text:
\begin{document}
\date{}

\title{Subinvariance in Leibniz Algebras}
\author{Kailash C. Misra, Ernie Stitzinger and Xingjian Yu}
\address{Department of Mathematics, North Carolina State University, Raleigh, NC 27695-8205}
\email{misra@ncsu.edu, stitz@ncsu.edu, xyu13@ncsu.edu }
\subjclass[2010]{17A32 , 17A60 }
\keywords{Leibniz Algebra, subinvariant subalgebras, radical and nilradical}
\thanks{KCM is partially supported by Simons Foundation grant \#  636482}

\begin{abstract}
Leibniz algebras are certain generalizations of Lie algebras. Motivated by the concept of subinvariance in group theory, Schenkman studied properties of subinvariant subalgebras of a Lie algebra. In this paper we define subinvariant subalgebras of Leibniz algebras and study their properties. It is shown that the signature results on subinvariance in Lie algebras have analogs for Leibniz algebras.

\end{abstract}

\maketitle
\bigskip
\section{Introduction}

Let $\A$ be a vector space over field $F$ equipped with a bilinear multiplication map $ \A \times \A \longrightarrow \A$ given by $(x, y) \longrightarrow xy$ for all $x, y \in \A$. For each $x \in \A$, define the left (resp. right) multiplication operator $L_x :\A \longrightarrow \A$ (resp. $R_x :\A \longrightarrow \A$ defined by $L_x(y) = xy$ (resp. $R_x(y) = yx$) for all $y \in \A$. The vector space $\A$ is a left (resp. right) Leibniz algebra if $L_x$ (resp. $R_x$) is a derivation for each $x \in \A$. Following Barnes \cite{B}, in this paper Leibniz algebras will always refer to finite dimensional left Leibniz algebras. Observe that a Lie algebra is a Leibniz algebra but not conversely. In particular the Leibniz algebra $\A$ has an abelian ideal ${\rm Leib}(\A)={\rm span}\{x^2= xx \mid x\in \A\}$. The ideal Leib$(\A)= \{0\}$ if and only if $\A$ is a Lie algebra. For a subalgebra $B$ of $\A$, the left (resp. right) centralizer of $B$ in $\A$ is defined by $Z^{\ell}_{\A}(B) = \{x \in \A \mid xa = 0 \ \ {\rm for} \, {\rm all} \ \ a \in B\}$ (resp. $Z^r_{\A}(B) = \{x \in \A \mid ax = 0 \ \  {\rm for} \, {\rm all} \ \ a \in B\}$). The cetralizer of $B$ in $\A$ is $Z_{\A}(B) = Z^{\ell}_{\A}(B) \cap Z^r_{\A}(B)$. The left (resp. right) center of $\A$ is $Z^{\ell}(\A) = Z^{\ell}_{\A}(\A)$. (resp. $Z^r(\A) = Z^r_{\A}(\A)$) and the center of $\A$ is $Z(\A) = Z^{\ell}(\A) \cap Z^r(\A)$. Note that ${\rm Leib}(\A) \subseteq Z^{\ell}(\A)$ and the center of $\A$ is $Z(\A) = Z^{\ell}(\A) \cap Z^r(\A)$. We denote $L(\A) = \{L_x \mid x \in\A\}$ which is a Lie algebra of linear transformations on $\A$. For a Leibniz algebra $\A$, we define the ideals $\A^{(1)} = \A = \A^{1}$, $\A^{(i)} = \A^{(i-1)} \A^{(i-1)}$ and $\A^{i} = \A\A^{i-1}$ for $i \in \mathbb{Z}_{\geq 2}$. The Leibniz algebra is said to be solvable (resp. nilpotent) if $\A^{(m)} = \{0\}$ (resp. $\A^{m} = \{0\}$) for some positive integer $m$. The maximal solvable (resp. nilpotent) ideal of $\A$ is called the radical (resp. nilradical) and denoted by rad$(\A)$ (resp. nilrad$(\A)$).

The concept of subinvariance was popularized for groups by Wielandt \cite{W} and for Lie algebras by Schenkman \cite{SE}. A subalgebra $B$ of the Leibniz algebra $\A$ is subinvariant in $\A$ if there exists a chain of subalgebras $B = B_t \subset B_{t-1} \subset \cdots \subset B_0 =A$ such that $B_i \vartriangleleft B_{i-1}$ for all $1 \leq i \leq t$. Such a series is called a normal series from $B$ to $A$. This normal series is a composition series from $B$ to $A$ if $B_i$ is a maximal ideal of $B_{i-1}$, $1 \leq i \leq t$ and in such case $B_{i-1}/B_i$ are called composition factors. As shown in  \cite{DMS}, the Leibniz algebra $\A$ is nilpotent if and only if all subalgebras are subinvariant in $\A$. In this paper we show that most important results on subinvariance of Lie algebras in \cite{SE} has Leibniz algebra analogs although techniques to obtain these results are different.

We organize this paper as follows. In Section 2, we give results on subinvariant subalgebras of Leibniz algebras over any field $F$. We define a function $f$ on a pair of subsets of a Leibniz algebra $\A$ and show that it has certain nice property which is used to prove several results through out the paper. Let $B$ be a subinvariant subalgebra of $\A$ and $B^{\omega}$ be the smallest member in the lower central series of $B$, then it is shown that $B^{\omega}$ is an ideal of $\A$. Also if $Z_{\A}(B) = \{0\}$, we show that $Z_{\A}(B^{\omega}) \subseteq B^{\omega}$. In Section 3, we collect the results on subinvariant subalgebras of Leibniz algebras over field $F$ of characteristic zero and show that the important results in this section fail over field of positive characteristic. We show that nilpotent (resp. solvable) subinvariant subalgebra of $\A$ is contained in $nilrad(\A)$ (resp. $rad(\A)$) and for any subinvariant subalgebra $B$ of $\A$, we show that $nilrad(B) = B \cap nilrad(\A)$ and
$rad(B) = B \cap rad(\A)$. Furthermore, it is shown that the subinvariant subalgebras of $\A$ contained in a subalgebra $B$ form a lattice under algebra generation and intersection, the latter of which is obvious. This lattice contains a unique maximal element which is an ideal in $B$, but not necessarily in $\A$. In particular, the collection of all nilpotent subinvariant subalgebras of $\A$ contained in $B$ is a sublattice with a unique maximal element. Finally in the last section we remark that under the assumption $Z^{\ell}(\A) = \{0\}$, the tower theorem for $\A$ follows trivially from the tower theorem for Lie algebra in \cite{SE}.

\bigskip
\section{General Results}
In this section we give some useful basic results on properties of subinvariant subalgebras of a Leibniz algebra $\A$ over field $F$. For a pair of subsets $B$ and $C$ of $\A$, we denote $f(B,C)$ to be the subalgebra of $\A$ containing $BC + CB$. We will denote $f(B,C) = BC + CB = f^1(B, C)$. For all $n \in \mathbb{Z}_{\geq 1}$, we define $f^{n+1}(B, C) = f(B, f^n(B, C))$ inductively. The following result on this function $f$ will be useful later.

\begin{lemma}\label{function}
Let $B$, $C$ be any two subsets of the Leibniz algebra $\A$. Then $f(B^n, C) \subseteq f^n(B, C)$ for all $n \in \mathbb{Z}_{\geq 1}$.

\end{lemma}
\begin{proof}
We prove by induction on $n$. Suppose $f(B^n, C) \subseteq f^n(B, C)$. Then $f(B^{n+1}, C) = B^{n+1}C + CB^{n+1} = (BB^n)C + C(BB^n) \subseteq B(B^nC) + B^n(BC) + (CB)B^n + B(CB^n) \subseteq Bf(B^n, C) + f(B^n, BC +CB) \subseteq Bf^n(B, C) + f^n(B, BC+CB) \subseteq f^{n+1}(B, C) + f^{n+1}(B, C) = f^{n+1}(B, C)$

\end{proof}

The following theorem is a Leibniz algebra analog of \cite[Theorem 1]{SE} and can be proved similarly.

\begin{thm}\label{basic}
Let $B, B'$ be subinvariant subalgebras of the Leibniz algebra $\A$ over field $F$, $C$ be a subinvariant subalgebra of $B$ and $K$ any subalgebra of $\A$.
Let $\phi : \A \longrightarrow \A'$ be a homomorphism from $\A$ to a Leibniz algebra $\A'$ over field $F$. Then the following statements hold.
\begin{enumerate}
\item $C$ is subinvariant in $\A$.
\item $B \cap B'$ is subinvariant in $\A$.
\item $K\cap B$ is subinvariant in $K$.
\item $\phi (B)$ is subinvariant in $\phi(\A)$.
\end{enumerate}

\end{thm}

For a subalgebra $B$ of the Leibniz algebra $\A$, define $B^{\omega} = \bigcap_{i=1}^{\infty}B^{i}$ which is an ideal of $B$.
The following result has been proved in \cite[Theorem 4.4]{B} using a different approach. As shown below it follows immediately
using the function $f$.

\begin{thm}\label{omega}
If $B$ is a subinvariant subalgebra of $\A$, then $B^{\omega}$ is an ideal of $\A$
\end{thm}

\begin{proof} Since $\A$ is finite dimensional, so is $B$. Hence $B^{\omega} =B^{s}$ for some $s \in \mathbb{Z}_{\geq 1}$ in the lower central series.
Also since $B$ is subinvariant in $\A$ we have a normal series $B = B_t \vartriangleleft B_{t-1} \vartriangleleft \cdots \vartriangleleft B_1 \vartriangleleft B_0 =A$. So we have a normal series of length $s+t$:
$$
B^{\omega}=B_{s+t-1} \vartriangleleft B_{s+t-2} \vartriangleleft \cdots \vartriangleleft B_t \vartriangleleft B_{t-1} \vartriangleleft \cdots \vartriangleleft B_0 =A.
$$
where we denote $B^j = B_{j+t-1}, 1\leq j \leq s$.
Observe that $f(B, \A) = B\A +\A B \subset B_1\A +\A B_1 \subseteq B_1$ since $B_1 \vartriangleleft \A$. Assume that $f^i (B, \A) \subseteq B_i$. Then
$f^{i+1}(B, \A) = f(B, f^i(B, \A)) \subseteq f(B, B_i) = BB_i + B_iB$. For $0 \leq i \leq t-1$, $BB_i + B_iB \subseteq B_{i+1}B_i + B_iB_{i+1} \subseteq B_{i+1}$ since $B_{i+1} \vartriangleleft B_i$. For $t \leq i \leq s+t-2$, we have $BB_i + B_iB = B_{i+1}$ by definition of lower central series. Thus we have $f^{s+t-1}(B, \A) \subseteq B_{s+t-1} = B^{\omega}$.
Now since $B^{\omega} =B^{s} = B_{s+t-1}$, by Lemma \ref{function} we obtain $B^{\omega}\A + \A B^{\omega} = f(B^{\omega}, \A) = f(B^s, \A)
= f(B^{s+t-1}, \A) \subset f^{s+t-1}(B, \A) \subseteq B^{\omega}$ implying that $B^{\omega}  \vartriangleleft \A$.

\end{proof}

\begin{cor}
If $B$ is a subinvariant subalgebra of $\A$ and $B^2 = B$, then $B$ is an ideal of $\A$.
\end{cor}
\begin{proof}
This follows from Theorem \ref{omega} since  $B^2 = B$ implies that $B^{\omega} = B$.
\end{proof}

We recall that a subalgebra $H$ of a Leibniz algebra $\A$ is a Cartan subalgebra if $H$ is nilpotent and
$N_{\A}(H) = \{x \in \A \mid xh, hx \in H \ \ {\rm for} \ \ {\rm all} \ \ h \in H\} = H$. The existence of a Cartan subalgebra of a finite dimensional Leibniz algebra $\A$ is shown in \cite{B}.

\begin{thm}\label{cartan}
Let $\A$ be a finite dimensional Leibniz algebra over field $F$. Then $\A = \A^{\omega} + H$ where $H$ is a Cartan subalgebra of $\A$.
\end{thm}
\begin{proof}
Let $H$ be a Cartan subalgebra of $\A$. Then $(H+\A^{\omega})/\A^{\omega}$ is a Cartan subalgebra of the nilpotent Leibniz algebra $\A/\A^{\omega}$.
Since a proper subalgebra of a nilpotent Leibniz algebra is contained in its normalizer, we have $(H+\A^{\omega})/\A^{\omega} = \A/\A^{\omega}$. This implies that
$H+ \A^{\omega} = \A$.
\end{proof}

The following result is an immediate consequence of Theorems \ref{omega} and \ref{cartan}.

\begin{cor}
Let $B$ be a subinvariant subalgebra of the finite dimensional Leibniz algebra $\A$. Then $B = B^{\omega} + H$ where $B^{\omega}$ is an ideal of $\A$ and
$H$ is a Cartan subalgebra of $B$.
\end{cor}

