TITLE: Spin and wavefunction of Excitons
QUESTION [1 upvotes]: My Bachelor thesis is all around Excitons (specifially transitions between excitons of different energies). During my work I often had trouble with the spin and the wavefunction of them. Is there maybe some good (free) literature about the theory of excitons ? I found some books in the internet but they werent for free. And my university hasnone.
1)
I read about "heavy holes" that can have spin of +-3/2. How can these be created ? And for example a 3/2 hole bonds with a 1/2 electron. Then the Exciton has spin 2. How would the wave function of this look like ? I know that in 3D the wavefunctions of hydrogen can be used for the wavefunctions of the exciton. And thats why I think if there is a spin 2 Exciton, it can only exist at higher Energies E(n) (since a hydrogen atom only can have spin 2 when it's n is 3 or higher). But on the other side, my professor now wrote me that there is also a $|2,1,-2\rangle$ (n=2,l=1,m=-2)
2)
$$|3,2,2\rangle \Rightarrow |1,0,0\rangle + 2 photons $$ is this possible ?
Or is just the following possible: $$|3,2,2\rangle \Rightarrow |2,1,1\rangle + 1 photon \Rightarrow (|1,0,0\rangle + 1 photon) + 1 photon $$
Or in Words:
If now there is an Exciton with spin 2 and n=3. Can it for example jump directly to n=1 with the emission of 2 photons (afterwards it will be an exciton with spin 0)? Or can it only First jump to n=1 with the emission of 1 photon (then it will be an Exciton with spin 1) and then to n=0 with the emission of another Photon (then it will be an Exciton with spin 0)

REPLY [0 votes]: Excitons are quasi-particles made by a bound electron-hole pair. In general one refers to exciton as spin conserving excitations. In case spin is a good quantum number, the spin of the state in which the hole is created in valence is the same as the one in which the electron goes in conduction. As a result the spin of the exciton is zero. Thinking to the exciton in terms of the hydroge atoms, this means that the excitonic wave-function is a singlet. This kind of excitons are the only ones which are optically bright at equilibrium, i.e. they can measured in the absorption spectrum of a material.
More in general one can speak about excitons for any bound particle-hole pair and thus also include pairs which have triplet character. These excitons are optically dark. The overall spin of the state is 1 and it is more correct to call this kind of states magnons.
If spin is not a good quantum number, then in general excitons and magnons can be mixed. For a theory paper you can have a look here: https://arxiv.org/abs/2103.02266
Instead, the nomenclature light hole and heavy hole is more related to the orbital angular momentum. Similarly to the spin case, one can think to the exciton orbital momentum in terms of the relative momentum between the electron and the hole. Similarly to the fact that an excitation must conserve the spin, there are selection rules for the orbital momentum for bright excitons. In general photons can change the orbital momentum of the electrons by one. To jump from l=2 to l=0 you need to photons. Now the intermediate state in the process can be real, but it can also be virtual (as it happens in non linear optics). In the case of a virtual intermediate state one can think in terms of a process where two photons are directly emitted.


TITLE: Integral involving exponential and power
QUESTION [1 upvotes]: I have bumped into the following integral, which Mathematica is apparently not able to solve. I have tried a couple of change of variables and a series expansion for $x$ close to $1$, but without much success. I am interested in a closed-form solution if existent: it doesn't look as harmful as it might turn out to be, though...
$$
I(c,\xi)=\int_1^\infty\mathrm{d}x\ \mathrm{e}^{-cx}x^{\frac{\xi -1}{\xi }} \left(x^{1/\xi }-1\right)^{\xi }\ ,
$$
for $c,\xi>0$ (and $\xi$ not an integer in general). Thanks for your help folks.

REPLY [0 votes]: The case $\xi=1/2$ can be solved in terms of modified Bessel- and Struvefunctions. For a proof rewrite $x=\cosh(t)$ so that
$$
I(c,\frac{1}2)=\int_0^{\infty}dte^{-c \cosh(t)}\frac{\sinh^2(t)}{\cosh(t)}=\int_0^{\infty}dte^{-c \cosh(t)}\cosh(t)-\int_0^{\infty}dte^{-c \cosh(t)}\frac{1}{\cosh(t)}
$$
According to DMLF 10.32.9 we can rewrite this as
$$
I(c,\frac{1}2)=K_1(c)+\int dc K_0(c)
$$
the last integral equals (10.43.1 of DMLF) a combination of modified Bessel functions $K_{\nu}(z)$ and modified Struve functions $L_{\mu
}(z)$. We obtain
$$
I(c,\frac{1}2)=K_1(c)+\frac{1}{2}\pi c (L_{-1
}(c)K_{0}(c)+L_{0
}(c)K_{1}(c))+C
$$
Because $I(\infty,\xi)=0$ and $K_1(\infty) =0$ as well as $\lim_{c\rightarrow\infty} c (L_{-1
}(c)K_{0}(c)+L_{0
}(c)K_{1}(c))=1$ we have $C=-\frac{\pi}{2}$ and finally
$$
I(c,\frac{1}2)=K_1(c)+\frac{1}{2}\pi c (L_{-1
}(c)K_{0}(c)+L_{0
}(c)K_{1}(c))-\frac{\pi}{2}
$$
I suspect that this method can't be generalized to other non-integer values of $\xi$ but maybe i'm wrong..


text:
\begin{document}
\thispagestyle{empty}
\maketitle
\begin{abstract}
	Forward invariance of a basin of attraction is often overlooked when using a Lyapunov stability theorem to prove local stability; even if the Lyapunov function decreases monotonically in a neighborhood of an equilibrium, the dynamic may escape from this neighborhood. In this note, we fix this gap by finding a smaller neighborhood that is forward invariant. This helps us to prove local stability more naturally without tracking each solution path. Similarly, we prove a transitivity theorem about basins of attractions without requiring forward invariance.

	\noindent
	{\it Keywords:} Lyapunov function, local stability, forward invariance, evolutionary dynamics,

\end{abstract}

\section{Introduction}
The idea of Lyapunov stability theorem or Lyapunov's direct method is intuitive: if we find a mapping (\textit{Lyapunov function}) from the current state of a dynamic to a real number such that i) the function attains a local minimum only at an equilibrium (possibly a set) and ii) its value decreases as long as the current state has not reached the equilibrium, then the equilibrium is stable under the dynamic. With this on hand, (we hope that) we do not have to identify a solution path; we just find a Lyapunov function and see how it behaves in the neighborhood–--in particular, the value and first-order derivatives at each point in the state space. So, we typically find a neighborhood where the decrease in the Lyapunov function is guaranteed, which call here a monotone decrease neighborhood, and expect this neighborhood to be a basin of attraction.

However, a basin of attraction must be forward invariant. (This does not matter for global stability, of course.) Precisely, in known versions of Lyapunov stability theorem (e.g. \citealt{Smirnov01DI}), the monotone decrease must be assured to hold on each solution path. Even if we find a monotone decrease neighborhood, a solution path may escape from this neighborhood and eventually the Lyapunov function may not decrease after the escape. This imposes an additional burden of proof, losing an appeal of the theorem to intuition since we eventually need to identify a solution path.
This is overlooked in practice; e.g. \cite{Sandholm10TELocalStbl} and \cite{ZusaiTBRD} on evolutionary dynamics in games, which we fix in this paper.

Similarly, we would expect transitivity of such basins of attractions. That is, if we find a Lyapunov function that decreases in $X_1$ and attains the minimum in $X_2$ and another that decreases in $X_2$ and attains the minimum in $X^\ast$, then we expect $X^\ast$ to be stable in $X_1$. Again, known versions of transitivity theorems as in \cite{Conley78_IsoInvSets} (see also \cite[Theorem 3]{OSTsBRD}) require $X_1$ to be forward invariant and $X_2$ to be forward and also strongly negative (i.e., backward) invariant.\footnote{Strong negative invariance of $X$ means that, if a solution path (starting at time $0$) visits $X$ at any positive time, then it must have started from $X$ at time 0.}

In applications to economics or game theory, we hope to find a Lyapunov function from economic intuition. Under an agent-based dynamic in a game or economic model, an aggregate of agents' possible gains from adjustment of their choices can be used as a candidate for a Lyapunov function once we find a neighborhood where an agent's revision of the choice incurs negative payoff externality to others' gains from further changes, as generally proven by \cite{ZusaiGains}. However, forward invariance needs more mathematical examination of the dynamic system, which may not be appealing to economic intuition.

In this paper, we reduce the burden of proof by showing that we can construct a forward invariant (smaller) neighborhood from a monotone decrease neighborhood. This fills the gap in applications, as in the papers mentioned above.
Further, this helps us to establish a transitivity theorem without requiring forward or negative invariance.

We consider a differential inclusion (a set-valued differential equation) and also an equilibrium set, not necessarily a point. This generalization is needed to cover evolutionary dynamics in games, since Nash equilibrium may constitute a (connected) set and also a transition may not be uniquely specified when there are multiple best responses.

\section{Definitions and theorems}
We consider an autonomous differential inclusion $\bcV$ such as
$$ \dot\x\in \bcV(\x)$$
on a compact metric $A$-dimensional real space $\cX\subset\R^A$ with $A<\infty$. $T\cX$ stands for the tangent space of $\cX$.\footnote{Below the statements of the definitions follow \cite{SandholmPopText}, a canonical reference book on evolutionary dynamics in games.} As a solution concept for the differential inclusion, we adopt a Carath\'{e}odory solution; that is, a solution path $\{\x_t\}_{t\ge 0}$ must be Lipschitz continuity at every $t\ge 0$ and also differentiable with derivative $\dot\x_t\in \bcV(\x_t)$ at almost every $t$.

Let $X^\ast$ be a nonempty closed set. We say $X^\ast$ is {\bf Lyapunov stable} under $\bcV$ if for any open neighborhood $O$ of $X^\ast$ there exists a neighborhood $O'$ of $A$ such that {\it every} solution path $\{\x^t\}_{t\ge 0}$ that starts from $O'$ remains in $O$. $X^\ast$ is {\bf attracting} if there is a neighborhood $O$ of $X^\ast$ such that {\it every} solution that starts in $O$ converges to $X^\ast$; $O$ is called a basin of attraction to $X^\ast$. If it is the entire space $\cX$, then we say $X^\ast$ is globally attracting. $X^\ast$ is {\bf asymptotically stable} if it is Lyapunov stable and attracting; it is globally asymptotically stable if it is Lyapunov stable and globally attracting.

