TITLE: Ideal of the twisted cubic
QUESTION [11 upvotes]: The twisted cubic is the image of the morphism $\phi : \mathbb{P}^1 \to \mathbb{P}^3 , (x:y) \mapsto (x^3:x^2 y:x y^2:y^3)$, it is given by $X = V(ad-bc,b^2-ac,c^2-bd)$. Now I would like to compute $I(X)$, which equals by the Nullstellensatz the radical of the ideal $I := (ad-bc,b^2-ac,c^2-bd) \subseteq k[a,b,c,d]$. I think that that $I$ is already a radical ideal, even a prime ideal. Namely, I suspect that
$$\phi^* : Q:=k[a,b,c,d]/I \to k[s,t] , a \mapsto s^3, b \mapsto s^2 t , c \mapsto s t^2 , d \mapsto t^3$$
is an injection. If this is true: How can we prove that? I've already tried to find a $k$-basis of the quotient, but this turned out to be a big mess. Even the representation of the quotient as a monoid algebra doesn't seem to help. Another idea is the following: A formal manipulation of generators and relations implies $Q_a \cong k[a,b]_a$. Thus it suffices to prove that $Q \to Q_a$ is injective, i.e. that $a$ is not a zero divisor.

REPLY [14 votes]: Here is a purely algebraic proof that $I(X)=I$.
It is of course sufficient to prove that $I(X) \subset I$ and for that it suffices to prove that every homogeneous polynomial $P(a,b,c,d)$
which vanishes on $X$ is in $I$.
Lemma
Any homogeneous  polynomial $P(a,b,c,d)\in k[a,b,c,d]$ can be written $$P(a,b,c,d)=R(a,d) +S(a,d)b+T(a,d)c+i(a,b,c,d)  $$  for some polynomials $R,S,T\in k[a,d]$
 and a polynomial $i\in I$
The easy proof is by induction on the degree of $P$ and I'll leave it to you.
Now  back to our problem. If now that homogeneous $P$ is in $ I(X)$ , we write it as in the lemma and  get by using that $P$ vanishes on $X$ that for all $(x:y)\in \mathbb P^1_k$   $$0=P(x^3,x^2y,xy^2,y^3)=R(x^3,y^3) +S(x^3,y^3)x^2y+T(x^3,y^3)xy^2+0 $$
By considering exponents modulo $3$ for $x$ and $y$, we see that no cancellation occurs, hence  that $R=S=T=0$ and thus  $P=i\in I$ as required.


TITLE: Straight-line embedding planar graph
QUESTION [5 upvotes]: I want to prove that every normal planar graph has a straight-line embedding.
First, I assume that the planar graph $G$ is maximal planar, i.e the number of edges is $3n-6$ for $|G|=|V(G)|\ge3$. If the graph is not maximal planar I simply add some edges until it is maximal. I also know that all faces form triangles.
Now I want to prove the result with induction with respect to $n$.
$n=3:$ This case is trivial, it is just a triangle.
$n-1\rightarrow n$. I assume I know that the embedding works for a graph with $n-1$ vertices. I do not know hot to conclude. Intuitively it is clear to me because it is nothing more than some triangles which I clue together.
Maybe you can help me.

REPLY [6 votes]: Note that all faces of maximal planar graph (including the outer one) are triangles.
Consider a maximal planar graph $G$ on $n$ vertices. It has vertex $v$ such that $\deg v \le 5$ (otherwise it woudl have at least $\frac{6n}2 = 3n > 3n - 6$ edges). Since all faces of $G$ are triangles, there is also cycle $C$ such that $V(C) = N(v)$. Let remove $v$ from $G$, add $|C| - 3$ chords into $C$ (if $|C| = 5$ chords should be adjacent) and get maximal planar graph $H$ on $n - 1$ vertices. By induction hypothesis it has a straight-line embedding. Now remove all edges from $E(H) \setminus E(G)$. If cycle $C$ is an inner face (of size at most 5) it is always possible to place $v$ inside this face and connect to its neigbours in $G$ by straight lines. If $C$ is the outer face then all (at most 2) edges from $E(H) \setminus E(G)$ are edges of outer face of embedding of $H$. Then it is possible to place $v$ on outer face of $H - (E(H) \setminus E(G))$ and connect it to all neighbours in $G$ by straight lines.
P. S. Both cases of placing $v$ are obvious when $\deg v = 3$ and $\deg v = 4$, and not hard to proove for $\deg v = 5$; ask in comments if needed.