The following is a Leibniz algebra analog of a well known Lie algebra result which we need to prove in the next theorem.
\begin{lemma}\label{ideal}
Let $\A$ be a nilpotent finite dimensional Leibniz algebra with center $Z(\A)$ and let $B$ be a nonzero ideal of $\A$. Then $B \cap Z(\A) \neq \{0\}$.
\end{lemma}
\begin{proof}
Since $\A$ is nilpotent its center $Z(\A) \neq \{0\}$. Also
$\A$ acts nilpotently on the ideal $B$ by both the left and right multiplications. Hence by Engle's theorem there is a  nonzero element $b \in B$ such that
$ab = 0 =ba$ for all $a \in \A$. Hence $b \in B\cap Z(\A)$ and $B\cap Z(\A) \neq \{0\}$.
\end{proof}
\begin{thm}\label{center}
Let $\A$ be a finite dimensional Leibniz algebra. If the centralizer $Z_{\A}(\A^{\omega})$ is not contained in $\A^{\omega}$, then the center
$Z(\A) \neq \{0\}$.
\end{thm}
\begin{proof}
By Theorem \ref{cartan}, $\A = \A^{\omega} + H$ where $H$ is a Cartan subalgebra of $\A$. By definition,
$C = Z_{\A}(\A^{\omega})$ is an ideal of $\A$. Consider the subalgebra $\A_1 = C + H$ of $\A$. Then by Theorem \ref{cartan} $\A_1 = \A_1^{\omega} +H_1$
where $ \A_1^{\omega}$ is an ideal of $\A$ and $H_1$ is a Cartan subalgebra of $\A_1$. Also $\A_1^{\omega} = (C+H)^{\omega} \subseteq C$ since $H$ is nilpotent.
\smallskip

If $C \subseteq \A_1^{\omega}$ then  $C \subseteq \A^{\omega}$ which is not the case by assumption. Hence $C \subsetneq \A_1^{\omega}$. Choose
$c \in C \setminus \A_1^{\omega}$. Since $C \subset \A_1 =  \A_1^{\omega} +H_1$, $c = a + h$ for some $a \in \A_1^{\omega}, h \in H_1$. If $h = 0$ then
$c = a \in \A_1^{\omega}$ which is a contradiction. Hence $0 \not= h = a -c \in C$ implying that $C \cap H_1 \not= \{0\}$.
\smallskip

Since $C$ is an ideal of $\A$, $C\cap H_1$ is an ideal of $H_1$. The center $Z_1 = Z(H_1) \neq \{0\}$ since $H_1$ is nilpotent. So by Lemma \ref{ideal}, we have $(C\cap H_1)\cap Z_1 \neq \{0\}$. Since $(C\cap H_1)\cap Z_1 \subseteq C\cap H_1 = Z_{\A}(\A^{\omega}) \cap Z(H_1)$, we have
$Z_{\A}(\A^{\omega}) \cap Z(H_1) \neq \{0\}$. Furthermore, $\A = \A^{\omega} + H \subseteq \A^{\omega} +\A_1 = \A^{\omega} +\A_1^{\omega} +H_1 = \A^{\omega} +H_1$. Hence $Z_{\A}(\A^{\omega}) \cap Z(H_1) \subset Z(\A)$ and $Z(\A)  \neq \{0\}$.
\end{proof}
\begin{thm}\label{sub-omega}
Let $B$ be a subinvariant subalgebra of $\A$. If $Z_{\A}(B) = \{0\}$, then $Z_{\A}(B^{\omega}) \subseteq B^{\omega}$.
\end{thm}
\begin{proof}
Since $Z_{\A}(B) = \{0\}$ implies $Z(B) = \{0\}$, by Theorem \ref{center} $Z_B(B^{\omega}) \subseteq B^{\omega}$. Suppose $Z_{\A}(B^{\omega}) \subsetneq B^{\omega}$. Then $Z_{\A}(B^{\omega}) \subsetneq B$ since $Z_{\A}(B^{\omega}) \subseteq B$ implies $Z_{\A}(B^{\omega}) = Z_{\A}(B^{\omega})\cap B = Z_B(B^{\omega})\subseteq B^{\omega}$. Since $B^{\omega}$ is an ideal of $\A$, and $Z_{\A}(B)$ is an ideal of
$\A$ we have $K = Z_{\A}(B^{\omega}) +B$  a subalgebra of $\A$ properly containing $B$. Since $B$ is a subinvariant subalgebra of $\A$, $B$ is a subinvariant subalgebra of $K$. Therefore, $B$ is an ideal of some subinvariant subalgebra $B_1$ of $K$ and $B^{\omega} \subseteq B_1^{\omega}$. Since  $Z_{\A}(B^{\omega}) \subsetneq B^{\omega}$, there exists $z \in  B_1\cap Z_{\A}(B^{\omega})$, and $z \not\in B$. There are four possibilities: Case $(1)$: $z^2 = 0$; Case $(2)$: $z^2 \not\in B$; Case $(3)$: $z^2 \in B\setminus B^{\omega}$; or  Case $(4)$: $z^2 \in B^{\omega}$.

Case $(1)$: If $z^2 = 0$, then  choose $C = {\rm span}\{z, B\}$ which is a subalgebra properly containing $B$ since $z \in B_1$ and  $B \vartriangleleft B_1$.
Clearly $B^{\omega} \subseteq  C^{\omega}$. Since $z^2 = 0$, $z \in B_1$, and $B \vartriangleleft B_1$, we have $(z + B)(z + B) \subseteq B^2 + (B \cap Z_{\A}(B^{\omega})) = B^2 + Z_{B}(B^{\omega}) \subseteq B^2 + B^{\omega} \subseteq B^2$ which implies that $C^2 = B^2$. Now  $(z + B)(z + B)(z + B) \subseteq (z + B) B^2 \subseteq B^3$ since as shown above $zB \subseteq B^2$ and  by Leibniz identity $zB^2 \subseteq  (zB)B + B(zB) \subseteq B^2B + BB^2 \subseteq B^3$. This implies that $C^3 = B^3$. Since $C^{\omega} = C^t$ for some $t \in \mathbb{Z}_{\geq 2}$, repeating this process we get $C^{\omega} = C^t = B^t = B^{\omega}$. Since $z \in C \cap Z_{\A}(B^{\omega}) = Z_C(C^{\omega})$ and $z \not\in C^{\omega} = B^{\omega} \subset B$, by Theorem \ref{center} we have $Z(C) \neq \{0\}$. Since $Z_{\A}(B) \supseteq Z_C(B) \supseteq Z_C(C) = Z(C)$, we get
$Z_{\A}(B) \neq \{0\}$ which is a contradiction.

Case $(2)$: If $z^2 \not\in B$, then $z^4 = (z^2)^2 = 0$ since $z^2 \in {\rm Leib}(\A)$ which is an abelian ideal of $\A$. Since $z^2 \in B_1$ and $B \vartriangleleft B_1$, we have $C = {\rm span}\{z^2, B\}$ a subalgebra. Clearly $B^{\omega} \subseteq  C^{\omega}$. Then as in Case $(1)$, we get $B^{\omega} = C^{\omega}$ which leads to a contradiction.

Case $(3)$: In this case $z^2 \in B$, but $z^2 \not\in B^{\omega}$. Since $z \in Z_{\A}(B^{\omega})$ which is an ideal in $\A$, $z^2 \in B \cap  Z_{\A}(B^{\omega}) = Z_B(B^{\omega}) \subseteq B^{\omega}$ which is a contradiction.

Case $(4)$: Suppose $z^2 \in B^{\omega}$. As in case $(1)$ consider the subalgebra $C = {\rm span}\{z, B\}$ properly containing $B$. Since $B^{\omega} \vartriangleleft B$, $(z + B)(z + B) \subseteq B^2 + B^{\omega} \subseteq B^2$. This implies $C^2 = B^2$. Repeating this process as in case $(1)$, we get $C^{\omega} = B^{\omega}$ which leads to a contradiction.

Hence we have $Z_{\A}(B^{\omega}) \subseteq B^{\omega}$ proving the theorem.
\end{proof}
\bigskip
\section{Characteristic zero case}
In this section all Leibniz algebras are over the field $F$ of characteristic zero. We extend some classic subinvariance results in Lie algebras to Leibniz algebras and show that most of them do not hold over field of finite characteristic. Let $\mathfrak{A}$ be an associative algebra over field $F$. An element $a \in \mathfrak{A}$ is nilpotent if $a^k = 0$ for some positive integer $k$. A subalgebra $\mathfrak{B}$ of the associative Lie algebra $\mathfrak{A}$ is nilpotent if there exist a positive integer $k$ such that every product of $k$ elements in $\mathfrak{B}$ is zero. The radical of $\mathfrak{A}$  denoted by $rad(\mathfrak{A})$ is the maximal nilpotent ideal of $\mathfrak{A}$. For a Lie algebra $L$ of linear transformations  on a finite dimensional vector space over field $F$, we denote $L^*$ to be the associative envelope of $L$. We recall the following interesting result from \cite[page 45]{Jacobson}.

\begin{thm}\cite[Corollary 2.5.2]{Jacobson}\label{associative}
Let $L$ be a Lie algebra of linear transformations on a finite dimensional vector space over field $F$ of characteristic zero. Then $L\cap rad(L^*)$ is the set of all nilpotent elements of $rad(L)$ and  $[rad(L), L] \subseteq rad(L^*)$.
\end{thm}

This leads to the following interesting result for Leibniz algebras over field of characteristic zero.

\begin{thm}\label{radical}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero. Then $\A R + R\A \subseteq N$ where $R = rad(\A)$ and $N = nilrad(\A)$.
\end{thm}
\begin{proof}
Consider the homomorphism $\phi : \A \longrightarrow L(\A)$ defined by $\phi (x) = L_x$ for all $x \in \A$. Then $\phi (R) \subseteq rad(L(\A))$. So by Theorem \ref{associative} we have $[\phi (R) , L(\A)] \subseteq [rad(L(\A) , L(\A)] \subseteq rad((L(\A))^*)$. Hence each element in
$[\phi (R) , L(\A)]$ is nilpotent in the associative envelope $L(\A)^*$. This implies that there exists positive integer $n$ such that for all $n$-tuple of elements $\{(y_i, x_i) \mid 1 \leq i \leq n\}$ in $R\times \A$ or $\A \times R$, we have $[L_{y_n}, L_{x_n}] \cdots  [L_{y_1}, L_{x_1}] = 0$  which implies $(y_nx_n) \cdots (y_1x_1)x = 0$ for all $x \in \A$. Therefore, we have $(\A R + R\A)^{n+1} = \{0\}$ which implies $\A R + R\A \subseteq N$.

\end{proof}

\begin{cor}\label{nilpotent}  Let $\A$ be a solvable Leibniz algebra over field $F$ of characteristic zero. Then $\A^2$ is nilpotent.
\end{cor}
\begin{proof} Since $\A$ is solvable, $R = rad(\A) = \A$. Hence by Theorem \ref{radical}, $\A^2 \subseteq nilrad(\A)$. Hence $\A^2$ is nilpotent.

\end{proof}

\begin{cor} Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero. Then $\A^2 \cap R = \A R + R\A$ where $R = rad(\A)$.
\end{cor}

\begin{proof}
By Levi's Theorem \cite[Theorem 1]{B1}, $\A = R + S$ where $S$ is a semisimple subalgebra of $\A$ and $S\cap R = \{0\}$. Then $\A^2 = (R + S)(R + S) =\A R + R\A + S^2$. Hence $\A^2 \cap R = \A R + R\A$ since $S^2 \cap R \subseteq S\cap R = \{0\}$.
\end{proof}

\begin{cor}\label{nilrad}  Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero. Then $N = nilrad(R)$ where $R = rad(\A)$, and $N = nilrad(\A)$.
\end{cor}

\begin{proof} Since $N$ is a nilpotent ideal of $R$, $N \subseteq nilrad(R)$. By Theorem \ref{radical} $\A (nilrad(R)) + (nilrad(R))\A \subseteq \A R + R\A \subseteq N$ implies that $nilrad(R)$ is a nilpotent ideal of $\A$. Hence $nilrad(R) \subseteq N$ giving $N = nilrad(R)$.

\end{proof}

\begin{thm}\label{solv-sub}
 Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $S$ be a solvable subinvariant subalgebra of $\A$. Then $S \subseteq R$ where $R = rad(\A)$.