\paragraph{Lyapunov stability theorem.}

\begin{thm}[Lyapunov stability theorem]\label{thm:Lyapunov_DI}
	Let $X^\ast$ be a non-empty closed set in a compact metric space $\cX$ with tangent space $T\cX$, and $X'$ be a neighborhood of $X^\ast$. Suppose that  continuous function $W:\cX\to\R$ and lower semicontinuous function $\tilde W:\cX\to\R$ satisfy (a) $W(\x)\ge 0$ and $\tilde W(\x)\le 0$ for all $\x\in X'$ and (b) $\cl X'\cap W^{-1}(0)= X'\cap\tilde W^{-1}(0)=X^\ast$. In addition, assume that $W$ is Lipschitz continuous in $\x\in X'$. i) If a differential inclusion $\bcV:\cX\to T\cX$ satisfies\footnote{$D$ denotes differentiation, so $DW(\x)=\tdif{W}{\x}(\x)=[\tpdif{W}{x_1}(\x),\ldots,\tpdif{W}{x_A}(\x)]$.}
	\begin{equation}
	DW(\x)\dot\x \le \tilde W(\x) \qquad \text{for any }\dot\x\in\bcV(\x)  \label{eq:LCondtn_DI}
	\end{equation}
	whenever $W$ is differentiable at $\x\in X'$, then $X^\ast$ is asymptotically stable under $\bcV$. ii) If $X'$ is forward invariant, i.e., every Carath\'{e}odory solution path $\{\x_t\}$ starting from $X'$ at time $0$ remains in $X'$ for all moments of time $t\in[0,\infty)$, then $X'$ is a basin of attraction to $X^\ast$.
\end{thm}
We call $W$ a \textit{Lyapunov function} and $\tilde W$ a \textit{decaying rate function}. Note that we allow for multiplicity of transition vectors, while requiring functions $W$ and $\tilde W$ to be well defined (the uniqueness of the values) as functions of state variable $\x$, independently of the choice of transition vector $\dot\x$ from $\bcV(\x)$.

In a standard Lyapunov stability theorem (e.g. \citet[\S5.5.3]{Robinson_DynSysEd2}) for a differential equation, a decaying rate function $\tilde W$ is not explicitly required while $\dot W$ is assumed to be (strictly) negative until $\x$ reaches the limit set $X^\ast$. The most significant difference is the requirement of lower semicontinuity of $\tilde W$. This assures the existence of a lower bound on the decaying rate $\dot W(\x)\le \bar w<0$ in a hypothetical case in which $\x$ remained out of an arbitrarily small neighborhood of $X^\ast$ for an arbitrarily long period of time. This excludes the possibility that $\x$ would stay there forever and guarantees convergence to  $X^\ast$ (not only Lyapunov stability, i.e., no asymptotic escape from $X^\ast$).

\citet[Theorem 7]{ZusaiTBRD} modifies the Lyapunov stability theorem for a differential inclusion in \citet[Theorem 8.2]{Smirnov01DI}. While the latter is applicable to a singleton of an equilibrium point, the former allows convergence to a set of equilibria.

\Cref{thm:Lyapunov_DI} in this paper relaxes assumptions in \citet[Theorem 7]{ZusaiTBRD}. The previous version imposes a stronger assumption than \eqref{eq:LCondtn_DI}: every Carath\'{e}odory solution $\{\x_t\}$ starting from $X'$ should satisfy
\begin{equation}
\dot W(\x_t) \le \tilde W(\x_t) \quad \text{for almost all }t\in[0,\infty). \label{eq:LCondtn_DI_tBRD}
\end{equation}
If $X'$ is forward invariant, then \eqref{eq:LCondtn_DI} implies this condition; thus, \cite{ZusaiTBRD} is straightforwardly applied and we can conclude that $X^\ast$ is asymptotically stable and $X'$ is a basin of attraction, as restated in part ii) of \Cref{thm:Lyapunov_DI}.\footnote{Once monotone decrease in $W$ is confirmed for any solution path as in \eqref{eq:LCondtn_DI_tBRD}, convergence to $W^{-1}(0)$ is obtained simply by using Gr\"{o}nwall's inequality.} In part i) of \Cref{thm:Lyapunov_DI} in the current version, we do not require forward invariance of $X'$; a solution trajectory may escape from $X'$ and thus \eqref{eq:LCondtn_DI_tBRD} may not be maintained. Thus, the current version weakens the assumption.

Besides, \eqref{eq:LCondtn_DI} is assumed for every point in the entire space $\cX$ and $\tilde W$ is assumed to be continuous. By checking the places in the proof where the definition of the domain for condition (i) in the theorem\footnote{In the notation of the current version, the condition reads as $W(\x)\ge 0$ and $\tilde W(\x)\le 0$ for all $\x\in\cX$. Thus it corresponds with condition (a) in \Cref{thm:Lyapunov_DI}.} and continuity of $\tilde W$ were used, one can easily find that it is innocuous to replace simplex $\Delta^A\subset\R^A$ with a closed subset $X^\ast$ of a compact metric space $\cX\subset\R^A$ and relax continuity of $\tilde W$  to lower semicontinuity.\footnote{Specifically, $\breve A$ in the proof \citep[p.25]{ZusaiTBRD} should be defined as a subset of $\cl X'$. The continuity of $\tilde W$ was used to assure the existence of the minimum of $\tilde W$ in $\breve A$; for this, lower semicontinuity is sufficient. Then, with the observation that forward invariance of $X'$ implies \eqref{eq:LCondtn_DI_tBRD}, the proof for the previous version applies to part ii) of the current version.}

\paragraph{Transivitity theorem.}
\begin{thm}[Transivitity theorem]\label{clm:Lyap_trans}
	Let $X_1\supset X_2 \supset X^\ast$ be three non-empty subsets of a compact metric space $\cX$; assume that $X^\ast$ is closed and $X_1$ is open. Suppose that two Lipschitz continuous functions $W_1,W_2:X_1\to\R$ and two lower semicontinuous functions $\tilde W_1,\tilde W_2:X_1\to\R$ satisfy the following assumptions: for any $\x\in X_1$,
	\begin{enumerate}[a)]
		\item i) $W_1(\x)\ge 0$, ii) $\tilde W_1(\x)\le 0$, and iii) $\cl X_1\cap {W_1}^{-1}(0)=\cl X_1\cap {\tilde W_1}^{-1}(0)=\cl X_2$;
		\item i) $W_2(\x)\ge 0$, ii) $\left[\x\in X_2 ~\Rightarrow~ \tilde W_2(\x)\le 0 \right]$, and iii) $\cl X_2\cap{W_2}^{-1}(0)=\cl X_2\cap{\tilde W_2}^{-1}(0)=X^\ast$;
		\item $\tilde W_1(\x)+\tilde W_2(\x)\le 0$.
	\end{enumerate}
	Furthermore, assume that
	$$ \text{a-iv) } DW_1(\x) \dot\x \le \tilde W_1(\x), \qquad  \text{b-iv) } DW_2(\x) \dot\x \le \tilde W_2(\x) \qquad\text{for any $\dot\x\in\bcV(\x)$},$$
	whenever $W_1$ and $W_2$ are differentiable at $\x\in X_1$. Then, $X^\ast$ is asymptotically stable under $\bcV$.
\end{thm}

Conditions a) imply that $W_1$ works as a Lyapunov function for local asymptotic stability of $X_2$ in (a subset of) $X_1$ and conditions b) imply that $W_2$ works as a Lyapunov function for local asymptotic stability of $X^\ast$ in (a subset of) $X_2$. So, we may jump to conclude that $X^\ast$ is asymptotically stable in $X_1$. If $X_1$ is indeed forward invariant and thus a basin of attraction to $X_2$ and $X_2$ is forward and also strongly negative invariant, then the transitivity theorem as in \citet[\S II.5.3.D]{Conley78_IsoInvSets} and \citet[Thorem 3]{OSTsBRD} guarantees asymptotic stability of $X^\ast$ with $X_1$ being a basin of attraction to $X^\ast$.

However, we do not assume invariance of these sets in our theorem. Even though \Cref{thm:Lyapunov_DI} eventually assures the existence of some forward invariant subset of $X_2$ from conditions b), the Lyapunov function $W_1$ in conditions a) guarantees convergence only to $X_2$, but not necessarily to this forward invariant subset. Furthermore, the standard transitivity theorem also requires strongly negative invariance of $X_2$, which may not be satisfied by the basin of attraction that we could find using \Cref{thm:Lyapunov_DI}.

The above theorem  avoids this issue by imposing condition c) to hold in the whole $X_1$, which we use to construct a Lyapunov function in $X_1$ to $X^\ast$. Then, we apply \Cref{thm:Lyapunov_DI} and thus the basin of attraction to $X^\ast$ is smaller than $X_1$. When current state $\x$ is in the interim subset $X_2$, conditions a) and b-ii) imply condition c). Condition c) deals with the case that $\x$ is still out of $X_2$ and thus $\tilde W_2(\x)$ may be positive (and thus $W_2$ may be increasing over time). Condition c) requires this to be suppressed by $\tilde W_1$, which must be negative in the entire $X_1$ by condition a-iii).

\paragraph{Applications in game theory.} Local stability of an equilibrium is one of the fundamental issues in game theory. A game may exhibit multiple Nash equilibria and thus each equilibrium may not be globally stable; thus, while we investigate global stability of the set of equilibria, we check local stability of each isolated equilibrium or each isolated connected set of equilibria.

A canonical condition to derive local stability under economically reasonable dynamics is negative payoff externality, specifically called \textit{self-defeating externality }by \citet{HofSand09JET_StableGames}. We regard \textit{a (population) game} as a mapping (payoff function) $\F$ from a distribution of strategies among (continuously many) agents $\x\in \Delta^A$ to a payoff vector $\bpi=\F(\x)\in\R^A$. Self-defeating externality boils down to negative semidefiniteness of $\z\cdot D\F(\x)\z$; a marginal deviation $\z$ in the strategy distribution from $\x$ triggers the change in the payoff vector, which is approximated as $D\F(\x)\z$. Self-defeating externality imposes a negative correlation between $\z$ and $D\F(\x)\z$; a strategy whose share increases by this deviation should face a decrease in its payoff and thus becomes less disadvantageous.

From this condition on $\F$, economists naturally expect agents to return to the equilibrium, while it needs to formulate how agents revise their choices of strategies in response to payoff changes. An evolutionary dynamic is a dynamic of the strategy distribution, constructed as a mean dynamic of agents each of whom revises its own strategy upon a receipt of a revision opportunity following a Poisson process \citep{SandholmPopText}. \cite{Sandholm10TELocalStbl} considers an equilibrium that is essentially characterized by self-defeating externality, called a regular evolutionary stable state and attempts to prove its local stability under several canonical classes of evolutionary dynamics, such as excess payoff dynamics and pairwise payoff comparison dynamics. Similarly, \cite{ZusaiTBRD} proves it for another class of dynamics, called tempered best response dynamics. These papers refer to a standard version of Lyapunov stability theorem as mentioned after \Cref{thm:Lyapunov_DI}, which requires monotone decrease in the Lyapunov function along with each solution path. However, these papers confirm its decrease only at each point in a neighborhood of a regular ESS, where the self-defeating externality holds. It is left unchecked whether this neighborhood is forward invariant. Our \Cref{thm:Lyapunov_DI} fixes this overlooked point.

