous space is automatically affine. The
consideration of this class of homogeneous spaces is motivated by
the following question: when does "the stabilizer of a point $x$
on an affine $G$-variety $X$ equals $H$" imply "the orbit $Gx$ is
closed" ?

For a reductive group $G$ a homogeneous space $G/H$ is an affine variety
if and only if the subgroup $H$ is reductive (Matsushima's Criterion).
Note that for an arbitrary affine algebraic group
$G$ a description of affine homogeneous spaces $G/H$
in group-theoretic terms for the pair $(G,H)$ is an open problem,
for more information see~\cite[Ch.2]{gr}.

For reductive $G$ a description of affinely closed homogeneous spaces
follows directly from the result due to D.~Luna~\cite{lu}:

\begin{theorem}\label{tl}
Let $G$ be a reductive group. A homogeneous space
$G/H$ is affinely closed if and only if the subgroup $H$
is reductive and has a finite index in its normalizer $N_G(H)$.
Moreover, if $G$ acts on an affine variety  $X$ and the stabilizer of a point
$x\in X$ contains a reductive subgroup $H$ such that the group
$N_G(H)/H$ is finite, then the orbit $Gx$ is closed.
\end{theorem}

For example, for a maximal torus $T$ of a reductive group $G$ the
Weyl group $W=N_G(T)/T$ is finite, hence $G/T$ is affinely closed.
If $\rho:H\to SL(V)$ is an irreducible representation of a
semisimple group, then $SL(V)/\rho(H)$ is affinely closed (by the
Schur Lemma, the group $N_{SL(V)}(\rho(H))/\rho(H)$ is finite).
Thus the class of affinely closed homogeneous spaces is wide.

In~\cite{ar}, affinely closed homogeneous spaces of a reductive group $G$
play a key role in a classification of affine $G$-algebras such that
any invariant subalgebra is finitely generated.
Characterizations of complex affinely closed homogeneous spaces
of reductive groups in terms of compact transformation groups
and invariant algebras on compact homogeneous spaces are given in
\cite{lat} and~\cite{gl}.

The aim of this paper is to generalize the result of D.~Luna to the case
of an arbitrary affine algebraic group $G$ (Theorem~\ref{tm}) and to obtain
a classification of affine $G$-algebras with finitely generated invariant
subalgebras (Theorem~\ref{tmm}). Note that a characteristic-free variant
of these results for a solvable group $G$ is given in~\cite{ten}.

\section{A description of affinely closed spaces}
Let us fix a Levi decomposition $G=LG^u$ of the group $G$ in a
semidirect product of a reductive subgroup $L$ and the unipotent
radical $G^u$. By $\phi$ denote the homomorphism $\phi: G \to
G/G^u$. We shall identify the image of $\phi$ with $L$. Put
$K=\phi(H)$.

\begin{theorem} \label{tm}
The following conditions are equivalent:

\ {\rm (1)}\ $G/H$ is affinely closed;

\ {\rm (2)}\ $L/K$ is affinely closed.
\end{theorem}

\begin{proof}
The subgroup $H$ is observable in $G$
if and only if the subgroup $K$ is observable in $L$ \cite{suh},
\cite[Th.7.3]{gr}.

\smallskip

Suppose that $L/K$ admits a non-trivial affine embedding. Then
there are an $L$-module $V$ and a vector $v\in V$ such that the
stabilizer $L_v$ equals $K$ and the orbit boundary $Y=Z\setminus
Lv$, where $Z=\overline{Lv}$, is nonempty. Let $I(Y)$ be the ideal
in $\kk[Z]$ defining the subvariety $Y$. Recall that for an action
of an algebraic group $G$ on an affine variety $X$ any element
$f\in\kk[X]$ belongs to a finite-dimensional invariant subspace,
or, equivalently, $\kk[X]$ is a sum of its finite-dimensional
$G$-submodules. Thus there exists an $L$-submodule $V_1\subset
I(Y)$ that generates $I(Y)$ as an ideal. The inclusion
$V_1\subset\kk[Z]$ defines $L$-equivariant morphism $\psi:Z\to
V_1^*$ and $\psi^{-1}(0)=Y$. Then $L$-equivariant morphism $\xi:
Z\to V_2=V_1^*\oplus (V\otimes V_1^*)$, $z\to (\psi(z),
z\otimes\psi(z))$ maps $Y$ to the origin and is injective on the
open orbit in $Z$. Hence we obtain an embedding of $L/K$ in an
$L$-module such that the closure of the image of this embedding
contains the origin. Put $v_2=\xi(v)$. By the Hilbert-Mumford
Criterion, there is a one-parameter subgroup $\lambda: \kk^* \to
L$ such that $\lim_{t\to 0} \lambda(t)v_2=0$. Consider the weight
decomposition $v_2=v_2^{(i_1)}+\dots+v_2^{(i_s)}$ of the vector
$v_2$, where $\lambda(t)v_2^{(i_k)}=t^{i_k}v_2^{(i_k)}$. Here all
$i_k$ are positive.

By the identification $G/G^u=L$, one may consider $V_2$ as a
$G$-module. Let $W$ be a finite-dimensional $G$-module with a
vector $w$ whose stabilizer equals $H$. Replacing the pair $(W,
w)$ by the pair $(W\oplus (W\otimes W), w+w\otimes w)$, one may
suppose that the orbit $Gw$ intersects the line $\kk w$ only at
the point $w$. For a sufficiently large $N$ in the $G$-module
$W\otimes V_2^{\otimes N}$ one has $\lim_{t\to 0} \lambda(t)
(w\otimes v_2^{\otimes N})=0$ ($\lambda(\kk^*)$ may be considered
as a subgroup of $G$). On the other hand, the stabilizer of
$w\otimes v_2^{\otimes N}$ coincides with $H$. This implies that
the space $G/H$ is not affinely closed.