\end{thm}
\begin{proof}
By Levi's Theorem \cite[Theorem 1]{B1}, $\A/R$ is semisimple. Since ${\rm Leib}(\A) \subseteq R$, $\A/R$ is a Lie algebra and since $S$ is a subinvariant subalgebra of $\A$, $(S+R)/R$ is a subinvariant subalgebra of $\A/R$. Hence by \cite[Lemma 4.2]{SE}, $(S+R)/R \subseteq rad(\A/R) = \{0\}$ which implies $S \subseteq R$.
\end{proof}

The following example is a particular case of the example due to Jacobson \cite[page 75]{Jacobson} and Seligman \cite[page 163]{S}, (also see \cite[Example 2.4]{Kristen}) for Lie algebras, hence Leibniz algebras over field $F$ of characteristic $p>2$ and shows that the above theorem does not hold over field of finite characteristic.

\begin{ex}\label{char-p}
Let $F$ be a field of characteristic $p>2$. Let $R$ be the commutative associative algebra over $F$ with basis $\{1, a, a^2, \cdots , a^{p-1}\}$ where $a^p = 0$. Then dim$(R) = p$. The maximal nilpotent ideal $N$ of $R$ has basis $\{a, a^2, \cdots , a^{p-1}\}$ and dim$(R/N) = 1$. Consider the simple Lie algebra $L = s\ell(2, F)$. Then $M = L\otimes R$ is a Lie algebra with $[x\otimes a^s, y\otimes a^t] = [x,y]\otimes a^{s+t}$ for all $x,y \in L$ and $0 \leq s, t \leq p-1$. Since $(L\otimes N)^p = L^p\otimes N^p = L\otimes 0 = 0$, $L\otimes N$ is a nilpotent (hence solvable) ideal of $M$. Since $M/(L\otimes N) = (L\otimes R)/(L\otimes N)$ is isomorphic to $L$, $L\otimes N$ is both the radical and nilradical of $M$.  Consider the derivation $\delta = 1\otimes \frac{d}{da}$ in $Der(M)$ defined by $\delta (x\otimes a^t) =t(x\otimes a^{t-1})$ for all $x\in L$ and $1\leq t\leq p-1$.
Consider the semidirect product Lie algebra (hence Leibniz algebra) $P = M \oplus {\rm span}\{\delta\}$ with the multiplication $[x+\delta, y+\delta] = [x, y] + \delta (y) - \delta (x)$ for all $x, y \in M$. The only nontrivial ideal of $P$ is $M$ which is not solvable. Hence
$rad(P) (= nilrad(P)) = \{0\}$. Since $L\otimes N \vartriangleleft M \vartriangleleft P$, $L\otimes N$ is a nilpotent, hence solvable subinvariant subalgebra of $P$. However, $L\otimes N \subsetneq rad(P) = \{0\}$. This shows that Theorem \ref{solv-sub} as well as \cite[Lemma 4.2]{SE} does not hold over field of finite characteristic.
\end{ex}

\begin{thm}\label{ideal-nil}
 Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero, $B \vartriangleleft \A$ , $R = rad(B)$, and $N = nilrad(B)$.
 Then $R$ and $N$ are ideals of $\A$ and $\A R + R\A \subseteq N$.
\end{thm}

\begin{proof}
If $B = \A$, then result holds by Theorem \ref{ideal}. So assume $B \subsetneq \A$ and choose an element $x \in \A \setminus B$. Let $C = <x>$ be the cyclic Leibniz algebra generated by $x$. Consider the Leibniz algebra $\bar{B} = B + C$ and denote $\bar{R} = rad (\bar{B})$. Since $B$ is an ideal of $\A$, it is also an ideal of $\bar{B}$. Hence $R$ is subinvariant in $\bar{B}$ and $R \subseteq \bar{R}$ by Theorem \ref{solv-sub}. This implies $R \subseteq \bar{R} \cap B$. Since  $\bar{R}$ and $B$ are ideals of $\bar{B}$, we get $ \bar{R} \cap B \subseteq R$. Hence $R = \bar{R} \cap B$ is an ideal of $\bar{B}$ implying that $xR + Rx \subseteq R$ for all $x \in \A \setminus B$. Since $ R \vartriangleleft B$ we get $R$ is an ideal of $\A$.

Since $R$ and $C$ are solvable,  the subalgebra $R_1 = R + C$ is solvable and $R_1^2$ is nilpotent by Corollary \ref{nilpotent}. Furthermore, $R_1R + RR_1 \subseteq R_1^2 \cap R$ and $R_1^2 \cap R$ is a nilpotent ideal of $R$. Hence $R_1R + RR_1 \subseteq nilrad(R)$. Since $nilrad(R) = nilrad(B)$ by Corollary \ref{nilrad} we have $xR + Rx \subseteq nilrad(R)$ for all $x \in \A$. Therefore, $N = nilrad(B) = nilrad(R)$ is an ideal of $\A$ and $\A R + R\A \subseteq N$.
\end{proof}

\begin{thm}\label{nil-sub}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $M$ be a nilpotent subinvariant subalgebra of $\A$. Then $M \subseteq nilrad(\A)$.
\end{thm}

\begin{proof}
Since $M$ is a subinvariant subalgebra of $\A$, we have a chain of subalgebras $M=M_t \subset M_{t-1} \subset \cdots \subset M_1 \subset M_0 = \A$ such that $M_i \vartriangleleft M_{i-1}, 1 \leq i \leq t$. Denote $N_i = nilrad(M_{i-1}), 1 \leq i \leq t$ which is a nilpotent ideal of $M_{i-1}$. Since $M = M_t$ is a nilpotent ideal of $M_{t-1}$, $M= M_t \subseteq N_t$. By Theorem \ref{ideal-nil}, $N_t$ is an ideal of $M_{t-2}$. Hence $N_t \subseteq N_{t-1}$ and by Theorem \ref{ideal-nil}, $N_{t-1}$ is an ideal of $M_{t-3}$. Continuing this process we get $M \subseteq N_t \subseteq N_{t-1} \subseteq \cdots \subseteq N_2 \subseteq N_1 = nilrad(M_0) = nilrad(\A)$.
\end{proof}

\begin{rmk} As seen in Example \ref{char-p}, $L\otimes N$ is a nilpotent subinvariant subalgebra of $P$, but $L\otimes N \subsetneq nilrad(P) = \{0\}$. Thus Theorem \ref{nil-sub} does not hold over field of finite characteristic.
\end{rmk}

\begin{thm}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $B$ be a subinvariant subalgebra of $\A$. Then $rad(B) = B\cap rad(\A)$ and $nilrad(B) = B\cap nilrad(\A)$.
\end{thm}

\begin{proof}
Since $B \cap rad(\A)$ (resp. $B \cap nilrad(\A)$) is a solvable (resp. nilpotent) ideal of $B$, we have $B\cap rad(\A) \subseteq rad(B)$ (resp. $B\cap nilrad(\A) \subseteq nilrad(B)$). Also $rad(B)$ (resp. $nilrad(B)$) is a solvable (resp. nilpotent) subinvariant subalgebra of $\A$ implies by Theorem \ref{solv-sub} (resp. Theorem \ref{nil-sub}) that $rad(B)\subseteq B\cap rad(\A)$ (resp. $nilrad(B) \subseteq B\cap nilrad(\A)$). Hence
$rad(B) = B\cap rad(\A)$ (resp. $nilrad(B) = B\cap nilrad(\A)$).
\end{proof}

\begin{lemma}\label{nil-subin}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $B, C$ be two nilpotent subinvariant subalgebras of $\A$. Then the subalgebra $D$ generated by $B$ and $C$ is a nilpotent subinvariant subalgebra of $\A$.
\end{lemma}

\begin{proof} Since $B$ and $C$ are nilpotent, by Theorem \ref{nil-sub}, they are contained in $nilrad(\A)$, hence $D \subseteq nilrad(\A)$. So by \cite[Theorem 4.16]{DMS}, $D$ is a subinvariant subalgebra of $nilrad(\A)$. Therefore, $D$ is a nilpotent subinvariant subalgebra of $\A$.
\end{proof}

\begin{thm}\label{subin}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $B, C$ be two subinvariant subalgebras of $\A$. Then the subalgebra $D$ generated by $B$ and $C$ is a subinvariant subalgebra of $\A$.
\end{thm}

\begin{proof}
If both $B$ and $C$ are nilpotent then the result follows by Lemma \ref{nil-subin}.
Suppose at least one of the subalgebras $B,  C$ is not nilpotent. Then $B^{\omega} \neq \{0\}$ and/or  $C^{\omega} \neq \{0\}$. Since $B$ and $C$ are subinvariant in $\A$, by Theorem \ref{omega} $B^{\omega}$ and $C^{\omega}$ are ideals of $\A$. Hence the subalgebra $K$ generated by
$\{B^{\omega}, C^{\omega}\}$ is an ideal of $\A$ and quotient algebras $(B+K)/K$ and $(C+K)/K$ are nilpotent in $\A/K$. Hence by Lemma \ref{nil-subin}, $(D+K)/K$ is subinvariant in $\A/K$. Since $D$ contains $K$, $D$ is subinvariant in $\A$.

\end{proof}

\begin{thm}\label{sub-sub}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $B$ be a subinvariant subalgebra of $\A$. Let $c \in \A$ and $\bar{B}$ be the smallest subalgebra containing $B$ such that $f(c, \bar{B}) \subseteq \bar{B}$. Then $\bar{B}$ is a subinvariant subalgebra of $\A$.
\end{thm}

\begin{proof}
If $c \in B$ then $\bar{B} = B$ and there is nothing to prove. So assume that $c \nin B$. Recall that $f^0(c, B) = B, f^{i+1}(c, B) = f(c, f^i(c, B))$ for all $i \in \mathbb{Z}_{i \geq 0}$. Consider the subalgebra $C = \sum_{i \geq 0}f^i(c, B)$. By definition  $B = f^0(c, B)  \subseteq C$. Since $cf^i(c, B) + f^i(c, B)c = f(c, f^i(c, B)) = f^{i+1}(c, B)$, we have $c C + Cc \subseteq C$ which implies $\bar{B} \subseteq C$. Also by definition $B \subseteq \bar{B}$ and $f(c, B) \subseteq f(c, \bar{B}) \subseteq \bar{B}$. Assume $f^i(c, B) \subseteq \bar{B}$. Then $f^{i+1}(c, B) = f(c, f^i(c, B)) \subseteq f(c, \bar{B}) \subseteq \bar{B}$. So by induction $f^i(c, B) \subseteq \bar{B}$ for all $i \geq 0$ which implies $C = \bar{B}$.

If $B$ is nilpotent, then $B \subseteq nilrad(\A)$. Since $nilrad(\A)$ is an ideal of $\A$, we have $f(c, B) = cB +Bc \subseteq nilrad(\A)$. Suppose $f^i(c, B) \subseteq nilrad(\A)$. Then $f^{i+1}(c, B) = f(c, f^i(c, B)) = cf^i(c, B) + f^i(c, B)c \subseteq nilrad(\A)$. So by induction $\bar{B} = \sum_{i \geq 0}f^i(c, B) \subseteq nilrad(\A)$. Hence by \cite[Theorem 4.16]{DMS}, $\bar{B}$ is a subinvariant subalgebra in $nilrad(\A)$, hence in $\A$.

Suppose $B$ is not nilpotent.Then by Theorem \ref{omega}, $B^{\omega} \neq \{0\}$ is an ideal of $\A$. Note that $B_1 = B/B^{\omega}$ is a nilpotent subalgebra of $\A_1 = \A/B^{\omega}$. Define the subalgebra $\bar{B}_1 = \bar{B}/B^{\omega}$. Then as before $\bar{B}_1 = \sum_{i \geq 0}((f^i(c, B) + B^{\omega})/B^{\omega} = \sum_{i \geq 0} f^i(c_1, B_1) \subseteq \A_1$ where $c_1 = c+B^{\omega} \in \A_1$. Hence $\bar{B}_1$ is a subinvariant subalgebra of $\A_1$ which implies $\bar{B}$ is a subinvariant subalgebra of $\A$ since $\bar{B} \supseteq B^{\omega}$.
\end{proof}

\begin{cor}\label{minimal}
Let $\A$ be a Leibniz algebra over field $F$ of characteristic zero and $B$ be a subalgebra. Let $\underline{B}$ be the subalgebra of $B$ generated by all subinvariant subalgebras of $\A$ contained in $B$. Then $\underline{B}$ is an ideal of $B$.
\end{cor}

\begin{proof} By definition $\underline{B}$ is a subinvariant subalgebra of $\A$ and $\underline{B} \subseteq B$. Let $c \in B$ and $B_1$ be the minimal subalgebra of $\A$ such that $\underline{B} \subseteq B_1$ and $f(c, B_1) \subseteq B_1$. Then by Theorem \ref{sub-sub}, $B_1$ is a subinvariant subalgebra of $\A$. By definition $B_1 \subseteq B$, hence $B_1 \subseteq \underline{B}$ which implies $B_1 = \underline{B}$. Therefore, for all $c \in B$, we have $c \underline{B} + \underline{B} c = f(c, \underline{B}) = f(c, B_1) \subseteq B_1 = \underline{B}$. Hence
$\underline{B}$ is an ideal of $B$.
\end{proof}

The following example shows that in Corollary \ref{minimal}, the subalgebra $\underline{B}$ is not necessarily an ideal of $\A$.

\begin{ex} Let $\A = {\rm span}\{x, y, z, t\}$ be  the Leibniz algebra over field $\mathbb{C}$ with nonzero multiplications $tx = x, ty = y, zy = x$. Then $B = {\rm span}\{z, t\}$ is an abelian self-normalizing subalgebra (hence a Cartan subalgebra) of $\A$. The subalgebra ${\rm span}\{z\}$ is a subinvariant subalgebra $\A$ since we have a chain of ideals ${\rm span}\{z\} \vartriangleleft {\rm span}\{z, x\} \vartriangleleft {\rm span}\{z, x, y\} \vartriangleleft \A$. Since $B$ is self-normalizing, $\underline{B} \neq B$. Hence $\underline{B} = {\rm span}\{z\}$ which is clearly an ideal of $B$, but not an ideal of $\A$.