In both the two papers, the Lyapunov function is decomposed to two parts; one is the aggregate of possible payoff gains for agents from revisions of strategies and another is the mass of agents who currently choose the strategies that are to be abandoned in the regular ESS. Thanks to this decomposition, we can apply \Cref{clm:Lyap_trans}. \cite{Zusai_dynStblESS} uses it to generalize their results to a broader class of economically natural dynamics.

\section{Proofs}
\subsection*{Proof of \Cref{thm:Lyapunov_DI}}

\begin{proof}
	Here we prove the difference in part i) from \citet[Theorem 7]{ZusaiTBRD}. For this, we focus on the case of $X'\subsetneq \cX$ and find a forward invariant subset of $X'$. Once we find it, any Carath\'{e}odory solution starting from the forward invariant subset remains there and thus satisfies \eqref{eq:LCondtn_DI_tBRD} as \eqref{eq:LCondtn_DI} holds for $\x=\x_t$ at each time $t\in\R_+$. Then, \citet[Theorem 7]{ZusaiTBRD} is applied and assures asymptotic stability of $X^\ast$ while having the forward invariant subset as a basin of attraction.

	First, construct a distance from point $\x\in\cX$ to $X^\ast$ based on the metric on $\cX$, say $d:\cX\times\cX\to\R_+$, by
	$$ d_\ast(\x) \coloneqq \min_{\x^\ast\in X^\ast} d(\x,\x^\ast).$$
	Since $X^\ast$ is a non-empty compact set and $d(\x,\x^\ast)$ is continuous in $\x^\ast$ when $\x$ is fixed, Weierstrass theorem assures the existence of the minimum in the above definition of $d_\ast(\x)$. This $d_\ast$ satisfies
	$$ d_\ast(\x)\ge 0; \qquad d_\ast(\x)=0 \ \Leftrightarrow\ \x\in X^\ast.$$

	Let $\bar d$ be the shortest distance from the complement of $X'$ to $X^\ast$:
	\begin{equation}
	\bar d \coloneqq \min_{\x\in \cX\setminus X'} d_\ast(\x). \label{eq:prf:Lyapunov_DI_dfnBarD}
	\end{equation}
	Maximum theorem guarantees continuity of $d_\ast:\cX\to\R_+$ by continuity of $d(\x,\x^\ast)$ in both $\x$ and $\x^\ast$. Besides, $\cX\setminus X'$ is a non-empty compact subset by $X'\subsetneq \cX$ and the openness of $X'$. Hence, the minimum in \eqref{eq:prf:Lyapunov_DI_dfnBarD} exists. It follows that
	\begin{equation}\label{eq:prf:Lyapunov_DI_barD}
	\bar d>0; \qquad d_\ast(\x)<\bar d \ \Rightarrow\ \x\in X'.
	\end{equation}

	\begin{figure}[t!]
		\centering
		\includegraphics[scale=0.6]{Fig/LyapunovDIprf_ForwardInv.pdf}
		\caption{Sets in the proof of \Cref{thm:Lyapunov_DI}. $X^\ast$ is the black area in the center and $X''$ is the light gray area. $\cl X'$ is the entire oval, with the outermost outline. $X'_0$ is the dark gray area, including the both boundaries.}
	\end{figure}

	Define set $X'_0\subset\cl X'$ by
	$$ X'_0 \coloneqq \cl X' ~\cap ~ {d_\ast}^{-1}([\bar d/2, \infty)).$$
	Since both $\cl X'$ and ${d_\ast}^{-1}([\bar d/2, \infty))$ are closed, $X'_0$ is closed and thus compact in $\cX$. It is not empty, as proven here. Suppose $X_0'=\emptyset$; then, any $\x\in\cX$ with $d_\ast(\x)\ge \bar d/2$ must be out of $\cl X'$. On the other hand, since $\cl X'$ is not empty, $X'$ has at least one boundary point $\x^0$; then, $d_\ast(\x^0)\ge \bar d$.\footnote{We can make a sequence converging to $\x^0$ from elements out of $X'$, whose distance from $X^\ast$ cannot be smaller than $\bar d$ by \eqref{eq:prf:Lyapunov_DI_dfnBarD}.} By the former statement, this implies $\x^0\notin\cl X'$ but it contradicts with $\x^0$ being on the boundary of $X'$; hence, $X'_0$ cannot be empty.

	Let $\bar w$ be the minimum of $W$ in $X'_0$;
	$$ \bar w \coloneqq \min_{\x\in X'_0} W(\x).$$
	Since $X'_0$ is compact and nonempty and $W$ is (Lipschitz) continuous, the minimum exists. Furthermore, it is positive; we have $X_0'\subset\cl X'$ by construction and $W(\x)\ge 0$ for all $\x\in\cl X'$ by condition (a) and continuity of $W$, while no element $\x\in X'_0$ belongs to $X^\ast$ since $d_\ast(\x)\ge \bar d>0$ for any $\x\in X'_0$. Because $X^\ast =\cl X'\cap W^{-1}(0)$ by condition (b) and $X'_0\subset \cl X'$, it implies $\x\in X'_0 \Rightarrow W(\x)>0$. Hence we have $\bar w>0$; by the definition of $\bar w$, we have
	\begin{equation}\label{eq:prf:Lyapunov_DI_barW}
	\underbrace{\left[ \x\in\cl X' \text{ and } d_\ast(\x) \ge \bar d/2\right]}_{\text{ i.e., }\x\in X'_0}  \ \Rightarrow\ W(\x)\ge \bar w.
	\end{equation}

	Define set $X''\subset X'$ by
	\begin{equation}
	X''=W^{-1}([0,\bar w/2)) \cap X'. \label{eq:prf:Lyapunov_DI_dfnX''}
	\end{equation}
	This set is an (open) neighborhood of $X^\ast$ by $X^\ast\subset X''$, since  $W=0$ at anywhere in $X^\ast$ and $X^\ast \subset X'$. Now we prove that $X''$ is wholly contained in set $d_\ast^{-1}([0,\bar d/2))$. Assume that there exists $\x\in X''$ such that $d_\ast(\x)\ge \bar d/2$. These jointly imply $W(\x)\ge \bar w$ by \eqref{eq:prf:Lyapunov_DI_barW} since $\x\in X''\subset X'\subset \cl X'$. However, this contradicts with $W(\x)\in [0,\bar w/2)$ for $\x$ to belong to $X''$. Hence, we have
	\begin{equation}
	\x\in X'' \ \Rightarrow\ d_\ast(\x)< \bar d/2. \label{eq:prf:Lyapunov_DI_X''}
	\end{equation}

	Now we prove $X''$ is forward invariant. To verify it by contradiction, assume that there is a Carath\'{e}odory solution trajectory $\{\x_t\}$ starting from $X''$ but escaping $X''$ at some moment of time:
	\begin{equation}\label{eq:prf:Lyapunov_DI_Hypo}
	\x_0\in X'', \qquad \text{ and }\x_T\notin X'' \text{ at some }T>0.
	\end{equation}
	The statement $\x_T\notin X''$ means $\x_T\notin X'$ or $W(\x_T)>\bar w/2$ by \eqref{eq:prf:Lyapunov_DI_dfnX''}. In the former case, we have $d_\ast(\x_T)\ge \bar d$ by \eqref{eq:prf:Lyapunov_DI_dfnBarD} while $d_\ast(\x_0)< \bar d/2$ by \eqref{eq:prf:Lyapunov_DI_X''}. By continuity of $d_\ast(\x)$ in $\x$ and of $\x_t$ in $t$ on a Carath\'{e}odory solution trajectory $\{\x_t\}$, $d_\ast(\x_t)$ is continuous in $t$; hence, there exists a moment of time $T'\in (0,T)$ such that $d_\ast(\x_{T'})=0.9 \bar d \in(0.5 \bar d, \bar d)\subset (d_\ast(\x_0), d_\ast(\x_T))$. At this point, $\x_{T'}\notin X''$ by \eqref{eq:prf:Lyapunov_DI_X''} while $\x_{T'}\in X'$ by \eqref{eq:prf:Lyapunov_DI_barD}; thus, $W(\x_{T'})\ge \bar w/2$ by \eqref{eq:prf:Lyapunov_DI_dfnX''}. Hence, the first case of escaping $X''$ implies the existence of $T'>0$ such that
	$$  W(\x_{T'})\ge \bar w/2 \qquad\textit{ and }\qquad \x_{T'}\in X'. $$
	In the second (but not the first) case, we have $W(\x_T)\ge \bar w/2$ but $\x_T\in X'$; that is, the above statement holds with $T'=T
	.$

	This implies the existence of $\bar T\in(0, T']$ such that
	\begin{equation}
	W(\x_{\bar T})\ge \bar w/2, \qquad\text{ and }\qquad \left[ \x_t\in X' \ \text{ for all }t< \bar T \right]. \label{eq:prf:Lyapunov_DI_barT}
	\end{equation}
	To prove it, assume $\x_{t'}\notin X'$ at some $t'< T'$, i.e., the negation of the latter condition with $\bar T=T'$; if there is no such $t'\le T'$, then it suggests that the claim \eqref{eq:prf:Lyapunov_DI_barT} holds at $\bar T=T'$ by the fact  $W(\x_{T'})\ge \bar w/2$.

