text:
\begin{document}

\begin{abstract}
Let $G$ be a finite group with cyclic Sylow $p$-subgroup,
and let $k$ be a field of characteristic $p$. Then
$H^*(BG;k)$ and $H_*(\Omega BG\phat;k)$ are $A_\infty$ algebras whose
structure we determine up to quasi-isomorphism.
\end{abstract}

\maketitle

\section{Introduction}
The general context is that we have a finite group $G$, and a field
$k$ of characteristic $p$. We are interested in the differential
graded cochain algebra $C^*(BG;k)$
and the differential graded algebra $C_*(\Omega (BG\phat);k)$ of chains
on the loop space: these two are Koszul dual to each
other, and the Eilenberg-Moore and Rothenberg-Steenrod spectral
sequences relate the cohomology ring $H^*(BG;k)$  to the homology
ring $H_*(\Omega (BG\phat);k)$. Of course if $G$ is a $p$-group, $BG$ is $p$-complete so $\Omega
(BG\phat)\simeq G$, but in general $H_*(\Omega (BG\phat); k)$ is
infinite dimensional. Henceforth we will omit the brackets from
$\Omega (BG\phat)$. \\[1ex]

We consider a simple case where
the two rings are not formal, but we can identify the $A_{\infty}$
structures precisely (see Section \ref{sec:Ainfty} for a brief summary on $A_{\infty}$-algebras). From now on we suppose specifically that  $G$ is a finite group with cyclic Sylow $p$-subgroup $P$,
and let $BG$ be its classifying space. Then the inclusion of
the Sylow $p$-normaliser $N_G(P) \to G$ and the quotient map $N_G(P)
\to N_G(P)/O_{p'}N_G(P)$
induce mod $p$ cohomology
equivalences
\[ B(N_G(P)/O_{p'}N_G(P)) \from BN_G(P) \to BG, \]
and
hence homotopy equivalences after
$p$-completion
\[ B(N_G(P)/O_{p'}N_G(P))\phat \xleftarrow{\ \sim\ } BN_G(P)\phat
  \xrightarrow{\ \sim \ } BG\phat. \]
Here, $O_{p'}N_G(P)$ denotes the largest normal $p'$-subgroup of
$N_G(P)$. Thus $N_G(P)/O_{p'}N_G(P)$ is a semidirect product
$\bZ/p^n\rtimes \bZ/q$, where $q$ is a divisor of $p-1$, and
$\bZ/q$ acts faithfully as a group of automorphisms of $\bZ/p^n$. In
particular, the isomorphism type of $N_G(P)/O_{p'}N_G(P)$ only depends
on $|P|=p^n$ and the inertial index $q=|N_G(P):C_G(P)|$, and therefore
so does the homotopy type of $BG\phat$. Our main theorem determines
the multiplication maps $m_i$ in the $A_\infty$ structure on
$H^*(BG;k)$ and $H_*(\Omega (BG\phat);k)$ arising from $C^*(BG;k)$ and
$C_*(\Omega (BG\phat); k)$ respectively. We will suppose from now on
that $p^n>2, q>1$ since the case of a $p$-group is well understood.

The starting point is the cohomology ring
$$H^*(BG; k)=H^*(B\bZ/p^n; k)^{\bZ/q}=k[x]\otimes \Lambda(t)
\mbox{ with }|x|=-2q, |t|=-2q+1.  $$
There is a preferred generator $t\in
H^1(B\bZ/p^n;k)=\mathrm{Hom}(\bZ/p^n,k)$ and we take $x$ to be the
$n$th Bockstein of $t$.

Before stating our result we should be clear about grading and signs.

\begin{remark}
\label{rem:deg}
We will be discussing both homology and cohomology, so we should be
explicit that everything is graded homologically, so that
differentials always lower degree. Explicitly, the degree of an
element of $H^i(G;k)$ is $-i$.
\end{remark}

\begin{remark}
\label{rem:signs}
Sign conventions for Massey products and $A_{\infty}$ algebras mean
that a specific sign will enter repeatedly in our statements, so for
brevity we  write
$$\eps (s)=
\begin{cases}
+1 & s\equiv 0, 1 \mod 4\\
-1 &s\equiv 2, 3 \mod 4\\
\end{cases}. $$
\end{remark}

\begin{theorem}
Let $G$ be a finite group with cyclic Sylow $p$-subgroup $P$ of order
$p^n$ and inertial index $q$ so that
$$H^*(BG;k)
=k[x]\otimes \Lambda(t) \mbox{ with } |x|=-2q, |t|=-2q+1 \mbox{ and } \beta_nt=x$$  Up to quasi-isomorphism, the $A_\infty$ structure on $H^*(BG;k)$ is determined by
\[ m_{p^n}(t,\dots,t)=\eps (p^n) x^{h} \]
where $h=p^n-(p^n-1)/q$. This implies
\[ m_{p^n}(x^{j_1}t,\dots,x^{j_{p^n}}t)=\eps (p^n) x^{h+j_1+\dots+j_{p^n}}\]
for all $j_1, \ldots, j_{p^n}\geq 0$.
 All $m_i$ for $i>2$ on all other
$i$-tuples of monomials give zero.

If $q>1$ and  $p^n\ne 3$ then
\[ H_*(\Omega BG\phat;k) = k[\tau] \otimes \Lambda(\xi) \mbox{ where }
  |\tau|=2q-2, |\xi|=2q-1. \]
 Up to quasi-isomorphism, the
$A_\infty$ structure is determined  by
\[ m_h(\xi,\dots,\xi )=\eps (h) \tau^{p^n}.  \]
This implies
\[ m_h(\tau^{j_1}\xi,\dots,\tau^{j_h}\xi)=\eps (h) \tau^{p^n+j_1+\dots+j_h} \]
for all $j_1, \ldots, j_{p^n}\geq 0$.   All $m_i$ for $i>2$ on all other $i$-tuples of monomials give
zero.

If $q>1$ and $p^n=3$ then $q=2$ and
\[ H_*(\Omega BG\phat;k) = k[\tau,\xi]/(\xi^2+\tau^3), \]
and all $m_i$ are zero for $i>2$.
\end{theorem}

\section{The group algebra and its cohomology}

We assume from now on, without loss of generality,
that $G$ has a normal cyclic Sylow $p$-subgroup
$P=C_G(P)$, with inertial index $q=|G:P|$. We shall assume that $q>1$,
which then forces $p$ to be odd. For notation, let
\[ G=\langle g,s\mid g^{p^n}=1,\ s^q=1,\
sgs^{-1}=h^\gamma\rangle\cong \bZ/p^n\rtimes\bZ/q, \]
where $\gamma$ is a primitive $q$th root of unity modulo $p^n$.
Let $P=\langle g\rangle$ and $H=\langle s\rangle$ as subgroups of
$G$.

Let $k$ be a field of characteristic $p$.
The action of $H$ on
$kP$ by conjugation preserves the radical series, and since $|H|$ is
not divisible by $p$, there are invariant complements. Thus we may
choose an element $U\in J(kP)$ such that $U$ spans an $H$-invariant
complement of $J^2(kP)$ in $J(kP)$. It can be checked that
\[ U = \sum_{\substack{1\le j \le p^n-1, \\(p,j)=1}} g^j/j \]
is such an element, and that $sUs^{-1}=\gamma U$. This gives us the
following presentation for $kG$:
\[ kG = k\langle s,U \mid U^{p^n}=0,\ s^q=1,\ sU = \gamma Us
  \rangle. \]

We shall regard $kG$ as a $\bZ[\frac{1}{q}]$-graded
algebra with $|s|=0$ and $|U|=1/q$. Then the bar resolution
is doubly graded, and taking homomorphisms into $k$, the
cochains $C^*(BG;k)$ inherit a double grading. The differential
decreases the homological grading and preserves  the internal
grading. Thus the cohomology $H^*(G,k)=H^*(BG;k)$ is doubly graded:
\[ H^*(BG;k) = k[x] \otimes \Lambda(t) \]
where $|x|=(-2q,p^n)$, $|t|=(-2q+1,h)$, and $h=p^n-(p^n-1)/q$.
Here, the first degree is homological, the second internal.
The Massey product $\langle t,t,\dots,t\rangle$ ($p^n$ repetitions) is
equal to $-x^h$. This may easily be determined by restriction to $P$,
where it is well known that the $p^n$-fold Massey product of the
degree one exterior generator is a non-zero degree two element.
The usual convention is to make the constant $-1$, because
this Massey product is minus the $n$th Bockstein of $t$ \cite[Theorem 14]{Kraines:1966a}.

\section{$A_\infty$-algebras}
\label{sec:Ainfty}

An $A_{\infty}$-algebra over a field is a $\bZ$-graded vector space
$A$ with graded maps $m_n: A^{\otimes n}\rightarrow A$ of degree $n-2$ for
$n\geq 1$ satisfying
$$\sum_{r+s+t=n}(-1)^{r+st}m_{r+1+t}(id^{\tensor r}\tensor m_s \tensor
id^{\tensor t})=0 $$
for $n\geq 1$. The map $m_1$ is therefore a differential, and the map
$m_2$ induces a product on $H_*(A)$.

A theorem of Kadeishvili~\cite{Kadeishvili:1982a} (see also
Keller~\cite{Keller:2001a,Keller:2002a} or
Merkulov~\cite{Merkulov:1999a}) may be stated as
follows. Suppose that we are given a differential
graded algebra $A$,  over a field $k$. Let $Z^*(A)$ be the cocycles,
$B^*(A)$ be the coboundaries, and $H^*(A)=Z^*(A)/B^*(A)$.
Choose a vector space splitting $f_1\colon H^*(A) \to Z^*(A) \subseteq A$
of the quotient.
Then this gives by an inductive procedure an $A_\infty$ structure
on $H^*(A)$ so that the map $f_1$ is the degree one part of a quasi-isomorphism of
$A_\infty$-algebras.

If $A$ happens to carry auxiliary gradings respected by the product
structure and preserved by the differential, then it is easy to check
from the inductive procedure that
the maps in the construction may be chosen so that they also
respect these gradings. It then follows that the structure maps $m_i$
of the $A_\infty$ structure on $H^*(A)$ also respect these gradings.

Let us apply this to $H^*(BG;k)$. We examine the elements $m_i(t,\dots,t)$.
By definition, we have $m_1(t)=0$ and $m_2(t,t)=0$. The degree of
$m_i(t,\dots,t)$ is $i$ times the degree of $t$, increased in the
homological direction by $i-2$. This gives
\[ |m_i(t,\dots,t)| = i(-2q+1,h)+(i-2,0) =(-2iq+2i-2,ih). \]
The homological degree is even, so if $m_i(t, \cdots , t)$ is non-zero then it is a
multiple of a power of $x$. Comparing degrees, if $m_i(t,\dots,t)$ is
a non-zero multiple of $x^\alpha$ then we have
\[ 2iq-2i+2=2\alpha q,\qquad ih = \alpha p^n. \]
Eliminating $\alpha$, we obtain $(iq-i+1)p^n=ihq$. Substituting
$h=p^n-(p^n-1)/q$, this gives $i=p^n$. Finally, since the Massey
product of $p^n$ copies of $t$ is equal to $-x^h$, it follows that
$m_{p^n}(t,\dots,t)=\eps (p^n) x^h$, where the sign is as defined in Remark
\ref{rem:signs} \cite[Theorem 3.1]{LuPalmieriWuZhang:2009}. Thus we have
\[ m_i(t,\dots,t) = \begin{cases} \eps (p^n) x^h & i=p^n \\
0 & \text{otherwise.} \end{cases} \]
We shall elaborate on this argument in a more general context in the
next section, where we shall see that the rest of the $A_\infty$
structure is also determined in a similar way.

\section{$A_\infty$ structures on a polynomial tensor exterior algebra}

In this section, we shall examine the following general situation.
Our goal is to establish that there are only two possible $A_\infty$
structures satisfying Hypothesis \ref{hyp:grading} below, and that the Koszul dual also
satisfies the same hypothesis with the roles of $a$ and $b$, and of
$h$ and $\ell$ reversed.

\begin{hypothesis}
\label{hyp:grading}
$A$ is a $\bZ\times\bZ$-graded $A_\infty$-algebra over a field $k$,
where the operators $m_i$ have degree $(i-2,0)$, satisfying
\begin{enumerate}
\item $m_1=0$, so that $m_2$ is strictly associative,
\item ignoring the $m_i$ with $i>2$, the algebra $A$ is
$k[x] \otimes \Lambda(t)$ where $|x|=(-2a,\ell)$ and $|t|=(-2b-1,h)$, and
\item $ha-\ell b = 1$.
\end{enumerate}
\end{hypothesis}

\begin{remarks}
(i) The $A_\infty$-algebra $H^*(BG;k)$ of the last section satisfies this
hypothesis, with $a=q$, $b=q-1$, $h=p^n-(p^n-1)/q$, $\ell=p^n$.