\end{ex}
\section{Remark on Tower Theorem}
In the fundamental work on subinvariance by Wielandt \cite{W} for groups and by Schenkman \cite{SE} for Lie algebras, there is an interesting result known as tower theorem. Let us recall the tower theorem for Lie algebras. Let $\g$ be a finite dimensional Lie algebra over field $F$ of characteristic zero. Assume that $\g$ has center zero. Then $\g$ is isomorphic to the ideal $\{ad_x \mid x \in \g\}$ in $Der(\g)$. Thus there is a natural embedding of $\g \subset Der(\g)$. Define $\g_0 = \g, \g_1 = Der(\g)$ and $\g_i = Der(\g_{i-1})$ for all $i \in \mathbb{Z}_{\geq 1}$. It is shown in \cite[Theorem 17]{SE} that there is a natural embedding of $\g_{i-1}$ in $\g_i$ and the tower of Lie algebras $\{\g = \g_0 \subset \g_1 \subset \g_2 \subset \cdots \}$ terminates in a limit Lie algebra $\g_{\infty}$ which is a complete Lie algebra and ${\rm dim}(\g_{\infty}) \leq (d+c)$, where $d = {\rm dim}(Der(\g^{\omega}))$ and $c$ is the dimension of the center of $\g^{\omega}$. Later Carles \cite{C} gave a description of each member $\g_i$ in the tower including the limit $\g_{\infty}$ and extended the tower theorem for Lie algebras with nontrivial center.

 Now let us consider the case of a Leibniz algebra $\A$ over field $F$ of characteristic zero. Assume that the left center $Z^{\ell}(\A) = \{0\}$. Then
 $\A$ is isomorphic to $L(\A) = \{L_x \mid x \in \A\} \subset Der(\A)$. Thus we have a natural embedding of $\A$ in $Der(\A)$. Now define $\A_0 = \A, \A_1 = Der(\A)$ and $\A_i = Der(\A_{i-1}$ for all $i \in \mathbb{Z}_{\geq 1}$. Since $Leib(\A) \subseteq Z^{\ell}(\A)$, we have $Leib(\A) = \{0\}$ which implies that $\A$ is indeed a Lie algebra and the tower theorem for Leibniz algebras follows by \cite[Theorem 17]{SE} for Lie algebra. Observe that by definition of complete Leibniz algebra in \cite{Kristen}, since the limit Leibniz algebra $\A_{\infty}$ is complete as a Lie algebra, it will be also complete as a Leibniz algebra. It would be an interesting problem to investigate whether there is any Leibniz algebra analog of Carles' \cite{C} Lie algebra results when $Z^{\ell}(\A) \neq \{0\}$ and $Leib(\A)\neq \{0\}$.


TITLE: Rewriting a sum of harmonic powers as a minimal polynomial
QUESTION [3 upvotes]: Revisiting one of my older questions, I've decided to try to tackle a simpler version of the problem, this time without the square root coefficients.

Let $x_0$ be a real number such that it satisfies the equation $$x_0+x_0^{1/2}+x_0^{1/3}+\cdots+x_0^{1/n}=1$$ for a natural number $n$. What is the minimal polynomial in $\mathbb{Z}[x]$?

Of course, this is possible by brute force: isolating the smallest power of $x$ then raising both sides by its reciprocal and repeating, but it becomes extremely tedious to do when $n$ is large. Also, this does not guarantee that the polynomial obtained is minimal.
This works fine for $n=1,2,3$. The minimal polynomials are, respectively, $$x-1,\quad x^2-3x+1,\quad x^5-8x^4+24x^3-21x^2+10x-1$$ and it may be interesting to note that the sign of the coefficients are alternating.
Is there an efficient way of doing this for the general case?

REPLY [1 votes]: Let $x=y^L$, where $L$ is the least common multiple of $2,3,\dots,n$. Then you have the two equations, $y^L-x=0,y^L+y^{L/2}+y^{L/3}+\cdots+y^{L/n}-1=0$. The resultant of these two polynomials will be a polynomial satisfied by $x$. I suspect, but am not sure I could prove, that it's the minimal polynomial.
The resultant can be computed as a determinant. Unfortunately, it's the determinant of a $2L\times2L$ matrix, but at least it's a very sparse matrix. I have a hunch the problem is inherently tedious and that there's no efficient way to do it.


text:
\vspace{-5pt}
\section{Conclusion}
\vspace{-5pt}
In this paper, we proposed a residual one-point feedback oracle for zeroth-order optimization, which estimates the gradient of the objective function using a single query of the function value at each iteration. When the function evaluation is noiseless, we showed that ZO using the proposed oracle can achieve the same iteration complexity as ZO using two-point oracles when the function is non-smooth. When the function is smooth, this complexity of ZO can be further improved. This is the first time that a one-point zeroth-order oracle is shown to match the performance of two-point oracles in ZO. In addition, we considered a more realistic scenario where the function evaluation is corrupted by noise. We showed that the convergence rate of ZO using the proposed oracle matches the best known results using one-point feedback or two-point feedback with uncontrollable data samples. We provided numerical experiments that showed that the proposed oracle outperforms the one-point oracle and is as effective as two-point feedback methods.

config: arxiv


TITLE: Algebra simplification of transfer function, help please!
QUESTION [0 upvotes]: I ran into problems trying to simplify this:
Given
$$ H(s)=\frac{0.9661s^4}{s^4+8.824S^3+44.86s^{2}+105.6s+254.2} $$
now subsitute $s=(\frac{1-z^{-1}}{1+z^{-1}})$ into $H(s)$ to get $G(z)$:
$$ G(z) = \frac{0.9661(\frac{1-z^{-1}}{1+z^{-1}})}{(\frac{1-z^{-1}}{1+z^{-1}})^{4}+8.824(\frac{1-z^{-1}}{1+z^{-1}})^{3}+44.86(\frac{1-z^{-1}}{1+z^{-1}})^{2}+105.6(\frac{1-z^{-1}}{1+z^{-1}})+254.2}$$
It's for a signal theory problem where a transformation is performed to get the digital transfer function from the s-domain, the answer according to my guide is:
$$G(z) = \frac{0.003-0.012z^{-1}+0.018z^{-2}-0.012z^{-3}+0.003z^{-4}}{1+2.808z^{-1}+3.294z^{-2}+1.1857z^{-3}+0.421z^{-4}}$$
But for the life of me I can't seem to get the same values, I also tried coding a simple MatLab program using the expand function but it doesn't return the answer in the form I need it to be.
Help would be SO greatly appreciated, especially if someone can advise what to do in Matlab!

REPLY [1 votes]: So, starting with the function
$$ H(s)=\frac{0.9661s^4}{s^4+8.824S^3+44.86s^2+105.6s+254.2} $$
we want to substitute $s=\frac{1-z^{-1}}{1+z^{-1}}=\frac{z-1}{z+1}$.
Now (finger crossed)
\begin{align*}H(z)&=\frac{0.9661\left(\frac{z-1}{z+1}\right)^4}{\left(\frac{z-1}{z+1}\right)^4+8.824\left(\frac{z-1}{z+1}\right)^3+44.86\left(\frac{z-1}{z+1}\right)^2+105.6\left(\frac{z-1}{z+1}\right)+254.2}\\
&=\frac{0.9661(z-1)^4}{(z-1)^4+8.824(z-1)^3(z+1)+44.86(z-1)^2(z+1)^2+105.6(z-1)(z+1)^3+254.2(z+1)^4}\\
&= \frac{0.00233085 z^4 - 0.0093234 z^3 + 0.0139851 z^2 - 0.0093234 z + 0.00233085}{z^4 + 2.91049 z^3 + 3.47777 z^2 + 1.97655 z + 0.447873}\\
&= \frac{0.00233085  - 0.0093234 z^{-1} + 0.0139851 z^{-2} - 0.0093234 z^{-3} + 0.00233085z^{-4}}{1 + 2.91049 z^{-1} + 3.47777 z^{-2} + 1.97655 z^{-3} + 0.447873z^{-4}}
\end{align*}
where I've used Wolframalpha in the third step.
(I honestly don't know what this exercise is about, but it makes no sense to me computing this by hand)


TITLE: Evaluate $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$
QUESTION [8 upvotes]: Let $A_0=\dfrac{3}{4}$, and $A_{n+1}=\dfrac{1+\sqrt{A_n}}{2}$ for all $n\geq0$.
How to find the value of  $\displaystyle\prod_{n=1}^\infty A_n$ ?
I don't have any idea.
Thank you.

REPLY [21 votes]: If we set $A_n = \cos^2(\theta_n)$, we get:
$$ A_{n+1} = \frac{1+\cos(\theta_n)}{2} = \cos^2\left(\frac{\theta_n}{2}\right),$$
and since $\theta_0=\frac{\pi}{6}$, induction gives:
$$ A_n = \cos^2\left(\frac{\pi}{6\cdot 2^n}\right)=\left(\frac{\sin\left(\frac{\pi}{6\cdot 2^{n-1}}\right)}{2\sin\left(\frac{\pi}{6\cdot 2^n}\right)}\right)^2 $$
from which it follows that:
$$ \prod_{n=1}^{+\infty}A_n = \left(\frac{3}{\pi}\right)^2 = \color{red}{\frac{9}{\pi^2}}.$$


TITLE: Can three stacks be implemented in one array, with O(1) push/pop time?
QUESTION [9 upvotes]: Two stacks can be efficiently implemented using one fixed sized array: stack #1 starts from the left end and grows to the right, and stack #2 starts from the right end and grows to the left. Is the same possible for three stacks?
More specifically, is it possible to implement three stacks given the following conditions:

You have a fixed size array that can hold N objects.
As long as the sum of the three stack sizes is < N, push() should not fail.
Both push() and pop() operations should take O(1) time.
In addition to the array, you can use only O(1) additional space.

Here are examples of solutions that do not satisfy these requirements:

Splitting the array into 3 fixed parts and using each part for a stack (violates 2).
Similar to the above but with movable boundaries between stacks (violates 3).
Simple linked-list based implementations (violates 4).

I'll accept non-trivial algorithms or impossibility proofs even if they don't meet all the conditions (1)-(4) exactly, for example, an algorithm where push/pop take O(1) amortized time, or where the additional memory is smaller than O(N), eg O(log N). Or an impossibility proof that shows that for example, accessing fewer than 5 elements of the array per push/pop is impossible.

REPLY [6 votes]: Fredman and Goldsmith showed in "Three Stacks" (Journal of Algorithms, 1994) that $\Theta(n^\varepsilon)$ bits of wasted space is achievable. It is also the minimum needed for arrays of size at least 16 quattuordecillion yottabytes. I described a simple algorithm wasting $\Theta(\sqrt{n})$ words of space in my StackOverflow answer to this question. As @dmitri-urbanowicz mentioned in the comments, this is basically just treating the array as $\sqrt{n}$ blocks of size $\sqrt{n}$, where each block is used for exactly one stack and contains a single pointer to the next block in that stack.


TITLE: For which values of $x$ are the terms of this sequence always positive?
QUESTION [0 upvotes]: Let the sequence $a_n$ be defined as such:
$$\begin{cases}a_0=1\\ a_1=x\\a_n=xa_{n-1}-a_{n-2},\ n \in \mathbb{N}\setminus\{1,2\} \end{cases}$$
I am looking for the $x$ for which the sequence is always positive.
For $x\ge 2$, the sequence is always positive, since it is the sequence of natural numbers for $x=2$.Then, a sequence for $x>2$ is always greater or equal to the sequence for which $x=2$.  I tried finding the limit $\lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}$, and since $x>1$ ($a_2=x^2-1$), this limit only exists for $x\ge 2$, as $L=\frac{x\pm \sqrt(x^2-4)}{2}$, so for $2>x>1$ the sequence can't converge. (Since this limit is always 1 for positive sequences that converge.)
Also, if for $2>x>1$ $\ \ \ a_n\le a_{n-1}\Leftrightarrow a_{n+1}\le(x-1)a_{n-1}\Rightarrow a_{n+1}<a_{n-1}$  for positive $a_n$. Meaning that the sequence has to be stricly increasing or at some point it will be negative, because it can't converge since the limit above (of the ratio of $a_{n+1}$ and $a_n$) doesn't exist. But now I'm stuck. How can I show that for all $1<x<2$ the sequence isn't strictly increasing? Or can you find some value for which it is? (I tested experimentally and it seems that for all values under 2 the sequence decreases at some point, but one can never be sure.)
P.S. I found a way to write the general expression of the sequence, but it didn't help me much.
$a_n = x^n - {n-1 \choose 1}x^{n-2}+{n-2 \choose 2}x^{n-4}-{n-3 \choose 3}x^{n-6}+\cdots  $

REPLY [1 votes]: The characteristic equation has two roots: $r=\frac{x+\sqrt{x^2-4}}{2}, s=\frac{x-\sqrt{x^2-4}}{2}$.
Note that $a_n-(r+s)a_{n-1}+rsa_{n-2}=0$
Case 1: $x>2$, $r, s$ are real and $r>s>0$.
$$
a_{n+1} - s a_n = r(a_n-sa_{n-1})
\Rightarrow a_{n+1} - s a_n =  r^n(a_1-sa_0) =r^n (x-s) =r^{n+1}\tag1\\
$$
$$a_{n+1} - r a_n = s(a_n-ra_{n-1})
\Rightarrow a_{n+1} - r a_n = s^n(a_1-ra_0)= s^n (x-r) = s^{n+1}\tag2
$$
(1)-(2),
$$a_n=\frac{r^{n+1}-s^{n+1} }{r-s} >0  \tag 3$$
(Note that $a_n=\frac{r^{n+1}-s^{n+1} }{r-s}$ as long as $r\neq s$, even if they are complex).
Case 2: $x=2$.
$$a_n-2a_{n-1}+a_{n-2}=0 \implies a_n-a_{n-1}=a_{n-1}-a_{n-2} \\
\implies a_n-a_{n-1}=a_1-a_0=1 \implies a_n=n$$
So $a_n>0, \forall n$.
Case 3: $x<2$, $r$ and $s$ are conjugate complex numbers.
Let $r=\exp\{i\theta\}, s=\exp\{i(-\theta)\}$ (since $rs=1$ we know $|r|=|s|=1$).
Now $$a_n=\frac{r^{n+1}-s^{n+1} }{r-s}=\frac{2\sin((n+1)\theta)}{2 \sin \theta} = \frac{\sin((n+1)\theta)}{\sin \theta}$$
Note that $\theta=\tan^{-1}\frac{4-x^2}{x} \in (0,\frac{\pi}{2}) \implies \lambda :=\frac{\theta}{\pi} \in (0,\frac 12)$.
If $a_n>0, \forall n$, then $\lfloor n \lambda \rfloor$ is always even. Let $k$ be the smallest integer such that $k\lambda \ge 2$, then $(k-1)\lambda < 1, \lambda = k\lambda - (k-1)\lambda > 2-1=1$, a contradiction.
So for some $n$, $a_n \leq 0$.
Conclusion: The sequence is always positive if and only if $x\geqslant 2$.