\smallskip

Conversely, suppose that $G/H$ admits a non-trivial affine
embedding. This embedding corresponds to a $G$-invariant
subalgebra $A\subset\kk[G/H]$ containing a non-trivial
$G$-invariant ideal $I$. Note that the algebra $\kk[L]$ may be
identified with the subalgebra in $\kk[G]$ of (left- or right-)
$G^u$-invariant functions, $\kk[G/H]$ is realized in $\kk[G]$ as
the subalgebra of right $H$-invariants, and $\kk[L/K]$ is the
subalgebra of left $G^u$-invariants in $\kk[G/H]$. Consider the
action of $G^u$ on the ideal $I$. By the Lie-Kolchin Theorem,
there is a non-zero $G^u$-invariant element in $I$. Thus the
subalgebra $A\cap\kk[L/K]$ contains the non-trivial $L$-invariant
ideal $I\cap\kk[L/K]$. If the space $L/K$ is affinely closed then
we get a contradiction with the following lemma.

\begin{lemma}\label{lv}
Let $L/K$ be an affinely closed space of a reductive group
$L$. Then any $L$-invariant subalgebra in $\kk[L/K]$ is finitely generated and
does not contain non-trivial $L$-invariant ideals.
\end{lemma}

\begin{proof}
Let $B\subset\kk[L/K]$ be a non-finitely generated
invariant subalgebra. For any chain $W_1\subset W_2\subset
W_3\subset\dots$ of finite-dimensional $L$-invariant submodules in
$\kk[L/K]$ with $\cup_{i=1}^{\infty} W_i=\kk[L/K]$, the chain of
subalgebras $B_1\subset B_2\subset B_3\subset\dots$ generated by
$W_i$ does not stabilize. Hence one may suppose that all
inclusions here are strict. Let $Z_i$ be the affine $L$-variety
corresponding the algebra $B_i$. The inclusion $B_i\subset
\kk[L/K]$ induces the dominant morphism $L/K\to Z_i$ and
Theorem~\ref{tl} implies that $Z_i=L/K_i$, $K\subset K_i$. But
$B_1\subset B_2\subset B_3\subset\dots$, and any $K_i$ is strictly
contained in $K_{i-1}$, a contradiction. This shows that $B$ is
finitely generated and, as proved above, $L$ acts transitively on
the affine variety $Z$ corresponding to $B$. But any non-trivial
$L$-invariant ideal in $B$ corresponds to a proper $L$-invariant
subvariety in $Z$.
\end{proof}

Theorem~\ref{tm} is proved.
\end{proof}

\begin{corollary}\label{c1}
Let $G/H$ be an affinely closed homogeneous space. Then for any affine
$G$-variety $X$ and a point $x\in X$ such that $Hx=x$, the orbit $Gx$
is closed.
\end{corollary}

\begin{proof}
The stabilizer $G_x$ is observable in $G$, hence $\phi(G_x)$ is
observable in $L$. The subgroup $\phi(G_x)$ contains $K=\phi(H)$,
and Theorems~\ref{tl} and~\ref{tm} imply that the space $L/\phi(G_x)$ is
affinely closed. By Theorem~\ref{tm}, the space $G/G_x$ is
affinely closed.
\end{proof}

 In particular, we get

\begin{corollary}\label{c2}
If $X$ is an affine $G$-variety and a point $x\in X$ is $T$-fixed,
where $T$ is a maximal torus of the group $G$, then the orbit $Gx$
is closed.
\end{corollary}

\section{$G$-algebras with finitely generated invariant
subalgebras}
Below {\it an affine algebra} over a field $\kk$ means
a finitely generated associative commutative $\kk$-algebra with
unit. Let $F$ be a subgroup of the automorphism group of an affine
algebra ${\mathcal A}$ and ${\rm rad}({\mathcal A})$ the set of nilpotent
elements of the algebra ${\mathcal A}$. Clearly, ${\rm rad}({\mathcal A})$
is an $F$-invariant ideal in ${\mathcal A}$.

\begin{lemma}\label{lr}
The following conditions are equivalent:

{\rm (1)}\ any $F$-invariant subalgebra in ${\mathcal A}$ is finitely generated;

{\rm (2)}\ any $F$-invariant subalgebra in ${\mathcal A}/{\rm rad}({\mathcal A})$
is finitely generated and $\dim {\rm rad}({\mathcal A})<\infty$.
\end{lemma}

\begin{proof}
Any finite-dimensional subspace in
${\rm rad}({\mathcal A})$ generates a finite-dimensional subalgebra in ${\mathcal A}$.
Hence if
$\dim {\rm rad}({\mathcal A})=\infty$, then the subalgebra generated by this
subspace is not finitely generated. On the other hand, the preimage in
${\mathcal A}$ of any non-finitely generated subalgebra in
${\mathcal A}/{\rm rad}({\mathcal A})$ is not finitely generated.

Conversely, suppose that (2) holds.
Then any subalgebra in ${\mathcal A}$ is generated by elements whose images
generate the image of this subalgebra in
${\mathcal A}/{\rm rad}({\mathcal A})$, and by a basis
of the radical of the subalgebra.
\end{proof}

By definition, a {\it $G$-algebra} is an affine algebra ${\mathcal A}$ with an
action (by automorphisms) of an algebraic group $G$ such that
any element $a\in {\mathcal A}$ is contained in a finite-dimensional
$G$-invariant subspace, where $G$ acts rationally.
Our aim is to describe all $G$-algebras such that any $G$-invariant
subalgebra is finitely generated.
By Lemma~\ref{lr}, we may assume that ${\rm rad}({\mathcal A})=0$.

Let $X=\Spec (\mathcal A)$ be the affine variety corresponding
to affine algebra ${\mathcal A}$ without nilpotents. To fix a
structure of $G$-algebra on ${\mathcal A}=\kk[X]$ is nothing else but to
fix an (algebraic) $G$-action on $X$. Define {\it the dimension}
$\dim{\mathcal A}$ of the algebra ${\mathcal A}$ as the dimension of the variety $X$.

\begin{lemma}\label{ld}
If $\dim{\mathcal A}\le 1$, then any subalgebra in ${\mathcal A}$ is finitely generated.
\end{lemma}

\begin{proof}
The case when $X$ is irreducible is considered
in~\cite[Prop.~2]{ar}. If $X=X_1\cup\dots\cup X_m$ is the decomposition on
irreducible components, then ${\mathcal A}$ is embedded into the direct sum of
the algebras $\kk[X_i]$, and any subalgebra in a summand is finitely generated.
Now it is easy to finish the proof by induction on $m$
considering the projection of ${\mathcal A}$ on $\kk[X_1]$.
\end{proof}

\begin{lemma}\label{lcom}
Suppose that $X=Z_1\cup Z_2$, where $Z_1$ and $Z_2$ are
closed invariant subvarieties. Then the following conditions are equivalent:

{\rm (1)}\ any invariant subalgebra in ${\mathcal A}$ is finitely generated;

{\rm (2)}\ any invariant subalgebra in $\kk[Z_1]$ and in $\kk[Z_2]$
is finitely generated.
\end{lemma}

\begin{proof}
If there is a non-finitely generated subalgebra in $\kk[Z_1]$,
then one may consider its preimage with respect to the restriction
homomorphism $\kk[X]\to\kk[Z_1]$. To prove the converse,
embed $\kk[X]$ in $\kk[Z_1]\oplus\kk[Z_2]$ and use the arguments
from the proof of the previous lemma.
\end{proof}

By Lemma~\ref{lcom}, one may assume that $G$ acts transitively
on the set of irreducible components of the variety $X$.