	By \eqref{eq:prf:Lyapunov_DI_barD}, the hypothesis $\x_{t'}\notin X'$ implies $d_\ast(\x_{t'})\ge \bar d$. Again, by continuity of $d_\ast(\x_t)$ in $t$, the set $\{t\le t' \mid  d_\ast(\x_t)\ge \bar d\}$ is closed and thus compact; by the fact $d_\ast(\x_0)<\bar d/2$, this implies the existence of the minimum $\bar T$ in this set and $\bar T>0$. That is, we have $d_\ast(\x_t)<\bar d$ for all $t< \bar T$ while $d_\ast(\x_{\bar T})=\bar d$. The former implies $\x_t\in X'$ for all $t<\bar T$ by \eqref{eq:prf:Lyapunov_DI_barD} and the latter implies $W(\x_{\bar T})\ge \bar w$ by $\x_{\bar T}=\lim_{t\to\bar T}x_t\in\cl X'$ and \eqref{eq:prf:Lyapunov_DI_barW}. Thus, the above claim \eqref{eq:prf:Lyapunov_DI_barT} holds at this $\bar T\in(0, T']$. Since condition (a) and \eqref{eq:LCondtn_DI} hold almost everywhere in $X'$, we have $\dot W(\x_\tau)\le \tilde W(\x_\tau)\le 0$ at almost all $\tau< \bar T$;\footnote{A Carath\'{e}odory solution trajectory is differentiable at almost all moments of time, though it may not be so at all moments.} thus, we have
	$$ W(\x_{\bar T})\le W(\x_0)+\int_0^{\bar T}  \tilde W(\x_\tau) d\tau\le W(\x_0).$$
	Since $W(\x_0)<\bar w/2$ by $\x_0\in X''$, we have $W(\x_{\bar T})<\bar w/2$ in \eqref{eq:prf:Lyapunov_DI_barT}. This contradicts with $W(\x_{\bar T})\ge \bar w/2.$

	Therefore, the hypothesis \eqref{eq:prf:Lyapunov_DI_Hypo} cannot hold: any Carath\'{e}odory solution trajectory $\{\x_t\}$ starting from $X''$ cannot escape $X''$ at any moment of time. That is, $X''$ is forward invariant.
\end{proof}

\subsection*{Proof of \Cref{clm:Lyap_trans}}
\begin{proof}
	Define a Lyapunov function $W:X_1\to\R$ and a decaying rate function $\tilde W:X_1\to\R$ by
	$$ W(\x) \coloneqq 2 W_1(\x)+W_2(\x), \qquad  \tilde W(\x) \coloneqq 2 \tilde W_1(\x)+\tilde W_2(\x) \qquad\text{ for each }\x\in X_1.$$
	Lipschitz continuity of $W_1$ and $W_2$ and lower semicontinuity of $\tilde W_1$ and $\tilde W_2$ are succeeded to those of $W$ and $\tilde W$,  respectively.
	It is immediate from assumptions a-i,iv), b-i,iv) and c) to see that
	\begin{align}
	W(\x)= 2 W_1(\x)+W_2(\x) \ge 0, \notag\\
	\tilde W(\x)= \tilde W_1(\x)+\{\tilde W_1(\x)+\tilde W_2(\x)\}\le 0, \notag\\
	DW(\x)\dot\x =2 DW_1(\x)\dot\x+ DW_2(\x)\dot\x \le 2 \tilde W_1(\x)+\tilde W_2(\x) =\tilde W(\x)
	\label{eq:prf_Lyap_trans_WLyap1}
	\end{align}
	for any $\x\in X_1, \dot\x\in\bcV(\x)$ (for the last equation assuming that $W_1$ and $W_2$ are differentiable at $\x$).

	Further, since $X^\ast\subset X_2$, it follows assumptions a-iii) and b-iii) that $W(\x)=\tilde W(\x)=0$ if $\x\in X^\ast$; thus $X^\ast$ is contained in $\cl X_1\cap W^{-1}(0)$ and $\cl X_1\cap \tilde W^{-1}(0)$ by $X^\ast\subset X_2\subset X_1\subset \cl X_1$. In contrary, assume $W(\x)=0$ at $\x\in\cl X_1$ first. By assumptions a-i) and b-i), it must be the case that $W_1(\x)=0$ and $W_2(\x)=0$. The former implies $\x\in\cl X_2$ by assumption a-iii). Together with this, the latter implies $\x\in X^\ast$ by assumption b-iii). Separately from this, now assume $\tilde W(\x)=0$ at $\x\in\cl X_1$. By assumptions a-ii) and c), it must be the case that $\tilde W_1(\x)=0$ and $\tilde W_1(\x)+\tilde W_2(\x)=0$.\footnote{Note that the latter condition alone cannot assure $\tilde W_2(\x)=0$, since $\tilde W_2(\x)$ could take a positive value unless $\x$ is in $X_2$.} The former implies $\x\in\cl X_2$ by assumption a-iii); besides, by plugging the former into the latter, we have $\tilde W_2(\x)=0$. These two statements jointly imply $\x\in X^\ast$ by assumption b-iii). In sum, we have verified
	\begin{equation}
	\cl X_1\cap W^{-1}(0)=\cl X_1\cap {\tilde W}^{-1}(0)=X^\ast. \label{eq:prf_Lyap_trans_WLyap2}
	\end{equation}
	Note that the first equality is due to the fact that $X^\ast\subset X_1$ and thus $X^\ast \cap \bd X_1=\emptyset$ since $X_1$ is open.

	We have verified all the assumptions in \Cref{thm:Lyapunov_DI}; therefore, $X^\ast$ is asymptotically stable. Notice that $X_1$ may not be forward invariant, but part i) of \Cref{thm:Lyapunov_DI} assures that we can make some subset of $X_1$ as a basin of attraction to $X^\ast$.
\end{proof}

\renewcommand\refname{References}
\bibliography{../DZbib}

\end{document}

config: arxiv


TITLE: Determine $y$ explicitly as a function of $x:$ $\ln(1+y)-\ln(y)=\ln(x^3+1)-2 \ln(x-1)$
QUESTION [1 upvotes]: I have taken the exp on both sides to give;
$$e^{\ln(1+y)}-e^{\ln y}=e^{\ln(x^3+1)}-e^{2\ln(x-1)}$$
$$1+y-y=x^3+1-(x-1)^2$$
Then simplified the LHS and factored the RHS to give;
$$y=(x-1)(x^2-2)$$
Is this correct, or have i gone too far down the rabbit hole?

REPLY [1 votes]: Taking the exponent is not linear:
$$a+b=c+d$$
Means:
$$\exp(a+b)=\exp(c+d) $$
$$\exp(a) \exp(b) =\exp(c) \exp(d) $$
My hint to you is to first combine the logarithms using the identities:
$$ \log(a)+\log(b) = \log(ab)$$
$$ \log(a)-\log(b) = \log(\frac{a}{b})$$
$$c \log(a) = \log(a^c) $$
If you use these, you can simplify the terms to two logs on each side and the apply the exponentiation.
You will get something like:
$$\ln\left(\frac{1+y}{y}\right)= \ln\left(\frac{x^3+1}{(x-1)^2}\right)$$
$$\ln\left(1+\frac{1}{y}\right)= \ln\left(\frac{x^3+1}{(x-1)^2}\right)$$
So:
$$1+\frac{1}{y} =\frac{x^3+1}{(x-1)^2}$$


TITLE: Entire even function with exponential type which strictly decreases and vanishes at infinities
QUESTION [2 upvotes]: I am looking for the function $f$ which satisfies the following conditions:

$f$ should be an even entire function, i.e. $f:\mathbb{C}\rightarrow \mathbb{C}, f(z)=f(-z)$.
$f$ should have a real value in both real and imaginary axis, i.e. $f(\mathbb{R}\cup i\mathbb{R})\subset \mathbb{R}$.
$f$ should monotonically decrease on real axis as $|x|\rightarrow 0$: in other words, $f'(x\in \mathbb{R}\setminus {0})\neq 0$.
$f(0)>0$ and $f$ should vanish on the infinity on real axis, i.e. $\lim_{x\in \mathbb{R}\rightarrow \infty }f(x)=0$.
$f$ should be an exponential type, i.e. $|f(z)|\leq A e^{c|z|}$ on large enough $|z|$ for some $A,c$.

The motivation of this question is to find the entire function with exponential type whose shape is similar with the Gaussian function $e^{-x^2}$. We can easilly see that gaussian function satisfies all the conditions except condition 5.
Another example I have tried is $J_0(z)$ or $\sin x/x$, which satisfies every conditions but condition 3 and 4. There are various examples which satisfies all the properties above but condition 1., for example, $(1/1+z^2)$.
Maybe such function $f$ exists but not in the closed form, for example, in the form of Taylor expansion: if so, then I want its rough form. Maybe such function never exists due to the property of exponential type functions: if so, then I want a rough proof of it.

REPLY [0 votes]: It seems that I myself showed the non-existence of such function. [Edit-Thanks to @conrad, Taking $a=0$ does not directly means that the function is exponential-type zero, showing that the argument below is flawed.]
Since $f$ is an entire exponential-type function, we may write
$$
f(z)=z^m e^{az+b} \prod_i (1-z_i/a_i)
$$
By scaling let $f(0)=1$. Then, with the fact that $f$ is even, $a=b=m=0$. This implies that
$$
f(z)=\prod_i (1-z_i/a_i)
$$
which now shows that $f$ is an exponential-type zero function. Now since Bernstein Inequality shows, for an exponential-type $\tau$ function, if $f$ is bounded on the real axis,
$$
|f'(x)|\leq \tau |f(x)|,\quad x\in \mathbb{R}
$$
but since $\tau=0$, $|f'(x)|=0$, showing $f=0$.


text:
\begin{document}
\maketitle \abstract
Let $\mathcal{A}$ be a unital associative algebra and $\mathcal{M}$ be an $\mathcal{A}$-bimodule.
A linear mapping $\varphi$ from $\mathcal{A}$ into an $\mathcal{A}$-bimodule $\mathcal{M}$ is called a Lie derivation if
$\varphi[A,B]=[\varphi(A),B]+[A,\varphi(B)]$ for each $A,B$ in $\mathcal{A}$,
and $\varphi$ is called a \emph{local Lie derivation} if for every $A$ in $\mathcal{A}$,
there exists a Lie derivation $\varphi_{A}$ (depending on $A$) from $\mathcal{A}$ into $\mathcal{M}$
such that $\varphi(A)=\varphi_{A}(A)$. In this paper, we prove that every local Lie derivation on von Neumann algebras is a Lie derivation;
and we show that if $\mathcal M$ is a type I von Neumann algebra with atomic lattice
of projections, then every local Lie derivation on $LS(\mathcal M)$ is a Lie derivation.

\
{\textbf{Keywords:}} Lie derivation, local Lie derivation, von Neumann algebra, locally measurable operator.

\
{\textbf{Mathematics Subject Classification(2010):}} 46L57; 47L35; 46L50

\
\section{Introduction}\

Let $\mathcal{A}$ be a unital associative algebra
over the complex field $\mathbb{C}$ and $\mathcal{M}$ be an $\mathcal{A}$-bimodule.
An linear mapping $\delta$ from $\mathcal{A}$ into $\mathcal{M}$ is
called a \emph{derivation} if $\delta(AB)=\delta(A)B+A\delta(B)$ for each $A$ and $B$ in $\mathcal{A}$.
In particular, a derivation $\delta_M$ defined by $\delta_{M}(A)=MA-AM$ for every $A$ in $\mathcal{A}$ is called an inner derivation,
where $M$ is a fixed element in $\mathcal{M}$.