TITLE: is linear combination unique for each vector?
QUESTION [1 upvotes]: each vector $\vec x$∈$R^n$ can be expressed as a sum
$$\vec x = \vec x_1\vec e_1+···+\vec x_n\vec e_n$$
Show that this expression is unique, that is, there does not exist other, different linear combination of vectors $$\vec e_1,\vec e_2,...,\vec e_n$$ which is equal to $\vec x$.
I've test vector $$\begin{bmatrix}16 \\32 \\\end{bmatrix} = 1\begin{bmatrix}1 \\1 \\\end{bmatrix} + 2\begin{bmatrix}8 \\16 \\\end{bmatrix}$$ $$\begin{bmatrix}16 \\32 \\\end{bmatrix} = 2\begin{bmatrix}2 \\8 \\\end{bmatrix} + 4\begin{bmatrix}3 \\4 \\\end{bmatrix}$$
and wrote it in such different ways. So I don't get what's happening here, is not this against uniqueness? Or if I am wrong then give me some hints how to prove it.

REPLY [0 votes]: This statement means that for a given basis of linearly independent vectors $\vec e_1, \dots, \vec e_n$ there is a unique decomposition of any vector $\vec x$ into a linear combination $\vec x = x_1\vec e_1 + \dots + x_n \vec e_n$. Note that $x_i$ are not vectors, but real numbers: coordinates of the vector $\vec x$ in the basis $\vec e_1, \dots, \vec e_n$.
In your example you considered two different bases, so of course the coordinates are different.


TITLE: prove that $x^4+y^4-4b^2xy \geq -2b^4$ for all $x,y \in \Bbb{R}$?
QUESTION [2 upvotes]: How would you prove that $$x^4+y^4-4b^2xy \geq -2b^4\;\;\;\;\;\;\;\; \forall x,y \in \Bbb{R}?$$
My idea was to show that the function $H(x,y)=x^4+y^4-4b^2xy+2b^4 \geq 0$ for all $x,y \in \Bbb{R}$ and to show that I am considering finding the absolute minimum of the function and show that it is equal to or greater then $0$ but im strugging to show this

REPLY [6 votes]: Since $x^4+y^4\geq 2x^2y^2$ we have:
$$x^4+y^4-4b^2xy +2b^4\geq 2x^2y^2 -4b^2xy +2b^4 =2 (xy-b^2)^2\geq 0$$


TITLE: Finding roots with mod and greatest integer
QUESTION [1 upvotes]: I have to find the roots to this equation
$$|x^2-[x]| =1$$
Where [x] is the greatest integer function and |a| is the modulus of a.
I don’t know how to go about it, hit and trial gave me nothing, help me out.

REPLY [0 votes]: Here is how I solved it:

Consider x in (k, k+1]
x = k+r
|x²-ceil(x)| = |(k+r)²-(k+1)| = 1
Case1: (k+r)²-(k+1) >= 0
|(k+r)²-(k+1)| = (k+r)²-k-1 = 1
(k²+2kr+r²)-k = 2
1r² +2kr +k²-k-2 = 0
r in (-2k +/- sqrt(4k²-4(k²-k-2))) / 2
    = -k +/- sqrt(k²-(k²-k-2))
    = -k +/- sqrt(k²-k²+k+2)
    = -k +/- sqrt(k+2)
    = {-k + sqrt(k+2), -k - sqrt(k+2)} = {r1, r2}
For real roots, k must be in {-2, -1, ...}
r must be within (0, 1]
    0 < r1:
        -k + sqrt(k+2) > 0
        sqrt(k+2) > k
        true for k <= 0, consider k > 0
        k+2 > k²
        1k² - 1k - 2 < 0
        (1 +/- sqrt(1 - 4*(-2))) / 2 = (1 +/- 3) / 2 = {-1, 2}
        k > 0 and 1k² - 1k - 2 < 0 is false for k → inf
        0 < k < 2
        k = 1
        k in {-2, -1, 0, 1}
    r1 <= 1:
        -k + sqrt(k+2) <= 1
        sqrt(k+2) <= 1+k
        false for k < -1
        k+2 <= (1+k)² = k²+2k+1
        0 <= k² + k - 1
        0 <= (-1)² + (-1) - 1 = -1, false
        0 <= (0)² + (0) - 1 = -1, false
        0 <= (1)² + (1) - 1 = 1, true
        k = 1
    r1 = -k + sqrt(k+2) = -1 + sqrt(1+2) = sqrt(3) - 1
    root candidate: 1 + r1 = 1 + sqrt(3) - 1 = sqrt(3)
    0 < r2:
        -k - sqrt(k+2) > 0
        k < -sqrt(k+2), false for k >= 0
        -1 < -sqrt(-1+2), 1 > sqrt(1), false
        -2 < -sqrt(-2+2), 2 > sqrt(0), true
        k = -2
    r2 <= 1:
        r2 = -k - sqrt(k+2) = 2 - sqrt(-2+2) = 2, false
Case2: (k+r)²-(k+1) < 0
|(k+r)²-(k+1)| = -(k+r)²+k+1 = 1
-(k²+2kr+r²)+k = 0
k²+2kr+r²-k = 0
1r² + 2kr + k²-k = 0
r in (-2k +/- sqrt(4k² - 4(k²-k))) / 2
    = -k +/- sqrt(k² - (k²-k))
    = -k +/- sqrt(k)
    = {-k - sqrt(k), -k + sqrt(k)}
For real roots, k must be in {0, 1, ...}
r must be within (0, 1]
    0 < r1 = -k - sqrt(k) is false for those k
    0 < r2:
        0 < -k + sqrt(k), k < sqrt(k), k² < k, k < 0, false
    No root candidates for Case2
The only root found is sqrt(3) when Case1 pertains, a simple test shows that
this is indeed the root:
|sqrt(3)²-ceil(sqrt(3))| = 1
|3-2| = 1
1 = 1


TITLE: Condition on integers $a,b$ such that $a^{1/b}$ is rational number
QUESTION [1 upvotes]: If $a, b\in \mathbb Z$ then what is condition on them such that $a^{1/b}\in \mathbb Q$?

I wanted to show above. My attempt:

Let $a^{1/b}=p/q$ then $a=p^b/q^b$ that is  $aq^b=p^b$...

I am not able to proceed further, any help will be appreciated.

REPLY [1 votes]: It's not enough to assume just $a^{1/b}=\frac{p}{q}$, you still need to assume (which makes sense) that $\frac{p}{q}$ is irreducible or $\gcd(p,q)=1$. But then $\gcd\left(p^b,q^b\right)=1$ (left as an exercise) and from Bezout, $\exists m,n \in\mathbb{Z}: mp^b+nq^b=1$. Combine this with $aq^b=p^b$ and the result is
$$maq^b+nq^b=1 \Rightarrow q^b(ma+n)=1 \Rightarrow q^b \mid 1$$
which is only possible when $q=1$ or $q=-1$. Let's assume $q=1$, then $a^{1/b}=\frac{p}{q}=p \Rightarrow \color{red}{a=p^b}$.

A few notes on why assuming $\frac{p}{q}$ is irreducible makes sense. If it's not irreducible, it can be made irreducible, via
$$\gcd(p,q)=d>1 \Rightarrow p=p_1\cdot d, q=q_1\cdot d, \color{red}{\gcd(p_1,q_1)=1} \Rightarrow\\
\frac{p}{q}=\frac{p_1\cdot d}{q_1\cdot d}=\frac{p_1}{q_1}$$
and we can use $\frac{p_1}{q_1}$ instead.


TITLE: Find the Cayley graph of the group $G = ( \mathbb{Z}/2 \mathbb{Z}) × ( \mathbb{Z}/2 \mathbb{Z})$ with generating set ${(1, 0),(0, 1)}$.
QUESTION [1 upvotes]: I am quite new to geometric group theory and can't really visualize the way the Cayley graph changes when you change the generating set. I found many possible graphs here but I don't really see how to construct the one I am looking for.

REPLY [0 votes]: Write out the four elements, clearly $(0,0)$ is connected to $(1,0),(0,1)$ and these are--in turn--each connected to $(0,0)$ and $(1,1)$ the first by adding themselves and the second by adding the other. And as each vertex has degree $2$ (Cayley graphs are regular graphs), that's all you need as $(1,1)$ already has both of it's connected vertices listed. So you end up with a square, the upper left corner being $(0,0)$, the two adjacent vertices $(1,0)$ and $(0,1)$, and the corner opposite $(0,0)$ is $(1,1)$.


TITLE: Van der Pol Equation: Undamped driven oscillator
QUESTION [0 upvotes]: From the van der pol equation
$$\ddot x - a(1-x^2)\dot x + x = b\cos(\omega t)$$
writing in two first order equations has the system:
$\dot x = y$,
$\dot y = a(1-x^2)y-x+b\cos(\omega t).$
In the undamped undriven case, (a=0 and b=0). I must show the general solution of equation :
$\dot y = a(1-x^2)y-x+b\cos(\omega t)$
is
$ x = \alpha \cos(t+\beta)$
and $y = -\alpha \sin(t+\beta).$
If a=0 and b=0, then $\dot y = -x$ which means $$\ddot x + x = 0.$$
Solving this differential equation gives roots at $i$ and $-i$, so general solution would be $$x(t) = A\cos(-t)+ B\sin(-t).$$
I'm not sure how from here you would get the disired solution of
$ x = \alpha \cos(t+\beta)$
and $y = -\alpha \sin(t+\beta),$ since only conditions given were a=0 and b=0 but no initial conditions of $x(t)$ were given.

REPLY [0 votes]: You can give the solution as
$$
x(t)=A\cos(t)+B\sin(t)
$$
or as
$$
x(t)=α\cos(t+β)=(α\cos(β))\cos(t)+(-α\sin(β))\sin(t).
$$
Now it should be obvious that the correspondence between the parameters is $(A,B)=(α\cos(β),-α\sin(β))$ or
$$
A-iB=αe^{iβ},
$$
that is, between the Cartesian and the polar representation of a point in the (complex) plane.

Note that for small positive $a$ and small $x$ the friction coefficient is negative, that is, the friction term is boosting the solution away from the zero equilibrium. For large $x$ the friction is positive, the solution is a cycle that spends only a short part of its period in the boosting region $|x|<1$. A small forcing term will thus give perturbations of the limit cycle, while a large forcing term will give rise to "chaotic" solutions. The quality of the "chaos" may also depend on the frequency $ω$.


TITLE: Proof of an upper bound of the product of binomial coefficients on a same row of Pascal triangle
QUESTION [3 upvotes]: We have to prove the following inequality:
$$\forall n \in \mathbb{N}, n \ge 2, \ \ \Pi_{k=0}^n \binom{n}{k} \le \left(\frac{2^n-2}{n-1}\right)^{n-1}$$
the progress where I am at, is visible in the following picture:

I'm especially confused, since the term (2*((n-1)/n)) has the limit zero when calculated to infinity and the other term diverges to infinity when calculated there.
so the first question would be, if these two terms are multiplied, which limit comes out as a result?
the second question would be simply how to continue on from there, or if you know a better way to approach this inequality proof.
I am very thankful for your helpful answers.

REPLY [4 votes]: By AM-GM inequality
$$
\left( {\prod\limits_{k = 0}^n \binom{n}{k} } \right)^{1/(n - 1)}  = \left( {\prod\limits_{k = 1}^{n - 1} \binom{n}{k} } \right)^{1/(n - 1)}  \le \frac{{\sum\nolimits_{k = 1}^{n - 1} \binom{n}{k} }}{{n - 1}} = \frac{{2^n  - 2}}{{n - 1}}
$$ provided $n\ge 2$.