Below we generalize a construction from~\cite{ar} to the case
of non-connnected groups and reducible varieties.
Let $Y$ be a closed subvariety of $X$.
Consider a subalgebra
$$
 {\mathcal A}(X,Y)=\{ f\in\kk[X] \mid f(y_1)=f(y_2) \ \forall\, y_1,y_2\in Y \}.
$$

\begin{lemma}\label{lk}
If $Y$ contains an irreducible component of positive dimension that
does not coincide with any irreducible component of $X$, then
${\mathcal A}(X,Y)$ is not finitely generated.
\end{lemma}

\begin{proof}
Note that ${\mathcal A}(X,Y)=\kk\oplus I(Y)$. If ${\mathcal
A}(X,Y)$ is finitely generated, then one may assume that
generators $f_1,\dots,f_k$ are in $I(Y)$. Any monom in
$f_1,\dots,f_k$ of degree $s$ is in $I(Y)^s$. Hence it is
sufficient to prove that for some $l$ the space $I(Y)/I(Y)^l$ is
infinite-dimensional.

Let $Y=Y_1\cup\dots\cup Y_n$ and $X=X_1\cup\dots\cup X_m$ be the
decompositions on irreducible components, and $Y_1\subset X_1$,
$Y_1\ne X_1$, $\dim Y_1>0$. Suppose that $f\in I(Y)$ and $f$ is
not identically zero on $X_1$. Let ${\mathcal O}_{X_1,Y_1}$ be the
local ring of the subvariety $Y_1$ in $X_1$ and ${\mathcal I}$ its
maximal ideal. By the Nakayama Lemma, $\cap_{i=1}^{\infty} {\mathcal
I}^i=0$, hence after restriction to $X_1$ the element $f$ belongs
to ${\mathcal I}^{l-1}\setminus {\mathcal I}^l$ for some $l\ge 2$. Let $W$
be a subspace in $\kk[X]$ complementary to $I(Y_1)$. Note that
$\dim Y_1>0$ implies $\dim W=\infty$. The subspace $fW$ may be
considered as an infinite-dimensional subspace in ${\mathcal I}^{l-1}$
with zero intersection with ${\mathcal I}^l$. Hence $fW$ determines an
infinite-dimensional subspace in $I(Y)/I(Y)^l$.
\end{proof}

We conclude that any invariant subvariety $Y$ satisfying the conditions
of Lemma~\ref{lk}, determines the non-finitely generated invariant subalgebra
${\mathcal A}(X,Y)$ in $\kk[X]$.

\smallskip

By $G^0$ denote the connected component of unit of a group $G$.

\begin{theorem}\label{tmm}
Let ${\mathcal A}$ be a $G$-algebra without nilpotents with the non-trivial
induced action of the subgroup $G^u$. The following conditions are
equivalent:

\smallskip

{\rm (1)}\ any $G$-invariant subalgebra in ${\mathcal A}$ is finitely generated;

{\rm (2)}\ any $G$-invariant subalgebra in ${\mathcal A}$ does not contain
non-trivial $G$-invariant ideals;

{\rm (3)}\ any $L$-invariant subalgebra in ${\mathcal A}^{G^u}$ does not contain
non-trivial $L$-invariant ideals;

{\rm (4)}\ ${\mathcal A}=\kk[G/H]$, where $G/H$ is an affinely closed
homogeneous space;

{\rm (5)}\ ${\mathcal A}^{G^u}=\kk[L/K]$, where $L/K$ is an affinely closed
homogeneous space.
\end{theorem}

\begin{proof}
$(1)\Rightarrow (4)$ \ {\it Step 1.}\ Suppose that
the action $G:X$ is not transitive. The closure $Y$ of a $G$-orbit
on $X$ is an invariant subvariety and we may apply Lemma~\ref{lk}
with the only exceptions $Y=X$ and $\dim Y=0$. Hence (1) implies
that $G^0$ acts on any component $X_i$ either with an open orbit and the
boundary of this orbit is a finite set of points, or trivially. In these cases
the action $G^u:X$ is trivial~\cite[Th.3]{po}, see
also~\cite[Prop.4]{ar}.

{\it Step 2.}\ Suppose that the action $G:X$ is transitive.
Then $X=G/H$. If
$G/H$ admits a non-trivial affine embedding $G/H\hookrightarrow X'$,
then $\kk[X']$ is an invariant subalgebra in ${\mathcal A}$. By Step 1,
this subalgebra contains a non-finitely generated invariant subalgebra.

\smallskip

$(2)\Rightarrow (4)$ \ The absence of non-trivial invariant ideals in
${\mathcal A}$ implies that $X=G/H$, and the absence of non-trivial invariant ideals
in invariant subalgebras implies that $G/H$ does not admit non-trivial
embeddings.

\smallskip

Proofs $(4)\Rightarrow (1)$ and $(4)\Rightarrow (2)$ are analogous
to the proof of Lemma~\ref{lv} (one should use
Corollary~\ref{c1}). By the same arguments we get $(5)\Rightarrow
(3)$.

\smallskip

We know that $\kk[G/H]^{G^u}=\kk[L/K]$. Theorem~\ref{tm}
implies $(4)\Rightarrow (5)$.