TITLE: Why do all "divisibility tricks" seem to use linear combinations, and are there any that don't?
QUESTION [21 upvotes]: When I say "divisibility trick" I mean "a recursive algorithm designed to show that, after multiple iterations, if the final output is a multiple of the desired number, then the original was also a multiple of the same number." Here's an example for a divisibility trick for 17.

Rewrite $n$ as $10q+r$, with $r<10$. Then, evaluate $|q-5r|$. Repeat this process until left with an easily factorable number.

Here's another.

Rewrite $n$ as $100q+r$, with $r<100$. Then, evaluate $|r-2q|$. Repeat this process until left with an easily factorable number.

Just to show these both work (or, at least, work for one particular number, let's try both on $31382$.

METHOD ONE:
$31382\rightarrow3128\rightarrow272\rightarrow17$, ergo $17\ |\ 31382$.

METHOD TWO:
$31382\rightarrow544\rightarrow34$, ergo $17\ |\ 31382$.

These divisibility tricks rely on breaking the number down into groups of digits and applying some linear operation to them. However, when we try to use a non-linear function, things seem to break down. For example, much as how method one here draws off the fact that $17\ |\ 51$ and the second relies off of $17\ |\ 119$, let's try to do something with $17\ |\ 34$. Namely:

Rewrite $n$ as $10q+r$, with $r<10$. Then, evaluate $|6q^2-5r|$. Repeat this process until left with an easily factorable number.

We can try this with $34$ and see, yes, $6(9)-5(4)=34$, so $17\ |\ 34$. But this fails for most numbers. For $51$, we have $51\rightarrow145$, and it diverges from there (also note that $17\not|\ 145$). Even with $17$, which is obviously a multiple of $17$, we have $17\rightarrow29$.
What separates the wheat from the chaff here, so to speak? Why is it that if we break down the digits of the multiple of some prime and make a linear relation around it, it seems to be true for all other multiples of the prime, but the same doesn't work for, say, a quadratic relation?

REPLY [12 votes]: We can invent "divisibility tricks" that don't have this property. For example:
"To test if $n$ is divisible by $3$, write $n$ as $10q+r$. Then, compute $q^3 + r^3$ and see if this is divisible by $3$."
This works because of Fermat's little theorem: for all integers $a$ and primes $p$, $a^p \equiv a \pmod p$. In particular, $q^3 \equiv q \pmod3$ and $r^3 \equiv r\pmod3$, so $q^3+r^3 \equiv q + r \equiv 10q+r \pmod 3$.
This is not a good trick! Not because it doesn't work, but because checking $q^3+r^3$ for divisibility by $3$ is probably harder than checking $n$ was.
In highly specific cases, we might be able to take advantage of this kind of trick, though. For example, if you are faced with trying to figure out if $10000256$ is divisible by $7$, it might help to realize that $10000256 = 10^7 + 2 \cdot 2^7$, so it is congruent to $10 + 2 \cdot 2 = 14$ modulo $7$.


TITLE: How to show $\operatorname{Hom}_A(M,N)$ is a finitely generated $A$-module?
QUESTION [0 upvotes]: Let $A$ be a commutative Noetherian ring with identity and $M,N$ two finitely generated $A$-modules.
How to show $\operatorname{Hom}_A(M,N)$ is a finitely generated $A$-module?

REPLY [4 votes]: Consider the surjection $A^n \to M$. This induces an injection $Hom_A(M,N) \to Hom_A(A^n, N) = N^n$. So this is a submodule of a finitely generated module and by the Noetherian condition, we are done.


text:
\begin{document}

\maketitle

\vspace{-3.5ex}
\thanks{ \centerline{\footnotesize
Department of Mathematics, Weizmann Institute}}
\thanks{\footnotesize\centerline{ Rehovot, 7610001, Israel}}
\thanks{\footnotesize \centerline{
e-mail: \texttt{victor.katsnelson@weizmann.ac.il, victorkatsnelson@gmail.com}}}