TITLE: Why do the terrestrial planets have many of the same gases in their atmospheres?
QUESTION [8 upvotes]: While the gases are present in very different abundances, and the atmospheres have very different pressures, Earth, Mars and Venus all have a similar cast of gases: nitrogen, argon, CO2, others in trace amounts.
I've asked an earth scientist friend and his short answer was "volcanoes." I'm afraid I don't know enough about solid earth processes or geochemistry to figure out where to find out more...where could I start?

REPLY [2 votes]: Star outgassing and supernovae determine concentrations of light elements in the universe.  Comet accretion and planetary outgassing determine what gets added.  Temperature and pressure plus energy inputs (solar UV) drive the chemistry.  Geochemistry modulates it.  Granite is rather reactive over time.


TITLE: Derivative of $\int_a^x \frac{1}{k^x} \: dk$
QUESTION [0 upvotes]: Is it possible to define a derivative for the function: $$\int_a^x \frac{1}{k^x} \: dk$$ (where a is a constant) and if so what would it be?

REPLY [0 votes]: Using the Leibniz rule,
$$\frac{d}{dx} \, \int_{a}^{b} f(x,t) \, dt = f(x,b) \, \partial_{x}b - f(x,a) \, \partial_{x}a + \int_{a}^{b} \partial_{x}f(x,t) \, dt$$
then
\begin{align}
\frac{d}{dx} \, \int_{a}^{x} \frac{dt}{t^{x}} &= \frac{1}{x^{x}} + \int_{a}^{x} \partial_{x} e^{-x \ln t} \, dt \\
&= \frac{1}{x^x} - \int_{a}^{x} \frac{\ln t}{t^{x}} \, dt
\end{align}
An alternate method may be used, that avoids the integral, or provides a value for a new integral. This is seen by the following.
Since, for $x \neq 1$,
$$\int_{a}^{x} \frac{dt}{t^x} = \left[ \frac{t^{1-x}}{1-x} \right]_{a}^{x} = \frac{x^{1-x} - a^{1-x}}{1-x}$$
then
$$\frac{d}{dx} \, \int_{a}^{x} \frac{dt}{t^{x}} = \frac{1}{(1-x)^{2}} \, \left[ x^{1-x} \, \left(\frac{1}{x} - 1 - (1-x) \, \ln x \right) + a^{1-x} \, (\ln a - 1) \right].$$
This result also leads to
$$\int_{a}^{x} \frac{\ln t}{t^{x}} \, dt = \frac{1}{(1-x)^{2}} \, \left[ (1-x) \, x^{1-x} \, \ln\left(\frac{x}{e}\right) - a^{1-x} \, \ln\left(\frac{a}{e}\right) \right].$$


TITLE: Showing the intersection/union of a cone is a cone
QUESTION [2 upvotes]: Defining a set $C \subset \mathbb{R}^n$ as a cone if for ever $x \in C$ and $\alpha \geq 0$
we have $\alpha x \in C$. ie they are closed under scalar multiplication.
How can I show that the intersection and unions of a cone is also a cone?
I am new to mathematical proofs so am not sure if I have the correct approach, but this is what I tried:
Let $C_1$ and $C_2$ be two cones.
The domain of $C_1 \cap C_2 $ = $C_1 \cap C_2$
Then $\forall $ $x\in C_1$, $y\in C_2$ and $\forall \alpha ,\beta \geq 0$
$ \alpha x + \beta y \in C_1 \cap C_2$
Hence the intersection of cones is a cone. Now I only showed this for the case of two intersections, but could I use a similar method to prove the intersection of n cones is a cone? Have I sufficiently proved this statement?
For the union I did dom$C_1 \cup C_2$ = $C_1 \cap C_2$ and did the same thing. IS this correct?

REPLY [2 votes]: First, show that the intersection $C_1\cap C_2$ is a cone: take $\alpha\ge0$, $x\in C_1\cap C_2$. Then $x\in C_1$ and $x\in C_2$, and since both are cones $\alpha x\in C_1$ and $\alpha x\in C_2$.
To prove that the intersection of $n$ cones is a cone, use induction with the above argument in the induction step.
The proof for union of cones is a cone, use a similar argument.


TITLE: Construct an example of set $A$ for which $A+A=A $ but $0∉cl(A)$
QUESTION [14 upvotes]: How to prove that convexity is necessary condition in this question? Need to construct an example of set $A$ for which $A+A=A$ but $0 \notin cl(A)$. From the linked question follows that $A$ must be non-convex.

REPLY [7 votes]: This is merely a distilled version of the answer by coffeemath (hence CW). Pick an irrational number $\alpha$ and let $A=\{(m,n)\in\mathbb Z^2:m\alpha+n>0\}$. This is a discrete set, it does not contain $(0,0)$, and obviously satisfies $A+A\subseteq A$. To prove the reverse inclusion $A\subseteq A+A$, note that for any $(m,n)\in A$ there exists $(m',n')$ such that $0<m'\alpha+n'<m\alpha+n$, and therefore $(m,n)=(m',n')+(m-m',n-n')\in A+A$.
This construction comes up in complex analysis, since the linear combinations of the functions $\{z^mw^n: (m,n)\in A\}$ form an algebra on the torus $|z|=|w|=1$ (here $z,w$ are complex). However, in this case one throws in $(0,0)$ (hence, the constant function) in order to have a unital algebra. See, e.g., Banach spaces of analytic functions by K. Hoffman, pp 54-57.


TITLE: If $Y_1=(X_1,...,X_l)$ then $\sigma(Y_1)=\sigma\left(\cup_{i=1}^l \sigma(X_i)\right)$.
QUESTION [1 upvotes]: $(\Omega, \mathcal A, P)$ probability space, $(\Omega_j, \mathcal A_j)$ measure space, $X_j : \Omega \rightarrow \Omega_j, ~ j =1, \ldots, n$ random variables. Furthermore
$\mathcal B_1 := \bigotimes_{j=1}^l \mathcal A_j, ~ \mathcal B_2 := \bigotimes_{j=l+1}^n \mathcal A_j, ~ \overline{\Omega_1} := \times_{j=1}^l \Omega_j, ~ \overline{\Omega_2} := \times_{j=l+1}^n \Omega_j, ~ \\ Y_1 := (X_1, \ldots, X_l), ~ Y_2 := (X_{l+1}, \ldots, X_n), ~ l \in \{1, \ldots, n-1 \}$.
I want to show that for independent $X_1, \ldots, X_n$ $Y_1$ and $Y_2$ are independent, too. Can somebody tell me why $\sigma(Y_1) := Y_1^{-1} (\mathcal B_1)$ is the same as $\sigma(\bigcup_{j=1}^l \sigma(X_j))$? I've already shown that the latter is independent from $\sigma(\bigcup_{j=l+1}^n \sigma(X_j))$, so it's just this step missing.
Thanks in advance!

REPLY [2 votes]: Note that $\mathcal{B}_1$ is generated by the set
$$\mathcal{G}=\{A_1\times \cdots \times A_j | A_i \in \mathcal{A}_i \text{ for } i\in\{1,..,l\}\}$$
therefore we have that $Y_1^{-1} (\mathcal{B}_1)=\sigma (Y_1^{-1}(\mathcal{G}))$ (see why here).
Now For any $G\in \mathcal{G}$, there exists $A_1\in\mathcal{A}_1,...,A_l\in\mathcal{A}_l$ such that $G=A_1\times \cdots \times A_l$. We see that
$$Y^{-1}_1(G)=\bigcap_{i=1}^lX_i^{-1}(A_i)\in \sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right),$$
since $X_i^{-1}(A_i)\in \sigma(X_i)$ for each $i\in\{1,...,l\}$. We have now shown that
$$
Y^{-1}_1(\mathcal{G}) \subset \sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right) \implies Y_1^{-1} (\mathcal{B}_1)=\sigma (Y_1^{-1}(G)) \subset\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right)
$$
For the converse we note that the bundled map $Y_1=(X_1,...,X_l)$ is $Y_1^{-1} (\mathcal{B}_1)/\mathcal{A}_1\otimes\cdots\otimes \mathcal{A}_l $-measurable, and this happens if and only if each coordinate map $X_i$ is $Y_1^{-1} (\mathcal{B}_1)/\mathcal{A}_i$ measurable for $i\in\{1,...,l\}$ (see why here). Thus we conclude that $Y_1^{-1} (\mathcal{B}_1)$ also makes all $X_1,...,X_l$ measurable. But since $\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right) $ is the smallest sigma-algebra making $X_1,...,X_l$ measurable we have that
$$
\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right)  \subset Y_1^{-1} (\mathcal{B}_1).
$$
We conclude that
$$
\sigma\left(\bigcup_{i=1}^l \sigma (X_i)\right)  = Y_1^{-1} (\mathcal{B}_1).
$$


TITLE: Confusion regarding momentum and energy
QUESTION [0 upvotes]: Can a body have momentum without energy?
Please don't use a lot of differentiation.
A little theoretical background in the answer would be appreciated.

REPLY [0 votes]: No it's not possible. Angular momentum is defined as cross product of radius vector and linear velocity. Angular momentum is only zero when the body doesn't possess any translatory or rotatory motion, in which energy can never be equal to zero


TITLE: Find an equation where all 'y' is always irrational for all integer values of x
QUESTION [1 upvotes]: Intuitively it appears to me that if $x$ is an integer, $y$ has to be an irrational number for the following equation.
$10y^2-10x-1 = 0$
Can someone prove me right or wrong?

REPLY [1 votes]: You are correct.
If $$10y^2-10x-1 = 0$$ and if $x$ is an integer then $y$ is irrational.
Note that if $y= \frac {m}{n}$  where $m$ and $n$ are relatively prime, then $$ 10 m^2=n^2(10x+1)$$
Obviously $10x+1$  is not a multiple of $10$ which makes $n^2$ to be a multiple o f $10$. Now if $n^2$ is a multiple of $10$ , $n$ is a multiple of $10$ thus $n^2$ is a multiple of $100$ which in turn makes $m $ to be a multiple of $10$
That contradicts the relatively prime assumption of $m$ and $n$.
Thus, $y$ is irrational.


text:
\section{Zeta functions and complex dimensions}
\label{sec:zeta-fns}

We will require certain mild hypotheses on the fractal string
\sL which gives the scaling of the spray \tiling. These
conditions are phrased as growth conditions on a zeta function
associated with \sL, within a suitable window, as defined just below.

\begin{defn}\label{def:scaling-zeta-function}
  For a fractal string $\sL = \{\scale_j\}_{j=1}^\iy$, the \emph{scaling zeta function} is given by
  \begin{equation}\label{eqn:scaling-zeta}
    \gzL(s) = \sum_{j=1}^\iy \scale_j^s
  \end{equation}
  for $s \in \bC$ with $\Re(s) > \abscissa$, where \abscissa is the abscissa of convergence of this series. (Compare with \cite[Def.\ 1.8]{FGCD}, where $\gzL$ is called the \emph{geometric zeta function} of $\sL$.)
  Recall that $\abscissa := \inf\{\ga \in \bR \suth \sum_{j=1}^\iy \scale_j^\ga < \iy\}$ and that \gzL is holomorphic (i.e., analytic) for $\Re s > \abscissa$. Henceforth, if $W \ci \bC$ contains $\{\Re s > \abscissa\}$ and \gzL has a meromorphic continuation (necessarily unique) to a connected open neighborhood of $W$, we abuse notation and continue to denote by \gzL its meromorphic extension. Under these assumptions, each pole $\gw \in W$ of \gzL is called a \emph{visible complex dimension} of \sL and

  the set of visible complex dimensions is written as
  \begin{equation}\label{comp-dim}
    \sD_{\sL}(W)=\{\pole \in W \suth \pole \mbox{ is a pole of } \gzL\}.
  \end{equation}

\end{defn}

Moreover, in the special case when \gzL has a meromorphic continuation to all of \bC, we may choose $W = \bC$ and then simply write $\sD_{\sL}:=\sD_{\sL}(\bC)$ and refer to $\sD_{\sL}$ as \emph{the complex dimensions of \sL}.