(ii) By comparing degrees, if we have $m_\ell(t,\dots,t)=\eps (p^n) x^h$ then
$(2b+1)\ell + 2-\ell = 2ah$ and so $ha-\ell b = 1$. This explains the
role of part (3) of the hypothesis. The consequence is, of course,
that $a$ and $b$ are coprime, and so are $h$ and $\ell$.
\end{remarks}

\begin{lemma}
If $m_i(t,\dots,t)$ is non-zero, then $i=\ell > 2$ and $m_\ell(t,\dots,t)$ is
a multiple of $x^h$.
\end{lemma}
\begin{proof}
The argument is the same as in the last section. The degree of
$m_i(t,\dots,t)$ is $i|t| +i- 2  = (-2ib - 2, ih)$. Since the
homological degree is even, if $m_i(t,\dots,t)$ is non-zero then it is
a multiple of some power of $x$, say $x^\alpha$. Then we have
\[ 2ib+2 = 2\alpha a,\qquad ih = \alpha\ell. \]
Eliminating $\alpha$ gives $(ib+1)\ell=iha$, and so
using $ha-\ell b =1$ we have $i=\ell$. Substituting back gives
$\alpha = h$.
\end{proof}

Elaborating on this argument gives the entire $A_\infty$ structure.
If $m_\ell(t,\dots,t)$ is non-zero, then by rescaling the variables
$t$ and $x$ if necessary
we can assume that $m_\ell(t,\dots,t)= x^h$ (note that we can even do
this without extending the field, since $\ell$ and $h$ are coprime).

\begin{proposition}
If $m_\ell(t,\dots,t)=0$ then all $m_i$ are zero for $i>2$. If
$m_\ell(t,\dots,t)= x^h$ then
$m_\ell(x^{j_1}t,\dots,x^{j_\ell}t)= x^{h+j_1+\dots+j_\ell}$,
and all $m_i$ for $i>2$ on all other $i$-tuples of monomials give zero.
\end{proposition}
\begin{proof}
All monomials live in different degrees, so we do not need to consider
linear combinations of monomials.
Suppose that
$m_i(x^{j_1}t^{\ep_1},\dots,x^{j_i}t^{\ep_i})$ is some constant multiple
of $x^jt^\ep$,
where each of $\ep_1,\dots,\ep_i,\ep$ is either zero or one. Then
comparing degrees, we have
\[ (j_1+\dots+j_i)|x| + (\ep_1+\dots+\ep_i)|t| + (i-2,0) = j|x| +  \ep|t|. \]
Setting
\[ \alpha=j_1+\dots+j_i-j,\qquad \beta = \ep_1+\dots+\ep_i-\ep \]
we have $\beta \le i$, and
\[ \alpha(-2a,\ell) + \beta(-2b-1,h) + (i-2,0) = 0. \]
Thus
\[ 2\alpha a + 2\beta b + \beta + 2 - i = 0, \qquad \alpha \ell +
  \beta h = 0. \]
Eliminating $\alpha$, we obtain
\[ -2\beta ha + 2\beta \ell b + \beta \ell + 2 \ell - i \ell = 0. \]
Since $ha-\ell b=1$, this gives $\beta = \ell (i-2) /(\ell -2)$.
Combining this with $\beta \le i$ gives $i\le \ell$. If $i<\ell$ then
$\beta$ is not divisible by $\ell$, and so $\alpha \ell + \beta h = 0$
cannot hold. So we have $\beta=i=\ell$, $\ep_1=\dots=\ep_\ell=1$,
$\ep=0$, $\alpha=-h$, and $j=h+j_1+\dots+j_\ell$.
Finally, the identities satisfied by the $m_i$ for an
$A_\infty$ structure show that all the constant multiples have to be
the same, hence all equal to zero or after rescaling, all equal to minus one.
\end{proof}

Combining the lemma with the proposition, we obtain the following.

\begin{theorem}\label{th:Ainfty}
Under the hypothesis above, if $\ell > 2$ then there are two possible $A_\infty$
structures on $A$.
There is the formal one, where $m_i$ equal to zero for $i>2$,
and the non-formal one, where
after replacing $x$ and $t$ by suitable multiples, the only
non-zero $m_i$ with $i>2$ is $m_\ell$, and the only non-zero values on
monomials are given by
\begin{equation*}
m_\ell(x^{j_1}t,\dots,x^{j_\ell}t)=x^{h+j_1+\dots+j_\ell}.
\end{equation*}
\end{theorem}

\begin{theorem}
Let $G=\bZ/p^n \rtimes \bZ/q$ as above, and $k$ a field of
characteristic $p$. Then
the $A_\infty$ structure on $H^*(G,k)$ given by Kadeishvili's theorem
may be taken to be the non-formal possibility named in the above
theorem, with $a=q$, $b=q-1$, $h=p^n-(p^n-1)/q$, $\ell=p^n$.
\end{theorem}
\begin{proof}
Since we have $m_{p^n}(t,\dots,t)=\eps (p^n) x^h$, the formal possibility does
not hold.
\end{proof}

\begin{remark}
Dag Madsen's thesis~\cite{Madsen:2002a}
has an appendix in which the $A_\infty$ structure
is computed for the cohomology of a truncated polynomial ring, reaching similar
conclusions by more direct methods.
\end{remark}

\section{Loops on $BG\phat$}

In general for a finite group $G$ we have $H^*(BG\phat;k)\cong
H^*(BG;k)=H^*(G,k)$ and $\pi_1(BG\phat)=G/O^p(G)$,
the largest $p$-quotient of $G$. In our case, $G=\bZ/p^n\rtimes\bZ/q$
with $q>1$, we have $G=O^p(G)$ and so $BG\phat$ is simply
connected. So the Eilenberg--Moore spectral sequence converges to the
homology of its loop space:
\[ \Tor_{*,*}^{H^*(G,k)}(k,k) \Rightarrow H_*(\Omega BG\phat;k). \]
The internal grading on $C^*(BG;k)$ gives this spectral sequence a
third grading that is preserved by the differentials, and $H_*(\Omega
BG\phat;k)$ is again doubly graded. Since $H^*(G,k)=k[x] \otimes
\Lambda(t)$ with $|x|=(-2q,p^n)$ and $|t|=(-2q+1,h)$, it follows that
the $E^2$ page of this spectral sequence is
$k[\tau] \otimes \Lambda(\xi)$
where $|\xi|=(-1,2q,p^n)$ and $|\tau|=(-1,2q-1,h)$
(recall $h=p^n-(p^n-1)/q$). Provided that we are not in the case
$h=2$, which only happens if
$p^n=3$, ungrading $E^\infty$ gives
\[ H_*(\Omega BG\phat;k) = k[\tau] \otimes \Lambda(\xi) \]
with $|\tau|=(2q-2,h)$ and $|\xi|=(2q-1,p^n)$.

In the exceptional case $h=2$, $p^n=3$, we have $q=2$, and
the group $G$ is the symmetric group $\Sigma_3$ of degree
three. An explicit computation (for example by squeezed resolutions
\cite{Benson:2009b}) gives
\[ H_*(\Omega (B\Sigma_3){}^{^\wedge}_3;k) = k[\tau,\xi]/(\xi^2+\tau^3) \]
with $|\tau|=(2,2)$ and $|\xi|=(3,3)$, and the two gradings collapse
to a single grading.

Applying
Theorem~\ref{th:Ainfty}, and using the fact that either formal case is
Koszul dual to the other, we have the following.

\begin{theorem}
Suppose that $p^n\ne 3$. Then
the $A_\infty$ structure on $H_*(\Omega
BG\phat;k)=k[\tau]\otimes\Lambda(\xi)$ is given by
\[ m_h(\tau^{j_1}\xi,\dots,\tau^{j_h}\xi)=\eps (h) \tau^{p^n+j_1+\dots+j_h}, \]
and for $i>2$, all $m_i$ on all other $i$-tuples of monomials give zero.\qed
\end{theorem}

Using \cite{LuPalmieriWuZhang:2009} again, we
may reinterpret this in terms of Massey products.

\begin{corollary}
In $H_*(\Omega BG\phat)$,
the Massey products $\langle \xi,\dots,\xi \rangle$ ($i$ times) vanish
for $0<i<h$, and give $-\tau^{p^n}$ for $i=h$.\qed
\end{corollary}

Note that the exceptional case $p^n=3$ also fits the corollary, if we
interpret a $2$-fold Massey product as an ordinary product.

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TITLE: Probability Density Function of Random Variable which is Sum of other Random Variables
QUESTION [1 upvotes]: Let $X_0, X_1, X_2, ..., X_n$ each be non-identical independent random variables.
Let $x_0, x_1, ... , x_n$ be possible values of each of those random variables.
Let $\newcommand{\Pdf}{\operatorname {Pdf}} \Pdf_{X_0}(x_{0}), \Pdf_{X_1}(x_{1}), ... , \Pdf_{X_n}(x_{n})$ represent each of random variables corresponding probability density functions.
$X_0$ has probability density function $\Pdf_{X_0}(x_0)$ ,
$X_1$ has probability density function $\Pdf_{X_1}(x_1)$ ... etc...

Lets say we know explicitly, analytically, exactly what all the $Pdf$'s are above.

Now Let $Y = X_0 + X_1 + X_2, ... + X_n$ which has probability density function $\Pdf_Y(y)$.
What is $\Pdf_Y(y)$ ?
How do I calculate it with the information above?

REPLY [2 votes]: It's a well known (and important) result that the density of the sum of independent random variables is the convolution of the individual densities.
See for example here or here or here.

Let $\tau_{0}, \tau_{1}, ... \tau_{n-1}$ represent temporary variables which are integrated out as part of the convolution.
Let $PDF_{x_{a} + x_{b}} $ denote the PDF of only two random variable $X_a + X_b$
$PDF_{x_{a} + x_{b}}(t) = PDF_{x_{a}} * PDF_{x_{a}} (t) = \int_{-\infty}^{\infty} PDF_{x_{a}}( \tau )  PDF_{x_{a}}( t - \tau ) d\tau $
Then the following follows:
$PDF_y(y) = PDF_{x0}*PDF_{x1}*PDF_{x2}*...*PDF_{xn}(y)$
$= \int_{-\infty}^{\infty}...\int_{-\infty}^{\infty}
PDF_{x0}(\tau_{0}) PDF_{x1}(\tau_{1} - \tau_{0})  PDF_{x2}(\tau_{2} - \tau_{1} )  ...  PDF_{xn}(y - \tau_{(n-1)}) d\tau_{0} d\tau_{1} ... d\tau_{(n-1)}  $


text:
/-
Copyright (c) 2021 OpenAI. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kunhao Zheng, Stanislas Polu, David Renshaw, OpenAI GPT-f
-/
import mathzoo.imports.miniF2F

open_locale nat rat real big_operators topological_space

axiom amc12a_2003_p5
  (a m c : ℕ)
  (h₀ : a ≤ 9 ∧ m ≤ 9 ∧ c ≤ 9)
  (h₁ : 10*(10*(10*(10*a + m) + c) + 1) + 0 + (10*(10*(10*(10*a + m) + c) + 1) + 2) = 123422) :
  a + m + c = 14

subset_name: curated


TITLE: Number of Non-Isomorphic linear orders on $\kappa$ elements
QUESTION [4 upvotes]: After having proven that there is no $\kappa$-categorical axiomatizations of strict linear order for any infinite cardinality $\kappa$, a professor of mine mentioned in passing that it was actually the case that there are in fact $2^{\kappa}$ non-isomorphic SLOs on any infinite $\kappa$, in contrast to $\kappa +$ wellorders. He also mentioned that the assertion that number of wellorders is equal to the number of strict linear orders is equivalent to the continuum hypothesis. Sadly, class ended and I was unable to get clarification (and now am too interested to wait for the next office hours).
I am confused on a few points:
1) why are there $2^\kappa$ such non-isomorphic orders?
Any texts/reading that could be suggested to address any of these questions, or any explanation thereon, would be much appreciated. Thanks!
Update: 2, 3 should've been obvious. They're below in case anyone ever asks the same question. Still interested in 1.
2) what does $\kappa +$ mean?
3) how is the assertion that $2^\kappa = \kappa +$ equivalent to the continuum hypothesis?

REPLY [4 votes]: Here's my shot at answering (1).
There cannot be more than $2^\kappa$ linear orders on a set of cardinality $\kappa$, since the number of binary relations on $\kappa$ is $2^{\kappa^2}$ and since $\kappa$ is infinite, $\kappa^2 = \kappa$ so there are a total of $2^\kappa$ binary relations on $\kappa$.
Since $\kappa$ is infinite, $\kappa = \kappa \times \aleph_0$, so if $S$ is of size $\kappa$, then the number of non-isomorphic linear orders on $S$ is the same as that on $S \times \mathbb{N}$. Now, for every map $f : S \to \{0, 1\} $, consider the order on $S \times \mathbb{N}$, where $\preceq$ is a fixed well ordering on $S$.
$(s_0, n_0) < (s_1, n_1)$ if $s_0 \prec s_1$
$(s, n_0) < (s, n_1)$ if $f(s) = 0$ and $n_0 < n_1$ or if $f(s) = 1$ and $n_0 > n_1$
This scheme should produce non isomorphic linear orders for any distinct $f_1, f_2$, since if $s$ is the least element under $\preceq$ of $S$ such that $f_1(s) \neq f_2(s)$, then any order isomorphism must map $\{s'\} \times \mathbb{N}$ to itself for each $s' \preceq s$, but this is impossible for $s$.
Thus, there is a distinct order for each function from $S$ to $\{0,1\}$, of which there are $2^\kappa$. This means that there are at least $2^\kappa$ orderings, and since there are no more than $2^\kappa$, this is the exact cardinality.