\thispagestyle{empty}

\vspace{2.0ex}
\begin{abstract}
\noindent
Let \(w_{T}(t)\) and \(w_{\Omega}(\omega)\) be functions defined for \(t\in\mathbb{R}\)
and \(\omega\in\mathbb{R}\) respectively, where \(\mathbb{R}=(-\infty,\infty)\). The functions \(w_{T}(t)\) and
\(w_{\Omega}(\omega)\) are assumed to be non-negative everywhere and bounded away from
zero outside some sets of finite Lebesgue measure. Moreover some regularity conditions are posed
on the these functions. Under these regularity conditions, the functions \(w_{T}\) and
\(w_{\Omega}\) are locally bounded and
 grow not faster than exponentially.
We associate the inner product space \(\mathcal{H}_{w_T,w_{\Omega}}\) with the
functions \(w_{T}\) and \(w_{\Omega}\). The space \(\mathcal{H}_{w_T,w_{\Omega}}\) consists of those
functions \(x\)  for which
\(\int\limits_{-\infty}^{\infty}|x(t)|^2w_{T}(t)dt+
\int\limits_{-\infty}^{\infty}|\hat{x}(\omega)|^2w_{\Omega}(\omega)d\omega=
\|x\|^2_{\mathcal{H}_{w_T,w_{\Omega}}}<\infty\), where \(\hat{x}\)
is the Fourier transform of the function \(x\).
We show that the system of Gaussians
\(\big\lbrace\exp(-\alpha(t-\tau)^2)\big\rbrace\), where \(\alpha\) runs over \(\mathbb{R}^{+}=(0,+\infty)\) and \(\tau\) runs over \(\mathbb{R}\),
is a complete system in the space \(\mathcal{H}_{w_T,w_{\Omega}}\).

\end{abstract}

\vspace{2.0ex}
\noindent
{\small\textbf{Mathematics Subject Classification\ (2010).}\ Primary 46E20, 46E35; \ Secondary
41A28, 42B10.}

\vspace{1.0ex}
\noindent
{\small\textbf{Keywords:}\
 Fourier transform,  Hilbert functional space, Sobolev space, Gaussians.}

\normalsize
\vspace{3.0ex}
\noindent
\textsf{Notation:}\\
\(\mathbb{R}\) \ \ - the real axis, i.e. the set of all real numbers.\\
\(\mathbb{R}_{+}\) - the set of all \emph{strictly} positive real numbers.\\
\(\mathbb{N}\) \ \ - the set of all natural numbers.\\
\(d\xi,\,dt,\,d\omega\) - \begin{minipage}[t]{0.8\linewidth}
the normalized Lebesgue measure on the real axis \(\mathbb{R}\) (if the variable on \(\mathbb{R}\) is denoted by \(\xi,\,t,\,\omega\) respectively.)
\end{minipage}
\section{The space \(\boldsymbol{\mathcal{H}_{w_T,w_{\Omega}}}\).}
\subsection{The space \(\boldsymbol{L^2\pmb{(}\pmb{\mathbb{R}}, w\pmb{(}\xi\pmb{)}d\xi\pmb{)}}\).}
Let \(w(\xi)\) be a function defined  for all \(\xi\in\mathbb{R}\). In what follow we always assume that \(w(\xi)\) satisfies the conditions:
\begin{equation}
\label{NNe}
0\leq{}w(\xi)<\infty,\ \ \forall\,\xi\in\mathbb{R}.
\end{equation}
\centerline{\textit{A function \(w(\xi)\) satisfying the condition  \eqref{NNe} is said to be \textsf{the weight function}.}}
We interpret the variable \(\xi\) either as the time variable \(t\), or  as the frequency
variable~\(\omega\).
\begin{definition}
\label{defL2}
We associate the space \(L^2(\mathbb{R},w(\xi)d\xi)\) with the weight function \(w(\xi)\).
This is the space of all functions \(x(\xi)\) which are defined \(d\xi\)-almost everywhere on \(\mathbb{R}\), take values in the set \(\mathbb{C}\) and satisfy the condition
\begin{subequations}
\label{DefL2}
\begin{gather}
\label{DefL2a}
\|x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)} <\infty,\\[-3.0ex]
\intertext{where}
\notag\\[-6.5ex]
\label{DefL2b}
\|x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}=\bigg(\int\limits_{\mathbb{R}}|x(\xi)|^2w(\xi)\,d\xi\bigg)^{1/2}.
\end{gather}
\end{subequations}
The functions \(x^{\prime}(\xi)\) and \(x^{\prime\prime}(\xi)\)
 such that \(\|x^{\prime}(\xi)-x^{\prime\prime}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0\)
 determine the same element of the space \(L^2(\mathbb{R},w(\xi)d\xi)\).
\end{definition}
\vspace{2.0ex}
\noindent
 The space \(L^2(\mathbb{R},w(\xi)d\xi)\) is an inner product vector space.
 If \(x_1,x_2\in\!L^2(\mathbb{R},w(\xi)d\xi)\), then the inner product
 \(\langle{}x_1,x_2\rangle_{L^2(\mathbb{R},w(\xi)d\xi)}\) is defined as
\begin{equation}
\label{InnPr}
\langle{}x_1,x_2\rangle_{L^2(\mathbb{R},w(\xi)d\xi)}=
\int\limits
_{\mathbb{R}}x_1(\xi)\overline{x_2(\xi)}w(\xi)d\xi\,.
\end{equation}
The inner product \eqref{InPr} generates the norm \eqref{DefL2b}:
\[\|x\|^2_{L^2(\mathbb{R},w(\xi)d\xi)}=\langle{}x,x\rangle_{L^2(\mathbb{R},w(\xi)d\xi)},
 \ \ \forall\,x\in{}L^2(\mathbb{R},w(\xi)d\xi).\]