\smallskip

$(3)\Rightarrow (2)$\ Let ${\mathcal B}$ be an invariant subalgebra in ${\mathcal A}$
and $I$ a non-trivial invariant ideal in ${\mathcal B}$. Then
$I\cap {\mathcal A}^{G^u}$ is a non-trivial invariant ideal in
${\mathcal B}\cap {\mathcal A}^{G^u}$.
This completes the proof of Theorem~\ref{tmm}.
\end{proof}

\begin{remark}
 1) \ The conditions of Theorem~\ref{tmm} are not
equivalent to the condition "any $L$-invariant subalgebra in ${\mathcal A}^{G^u}$
is finitely generated": one may consider $G=G^u=(\kk, +)$ acting
on $\kk[x,y]$ by the formula $(a, f(x,y)) \to f(x+ay,y)$.

\smallskip

\ 2)\ The implication $(1)\Rightarrow (5)$ is incorrect if
${\mathcal A}={\mathcal A}^{G^u}$, see Lemma~\ref{ld} and~\cite{ar}.

\smallskip

\ 3) \ The restriction ${\mathcal A}\ne{\mathcal A}^{G^u}$ is natural
because the case of reductive group actions was studied
in~\cite{ar} under the assumptions that $G$ is connected and $X$
is irreducible. But Lemma~\ref{lk} and above arguments show that
if a reductive $G$ acts transitively on the set of irreducible
components of $X$, then any invariant subalgebra in $\kk[X]$ is
finitely generated if and only if either the $G^0$-algebras
$\kk[X_i]$ for any irreducible component $X_i$
have (in the terminology of~\cite{ar}) type C or HV, or
$X=G/H$ and $G/H$ is affinely closed (in this case, the
$G^0$-algebras $\kk[X_i]$ may not have type N, see example 5)
below). Surprisingly, the restriction ${\mathcal A}\ne{\mathcal A}^{G^u}$
simplifies the main results.
\end{remark}

\begin{corollary}
Let $G/H$ be a quasi-affine homogeneous space. Suppose that $H$
does not contain $G^u$. Then either $G/H$ is affinely closed or there are
infinitely many pairwise nonisomorphic affine embeddings
$G/H\hookrightarrow X_i$ and a sequence of dominant equivariant morphisms:
$$
X_1\gets X_2\gets X_3\gets \dots \ \ .
$$
\end{corollary}

\begin{proof}
Let $G/H\hookrightarrow X$ be a non-trivial affine embedding
and $Y$ the complement to the open orbit in $X$.
The algerba ${\mathcal A}(X,Y)$ is not finitely generated.
Let $p_0\in I(Y)$ be a non-zero element and
$f_1,\dots,f_s$ a set of generators of $\kk[X]$. Put
$p_i=p_0f_i$, $i=1,\dots,s$ and extend the set $p_0,p_1,\dots,p_s$
to an (infinite) generating set
$p_0,p_1,\dots,p_s,h_1,h_2,\dots$ of the algebra ${\mathcal A}(X,Y)$.
The affine $G$-varieties $X_i$ corresponding to the algebras
$$
{\mathcal B}_i=\kk[<Gp_0,Gp_1,\dots,Gp_s,Gh_1,\dots,Gh_i>] \subset
{\mathcal A}(X,Y) \subset \kk[X] \subset \kk[G/H],
$$
\noindent define embeddings
$G/H\hookrightarrow X_i$ and the inclusions of algebras determine
the desired chain of dominant morphisms. In the sequence
$X_i$ there is a subsequence consisting of pairwise nonisomorphic
embeddings.
(By definition, an isomorphish of two embeddings sends the base point
to the base point and is the unique equivariant morphism
identical on the open orbit.)
\end{proof}

\section{Some remarks on affine embeddings of homogeneous spaces for
non-reductive groups}
In this section we consider elementary
examples that provide negative answers to some natural questions.

\smallskip

Let $V$ be a finite-dimensional $G$-module and $v\in V$.

\smallskip

1) {\it The orbit $Gv$ is closed, but the orbit $Lv$ is not closed.}\
Consider
$$
  V=\kk^2,\  v=(1,1), \
G=\left\{\left( \begin{array}{cc}
               t & a \\
               0 & 1
              \end{array} \right)\right\},\
L=\left\{\left( \begin{array}{cc}
               t & 0 \\
               0 & 1
              \end{array} \right)\right\}.
$$

\smallskip

2) {\it The orbit $Lv$ is closed, but the orbit $Gv$ is not
closed.} Consider $$
  V=\kk^2,\  v=(1,1), \
G=\left\{\left( \begin{array}{cc}
               t & a \\
               0 & t^{-1}
              \end{array} \right)\right\},\
L=\left\{\left( \begin{array}{cc}
               t & 0 \\
               0 & t^{-1}
              \end{array} \right)\right\}.
$$

\smallskip

3) {\it The space $G/H$ is affinely closed does not imply that
all $L$-orbits on $G/H$ are closed.}\ Consider
$$
G=\left\{\left( \begin{array}{cc}
               t & a \\
               0 & t^{-1}
              \end{array} \right)\right\},\,
H=L=\left\{\left( \begin{array}{cc}
               t & 0 \\
               0 & t^{-1}
              \end{array} \right)\right\},\,
x=\left( \begin{array}{cc}
          1 & 1 \\
          0 & 1
          \end{array} \right)
$$
$$
\Rightarrow \overline{LxH}\ne LxH.
$$

\smallskip

4) {\it If $G^0/H^0$ is affinely closed then
$G/H$ is affinely closed, but the converse is not true.}
\ One may take $G=SL(2)$ with any finite non-Abelian subgroup $H$.

\smallskip

5) {\it If $G^0/(H\cap G^0)$ is affinely closed then
$G/H$ is affinely closed, but the converse is not true.} \
Consider
$$
  G=N_{SL(2)}T, \ \ H=<
\left( \begin{array}{cc}
        0 & 1 \\
       -1 & 0
      \end{array} \right) >.
$$

\smallskip

6) {\it The condition $0\in\overline{Gv}$ does not imply that $0$
may be obtained as the limit of a one-parameter subgroup in $G$.}\
The corresponding example for a solvable
$G$ is given in~\cite[sec.11]{bir}.

\smallskip

An important open problem is to characterize affinely closed spaces
(for both reductive and non-reductive groups) over an algebraically
closed field of positive characteristic. In particular, it is not known
do we have here
Corollary~\ref{c1}. Some results in this direction
may be found in~\cite[sec.8]{ar}.