In \cite{sakai}, S. Sakai proves that every derivation on von Neumann algebras is an inner derivation.
In \cite{christensen}, E. Christensen shows that every derivation on nest algebras on a Hilbert space $\mathcal H$
is an inner derivation.
For more information on derivations and inner derivations, we refer to \cite{DuZhang,DuZhang1,hejun}.

In \cite{R. Kadison,D. Larson}, R. Kadison and D. Larson introduce the concept of local derivations.
A linear mapping $\delta$ from $\mathcal{A}$ into
$\mathcal{M}$ is called a \emph{local derivation} if for every $A$ in $\mathcal{A}$,
there exists a derivation $\delta_{A}$ (depending on $A$) from $\mathcal{A}$ into $\mathcal{M}$
such that $\delta(A)=\delta_{A}(A)$.

In \cite{R. Kadison}, R. Kadison proves that every continuous local
derivation from a von Neumann algebra into its dual Banach
module is a derivation. In \cite{D. Larson}, D. Larson and A. Sourour prove that
if $X$ is a Banach space, then every local derivation on $B(X)$ is a derivation.
In \cite{B. Johnson}, B. Jonson shows that every local derivation from a $C^*$-algebra
into its Banach bimodule is a derivation. In \cite{HadwinLi1, HadwinLi2}, D. Hadwin and J. Li
characterize local derivations on non self-adjoint operator algebras such as nest algebras and CDCSL algebras.

A linear mapping $\varphi$ from $\mathcal{A}$ into an $\mathcal{A}$-bimodule $\mathcal{M}$ is called a Lie derivation if
$\varphi[A,B]=[\varphi(A),B]+[A,\varphi(B)]$ for each $A$ and $B$ in $\mathcal{A}$, where $[A,B]=AB-BA$ is the usual Lie product.
A Lie derivation $\varphi$ is said to be standard if it can be decomposed as $\varphi=\delta+\tau$, where $\delta$ is a derivation from
$\mathcal{A}$ into $\mathcal{M}$ and $\tau$ is a linear mapping from $\mathcal{A}$ into $\mathcal{Z}(\mathcal{A},\mathcal{M})$ such that $\tau[A,B]=0$ for
each $A$ and $B$ in $\mathcal{A}$, where $\mathcal{Z}(\mathcal{A},\mathcal{M})=\{M\in\mathcal{M}:AM=MA~\mbox{for every}~A~\mbox{in}~\mathcal{A}\}$.

In \cite{Johnson1}, B. Johnson proves that every continuous Lie derivation from a $C^{*}$-algebra into its Banach bimodule is standard.
In \cite{M. Mathieu}, M. Mathieu and A. Villena prove that every Lie derivation on a $C^*$-algebra is standard.
In \cite{W. Cheung}, W. Cheung characterizes Lie derivations on triangular algebras. In \cite{F. Lu3}, F. Lu proves that every Lie derivation on a
completely distributed commutative subspace lattice algebra is standard. In \cite{D. Benkovic}, D. Benkovi$\check{c}$ proves that every Lie derivation
on matrix algebra $M_n(\mathcal A)$ is standard, where $n\geq2$ and $\mathcal A$ is a 2-torsion free unital algebra.

Similar to local derivations, In \cite{L. Chen}, L. Chen, F. Lu and T. Wang introduce the concept of local Lie derivations.
A linear mapping $\varphi$ from $\mathcal{A}$ into
$\mathcal{M}$ is called a \emph{local Lie derivation} if for every $A$ in $\mathcal{A}$,
there exists a Lie derivation $\varphi_{A}$ (depending on $A$) from $\mathcal{A}$ into $\mathcal{M}$
such that $\varphi(A)=\varphi_{A}(A)$.

In \cite{L. Chen}, L. Chen, F. Lu and T. Wang prove that every local Lie derivation on $B(X)$ is a Lie derivation, where $X$ is a Banach space of dimension exceeding 2.
In \cite{L. Chen2},  L. Chen and F. Lu prove that every local Lie derivation on nest algebras is a Lie derivation.
In \cite{liudan}, D. Liu and J. Zhang prove that under certain conditions, every local Lie derivation on triangular algebras is a Lie derivation.
In \cite{hejun1}, J. He, J. Li, G. An and W. Huang prove that every local Lie derivation on some algebras such as finite von Neumann algebras, nest algebras, Jiang-Su algebra and UHF algebras
is a Lie derivation.

Compare with the characterizations of derivations on Banach algebras, investigation of
derivations on  unbounded operator algebras begin much later.

In \cite{Segal}, I. Segal studies the theory of noncommutative integration,
and introduces various classes of non-trivial $*$-algebras of unbounded
operators. In this paper, we mainly consider
the $*$-algebra $S(\mathcal M)$ of all measurable operators
and the $*$-algebra $LS(\mathcal M)$ of all locally measurable operators
affiliated with a von Neumann algebra $\mathcal M$.
In \cite{Segal}, I. Segal shows that the algebraic and topological properties
of the measurable operators algebra $S(\mathcal M)$ are similar to the von Neumann algebra $\mathcal M$.
If $\mathcal M$ is a commutative von Neumann algebra,
then $\mathcal M$ is $*$-isomorphic to the algebra
$L^\infty(\Omega,\Sigma,\mu)$ of all essentially bounded measurable
complex functions on a measure space $(\Omega,\Sigma,\mu)$; and
$S(\mathcal M)$ is $*$-isomorphic to the algebra $L^0(\Omega,\Sigma,\mu)$
of all measurable almost everywhere finite complex-valued functions on
$(\Omega,\Sigma,\mu)$. In \cite{Ber1}, A. Ber, V. Chilin and F. Sukochev show that there exists a derivation on $L^0(0,1)$
is not an inner derivation, and the derivation is discontinuous in the measure topology.
This result means that the properties of derivations on $S(\mathcal M)$
are different from the derivations
on $\mathcal M$.

In \cite{Albeverio1, Albeverio2}, Albeverio, Ayupov and Kudaybergenov study
the properties of derivations on various classes of measurable algebras.
If $\mathcal M$ is a type I von Neumann algebra,
in \cite{Albeverio1}, the authors prove that every derivation
on $LS(\mathcal M)$ is an inner derivation if and only if it is $\mathcal Z(\mathcal M)$ linear;
in \cite{Albeverio2}, the authors give the decomposition form of derivations on $S(\mathcal M)$
and $LS(\mathcal M)$; they also prove that if $\mathcal M$ is a type $\mathrm{I}_\infty$ von Neumann algebra,
then every derivation on $S(\mathcal M)$ or $LS(\mathcal M)$ is an inner derivation.
If $\mathcal M$ is a properly infinite von Neumann algebra,
in \cite{Ber3}, A. Ber, V. Chilin and F. Sukochev prove that
every derivation on $LS(\mathcal M)$ is continuous with respect
to the local measure topology $t(\mathcal M)$; and in \cite{Ber2}, the authors
show that every derivation on $LS(\mathcal M)$ is an inner derivation.
In \cite{Albeverio3}, S. Albeverio and S. Ayupov give a characterization of local derivations
on $S(\mathcal M)$, where $\mathcal M$ is an abelian von Neumann algebra.
In \cite{D. Hadwin}, D. Hadwin and J. Li prove that if $\mathcal M$ is a von Neumann
algebra without abelian direct summands, then every local derivation
on $LS(\mathcal M)$ or $S(\mathcal M)$ is a derivation.
In \cite{V. Chilin}, V. Chilin and I. Juraev  show that every Lie derivation on
$LS(\mathcal M)$ or $S(\mathcal M)$ is standard.

This paper is organized as follows. In Section 2, we recall the definitions of algebras of measurable operators and local measurable operators.

In Section 3, we generalize the Corollary 3.2 in \cite{hejun1} and prove that every local Lie derivation
on von Neumann algebras is a Lie derivation.

In Section 4, we prove that if $\mathcal M$ is a type I von Neumann algebra with an atomic lattice
of projections, then every local Lie derivation on $LS(\mathcal M)$ is a Lie derivation.

\section{Preliminaries}\

Let $\mathcal H$ be a complex Hilbert space
and $B(\mathcal H)$ be the algebra of all bounded linear operators on $\mathcal H$.
Suppose that $\mathcal{M}$ is a von Neumann algebra on
$\mathcal{H}$ and
$\mathcal Z(\mathcal M)=\mathcal M\cap\mathcal M'$
is the center of $\mathcal M$, where
$$\mathcal M'=\{a\in B(\mathcal H):ab=ba~\mathrm{for~every}~b~\mathrm{in}~\mathcal M\}.$$
Denote by
$\mathcal P(\mathcal M)=\{p\in\mathcal M:p=p^*=p^2\}$
the lattice of all projections in $\mathcal M$ and by $\mathcal P_{fin}(\mathcal M)$
the set of all finite projections in $\mathcal M$. For each $p$ and $q$
in $\mathcal P(\mathcal M)$, if we define the
inclusion relation $p\subset q$ by $p\leq q$, then
$\mathcal P(\mathcal M)$ is a complete lattice. Suppose that
$\{p_l\}_{l\in\lambda}$ is a family of projections in $\mathcal M$, we denote
$${\sup_{{l\in\lambda}}}p_l=\overline{\bigcup_{l\in\lambda} p_l\mathcal H}~~\mathrm{and}~~
{\inf_{{l\in\lambda}}}p_l=\bigcap_{l\in\lambda} p_l\mathcal H.$$
If $\{p_l\}_{l\in\lambda}$
is an orthogonal family of projections in $\mathcal M$, then we have that
$${\sup_{{l\in\lambda}}}p_l=\sum_{l\in\lambda}p_l.$$

Let $x$ be a closed densely defined linear operator on $\mathcal H$
with the domain $\mathcal D(x)$, where $\mathcal D(x)$ is a linear subspace of $\mathcal H$.
$x$ is said to be \emph{affiliated} with $\mathcal M$,
denote by $x\eta\mathcal M$, if
$u^*xu=x$ for every unitary element $u$ in $\mathcal{M}'$.

A linear operator affiliated with $\mathcal M$
is said to be \emph{measurable} with respect to
$\mathcal M$, if there exists a sequence $\{p_n\}_{n=1}^{\infty}\subset\mathcal P(\mathcal M)$ such that $p_n\uparrow1$,
$p_n(\mathcal H)\subset\mathcal D(x)$ and $p_n^{\bot}=1-p_n\in \mathcal P_{fin}(\mathcal M)$ for every $n\in\mathbb{N}$,
where $\mathbb{N}$ is the set of all natural numbers. Denote by $S(\mathcal M)$ the set of all measurable operators
affiliated with the von Neumann algebra $\mathcal M$.