 \subsection{Fourier transform.}
For a complex-valued function \(x\) defined on \(\mathbb{R}\), its Fourier transform \(\hat{x}\) is
\begin{subequations}
\label{FoT}
\begin{equation}
\label{DiFoT}
\hat{x}(\omega)=\int\limits_{\mathbb{R}}x(t)e^{-2\pi{}i\omega{}t}dt\,,
\quad\omega\in\mathbb{R}.
\end{equation}
For a complex-valued function \(y\) defined on \(\mathbb{R}\), its inverse Fourier transform \(\check{y}\) is
\begin{equation}
\label{InFoT}
\check{y}(t)=\int\limits_{\mathbb{R}}y(\omega)e^{2\pi{}it\omega{}}d\omega\,,
\quad{t}\in\mathbb{R}.
\end{equation}
The transforms \eqref{DiFoT} and \eqref{InFoT} are mutually inverse:
\begin{equation}
\label{MuInv}
\textup{if } y=\hat{x}, \textup{ than } x=\check{y}.
\end{equation}
\end{subequations}
The Fourier transform and the inverse Fourier transform can be considered as operators \(\mathscr{F}\) and \(\mathscr{F}^{-1}\) acting in \(L^2(\mathbb{R})\):
\begin{subequations}
\label{FoOp}
\begin{align}
\label{DiFoOp}
\mathscr{F}x&=\hat{x}\,,\\
\mathscr{F}^{-1}y&=\check{y}\,.
\label{MuInV}
\end{align}
\end{subequations}
The Fourier operator \(\mathscr{F}\) is  well defined for \(x\in{}L^2(\mathbb{R},dt)\) and maps
the space \(L^2(\mathbb{R},dt)\) \emph{onto} the space \(L^2(\mathbb{R},d\omega)\) isometrically: the Parseval equality
\begin{equation}
\label{Par}
\|x\|^2_{L^2(\mathbb{R},dt)}=
\|\hat{x}\|^2_{L^2(\mathbb{R},d\omega)}, \ \ \forall\,x\in{}L^2(\mathbb{R},dt)
\end{equation}
holds.

The equalities \eqref{FoOp} can be expressed as
\begin{equation}
\label{MuIn}
\mathscr{F}^{-1}\mathscr{F}=\mathscr{I}_{T},\quad \mathscr{F}\,\mathscr{F}^{-1}=\mathscr{I}_{\Omega},
\end{equation}
where \(\mathscr{I}_{T}\) and \(\mathscr{I}_{\Omega}\) are the identity operators
in the spaces \(L^2(\mathbb{R},dt)\) and \(L^2(\mathbb{R},d\omega)\)
respectively.
\subsection{Definition of the space \(\boldsymbol{\mathcal{H}_{w_T,w_{\Omega}}}\).}
Let \(w_{T}(t)\) and \(w_{\Omega}(\omega)\) be two weight functions defined for \(t\in\mathbb{R}\)
and \(\omega\in\mathbb{R}\) respectively.
We associate the spaces \(L^2(\mathbb{R},w_T(t)dt)\) and \(L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)\) with these weight functions.

 \vspace{2.0ex}
 \noindent
 \begin{definition}
 \label{DMS}
 \begin{subequations}
  Let weight functions \(w_{T}(t)\) and \(w_{\Omega}(\omega)\) be given.\\
   \textsf{The space} \(\boldsymbol{\mathcal{H}_{w_T,w_{\Omega}}}\) is defined as the set of all those
  \(x(t)\in{}L^2(\mathbb{R}, dt)\) for which
  \begin{gather}
  x(t)\in{}L^2(\mathbb{R},w_T(t)dt) \textup{ \ and \ } \hat{x}(\omega)\in{}L^2(\mathbb{R},w_{\Omega}(\omega)d\omega),
  \end{gather}
  where \(\hat{x}(\omega\) is the Fourier transform of the function \(x(t)\): \(\hat{x}=\mathscr{F}x\).