TITLE: theorem of Mittag-Leffler
QUESTION [2 upvotes]: I want to find a meromorphic function that has poles exactly in the natural numbers and has the principal part $\frac{1}{z-n}.$
Therefore I used the theorem of Mittag-Leffler
and I received $\sum_{n=1}^{\infty}(\frac{1}{z-n}+\frac{1}{n}).$ Is this correct?
And unfortunately I don't quite understand how exactly I can read the principle part..
Thanks for your help!!

REPLY [2 votes]: Not sure what you mean by "principle part". I think you mean the Laurent series, which is going to be different for each singularity of a function.
I think you are mixing up a Laurent series vs a rational series. A Laurent series is of the form:
$$f(z) = \sum_{k=-\infty}^{\infty} a_k(z-z_0)^k $$
Notice how the series is centered at $z_0$, which is a singularity to the meromorphic function. A rational series is
$$ f(z) = \sum_{k=0}^\infty \frac{a_k}{z - b_k}$$
Notice that there is no "center". This is not a Laurent series, does not have an annulus of convergence and is instead governed by different theorems. However, you can use any sequence of convergent analytic functions to discover properties about the Laurent series.
For example, if I wanted coefficient $a_{-1}$ in the Laurent series centered at $n$, for the function defined by the rational series
$$f(z) = \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n}$$
Then all I have to do is integrate around a small compact contour $C$ that is centered at $n$, canceling out those terms that integrate to zero:
$$\begin{align} \int_C \sum_{k=-\infty}^{\infty} a_k (z-n)^k &= \int_C \sum_{k=1}^\infty \frac{1}{z-n}-\frac{1}{n} \\ a_{-1}\cdot 2\pi i &= \int_C \frac{1}{z-n}  \\ a_{-1}\cdot 2\pi i &= 2\pi i \\ a_{-1} &= 1 \end{align}$$
So now we know that the Laurent series centered at $n$ has a principle coefficient of $1$. Again, note there is a different Laurent series for each singularity.


text:
\begin{document}

\title{ Anytime Proximity Moving Horizon Estimation:\\ Stability and Regret }

\author{Meriem~Gharbi,
        Bahman~Gharesifard,~\IEEEmembership{Member,~IEEE}
        and~Christian~Ebenbauer,~\IEEEmembership{Member,~IEEE}

\thanks{This work was supported by the Deutsche Forschungsgemeinschaft (Grant EB 425/6-1). Bahman Gharesifard is thankful to the generous support of the Alexander von Humboldt Foundation.
}
\thanks{M. Gharbi and C. Ebenbauer are with Institute for Systems Theory and Automatic Control,
University of Stuttgart, 70550 Stuttgart, Germany (e-mail:
meriem.gharbi@ist.uni-stuttgart.de; ce@ist.uni-stuttgart.de).}
\thanks{B. Gharesifard is with the Department of Mathematics and Statistics, Queen's University, Kingston, ON K7L3N6, Canada (e-mail: bahman@mast.queensu.ca).}
}

\markboth{Journal of \LaTeX\ Class Files,~Vol.~ , No.~ , }
{Gharbi \MakeLowercase{\textit{et al.}}: Anytime Proximity Moving Horizon Estimation: Stability and Regret}

\maketitle

\begin{abstract}
In this paper, we address the efficient implementation of moving horizon state estimation of constrained discrete-time  linear systems. We propose a novel iteration scheme which employs a proximity-based formulation of the underlying optimization algorithm and reduces computational effort by performing only a limited number of  optimization iterations each time a new measurement is received.  We outline conditions under which global exponential stability of the underlying estimation errors is ensured.  Performance guarantees  of the iteration scheme in terms of regret upper bounds are also established.  A combined result shows that both exponential stability and a sublinear regret which can be rendered smaller by increasing the number of optimization iterations can be guaranteed. The stability and regret results of the proposed estimator are showcased through numerical simulations.
\end{abstract}

\IEEEpeerreviewmaketitle

\input{intro}
\input{setup}
\input{algorithm}
\input{stability}
\input{regret}
\input{example}
\input{conc}

\bibliography{bib}
\bibliographystyle{unsrt}

\renewcommand{\thesubsection}{\Alph{subsection}}
\section*{Appendix}
 \input{App}

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi

\begin{IEEEbiography}[{\includegraphics[width=1in,height=1.25in,clip,keepaspectratio]{Gharbi_Foto}}]{Meriem Gharbi} received the B.Sc. and M.Sc. degrees in Engineering Cybernetics from
the University of Stuttgart, Germany, in 2014 and 2017, respectively.
Since October 2017, she is a Ph.D. student at the Institute for System Theory and
Automatic Control, University of Stuttgart, Germany. \\
Her current research interests lie in the area of optimization-based state estimation with a focus on moving horizon estimation.

\end{IEEEbiography}

\begin{IEEEbiography}[{\includegraphics[width=1in,height=1.25in,clip,keepaspectratio]{Gharesifard}}]{Bahman Gharesifard}  received the B.Sc. degree in Mechanical  Engineering, in 2002, and the M.Sc. degree in Control and Dynamics,  in 2005, from Shiraz University, Iran.  He received his Ph.D.
in Mathematics from Queen's University, Canada, in 2009.
He is an Associate Professor with the Department of Mathematics and Statistics at Queen's University. He held postdoctoral positions with the Department of Mechanical and Aerospace Engineering at University of California, San Diego 2009-2012 and with the Coordinated Science Laboratory at the University of Illinois at Urbana-Champaign from 2012-2013. He held a visiting faculty position at the Institute for Systems Theory and Automatic Control at the University of Stuttgart in summer of 2016, and
was an Alexander von Humboldt research fellow at the same institute in 2019-2020. He received the 2019 CAIMS-PIMS Early Career Award, jointly awarded by the Canadian Applied \& Industrial Math Society and the Pacific Institute for the Mathematical Sciences, a Humboldt research fellowship for experienced researchers from the Alexander von Humboldt Foundation in 2019, and an NSERC Discovery Accelerator Supplement in 2019. He was a finalist (as an advisor) for the Best Student Paper Award at the American Control Conference in 2017. He received the Engineering and Applied Science First Year Instructor Teaching Award in 2015 and 2017, and was shortlisted for a Frank Knox Award for Excellence in Teaching in 2014. He serves on the Conference Editorial Board of the IEEE Control Systems Society, and as an
Associate editor for the IEEE Control System Letters.\\ His research interests include systems and controls, distributed control, distributed optimization and learning, geometric control theory, social and economic networks, game theory, geometric mechanics, and applied Riemannian geometry.

