text:
\begin{document}

\title{The competition-common enemy graphs
of digraphs satisfying Conditions $C(p)$ and $C'(p)$}

\author{\begin{tabular}{c}
{\sc Yoshio SANO}
\thanks{This work was supported by Priority Research Centers Program
through the National Research Foundation of Korea (NRF)
funded by the Ministry of Education, Science and Technology
(2009-0094069).}\\
\\
Pohang Mathematics Institute \\
POSTECH, Pohang 790-784, Korea\\
{\tt ysano@postech.ac.kr}
\end{tabular} }

\date{}

\maketitle

\begin{abstract}
S. -R. Kim and F. S. Roberts (2002) introduced
the following conditions $C(p)$ and $C'(p)$ for digraphs
as generalizations of the condition for digraphs to be semiorders.
The condition $C(p)$ (resp. $C'(p)$) is:
For any set $S$ of $p$ vertices in $D$,
there exists $x \in S$ such that
$N^+_D(x) \subseteq N^+_D(y)$
(resp. $N^-_D(x) \subseteq N^-_D(y)$)
for all $y \in S$,
where $N^+_D(x)$ (resp. $N^-_D(x)$)
is the set of out-neighbors (resp. in-neighbors) of $x$ in $D$.
The {\it competition graph}
of a digraph $D$ is the (simple undirected) graph
which has the same vertex set as $D$
and has an edge between two distinct vertices $x$ and $y$
if $N^+_D(x) \cap N^+_D(y) \neq \emptyset$.
Kim and Roberts characterized the competition graphs of digraphs which
satisfy Condition $C(p)$.

The {\it competition-common enemy graph}
of a digraph $D$ is the graph
which has the same vertex set as $D$
and has an edge between two distinct vertices $x$ and $y$
if it holds that both
$N^+_D(x) \cap N^+_D(y) \neq \emptyset$
and
$N^-_D(x) \cap N^-_D(y) \neq \emptyset$.
In this note, we characterize the competition-common enemy graphs
of digraphs satisfying Conditions $C(p)$ and $C'(p)$.
\end{abstract}

\noindent
{\bf Keywords:}
competition-common enemy graph;
semiorder;
interval order;
Condition $C(p)$

\newpage

\section{Introduction}

J. E. Cohen \cite{Cohen} introduced the notion of
a competition graph in 1968 in connection with a problem in ecology.
The {\it competition graph} $C(D)$ of a digraph $D$
is the (simple undirected) graph $G=(V,E)$ which
has the same vertex set as $D$
and has an edge between two distinct vertices
$x$ and $y$ if and only if $N^+_D(x) \cap N^+_D(y) \neq \emptyset$,
where $N^+_D(x) := \{v \in V(D) \mid (x,v) \in A(D) \}$
is the set of out-neighbors of $x$ in $D$.
It has been one of important research problems in the study of competition
graphs to characterize the competition graphs of digraphs satisfying
some conditions.

\begin{Defi}
{\rm
A digraph $D=(V,A)$ is called a
{\it semiorder} if there exist a real-valued function
$f:V \to \bR$ on the set $V$ and a positive real number $\delta \in \bR$
such that $(x,y) \in A$ if and only if $f(x) > f(y) + \delta$.

A digraph $D=(V,A)$ is called an
{\it interval order} if there exists
an assignment $J:V \to 2^{\bR}$
of a closed real interval $J(x) \subset \bR$
to each vertex $x \in V$
such that $(x,y) \in A$ if and only if $\min J(x) > \max J(y)$.
}
\qed
\end{Defi}

Kim and Roberts characterized the competition graphs of semiorders
and interval orders as follows:

\begin{Thm}[\cite{KimRoberts}]\label{thm:KR}
Let $G$ be a graph. Then the following are equivalent.
\begin{itemize}
\item[{\rm (a)}]
$G$ is the competition graph of a semiorder,
\item[{\rm (b)}]
$G$ is the competition graph of an interval order,
\item[{\rm (c)}]
$G = K_r \cup I_q$ where if $r \geq 2$ then $q \geq 1$.
\qed
\end{itemize}
\end{Thm}

\noindent
Moreover,
Kim and Roberts \cite{KimRoberts} introduced
some conditions, which are called Condition $C(p)$ and Condition $C'(p)$,
for digraphs
as generalizations of the condition for digraphs to be semiorders,
and they gave a characterization of the competition graphs of digraphs
satisfying Condition $C(p)$ to show Theorem \ref{thm:KR} as its corollary.

D. D. Scott \cite{Scott}
introduced the {\it competition-common enemy graph} of a digraph in 1987
as a variant of competition graph.
The {\it competition-common enemy graph}
of a digraph $D$ is the graph
which has the same vertex set as $D$
and has an edge between two distinct vertices $x$ and $y$
if it holds that both
$N^+_D(x) \cap N^+_D(y) \neq \emptyset$
and
$N^-_D(x) \cap N^-_D(y) \neq \emptyset$, where
$N^-_D(x) := \{v \in V(D) \mid (v,x) \in A(D) \}$
is the set of in-neighbors of $x$ in $D$.

\begin{figure}[h]
\begin{center}
 \resizebox{10cm}{!}{
  \includegraphics{interval.eps}
 }
\end{center}
\caption{An interval order $D$, the competition graph $C(D)$, and
the competition-common enemy graph $CCE(D)$}
\label{fig:interval}
\end{figure}

In this note, we characterize
the competition-common enemy graphs of semiorders
and interval orders as follows:

\begin{Thm}\label{thm:main0}
Let $G$ be a graph. Then the following are equivalent.
\begin{itemize}
\item[{\rm (a)}]
$G$ is the competition-common enemy graph
of a semiorder,
\item[{\rm (b)}]
$G$ is the competition-common enemy graph
of an interval order,
\item[{\rm (c)}]
$G = K_r \cup I_q$ where if $r \geq 2$ then $q \geq 2$.
\qed
\end{itemize}
\end{Thm}

\noindent
Furthermore,
we also characterize the competition-common enemy graphs
of digraphs satisfying Conditions $C(p)$ and $C'(p)$.

\section{Main Results}

\subsection{Conditions $C(p)$ and $C'(p)$}

\begin{Defi}
{\rm
Let $D$ be a digraph.
For a set $S$ of vertices in $D$, we define the following:
\begin{eqnarray*}
\cF^+_D(S) &:=&
\{x \in S \mid N^+_D(x) \subseteq N^+_D(y) \text{ for all } y \in S \}, \\
\cF^-_D(S) &:=&
\{x \in S \mid N^-_D(x) \subseteq N^-_D(y) \text{ for all } y \in S \}, \\
\cH^+_D(S) &:=&
\{x \in S \mid N^+_D(x) \supseteq N^+_D(y) \text{ for all } y \in S \}, \\
\cH^-_D(S) &:=&
\{x \in S \mid N^-_D(x) \supseteq N^-_D(y) \text{ for all } y \in S \}.
\end{eqnarray*}
(Note that, in \cite{KimRoberts},
an element in $\cF^+_D(S)$ is called a {\it foot} of $S$ and
an element in $\cH^+_D(S)$ is called a {\it head} of $S$.)

\begin{figure}[h]
\begin{center}
 \resizebox{10cm}{!}{
  \includegraphics{FS-HS.eps}
 }
\end{center}
\caption{Elements $x$ in $\cF^+_D(S)$, $\cF^-_D(S)$, $\cH^+_D(S)$,
and $\cH^-_D(S)$}
\label{fig:FS-HS}
\end{figure}

Let $p$ be a positive integer with $p \geq 2$.
We say that {\it $D$ satisfies Condition $C(p)$}
(resp. {\it Condition $C'(p)$}, {\it Condition $C^*(p)$},
{\it Condition ${C^*}'(p)$})
if the set $\cF^+_D(S)$ (resp. $\cF^-_D(S)$ $\cH^+_D(S)$ $\cH^-_D(S)$)
is not empty
for any set $S$ of $p$ vertices in the digraph $D$.
}
\end{Defi}

\begin{Prop}[\cite{KimRoberts}]\label{prop:C(p)}
Let $2 \leq p < q$.
If a digraph $D$ satisfies Condition $C(p)$,
then the digraph $D$ also satisfies Condition $C(q)$.
\end{Prop}

\begin{Lem}\label{lem:FD}
Let $D$ be a digraph and $T, U$ be sets of vertices in $D$.
If $\cF^-_D(T) \cap U \neq \emptyset$,
then $\cF^-_D(U) \subseteq \cF^-_D(T \cup U)$.
\end{Lem}

\begin{proof}
Take $t \in \cF^-_D(T) \cap U$.
Then $N^-_D(t) \subseteq N^-_D(t')$ for any $t' \in T \setminus U$.
If $\cF^-_D(U)$ is empty, then the lemma trivially holds.
So we assume that $\cF^-_D(U) \neq \emptyset$.
Take any $u \in \cF^-_D(U)$.
Then $N^-_D(u) \subseteq N^-_D(u')$ for any $u' \in U$.
Since $t \in U$, we have $N^-_D(u) \subseteq N^-_D(t)$.
Therefore, $N^-_D(u) \subseteq
N^-_D(t')$ for any $t' \in T \setminus U$.
Thus $N^-_D(u) \subseteq
N^-_D(s)$ for any $s \in (T \setminus U) \cup U= T \cup U$.
Hence the lemma holds.
\end{proof}

\begin{Prop}\label{prop:C'(p)}
Let $2 \leq p < q$.
If a digraph $D$ satisfies Condition $C'(p)$,
then the digraph $D$ also satisfies Condition $C'(q)$.
\end{Prop}

\begin{proof}
It is enough to show that
$D$ satisfies Condition $C'(p+1)$.
Let $S$ be any set of $p+1$ vertices of $D$, and
let $T$ be a subset of $S$ with $|T|=p$.
Then $\cF^-_D(T) \neq \emptyset$
since $D$ satisfies Condition $C'(p)$.
Take an element $x$ in $\cF^-_D(T)$.
Let $U$ be a subset of $S$ such that $|U|=p$ and $x \in U$.
Since $p \geq 2$, it holds that $T \cup U = S$.
By Lemma \ref{lem:FD},
we have $\cF^-_D(U) \subseteq \cF^-_D(T \cup U) = \cF^-_D(S)$.
Since $D$ satisfies Condition $C'(p)$,
$\cF^-_D(U) \neq \emptyset$.
Thus $\cF^-_D(S)$ is not empty.
\end{proof}