In practice, $W$ will be a window (the part of $\bC$ to the right of a screen $S$) as in Definition \ref{def:screen}, just below.
The following three definitions are excerpted from \cite[Section~5.3]{FGCD}.

\begin{defn}
  \label{def:screen}
  Let $S:\bR \to (-\iy, \abscissa]$ be a bounded Lipschitz continuous function. Then the \emph{screen}
  is $S = \{S(t) + \ii t \suth t \in \bR\}$, the graph of a function with
  the axes interchanged. Here and henceforth, we denote the imaginary unit by $\ii := \sqrt{-1}$. We let
  \linenopax
  \begin{align}
    \label{eqn:infS}
    \inf S &:= \inf\nolimits_{t \in \bR} S(t) = \inf\{\Re s \suth s \in S\}, \text{ and} \\
    \label{eqn:supS}
    \sup S &:= \sup\nolimits_{t \in \bR} S(t) = \sup\{\Re s \suth s \in S\}.
  \end{align}
  The screen is thus a vertical contour in \bC. The region to the
  right of the screen is the set $W$, called the \emph{window}:
  \linenopax
  \begin{align}
    \label{eqn:window}
    W &:= \{z \in \bC \suth \Re z \geq S(\Im z)\}.
  \end{align}
  For a given string \sL, we always choose $S$ to avoid $\sD_{\sL}$ and such that \gzL can be meromorphic{ally} continued to an open neighborhood of $W$. We also assume that $\sup S \leq \abscissa$, that is, $S(t) \leq \abscissa$ for every $t \in \bR$.
\end{defn}

\begin{defn}
  \label{def:languid}
  The fractal string \sL is said to be \emph{languid} if its
  associated zeta function \gzL satisfies certain horizontal and
  vertical growth conditions. Specifically, let
  $\{T_n\}_{n \in \bZ}$ be a sequence in \bR such that $T_{-n} < 0 < T_n$
  for $n \geq 1$, and
  \linenopax
  \begin{align}
    \label{eqn:Tn seq conds}
    \lim_{n \to \iy} T_n = \iy,
    \lim_{n \to \iy} T_{-n} = -\iy, \text{ and }
    \lim_{n \to \iy} \frac{T_n}{|T_{-n}|} = 1.
  \end{align}
  For \sL to be languid, there must exist constants $\languidOrder \in \bR$ and $c>0$, and a sequence $\{T_n\}$ as described in \eqref{eqn:Tn seq conds}, such that: \\
  \vstr
  For all $n \in \bZ$ and all $\ga \geq S(T_n)$,
  \linenopax
  \begin{align}
    \label{eqn:L1} \tag*{\textbf{L1}}
    |\gzL(\ga + \ii T_n)| \leq c \cdot \left(|T_n| + 1\right)^\languidOrder,
  \end{align}

  and for all $t \in \bR$, $|t| \geq 1$,
  \linenopax
  \begin{align}
    \label{eqn:L2} \tag*{\textbf{L2}}
    |\gzL(S(t) + \ii t)| \leq c \cdot |t|^\languidOrder.
  \end{align}
  In this case, \sL is said to be \emph{languid of order \languidOrder}.
\end{defn}

\begin{defn}
  \label{def:strongly-languid}
  The fractal string \sL is said to be \emph{strongly languid} of order \languidOrder and with constant $A$ iff it satisfies \textbf{\ref{eqn:L1}} and the following condition \textbf{\ref{eqn:L2'}}, which is clearly stronger than \textbf{\ref{eqn:L2}}:\\
  \vstr
  There exists a sequence of screens $S_m$ for $m \geq 1$,
  $t \in \bR$, with $\sup S_m \to -\iy$ as
  $m \to \iy$, and with a uniform Lipschitz bound, for which there exist constants $\languidOrder \in \bR$ and $A,c>0$ such that
  \linenopax
  \begin{align}
    \label{eqn:L2'}  \tag*{\textbf{L2$'$}}
    |\gzL(S_m(t) + \ii t)| \leq c \cdot A^{|S_m(t)|}(|t|+1)^\languidOrder,
  \end{align}
  for all $t \in \bR$ and $m \geq 1$.
\end{defn}

By saying ``\gzL is languid'', we mean just that \sL is languid.
In the rest of this paper, \tiling is assumed to be a fractal spray with a generator $\gen \ci \bRd$, scaled by the fractal string $\sL = \{\scale_j\}_{j=1}^\iy$ with $\ell_1=1$ as in Remark~\ref{rem:renormalize}. The tubular zeta function first appeared in \cite{TFCD}, but we need to modify the definition in order to extend it to the case when the generators are not monophase; thus, the following definition is new.

\begin{defn}\label{def:volume-zeta-fn}
  The \emph{tubular zeta function} \gzT of the fractal spray \tiling is defined by
  \begin{equation}\label{eqn:def:volume-zeta}
    \gzT(\ge,s)
    = \ge^{d-s} \sum_{j=1}^\iy \scale_j^s \left(
      \sum_{k=0}^{d}\frac{\genir^{s-k}}{s-k} \crv_k(\gen,\scale_j^{-1}\ge) - \frac{\genir^{s-d}}{s-d}\gl_d(G)\right)
  \end{equation}
  for every $\ge\in(0,\iy)$ and for each $s \in \bC$ such that the sum converges absolutely. As in Definition~\ref{def:scaling-zeta-function}, we will henceforth abuse notation and use $\gzT(\ge,s)$ to mean a meromorphic extension of the function defined by the formula \eqref{eqn:def:volume-zeta}, as convenient.
\end{defn}

Note that by \eqref{eqn:leb-coeff-rel}, \eqref{eqn:scaling-zeta}, and Proposition~\ref{cor:V(Gj,ge)}, for $\ge \geq \genir$, one has
\linenopax
\begin{align}\label{eqn:gzT-for-big-epsilon}
  \gzT(\ge,s)
    &= \ge^{d-s} \gzL(s) \left(
      \sum_{k=0}^{d}\frac{\genir^{s-k} \crv_k(\gen)}{s-k}- \frac{\genir^{s-d} \gl_d(G)}{s-d}\right)
    = \ge^{d-s} \gzL(s) \sum_{k=0}^{d-1}\frac{\genir^{s-k} \crv_k(\gen) (d-k)}{(s-k)(d-s)}.
\end{align}
Since $\gzL(s)$ is a Dirichlet series, it has an \emph{abscissa of convergence}: there is a unique number $\abscissa \in [-\iy,\iy]$ such that $\gzL(s)$ converges absolutely for $s$ with $\Re s > \abscissa$ and diverges for $s$ with $\Re s < \abscissa$. The abscissa of convergence is thus analogous to the radius of convergence of a power series.

  Note that in all reasonable situations we have $0 \leq \abscissa \leq d$. Indeed, $\abscissa \geq 0$ follows immediately from the non-finiteness of the fractal string (assumed in Definition~\ref{def:fractal_string}), and $\abscissa \leq d$ follows if one requires that the generated fractal spray $\sT$ have finite total volume. (Note that for fractal sprays with infinite total volume, the question for a (global) tube formula does not make sense.)

For the scaling zeta functions of the self-similar tilings discussed in Section~\ref{sec:Self-similar-tilings}, one has $0 < \abscissa < d$, and
\abscissa coincides with the Minkowski dimension, Hausdorff dimension, and similarity dimension of the associated self-similar set; for a precise statement, please see \cite[Section~4.3]{TFCD} and Section~\ref{sec:Self-similar-tilings} below (especially Remark~\ref{rem:incarnations-of-D}).

Note that the tubular zeta function \gzT may be viewed as a generating function for the geometry of the fractal spray \tiling.

Proposition~\ref{prop:poles} clarifies the relation between the scaling zeta function \gzL and the tubular zeta function \gzT of a fractal spray \tiling. It also motivates and justifies the definition of complex dimensions of fractal sprays. The intended application of Proposition~\ref{prop:poles} is with \gW as a suitable open  neighborhood of a window $W$ for the scaling zeta function \gzL of \tiling, as in Definition~\ref{def:screen}. (Proposition~\ref{prop:poles} is extended significantly in Theorem~\ref{thm:Meromorphic-continuation-of-gzT} and Lemma~\ref{thm:residues-of-gzT}.)

\begin{prop} \label{prop:poles}
  If \abscissa is the abscissa of convergence of $\gzL$, then for all $\ge > 0$, the series in \eqref{eqn:def:volume-zeta} defining $\gzT(\ge,s)$ converges absolutely for any fixed $s \in \bC \less \{0,1,\dots,d\}$ with $\Re s > \abscissa$. More generally, suppose $\gzL$ is meromorphic in a connected open set \gW containing $\{\Re s > \abscissa\}$. Then for all $\ge > 0$, the function $\gzT(\ge, \mydot)$ is meromorphic in \gW and each pole $\gw \in \gW$ of $\gzT(\ge, \mydot)$ is a pole of \gzL or belongs to the set $\{0,1,\ldots,d\}$.
\end{prop}
\begin{proof}
  Fix $\ge>0$.
  Upon expanding \eqref{eqn:def:volume-zeta} of Definition~\ref{def:volume-zeta-fn} and interchanging the sums, the tubular zeta function becomes
  \linenopax
  \begin{align}\label{eqn:zeta-expanded}
    \gzT(\ge,s)
    &= \sum_{k=0}^{d}\frac{ \ge^{d-s} \genir^{s-k} }{s-k}\sum_{j=1}^\iy \scale_j^s\crv_k(\gen,\scale_j^{-1}\ge) - \frac{\ge^{d-s} \genir^{s-d}}{s-d} \gzL(s) \gl_d(G).
  \end{align}
  It is clear that the second term on the right-hand side of \eqref{eqn:zeta-expanded}  is convergent for $s$ as in the hypotheses, so it remains to check that the first term is similarly convergent for each $k$.

  Since \ge is fixed and $\scale_j \searrow 0$, define $J=J(\ge)$ to be the index of the last scale greater than \ge:
  \linenopax
  \begin{align}\label{eqn:J(eps)}
    J(\ge) := \max\{j \geq 1 \suth \scale_j^{-1} \ge < \genir \} \vee 0.
  \end{align}
At the end of \eqref{eqn:J(eps)}, ``$\vee 0$'' indicates that $J(\ge) = 0$ for $\ge \geq \scale_1 \genir$.

  \footnote{By convention, $\sup \es = -\iy$. Thus the maximum $\vee 0$ is included so that $J(\ge) = 0$ when $\ge \geq \scale_1 \genir$.}

Now $\crv_k(\gen,\scale_j^{-1}\ge) = \crv_k(\gen)$ for all $j > J$, by \eqref{eqn:coeff-ext}, and so
  \linenopax
  \begin{align}\label{eqn:protosplit}
    \sum_{j=1}^\iy \scale_j^s \crv_k(\gen,\scale_j^{-1}\ge)
    = \sum_{j=1}^{J} \scale_j^s \crv_k(\gen,\scale_j^{-1}\ge)
     + \crv_k(\gen) \sum_{j=J+1}^{\iy} \scale_j^s.
  \end{align}
  Observe that the first sum on the right-hand side of \eqref{eqn:protosplit} is entire, as a finite sum of the entire functions $c_x x^s$, and that the second sum on the right-hand side of \eqref{eqn:protosplit} converges absolutely exactly where $\gzL$ does; that is, for $\Re s > \abscissa$.

  Justification of the claims of meromorphicity are obtained by parallel reasoning; the decomposition \eqref{eqn:protosplit} shows that, except possibly for $\{0,1,\dots,d\}$, the tubular zeta function is meromorphic precisely where the scaling zeta function is.
\end{proof}

\begin{defn}[Complex dimensions]\label{def:complex-dimn}

  The set
\[
\sD_{\sT} = \sD_{\sL} \cup \{0,1,\dots,d\}
\]
of (potential) poles of \gzT is called the set of \emph{complex dimensions} of $\sT$. Let $W \ci \bC$ be a window for \gzL as in Definition~\ref{def:screen}, so that \gzL is meromorphic in an open neighborhood of $W$.
  (Proposition~\ref{prop:poles} thus implies that for each fixed $\ge > 0$, the function $\zeta_\sT(\ge,\mydot)$ is also meromorphic in an open neighborhood of $W$.)
  Then $\sD_\tiling(W) := \sD_\tiling \cap W$ is called the set of \emph{visible complex dimensions} of $\sL$ in $W$, in parallel with \eqref{comp-dim}. We refer to $\sD_\sL$ as the \emph{scaling} complex dimensions and $\{0,1,\dots,d\}$ as the \emph{integer} complex dimensions of \tiling.

\end{defn}


TITLE: Sub- and superharmonic functions
QUESTION [0 upvotes]: Let $\Omega \subset \mathbb{R}^n$ and $f\in C^2(\Omega)$ such that $\Delta f(x)\leq 0$ for some $x\in\Omega$. Thus, can we derive:
$f(x)\geq \frac{1}{|B_r(x)|}\displaystyle\int_{B_r(x)}f(y)\,dy $
where $B_r(x)\Subset \Omega$? Or is there really $\Delta f(x)\leq 0$ for all $x\in\Omega$ needed?
Thanks.

REPLY [1 votes]: It's well known that if $f$ is subharmonic and smooth then $\Delta f\le0$. Perhaps less well known is the fact that the mean-value inequality at a single point implies the inequality for the Laplacian at that point. This is clear if you know the result expressing the Laplacian in terms of spherical averages; I don't recall the result so I'll give the proof first: If $f\in C^2(\Bbb R^2)$ then Taylor's theorem says
$$f(x,y)=f(0,0)+Ax+By+Cxy+Dx^2+Ey^2+o(x^2+y^2).$$Note that $$\Delta f(0,0)=2D+2E$$while $$\frac1{\pi r^2}\int_{D(0,r)}f(x,y)\,dxdy=f(0,0)+\frac12(D+E)+o(1).$$So

$\frac1{\pi r^2}\int_{D(0,r)}f(x,y)\,dxdy\le f(0,0)$ for all $r>0$ implies $\Delta f(0,0)\le0$.