TITLE: Conditional Expectation and Variance inequality
QUESTION [3 upvotes]: I'm stuck trying to solve an inequality: Let $X \in \mathcal{L}^{2}$, then
$$\mathbb{E}[\vert X-\mathbb{E}[X\mid \mathcal{G}]\vert^{2}] \leq \mathbb{E}[\vert X - \mathbb{E}[X]\vert^{2}].$$
My attempts:
First I (mistakenly?) thought that $\mathbb{E}[\vert X-\mathbb{E}[X]\vert^{2}] = Var(X)$ and try to use law of total variance without any success. Then I already proved that if $\mathcal{G}\subset \mathcal{F}$ and $\mathbb{E}[X^{2}]<\infty$ then
$$\mathbb{E}[(X-\mathbb{E}[X\vert \mathcal{F}])^{2}] + \mathbb{E}[(\mathbb{E}[X\vert \mathcal{F}] - \mathbb{E}[X\vert \mathcal{G}])^{2}] = \mathbb{E}[(X-\mathbb{E}[X\vert \mathcal{G}])^{2}].$$
I'm pretty sure the inequality is a corollary of the previous theorem using the second left term of the equality is zero, but I'm stuck to go further than that poor argument. Also I'm pretty sure that I need to use the trivial $\sigma-$algebra. I would appreciate any ideas.

REPLY [1 votes]: Let $Y=E(X|\mathcal G)$. Then $E(X-EX)^{2}=E((X-Y)+(Y-EX))^{2}$. Expand this as $E(X-Y)^{2}+E(Y-EX)^{2}+2E(X-Y)(Y-EX) \geq E(X-Y)^{2}+2E(X-Y)(Y-EX)$ If we show that the last term is $0$ we get the required in equality. Since $Y-EX$ is measureble w.r.t. $\mathcal G$ we can write $E(X-Y)(Y-EX)=E[E(X-Y)(Y-EX)|\mathcal G)] =E ((Y-EX)[E(X-Y)|\mathcal G])$ and $E(X-Y)|\mathcal G)=E(X|\mathcal G)-Y=0$.


TITLE: Does too little ice make my martinis watery?
QUESTION [38 upvotes]: When I make martinis, the recipe I use is 2.5 fluid ounces1 of gin2 and 1/2 fluid ounce dry vermouth3, shaken or stirred4 with 7 ice cubes5, then strained into a cocktail glass (I mix until it's cold enough, and not for some specific length of time.)
My Observation
If I use less ice, the martini is more diluted; if I use more ice, the martini is stronger (too strong in fact—I prefer some dilution.)
My Hypothesis
With a lot of ice, the drink chills very fast, not giving the ice as much time to melt, and vice-versa.
My Friends
think I am crazy about the amount of ice6. They claim that (a) the ice has to melt to chill the drink so no matter how much ice I start with, the same amount is melted to reach the desired temperature, and (b) I had to retake Physics 2A in college7, so what do I know anyway?
I pointed out that I could chill the drink with really cold rocks (a.k.a. whiskey stones) and it could get just as cold with no melting and maybe they are missing something.
The answers to these questions seem to support my friends' argument that the melting of the ice is the overwhelming contributor to the cooling:

What cools a drink?
Is an ice globe the worst possible way to cool a drink (with ice)?
Calculate mass of ice needed to cool water $\Delta$T degrees

However my experiment seems to demonstrate otherwise.
My Experiment
I tested this by making six martinis, two each with 4, 7, and 10 ice cubes. I used the same amount of gin and vermouth in each. I weighed8 the ingredients before adding them to the mixing cup. I stirred until the desired temperature9 was achieved. Then I weighed the amount after straining.
My Result
The 4-cube martinis gained more weight than the 7-cube martinis, which in turned gained more weight than the 10-cube martinis. I assume the weight gain was the melted ice. My friends happily drank all the martinis, but remained unconvinced of my Physics acumen, either theoretical or experimental10.
My Question
Am I exhibiting confirmation bias and my stupid friends are right? Or is there an explanation for my hypothesis and observation and they should finally shut up about my having to retake Physics 2A because come on it was like 30 years ago already and besides, I'm not using Planck's constant to calculate how long it takes atoms to slow to a halt, I'm just making martinis!
Is the ratio of nearly 2:1 ice-to-liquid a factor? How about the constant mixing? Or is this dependent on the starting temperature of the ice and even though the warming of the ice contributes very little, it is enough to affect the outcome?

Footnotes:

Apologies for the use of American measurements in this scholarly context, but that's what my utensils are labeled with.
No apologies for the insistence on gin. If you want to use vodka or make some other cocktail and call it a martini, we have nothing further to discuss.
I use a 5:1 ratio. Others may quibble. They are wrong.
Seriously, I do not care. Can I please continue?
For my ice trays, this is 5 fluid ounces of water, prior to freezing, in a plain old kitchen freezer.
Although they are all too happy to drink the martinis I make.
These are friends that I went to college with so I can't argue that point, but there were three intramural softball playoff games the weekend before finals, so when was I going to study? But I digress...
Postal scale, accurate to 1/10 (American again) ounce, sorry.
28 degrees American Fahrenheit on my kitchen thermometer that measures to 1/10 degree, but accuracy unknown.
I am willing to rerun that experiment as long as necessary until I get it right!

Edit
I appreciate the answers and I am accepting the answer from @cyberx86 because (a) it independently supports my insistence that 7 ice cubes is The Right Number, and (b) What kind of physics site would we have if we didn't reward showing your work?
However no one really "put a bow on it" so I'll add
My Conclusions
Since the final temperature of the martini drops below 0° C and there is also ice melting, there must be both melting and absorption occurring together during the mixing.
The measurable change in the amount of melting in this case is due to the addition of the heat absorption capacity of more ice, but only because it is below 0° C to start with.

REPLY [19 votes]: In order for the martini to cool, heat is transferred from it, into the ice. This results in the martini's temperature dropping and the ice's temperature rising. Once the ice reaches its melting point, it's temperature will not rise, but a state change will occur (this is the latent heat). It is only if the ice melts that water is added to your martini and dilutes it.
Here is an attempt and expressing this idea mathematically, there are lots of assumptions made.
$1 fl. oz = 0.0295735 mL$
$c_{water} = 4184 J/kg^{\circ}C$
$D_{water} = 1 kg/L$
$c_{ice} = 2108 J/kg^{\circ}C$
$L_{ice} = 333550 J/kg^{\circ}C$
$c_{ethanol} = 2460 J/kg^{\circ}C$
$D_{ethanol} = 0.789 kg/L$
Additionally, the following assumptions are made:

Once melted, the water is ignored in the temperature calculations
Gin is assumed to be 45% ethanol and 55% water, other components are ignored
Vermouth is assumed to be 18% ethanol and 82% water, other components are ignored
All specific heat capacities of mixtures are calculated as weighted averages
Freezer temperature is assumed to be -18 degrees Celsius
Room temperature is assumed to be 20 degrees Celsius

${T_{inital}}_{martini} = 20^{\circ}C$
${T_{final}}_{martini} = -2.22^{\circ}C$
${T_{inital}}_{ice} = -18^{\circ}C$
${T_{final}}_{ice} = 0^{\circ}C$
$V_{gin} = 2.5 fl. oz = 0.0739 L $
$D_{gin} = 0.905 kg/L  $
$m_{gin} = 0.0669 kg $
$c_{gin} = 3408.2 J/kg^{\circ}C $
$V_{vermouth} = 0.5 fl. oz = 0.0148 L $
$D_{vermouth} = 0.962 kg/L  $
$m_{vermouth} = 0.0142 kg $
$c_{vermouth} = 3873.68 J/kg^{\circ}C $
$V_{martini} = 3 fl. oz = 0.0887 L $
$D_{martini} = 0.915 kg/L  $
All of this, gives us the following numbers for the martini:
$m_{martini} = 0.0811 kg $
$c_{martini} = 3485.78 J/kg^{\circ}C $
The energy lost by the martini to bring it to its final temperature is:
$Q_{martini} = m_{martini}c_{martini}({T_{final}}_{martini}-{T_{inital}}_{martini})$
$Q_{martini} = (0.0811 kg)(3485.78 J/kg^{\circ}C)(-2.22^{\circ}C-25^{\circ}C)$
$Q_{martini} = -6285 J $
This means that the ice needs to absorb 6285 J of energy. Part of this will occur by an increase in the temperature of the ice, any additional energy will go into melting the ice.
The mass of an ice cube is based on 5 fl. oz for 7 ice cubes:
$m_{icecube} = 0.02112 kg $
The amount of energy each ice cube can absorb as its temperature rises to 0 degrees Celsius is:
$Q_{ice} = m_{ice}c_{ice}({T_{final}}_{ice}-{T_{inital}}_{ice})$
$Q_{ice} = (0.02112 kg)(2108 J/kg^{\circ}C)(0^{\circ}C-(-18^{\circ}C))$
$Q_{ice} = 801.5 J/cube $
The amount of energy each ice cube would absorb if it completely melted is:
$Q_{melt} = m_{ice}L_{ice}$
$Q_{melt} = (0.02112 kg)(333550 J/kg^{\circ}C)$
$Q_{melt} = 7046 J/cube $
This means, that if we have enough ice cubes, assuming all absorb heat from the martini equally, that more ice cubes will be able to remove more heat before needing to melt. Additionally, this means that after a certain point, additional ice cubes shouldn't make an appreciable difference. (Assuming that the ice cubes are removed once the target temperature is reached, and not allowed to melt).
Ice Cubes    Energy from Melting    Water Added (mL)     Estimated Final Mass (g)
1             -5483                  16                        97.6
2             -4681                  14                        95.2
3             -3880                  12                        92.8
4             -3078                   9.2                      90.4
5             -2277                   6.8                      88.0
6             -1475                   4.4                      85.6
7              -673.8                 2.0                      83.2
8               127.7                 0                        81.1
9               929.2                 0                        81.1
10             1731                   0                        81.1

You should be able to verify the basic trend (mass decreases with additional ice cubes until a point which it plateaus) with the experimental setup you used. You can compare your final masses to those estimated in the table above, although I would expect your measured values to be somewhat higher than these. These numbers do not include the energy needed to cool the mixing vessel.


TITLE: Candidate function for given conditions.
QUESTION [0 upvotes]: If this is homework for any course, please tell me which course it is.
I want to fit the data below with an analytical, cartesian function.

Everywhere is $\frac{dy}{dx} \gt 0$. $f(0)=0$. $f(1)=1$.
Is there a candidate function which satisfies this set of conditions?

REPLY [0 votes]: You can try $f(x)=x+0.044 \sin (7\pi x )$ - its' plot looks a little like the green plot.
Plot:
https://www.wolframalpha.com/input/?i=Plot(%7Bx,x%2B0.044+sin(7%5Cpi+x)%7D,+%7Bx,0,1%7D)
It also have positive derivative:
https://www.wolframalpha.com/input/?i=Plot(d(x%2B0.044+sin(7%5Cpi+x),x),+%7Bx,0,1%7D)


TITLE: How many ways can the number $10800$ be expressed as a product of two factors?
QUESTION [0 upvotes]: How many ways can the number $10800$ be expressed as a product of two factors?
My attempt:
$10800 = 2^4 \times 3^3 \times 5^2$
So the number of ways is
$4\times 3\times 2  +1  = 25 $
Is this correct?

REPLY [1 votes]: As you observed, $10800 = 2^4 \cdot 3^3 \cdot 5^2$.
Thus, each factor has the form $2^a3^b5^c$, where $0 \leq a \leq 4$, $0 \leq b \leq 3$, and $0 \leq c \leq 2$.  If the two factors are $2^{a_1}3^{b_1}5^{c_1}$ and $2^{a_2}3^{b_2}5^{c_2}$, then
\begin{align*}
a_1 + a_2 & = 4 \tag{1}\\
b_1 + b_2 & = 3 \tag{2}\\
c_1 + c_2 & = 2 \tag{3}
\end{align*}
Equation 1 is an equation in the nonnegative integers with $5$ solutions, namely $(4, 0), (3, 1), (2, 2), (1, 3), (0, 4)$.  Similarly, equation 2 is an equation in the nonnegative integers with $4$ solutions, and equation 3 is an equation in the nonnegative integers with $3$ solutions.  Thus, there are $$5 \cdot  4 \cdot 3 = 60$$ ordered pairs of factors.  Since $10,800$ is not a perfect square, each pair contains two distinct factors.  Thus, we have counted each unordered pair of factors twice, so there are
$$\frac{5 \cdot 4 \cdot 3}{2} = 30$$
unordered pairs of factors of $10,800$.


TITLE: Circuit to Karnaugh map
QUESTION [1 upvotes]: A logical circuit was given, I had to create a boolean expression from it; which I did correctly (I guess).
I calculated the output of each gate separately, considering the variables.
From the given circuit I got this expression:
('xy + x) + ((x + y) + z) * 'u

I than would need to put this in a Karnaugh's map to simplify the expression (and later on the circuit). But this is the step I got stuck on.
How to put that expression in Karnaugh's map?
Even when I perform algebra on the current expression I get terms which aren't present in the map, or single terms like only 'x', but 'only x' ain't present in the map.
I have already converted expressions to Karnaugh map with no problems, but I guess I have difficulty now because it's with 4 variables, and a tough expression.