For a graph $G$, we denote the set of all isolated vertices in $G$
by $I_G$. Then the graph $G - I_G$ is the union of the nontrivial
connected components of $G$.

\begin{Prop}\label{prop:clique}
Let $G$ be the competition-common enemy graph of
a digraph $D$ which satisfies
Conditions $C(p)$ and $C'(p)$ for some $p \geq 2$.
Suppose that $G-I_G$ has at least $p$ vertices.
Then $G-I_G$ is a clique of $G$.
\end{Prop}

\begin{proof}
Take any two vertices $a$ and $b$ in $G-I_G$.
Then $a$ and $b$ are not isolated.
Let $S$ be a set of $p$ vertices in $G-I_G$
containing the vertices $a$ and $b$.
Since $D$ satisfies Conditions $C(p)$ and $C'(p)$,
there exist $x \in \cF^+_D(S)$ and $y \in \cF^-_D(S)$.
Note that $x$ and $y$ are not isolated vertices.
Take $u \in N^+_D(x)$ and $v \in N^-_D(y)$.
By Condition $C(p)$, we have $u \in N^+_D(a) \cap N^+_D(b)$.
By Condition $C'(p)$, we have $v \in N^-_D(a) \cap N^-_D(b)$.
Therefore $a$ and $b$ are adjacent in $G-I_G$.
Hence the proposition holds.
\end{proof}

\subsection{Classifiation}

\begin{Thm}\label{thm:loopless}
Let $G$ be a graph and $p \geq 2$.
Suppose that $G-I_G$ has at least $p$ vertices.
Then $G$ is the competition-common enemy graph of
a loopless digraph satisfying Conditions $C(p)$ and $C'(p)$
if and only if
$G = K_r \cup I_q$ with $r \geq p$ and $q \geq 2$.
\end{Thm}

\begin{proof}
Fisrt, we show the ``only if" part.
Let $G$ be the competition-common enemy graph of
a loopless digraph $D$ satisfying Conditions $C(p)$ and $C'(p)$.
Proposition \ref{prop:clique} shows
that $G = K_r \cup I_q$ with $r \geq p$ and $q \geq 0$.
Suppose that $q = 0$ or $q=1$.
Since $r \geq p$,
by Propositions \ref{prop:C(p)} and \ref{prop:C'(p)},
$D$ satisfies Conditions $C(r)$ and $C'(r)$.
Let $x \in \cF^+_D(S)$ and $y \in \cF^-_D(S)$ where $S:=V(G-I_G)$.
Since $x$ and $y$ are not isolated in $G$,
we have $N^+_D(x) \neq \emptyset$ and $N^-_D(y) \neq \emptyset$.
Let $u \in N^+_D(x)$ and $v \in N^-_D(y)$.
If $u=v$, then $(s,u) \in A(D)$ and $(u,s) \in A(D)$ for any $s \in S$.
Let $S'$ be a set of $p$ vertices containing the vertex $u$.
Note that $S' \setminus\{u\} \subseteq S$ since $q \leq 1$.
By Condition $C(p)$, $\cF^+_D(S') \neq \emptyset$.
If $u \in \cF^+_D(S')$, then
we have $s \in N^+_D(u) \subseteq N^+_D(s)$ for $s \in S' \setminus\{u\}$,
i.e., $(s,s) \in A(D)$,
which contradicts that $D$ is loopless.
If $s \in \cF^+_D(S')$ for some $s \in S' \setminus\{u\}$, then
we have $u \in N^+_D(s) \subseteq N^+_D(u)$, i.e., $(u,u) \in A(D)$,
which contradicts that $D$ is loopless.
Therefore $u$ and $v$ must be distinct.
Since $q \leq 1$, at least one of $u$ and $v$ is in $S=V(G-I_G)$.
If $u \in S$, then
we have $u \in N^+_D(x) \subseteq N^+_D(u)$, i.e., $(u,u) \in A(D)$.
If $v \in S$, then
we have $v \in N^-_D(y) \subseteq N^-_D(v)$, i.e., $(v,v) \in A(D)$.
In any case, we reach a contradiction.
Thus we have $q \geq 2$.

Second, we show the ``if" part.
Let $G = K_r \cup I_q$ with $r \geq p$ and $q \geq 2$.
Let $a$ and $b$ be distinct vertices in $I_q$.
We define a digraph $D$ by
$V(D):=V(G)$ and
$A(D):=
\{(a, x) \mid x \in V(K_r) \}
\cup
\{(x, b) \mid x \in V(K_r) \}
\cup \{(a,b)\}$.
Then $D$ is loopless, $D$ satisfies Conditions
$C(p)$ and $C'(p)$, and
the competition-common enemy graph of $D$ is equal to $G$.
\end{proof}

The {\it double competition number} $dk(G)$ of a graph $G$
is the minimum number $k$ such that $G$ with $k$ new isolated vertices
is the competition-common enemy graph of an acyclic digraph.

\begin{Thm}\label{thm:acyclic}
Let $G$ be a graph and $p \geq 2$.
If $G$ is the competition-common enemy graph of an
acyclic digraph $D$ satisfying Conditions $C(p)$
and $C'(p)$, then
$G$ is one of the following graphs:
\begin{itemize}
\item[{\rm (a)}]
$I_q$ $(q \geq 1)$,
\item[{\rm (b)}]
$K_r \cup I_q$ $(r \geq p$, $q \geq 2)$,
\item[{\rm (c)}]
$H \cup I_q$ where $|V(H)| < p$, $I_H=\emptyset$, and
$q \geq dk(H)$.
\end{itemize}
\end{Thm}

\begin{proof}
Let $G$ be the competition-common enemy graph of
an acyclic digraph $D$ satisfying
Conditions $C(p)$ and $C'(p)$.
If there is no nontrivial connected component in $G$, then (a) holds.
Let $H$ be the union of all nontrivial connected components of $G$.
Then we have $G = H \cup I_q$ with $q \geq 0$ and $I_H=\emptyset$.
If $H$ has at least $p$ vertices,
then it follows from Theorem \ref{thm:loopless} that
$G=K_r \cup I_q$ with $r \geq p$ and $q \geq 2$, i.e., (b) holds.
Suppose that
the number of the vertices of $H$ is less than $p$.
Since $G$ is the competition-common enemy graph
of an acyclic digraph $D$,
the double competition number
$dk(G)$ of $G$ is equal to $0$.
Therefore, there must be at least $dk(H)$ vertices in $I_q$.
Hence (c) holds.
\end{proof}

\subsection{Proof of Theorem \ref{thm:main0}}

\begin{proof}[Proof of Theorem \ref{thm:main0}]

(a) $\Rightarrow$ (b):
Since semiorders are a special case of interval orders
where every interval has the same length,
(a) implies (b).

(b) $\Rightarrow$ (c):
We can easily check that any
interval order satisfies Conditions
$C(2)$ and $C'(2)$.
By Theorem \ref{thm:acyclic} with $p=2$,
we can conclude that if (b) then (c).

(c) $\Rightarrow$ (a):
Suppose that
$G = I_q$ $(q \geq 1)$ or
$G = K_r \cup I_q$ $(r \geq 2$, $q \geq 2)$.
When $G = I_q$,
we let $f_1(x):=0$ for any $x \in V(G)$ and let $\delta_1:=1$.
Then $G$ is the competition-common enemy graph of the semiorder
defined by $f_1$ and $\delta_1$.
When $G = K_r \cup I_q$,
we take a vertex $a$ in $I_q$,
let $f_2(x):=0$ for any $x \in V(K_r)$,
$f_2(a):=2$, and
$f_2(b):=-2$ for any $b \in V(I_q) \setminus \{a\}$,
and let $\delta_2:=1$.
Then $G$ is the competition-common enemy graph of the semiorder
defined by $f_2$ and $\delta_2$.

Hence Theorem \ref{thm:main0} holds.
\end{proof}

\section{Concluding Remarks}

In this section,
we present some problems for further study.