TITLE: In kinetic theory of gases, how do we figure out the forces exerted by walls on molecules?
QUESTION [0 upvotes]: This is a cube with each side having a length of $l$ containing a certain amount of molecules in it.
Let's say a molecule of mass m is traveling in the $y$-direction with the velocity $v$ and hits the wall 1. Obviously, just before the molecule hits the wall 1, its momentum is $mv$. After hitting the wall 1, the direction of the molecule will be reversed, and its velocity and momentum will be $-v$ and $-mv$, respectively. So the change in momentum is $2mv$. So far, so good.
Now, let's say the time the molecule is in contact with the wall 1 is $t$. In other words, the wall 1 applies a certain force on the molecule for the time $t$, so to get the force the wall
1 exerts on the molecule, we simply divide the change in momentum $2mv$ by $t$.
However, this is not what is done in the proofs I have seen.
As the distance between wall 1 and wall 2 is $l$, and the velocity of the molecule is $v$, after hitting the wall 1, the time it takes for the molecule to travel back to the wall 2 and come back and hit the wall 1 again is, $t_1 = \frac{2l}{v}$.
And they calculate the force the wall 1 exerts on the molecule by dividing the change in momentum $2mv$ by $t_1$.
I do not understand it. Someone please help.

REPLY [4 votes]: It's from the impulse formula $force \times time = \delta mv$
Where $\delta mv$ stands for change in momentum.  They consider it over 1 second, so the time used is 1 second, not the actual time of contact of the molecule with the wall.  There are many collisions of the molecule with wall 1 and 2 in one second. The time between collisions is $\frac{l}v$, and the number of collisions per second is $\frac{v}l$, and the number of collisions with wall 1 per second is $\frac{v}{2l}$.   So the  formula above becomes
$force \times time = 2mv\times\frac{v}{2l}$  with $time =1$
so the force on wall 1 is $force = \frac{mv^2}{l}$


TITLE: Given N1 (time to termination) for an absorbing markov chain find a possible P
QUESTION [0 upvotes]: I am working with an absorbing markov chain as described here: https://en.wikipedia.org/wiki/Absorbing_Markov_chain
Normally you would have a Matrix:
$$P=\begin{pmatrix}Q & R\\ 0 & I_r\end{pmatrix}$$
where $Q$ is a t-by-t matrix, $R$ is a nonzero t-by-r matrix, $0$ is an r-by-t zero matrix, and $I_r$ is the r-by-r identity matrix. Thus, $Q$ describes the probability of transitioning from some transient state to another while $R$ describes the probability of transitioning from some transient state to some absorbing state.
From here you would compute things like $N$ (a t-by-t matrix where each cell is the expected time spent in state j when starting from state i before absorption)
And $N1$ which is a sum  over the rows of $N$ and is the expected time to termination from  every initial state i
My Question is this:
Given $N1$ can a plausible matrix $P$ be discovered that would result in $N1$?
If not can $P$ be computed given $N$?

REPLY [1 votes]: We assume that $N_{ij}$ for transient $i,j$ is   the expected time (including time $0$) spent in state $j$ when starting from state $i$,  before absorption. Extend $N$ to be a square matrix of side $t+r$ as follows: If $i$ is absorbing, let $N_{ii}=1$. If $i,j$ are distinct and at least one of them is absorbing, let  $N_{ij}=0$.
If $i,j$ are both transient,
then $(PN)_{ij}=\sum_{\ell}p_{i \ell}N_{\ell j}$ is is   the expected time (NOT including time $0$) spent in state $j$ when starting from state $i$,  before absorption. Thus in this case,
$$(N-PN)_{ij}=N_{ij}-\sum_{\ell}p_{i \ell}N_{\ell j}=I_{ij} \,.$$
By inspection, this identity also holds if at least one of $i,j$ is absorbing, so it holds  for all $i,j$. Thus
$(I-P)N=I$ so $P=I-N^{-1}$.


text:
\section{H-Cobordism Theorem}
Tags: Topology

\begin{theorem}
Let $X^n, Y^n$ be two [[Definition:Simply Connected|simply connected]] manifolds.
Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an [[Definition:Cobordism|h-cobordism]] between $X$ and $Y$.
Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a [[Definition:Diffeomorphism|diffeomorphism]].
In particular, $X$ and $Y$ are [[Definition:Diffeomorphism|diffeomorphic]].
\end{theorem}

\begin{proof}
{{ProofWanted|In progress}}
[[Category:Topology]]
2guptkt0e7yhebs6rugk4vfd5rl5e1l
\end{proof}

config: wiki


TITLE: Will this equation with $\sin$ and $\arcsin$ cancel?
QUESTION [0 upvotes]: It can be said that $\arcsin(\sin(x))= x$ are inverses if $x \lt 2\pi$.
Can it also be generalized so that $\arcsin(\sin(d\cdot x))= d\cdot x$ if $x \lt 2\pi\cdot d$ for a constant $d$?

REPLY [0 votes]: You are incorrect about the domain of arcsine. Arcsine has a domain of values $y$ where $-1 \leq y \leq 1$, which means if we take $y = \sin(x)$ our values of $x$ are restricted by $-\pi/2 \leq x \leq \pi/2$. So as long as your constant $d$ is such that $zd = x$ where $x \in [-\pi/2, \pi/2]$ then we do know $\arcsin(\sin(dz)) = dz$. To say that this idea is "generalized" is also not true though, since it requires us to introduce a constant $d$ in addition to our variable $x$. The true generalization is that $\arcsin(\sin(x)) = x$ for $x \in [-\pi/2, \pi/2]$. It is trivial that we can write $x$ as a product of another variable and a constant. Choose any $c \neq 0$ and set $z = \frac{x}{c}$ to see what I mean.


TITLE: A closed linear functional must be bounded？
QUESTION [2 upvotes]: it‘s a problem from salamon’s functional analysis

Let $X$ be a real Banach space, Let $Y\subset X$ be a linear subspace, and let $T:Y\to \mathbb{R} $ be a linear functional. show that $T$ is closeable if and only if $T$ is bounded.

The definition of closeable:

Let $X$ and $Y$ be Banach spaces, let $dom(A)\subset X$ be a linear subspace, and let $A:dom(A)\to Y$ be a linear operator.  The operator $A$ is called closeable if there exists a closed linear operator $A’: dom(A’)\to Y$ on a linear subspace $dom(A’)\subset X$ such that
$$dom(A)\subset dom(A’), A’x=Ax, for~all~x\in dom(A)$$

If $T$ is bounded, we can easily show that $T$ is closeable by Hahn-Banach theorem and closed graph theorem.
However, conversely, I don’t know how to deal with it. If $T’$ is a closed linear functional which extends $T$, we cannot get the boundedness of $T’$ since the domain of $T’$ may not be closed, and hence we cannot use closed graph theorem. It seems that we should prove that a closed linear functional must be bounded, but I don’t know how to do it.
I do hope someone can help me.Thanks!
Update:
I have some ideas although I’m not sure whether it is right.
Claim: Suppose that $X$ is a normed vector space but not complete, and $T:X\to \mathbb{R}$ is a closed linear functional.We can show that $T$ must be bounded.
Proof: If $T$ is unbounded, then $Ker T$ is a dense subspace of $X$. Chose $y\in X\backslash Ker(T)$, and $\left\{ x_n \right\} \subset Ker\left( T \right) $ such that $x_n\to y$.
Note that $Tx_n\to 0$,$x_n\to y$ implies that $Ty=0 $ since $T$ is closed, so it contradicts with the assumtion that $y\notin Ker(T)$.

REPLY [1 votes]: You don't need to argue by contradiction, your argument is a direct one.
Indeed, let $\{x_n\}\subset\ker T$ be Cauchy. Since $X$ is Banach, there exists $y\in X$ with $y=\lim x_n$. The sequence $\{(x_n,0)\}$ is a Cauchy sequence in the graph of $T$. As this is closed and $X\times\mathbb R$ is Banach, it is complete; so $(y,0)$ is in the graph of $T$, i.e. $Ty=0$. This shows that $\ker T$ is closed, and so $T$ is bounded (this last implication is where one uses that $T$ is a functional).


TITLE: Prove convergence of random variables
QUESTION [0 upvotes]: I have a problem with following task:
We have $X_1,X_2,...$ independent random variables, that almost surely $X_1\ge X_2\ge...\ge 0$. Prove that if $X_n\rightarrow0$ converges in probability then $X_n\rightarrow0$ converges almost surely. I am hitting a wall with this one. I will be very glad for as simple as possible explanation.

REPLY [1 votes]: Let $u_i$ be the essential supremum of $X_i$, i.e. the least constant $c$ such that $X_i \le c$ a.s.  If $X_{i-1}$ and $X_i$ are independent but $X_{i-1} \ge X_i$ a.s., then $c_i$ must exist and $X_{i-1} \ge c_i \ge X_i$ a.s.  Show that
if $X_i \to 0$ in probability, $c_i \to 0$.


TITLE: Show that $f(x)=x²$ is a Regulated function
QUESTION [0 upvotes]: Show that $f(x)=x^2$, $x\in[0,1]$ is a Regulated function by giving a Step function $g_n$ that converges uniformly to $f$.
Show that $\|g_n-f\| \to 0$

REPLY [0 votes]: For every $n$ define:
$$ g_n(x)=\sum\limits_{k=1}^n f\left( \frac{k}{n} \right) \cdot \mathbf{1}_{ \Big[\frac{k-1}{n}, \frac{k}{n}\Big) }+f(1)\cdot \mathbf{1}_{\{ 1\} } $$
Where $\mathbf{1}_{E}$ is the indicator function of $E$. This function is uniformly continuous, i.e for all $\epsilon>0$ there exists $\delta_\epsilon>0$, such that $\vert f(x)-f(y)\vert<\epsilon$ if $\vert x-y\vert<\delta_\epsilon$. For $n$ large enough, $\frac{1}{n}<\delta_\epsilon$ and:
$$ \Big\vert f(x)-g_n(x) \Big\vert \overset{\text{if} \; x\in \big[\frac{k-1}{n}, \frac{k}{n}\big) }{=} \Big\vert f(x)-f\left( \frac{k}{n} \right) \Big\vert< \epsilon $$
Since every $x\neq 1$ is in such an interval, this shows that $\Vert f-g_n\Vert_\infty\rightarrow 0$.
This approximation has the benefit of working for any continuous function on closed bounded interval.


TITLE: Evaluation of a digamma series involving golden-ratio
QUESTION [2 upvotes]: Let $\varphi =\frac{1}{2} \left(\sqrt{5}+1\right), a=\tan \left(\frac{\sqrt{5} \pi }{2}\right)$, then how can one prove
$$\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+\varphi)-\psi ^{(0)}\left(n-\frac{1}{\varphi}\right)}{n^2+n-1}=\frac{\pi ^2 a^2}{\sqrt{5}}+\frac{4 \pi  a}{5}+\frac{\pi ^2}{2 \sqrt{5}}$$
Note that $n^2+n-1=(n+\varphi) \left(n-\frac{1}{\varphi}\right)$. Maybe we should consider the generalized sum i.e. $\sum _{n=1}^{\infty } \frac{\psi ^{(0)}(n+t)-\psi ^{(0)}(n+s)}{(n+s) (n+t)}$? Any help will be appreciated.

REPLY [2 votes]: We can use the representation
\begin{equation}
\psi(a)-\psi(b)=(a-b)\sum_{p=0}^\infty\frac{1}{(p+a)(p+b)}
\end{equation}
to express the series
\begin{align}
S(s,t)&=\sum _{n=1}^{\infty } \frac{\psi (n+t)-\psi (n+s)}{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{p=0}^\infty\frac{1}{(p+n+t)(p+n+s)} \frac1{(n+s) (n+t)}\\
&=(t-s)\sum _{n=1}^{\infty }\sum_{k=n}^\infty\frac{1}{(k+t)(k+s)} \frac1{(n+s) (n+t)}
\end{align}
The case $t=\phi, s=-1/\phi$ is explicitely solved in this question.
For the general case, as shown in the answers, and with $f(n)=1/\left((n+s) (n+t)  \right)$, we have
\begin{equation}
\left[\sum_{n=1}^\infty f(n)\right]^2=2\sum_{n=1}^\infty \sum_{k=n}^{\infty} f(n)f(k) - \sum_{n=1}^{\infty}f(n)^2
\end{equation}
then,
\begin{equation}
S(s,t)=\frac{t-s}{2}\left[\sum _{n=1}^{\infty }\frac1{(n+s) (n+t)}\right]^2+\frac{t-s}{2}\sum _{n=1}^{\infty }\frac1{(n+s)^2 (n+t)^2}
\end{equation}
With
\begin{equation}
\frac1{(n+s)^2 (n+t)^2}=\frac{1}{(t-s)^2}\left[\frac{1}{(n+s)^2}+\frac{1}{(n+t)^2}\right]+\frac{2}{(t-s)^3}\left[\frac{1}{n+t}-\frac{1}{n+s}\right]
\end{equation}
and using again the representations in terms of the polygamma functions given above, we obtain
\begin{align}
S(s,t)=\frac{1}{2(t-s)}&\left[\psi(t+1)-\psi(s+1)\right]^2\\
&+\frac{1}{2(t-s)}\left[\psi^{(1)}(t+1)+\psi^{(1)}(s+1)\right]\\
&-\frac{1}{(t-s)^2}\left[\psi(t+1)-\psi(s+1)\right]
\end{align}