A linear operator affiliated with $\mathcal M$
 is said to be \emph{locally measurable} with respect to
$\mathcal M$, if there exists a sequence $\{z_n\}_{n=1}^{\infty}\subset\mathcal P(\mathcal Z(\mathcal M))$
such that $z_n\uparrow1$ and $z_nx\in S(\mathcal M)$ for every $n\in\mathbb{N}$.
Denote by $LS(\mathcal M)$ the set of all locally measurable operators
affiliated with the von Neumann algebra $\mathcal M$.

In \cite{M. Muratov}, Muratov and Chilin prove that $S(\mathcal M)$ and $LS(\mathcal M)$
are both unital $*$-algebras and $\mathcal M\subset S(\mathcal M)\subset LS(\mathcal M)$;
the authors also show that if $\mathcal M$ is a finite von Neumann algebra
or $\mathrm{dim}(\mathcal Z(\mathcal M))<\infty$, then
$S(\mathcal M)=LS(\mathcal M)$; if $\mathcal M$ is a type III von Neumann algebra and
$\mathrm{dim}(\mathcal Z(\mathcal M))=\infty$, then $S(\mathcal M)=\mathcal M$ and
$LS(\mathcal M)\neq\mathcal M$.

\section{Local Lie derivations on von Neumann algebras}\

In this section, we consider local Lie derivations on von Neumann algebras. To prove
our main theorem, we need the following lemma.

\begin{lemma}\label{01}
Let $\mathcal{A}_1$ and $\mathcal{A}_2$ be two unital algebras and $\mathcal{A}=\mathcal{A}_1\bigoplus\mathcal{A}_2$.
If the following five conditions hold:\\
$(1)$ each Lie derivation on $\mathcal{A}$ is standard;\\
$(2)$ each derivation on $\mathcal{A}$ is inner;\\
$(3)$ each local derivation on $\mathcal{A}$ is a derivation;\\
$(4)$ $\mathcal{Z}(\mathcal{A}_1)\cap [\mathcal{A}_1,\mathcal{A}_1]=\{0\}$;\\
$(5)$ $\mathcal{A}_2=[\mathcal{A}_2,\mathcal{A}_2]$,\\
then every local Lie derivation on $\mathcal{A}$ is a Lie derivation.
\end{lemma}
\begin{proof}
Denote the units of $\mathcal{A}$, $\mathcal{A}_1$ and $\mathcal{A}_2$ by $I$, $P$ and $Q$, respectively.
For each $A$ in $\mathcal A$, we have that $A=PA+QA=A_1+A_2$, where $A_i\in\mathcal{A}_i,~i=1,2$.

In the following we suppose that $\varphi$ is a local Lie derivation on $\mathcal{A}$.

By the definition of local Lie derivation, we know that for every $A_1$ in $\mathcal{A}_1$,
there exists a Lie derivation $\varphi_{A_1}$ on $\mathcal{A}$ such that $\varphi(A_1)=\varphi_{A_1}(A_1)$.
Since $\varphi_{A_1}$ is standard and each derivation on $\mathcal{A}$ is inner, we can obtain that
$$\varphi(A_1)=\varphi_{A_1}(A_1)=\delta_{A_1}(A_1)+\tau_{A_1}(A_1)=[A_1,T_{A_1}]+P\tau_{A_1}(A_1)+Q\tau_{A_1}(A_1),$$
where $\delta_{A_1}$ is a derivation on $\mathcal{A}$, $T_{A_1}$ is an element in $\mathcal{A}$, and $\tau_{A_1}$ is a linear mapping from
$\mathcal{A}$ into $\mathcal{Z}(\mathcal{A})$ such that $\tau_{A_1}([\mathcal{A},\mathcal{A}])=0$.

It means that $\varphi$ has a decomposition at $A_1$. Next we show that the decomposition at $A_1$
is unique. Assume there is another decomposition at $A_1$, that is
$$\varphi(A_1)=\varphi^{'}_{A_1}(A_1)=\delta^{'}_{A_1}(A_1)+\tau^{'}_{A_1}(A_1)=[A_1,T^{'}_{A_1}]+P\tau^{'}_{A_1}(A_1)+Q\tau^{'}_{A_1}(A_1),$$
where $\delta^{'}_{A_1}$ is a derivation on $\mathcal{A}$, $T^{'}_{A_1}$ is an element in $\mathcal{A}$ and $\tau^{'}_{A_1}$ is a linear mapping from
$\mathcal{A}$ into $\mathcal{Z}(\mathcal{A})$ such that $\tau^{'}_{A_1}([\mathcal{A},\mathcal{A}])=0$.

Then we have that
$$[A_1,T_{A_1}]+P\tau_{A_1}(A_1)+Q\tau_{A_1}(A_1)=[A_1,T^{'}_{A_1}]+P\tau^{'}_{A_1}(A_1)+Q\tau^{'}_{A_1}(A_1).$$
Thus
$$[A_1,T_{A_1}]-[A_1,T^{'}_{A_1}]=P\tau^{'}_{A_1}(A_1)-P\tau_{A_1}(A_1)+Q\tau^{'}_{A_1}(A_1)-Q\tau_{A_1}(A_1).$$

Since $[A_1,T_{A_1}]-[A_1,T^{'}_{A_1}]$ and $P\tau^{'}_{A_1}(A_1)-P\tau_{A_1}(A_1)\in\mathcal{A}_1$ belong to $\mathcal{A}_1$,
and $Q\tau^{'}_{A_1}(A_1)-Q\tau_{A_1}(A_1)$ belongs to $\mathcal{A}_2$,
we have that $Q\tau^{'}_{A_1}(A_1)-Q\tau_{A_1}(A_1)=0$. Moreover, we can obtain that
$$[A_1,T_{A_1}]-[A_1,T^{'}_{A_1}]=[A_1,PT_{A_1}]-[A_1,PT^{'}_{A_1}]\in[\mathcal{A}_1,\mathcal{A}_1],$$
and
$$P\tau^{'}_{A_1}(A_1)-P\tau_{A_1}(A_1)\in\mathcal{Z}(\mathcal{A}_1).$$

By condition (4), it follows that
$[A_1,T_{A_1}]-[A_1,T^{'}_{A_1}]=P\tau^{'}_{A_1}(A_1)-P\tau_{A_1}(A_1)=0.$
It implies that $\delta_{A_1}(A_1)=\delta^{'}_{A_1}(A_1)$ and $\tau_{A_1}(A_1)=\tau^{'}_{A_1}(A_1)$.
Hence the decomposition is unique.

Now we have $\varphi|_{\mathcal{A}_1}=\delta_1+\tau_1$, where $\delta_1$ is a mapping from $\mathcal{A}_1$ into $\mathcal{A}_1$
such that
$\delta_1(A_1)=[A_1,S_{A_1}]$ for some element $S_{A_1}$ in $\mathcal{A}_1$, and $\tau_1$ is a
mapping from $\mathcal{A}_1$ into $\mathcal{Z}(\mathcal{A})$ such that $\tau_1([\mathcal{A}_1,\mathcal{A}_1])=0$.

Next we prove that $\delta_1$ and $\tau_1$ are linear mappings. For each $A_1$ and $B_1$ in $\mathcal{A}_1$, we have that
$$\varphi(A_1)=\delta_1(A_1)+\tau_1(A_1)=[A_1,S_{A_1}]+\tau_1(A_1),$$
$$\varphi(B_1)=\delta_1(B_1)+\tau_1(B_1)=[B_1,S_{B_1}]+\tau_1(B_1),$$
and
$$\varphi(A_1+B_1)=\delta_1(A_1+B_1)+\tau_1(A_1+B_1)=[A_1+B_1,S_{A_1+B_1}]+\tau_1(A_1+B_1).$$
Since $\varphi$ is additive, through a discussion similar to that before, it implies that
$$[A_1+B_1,S_{A_1+B_1}]=[A_1,S_{A_1}]+[B_1,S_{B_1}]$$
and
$$\tau_1(A_1+B_1)=\tau_1(A_1)+\tau_1(B_1).$$
It means that $\delta_1$ and $\tau_1$ are additive mappings. Using the same technique,
we can prove that $\delta_1$ and $\tau_1$ are homogeneous. Hence $\delta_1$ and $\tau_1$ are linear mappings.

For every $A_2$ in $\mathcal{A}_2$, we have that
$$\varphi(A_2)=\varphi_{A_2}(A_2)=\delta_{A_2}(A_2)+\tau_{A_2}(A_2)=[A_2,T_{A_2}]+\tau_{A_2}(A_2),$$
where $\delta_{A_2}$ is a derivation on $\mathcal{A}$, $T_{A_2}$ is an element in $\mathcal{A}$ and $\tau_{A_2}$ is a linear mapping from
$\mathcal{A}$ into $\mathcal{Z}(\mathcal{A})$ such that $\tau_{A_2}([\mathcal{A},\mathcal{A}])=0$.
By condition (5), we have that
$\tau_{A_2}(A_2)=0$. Thus $\varphi(A_2)=[A_2,T_{A_2}]=[A_2,QT_{A_2}].$

Let $\varphi|_{\mathcal{A}_2}=\delta_2$. Then we have $\delta_2(A_2)=[A_2,S_{A_2}]$ for some element $S_{A_2}$ in $\mathcal{A}_2$.
And obviously, $\delta_2$ is linear.

Define two linear mappings as follows:
$$\delta(A)=\delta_1(A_1)+\delta_2(A_2),~~
\tau(A)=\tau_1(A_1),$$
for all $A=A_1+A_2\in\mathcal{A}$.
By the previous discussion, $\tau$ is a linear mapping from
$\mathcal{A}$ into $\mathcal{Z}(\mathcal{A})$ such that $\tau([\mathcal{A},\mathcal{A}])=0$.
In addition,
$$\delta(A)=\delta_1(A_1)+\delta_2(A_2)=[A_1,S_{A_1}]+[A_2,S_{A_2}]=[A_1+A_2,S_{A_1}+S_{A_2}]=[A,S_{A_1}+S_{A_2}].$$
It means that $\delta$ is a local derivation. By condition (3), $\delta$ is a derivation.
Notice that
$$\varphi(A)=\varphi(A_1)+\varphi(A_2)=\delta_1(A_1)+\tau_1(A_1)+\delta_2(A_2)=\delta(A)+\tau(A).$$
Hence $\varphi$ is a standard Lie derivation.
\end{proof}

By Lemma 3.1, we have the following result.

\begin{theorem}\label{02}
Every local Lie derivation on a von Neumann algebra is a Lie derivation.
\end{theorem}
\begin{proof}
Let $\mathcal A$ be a von Neumann algebra. It is well known that $\mathcal{A}=\mathcal{A}_1\bigoplus\mathcal{A}_2$,
where $\mathcal{A}_1$ is a finite von Neumann algebra, and $\mathcal{A}_2$ is a proper infinite von Neumann algebra.