In Theorem \ref{thm:loopless},
we gave a characterization of the competition common-enemy
graphs $G$ of digraphs satisfying Conditions $C(p)$ and $C'(p)$
if the number of the vertices of $G-I_G$ is at least $p$.

\begin{Prob}
Characterize
the competition-common enemy graphs $G$
of digraphs satisfying Conditions
$C(p)$ and $C'(p)$ when
the number of the vertices of $G-I_G$ is less than $p$.
\end{Prob}

In this note, we didn't consider Conditions $C^*(p)$ and ${C^*}'(p)$.

\begin{Prob}
Characterize
the competition-common enemy graphs of digraphs satisfying Conditions
$C^*(p)$ and ${C^*}'(p)$.
\end{Prob}

Niche graphs are another variant of competition graphs
and were introduced by
C. Cable, K. F. Jones, J.R. Lundgren, and S. Seager \cite{CJLS}.
The {\it niche graph} of a digraph $D$
is the graph which has the same vertex set as $D$
and has an edge between two distinct vertices $x$ and $y$
if
$N^+_D(x) \cap N^+_D(y) \neq \emptyset$
or
$N^-_D(x) \cap N^-_D(y) \neq \emptyset$.

\begin{Prob}
What are the niche graphs of digraphs satisfying Conditions
$C(p)$, $C'(p)$, $C^*(p)$, or ${C^*}'(p)$?
\end{Prob}


TITLE: On an injective ring homomorphism from the ring of continuous functions to the ring of differentiable functions
QUESTION [4 upvotes]: Let $\phi : C \to D$ be an injective ring homomorphism such that $\phi(1)=1$, where $1$ denotes the constant function $1$ and $C$, $D$ are the rings of continuous, respectively differentiable functions on $\mathbb{R}$. If $f \in C ,g \in D$ are such that $\phi(f)=g$, and $t_0\in \mathbb R$, is it true that $\phi (f-g(t_0))=g-g(t_0)$ ? This is used on page 944, Problem 11617. And why if the image of $\phi $ consists of only constant functions, then $\phi$ is not injective ? Please help. Thanks in advance.

REPLY [5 votes]: For the first part note that a standard argument shows that $\phi(q)=q$ where $q \in \mathbb{Q}$ denotes the constant function $q$. If you show that this also holds for irrational numbers you are done for the first part. I'll give a sketch of the proof.
Let $C=C(\mathbb{R})$ and $D=D(\mathbb{R})$.
Define $i: \mathbb{R} \to C(\mathbb{R})$ by $i(x) = c_x$, where $c_x$ is the constantly equal to $x$ function.
For every $x_0$ in $\mathbb{R}$ define $e_{x_0}: D(\mathbb{R}) \to \mathbb{R}$ the evaluation morphism at $x_0$.
Take $x_0$ in $\mathbb{R}$ and prove that $\psi = ev_{x_0} \circ \phi \circ i :\mathbb{R} \to \mathbb{R}$ is a ring homomorphism, which sends positive numbers to positive numbers. It may help to observe that $c_x = c_{\sqrt{x}} \cdot c_{\sqrt{x}}$ for $x > 0$.
Now $\psi:\mathbb{R} \to \mathbb{R}$ is a strictly increasing ring homomorphism which agrees with the identity on $\mathbb{Q}$. Therefore, $\psi$ is the identity. And because $x_0$ was arbitrary, then $\phi \circ i :\mathbb{R} \to D(\mathbb{R})$ maps $x \mapsto c_x$ for every $x \in \mathbb{R}$. That is, $\phi(c_x)=c_x$ for every $x\in\mathbb{R}$.
For the second part suppose that the image of $\phi$ consists only of constant functions. Take $f = \chi_{[0;1]}$ and $g = \chi_{[2;3]}$. Clearly, $f\cdot g \equiv 0$. So $0 = \phi(f\cdot g) = \phi(f)\phi(g)$. Because the image of $\phi$ consists only of constant functions, then $\phi(f) =0$ or $\phi(g)=0$. Because $f,g \neq 0$ this implies that $\phi$ is not injective, contradicting the hypothesis of $\phi$ being injective. Then, for injective $\phi$, the image of $\phi$ can't consist only of constant functions.


TITLE: Weird relation between bitwise 'AND' and 'OR' Operations.
QUESTION [3 upvotes]: I was trying to plot some boolean operations data using Matplotlib Python. I noticed some weird behavior of 'and'(&) and 'or'(|) operations on the same data pair.
Suppose:
A=50(0011 0010)
B=60(0011 1100)

A&B = 48(0011 0000)
A|B = 62(0011 1110)

Notice how both of them have the same distance(2) from the bigger and lower number. In other words- have the same distance(7) from the arithmetic average of A and B(55). Another example:
A=96(0110 0000)
B=82(0101 0010)

A&B = 64(0100 0000)
A|B = 114(0111 0010)

It is the same, A&B is 18 lower than B, and A|B is 18 higher than A. both are at the same distance from the mean of A and B(89) which is 25.
Mathematically:
$$
 |(A.B - \mu(A,B))| =  |(A+B - \mu(A,B))|
$$
It is almost true for all positive numbers(some weird problem arises if the lower bound gets under 0)
My First question is, why do they behave this way? Is this connected to De Morgan's theorem?
I tried to see if there is a relation between the distance that those operations have from the actual number(A and B) and the distance of A and B. I found this-
variable A = 10000, 10000, 10000, 10000,.....,10000
variable B = 10000, 10001, 10002, 10003,.....,15000
so on x-axis, I put B-A(0,1,2,3,4,....,5000).
On the y-axis, I put (A|B - B) (which is equal to (A - A&B))
The result was like this:
Matplotlib data of 5000 distances
Now my second question is, is it possible to get a function for this relation
Something like-
$$
f(y) = x^2 + sin(x)....
$$
If Anybody still reading, the plot for the distance of A&B and A|B from the arithmetic mean of A and B looks like this:
Distance from the mean of A and B plot
Same plot as before if A and B started from 20000, 20000. and plotted 10000 points instead of 5000.
10000 data points with A=20000, B starts from 20000
Thank you for your time:)

REPLY [3 votes]: Consider numbers with 1 bit, then (with hopefully obvious terminology):
$$A=0011$$
$$B=0101$$
$$A\&B=0001$$
$$A|B=0111$$
$$A\&B + A|B=0112$$
$$A+B = 0112$$
You should be able to see from there that the relationship you mention is true for any combination of single bits and so will also be true for any combination of bits.


TITLE: "Trace" of a distribution?
QUESTION [0 upvotes]: In quantum field theory we have to sometimes take the "trace" of a distribution $M(x,y)$, $\text{tr}M\sim\int dx M(x,x)$. This happens for instance when we try to expand the determinant of a Dirac operator $\det{\not}D=\exp\text{tr}\log{\not}D.$
These traces are not well defined mathematically, because they are divergent. A simple example would be looking at the distribution $f(x)\delta(x-y)$, whose trace is taken to be $\delta(0)\int dx f(x)$. Somehow, physicists are able to get away with these infinite $\delta(0)$ factors and still get something sensible. In the cases I have seen, these factors are usually assumed to be related to the volume of space, taken to be infinite.
Since these objects are not of trace class, these calculations are not guaranteed to make sense. A trace is supposed to have the same value regardless of what basis you calculate it in, so is the trace of $M(x,y)$ in position space the same as it is in momentum space?
For that matter, how does one take the trace of a distribution like $f(x)\partial_{\mu}\delta(x-y)$, or an arbitrary number of derivatives acting on a delta function?
Mathematicians seem to think there is no way of doing it, but physicists have found ways of getting sensible results! There must be some way of defining it properly, right?

REPLY [0 votes]: Yes, in some cases the trace is related to the volume of the space. However, it would be more appropriate to say that it is related to the cardinality of the space, in other words, the number of modes in the space. People usually get around the problem of infinite cardinality by reducing the number of modes to a finite number. This is done with the aid of some form of regularization. The quantities that would have been infinite then become finite and one can proceed to perform the calculation. In the final result, these quantities would cancel out if the result represents a physical quantity. If they don't then it usually means that there is something wrong. After they cancelled, one can undo the regularization that would have made these quantities infinite. Since they are gone, the result remains finite.
What all this really means is that even if you are working in a space where some of these traces would diverge, the results of calculations for physical quantities that one would for instance get from measurements would always be finite. The question is just how to perform the calculation with these infinities so that they would cancel. The reason why it is difficult to work with infinities is because they obey cardinal arithmetic. Regularization converts them into ordinal numbers so that one can handle them in the usual way with ordinal arithmetic. Then, after they cancelled out one can take the necessary limit to remove the regularization.
Hope it helps.


TITLE: Inequality with Algebra $\sup_{x,y,z\in A}\left(\left(13x+\frac{5}{x}\right)+\left(13y+\frac{5}{y}\right)+\left(13z+\frac{5}{z}\right)\right)= 63$
QUESTION [2 upvotes]: let $A=\{(x,y,z)|x,y,z\in [\frac{1}{2},2],xyz=1\}$
Maybe have
$$f=\sup_{x,y,z\in A}\left(\left(13x+\dfrac{5}{x}\right)+\left(13y+\dfrac{5}{y}\right)+\left(13z+\dfrac{5}{z}\right)\right)=  63?$$
Here the maximum of $f$ is $63$,reached at the point $(x,y,z)$ on the space surface $xyz=1$,and  $x=2,y=\dfrac{1}{2},z=1$
I attempted a proof by AM-GM inequality,But I am not able to solve the upper part,   this inequality when $x=2,y=\dfrac{1}{2},z=1$,then $f=63$
This problem  can use from space Analysis geometry? ,take $y=\dfrac{15}{3x}$

REPLY [0 votes]: Start with any $(x,y,z)\in A$.  If $x\leq1$, then replace $y$ and $z$ by $2$ and $1/(2x)$.  Does that increase the sum?