 \textup{Equivalently}\\
  \textsf{The space} \(\boldsymbol{\mathcal{H}_{w_T,w_{\Omega}}}\) is defined as the set of all those \(y(\omega)\in{}L^2(\mathbb{R}, d\omega)\) for which
  \begin{gather}
  y(\omega)\in{}L^2(\mathbb{R},w_{\Omega}(\omega)d\omega) \textup{ and } \check{y}(t)\in{}L^2(\mathbb{R},w_{T}(t)dt),
  \end{gather}
  \end{subequations}
  where \(\check{y}(t)\) is the inverse Fourier transform of the function \(y(\omega)\):
  \(\check{y}=\mathscr{F}^{-1}y\).
 \end{definition}

 The space \(\mathcal{H}_{w_T,w_{\Omega}}\) is an inner product space.
 If \(x_1,\,x_2\in\mathcal{H}_{w_T,w_{\Omega}}\), then the inner product
 \(\langle{}x_1,x_2\rangle_{\mathcal{H}_{w_T,w_{\Omega}}}\) is defined as
 \begin{equation}
 \label{InPr}
 \langle{}x_1,x_2\rangle_{\mathcal{H}_{w_T,w_{\Omega}}}
 =\langle{}x_1,x_2\rangle_{L^2(\mathbb{R},w_T(t)dt)}
 +\langle{}\widehat{x_1},\widehat{x_2}\rangle_{L^2(\mathbb{R},w_{\Omega}(\omega)dt)}\,.
 \end{equation}
In particular, the expression \(\|x\|^{2}_{\mathcal{H}_{w_T,w_{\Omega}}}\) for the square of norm of \(x\in\mathcal{H}_{w_T,w_{\Omega}}\) is
\begin{equation}
\label{IOW}
\|x\|^{2}_{\mathcal{H}_{w_T,w_{\Omega}}}=\|x\|^{2}_{L^2(\mathbb{R},w_T(t)dt)}+
\|\hat{x}\|^{2}_{L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)}.
\end{equation}
 \begin{remark}
 If the weight functions \(\,w_{T}(t)\), \(w_{\Omega}(\omega)\) grow very fast as
\(|t|\to\infty,\,|\omega|\to\infty\), then the space \(\mathcal{H}_{w_{T},w_{\Omega}}\) may consist of  the identically zero function only.

For example, if \(w_{T}(t)=\exp(\alpha{}t^2),\,w_{\Omega}(\omega)=\exp(\beta{}\omega^2)\), where
\(\alpha>0,\,\beta>0\), and \(\alpha\beta>4\pi^2\), then the space
\(\mathcal{H}_{w_{T},w_{\Omega}}\) contains
only the identically zero function. This statement is a version of Hardy's Theorem\,\footnote{
Concerning this theorem, we  refer to \cite[Sec.\,3.2]{DyMcK}.}.
 \end{remark}
 \noindent
\textbf{EXAMPLE}.
Let
\vspace{-0.5ex}
\begin{equation}
\label{SWF}
w_{T}(t)\equiv1 \ \textup{ and } \ w_{\Omega}(\omega)=(2\pi\omega)^{2n}.
\end{equation}
For \(x(t)\in{}L^2(\mathbb{R},dt)\),
the condition \(\int\limits_{\mathbb{R}}|\hat{x}(\omega)|^2{\omega}^{2n}d\omega<\infty\)
means that the \(n\)-th derivative \(x^{(n)}(t)\) of the function \(x(t)\), considered as a \textsf{distribution} originally, is actually a \textsf{function} from \(L^2(\mathbb{R},dt)\). Moreover
\begin{equation*}
\int\limits_{\mathbb{R}}\big|\hat{x}(\omega)\big|^2w_{\Omega}(\omega)d\omega=
\int\limits_{\mathbb{R}}\big|x^{(n)}(t)\big|^2dt.
\end{equation*}
Thus for the weight functions of the form \eqref{SWF},
\begin{equation*}
\|x\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}}=\int\limits_{\mathbb{R}}|x(t)|^2dt+
\int\limits_{\mathbb{R}}\big|x^{(n)}(t)\big|^2dt.
\end{equation*}
So for the weight functions of the form \eqref{SWF}, the space \(\mathcal{H}_{w_{T},w_{\Omega}}\)
is the Sobolev space \(W_{\;2}^{(n)}\).
 \subsection{The non-degeneracy condition.}
 \begin{definition} \
 \label{DeCo}
 The weight functions \(w_{T}(t)\)  \textsf{satisfy
 the non-de\-generacy condition} if there exist  number
 \(\varepsilon_T>0\)  such that the set
 \(E_{T}=\lbrace{}t:\,w_{T}(t)<\varepsilon_{T}\rbrace\)
 is of finite Lebesgue measure:
 \(\textup{mes}\,E_{T}<\infty\).