By \cite[Theorem 1.1]{M. Mathieu}, we know that every Lie derivation on $\mathcal{A}$ is standard,
by \cite[Theorem 1]{sakai}, we know that every derivation on $\mathcal{A}$ is inner, and by
\cite[Theorem 5.3]{B. Johnson}, we know that every local derivation on $\mathcal{A}$ is a derivation.
Since $\mathcal{A}_2$ is a proper infinite von Neumann algebra, we known that $\mathcal{A}_2=[\mathcal{A}_2,\mathcal{A}_2]$(see in \cite{sunouchi}).

Hence it is sufficient to prove that $\mathcal{Z}(\mathcal{A}_1)\cap [\mathcal{A}_1,\mathcal{A}_1]=\{0\}$.
Since $\mathcal{A}_1$ is finite and by \cite[Theorem 8.2.8]{R.Kadison J. Ringrose}, it follows that there is a center-valued trace $\tau$ on $\mathcal{A}_1$
such that $\tau(Z)=Z$ for every $Z$ in $\mathcal{Z}(\mathcal{A}_1)$ and $\tau([A,B])=0$ for each $A$ and $B$ in $\mathcal{A}_1$.
Suppose that $A\in\mathcal{Z}(\mathcal{A}_1)\cap [\mathcal{A}_1,\mathcal{A}_1]$,
then we have that $\tau(A)=A$ and $\tau(A)=0$. it implies that $A=0$.

By Lemma \ref{01}, we know that every local Lie derivation on a von Neumann algebra is a Lie derivation.
\end{proof}

\section{Local Lie derivations on algebras of locally measurable operators}\

In this section, we mainly consider local Lie derivations on algebras
of all locally measurable operators affiliated with a type I von Neumann algebra.
To prove the main result, we need the following lemmas.

\begin{lemma}\label{03}
Suppose that $\mathcal{A}$ is a commutative unital algebra and $\mathcal{J}=M_n(\mathcal{A})$.
Then $\mathcal{Z}(\mathcal{J})\cap[\mathcal{J},\mathcal{J}]=\{0\}$
\end{lemma}
\begin{proof}
Let $\{e_{i,j}\}_{i,j=1}^{n}$ be the system of matrix units in $M_n(\mathcal{A})$.
Then for every element $A$ in $\mathcal{J}$, we have that $A=\sum_{i,j=1}^{n}a_{ij}e_{ij}$,
where $a_{ij}\in\mathcal{A}$.

Define a linear mapping $\tau$ from $\mathcal{J}$ into $\mathcal{A}$ by
$\tau(A)=\sum_{i=1}^{n}a_{ii}$ for every $A=\sum_{i,j=1}^{n}a_{ij}e_{ij}\in\mathcal{J}$.
Since $\mathcal{A}$ is  commutative, it is not difficult to verify that $\tau([A,B])=0$ for each $A$ and $B$ in $\mathcal{J}$.

It should be noticed that $\mathcal{Z}(\mathcal{J})=\{A:~A=\sum_{i=1}^{n}ae_{ii},~a\in\mathcal{A}\}$.
Suppose that $A=\sum_{i=1}^{n}ae_{ii}$ is an element in $\mathcal{Z}(\mathcal{J})\cap[\mathcal{J},\mathcal{J}]$,
then by the definition of $\tau$, we have that $\tau(A)=na$ and $\tau(A)=0$.
It implies that $A=0$.
\end{proof}

\begin{lemma}\label{04}
Suppose that $\mathcal{A}=\prod_{i\in\Lambda}\mathcal{A}_i$.
If $\mathcal{Z}(\mathcal{A}_i)\cap [\mathcal{A}_i,\mathcal{A}_i]=\{0\}$ for every $i\in\Lambda$,\\
then we have that $\mathcal{Z}(\mathcal{A})\cap [\mathcal{A},\mathcal{A}]=\{0\}$.
\end{lemma}
\begin{proof}
Let $A=\{a_i\}_{i\in\Lambda}$ be an element in $\mathcal{Z}(\mathcal{A})\cap [\mathcal{A},\mathcal{A}]$.
Then for every $i\in\Lambda$, we have that $a_i\in\mathcal{Z}(\mathcal{A}_i)\cap [\mathcal{A}_i,\mathcal{A}_i]$.
By assumption, it follows that $a_i=0$. Hence $A=0$.
\end{proof}

\begin{lemma}\label{05}
Suppose that $\mathcal{M}$ is a type $\mathrm{I}_{\infty}$ von Neumann algebra. Then $LS(\mathcal{M})=[LS(\mathcal{M}),LS(\mathcal{M})]$.
\end{lemma}
\begin{proof}
By \cite{M. Muratov1}, we know that for every $x$ in $LS(\mathcal{M})$, there exists a sequence $\{z_n\}$ of mutually orthogonal central projections in $\mathcal{M}$ with $\sum_{n=1}^{\infty}z_n=I$,
such that $x=\sum_{n=1}^{\infty}z_nx$, and $z_nx\in\mathcal{M}$ for every $n\in\mathbb{N}$.
Since $\mathcal{M}$ is a proper infinite von Neumann algebra, it is well known that $\mathcal{M}=[\mathcal{M},\mathcal{M}]$.
Thus we have that $z_nx=\sum_{i=1}^{k}[a^{n}_i,b^{n}_i]$, where $a^{n}_i,b^{n}_i\in\mathcal{M}$ for each $n$ and $i$.

Set $s_i=\sum_{n=1}^{\infty}z_na^{n}_i$ and $t_i=\sum_{n=1}^{\infty}z_nb^{n}_i$.
By the definition of locally measurable operators, it is easy to show that $s_i$ and $t_i$ are two elements in $LS(\mathcal{M})$.

Since that $\{z_n\}$ are mutually orthogonal central projections, we can obtain that
$$[s_i,t_i]=[\sum_{n=1}^{\infty}z_na^{n}_i,\sum_{n=1}^{\infty}z_nb^{n}_i]=\sum_{n=1}^{\infty}z_n[a^{n}_i,b^{n}_i],$$
moreover, we have that
$$\sum_{i=1}^{k}[s_i,t_i]=\sum_{i=1}^{k}\sum_{n=1}^{\infty}z_n[a^{n}_i,b^{n}_i]=\sum_{n=1}^{\infty}z_n(\sum_{i=1}^{k}[a^{n}_i,b^{n}_i])
=\sum_{n=1}^{\infty}z_nx=x.$$
It follows that $x\in[LS(\mathcal{M}),LS(\mathcal{M})]$.
\end{proof}

In the following we show the main result of this section.

\begin{theorem}\label{06}
Suppose that $\mathcal{M}$ is a type $\mathrm{I}$ von Neumann algebra with an atomic lattice of projections.
Then every local Lie derivation from $LS(\mathcal{M})$ into itself is a Lie derivation.
\end{theorem}
\begin{proof}
By \cite[Theorem 6.5.2]{R.Kadison J. Ringrose}, we know that $\mathcal{M}=\mathcal{M}_1\bigoplus\mathcal{M}_2$,
where $\mathcal{M}_1$ is a type $\mathrm{I}_{finite}$ von Neumann algebra
and $\mathcal{M}_2$ is a type $\mathrm{I}_{\infty}$ von Neumann algebra.
Hence by \cite[Proposition 1.1]{Albeverio2}, we have that $LS(\mathcal{M})\cong LS(\mathcal{M}_1)\bigoplus LS(\mathcal{M}_2)$.

In the following we will verify the conditions (1) to (5) in Lemma \ref{01} one by one.

By \cite[Theorem 1]{V. Chilin}, we know that every Lie derivation on $LS(\mathcal{M})$ is standard;
by \cite[Corollary 5,12]{Albeverio2}, we know that every derivation on $LS(\mathcal{M})$ is inner for a von Neumann
algebra with atomic lattice of projections.

It is proved in \cite{HadwinLi1} that every local derivation on $LS(\mathcal{M})$ is a derivation
for a von Neumann algebra without abelian direct summands.
While for an abelian von Neumann algebra with atomic lattice of projections,
by \cite[Theorem 3.8]{Albeverio3} we know that every local derivation on $LS(\mathcal{M})$ is a derivation.
Associated the two results, we can obtain each local derivation on $LS(\mathcal{M})$ is a derivation
for a von Neumann algebra with atomic lattice of projections.

Since $\mathcal{M}_1$ is a type $\mathrm{I}_{finite}$ von Neumann algebra,
we know that $\mathcal{M}_1=\bigoplus_{n=1}^{\infty}\mathcal{A}_n$,
where each $\mathcal{A}_n$ is a homogenous type $\mathrm{I}_n$ von Neumann algebra.
Hence $LS(\mathcal{M}_1)\cong\prod_{n=1}^{\infty}LS(\mathcal{A}_n)$.
Since $\mathcal{A}_n$ is a homogenous type $\mathrm{I}_n$ von Neumann algebra,
by \cite{Albeverio2} we know that $LS(\mathcal{A}_n)\cong M_n(\mathcal{Z}(LS(\mathcal{A}_n)))$.
By Lemmas \ref{03} and \ref{04}, we know that the condition (4) in Lemma \ref{01} holds.
And by Lemma \ref{05}, the condition (5) in Lemma \ref{01} holds.
\end{proof}

\bibliographystyle{amsplain}


TITLE: Determine the remainder
QUESTION [0 upvotes]: Determine the remainder when $2^{2018}$ is divided by $55$

REPLY [1 votes]: As Carmichael Function $\lambda(55)=20$ and $(55,2)=1$ and $2018\equiv-2\pmod{20}$
$$2^{2018}\equiv2^{-2}\pmod{55}$$
Now as $14\cdot4\equiv1\pmod{55},4^{-1}\equiv14$


TITLE: First derivative of $e^{-2x}/(1-x)$
QUESTION [0 upvotes]: Could someone please help me with these derivatives?
I have to find the first derivative of
$$f(x) = \frac{e^{-2x}}{1-x}.$$
Then I have to find the second derivative of that.
For the first derivative I get
$$ f'(x) = \frac{-2e^{-2x}}{(1-x)^2} $$
The answer book disagrees with me though. What am I doing wrong?