REPLY [0 votes]: $p=x+y,xy=q, \dfrac{1}{2} \le z \le 2 \implies \dfrac{1}{2} \le q \le 2 $
$f=13p+\dfrac{5p+13}{q}+5q$
consider $\dfrac{A}{q}+Bq$ will get max at two ends, so $f(p,q)$ will get max at $q=2$ or $q= \dfrac{1}{2}$
when $q=2 ,y=\dfrac{2}{x}$, put it in $f$,it become $f(x)= \dfrac{A}{x}+Bx+C$,again, $f(x)$ will get max at two ends,so you will have $x=2$ or $x=\dfrac{1}{2}$ to check which one is bigger.


TITLE: Integral on surface
QUESTION [3 upvotes]: Please help me calculate integral on surface.
$$\iint\!(2-y)\,dS $$
$$ 0\leq z\leq 1;\,  y =1 $$
I can't understand what should I do with '$z$' coordinate. I assume I should do something with it, not to leave blank.
Maybe like this?
$$= \iint\!dy dz  = 1?$$

REPLY [3 votes]: Most likely there is part of a surface $S$ which is bounded between $0\leq z\leq 1$ and $y=1$.
If the integral is on the surface $y=1$ and bounded between $0\leq z\leq 1$, then this surface is an infinite strip parallel to the $x$-axis, there is no way the integral is $1$ unless you have a bound on $x$ as well.
If it is the former case, what you wanna do is to look for a parametrization of $S$:
$$
S: \mathbf{\Phi}(u,v) = \big(x(u,v),y(u,v),z(u,v)\big),
$$
for $u\in [a,b]$, $v\in [c,d]$, then
$$
\iint_S (2-y)\,dS = \int^d_c\int^b_a \big(2-y(u,v)\big)\,\left|\frac{\partial \mathbf{\Phi}}{\partial u}\times \frac{\partial \mathbf{\Phi}}{\partial v}\right|\,dudv.
$$

EDIT: for the $z = 1$ case, you certain have:
$$
\iint_S (2-z)\,dS = \int^{1}_0\color{red}{\int^{b}_{a}} (2-z)\sqrt{1 + (z_x)^2 + (z_y)^2} \,\color{red}{dx} dy  = \color{red}{\int^{b}_{a}dx}\int^{1}_0(2-1)dy.
$$
Like I said in the original answer, you have to have a bound on $x$, otherwise the surface is infinite.


TITLE: Dual groups in Fourier analysis
QUESTION [3 upvotes]: For finite abelian groups and the additive group of real numbers, the dual group is isomorphic to itself.
But for the unit circle in the complex plane, the dual group is isomorphic to $\mathbb{Z}$.
I know why it should, but I don't see where does this difference come from. Is there any topological concern(like the fundamental group, etc..) in this regard or any other relative algebraic reasons like the degree of a map?
If there is any relative notion, how does it affect to the Dual group of an abelian group?

REPLY [3 votes]: Really you should treat self-dual groups as "special" and non self-dual groups as "general." You get lucky in some cases, but there's no reason to think characters on a group should be isomorphic to the group itself a priori.
One thing that's worth noting is that the dual of a compact group is a discrete group, and vice versa. So the dual of, for example, $S^1$ with the usual topology could not be $S^1$ because it's not discrete. Similarly for $\Bbb Z$. In this way you can see finite groups are dual to finite groups thanks to the fact that compact + discrete implies finite--even aside from the canonical embedding of $G$ into $\widehat{\widehat{G}}$. For $\Bbb R$ you get lucky because of the specific nature of continuous homomorphisms into $S^1$ from $\Bbb R$ and the fact that the only closed subgroups of $\Bbb R$ are $\Bbb R$ and $\alpha\Bbb Z$ for some $\alpha\in\Bbb R$, in particular the kernel of the homomorphism into $S^1$ is such a closed subgroup--being the inverse image of the closed set $\{1\}\subseteq S^1$.


TITLE: Lognormal distribution inverse equivalent
QUESTION [4 upvotes]: In Lognormal distribution if the random variable X is log-normally distributed, then Y = ln(X) has a normal distribution.
Is there inverse equivalent to lognormal distribution where Y = exp(X) has a normal distribution?

REPLY [5 votes]: This is easily shown to be impossible for any real-valued random variable $X$:  if we require $$Y = e^X \sim \operatorname{Normal}(\mu,\sigma^2),$$ then there must be some value of $X$ for which $e^X < 0$.  But this is impossible if $X \in \mathbb R$.


TITLE: What are all the stationary and pointwise independent random processes?
QUESTION [6 upvotes]: In the 60's, I. Gel'fand introduced the concept of generalized stochastic processes (Ch. III, Vol. 4 of his work on Generalized functions). For a generalized stochastic process $\Phi$, he defines the concepts of stationarity ($\Phi(\varphi)$ and $\Phi(\varphi(\cdot - t_0))$ have the same law)  and of independence at every point (the random variable $\Phi(\varphi_1)$ and $\Phi(\varphi_2)$ are independent if $\varphi_1$ and $\varphi_2$ have disjoint supports).
Gel'fand especially introduces the complete class of Lévy white noises as generalized stochastic processes with characteristic functional of the form
$$L(\varphi) =\exp\left( \int f(\varphi(t)) \mathrm{d}t \right),$$
with $f$ a function that has a L\'evy-Khintchine representation.
Obviously, white noises are not the only stationary and independent at every point processes (ex: the weak derivative of a white noise). I am interested by a characterization of stationary and independent at every point processes. Especially, Gel'fand conjectured the following result.
Conjecture:
If a generalized stochastic process $\Phi$ is stationary and independent at every point, then the characteristic function $L$ of $\Phi$ has the form
$$L(\varphi) =\exp\left( \int f(\varphi(t),\varphi^{(1)}(t),\cdots, \varphi^{(n)}(t)) \mathrm{d}t \right),$$
with $f$ a continuous function from $\mathbb{R}^{n+1}$ to $\mathbb{C}$ with $f(0)=0$.
Is that result true? In order to express a kind of reciprocal result, can we characterize the functions $f$ such that the previous functional is a characteristic functional (main problem: its positive-definiteness)?
Some people are extensively studied the positive-definiteness of functionals but wasn't able to find references that are clearly answering my question.
Thanks for your attention.

REPLY [2 votes]: I would suggest the reference [1], where J.N. Pandey addresses this problem.
[1] J.N. Pandey, ``On the positive definiteness of a functional'', Canad. Math. Bull., Vol 22 (2), 1979


TITLE: invertibility of a matrix with a Gaussian perturbation
QUESTION [4 upvotes]: Suppose that $A$ is an arbitrary fixed $n\times n$ matrix and $G$ a random $n\times n$ matrix with i.i.d. $N(0,1)$ entries. Is there a simple proof that $A+G$ is invertible with probability 1?
What if $G$ is a random Wigner matrix (symmetric, upper diagonal entries are i.i.d. $N(0,1)$)? Is $A+G$ still invertible with probability 1? Is there a simple proof?

REPLY [4 votes]: We  use the idea suggested by Alekk.
Let $A_{i,j}$ the entries of $A$. Then in the first case, the entries of $M:=A+G$ are Gaussian independent random variables, that is, $M_{i,j}\sim N(A_{i,j},1)$. Denote $N$ the set of elements of $\Bbb R^{n^2}$ such that the matrix of generic term $x_{i,j}$ is not invertible. This set has null Lebesgue measure in $\Bbb R^{n^2}$ as it's the zeros of a polynomial. By independence, the family $(M_{i,j},i,j\in [n])$ is Gaussian, so
$$p:=P(A+G\mbox{ is not invertible})=P((M_{i,j})_{i,j=1}^n\in S).$$
As the law of $(M_{i,j},i,j\in [n])$ is absolutely continuous with respect to Lebesgue measure in $\Bbb R^{n^2}$, we conclude that $p=0$.
In the second case, write $\det(A+G)$ as a polynomial of the $G_{i,j},i\leqslant j$, and use the fact that $(G_{i,j},1\leqslant i\leqslant j\leqslant n)$ is Gaussian to conclude in the same way as in the first case.

The result doesn't depend on the choice of the deterministic matrix $A$.
We don't need i.i.d.ness, just the fact that $(G_{i,j},i,j\in [n])$ is Gaussian in the first case, $(G_{i,j},i\leqslant j,i,j\in [n])$ in the second.
We can have a more general result when the law of $(A_{i,j}+G_{i,j})_{i,j\in [n]}$ is absolutely continuous with respect to Lebesgue measure on $\Bbb R^{n^2}$.