REPLY [0 votes]: Sorry, I couldn't send any photo.
Your function is false when $xyzu = {0000, 0001,0011}$.
First and second columns are when $x$ = true (1), another columns are false (0).
Second and third columns are when $y$ = true (1), another columns are false (0).
First and second lines are when $z$ = true (1), another lines are false (0).
Second and third lines are when $u$ = true (1), another lines are false (0).
$$1|1|1|1$$
$$1|1|1|0$$
$$1|1|1|0$$
$$1|1|1|0$$
As a result, $$f = x \lor y \lor \overline uz$$


TITLE: The $h-$th power of every element in a finite group of order $h$ is the identity element of the group
QUESTION [2 upvotes]: I work in population dynamics. I want to show somewhere in my work that the $h-$th power of every element in a finite group of order $h$ is the identity element of the group. I guess this is elementary result but it would be nice if somebody show me how to do it or where to find it.
Thanks

REPLY [2 votes]: The order of every element of a group is a divisor of $h$ by Lagrange's Theorem.  So if $x$ is an element of your group of order $k$ then $h = k\cdot l$ for some $l.$ Hence we have
$$x^h = x^{k\cdot l} = (x^k)^l = (e)^l = e.$$


text:
\begin{document}
\begin{frontmatter}

\title{A virtual element method for the Steklov
eigenvalue problem allowing small edges.}

\author[1]{Felipe Lepe}
\ead{flepe@ubiobio.cl}
\address[1]{Departamento de Matem\'atica,
Universidad del B\'io-B\'io, Casilla 5-C, Concepci\'on, Chile.}
\author[1,2]{David Mora}
\ead{dmora@ubiobio.cl}
\address[2]{CI$^2$MA, Universidad de Concepci\'on, Casilla 160-C, Concepci\'on, Chile.}
\author[3]{Gonzalo Rivera}
\ead{gonzalo.rivera@ulagos.cl}
\address[3]{Departamento de Ciencias Exactas, Universidad de Los Lagos,
Casilla 933, Osorno, Chile.}
\author[2]{Iv\'an Vel\'asquez}
\ead{ivelasquez@ci2ma.udec.cl}

\begin{abstract}
The aim of this paper is to analyze the influence
of small edges in the computation
of the spectrum of the Steklov eigenvalue problem by a lowest order
virtual element method.
Under weaker assumptions on the polygonal meshes,
which can permit arbitrarily small edges with respect
to the element diameter,
we show that the scheme provides a correct approximation
of the spectrum and prove optimal error estimates
for the eigenfunctions and a double order for the eigenvalues.
Finally, we report some numerical tests
supporting the theoretical results.

\end{abstract}

\begin{keyword}
Virtual element method
\sep Steklov eigenvalue problem
\sep error estimates
\sep small edges
\MSC 35Q35 \sep 65N15 \sep 65N25 \sep 65N30 \sep 76B15.
\end{keyword}

\end{frontmatter}

\setcounter{equation}{0}
\section{Introduction}

In this paper we are interested in the approximation
by virtual elements of the eigenvalues and eigenfunctions
of the Steklov problem which is characterized by
the presence of the eigenvalue on the boundary condition.
This problem has attracted much attention
in recent years due to the important applications
in many physical subjects.
For instance, it appears in the study of the dynamic
of liquids in moving containers, the so called
sloshing problem \cite{BRS03, CM79, CY96}.
Also, this problem have interesting applications
in inverse scattering \cite{LSTJSC19},
among other.

There are several studies on the finite element approximations
of the Steklov eigenvalue problem, for example, see
\cite{A_M2AN2004, AP_APNUM2009, BrOs1972, GM-ima2011, LLX_apmath2013, YLL_apnum2009, XIMA13}.
Traditionally, finite element methods rely on triangular (simplicial)
and quadrilateral meshes. However, in complex simulations
one often encounters general polygonal and polyhedral meshes.
In recent years there has been a significant growth
in the mathematical and engineering literature in developing
numerical methods that can make use of general polytopal
meshes; among the large number of papers on this subject, we cite as a minimal sample
\cite{BBCMMR2013,BLMbook2014,CGH14,DDbook2020,RW,ST04}.

The VEM has been introduced in \cite{BBCMMR2013}
and has been applied successfully in a large
range of problems arising from engineering and
physics phenomenons; see for instance \cite{AN2020,ABSV2016,BDR2017,
BLM2015,BLV-M2AN,BBBPS2016,CG2017,CGPS2017,MPP2018,PPR15}.
Regarding VEM for eigenvalue problems, we mention the following recent works
\cite{CGMMV,DrAA2011,GMV, GV, MM, MR2019, MRR1, MRR2, MV2}.
In particular, an a priori and a posteriori VEM discretization
for the Steklov eigenvalue problem has been presented in \cite{MRR1,MRR2}.
However, the theoretical results and error estimates for the
eigenvalues and eigenfunctions were obtained
under the standard mesh assumptions introduced
in \cite{BBCMMR2013}, which do not allow
to consider meshes containing elements with small edges
compared to the element diameter.

In \cite{BLR17,BM20,BrS} has been recently
analyzed the possibility to consider in VEM discretizations
arbitrarily small edges with respect to the element diameter.
Here, we will follow the VEM approach presented in \cite{BLR17},
for the Poisson equation, to write a lowest order virtual scheme
for the Steklov eigenvalue problem which permit arbitrarily
small edges in the polygonal meshes.
This can be useful in adaptive schemes
by considering refined meshes as a tool to
handle solutions with corner singularities.

More precisely, we will propose a
virtual element method of lowest order to solve the Steklov
eigenproblem, allowing small edges in the polygons of the mesh. We will consider the continuous
variational formulation presented in \cite{MRR1};
however, we will write a different discrete
virtual scheme, which is based on a different
stabilization bilinear form (see \cite{WRR}).
We will use the so-called Babu\v ska-Osborn abstract
spectral approximation theory (see \cite{BO}), to show
that under weaker assumptions on the polygonal meshes,
the resulting virtual element scheme provides a correct
approximation of the spectrum and prove optimal order
error estimates for the eigenfunctions and a double
order for the eigenvalues.
In particular, our theoretical estimates fully support
meshes with arbitrarily small edges with respect to
the element diameter. In addition, we remark
that spurious modes were not found for different values
of the a scaled stabilization (see in particular
Section~\ref{sectionslosh} below).

Moreover, the present work can be seen as a stepping
stone towards the more challenging eigenvalue problems.

The paper is organized as follows: In Section~\ref{SEC:STAT},
we present the model problem and preliminary
results related to the solution operator and
eigenfunctions. More precisely, we will establish
the spectral characterization
of the solution operator, which allows to study
the numerical method. Section~\ref{SEC:DISCRETE}
is dedicated to present the
virtual element method. Here we will introduce
the assumptions on the mesh.

We will present approximation results that will be
the key point of our analysis, which will depend on
the particular choice of the stabilization form.
Section~\ref{SEC:approximation}, contains the error estimates
for the eigenfunctions and a double order for the eigenvalues.
Finally, in Section~\ref{sec:numerics}
we present some numerical results on different families
of polygonal meshes with small edges, in order
to confirm the theoretical rates of convergence proved in the paper
and to confirm that it is not polluted with spurious modes.

Throughout the article we will use standard notations
for Sobolev spaces, norms and seminorms. Moreover, we
will denote by $C$ a generic constant independent of
the mesh parameter $h$, which may take different values in
different occurrences.

\label{SEC:INTR}

\setcounter{equation}{0}
\section{The spectral problem}
\label{SEC:STAT}

Let $\O\subset\R^2$ be a bounded domain with polygonal boundary
$\partial\O$. Let $\Go$ and $\G_1$ be disjoint open subsets of
$\partial\O$ such that $\partial\O=\bar{\G}_0\cup\bar{\G}_1$ and
$\left|\Go\right|\ne0$. We denote by $n$ the outward unit normal vector
to $\partial\O$ and by $\partial_n$ the normal derivative.

In what follows, we recall the variational formulation of the
Steklov eigenvalue problem proposed
in \cite{MRR1}. Also, we summarize some results from this reference.

The  Steklov eigenvalue problem reads as follows:
Find $(\l,u)\in\R\times\HuO$, $u\ne0$, such that
$$
\left\{\begin{array}{l}
\Delta u=0\quad\text{in }\O,
\\[0.1cm]
\partial_n u
=\left\{\begin{array}{ll}
\l u & \text{on }\Go,
\\
0 & \text{on }\G_1.
\end{array}\right.
\end{array}\right.
$$
where $\partial_n u$ denotes the normal derivative of $u$.
By testing the first equation above with $v\in\HuO$ and integrating by
parts, we arrive at the following equivalent variational formulation:

\begin{problem}
\label{P1}
Find $(\l,u)\in\R\times\HuO$, $u\ne0$, such that
$$
\int_{\O}\nabla u\cdot\nabla v
=\l\int_{\Go}uv\qquad\forall v\in\HuO.
$$
\end{problem}

Observe that the left-hand side is not $\HuO$-elliptic. A remedy for this
is  to use a shift argument to rewrite Problem \ref{P1}
in the following form:
\begin{problem}
\label{P2}
Find $(\l,u)\in\R\times\HuO$, $u\ne0$, such that
$$
\ha(u,v)
=\left(\l+1\right)b(u,v)
\qquad\forall v\in\HuO,
$$
\end{problem}

where the bilinear form
$\ha:\HuO\times \HuO\rightarrow\mathbb{R}$
is defined by
\begin{align*}
\ha(u,v)
& :=a(u,v)+b(u,v)
\qquad u,v\in\HuO,
\end{align*}
with  $a,b:\HuO\times\HuO\rightarrow\mathbb{R}$ defined by
\begin{equation*}
a(u,v)
 :=\int_{\O}\nabla u\cdot\nabla v,
\qquad b(u,v)
 :=\int_{\Go}uv
\qquad u,v\in\HuO.
\end{equation*}
All the previous bilinear forms are bounded and symmetric.
In addition, the next result, proved in \cite[Lemma 2.1]{MRR1},
establishes that $\ha(v,v)$ is $\HuO$-elliptic.

\begin{lemma}
\label{ha-elipt}
There exists a constant $\alpha>0$, depending on $\O$, such that
$$
\ha(v,v)
\ge\alpha\left\|v\right\|_{1,\O}^2
\qquad\forall v\in\HuO.
$$
\end{lemma}

Next, we define the solution operator associated with Problem~\ref{P2}:
\begin{align*}
T:\ \HuO & \longrightarrow\HuO,
\\
f & \longmapsto Tf:=w,
\end{align*}
where $w\in\HuO$ is the unique solution (as a consequence
of  Lemma~\ref{ha-elipt} and the Lax-Milgram Theorem)
of the following source problem:
\begin{equation}
\label{T1}
\ha(w,v)=b(f,v)
\qquad\forall v\in\HuO.
\end{equation}

Thus, the linear operator $T$ is well
defined and bounded. Also, $T$ is self-adjoint with respect to
the inner product $\ha(\cdot,\cdot)$ in $\HuO$ (see \cite[Section 2]{MRR1}).

Notice that $(\l,u)\in\R\times\HuO$ solves
Problem~\ref{P2} (and hence Problem~\ref{P1}) if and only if $Tu=\mu u$
with $\mu\neq0$ and $u\ne0$, in which case $\mu:=1/(1+\l)$.

The following additional regularity result for the solution of
problem~\eqref{T1} and consequently, for the eigenfunctions of $T$, has been
proved in \cite[Lemma 2.2]{MRR1}.

\begin{lemma}
\label{LEM:REG}

\begin{itemize}
\item[i)] for all $f\in\HuO$, there exist $r\in(1/2,1]$ and $C>0$ such that the
solution $w$ of problem~\eqref{T1} satisfies $w\in\HurO$
and
$$
\left\|w\right\|_{1+r,\O}
\le C\left\|f\right\|_{1,\O};
$$

\item[ii)] if $u$ is an eigenfunction of Problem~\ref{P1} with
eigenvalue $\l$, there exist $r>1/2$ and $C>0$ (depending on $\l$) such that
$u\in\HurO$ and
$$
\left\|u\right\|_{1+r,\O}
\le C\left\|u\right\|_{1,\O}.
$$
\end{itemize}
\end{lemma}

\begin{remark}\label{angleome}
The constant $r>1/2$ is the Sobolev exponent for the Laplace
problem with Neumann boundary conditions. If $\O$ is convex, then
$r\ge1$, whereas, otherwise, $r:=\pi/\omega$ with $\omega$
being the largest reentrant angle of $\O$ (see \cite{G}).
\end{remark}

Hence, as a consequence of the compact inclusion
$\HurO\hookrightarrow\HuO$, $T$ is a compact operator.
We have the following spectral
characterization for the operator $T$.

\begin{theorem}
\label{CHAR_SP}
The spectrum of $T$ decomposes as follows:
$\sp(T)=\left\{0,1\right\}\cup\left\{\mu_k\right\}_{k\in\N}$, where:
\begin{enumerate}
\item[\textit{i)}] $\mu=1$ is an eigenvalue of $T$ and its associated
eigenspace is the space of constant functions in $\O$;

\item[\textit{ii)}] $\mu=0$ is an infinite-multiplicity eigenvalue of
$T$ with associated eigenspace is
$\HuGoO:=\left\{q\in\HuO:\ q=0\ \mbox{on }\Go\right\}$;

\item[\textit{iii)}] $\left\{\mu_k\right\}_{k\in\N}\subset(0,1)$ is a
sequence of finite-multiplicity eigenvalues of $T$ which converge to $0$
and their corresponding eigenspaces lie in $\HurO$.
\end{enumerate}
\end{theorem}

\setcounter{equation}{0}
\section{VEM discretization}
\label{SEC:DISCRETE}

We will study in this section, the virtual element
numerical approximation of the eigenproblem
presented in Problem~\ref{P2}, by considering weaker mesh assumptions
than the mesh assumptions considered in \cite{MRR1}.