 The weight functions \(w_{\Omega}(\omega)\)  \textsf{satisfy
 the non-de\-generacy condition} if there exist  number
 \(\varepsilon_{\Omega}>0\)  such that the set
 \(E_{\Omega}=\lbrace{}t:\,w_{\Omega}(\omega)<\varepsilon_{\Omega}\rbrace\)
 is of finite Lebesgue measure:
 \(\textup{mes}\,E_{\Omega}<\infty\).
 \end{definition}
\begin{theorem} \
If both weight functions \(w_{T}\) and \(w_{\Omega}\) satisfy the non-degeneracy conditions,
then the the inequality
\begin{equation}
\label{MaNo}
\|x\|^2_{L^2(\mathbb{R},dt)}\leq{}B\|x\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}}
\end{equation}
holds for every  \(x\in\mathcal{H}_{w_{T},w_{\Omega}}\), where \(B<\infty\) is
a constant which does not depend on \(x\).
\end{theorem}
\begin{proof}
In \cite{Naz} F.\,Nazarov proved the remarkable inequality
\begin{equation}
\label{NazIn}
\int\limits_{\mathbb{R}}|x(t)|^2dt\leq{}A\exp\{A\cdot\textup{mes}\,E\cdot\textup{mes}\,F\,\}
\Big(\int\limits_{\mathbb{R}\setminus\,E}|x(t)|^2dt+
\int\limits_{\mathbb{R}\setminus\,F}|\hat{x}(\omega)|^2d\omega\Big)\,,
\end{equation}
where \(x\) is an arbitrary function from \(L^2(\mathbb{R})\),
\(\hat{x}\) is the Fourier transform of the function \(x\),
\(E\) and \(F\) are
arbitrary subsets of \(\mathbb{R}\), and \(A\), \(0<A<\infty\), is an absolute constant.
(The value \(A\) does not depend on \(x,\,E,\,F\).) We apply the Nazarov inequality
\eqref{NazIn} to the sets \(E=E_T\),\,\(F=E_\Omega\)
which appear in Definition \ref{DeCo}
and to  an arbitrary function  \(x\in\mathcal{H}_{w_{T},w_{\Omega}}\).
Since \(\varepsilon_{T}\leq{}w_{T}(t)\) for \(t\in\mathbb{R}\setminus{}E_T\)
and \(\varepsilon_{\Omega}\leq{}w_{\Omega}(\omega)\) for
\(\omega\in\mathbb{R}\setminus{}E_{\Omega}\), the inequalities
\begin{equation*}
\int\limits_{\mathbb{R}\setminus{}E_{T}}|x(t)|^2dt\leq
\frac{1}{\varepsilon_T}\int\limits_{\mathbb{R}}|x(t)|^2w_T(t)dt,\
\int\limits_{\mathbb{R}\setminus{}E_{\Omega}}
|\hat{x}(\omega)|^2d\omega\leq
\frac{1}{\varepsilon_F}\int\limits_{\mathbb{R}}|\hat{x}(\omega)|^2w_{\Omega}(\omega)d\omega\,.
\end{equation*}
hold.
Combine these inequalities with \eqref{NazIn}, we come to the inequality
\eqref{MaNo}.
\end{proof}
\begin{corollary}
\label{Cor}
For every \(x\in\mathcal{H}_{w_{T},w_{\Omega}}\), the inequalities
\begin{subequations}
\label{CorET}
\begin{align}
\|x\|^2_{L^2(\mathbb{R},(1+w_{T}(t))dt)}\leq&(B+1)\|x\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}},\\
\label{CorEO}
\|\hat{x}\|^2_{L^2(\mathbb{R},(1+w_{\Omega}(\omega))d\omega)}\leq&(B+1)\|x\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}}
\end{align}
\end{subequations}
hold, where \(B\) is the same constant that in \eqref{MaNo}.
\end{corollary}
\begin{theorem}
\label{Com}
If both weight functions \(w_{T}(t)\) and \(w_{\Omega}(\omega)\) satisfy the non-degeneracy conditions, then the space \(\mathcal{H}_{w_{T},w_{\Omega}}\) provided by the norm \eqref{IOW} is complete.
\end{theorem}
\begin{proof} Let \(\lbrace{}x_n\rbrace_n\) be a Cauchy sequence of elements from
\(\mathcal{H}_{w_{T},w_{\Omega}}\), that is
\begin{equation*}
\|x_n-x_m\|_{\mathcal{H}_{w_{T},w_{\Omega}}}\to0 \ \textup{ as } m\to\infty,\,n\to\infty.
\end{equation*}
From the inequality \eqref{MaNo} and from the Parseval equality it follows that
\begin{equation*}
\|x_n-x_m\|_{L^2(\mathbb{R},dt)}\to0 \ \textup{ and } \|\hat{x}_n-\hat{x}_m\|_{L^2(\mathbb{R},d\omega)}\to0
\textup{ as } m\to\infty,\,n\to\infty.
\end{equation*}
Since the space \(L^2(\mathbb{R},dt)\) is complete, there exists \(x\in{}L^2(\mathbb{R},dt)\)
such that
\begin{equation*}
\|x_n-x\|_{L^2(\mathbb{R},dt)}\to0 \ \textup{ as } n\to\infty.
\end{equation*}
In view of the Parseval equality,
\begin{equation*}
\|\hat{x}_n-\hat{x}\|_{L^2(\mathbb{R},d\omega)}\to0 \ \textup{ as } n\to\infty.
\end{equation*}
According to well known results from the measure theory\footnote{See for example
\cite[Sections 21,22]{Hal}}, we can select an increasing subsequence \(\lbrace{}n_k\rbrace\)
of natural numbers such that
\begin{subequations}
\label{CAE}
\begin{align}
\label{CAET}
x_{n_k}(t)\to x(t) \ &\textup{ as \(k\to\infty \ dt\)-almost everywhere on } \mathbb{R},\\
\label{CAEO}
 \hat{x}_{n_k}(\omega)\to\hat{x}(\omega) \ &\textup{ as
 \(k\to\infty \ d\omega\)-almost everywhere on } \mathbb{R}.
\end{align}
\end{subequations}
Since a Cauchy sequence is bounded,
\begin{equation*}
\int\limits_{\mathbb{R}}|x_{n_{k}}(t)|^2w_{T}(t)dt+
\int\limits_{\mathbb{R}}|\hat{x}_{n_{k}}(\omega)|^2w_{\Omega}(\omega)d\omega\leq{}C<\infty,
\ \ k=1,2,\,\ldots\,,
\end{equation*}
where \(C\) does not depend on \(k\). Using \eqref{CAE} and Fatou's Lemma, we conclude
that
\begin{equation*}
\int\limits_{\mathbb{R}}|x(t)|^2w_{T}(t)dt+
\int\limits_{\mathbb{R}}|\hat{x}(\omega)|^2w_{\Omega}(\omega)d\omega\leq{}C.
\end{equation*}
Thus \(x\in\mathcal{H}_{w_{T},w_{\Omega}}\).