TITLE: Difference between interior and set of accumulation points
QUESTION [2 upvotes]: I don't understand the difference between the interior of a set, and the set of all its accumulation points.
My understanding of an accumulation point is any point in a set which has an epsilon neighborhood around it, which is contained in the set- not necessarily implying that the accumulation point itself is in the set.
From what I can gather, the interior is identical. Can someone explain the difference?

REPLY [0 votes]: I assume that you're working with subsets of the real line with the usual topology.
The definition you gave for accumulation point is actually the definition for interior point, and it does imply that the point lies in the set. Indeed, an $\epsilon$-neighborhood of a point $x$ is an interval of the form $(x-\epsilon,x+\epsilon)$ with $\epsilon>0,$ and $x$ readily lies in any such interval, so $x$ lies in any set that contains such an interval.
We say that a point $x$ is an accumulation point of a set $A$ if every $\epsilon$-neighborhood of $x$ intersects $A$ at a point distinct from $x$. This does not mean that every epsilon neighborhood of $x$ is contained in $A$--it's possible that no $\epsilon$-neighborhood of $x$ is contained in $A,$ which means that accumulation points need not be interior points.
For an example to see that the interior and set of accumulation points are not the same, consider $[0,2]\setminus\{1\}$. The interior of the set is $(0,1)\cup(1,2),$ but the set of accumulation points is $[0,2]$. The point $1$ is an example of a non-interior accumulation point.
For another example, consider the set $\{0\}\cup\{\frac1n:n\in\Bbb Z^+\}.$ No point of this set has a neighborhood contained in the set, so the interior is empty, but it can be shown that $0$ is the unique accumulation point. So, even if an accumulation point is in the set, it need not be an interior point.


TITLE: Proving $\lim_{n\to\infty}\frac{\ln(n+1)}{3^{n^2-n}}=0$
QUESTION [0 upvotes]: I'm trying to prove that $\lim_{n\to\infty}\frac{\ln(n+1)}{3^{n^2-n}}=0$.
I prefer not to convert it to functions and use the L'hopital rule and without the taylor series.
I can't seem to think of a simple way which proves it. How should I prove it?

REPLY [3 votes]: You can use comparison. For instance, for any $n\geq1$ we have $\ln(n+1)\leq n$ and $3^{n^2-n}\geq n^2$. This gives
$$
0\leq \frac{\ln(n+1)}{3^{n^2-n}}\leq \frac{n}{n^2}
$$
After a very simple limit calculation on the leftmost and rightmost expressions we conclude by the squeeze theorem that the middle expression must indeed converge to $0$.


TITLE: How to show that the following definitions of Lebesgue measurable sets are equivalent.
QUESTION [0 upvotes]: I am a graduate student of Mathematics and we have measure theory in this semester.Our instructor is not teaching very well so I am self-studying measure theory.The approach I am following defines Lebesgue measurable sets as follows:
Let $m^*$ be an outer measure on $\mathbb R$ defined by $m^*(A)=\inf\{\sum\limits_{n=1}^\infty \ell(I_n):I_1,I_2,...$ are open intervals such that $A\subset \bigcup\limits_{n=1}^\infty I_n\}$,then call the collection of the $m^*$-measurable sets as $\mathcal M$.By Caratheodory's theorem we know that $\mathcal M$ is a $\sigma$-algebra and $m^*$ is a measure when restricted to $\mathcal M$.
Now with this definition how can I show that the following are equivalent:
$1)$ $A\subset \mathbb R$ is such that for any $E\subset \mathbb R$, $m^*(E)=m^*(E\cap A)+m^*(E\cap A^c)$.
$2)$ $\exists$ a Borel set $B\subset A$ such that $m^*(A\setminus B)=0$.
$3)$ For each $\epsilon>0,\exists G_\epsilon\supset A$ open such that $|G_\epsilon\setminus A|<\epsilon$.
I am studying measure theory from several books and each of them follow a different approach,so I need some help to connect the different approaches.Can someone help me?

REPLY [1 votes]: Firstly, I think that the easiest order of implications is $1) \implies 3)$, then $3) \implies 2)$, then $2)\implies 1)$. Before drafting a full answer, I will give you the sketch of each argument.
For $1) \implies 3)$, you essentially want to appeal to the definition of the outer measure. You can find a collection of intervals $(I_i)_i$ which covers $A$ and for which $m^*(A) \leq m^*\big(\bigcup_i I_i\big) < m^*(A) + \varepsilon$ by definition of the infimum. Then the set $G_\varepsilon = \bigcup_i I_i$ is the one you want, and you use the measurability of $A$ to argue that the desired inequality holds.
Side remark: almost all of these arguments are most direct to show in the case where $m^*(A) < \infty$; to handle the case where $m^*(A) = \infty$ what I believe you want to do is apply the result to each of the sets $A_n = [-n,n] \cap A$ and then pass to the limit.
For $3) \implies 2)$, you can find an open set $G_\varepsilon$ for each $\varepsilon > 0$ satisfying the hypothesis; choose $B = \bigcap_\varepsilon G_\varepsilon$ to be your borel set, and argue the result holds by monotonicity.
For $2) \implies 1)$, first prove the lemma that all borel sets are measurable (if the open intervals are measurable then the sigma algebra generated by the open intervals, i.e. the borel algebra, is contained in the sigma algebra of measurable sets). Then, the lemma that sets of outer measure zero are measurable. Finally, stitch these together to get the desired equation for $A$ to hold.

$1) \implies 3)$
Suppose $A$ is a set such that $\forall E \ \big( m^*(E) = m^*(E \cap A) + m^*(E \cap A^c)\big)$. In the first case, suppose $m^*(A) < \infty$ and let $\varepsilon > 0$.
By definition of the infimum, one can find a collection of intervals $(I_i)_i$ such that $A \subset \bigcup_i I_i$ and
$$m^*(A) \leq \sum_i \ell(I_i) < m^*(A) + \varepsilon.$$
Let $G = \bigcup_i I_i$; we will show that $m^*(G \setminus A) < \varepsilon$. First, apply the hypothesis $1)$ with the choice $E = G$. Then we conclude
$$m^*(G) = m^*(G \cap A) + m^*(G \cap A^c) = m^*(A) + m^*(G \setminus A)$$
by the fact that $A \subset G$. Next, we have the following estimation by the countable subadditivity of the outer measure.
\begin{align*}
m^*(G) &= m^*\big(\bigcup_i I_i\big)\\
&\leq \sum_i m^*(I_i)\\
&< m^*(A) + \varepsilon
\end{align*}
The above also assumes that $m^*(I) = \ell(I)$ for an interval $I$.
We conclude that $m^*(G \setminus A) = m^*(G) - m^*(A) < \varepsilon$. Observe the necessity of the hypothesis $m^*(A) < \infty$ in this argument, else the difference is undefined.
In the second case, suppose $m^*(A) = \infty$ and again let $\varepsilon > 0$. For each natural $n > 0$, let $A_n = [-n,n] \cap A$. Observe that for each $n$, we have $m^*(A_n) < \infty$ in which case (by the above argument) there is an open set $G_n$ such that $A_n \subset G_n$ and $m^*(G_n \setminus A_n) < \varepsilon / 2^n$. Observe the scaling of $\varepsilon$, which hints that in the end we will sum over all of these differences.
Indeed, let $G = \bigcup_n G_n$; we argue that this is the desired $G$. First, we have the lemma
$$\Big( \bigcup_n G_n \Big) \setminus \Big(\bigcup_n A_n\Big) \subset \bigcup_n \big(G_n \setminus A_n\big).$$
If a point $x$ belongs to the LHS, then $x \in G_m$ for some $m$ while $x\notin A_n$ for any $n$. In particular, $x \notin A_m$ and therefore $x \in G_m \setminus A_m$, hence $x$ belongs to the RHS. The inclusion follows.
In light of this, we have the following estimation, again relying on the (monotonicity and) countable subadditivity of the outer measure.
\begin{align*}
m^*(G \setminus A) &= m^* \Big(\Big( \bigcup_n G_n \Big) \setminus \Big(\bigcup_n A_n\Big)\Big) \\
&\leq m^*\Big(\bigcup_n \big(G_n \setminus A_n\big)\Big)\\
&\leq \sum_n m^*(G_n \setminus A_n)\\
&< \sum_n \varepsilon /2^n\\
&= \varepsilon
\end{align*}
This concludes the first implication.