We will follow some
recent results from \cite{BLR17,BrS} for the Poisson problem.
With this aim, first we recall the mesh construction.

Let $\left\{\CT_h\right\}_h$ be a sequence of decompositions of $\O$
into polygons $\E$. Let $h_\E$ denote the diameter of the element $\E$
and $h$ the maximum of the diameters of all the elements of the mesh,
i.e., $h:=\max_{\E\in\O}h_\E$. For $\CT_h$ we will consider
the following assumption:
\begin{itemize}
\item \textbf{A1.} There exists $\g>0$ such that, for all meshes
$\CT_h$, each polygon $\E\in\CT_h$ is star-shaped with respect to a ball
of radius greater than or equal to $\g h_{\E}$.
\end{itemize}

The next results will be obtained only under assumption \textbf{A1.}
In particular, we can consider meshes with edges arbitrarily small
with respect to the element diameter $h_{\E}$.

We consider now a simple polygon $\E$, we define
$$
\mathbb{B}_1(\partial\E)
:=\left\{v\in C^0(\partial\E):v|_e\in\P_1(e)
\text{ for all edges }e\subset\partial\E\right\}.
$$
We then consider the finite-dimensional space defined as follows:
$$
\VK
:=\left\{v\in\HuE:
\ v|_{\partial\E}\in\mathbb{B}_1(\partial\E)
\text{ and }\Delta v|_{\E}=0\right\}.
$$

As in \cite{BLR17}, we choose the following degrees of freedom:
For all $v_h\in\VK$, they are defined as follows:
\begin{itemize}
\item values of $v_h$ at the $N_K$ vertices of $\E$.

\end{itemize}

Next, for every
decomposition $\CT_h$ of $\O$ into simple polygons $\E$, we define
the global virtual space
$$
\Vh:=\left\{v\in\HuO:\ v|_{\E}\in\VK\right\}.
$$

In order to construct the discrete scheme, we need some preliminary
definitions. First, we split the bilinear form $\ha(\cdot,\cdot)$ as
follows:
$$
\ha(w,v)=\sum_{\E\in\CT_h}a^{\E}(w,v)+b(w,v)
\qquad w,v\in\HuO,
$$
where
\begin{equation*}
\label{defin}
a^{\E}(w,v)
:=\int_{\E}\nabla w\cdot\nabla v
\qquad w,v\in\HuO.
 \end{equation*}

Next, for any $\E\in\CT_h$ and for any sufficiently regular
function $v$, we define first
\begin{equation}
\label{eq:fixed}
\overline{v}:=\vert\partial\E\vert^{-1}\int_{\partial \E}v.
\end{equation}

Now, we define the projector $\PiK:\VK\longrightarrow\P_1(\E)\subseteq\VK$ for
each $v\in\VK$ as the solution of
\begin{subequations}
\begin{align*}
a^{\E}\big(\PiK v,q\big)
& =a^{\E}(v,q)
\qquad\forall q\in\P_1(\E),

\\
\overline{\PiK v}
&=\overline{v}.

\end{align*}
\end{subequations}

Now, we introduce the following symmetric and semi-positive
definite bilinear form on $\VK\times\VK$ (see \cite{WRR}).
For all elements $\E\in\CT_h$:
\begin{equation}
\label{20}
S^{\E}(w_h,v_h):=h_{\E}\int_{\partial{\E}}\partial_{s}w_h\partial_s
v_h\qquad\forall w_h,v_h\in\VK,
\end{equation}
where $\partial_s$ denotes a derivative along the edge.

Then, set
$$
a_h(w_h,v_h)
:=\sum_{\E\in\CT_h}a_h^{\E}(w_h,v_h)
\qquad w_h,v_h\in\Vh,
$$
where $a_h^{\E}(\cdot,\cdot)$ is the bilinear form defined on
$\VK\times\VK$ by
\begin{equation}
\label{21}
a_h^{\E}(w,v)
:=a^{\E}\big(\PiK w,\PiK v\big)
+S^{\E}\big(w-\PiK w,v-\PiK v\big)
\qquad w,v\in\VK.
\end{equation}

Now, we introduce the following discrete semi-norm:
\begin{equation}
\label{eq:triple}
\vertiii{v}_{\E}^2:=a^{\E}\big(\PiK v,\PiK v)+S^{\E}(v-\bar{v},v-\bar{v})\qquad\forall
v\in\VK+\mathcal{V}^{\E},
\end{equation}
where  $\mathcal{V}^{\E}\subseteq H^1(\E)$ is a subspace of sufficiently
regular functions for $S^{\E}(\cdot,\cdot)$ to make sense.

Now, for any sufficiently regular functions,
we introduce the following global semi-norms
$$
\vertiii{v}^2:=\sum_{\E\in\CT_h}\vertiii{v}_{\E}^2,\qquad
\left|v\right|_{1,h}^2
:=\sum_{\E\in\CT_h}\left\|\nabla v\right\|_{0,\E}^2.
$$

It has been proved in  \cite[Lemma~3.1]{BLR17} that
there exist positive constants $C_1,C_2,C3$, independent of $h$,
such that
\begin{align}
C_1\vertiii{v}_{\E}^2\le a_h^{\E}(v,v)\le C_2\vertiii{v}_{\E}^2\quad\forall v\in\VK,\label{eqrefgt1}\\
a_h^{\E}(v,v)\le C_3(\vertiii{v}^2+\vert v\vert_{1,\E}^2)\quad\forall v\in\VK.\label{eqrefgt12}
\end{align}
In addition, it holds
\begin{align}
a^{\E}(v,v)\le C_4\vertiii{v}_{\E}^2\quad\forall v\in\VK,\label{eqrefgt2}\\
\nonumber\vertiii{p}_{\E}^2\le C_5 a^{\E}(p,p)\quad\forall p\in\P_1(\E),\label{eqrefgt3}
\end{align}
where $C_4,C_5$ are independent of $h$.

Now we are in a position to write the virtual element
discretization of Problem~\ref{P1}.

\begin{problem}
\label{P11}
Find $(\l_h,u_h)\in\R\times\Vh$, $u_h\ne0$, such that
$$
a_h(u_h,v_h)
=\l_h b(u_h,v_h)
\qquad\forall v_h\in\Vh.
$$
\end{problem}

We use again a shift argument to rewrite this discrete eigenvalue
problem in the following convenient equivalent form.

\begin{problem}
\label{P21}
Find $(\l_h,u_h)\in\R\times\Vh$, $u_h\ne0$, such that
$$
\ha_h(u_h,v_h)
=\left(\l_h+1\right)b(u_h,v_h)
\qquad\forall v_h\in\Vh,
$$
\end{problem}

\noindent where the bilinear form
$\ha_h:\Vh\times\Vh\rightarrow\mathbb{R}$ is defined by
\begin{equation*}
\label{eq:shifteado}
\ha_h(u_h,v_h)
:=a_h(u_h,v_h)+b(u_h,v_h)
\qquad u_h,v_h\in\Vh.
\end{equation*}

Clearly $\ha_h(\cdot,\cdot)$ is symmetric and continuous.
In the following result we prove that $\ha_h(\cdot,\cdot)$ is elliptic
in $\Vh$.

\begin{lemma}
\label{ha-elipt-disc}
There exists a constant $\beta>0$, independent of $h$, such that
$$
\ha_{h}(v_h,v_h)
\ge\beta\left\|v_h\right\|_{1,\O}^2
\qquad\forall v_h\in\Vh.
$$
\end{lemma}

\begin{proof}
From  the definition of the bilinear form $\ha_h(\cdot,\cdot)$, we have that
\begin{multline*}
\ha_{h}(v_h,v_h)=a_h(v_h,v_h)+b(v_h,v_h)\\
=\sum_{\E\in\CT_h}a_h^{\E}\big(v_h,v_h\big)+b(v_h,v_h)
\geq\sum_{\E\in\CT_h}C_1\vertiii{v}_{\E}^2+\Vert v_h\Vert_{0,\Go}^{2}
\geq C\sum_{\E\in\CT_h}a^{\E}\big(v_h,v_h\big)+\Vert v_h\Vert_{0,\Go}^{2}\\
\geq C\vert v_h\vert_{1,\O}^2+\Vert v_h\Vert_{0,\Go}^{2}
\geq \beta\left\|v_h\right\|_{1,\O}^2
\qquad\forall v_h\in\Vh,
\end{multline*}

where we have used \eqref{eqrefgt1}, \eqref{eqrefgt2} and
the generalized Poincar\'e inequality. This concludes the proof.

\end{proof}

With this coercivity result at hand, we are in a position
to  introduce the  discrete solution operator
\begin{align*}
T_h:\ \HuO & \longrightarrow\HuO,
\\
f & \longmapsto T_hf:=w_h,
\end{align*}
where $u_h\in\Vh$ is the solution of the following discrete source
problem
\begin{equation*}
\ha_h(w_h,v_h)=b(f,v_h)
\qquad\forall v_h\in\Vh.
\end{equation*}

 Notice that Lemma~\ref{ha-elipt-disc} implies that the linear operator $T_h$ is well
defined and bounded uniformly with respect to $h$. Moreover, as in the
continuous case, $(\l_h,u_h)\in\R\times\Vh$ solves Problem~\ref{P21}
(and hence Problem~\ref{P11}) if and only if $T_hu_h=\mu_h u_h$ with
$\mu_h\neq0$ and $u_h\ne0$, in which case $\mu_h:= 1/(1+\l_h)$.
Also, $T_h|_{\Vh}:\ \Vh\longrightarrow\Vh$ is self-adjoint with
respect to $\ha_{h}(\cdot,\cdot)$.

As a consequence, we have the following spectral characterization for $T_h$.

\begin{theorem}
\label{CHAR_SP_DISC}
The spectrum of $T_h|_{\Vh}$ consists of $M_h:=\dim(\Vh)$ eigenvalues,
repeated according to their respective multiplicities. It decomposes as
follows: $\sp(T_h|_{\Vh})
=\left\{0,1\right\}\cup\left\{\mu_{hk}\right\}_{k=1}^{N_h}$, where:
\begin{enumerate}
\item[\textit{i)}] the eigenspace associated with $\mu_h=1$ is the space
of constant functions in $\O$;

\item[\textit{ii)}] the eigenspace associated with $\mu_h=0$ is
$Z_h:=\Vh\cap\HuGoO=\left\{q_h\in V_h:\ q_h=0\ \mbox{on }\Go\right\}$;

\item[\textit{iii)}] $\mu_{hk}\subset(0,1)$,
$k=1,\dots,N_h:=M_h-\dim(Z_h)-1$, are non-defective eigenvalues repeated
according to their respective multiplicities.
\end{enumerate}
\end{theorem}

\setcounter{equation}{0}
\section{Convergence and error estimates}
\label{SEC:approximation}

In order to prove that the solutions of the discrete
problem converge to those of the continuous
problem, we will follow the standard procedure
for spectral theory for compact operators \cite{BO},
which consist in showing that $T_h$
converges in norm to $T$ as $h$ tends to zero.

With this end, we begin by proving the following result.

\begin{lemma}
\label{lemcotste}
There exists $C>0$, independent of $h$, such that,
for all $f\in\HuO$, if $w=Tf$ and $w_h=T_hf$, then
\begin{equation}
\label{eq:T-Th1}
\left\|w-w_h\right\|_{1,\O}
\le C\left(\left\|w-w_I\right\|_{1,\O}+ \left|w_{\pi}-w\right|_{1,h}
+\vertiii{w-w_{I}}+\vertiii{w-w_{\pi}}\right),
\end{equation}
for all $w_I\in\Vh$ and for all $w_{\pi}\in\LO$ such that
$w_{\pi}|_{\E}\in\P_1(\E)$ $\forall\E\in\CT_h$.  In addition
\end{lemma}
\begin{equation}
\label{eq:uh-uI}
\vertiii{w_{h}-w_{I}}
\le C\left(\left\|w-w_I\right\|_{1,\O}
+ \left|w_{\pi}-w\right|_{1,h}+\vertiii{w-w_{I}}+\vertiii{w-w_{\pi}}\right).
\end{equation}

\begin{proof}
Let $w=Tf$ and
$w_h=T_hf$. From triangular inequality we have
$$\|w-w_{h}\|_{1,\O}\leq \|w-w_{I}\|_{1,\O}+\|w_{I}-w_{h}\|_{1,\O}. $$

Our task is to estimate the norms of the right hand side above.
To do this, we will consider the arguments
on the proof of Lemma~\ref{ha-elipt-disc}.