\end{IEEEbiography}

\begin{IEEEbiography}[{\includegraphics[width=1in,height=1.25in,clip,keepaspectratio]{ebenbauer13a}}]{Christian Ebenbauer} received the M.S. degree (Dipl.-Ing.) in Telematics (electrical engineering and computer science) from Graz University of Technology, Graz, Austria, in 2000 and the Ph.D. (Dr.-Ing.) in Mechanical Engineering from the University of Stuttgart, Germany, in 2005. After having completed his Ph.D., he was a Postdoctoral Associate and an Erwin Schr{\"o}dinger Fellow at the Laboratory for Information and Decision Systems, Massachusetts Institute of Technology, Cambridge, MA, USA. Since April 2009, he is a Full Professor of Computations in Control at the Institute for Systems Theory and Automatic Control, University of Stuttgart.\\
His research interests lie in the areas of dynamical systems, control theory, optimization, and computation. Recent research projects focus on system-theoretic approaches to optimization algorithms, extremum seeking control, and MPC algorithms.

\end{IEEEbiography}

\end{document}


TITLE: Quasiconvexity of the objective function
QUESTION [0 upvotes]: I´m having a lot of trouble showing if the objective function is quasiconvex or not. I already tried with the bordered Hessian but it gets really messy :-/. Maybe someone can help me with this! Thanks in advance!
The problem: $$max f(x,y)=x^{1/2}*y^{1/3}$$
subject to:$$3x+4y ≤25$$

REPLY [1 votes]: This is not a quasiconvex function but a strictly quasiconcave function. This is a Cobb-Douglas function with $\alpha = 1/2$ and $\beta = 1/3$. Thus $\alpha + \beta = 5/6$ which is less than 1. Thus the Cobb-Douglas function $x^{1/2}*y^{1/3}$ is concave and therefore it is also quasiconcave. The fact that it is strictly quasiconcave can be shown by solving the Bordered Hessian.


TITLE: Restrictions that comes from: If $(x, y)$ is a solution then $(-y, -x)$ is a solution too
QUESTION [2 upvotes]: If we have an equation like $f(x, y)=0$. And there is symmetry such that if $(x, y)$ is a solution then $(-y, -x)$ is a solution too, then what  kind of assumptions on $x$, $y$ can be made without loss of generality? Or what kind of maps can provide an easier equation?
I know that if $(x, y)$ is a solution and then $(-x, -y)$ is a solution too then we can solve for positive values only. Can you help me about above question?

REPLY [6 votes]: The map $(x,y) \mapsto (-y,-x)$ is a involution (i.e. if you apply it to itself you get the identity map) and it gives you a correspondance between the two regions
$$\{ (x,y) : x+y <0 \}$$ and $$\{ (x,y) : x+y >0 \}$$
while it has as fixed points the line $x+y=0$. So, you have a symmetry with respect to this line, and you can consider WLOG $$x+y \ge 0$$


TITLE: When integrating $x^2((3-x^3)^2)$ using $u$ substitution, why does $u$ have to be $x^3$ and not $3-x^3$?
QUESTION [4 upvotes]: Title. I'm a high student self studying calculus and was wondering why the $U$ in the $U$ substitution in this case is $x^3$ and not $3-x^3$ since they both seem to work. Any help will be appreciated!

REPLY [1 votes]: Both the sustitutions will work here.But for simplicity take u=x^3 .You can take any one of them as u. By the way , what are the limits ?


TITLE: Solving Special Function Equations Using Lie Symmetries
QUESTION [3 upvotes]: The lie group + representation theory approach to special functions & how they solve the ode's arising in physics is absolutely amazing. I've given an example of it's power below on Bessel's equation.
Kaufman's article describes algebraic methods for dealing with Hermite, Legendre & Associated Legendre. Can we take the other special functions mentioned in this paper, obtainable as linear combinations of the conformal symmetries of the Laplacian (expressed as lie algebra elements), and obtain their solution analogously to how Bessel is solved below? I believe it's something like a geometric interpretation of Weisner's method.
Bessel's equation seems to be saying: find a function in the plane such that when we shift it right, then shift it back left again, all locally (i.e. differentially) in polar coordinates, we get the same function:

(c.f. Killingbeck, Mathematical Technique's and Applications, sec. 8.21).
The idea is to take Bessel's equation, factor it, add an extra variable to make the factors parameter independent so that they become elements of a lie algebra, identify the meaning of those factors, in this case notice the Lie algebra factors are translations in Polar coordinates, and realize it's just a differential expression of a symmetry.
Bessel's equation arises from $LRv = v$ when you express $L$ & $R$ in polar coordinates. It makes sense to express them in polar coordinates since Bessel arises from separating the Laplacian assuming cylindrical symmetry, and the $LRv = v$ assumption (not $LRv = w$) is motivated by the symmetry of the Laplacian.
Using this idea we can, for some reason, actually solve Bessel's equation with a picture!:

We just want to shift $\mathcal{J}_n(r,\phi)$ in the x-direction using the operator $e^{a\tfrac{\partial}{\partial x}}$ expressed in polar coordinates: $e^{\tfrac{a}{2}(\mathcal{L}-\mathcal{R})}$ and realize it will be equal to $\mathcal{J}_n(r',\phi')$:

So the last line comes from dragging all this to the origin and putting it along the x-axis, here we see the geometric meaning of Bessel functions!
Hypergeometric is supposed to be related to $SL(2,R)$ symmetries, Bessel to translational planar symmetries, the Gamma function related to linear symmetries $y = ax + b$, etc... Is there an easy unified geometric exposition on how to deal with these babies?
It'd be great to understand the other equations, their formulation and solution, with a geometric interpretation like this one.
References:

Killingbeck, Mathematical Technique's and Applications, sec. 8.21
Vilenkin, Representation of Lie Groups and Special Functions Vol. 1
Vilenkin, Special Functions and Theory of Group Representations
Miller, Lie Theory and Special Functions
Kaufman, Special Functions of Mathematical Physics from the Viewpoint of Lie
Algebra

REPLY [2 votes]: You are correct in your interpretation that Weisner's method is geometric in nature: it is a method for finding generating functions for special functions using representation theory of Lie groups and Lie algebras. And as you know, Lie groups play an enormous role in modern geometry, on several different levels. Lie groups are smooth differentiable manifolds, that is, roughly, a type of manifold that is locally similar enough to a linear space to allow one to do calculus.
Lie and others showed that the most important equations for special functions and orthogonal polynomials tend to arise from group theoretical symmetries. The initial application that Lie had in mind was to the theory of differential equations: on the model of Galois theory and polynomial equations, the driving conception was of a theory capable of unifying, by the study of symmetry, the whole area of ordinary differential equations. However, the hope that Lie Theory would unify the entire field of ordinary differential equations was not fulfilled.
Thus, it is correct that you can take

the other special functions mentioned in this paper, obtainable as linear combinations of the conformal symmetries of the Laplacian (expressed as lie algebra elements), and obtain their solution analogously to how Bessel is solved below.

because of how these functions are defined.
However, there is no general method because this is not valid in general for arbitrary ordinary differential equations.