$2) \implies 1)$
I shall preface this section with the observation that this proof is a bit of a headache, not to understand but to come up with, once you've forgotten the order in which things must be done. Drawing pictures of $A$, $B$ and $E$ venn-diagram style helps immensely to determine which set must be split by which other set.
I shall leave it to you to cover the lemmas that each Borel set is measurable (because each open interval is measurable, plus sigma-algebra-generating tricks), as is each set with outer measure zero.
So let $A \subset \mathbb R$ and suppose $B \subset A$ is a Borel set such that $m^*(A \setminus B) = 0$. Observe that both $B$ and $A \setminus B$ are therefore measurable. Let $E \subset \mathbb R$ be arbitrary and apply the measurability of $B$; we have that
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c).$$
We then claim that $m^*(E \cap A) = m^*(E \cap B)$ and $m^*(E \cap A^c) = m^*(E \cap B^c)$. For the first, apply the measurability of the set $B$ to the set $E \cap A$; we have
$$m^*(E \cap A) = m^*(E \cap A \cap B) + m^*(E \cap A \cap B^c) = m^*(E \cap B) + 0.$$
The latter equality follows from the fact that $B \subset A$ together with $(E \cap A \cap B^c) \subset (A \setminus B)$ which hence has outer measure zero by hypothesis + monotonicity.
For the last step, we apply the measurability of $A \setminus B$ to the set $E \cap B^c$. We obtain
$$m^*(E \cap B^c) = m^*(E \cap B^c \cap (A\setminus B)) + m^*(E \cap B^c \cap (A\setminus B)^c) = 0 + m^*(E \cap A^c).$$
The latter equality follows again from hypothesis + monotonicity, together with set equality which can be argued by chasing the logic out of a venn diagram. Specifically,
$$E \cap B^c \cap (A\setminus B)^c = E \cap A^c.$$
We finally conclude $1)$. This proof may also be (carefully) adapted to the dual case, where $B$ is a borel set such that $A \subset B$ and $m^*(B \setminus A) = 0$.
This concludes the second implication.

$3) \implies 2)$
As I mentioned in the comments, I think it would be generally easier to replace condition $2)$ with the dual statement (that there exists a Borel set $B \supset A$ such that $m^*(B \setminus A) = 0$), prove that these three are equivalent, then recover the inner Borel approximation in a separate argument. Nevertheless, as these are all ultimately equivalent, it is theoretically possible to go directly from $3)$ to $2)$ in its current form, and so here is an argument of that.
Firstly, for each $n$ there is an open set $G_n \supset A$ such that $m^*(G_n \setminus A) < 1/n$. Let $B = \bigcap_n G_n$, which is Borel. We have $A \subset B$ since $A \subset G_n$ for every $n$, and therefore by monotonicity we have
$$0 \leq m^*(B \setminus A) \leq m^*(G_n \setminus A) < 1/n$$
for every $n$, hence $m^*(B \setminus A) = 0$ by the squeeze theorem. This is the inverse of what we aim to prove, but here is how the gap may be bridged: we shall first use the set $B$ to argue that $A$ satisfies $1)$ by identical argument as $2) \implies 1)$ above, then hence $A^c$ satisfies $3)$ and the above construction can be repeated to finally produce the desired Borel set $B'' \subset A$.
Indeed, we repeat that any Borel set and any set of outer measure zero is measurable, so $B$ and $B \setminus A$ are measurable. Let $E \subset \mathbb R$ be arbitrary; by the measurability of $B$ we have
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c).$$
Then we apply the measurability of $B \setminus A$ to the set $E \cap B$ and deduce
$$m^*(E \cap B) = m^*(E \cap B \cap (B \setminus A)) + m^*(E \cap B \cap (B \setminus A)^c) = 0 + m^*(E \cap A).$$
Here we have used that $E \cap B \cap (B \setminus A) \subset B \setminus A$ and therefore has outer measure zero by monotonicity. We have also used the fact that $E \cap B \cap (B \setminus A)^c = E \cap A$ which can be easily proven under the assumption that $A \subset B$.
Indeed, if $x$ is a point belonging to $E \cap B \cap (B\setminus A)^c$ then we have $x \in E$ and $x \in B$ and $X \notin B \setminus A$, the third statement being equivalent to $x \notin B$ or $x \in A$. Since $x \in B$ is already known, then we must have $x \in A$ and hence $x \in E \cap A$. This demonstrates that $E \cap B \cap (B\setminus A)^c \subset E \cap A$; the reverse inclusion is checked similarly.
We then apply the measurability of $B$ to the set $E \cap A^c$ and deduce
$$m^*(E \cap A^c) = m^*(E \cap A^c \cap B) + m^*(E \cap A^c \cap B^c) = 0 + m^*(E \cap B^c).$$
Here again we use $E \cap A^c \cap B \subset A^c \cap B = B \setminus A$ which has outer measure zero, as well as $E \cap A^c \cap B^c = E \cap B^c$. This follows because $A \subset B$ implies $B^c \subset A^c$, in which case $A^c \cap B^c = B^c$.
We have therefore shown that $m^*(E \cap B) = m^*(E \cap A)$ and $m^*(E \cap B^c) = m^*(E \cap A^c)$, hence
$$m^*(E) = m^*(E \cap B) + m^*(E \cap B^c) = m^*(E \cap A) + m^*(E \cap A^c).$$
It therefore follows that $A$ satisfies $1)$.
By symmetry in $1)$ we have that $A^c$ satisfies $1)$ as well. Then by earlier argument, $A^c$ satisfies $3)$. Finally, repeat the construction at the beginning to find a Borel set $B' \supset A^c$ such that $m^*(B' \setminus A^c) = 0$, and let $B'' = (B')^c$. This is the desired Borel set for the condition $2)$.
Firstly, $B'' \subset A$ is clear by definition. Then, one may unpack complements to conclude the set equality $A\setminus B'' = B' \setminus A^c$, in which case
$$m^*(A \setminus B'') = m^*(B' \setminus A) = 0.$$
This completes the theorem.


TITLE: Understanding hollow waveguides. What is $\lambda_g$ on the figure?
QUESTION [2 upvotes]: I am trying to understand the field distribution in waveguides.
On the figure is $HE_{12}$ mode of hollow fiber.
What is the $\lambda_g$ ? Is it $\lambda_g$ = $\lambda_0$ / $n_{eff}$.
Does the z field component repeat (circles) many times (L/$\lambda_g$) over the length L of the fiber ?

REPLY [2 votes]: $\lambda_g$ is the wavelength inside the waveguide. Yes, that pattern repeats itself all along the waveguide - and it propagates at a velocity that is less than the velocity of electromagnetic waves in free space.


TITLE: Leibniz rule for Pseudo-differential operators of negative order
QUESTION [2 upvotes]: Does anyone know of some good references for a fractional Leibniz rule for pseudo-differential operators of negative order? As a specific example, I would like to compute $\partial_{x}^{-1}(uv)$, assuming $\xi \neq 0$ in phase space and mean-zero $u,v$.

REPLY [2 votes]: Let $P$ be a pseudodifferential operator with symbol $p(x,\xi)$ belonging to $S^m_{1,0}$ and let $a(x)$ be a $C^\infty$ function. Then the operator $Pa$ defined by $Pa u=P(au)$ is a pseudodifferential operator of order $m$ with symbol $q(x,\xi)$ such that
$$
q-\sum_{\vert \alpha\vert\le N} \frac{1}{i^{\vert \alpha\vert}\alpha !}(\partial_\xi^\alpha p)(x,\xi)
(\partial_x^\alpha a)(x)\in S^{m-N-1}_{1,0},
$$
which provides an asymptotic expansion of the symbol of the operator Pa, following simply from the composition formula.


text:
\section{Benchmark FSI2: flow-induced vibrations}\label{chap:testFSI}
In this section, we perform the FSI simulation described in Section \ref{chap:benchmark}. From several simulation scenarios proposed in \cite{turek2006proposal}, we choose a scenario titled FSI2 since it corresponds to the largest magnitude of beam displacement and mesh deformation. Using this FSI simulation, we test and compare different mesh deformation techniques (MDTs) introduced in Section \ref{chap:methods}. We use stiffening degree $\chi=2.5$ for all MDTs but BE and IBE. To them, we apply no local stiffening.

For the analysis, we refine the geometry parametrization five times using uniform $h$-refinement and perform the simulation for 15 seconds with a time step $\Delta t=0.0025$ s. For time integration, we use the Newmark method \cite{wriggers2008nonlinear} with $\beta=0.5$ and $\gamma=1$ for the structure and the IMEX scheme \cite{john2016finite} with $\theta=0.5$ and no stabilization for the fluid. We achieve the coupling of fluid and structure by means of the partitioned Fluid-Dirichlet-Structure-Neumann algorithm \cite{dorfel2011fluid}.

The FSI2 scenario is characterized by the following parameters: fluid density $\rho_f=10^3$ kg$\cdot$m$^{-3}$, fluid kinetic viscosity $\nu_f=10^{-3}$ m$^2\cdot$s$^{-1}$, maximum inflow velocity $v_\text{max}=1.5$ m$\cdot$s$^{-1}$, structure density $\rho_s = 10^4$ kg$\cdot$m$^{-3}$, structure Young's modulus $E=1.4\times10^6$ kg$\cdot$m$^{-1}\cdot$s$^{-2}$, structure Poisson's ratio $\nu_s=0.4$, gravitational acceleration $\f{g}=(0,0)^T$ m$\cdot$s$^{-2}$ and mesh Poisson's ratio $\nu_a=0.3$ (where applicable). The corresponding Reynolds number is $Re = 100$, which results in an unstable flow and development of vortex shedding. Alternating downward and upward forces exerted on the structure by the fluid lead to oscillations of the beam which grow in magnitude until they reach a fully periodic regime. Figure \ref{fig:FSI2velocity} illustrates typical fluid velocity field and beam deformation when the oscillations are fully developed.