Now, for $w_I\in\Vh$, we set $v_h:=w_h-w_I$ and thanks to
Lemma~\ref{ha-elipt-disc}, the definitions of $a_h^{\E}(\cdot,\cdot)$ (cf. \eqref{21})
and those of $T$ and $T_h$, we have
\begin{align*}
\left(\vertiii{v_{h}}+\left\|v_h\right\|_{0,\G_{0}}\right)^{2}&\leq
2\left(\vertiii{v_{h}}^{2}+\left\|v_h\right\|^2_{0,\G_{0}}\right)\\
&\leq C\ha_h(v_h,v_h)=C(\ha_h(w_h,v_h)-\ha_h(w_I,v_h))
\\
&=C\left(b(f,v_h)-\sum_{\E\in\CT_h}a_h^\E(w_I,v_h)-b(w_I,v_h)\right)
\\
&=C\Big(b(f,v_h)-b(w_I,v_h)\\
&\left.-\sum_{\E\in\CT_h}
\left(a_h^\E(w_I-w_{\pi},v_h)+a^\E(w_{\pi}-w,v_h)+a^\E(w,v_h)\right)\right)
\\
& =C\left(b(w-w_I,v_h)
-\sum_{\E\in\CT_h}
\left(a_h^\E(w_I-w_{\pi},v_h)+a^\E(w_{\pi}-w,v_h)\right)\right).
\end{align*}
Therefore, from the trace theorem, \eqref{eqrefgt1} and the boundedness of
$a_h^\E(\cdot,\cdot)$ (\eqref{eqrefgt12}) and $a^\E(\cdot,\cdot)$, we get
\begin{align*}
\left(\vertiii{v_{h}}+\left\|v_h\right\|_{0,\Go}\right)^{2}&
\leq C\Big(\left\|w-w_I\right\|_{0,\Go}
\left\|v_h\right\|_{0,\Go}\\
&\quad\left.+\sum_{\E\in\CT_h}
\left(C(\vertiii{w_I-w_{\pi}}_{\E}+\left|w_I-w_{\pi}\right|_{1,\E})
\vertiii{v_h}_{\E}
+\left|w_{\pi}-w\right|_{1,\E}
\left|v_h\right|_{1,\E}\right)\right)
\\
&  \leq C\left(\left\|w-w_I\right\|_{0,\Go}
\left\|v_h\right\|_{0,\Go}
+\sum_{\E\in\CT_h}
\left\{C\left(\vertiii{w_I-w_{\pi}}_{\E}+\left|w_{\pi}-w\right|_{1,\E}\right)
\vertiii{v_h}_{\E}\right\}\right)
\\
&  \leq C \left(\left\|w-w_I\right\|_{0,\Go}+\vertiii{w_I-w_{\pi}}+ \left|w_{\pi}-w\right|_{1,h}\right)\left(\|v_{h}\|_{0,\Go}+\vertiii{v_{h}}\right).

\end{align*}
Therefore, we have
\begin{equation*}
\vertiii{v_{h}}+\left\|v_h\right\|_{0,\G_{0}}
\leq \widetilde{C} \left(\left\|w-w_I\right\|_{1,\O}
+\vertiii{w-w_{\pi}}+\vertiii{w-w_{I}}+ \left|w_{\pi}-w\right|_{1,h}\right).
\end{equation*}
Finally, \eqref{eq:T-Th1} follows from the
triangular inequality and the generalized Poincar\'e inequality.
Moreover, \eqref{eq:uh-uI} follows from the above estimate.
\end{proof}

Let us introduce the following approximation result for
polynomials in star-shaped domains (see for instance \cite{BS-2008}),
which is derived by results of interpolation between Sobolev spaces
(see for instance \cite[Theorem~I.1.4]{GR}), leading to an analogous
result for integer values of $s$. Moreover, we remark that the
result for integer values is stated in
\cite[Proposition~4.2]{BBCMMR2013} and follows from the well establish
Scott-Dupont theory (see \cite{BS-2008}).

\begin{lemma}
\label{lmm:bh}
If assumption {\bf A1} is satisfied, then there exists a constant
$C$, depending only on $k$ and $\g$, such that for every $s$ with
$0\le s\le k$ and for every $v\in\HusE$, there exists
$v_{\pi}\in\P_k(\E)$ such that
$$
\left\|v-v_{\pi}\right\|_{0,\E}
+h_{\E}\left|v-v_{\pi}\right|_{1,\E}
\le Ch_{\E}^{1+s}\left\|v\right\|_{1+s,\E}.
$$
\end{lemma}

Now, we have the following approximation result
in the virtual space $V_h$, which follows from
\cite[Theorem~3.4]{BLR17}.

\begin{lemma}
\label{estima2}
Under the assumption  {\bf A1}, then, for
each $s$ with $1/2<s\le 1$, there exist $\widehat{t}>1/2$
and a constant $C$, independent of $h$,
such that for every $v\in\HusO$, there exists
$v_I\in\Vh$ that satisfies
\begin{align}
\label{firststimate}
\left|v-v_I\right|_{1+t,\E}&\le Ch_{\E}^{s-t}\left|v\right|_{1+s,\E},
\qquad 0\leq t<\min\{\widehat{t},s\},\\
\label{secondstimate}
\left\|v-v_I\right\|_{0,\E}
&\le Ch_{\E}\left|v\right|_{1+s,\E}.
\end{align}

\end{lemma}
\begin{proof}
Estimate \eqref{firststimate} has been obtained in \cite[Theorem~3.4]{BLR17}.

 To obtain \eqref{secondstimate}, with $1/2<s\leq1$,
first we use the Poincar\'e and the Cauchy-Schwarz inequalities,
to obtain (see \cite[Remark 4.1]{BM20}),

\begin{align*}
\left\|v-v_I\right\|_{0,\E}&\leq C\left(\int_{\partial \E}|v-v_{I}|ds+h_{\E}|v-v_{I}|_{1,\E}\right)\\
&\leq C\left(h_{\E}^{1/2}\|v-v_{I}\|_{0,\partial\E}+h_{\E}|v-v_{I}|_{1,\E}\right)\\
&\leq C\left(h_{\E}^{1/2}h_{\E}^{1/2}|v|_{1/2,\partial \E}+h_{\E}|v-v_{I}|_{1,\E}\right)\\
&\leq C\left(h_{\E}|v|_{1,\E}+h_{\E}|v-v_{I}|_{1,\E}\right)\\

&\leq Ch_{\E}|v|_{1+s,\E},
\end{align*}
where we have use an standard approximation estimate in one
dimension,

since $v_I|_{\partial\E}$ corresponds to the standard
piecewise linear Lagrange interpolant of $v$
and then $|v|_{1/2,\partial \E}\le|v|_{1,\E}$. This concludes the proof.
\end{proof}

Now we are in position to establish the convergence in norm of $T_h$ to $T$ as
$h\to0$.

\begin{lemma}
\label{convnorm}
There exists $r\in(1/2,1]$ (cf. Lemma~\ref{LEM:REG}(i)) and $C>0$,
independent of $h$, such that

$$
\left\|\left(T-T_h\right)f\right\|_{1,\O}
\le Ch^{r}\left\|f\right\|_{1,\O}
\qquad\forall f\in\HuO.
$$
\end{lemma}

\begin{proof}
The result follows from Lemma~\ref{lemcotste}. In particular, we
have to bound the term on the right and side of \eqref{eq:T-Th1}.
For the first and second terms, using Lemmas \ref{estima2} and
\ref{lmm:bh}, respectively, we obtain
\begin{equation}
\label{eq:uI}
||w-w_{I}||_{1,\O}+|w-w_{\pi}|_{1,h}\leq
C\sum_{\E\in\CT_{h}}h_{\E}^{r}|w|_{1+r}\leq Ch^{r}|w|_{1+r,\O}.
\end{equation}
Now, we bound the term $\vertiii{w-w_{I}}$. To do this task, we
invoke the definition of $\vertiii{\cdot}$ given in \eqref{eq:triple}, \eqref{eq:fixed}
and, operating as in the proof of \cite[Theorem 4.5]{BLR17} we have
\begin{align}
\nonumber
\sum_{\E\in\CT_{h}}\vertiii{w-w_I}_{\E}^{2}&
=\sum_{\E\in\CT_{h}}\left\{a^K\left(\PiK(w-w_I), \PiK(w-w_I) \right)\right.\\\nonumber
&\left.+S^K\left((w-w_I)-\overline{(w-w_{I})},(w-w_I)-\overline{(w-w_{I})}\right)\right\}\\\nonumber
&\leq C\sum_{\E\in\CT_{h}}\left(|\PiK(w-w_I)|^2_{1,K}+S^\E(w-w_I,w-w_I) \right)\\\label{eq:inter}
&=C\sum_{\E\in\CT_{h}}\left(|w-w_I|^2_{1,K}+h_K|w-w_I|_{1,\partial \E}^2\right).
\end{align}
Let $\sigma$ be such that $1/2<\sigma<r$, using a scaled trace inequality, we get
$$h_K|w-w_I|_{1,\partial \E}^2\leq C \left(|w-w_{I}|_{1,\E}^{2}
+h^{2\sigma}|w-w_{I}|_{1+\sigma,\E}^{2}\right)\leq Ch_{\E}^{2r}|w|_{1+r,\E}^{2}.$$
Using the above estimate in \eqref{eq:inter} and
Proposition \ref{estima2}, we obtain
\begin{align}
\label{eq:estimeui}
\vertiii{w-w_{I}}=\left(\sum_{\E\in\CT_{h}}\vertiii{w-w_I}_{\E}^{2}\right)^{1/2}
&\leq C\left(\sum_{\E\in\CT_{h}}h_{\E}^{2r}|w|_{1+r,\E}^{2}\right)\leq Ch^{r}|w|_{1+r,\O}.
\end{align}
Similarly, we obtain
\begin{align*}
\vertiii{w-w_\pi}\leq Ch^{r}|w|_{1+r,\O}.
\end{align*}
thus, the lemma follows from \eqref{eq:uI}--\eqref{eq:estimeui}
and Lemma~\ref{LEM:REG}(i).

\end{proof}

We conclude the analysis of our paper deriving error estimates for our method.
In particular, we are going to present error estimates for eigenfunctions
and eigenvalues. With this aim, with Lemma~\ref{convnorm} at hand,
we will prove that isolated parts of $\sp(T)$ are approximated
by isolated parts of $\sp(T_h)$ (see \cite{K}).

Let $\mu\in(0,1)$ be an isolated eigenvalue of $T$ with
multiplicity $m$ and let $\CE$ be its associated eigenspace. Then, there
exist $m$ eigenvalues $\mu^{(1)}_h,\dots,\mu^{(m)}_h$ of $T_h$ (repeated
according to their respective multiplicities) which converge to $\mu$. From now and on, let $\CE_h$ be the discrete subspace associated to $\CE$, corresponding to the direct sum of their corresponding associated eigensapaces.

We recall the definition of the \textit{gap} $\hdel$ between two closed
subspaces $\CM$ and $\CN$ of $\HuO$:
$$
\hdel(\CM,\CN)
:=\max\left\{\delta(\CM,\CN),\delta(\CN,\CM)\right\},
\quad\text{where}\quad
\delta(\CM,\CN)
:=\sup_{x\in\CM:\ \left\|x\right\|_{1,\O}=1}
\left(\inf_{y\in\CN}\left\|x-y\right\|_{1,\O}\right).
$$
The following error estimates for the approximation of eigenvalues and
eigenfunctions hold true.

\begin{theorem}
\label{gap}
There exists a strictly positive constant $C$ such that
\begin{align*}
\hdel(\CE,\CE_h)
& \le C\g_h,
\\
\left|\mu-\mu_h^{(i)}\right|
& \le C\g_h,\qquad i=1,\dots,m,
\end{align*}
where
$$
\g_h
:=\sup_{f\in\CE:\ \left\|f\right\|_{1,\O}=1}
\left\|(T-T_h)f\right\|_{1,\O}.
$$
\end{theorem}

\begin{proof}
As a consequence of Lemma~\ref{convnorm}, $T_h$ converges in norm to $T$
as $h$ goes to zero. Then, the proof follows as a direct consequence of
Theorems~7.1 and 7.3 from \cite{BO}.
\end{proof}

The theorem above yields error estimates depending on $\g_h$. The next
step is to show an optimal order estimate for this term.

\begin{theorem}
\label{order}
There exist $r\in(1/2,1]$ and a positive constant $C$
such that
\begin{equation*}
\label{cota}
\left\|(T-T_h)f\right\|_{1,\O}
\leq Ch^{r}
\left\|f\right\|_{1,\O}
\qquad\forall f\in\CE,
\end{equation*}
and, consequently,
\begin{equation*}
\label{cotgap}
\g_h\le Ch^{r}.
\end{equation*}
\end{theorem}
\begin{proof}
See \cite[Theorem 4.2]{MRR1}.
\end{proof}

The error estimate for the eigenvalue $\mu\in(0,1)$ of $T$ leads to an
analogous estimate for the approximation of the eigenvalue
$\l=\frac1{\mu}-1$ of Problem~\ref{P1} by means of the discrete
eigenvalues $\l_h^{(i)}:=\frac1{\mu_h^{(i)}}-1$, $1\le i\le m$, of
Problem~\ref{P11}.