Given \(\varepsilon>0\), we choose \(K(\varepsilon)<\infty\) such that
\begin{equation*}
\Big(k>K(\varepsilon), l>K(\varepsilon)\Big)\Rightarrow
\big\|x_{n_k}-x_{n_l}\big\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}}\leq\varepsilon^2,
\end{equation*}
i.e.
\begin{equation*}
\int\limits_{\mathbb{R}}\big|x_{n_k}(t)-x_{n_l}(t)\big|^2w_{T}(t)dt +
\int\limits_{\mathbb{R}}\big|\widehat{x_{n_k}}(\omega)-\widehat{x_{n_l}}(\omega)\big|^2
w_{\Omega}(\omega)d\omega\leq\varepsilon^2.
\end{equation*}
Using \eqref{CAE}, we pass to the limit as \(l\to\infty\) in the last inequality.
By Fatou's Lemma, we obtain
\begin{equation*}
\int\limits_{\mathbb{R}}\big|x_{n_k}(t)-x(t)\big|^2w_{T}(t)dt +
\int\limits_{\mathbb{R}}\big|\widehat{x_{n_k}}(\omega)-\widehat{x}(\omega)\big|^2
w_{\Omega}(\omega)d\omega\leq\varepsilon^2,
\end{equation*}
i.e.
\begin{equation*}
\Big(k\geq{}K(\varepsilon)\Big)\Rightarrow
\big\|x_{n_k}-x\big\|^2_{\mathcal{H}_{w_{T},w_{\Omega}}}\leq\varepsilon^2.
\end{equation*}
\end{proof}

\subsection{The regularity condition.}
In what follows we impose some regularity condition on the weight function \(w(\xi)\).
In particular, this regularity condition ensure  that the shift operator \(\mathscr{T}_h\)
is a bounded operator in the space \(L^2(\mathbb{R},w(\xi)d\xi)\) for each \(h\in\mathbb{R}\).