To assess the simulation accuracy, we study the following quantities of interest when the beam oscillations are fully developed : $x$- and $y$-displacement components $u_s^x(A)$ and $u_s^y(A)$ of the point $A$ located in the middle of the beam right end; and drag and lift forces $F_D$ and $F_N$ exerted on the structure by the fluid which are defined as
\begin{equation}
(F_D,F_N) = \int\displaylimits_{\Sigma(t)}\pmb{\sigma}_f(\f{v}_f,p_f)\cdot\f{n}d\f{s}.
\end{equation}
Here, $\Sigma(t)$ is the entire boundary of the submerged solid, including the rigid disk and the flexible beam, at time $t$. Since we expect the quantities of interest to behave periodically, we report them in terms of their mean $\big((*)_\text{max}+(*)_\text{min}\big)/2$, amplitude $\big((*)_\text{max}-(*)_\text{min}\big)/2$ and frequency. In Table \ref{table:FSI2res}, we compare the simulation results obtained with the TINE technique against the reference results from \cite{turek2006proposal}.  Overall, our results seem to undershoot the reference values by about 5$\%$, which can be expected since we use much fewer degrees of freedom and a larger time step than the reference simulation. Despite this discrepancy, we are more than capable of reproducing a qualitatively correct behavior of the system and can use it to study the MDTs. Figures \ref{fig:FSI2_lift} and \ref{fig:FSI2_disp} illustrate behavior of the lift, drag and beam displacement.

\begin{figure*}
	\centering
	\includegraphics[height=3.2cm]{images/fig10a.png}
	\includegraphics[height=3.2cm]{images/fig10b.png}\caption{Benchmark FSI2: fully developed oscillation regime. Fluid velocity field for the maximal upward und downward beam deflection.}
	\label{fig:FSI2velocity}
\end{figure*}

\begin{figure*}
	\centering
	\includegraphics[height=3.5cm]{images/fig11a.pdf}
	\includegraphics[height=3.5cm]{images/fig11b.pdf}
	\caption{Benchmark FSI2: drag and lift behavior with different MDTs.}
	\label{fig:FSI2_lift}
\end{figure*}

\begin{figure*}
	\centering
	\includegraphics[height=3.5cm]{images/fig12a.pdf}
	\includegraphics[height=3.5cm]{images/fig12b.pdf}
	\caption{Benchmark FSI2: displacement of the middle point of the beam right end with different MDTs.}
	\label{fig:FSI2_disp}
\end{figure*}

\begin{table*}
	\centering
	\caption{Benchmark FSI2: simulation results with TINE ($\Delta t$=0.0025s, 44122 DoFs) vs reference results ($\Delta t$=0.001s, 304128 DoFs).}
	\begin{tabular}{ l c c c c c}
		\hline\noalign{\smallskip}
		& $F_D$ [$N$]&$F_L$ [$N$] & $u_s^x(A)$ [$10^{-3}m$]&$u_s^y(A)$ [$10^{-3}m$] & Fr [$1/s$] \\
		\noalign{\smallskip}\hline\noalign{\smallskip}
		Results &  203.19$\pm$63.88 & 1.22$\pm$223.35 & -12.75$\pm$11.37 & 1.24$\pm$74.55 & 1.94\\
		Reference &208.83$\pm$73.75& 0.88$\pm$234.2 &-14.58$\pm$12.44& 1.23$\pm$80.6 &2.0 \\
		\noalign{\smallskip}\hline
	\end{tabular}

	\label{table:FSI2res}
\end{table*}

Let us now focus on the fluid mesh deformation. When performing the FSI simulation, we apply each of the seven MDTs (HE, IHE, BE, IBE, LE, ILE and TINE) and study how a particular MDT handles mesh deformations occurring during the simulation. Figure \ref{fig:FSI2success} illustrates what portion of the simulation can be completed using different MDTs. As we can observe, simulations with the HE, IHE and IBE techniques had to be terminated before they could reach the end. All three techniques have failed to maintain bijectivity of the ALE mapping; however, different reasons have led to the failure. In the case of HE and IHE, the simulations have stopped at the ninth second---when the oscillations in the system start to develop. As the applied mesh deformation grows, the HE and IHE techniques fail due to their intrinsic inability to handle large deformations.

On the other hand, the IBE technique is able to handle mesh deformations occurring in the simulation but suffers from the accumulated deformation effect described in Section \ref{chap:testALE}. As a result, the simulation fails at the 14th second. Figure \ref{fig:FSI2_ALEnorm} depicts behavior of the ALE norm for all MDTs. We can observe that the ILE technique suffers from even stronger accumulated deformation than IBE but still manages to maintain a bijective ALE mapping until the simulation end. Unfortunately, the highly distorted mesh produced by the ILE technique (see Figure \ref{fig:FSI2meshILE}) affects the simulation results, see Figures \ref{fig:FSI2_lift} and \ref{fig:FSI2_disp}. Instead of a stable periodic behavior in the fully developed oscillation regime, we observe signs of damping. We observe a similar effect when using the IBE technique: the accumulated mesh deformation results in spurious amplification of the oscillations.

\begin{figure*}
	\centering
	\includegraphics[height=3.3cm]{images/fig13.pdf}
	\caption{Benchmark FSI2: portion of simulation completed by different MDTs before the ALE mapping becomes invalid.}
	\label{fig:FSI2success}
\end{figure*}

\begin{figure*}
	\centering
	\includegraphics[height=3.3cm]{images/fig14.pdf}
	\caption{Benchmark FSI2: ALE displacement norm with different MDTs.}
	\label{fig:FSI2_ALEnorm}
\end{figure*}

\begin{figure}
	\centering
	\includegraphics[height=6cm]{images/fig15a.png}\\
	\vspace{0.2cm}
	\includegraphics[height=6cm]{images/fig15b.png}
	\caption{Benchmark FSI2: accumulated distortion of the fluid mesh during the last oscillation period with the ILE technique.}
	\label{fig:FSI2meshILE}
\end{figure}

\begin{figure}
	\centering
	\includegraphics[height=6cm]{images/fig16a.png}\\
	\vspace{0.2cm}
	\includegraphics[height=6cm]{images/fig16b.png}
	\caption{Benchmark FSI2: fluid mesh with during the last oscillation period with the TINE technique.}
	\label{fig:FSI2meshTINE}
\end{figure}

Of the seven MDTs considered in this work, only three---BE, LE and TINE---were able to handle mesh deformations occurring in the simulation and maintain high mesh quality until the simulation end. Most importantly, the BE, LE and TINE techniques have demonstrated no signs of accumulated distortion. Using these techniques, we were able to reproduce a stable periodic behavior of the system and correct simulation results. Figure \ref{fig:FSI2meshTINE} depicts the fluid mesh at the end of the simulation deformed using the TINE technique. Of course, the BE, LE and TINE techniques differ a lot in terms of their computational cost. However, since construction of the ALE mapping corresponds only to a small portion of a total computational effort required to perform an FSI simulation, the choice of MDT does not affect the total computational cost too much. Table \ref{table:FSItimeComparison} compares computational cost of  BE, LE and TINE against the ILE technique which is often considered a default option in the FSI community.

\begin{table}
	\centering
	\caption{Benchmark FSI2: computational time comparison for MDTs which successfully completed the simulation. The ILE technique time is used as a reference point for comparison.}
	\begin{tabular}{l c c}
		\hline\noalign{\smallskip}
		& ALE time &  Total time\\
		\noalign{\smallskip}\hline\noalign{\smallskip}
		BE   & 2h10m (-48\%) & 23h55m (-7.3\%)\\
		LE & 1h39m (-60\%) & 23h20m (-9.6\%) \\
		ILE  & 4h7m (+0\%) & 25h48m (+0.0\%)\\
		TINE & 4h52m (+18\%) & 26h27m (+2.5\%) \\
		\noalign{\smallskip}\hline
	\end{tabular}
	\label{table:FSItimeComparison}
\end{table}

config: arxiv


TITLE: null,row and column space can have different bases
QUESTION [0 upvotes]: When solving examples that ask me to find a basis for the null, row or columnspace for a matrix I often get a different basis than what I find in the answer sheet. I know you can come up with many different bases for any given vector space so my answers can still be correct, I just have to make sure I made no mistakes. What I'm unsure about is if the basis I come up with can be composed of a different number of elements than what the answer sheet shows in it's solution and still be a correct answer.

REPLY [1 votes]: No, that cannot happen. Any two bases of a vector space have the same number of elements. That's a standard Linear Algebra theorem.