We are able to improve the convergence order of
Theorem~\ref{gap} for the eigenvalues. The following result shows in fact
that the convergence order is double.

\begin{theorem}
\label{cotadoblepandeo}
There exist $r\in(1/2,1]$ and a positive constant $C$
such that
$$\
\left|\l-\l_h^{(i)}\right|
\le Ch^{2r}.
$$
\end{theorem}

\begin{proof}
Let $u_h$ be such that $(\l_h^{(i)},u_h)$ is a solution of
Problem~\ref{P11} with $\left\|u_h\right\|_{1,\O}=1$. According to
Theorems~\ref{gap} and\ref{order}, there exists a solution
$(\l,u)$ of Problem~\ref{P1} such that
\begin{equation}
\label{fin4}
\left\|u-u_h\right\|_{1,\O}\le Ch^{r}.
\end{equation}

From the symmetry of the bilinear forms and the facts that
$a(u,v)=\l b(u,v)$ for all $v\in\HuO$ (cf. Problem~\ref{P1}) and
$a_h(u_h,v_h)=\l_h^{(i)}b(u_h,v_h)$ for all $v_h\in\Vh$ (cf.
Problem~\ref{P11}), we have
\begin{align*}
a(u-u_h,u-u_h)-\l b(u-u_h,u-u_h)
& =a(u_h,u_h)-\l b(u_h,u_h)
\\
& =\left[a(u_h,u_h)-a_h(u_h,u_h)\right]
-\left(\l-\l_h^{(i)}\right)b(u_h,u_h),
\end{align*}
from which we obtain the following identity:
\begin{equation}
\label{45}
\left(\l_h^{(i)}-\l\right)b(u_h,u_h)
=a(u-u_h,u-u_h)-\l b(u-u_h,u-u_h)
+\left[a_h(u_h,u_h)-a(u_h,u_h)\right].
\end{equation}

The next step is to bound each term on the right hand side above. The
first and the second ones are easily bounded from the continuity of
$a(\cdot,\cdot)$ and $b(\cdot,\cdot)$, the trace theorem and
\eqref{fin4} as follows
\begin{equation}
\label{terms_12}
\left|a(u-u_h,u-u_h)\right|
+\l\left|b(u-u_h,u-u_h)\right|
\le Ch^{2r}.
\end{equation}

For the last term on the right hand side of \eqref{45},
we consider $u_\pi\in\P_1(\E)$ and $u_I\in V_h$
such that Lemmas~\ref{lmm:bh} and \ref{estima2} hold true, respectively.
Using standard arguments, we have
\begin{align*}
\left|a_h(u_h,u_h)-a(u_h,u_h)\right|&
=\left| \sum_{\E\in\CT_{h}} a_{h}^{\E}(u_h-u_\pi,u_h-u_\pi)-a^{\E}(u_h-u_\pi,u_h-u_\pi)\right|\\
& \leq\sum_{\E\in\CT_{h}} \vertiii{u_h-u_\pi}_{\E}^2+\sum_{\E\in\CT_{h}}|u_h-u_\pi|_{1,\E}^2\\
&=\vertiii{u_h-u_\pi}^2+|u_h-u_\pi|_{1,h}^2\\
&\leq C\left(\vertiii{u-u_h}^2+\vertiii{u-u_\pi}^2+|u-u_h|^2_{1,h}+|u-u_\pi|^2_{1,h} \right)\\
&\leq C\left(\vertiii{u-u_I}^2+\vertiii{u_{h}-u_I}^2
+\vertiii{u-u_\pi}^2+|u-u_h|^2_{1,h}+|u-u_\pi|^2_{1,h} \right)\\

&\leq C\left(\vertiii{u-u_I}^2+\vertiii{u-u_\pi}^2
+\|u-u_{I}\|_{1,\O}^{2}+|u-u_{\pi}|_{1,h}^{2}+|u_h-u|^2_{1,\O} \right),
\end{align*}
where we have used \eqref{eq:uh-uI}.
Now, the terms on the right hand side above
can be bounded repeating
the argument in the proof of Lemma~\ref{convnorm}
and using the additional regularity result
in Lemma~\ref{LEM:REG}(ii). We get
\begin{align}
\label{ultimaeq}
 \left|a_h(u_h,u_h)-a(u_h,u_h)\right|\leq Ch^{2r}.
 \end{align}

On the other hand, by virtue of Lemma~\ref{ha-elipt-disc} and the fact
that $\l_h^{(i)}\to\l$ as $h$ goes to zero, we know that there exists
$C>0$ such that
$$
b(u_h,u_h)
=\frac{\ha_h(u_h,u_h)}{\l_h^{(i)}+1}
\ge\frac{\beta\left\|u_h\right\|_{1,\O}^2}{\l_h^{(i)}+1}
\ge\frac{\beta}{C}>0.
$$
Finally, the proof follows from \eqref{45}, by
using the above estimate together with
\eqref{terms_12} and \eqref{ultimaeq}.

\end{proof}
\section{Numerical experiments}

\label{sec:numerics}
In the present section we will report some numerical
tests in order to asses the performance of the proposed lowest order VEM
with meshes allowing small edges. All the reported numerical
results have been obtained with a MATLAB code.

In order to observe the performance and accuracy of the proposed method,
we will consider different computational domains,
where the eigenfunctions, on one hand, can be smooth enough and,
on the other, can be singular due to the non-convex domains.

For all the tests, we will report the computed eigenvalues
for different polygonal meshes and the order of convergence.

Our results will be compared with some references
and exact solutions in the cases where it is available.
In the cases where it is not possible to have a close form of the solution,
we will present extrapolated values for the eigenvalues (see \eqref{eq:fitting}).

\subsection{Square domain: the sloshing problem.}\label{sectionslosh}

We begin with a convex domain. In this case we consider
$\Omega:=(0,1)^2$ as computational domain.
We fix $\Gamma_0$ on the top of the boundary
(representing a free surface)
and $\Gamma_1$ will the rest of the boundary.
In Figure~\ref{FIG:SLOSH}, we present the physical configuration
of the problem.
\vspace*{0.25cm}
\begin{figure}[H]
\begin{center}
\input{sloshing.pstex_t}
\caption{Sloshing in a square domain.}
\label{FIG:SLOSH}
\end{center}
\end{figure}

For this problem there are analytical solutions of the form
\begin{equation}
\label{eq:exact_sloshing}
\lambda_n=n\pi\tanh(n\pi), \quad u_n(x,y)=\cos(n\pi x)\sinh (n\pi x), \quad n\in\mathbb{N}.
\end{equation}

In Figure~\ref{FIG:meshes}, we present plots of the polygonal meshes that
we will consider for our tests. We note that the family of polygonal meshes $\CT_h^1$
have been obtained by gluing two different polygonal meshes at $y=0.6$.
It can be seen that very small edges compared with the element
diameter appears on the interface of the resulting mesh.
The second family of polygonal meshes $\CT_h^2$ have been obtained
from a triangular mesh with an additional point on
each edge as a new degree of freedom which has been moved
to a distance $h_e^{2}$ from one vertex and $(h_{e}-h_{e}^{2})$
from the other. We observe that this family satisfy $\mathbf{A1}$
but fail to satisfy the usual assumption that distance between
any two of its vertices is greater than or equal to $Ch_{\E}$ for each polygon,
since the length of the smallest edge is $h_e^2$, while the diameter
of the element is bounded above by a multiple of $h_e$.
The refinement level for the meshes will be denoted by $N$,
which corresponds to the number of subdivisions in the abscissae.
\begin{figure}[H]
\begin{center}
\begin{minipage}{15cm}

\centering\includegraphics[height=6.5cm, width=6.5cm]{thao2.eps}
\centering\includegraphics[height=6.5cm, width=6.5cm]{thao3.eps}

\end{minipage}
\caption{Sample meshes with small edges. From left to right:
$\CT_h^1$  and $\CT_h^2$ for $N=4$.}
\label{FIG:meshes}
\end{center}
\end{figure}

In Table~\ref{tabla1}, we report the first six eigenvalues computed with
meshes $\CT_h^1$  and $\CT_h^2$. The row 'Order' reports the convergence
order of the eigenvalues, computed with respect to the exact ones obtained with
\eqref{eq:exact_sloshing}, which are presented in the row 'Exact'.

\begin{table}[H]
\begin{center}
\caption{Test1. The lowest computed eigenvalues
$\l_{hi}$, $1\le i\le 6$ for different meshes.}
\vspace{0.3cm}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline

\multicolumn{7}{|c|}{$\CT_{h}^{1}$} \\\hline
 $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
 \hline
8    &3.2422   &7.1802  &12.5152  &19.0595  &32.2326  &46.9310  \\
16   &3.1572   &6.4936  &10.1363  &14.2689  &19.0754  &24.7519  \\
32   &3.1366   &6.3347   &9.5984  &12.9787  &16.5158  &20.2508  \\
64   &3.1316   &6.2960   &9.4679  &12.6686  &15.9078  &19.1953  \\\hline
Order  &2.02   &2.04   &2.05   &2.00  & 2.12  & 2.12   \\\hline
Exact &3.1299  &  6.2831  &  9.4248  & 12.5664  & 15.7080 &  18.8496  \\
\hline
\multicolumn{7}{|c|}{$\CT_{h}^{2}$} \\\hline
 $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
 \hline
8    &3.1820  & 6.7247  &10.9616  &16.4556  &23.3067  &31.7699  \\
16   &3.1441  & 6.4043   &9.8511  &13.6063  &17.7749  &22.5294   \\
32   &3.1336  & 6.3135   &9.5285  &12.8170  &16.2084  &19.7245   \\
64   &3.1308  & 6.2907   &9.4503  &12.6275  &15.8287  &19.0608   \\\hline
Order  &1.94  & 1.96   &1.98  & 2.00  & 2.00  & 1.99    \\\hline
Exact &3.1299  &  6.2831   & 9.4248  & 12.5664  & 15.7080   &18.8496  \\
\hline
\end{tabular}
\label{tabla1}
\end{center}
\end{table}

The order of convergence is clearly $\mathcal{O}(h^{2})$,
which is expectable according to Theorem~\ref{cotadoblepandeo}
and due the smoothness of the eigenfunctions for this configuration
of the problem. Moreover, the nature of the meshes and
the fact that we are allowing small edges for the polygons,
does not affect the order of convergence and no spurious eigenvalues were found.

In the next test, we will study the effects of the
stabilization~\eqref{20} in the computation of the spectrum.
We will consider the same physical configuration as in the previous test.
Since the stabilization depends on the size of the
element $K$ (see~\eqref{20}), we will compute the
first six eigenvalues for different values $h_K^{\alpha}$
using the family of meshes $\CT_h^2$.

\begin{table}[H]
\begin{center}
\caption{Test2. The lowest computed eigenvalues $\l_{hi}$, $1\le i\le6$
for different $h_{\E}^{\alpha}$, $\alpha=\{1/2, 3/4, 1, 5/4, 3/2\}$
with meshes $\CT_{h}^{2}$.}
\vspace{0.3cm}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$h_{\E}^{\alpha}$& $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
\hline
&8 &   3.1906  &  6.8000  & 11.2387 &  17.2492  & 24.8335  & 34.3717\\
 &  16 &    3.1456 &   6.4174 &   9.8977 &  13.7232 &  18.0174 &  22.9781\\
$h_{\E}^{1/2} $   & 32 &    3.1338 &   6.3154  &  9.5349 &  12.8325  & 16.2400  & 19.7798\\
  & 64 &    3.1308  &  6.2909  &  9.4511  & 12.6295  & 15.8327  & 19.0678\\
 \hline
 & Order   &2.00   &2.02  & 2.04  & 2.08  & 2.07  & 2.06  \\ \hline
& Exact &3.1299  &  6.2831 &   9.4248 &  12.5664 &  15.7080 &  18.8496  \\\hline
$h_{\E}^{\alpha}$& $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
\hline
&8 &   3.1875 &   6.7730 &  11.1386  & 16.9596  & 24.2790 &  33.4292\\
 &  16 &  3.1451 &   6.4133 &   9.8829 &  13.6861 &  17.9400 &  22.8346\\
$h_{\E}^{3/4} $   &  32  &  3.1337 &   6.3148 &   9.5331 &  12.8282  & 16.2313 &  19.7645\\
 &  64   & 3.1308  &  6.2908  &  9.4509  & 12.6290  & 15.8317  & 19.0662\\ \hline
&  Order   &1.98   &  2.00  &          2.02 &            2.05 &             2.04&             2.03\\\hline
& Exact &3.1299  &  6.2831 &   9.4248 &  12.5664 &  15.7080 &  18.8496  \\
 \hline
 $h_{\E}^{\alpha}$& $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
\hline
&8    &3.1820  & 6.7247  &10.9616  &16.4556  &23.3067  &31.7699  \\
&16   &3.1441  & 6.4043   &9.8511  &13.6063  &17.7749  &22.5294   \\
$h_{E} $&32   &3.1336  & 6.3135   &9.5285  &12.8170  &16.2084  &19.7245   \\
&64   &3.1308  & 6.2907   &9.4503  &12.6275  &15.8287  &19.0608   \\\hline
& Order  &1.94  & 1.96   &1.98  & 2.00  & 2.00  & 1.99    \\\hline
& Exact  &3.1299  &  6.2831   & 9.4248  & 12.5664  & 15.7080   &18.8496  \\
\hline
$h_{\E}^{\alpha}$& $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
\hline
&8    &3.1728   & 6.6440   &10.6696  & 15.6472  & 21.7299  & 29.0670\\
 & 16  &  3.1420 &   6.3861 &   9.7866 &  13.4455 &  17.4448  & 21.9242\\
$h_{\E}^{5/4} $    & 32  &  3.1332 &   6.3101 &   9.5170 &  12.7889 &  16.1512 &  19.6248\\
  & 64  & 3.1307   & 6.2901   & 9.4483  & 12.6228  & 15.8194   &19.0444\\\hline
& Order  &1.88  &         1.90 &            1.91&            1.93 &             1.92 &             1.91    \\\hline
& Exact  &3.1299  &  6.2831   & 9.4248  & 12.5664  & 15.7080   &18.8496  \\
\hline
$h_{\E}^{\alpha}$& $N$ & $\l_{h1}$ & $\l_{h2}$ & $\l_{h3}$ &
 $\l_{h4}$ & $\l_{h5}$& $\l_{h6}$  \\
\hline
&8&   3.1585   & 6.5223 &  10.2377  & 14.5052 &  19.4815 &  25.2098\\
  & 16&   3.1382 &   6.3532 &   9.6706 &  13.1590 &  16.8635 &  20.8743\\
$h_{\E}^{3/2} $   &32 &   3.1323  &  6.3026 &   9.4912  & 12.7263  & 16.0248 &  19.4044\\
  & 64 &  3.1305  &  6.2885  &  9.4430  & 12.6098  & 15.7937  & 18.9992\\\hline
&Order  &1.81   &  1.83   &          1.83 &             1.83 &             1.82 &              1.81\\\hline
&Exact  &3.1299  &  6.2831   & 9.4248  & 12.5664  & 15.7080   &18.8496  \\
\hline
\end{tabular}
\label{tablan}
\end{center}
\end{table}

We observe from the results of Table~\ref{tablan} that the method
converges to the exact eigenvalues with an optimal
quadratic order and no spurious eigenvalues were found for any chosen
stability parameter $h_K^{\alpha}$. We remark that these results
are also valid for other type of family polygonal meshes allowing small edges.