REPLY [1 votes]: Let  $\displaystyle f(x)=\frac{u(x)}{v(x)}$, where $u=e^{-2x} \Rightarrow u'=-2e^{-2x} $ and $v=1-x \Rightarrow v'=-1$.
Now, as $\displaystyle\left(\frac{u}{v}\right)'=\frac{vu'-uv'}{v^2}$, then
$$f'(x) =\frac{(1-x)(-2e^{-2x})-e^{-2x}(-1)}{(1-x)^2}=\frac{-2e^{-2x}+2xe^{-2x}+e^{-2x}}{(1-x)^2}=\frac{(2x-1)e^{-2x}}{(1-x)^2}.$$


TITLE: Principal value integral
QUESTION [10 upvotes]: I am reading A. Zee, QFT in a nutshell, and in appendix 1 he has:

Meanwhile the principal value integral is defined by:
  $$\int dx\,{\cal P}{1\over x}f(x)~=~ \lim_{\epsilon \rightarrow 0} \int dx\, {x\over x^2+\epsilon^2}f(x)$$

Please can someone explain to me why this is the case? As I understood it the principal value integral is rather defined as
$$\int_a^b dx\,{\cal P}{1\over x}f(x)~=~ \lim_{\epsilon \rightarrow 0^+} \int_a^{-\epsilon} dx\, {1\over x}f(x)+\lim_{\epsilon \rightarrow 0^+} \int_{\epsilon}^b dx\, {1\over x}f(x),$$
where $a<0<b$. But as far as I can see these two definitions are not equivalent.

REPLY [1 votes]: There are two issues here which recur frequently on this forum--- which functions can be regarded as distributions and what is the nature of convergence in the sense of distributions?  Perhaps a mathematician's take might be of interest.  It is standard that locally integrable functions can be regarded as distributions in a natural way, less so that the same applies to  meromorphic functions (in the OP $\frac 1x$).  The method is the same in both cases but is particularly simple in the one in hand.   The function $\ln  |x|$ is locally integrable and so a distribution.  It has a distributional derivative and it is natural to define the distribution $\frac 1x$ to be this derivative.  One can show quite easily that it then has all of the properties that one expects (including those of the approach using principal parts).
As to convergence, I will forego the precise definition.  For almost all practical purposes, it suffices to know 2 facts. Firstly, if a  sequence of functions converges locally in the $L^1$-norm, then it converges in the sense of distributions and, secondly, we can always differentiate such a convergent family and convergence is retained (the dream theorem of every calculus student).
After this preparation, the question in hand has a one line proof.  The family $\frac 12 \ln (x^2+\epsilon^2)$ converges to $\ln|x|$ in the above sense and differentiating provides the required formula.
All the facts about distributions used here can be found in the monograph of L. Schwartz but I can recommend the elementary approach of J. Sebastiao e Silva which is expounded in "An Introduction to the Theory of Distributions" by Campos Ferreira. This uses only the tools of elementary one-dimensional calculus.


TITLE: Twisted Day convolution
QUESTION [5 upvotes]: Has anyone studied a version of Day convolution for an enriched presheaf category $V^{A^{\mathrm{op}}}$ where the monoidal structure of $V$ is "twisted" on one side by an action of $A$?  I'm thinking of modifying the standard formula
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c) \otimes A(b\otimes c,a)$$
to something like
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c)^b \otimes A(b\otimes c,a)$$
where $G(c)^b$ denotes an action of $b\in A$ on $G(c)\in V$.

REPLY [3 votes]: The following is a proof that the twisted monoidal structure $\otimes^\rho$ is a convolution on $[A, V]$, where $A$ is a monoidal category, $V$ the cosmos on which $A$ is enriched, and $\rho\colon A\times V\to V$ a monoidal action (in the sense that $(v^a)^b \cong v^{a\cdot b}$, $v^i \cong v$ where $i$ is the monoidal identity of $A$, and $(v\otimes w)^a \cong v^a \otimes w^a$).
To make things work, I must assume that

Every object $a$ of the monoidal structure $(A, \cdot, i)$ has a monoidal inverse $a^{-1}$;
The functor $(-)^a \colon V\to V$ is cocontinuous.

This is somehow restricive, but since we know (coends, Prop. A.3) that (strong) promonoidal structure in $C$ correspond bijectively with convolution structures on $[A, V]$, there must exist a convolution product such that
$$
F\otimes^\rho G \cong \int^{xy} Fx \otimes Gy \otimes P_\rho(x,y;-)
$$
so that the action $(x,y)\mapsto (Gy)^x$ is "absorbed" into the promonoidal functor $P\colon A^\text{op}\times A^\text{op} \times A \to V$.
I'm not able to find the $P_\rho$, but I'm able to find sufficient assumptions (those above) to ensure associativity. As I mentioned in the comment, I come up with a "twisted ninja Yoneda lemma" (coends, Prop. 2.1), and I learned something more about coend-nonsense, so thank you (-:
Let's start the coend-juggling:
\begin{align}
[(F \otimes^\rho G)\otimes^\rho H](a) &= \int^{bc} (F\otimes^\rho G)(b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes A(x\cdot y, b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes \left( \int^b (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \right) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes (Hc)^{x \cdot y}\otimes A((x\cdot y)\cdot c, a)
\end{align}
where I applied the ninja Yoneda lemma to the functor $S(b) = (Hc)^b \otimes A(b\cdot c, a)$ ($S$ results as the composition $b\overset{\Delta}\mapsto (b,b) \overset{\hat S}\mapsto (Hc)^b \otimes A(b\cdot c, a)$; this type of dependence is essential, and implicit in the definition). We now have to manipulate the other parenthesization of $F,G,H$: to obtain the same object, we need the assumptions of cocompleteness of the action and invertibility:
\begin{align}
[F \otimes^\rho (G\otimes^\rho H)](a) & \cong \int^{mn} Fm \otimes ((G\otimes^\rho H)(n))^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes \Big( Gz\otimes (Ht)^z \otimes A(z\cdot t, n)\Big)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(z\cdot t, n)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n)^m \otimes A(m\cdot n, a) \right)\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n) \otimes A(m\cdot n, a)^{m^{-1}} \right)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \Big( A(m\cdot (z\cdot t), a)^{m^{-1}} \Big)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(m\cdot (z\cdot t), a)
\end{align}
where I applied the assumptions, and the ninja Yoneda lemma.
The same manipulations show that the presheaf $\hom(i,-)$ is the monoidal unit.
\begin{align}
F \otimes^\rho I & = \int^{bc} Fb \otimes A(i,c)^b \otimes A(b\cdot c ,a ) \\
& \cong \int^b Fb \otimes \left( \int^c A(i,c) \otimes A(b\cdot c, a)^{b^{-1}} \right)^b \\
&\cong \cdots
\end{align}


TITLE: $y^2 - x^3$ not an embedded submanifold
QUESTION [4 upvotes]: How can I show that the cuspidal cubic $y^2 = x^3$ is not an embedded submanifold of $\Bbb{R}^2$? By embedded submanifold I mean a topological manifold in the subspace topology equipped with a smooth structure such that the inclusion of the curve into $\Bbb{R}^2$ is a smooth embedding. I don't even know where to start please help me. All the usual tricks I know of removing a point from a curve and see what happens don't work. How can I extract out information about the cusp to conclude it is not? Also can I put a smooth structure on it so it is an immersed submanifold? THankz.

REPLY [4 votes]: Suppose the cuspidal cubic $A=\{(x,y)\in\Bbb R^2|y^2=x^3\}$  is an embedded submanifold of $\Bbb R^2$, and let the inclusion map $i:A\hookrightarrow \Bbb R^2$ denote the smooth embedding. For $A$, there is a smooth chart $(U,\varphi)$ containing $(0,0)$ such that $\varphi(U)=(-\epsilon,\epsilon)$ and $\varphi(0,0)=0$, then $\varphi^{-1}: (-\epsilon,\epsilon)\to U$ is a diffeomorphism, so $i\circ \varphi^{-1}$ is a smooth map. Let $\pi:\Bbb R^2\to \Bbb R$ denote the projection $(x,y)\mapsto y$, we define $f:=\pi\circ i\circ \varphi^{-1}$, then $(i\circ \varphi^{-1})(t)=(f^{\frac{2}{3}}(t),f(t))$, because $f(0)=0$, $f^{\frac{2}{3}}(t)$ is not differentiable at $t=0$, which is a contradiction.


TITLE: PDF of sum of random variables (with uniform distribution)
QUESTION [2 upvotes]: How can I solve this:
Random variables $X,Y$ ~ Unif$(0, 1)$ are independent. Calculate the probability density function of sum $X + 3Y$.
I couldn't find a sum for uniformally  distributed random variables. I assume I have to go straight to the PDF and solve it that way.

REPLY [2 votes]: Easy Understanding of Convolution  The best way to understand convolution is given in the article in the link,using that
I am going to solve the above problem and hence you could follow the same for any similar problem such as this with not too much confusion.
$Z = X+ 3Y$ where X and Y are U(0,1).
I am going to define a new variable W where W is distributed according to U(0,3)
Thus $Z = X + 3Y = X+ W$ where X is U(0,1) and W is U(0,3).
Now I am going define the bounds
$t_{X_0} = 0$
$t_{X_1} = 1$
$t_{W_0} = 0$
$t_{W_1} = 3$
Thus $$f_Z(z) = 0, z \le t_{X_0}+t_{W_0} ,$$
$$f_Z(z) = \int_{max(t_{W_0}, t-t_{X_1})}^{min(t_{W_1}, t-t_{X_0})} f_W(w)f_X(z-w)dw,    \text{     }     t_{X_0}+t_{W_0} \le z \le t_{X_1}+t_{W_1},$$
$$f_Z(z) = 0, z \ge t_{X_1}+t_{W_1} ,$$
These translate to the following:
$$f_Z(z) = 0, z \le 0 ,$$
$$f_Z(z) = \int_{max(0, z-1)}^{min(3, z)} f_W(w)f_X(z-w)dw,    \text{     }     0\le z \le 4,$$
$$f_Z(z) = 0, z \ge 4 ,$$
$f_W(w) = \frac{1}{3}$ as $W$ is $U(0,3)$.
$f_X(x) = 1 $ as $X$ is $U(0,1)$,
The middle one needs to be split into three intervals, and they are a) $0\le z\le 1$, b) $1\le z\le 3$, and  c) $3\le z\le 4$.
Thus
$f_Z(z) = \int_{0}^{z}\frac{1}{3}dw = \frac{z}{3}$, $0\le z\le 1$
$f_Z(z) = \int_{z-1}^{z}\frac{1}{3}dw = \frac{1}{3}$, $1\le z\le 3$
$f_Z(z) = \int_{z-1}^{3}\frac{1}{3}dw = \frac{4-z}{3}$, $3\le z\le 4$
Sanity check is to find if $\int_{0}^{4} f_Z(z) = 1$ which it is in this case and hence the solution.
Goodluck