TITLE: Point J(−2,5) is 1 unit away from a line L that passes through point K(−3,0). Find the line L.
QUESTION [0 upvotes]: how do i find the equation of the L line given its distance from point J which is 1 unit away from it. there could be two Lines to this equation but I don't know how to find it

REPLY [0 votes]: Let the normal to the line be the unit vector $(\cos \theta, \sin \theta)$
Since $K(-3,0)$ is on the line, then the equaiton of the line is
$ \cos \theta (x + 3) + \sin \theta y = 0 $
Since $J(-2, 5)$ is $1 $ unit away from the line then,
$ 1 = |\cos \theta + 5 \sin \theta |$
Define $\phi = \text{ATAN2}(1, 5) = \cos^{-1} \dfrac{1}{\sqrt{26}}$ , then
$1 = \sqrt{26} | \cos(\theta - \phi ) |$
From which,
$ \cos(\theta - \phi) = \pm \dfrac{1}{\sqrt{26}} $
Hence,
$\theta = \phi + \cos^{-1} \pm \dfrac{1}{\sqrt{26}} = 2 \phi , \pi $
If $\theta = \pi$ then the equation of the line is: $-(x + 3) = 0$, i.e. $\boxed{x = -3}$
If $\theta = 2 \phi$, then the equation of the line is $\cos 2 \phi (x + 3) + \sin 2 \phi = 0 $
Now, $\cos 2 \phi = 2 \cos^2 \phi - 1 = \dfrac{2}{26} - 1 = -\dfrac{12}{13}$
And $\sin 2 \phi = \dfrac{5}{13} $
Hence, the equation is: $\boxed{-12 (x + 3) + 5 y = 13}$


TITLE: Are photons emitted by a magnet?
QUESTION [6 upvotes]: If you put a photon detector near a magnet (with the magnetic field static in time), is there some probability that the photon detector will detect a photon?
Does QFT not predict that a photon could be detected? If we have a uniform static classical $\vec{B}$ field in the $z$ direction, e.g. $A_{classical}^{\mu}=(0,-By,0,0)$ and we couple electrons to this classical field by adding the term $H_{interaction}=\bar{\psi}\gamma_{\mu}\psi A^{\mu}_{classical}$ to the QED Lagrangian, then there are then Feynman diagrams which seem to indicate that the classical field can produce real photons. (e.g. the Feynman diagram where the classical source emits a photon which fluctuates into an $e^{+}e^{-}$ bubble with a real photon emitted off of the bubble).

REPLY [6 votes]: Here is an argument based on spacetime symmetry and the fact that U(1) theory is abelian (therefore photons don't interact with each other) to show that the probability is zero.
Suppose we start from a uniform magnetic field in the $z$ direction. Since there is no preferred direction in the $xy$-plane, the final state photon must be emitted along the $z$ direction. The final state magnetic field must be uniform along the $z$ direction as well, due to the same argument.
Therefore if the probability is non-zero, a uniform magnetic field will continue to decay spontaneously until its strength approaches zero (as we approach $T=\infty$), and produce a collection of photons. Therefore the end state at $T=\infty$ will simply be a collection of photons, ie. $a^\dagger(\omega_1)\cdots a^\dagger(\omega_n)|0\rangle$.
Now QED is time symmetric. Therefore we can start from the final collection of photons and evolve backwards in time. However since in U(1) theories gauge particles do not interact with each other, we simply have a collection of free particles --- the interaction Hamiltonian is zero, and the initial state must be a collection of free particles as well. QED.
EDIT: Before anybody starts babbling about "non-linear" optics and "four-photon scattering", let me point out that the current experimentally accessible field intensity is well below the Schwinger limit [1], at which point we expect new physics to come out. Our current experimentally verified theory contains only linear photon interactions. As theorists tend to get carried away with formalism we should always remind ourselves that physics is ultimately an experimental science, and any theoretical formalism is ultimately an effective theory.
[1] https://en.wikipedia.org/wiki/Schwinger_limit


TITLE: Stackelberg Competition Matlab Problem
QUESTION [0 upvotes]: This is an extension to the problem I asked in a previous question which can be seen here, Economic price optimisation problem.
I have programmed a function in MATLAB to solve the Stackelberg competition problem set up in the question linked above. The problem has to be solved numerically which is why I decided to use MATLAB. The profit functions for the two Firms in competition are below, where the subscript number denotes the firm.
$$\Pi_1(p_1,p_2)=\frac{25p_1}{(p_1-0.6p_2)^{1.1}}-0.0815[\frac{25}{(p_1-0.6p_2)^{1.1}}]^{1.9}$$
$$\Pi_2(p_1,p_2)=\frac{25p_2}{(p_2-0.6p_1)^{1.1}}-0.0645[\frac{25}{(p_2-0.6p_1)^{1.1}}]^{1.9}$$
The code for the function I programmed to solve this can be seen below.
function y = OptimalPrices(p1, epsilon)
% Function finds optimal prices p1 and p2 for Stackelberg competition

% initial value

x = 1;
p1_guess = p1;

while p1_guess < 10;
    left_init = 0.5314*p1_guess + 0.7368; %left limit has to increase with the p1 guess
    p2 = MaxPi2(p1_guess, left_init, 6, 0.00001); %calculates p2 value
    Pi1 = f1_prof(p1_guess, p2); %finds Firm 1 profit using the p1 and p2 values
    p1vals(x) = p1_guess; %stores p1 value in a vector
    p2vals(x) = p2; %stores p2 value in a vector
    Pi1vals(x) = Pi1; %stores Pi1 value in a vector

    if imag(Pi1vals(x))>0 %if result is complex, sets equal to 0
        Pi1vals(x) = 0;
    end

    %increments
    x = x+1;
    p1_guess = p1_guess + epsilon;

end

[Pi1max, pos] = max(Pi1vals); %finds biggest Firm 1 profit and its position in vector
p1opt = p1vals(pos);
p2opt = p2vals(pos);

% Display results
p1opt
p2opt

Firm1_Profit = f1_prof(p1opt, p2opt)
Firm2_Profit = f2_prof(p1opt, p2opt)

end

As you may have noticed, nested within this function is another function titled $MaxPi2$. This solves the maximisation problem, $max _{p_2} \Pi_2(p_1,p_2)$, using the Bisection Method. The code for this can be seen below.
function p2 = MaxPi2(p1,l,r,epsilon)
% Finds value for p2 that maximises Pi2, using Bisection Method
% Looking for when the derivative of Firm 2's profit function is 0

% 'l' is the left end of the interval; 'r' is the right end
% epsilon is the precision parameter

% values of function at the ends
fl = f2_prof_prime(p1, l);
fr = f2_prof_prime(p1, r);

if fl*fr>0  % check that solution within the interval
    p2 = 0;
else
    p2 = (l+r)/2;
    dist = (r-l)/(1+abs(p2));
    % iterations
    while dist>=epsilon % while the length is large
        f = f2_prof_prime(p1, p2);
        if fl*f<0       % left half of interval
            r = p2;
            fr = f;
        else            % right half of interval
            l = p2;
            fl = f;
        end
        p2 = (l+r)/2;
        dist = (r-l)/(1+abs(p2));
    end

end
end

If I run the function $OptimalPrices(0, 0.0001)$, I get the following results, where $p_1$ and $p_2$ are the optimal prices set by the firms in this Stackelberg competition.
$$p_1=3.6150$$
$$p_2=3.1521$$
$$\Pi_1(p_1,p_2)=37.8202$$
$$\Pi_2(p_1,p_2)=50.0166$$
These results clearly show that Firm 2 will make a higher profit overall. Now, I was taught the "First Mover Advantage", that is, the Firm that sets its price first will make higher profits overall. I would just like to know if that is always the case, since although Firm 1 moves first in this competition, Firm 2 is making more profit? Does that mean my results are wrong somewhere? Or would Firm 2 actually make more profits in this case?

REPLY [0 votes]: In the Stackelberg competition where firms set price sequentially, it is not rare for the second firm to have a better payoff if the demand functions are not symmetric. In your case firm 2 has lower total costs and therefore, it is likely it will have higher total profits.


TITLE: Simultaneous Equations with a Modulus
QUESTION [3 upvotes]: I must confess I'm finding it pretty hard to get my head round these equations, much less solve them. So any help would be enormously appreciated. I'd appreciate if you'd show the logic of how to solve these, and keep it simple without explicitly mentioning the Extended Euclidean what have you, and all this other stuff. (I've seen it mentioned in other answers, but it's impenetrable to me). I'm only interested in the step-by-step process, and how it works, rather than just quoting formulaic methods. So please keep it simple and jargon-free. Thank you. Equations are:
$$
\begin{cases}
z_1 (8z_2 +  2) = 3 \mod 11\\
z_2 (4z_1 + 11) = 5 \mod 13
\end{cases}
$$
I'm trying to solve for $z_1$ and $z_2$ modulo whatever.

REPLY [1 votes]: So you will have $10$ answer pairs for the first equation.
Each of these is obtained by getting the inverse of some chosen value of $z_1$ and then solving for
$(8z_2 + 2) \equiv 3z_1^{-1} \bmod 11$ which calculates out to $z_2\equiv 7(3z_1^{-1}-2) \bmod 11$ since $8^{-1}\equiv 7 \bmod 11$.
You know in advance that $z_1\equiv 0 \bmod 11$ is not going to give a solution. Note that $z_2\equiv 0$ can be a solution though; in fact $(z_1,z_2)\equiv (7,0)\bmod 11$ is a solution for the first equation.
So for example with $z_1\equiv 6$, we know $z_1^{-1}\equiv 2$ so $z_2 \equiv 7(3\cdot 2 -2) \equiv 28\equiv 6 \bmod 11$.
Similarly you will have $12$ answer pairs for the second equation. Here of course $z_2\equiv 0 \bmod 13$ cannot be a solution, but this time $z_1\equiv 0$ is feasible and $(z_1,z_2)\equiv (0,4)\bmod 13$ satisfies the requirement.
Combining every solution to the first equation with every solution to the second through the Chinese Remainder Theorem, you will have $120$ answer pairs $\bmod 143$.
An example: combining $(6,6)\bmod 11$ with $(0,4)\bmod 13$ gives $(z_1,z_2)\equiv(39,17)\bmod 143$ as one of the answer pairs.