TITLE: Can $\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$ be simplified?
QUESTION [2 upvotes]: I'm trying to compute a complicated thing, and I end up with terms like:
$$\sum_{i=a}^{b}p^{i}\binom{i}{a}\binom{b}{i}$$
$a$ and $b$ are nonnegative integers, $0<p<1$.
I don't see a way to make this simpler. Am I missing something or is this the simplest thing that I'm going to get?

REPLY [1 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{i = a}^{b}p^{i}{i \choose a}{b \choose i} & =
\sum_{i = 0}^{b}{b \choose i}p^{i}\bracks{z^{a}}\pars{1 + z}^{i} =
\bracks{z^{a}}\sum_{i = 0}^{b}{b \choose i}\pars{p + pz}^{i} \\[5mm] & =
\bracks{z^{a}}\bracks{1 + \pars{p + pz}}^{\, b}
\\[5mm] & =
\pars{1 + p}^{b}\bracks{z^{a}}\bracks{1 + {p \over 1 + p}\,z}^{\, b} \\[5mm] &
\pars{1 + p}^{b}{b \choose a}\pars{p \over 1 + p}^{a}
\\[5mm] & =
\bbx{{p^{a} \over \pars{1 + p}^{a - b}}\,{b \choose a}} \\ &
\end{align}


TITLE: f is uniformly continuous if for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.
QUESTION [1 upvotes]: Let f : R → R be a given function. The following property will ensure that f is uniformly continuous:
for all $x$ and $y ∈ \mathbb{R}$, $|f(x) − f(y)| ≤ |x−y|^\frac{1}{2}$.
My attempt :
From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$
Now let $\epsilon \gt 0$ be given. Then
$$
 |x−y|^\frac{1}{2} \lt \epsilon \\
\implies
 |x-y| \lt {\epsilon}^2 = \delta \\
\implies |f(x)-f(y)| \lt \epsilon
.$$
Therefore the function is uniformly continuous.

REPLY [0 votes]: From definition a function $f$ is uniformly continuous if for a arbitrary $\epsilon \gt 0$ there exists a $\delta \gt 0$ such that for any two points $x,y$ in $\mathbb{R}$,
$$ |x-y| \lt \delta \implies |f(x)-f(y)| \lt \epsilon
$$
Now let $\epsilon \gt 0$ be given. Then let $$\delta = {\epsilon}^2.$$ Then
$$ |x-y| \lt {\epsilon}^2 = \delta \\
\implies |x−y|^\frac{1}{2} \lt \epsilon \\
\implies |f(x)-f(y)| \lt \epsilon
.$$
Therefore the function is uniformly continuous.


TITLE: An elementary proof for a bound on $x \log x$
QUESTION [5 upvotes]: During one of our information theory classes, the Professor used the following bound to prove a result.
For any $x,y \in (0,1)$, $x \neq y$, show that
$$|x \log(x) - y \log(y) | \leq |x-y|\log \left(\frac{1}{|x-y|}\right)$$
Note that RHS is positive when put this way. Also note that base of the logarithm is immaterial here(I take base $e$ to simplify calculations). Looking at $|x-y|$, I thought of using mean value theorem. W.L.O.G let $x > y$. Then by applying mean value theorem to $u \log u$, we have
$$x \log(x) - y \log(y) = (x-y)(1 + \log z) = (x-y)(\log ez)$$
for some $z \in (y,x)$. Now I do not know how to show (after removing logs).
$$ez \leq \left(\frac{1}{x-y}\right)$$
I'd like to know if my approach has any promise as well as any alternate ways to get this. As always any hints are appreciated.
Edit: Looks like I made an error while posting this problem, apparently the result also required that $|x-y| \le 0.5$. Kindly note this.

REPLY [1 votes]: From what I tried, the mean value approach will not work. So I found another approach using the method of secants. Given a function $f$, define for any two real numbers $b$ and $a$, $b \ne a$
$$R(b,a) = \frac{f(b)-f(a)}{b-a}$$
Now for a convex function (you need nothing more than the definition), we can show that for $a < b < c$
$$R(c,b)\ge R(c,a) \ge R(b,a)$$
We can further deduce for convex functions that if $a< b$, $c < d$, $a < c$ and $b<d$,
$$R(b,a) \le R(d,c)$$
Flip all inequalities for concave functions. Now in my problem, let $f(u) = -u\log(u)$. Set $f(0)=0$ here. Note that this is a concave function. Now, assuming $x > y$ in the original question, let $x-y = \tau$. Thus we get by putting $a=0$, $b=\tau$, $c=y$ and $d=x=y+\tau$, that
$$R(x,y) \leq R(\tau,0)\Rightarrow \frac{f(x)-f(y)}{x-y} \leq \frac{f(\tau)}{\tau} \\
\Rightarrow f(x) -f(y) \leq f(\tau)$$
Similarly noting that $f(1)=0$ and putting $a=y$, $b=x=y+\tau$, $c=1-\tau$, $d=1$, we get
$$f(x) -f(y) \geq -f(1-\tau)$$
Thus, noting that $f(u) \geq 0$ for $u\in(0,1]$, we get
$$|f(x) -f(y)| \leq \max\{f(\tau),f(1-\tau)\}$$
Since $\tau \leq 1/2$, $f(\tau) \geq f(1-\tau)$ which completes the proof. $\blacksquare$


TITLE: Finite Generated Abelian Torsion-Free Group is a Free Abelian Group
QUESTION [5 upvotes]: I am trying to prove that every Finite Generated Abelian Torsion-Free Group is a Free Abelian Group. In order to do this, I am trying to show that if $\{x_1, \dots, x_n\}$ is a minimal generator of the group and if $n_1x_1 + \dots n_n x_n=0$ then $n_1=\dots = n_n=0$. I'm stuck here.