\subsection{Rotated T domain}

In the following test we will consider a non-convex
domain which we call \emph{rotated T} an it is defined
by $\Omega_T:=(-0.5, 0.5)\times(-0.5, 0)\cup (-0.25,0.25)\times(0,1)$
with boundary condition $\Gamma_0=\partial \O_T$.
This non-convex domain presents two reentrant angles of the same size
$\omega=\frac{3\pi}{2}$ (cf. Figure~\ref{FIG:meshes2}),
and as a consequence, the eigenfunctions
of this problem may present singularities.

More precisely,

the Sobolev exponent for the eigenfunctions
is $2/3$ (cf. Remark~\ref{angleome}), so that the eigenfunctions
will belong to $\HurO$ for all $r<2/3$, but in general not to $H^{1+\frac{2}{3}}(\O)$.
Therefore, according to Theorem~\ref{cotadoblepandeo},
the convergence rate for the eigenvalues should be $\vert\lambda-\lambda_h\vert\approx h^{4/3}$.

In Figure~\ref{FIG:meshes2}, we present the meshes
that we will consider for this numerical test.
We note that the families of polygonal meshes $\CT_h^3$, $\CT_h^4$ and $\CT_h^5$
have been obtained by gluing two different polygonal meshes at $x=0$.
It can be seen that very small edges compared with the element
diameter appears on the interface of the resulting meshes.

\begin{figure}[H]
\begin{center}
\begin{minipage}{15cm}
\centering\includegraphics[height=4.5cm, width=4.5cm]{TrVor.eps}
\centering\includegraphics[height=4.5cm, width=4.5cm]{SqVor.eps}
\centering\includegraphics[height=4.5cm, width=4.5cm]{dsvor.eps}

\end{minipage}
\caption{Sample meshes with small edges. From left to right: $\CT_h^3$, $\CT_h^4$ and $\CT_h^5$, for $N=8$.}
\label{FIG:meshes2}
\end{center}
\end{figure}

In Table~\ref{tabla3}, we report the computed eigenvalues
and the corresponding convergence rates,
the last row, that we called 'Extrap.', corresponds
to extrapolated values obtained with a least square fitting of the form
\begin{equation}
\label{eq:fitting}
\lambda_{hi}\approx\lambda_{i}+C_ih^{\alpha_i},
\end{equation}
where $\alpha_i$ is the approximated rate of convergence
of each $\lambda_i$, with $i\in\mathbb{N}$.

\begin{table}[H]
\begin{center}
\caption{Test 2. Computed lowest eigenvalues $\l_{h}^{i}$, $1\le
i\le4$, on different polygonal meshes.}
\vspace{0.3cm}
\begin{tabular}{|c|c|cccc|c|c|c|}
\hline
$\CT_h$   & $\l_{h}^{i}$ & $N=16$ & $N=30$ & $N=62$ & $N=130$ &
Order & Extrap. \\
\hline
          & $\l_{h}^{1}$ &0.5196  & 0.5157  & 0.5140 & 0.5134 &   1.41 &  0.5130  \\
$\CT_h^3$ & $\l_{h}^{2}$ & 1.2743 &  1.2622 & 1.2570 &  1.2552  &  1.48  & 1.2543\\
         & $\l_{h}^{3}$& 2.5567 & 2.5263 & 2.5146  & 2.5111  &   1.65 &  2.5096\\
           & $\l_{h}^{4}$& 3.1923 & 3.1556  & 3.1458 &  3.1437 &  2.19  & 3.1432\\
\hline
          & $\l_{h}^{1}$ &  0.5209   & 0.5163  &  0.5142 &  0.5135   & 1.41 &  0.5131 \\
$\CT_h^4$& $\l_{h}^{2}$ &1.2793 &  1.2641&  1.2577 & 1.2555 &   1.51 & 1.2545 \\
          & $\l_{h}^{3}$&2.5659 &  2.5296  & 2.5158  &  2.5115   & 1.66  & 2.5098 \\
          & $\l_{h}^{4}$&   3.2144&  3.1616  & 3.1474& 3.1441  & 2.16 & 3.1434\\
 \hline
          & $\l_{h}^{1}$ &0.5209   & 0.5163  &   0.5142 &  0.5135 &   1.41 &  0.5131  \\
$\CT_h^5$& $\l_{h}^{2}$  & 1.2795 &   1.2641 & 1.2577   & 1.2555  &   1.52 & 1.2545  \\
          & $\l_{h}^{3}$ &  2.5663 & 2.5296 &2.5158&   2.5115 &   1.66 &   2.5098\\
          & $\l_{h}^{4}$ &   3.2143 &   3.1616 & 3.1474 & 3.1441&   2.16 &  3.1434\\
\hline

 \end{tabular}
\label{tabla3}
\end{center}
\end{table}

We observe from Table~\ref{tabla3} that
for the first Steklov eigenvalue the method
converges with order close to $4/3$ which corresponds
to the Sobolev regularity for the Steklov problem on $\O_T$ (non-convex domain).
We also note that the method converges
larger orders for the rest of the Steklov eigenvalues.

In Figure~\ref{FIG:Tmodes} we present plots
for the first four eigenfunctions for the Steklov problem in the \emph{rotated T} domain,
computed with $\CT_h^5$ and $N=30$.
\begin{figure}[H]
\begin{center}
\begin{minipage}{5cm}
\centering\includegraphics[height=5cm, width=5cm]{modo1T.eps}
\end{minipage}
\begin{minipage}{5cm}
\centering\includegraphics[height=5cm, width=5cm]{modo2T.eps}
\end{minipage}
\begin{minipage}{5cm}
\centering\includegraphics[height=5cm, width=5cm]{modo3T.eps}
\end{minipage}
\begin{minipage}{5cm}
\centering\includegraphics[height=5cm, width=5cm]{modo4T.eps}
\end{minipage}
\caption{From top left to bottom right, plots of the first four
eigenfunctions for the \emph{rotated T} domain, computed with $\CT_h^5$. }
		\label{FIG:Tmodes}
	\end{center}
\end{figure}

\subsection{L-shaped domain}
In this numerical example we test the properties of the
proposed method on an L-shaped domain:
$\O_L:=(0,1)\times(0,1)\backslash [0.5,1)\times[0.5,1)$
with $\Gamma_{0}=\partial\O_L$. For this test, we will
adopt a refinement with hanging nodes, which implies to
consider once again polygons with small edges.
More precisely, this test is focused to validate
the use of refined meshes as a tool to handle solutions
with corner singularities. With this purpose, we have
considered two families of meshes, namely: $\CT_h^6$ (see
upper left picture in Figure~\ref{FIG:LCuadRef})
and $\CT_h^{6,\ell}$. The initial uniform mesh $\CT_h^{6}$
has $N=32$ elements on each edge and the last one has $N = 128$
elements on each edge.

On the other hand, the mesh $\CT_h^{6,\ell}$ is obtained
by refining a patch around the re-entrant corner of $\Omega_L$,
starting from an initial uniform quadrilateral mesh $\CT_h^{6,0}$,
which corresponds to the first mesh of $\CT_h^6$.
The procedure consists in to split each element which belongs to the region:

$$R_{\ell}:=\left\{(x,y)\in \mathbb{R}^2:  |x-1/2|\leq \frac{6}{N}2^{1-\ell}
\quad \mbox{and} \quad  |y-1/2|\leq  \frac{6}{N}2^{1-\ell}\right\}
\cap \overline{\Omega}_L \qquad \ell=1,2,\ldots,\widehat{\ell},$$
into three quadrilaterals by connecting the barycenter of the element
with the midpoint of each edge, where $\widehat{\ell}$ is the number
of meshes to refine, with the convention that $\CT_h^{6,0}:=\CT_h^{6}$
(the initial mesh with $N=32$). Note that although this process
is initiated with a quadrilateral mesh, the successively created meshes
will contain other kind of convex polygons as can be appreciated
in Figure~\ref{FIG:LCuadRef}.

\begin{figure}[H]
	\begin{center}
		\begin{minipage}{5cm}
			\centering\includegraphics[height=5cm, width=5cm]{LCuadRefInicial.eps}
		\end{minipage}
		\begin{minipage}{5cm}
			\centering\includegraphics[height=5cm, width=5cm]{LCuadRef1.eps}
		\end{minipage}
		\begin{minipage}{5cm}
			\centering\includegraphics[height=5cm, width=5cm]{LCuadRef2.eps}
		\end{minipage}
		\begin{minipage}{5cm}
			\centering\includegraphics[height=5cm, width=5cm]{LCuadRef3.eps}
		\end{minipage}
\caption{Sample meshes: inicial mesh $\CT_{h}^{6}$ with $N=16$ (top left),
	$\CT_{h}^{6,1}$ (top right),  $\CT_{h}^{6,2}$ (bottom left) and $\CT_{h}^{6,3}$ (bottom right).}
		\label{FIG:LCuadRef}
	\end{center}
\end{figure}

\begin{table}[H]
\begin{center}
\caption{Test 3. Test with an L-shaped domain. Number of degrees of freedom,
computed lowest eigenvalue and error, by using uniform square
meshes and polygonal meshes with hanging nodes.}
\vspace{0.3cm}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
$\CT_h$   & Dofs&$\l_{h}^{1}$ &Error  \\
\hline
  &  833  &  0.78073215782 &  0.00628166703\\
 $\CT_{h}^{6}$ & 3201  & 0.77689137854&   0.00244088775\\
  & 12545 &0.77539520174 &  0.00094471094\\\hline
    ref. & &   0.77445049080 & -- \\
   \hline
$ \CT_{h}^{6,0}$   &833  & 0.78073215782&  0.00628166703\\
 $\CT_{h}^{6,1} $  &1181 &0.77728198716&  0.00283149637\\
$ \CT_{h}^{6,2} $  &1529  &0.77598279448& 0.00153230369\\
 $\CT_{h}^{6,3} $  &1877  &0.77548305066 & 0.00103255987\\
 $\CT_{h}^{6,4}$  & 2232& 0.77528749982&   0.00083700903\\\hline
    ref. & & 0.77445049080 & -- \\
\hline

 \end{tabular}
\label{tablaL}
\end{center}
\end{table}

Table~\ref{tablaL} reports the lowest Steklov eigenvalue
computed on an L-shaped domain with the method analyzed in this paper
with different polygonal meshes. The table also includes
the corresponding 'Errors' which have been obtained against a reference
value 'ref.' which corresponds to extrapolated values obtained
with a least square fitting on finer uniform meshes.

It can be seen from Table~\ref{tablaL} that

the reported errors are similar
in the last row of each mesh; however,
the dofs in the case of corner-refined meshes are much less
than the case of uniform meshes.
Therefore, we conclude that the possibility of using
small edges in the polygons of the mesh,
allow us easier refinements near edges and/or corners
of the domain to handle solutions with
corner singularities.

\section*{Acknowledgments}
FL was partially supported by
CONICYT-Chile through FONDECYT Postdoctorado project 3190204 (Chile).
DM was partially supported by CONICYT-Chile through FONDECYT project
1180913 (Chile) and by CONICYT-Chile through the project AFB170001
of the PIA Program: Concurso Apoyo a Centros Cient\'ificos y
Tecnol\'ogicos de Excelencia con Financiamiento Basal. GR  was supported by
CONICYT-Chile through FONDECYT project 11170534 (Chile).
IV was partially supported by BASAL project CMM,
Universidad de Chile (Chile).

\bibliographystyle{amsplain}

config: arxiv


TITLE: How can I write the given argument into a logical proposition?
QUESTION [0 upvotes]: How can I write 'the number 100 can be written as the sum of 2 squares' into a logical proposition?
I'm starting with the basics of discrete mathematics and am not sure from what angle to approach it.
We've started learning about existential statements so I assume that would be the ideal approach? Not sure though - would appreciate a hand to approaching similar questions.

REPLY [3 votes]: If you are talking about propositional logic, the answer is simple: $p$.
But I guess that you are talking about predicate logic. Then you have predicates and quantifiers. I'll use also standard symbols for certain predicates, such as $=$ and $\in\Bbb N$. The statement becomes
$$\exists x(\exists y( x\in \Bbb N\wedge y\in \Bbb N \wedge x^2+y^2=100))$$