REPLY [4 votes]: Woo-hoo! Here's the proof (without the structure theorem) that I was trying for yesterday.
Notation: If $w=(w_1,\dots,w_n)\in\Bbb Z^n$ we will write $$\gcd(w)=\gcd(w_1,\dots,w_n).$$ If $x_1,\dots,x_n$ is your minimal generating set we will write $$x=(x_1,\dots,x_n).$$The reason being that now if $w\in\Bbb Z^n$ we will want to write $$w\cdot x=w_1x_1+\dots+w_nx_n$$in order to make various things readable.
Here's the theorem I conjectured yesterday (based on the case $n=2$):
Theorem Suppose $m\in\Bbb Z^n$ and $\gcd(m)=1$. There exists a set $v_1,\dots, v_n$ of generators for $\Bbb Z^n$ with $m=v_1$.
You can find a proof of that at Generating Sets for Subgroups of $(\Bbb Z^n,+)$.. The theorem makes the result you want trivial:
Suppose $x_1,\dots,x_n$ is a minimal set of generators for your torsion-free abelian group. Suppose that there exists $m\in\Bbb Z^n$ with $m\ne0$ and $$m\cdot x=0.$$
(See above for the notation.)
As noted previously, if there exist $d\in\Bbb Z$ and $m'\in\Bbb Z^n$ with $m=dm'$ then $d(m'\cdot x)=0$ and hence $m'\cdot x=0$. So we may assume $\gcd(m)=1$. Let $v_1,\dots,v_n$ be a set of generators for $\Bbb Z^n$ with $v_1=m$.
Now if $w\in\Bbb Z^n$ there exist integers $k_j$ with $$w=k_1m+k_2v_2+\dots+k_nv_n.$$Hence $$w\cdot x=k_2(v_2\cdot x)+\dots+k_n(v_n\cdot x).$$So $v_2\cdot x,\dots,v_n\cdot x$ generate your group, contradicting the minimality of $m$.


TITLE: Orthogonal Skew Symmetric matrices are orthogonally conjugate
QUESTION [0 upvotes]: Let $X,Y$ be orthogonal skew symmetric matrices. Are $X$ and $Y$ orthogonally conjugate?
I.E. there exists $U \in O(n)$ such that $UX = YU$?
By orthogonality and skew symmetry, $XX^T = X(-X) = I$.
Also $YY^T = Y(-Y) = I$.
But I cannot go further.

REPLY [1 votes]: I assume the field is $\mathbb R$.
A straightforward way to do this is treat some arbitrary orthogonal skew-symmetric matrix $A$ as representing a skew-symmetric form, i.e. $\langle \mathbf x, \mathbf y\rangle:=\mathbf x^TA\mathbf y$. Note that skew-symmetry implies $\langle \mathbf x, \mathbf x\rangle=0$ for all $\mathbf x$.  Technically you just need to show that that such an $A$ always has a 2-dimensional $A$-invariant subspace and the result follows by induction on dimension.
In particular select some arbitrary $\mathbf v_2$ with length 1 (under the 2 norm). Consider the matrix $\mathbf v_2^TA$ which has rank 1 hence nullity of $n-1$ and this nullspace includes $\mathbf v_2$.  Build an orthonormal (w/ respect to dot product) basis for $\ker \mathbf v_2^TA$ as $ V'=\bigg[\begin{array}{c|c|c|c}
\mathbf v_2 &\mathbf v_3 &\cdots & \mathbf v_{n}
\end{array}\bigg]$.  Now select $\mathbf v_1$ via extending to orthogonal matrix $ V:=\bigg[\begin{array}{c|c|c|c}
\mathbf v_1 & \mathbf v_2 &\mathbf v_3 &\cdots & \mathbf v_{n}
\end{array}\bigg]$ and $\big\Vert \mathbf v_2^TAV\big\Vert_F=1\implies\langle \mathbf v_1, \mathbf v_2\rangle=\pm 1$ and we may assume $\langle\mathbf v_1, \mathbf v_2\rangle=1$ (via re-scaling $\mathbf v_1$ by $-1$ if needed).
Then
$V^TAV = \begin{bmatrix}  \langle \mathbf v_1, \mathbf v_1\rangle &\langle \mathbf v_2, \mathbf v_1\rangle &\mathbf x_1^T \\  \langle \mathbf v_1, \mathbf v_2\rangle &\langle \mathbf v_2, \mathbf v_2\rangle & \mathbf x_2^T \\ \mathbf y_1 & \mathbf y_2 & *\end{bmatrix}= \begin{bmatrix}  0 &-1 &\mathbf x_1^T \\  1 &0 & \mathbf x_2^T \\ -\mathbf x_1 & -\mathbf x_2 & A'\end{bmatrix}= \begin{bmatrix}  0 &-1 &\mathbf 0^T \\  1 &0 & \mathbf 0^T \\ \mathbf 0 & \mathbf 0 &  A'\end{bmatrix}$
where the middle equality follows by skew-symmetry and the final equality follows because $(V^TAV)\in O_n(\mathbb R)$ so all of its rows and columns have length one.  Note that $A'$ is real orthogonal and skew symmetric hence by induction hypothesis where is some $Q\in O_n(\mathbb R)$ of the form $Q:=\begin{bmatrix}  I_2 &\mathbf 0  \\  \mathbf 0 &  Q'\end{bmatrix}$ such that $(Q')^TA'Q'$ is block diagonal with each block given by $J:=\begin{bmatrix}  0 &-1 \\  1 &0 \end{bmatrix}$
Conclude: an arbitrary skew symmetric orthogonal matrix satisfies
$U^TAU = \begin{bmatrix}  J &\mathbf 0 &\dots &\mathbf 0\\\mathbf 0&J &\dots &\mathbf 0\\  \vdots  &  \vdots&\ddots&\vdots\\ \mathbf 0&\mathbf 0&\dots &J\end{bmatrix}=\mathbf J$ for some $U\in O_n(\mathbb R)$ thus in the original post both $X$ and $Y$ are othogonally similar to $\mathbf J$ hence they are orthogonally similar to each other.