The regularity condition is formulated un terms of the value \(M_{w}(\delta)\) which can be
interpreted as the multiplicative modulus  of continuity (m.m.c.) of the weight function \(w(\xi)\).
\begin{definition}
\label{DMMC}
Let \(w(\xi)\) be a weight function satisfying the condition
\begin{equation}
\label{StriPos}
w(\xi)>0, \ \ \forall\,\xi\in\mathbb{R}.
\end{equation}
 \textsf{The  multiplicative modulus  of continuity} \(M_{w}(\delta)\) of the weight function \(w(\xi)\)
is defined as
\begin{gather}
\label{RegVa}
M_{w}(\delta)\stackrel{\textup{def}}{=}
\sup\limits_{\substack{\xi^{\prime}\in\mathbb{R},
\,\xi^{\prime\prime}\in\mathbb{R}\\|\xi^{\prime}-\xi^{\prime\prime}|\leq\delta}}
\frac{w(\xi^{\prime})}{w(\xi^{\prime\prime})}\,,\hspace*{3.0ex}0\leq\delta<\infty.
\end{gather}
The function \(M_{w}(\delta)\) may take the value \(\infty\):
\(M_{w}(\delta)\leq\infty\) for \(\delta\in[0,\infty)\).
\end{definition}
\begin{remark}
Without condition
\eqref{StriPos}, the value \eqref{RegVa} may be not well defined: considering the ratio
\(\frac{w(\xi^{\prime})}{w(\xi^{\prime\prime})}\), we may come to the uncertainty of the form~
\(\frac{0}{0}\).
\end{remark}
\begin{remark}
\label{winv}
Under the condition \eqref{StriPos},
the function \(w^{-1}(\xi)=1/w(\xi)\) also is a weight function.
From \eqref{RegVa} it is clear that  multiplicative modules  of continuity of these weight functions
coincide:
\begin{equation}
\label{CoInc}
M_{w}(\delta)=M_{w^{-1}}(\delta), \ \ \forall\,\delta\in[0,\infty).
\end{equation}
\end{remark}
\begin{remark}
The function \(M_w(\delta)\) is not necessarily continuous with respect to
 \(\delta\).
\end{remark}
\begin{lemma}
\label{PMMC}
For any\footnote{To exclude division by zero, we must assume that the condition \eqref{StriPos} holds.} weight function \(w(\xi)\), its multiplicative modulus  of continuity \(M_{w}(\delta)\)
possesses the properties
\begin{enumerate}
\item[\textup{1}.] {\ }\\[-5.5ex]
\begin{equation}
\label{MMCz}
M_{w}(0)=1;
\end{equation}
\item[\textup{2}.] The function \(M_{w}(\delta)\) increases with respect to \(\delta\):
\begin{gather}
\label{Mon}
\text{If \(0\leq\delta^{\prime}<\delta^{\prime\prime}<\infty\), then
\(M_w(\delta^{\prime})\leq{M_w(\delta^{\prime\prime})}\).}\\[-2.0ex]
\intertext{In particular,}
\notag\\[-6.0ex]
 1\leq{}M_w(\delta)\,, \ \ \forall\,\delta\in\mathbb{R}_{+}.
\end{gather}
\item[\textup{3}.] The functions \(M_w\) is submultiplicative:
\begin{gather}
\label{SuMuPr}
M_w(\delta_1+\delta_2)\leq{}M_w(\delta_1)\cdot{}M_w(\delta_2), \ \ \forall\,\delta_1\in\mathbb{R}_{+},\,\forall\,\delta_2\in\mathbb{R}_{+};\\[-1.5ex]
\intertext{In particular,}
\notag\\[-6.0ex]
 \label{SubSp}
 M_w(n\delta)\leq{}(M_w(\delta))^n,\ \
\forall\,n\in\mathbb{N},\,\forall\,\delta\in\mathbb{R}_{+}.
\end{gather}
\end{enumerate}
\end{lemma}

From \eqref{Mon} and \eqref{SubSp} it follows that only two (mutually exclusive) possibilities.
can happen:

\centerline{Either \(M_{w}(\delta)<\infty \ \forall\,\delta\in\mathbb{R}_{+}\), or
\(M_{w}(\delta)=\infty \ \forall\,\delta\in\mathbb{R}_{+}\).}

\begin{definition}
\label{DefReCo}
\textsf{The weight functions \(w(\xi)\)  satisfies the regularity condition} if
\begin{subequations}
\label{FRC}
\begin{gather}
\label{FRC1}
M_{w}(\delta)<\infty, \ \ \forall\,\delta\in\mathbb{R}_{+},\\[-2.5ex]
\intertext{or, what is the same,}
\notag\\[-6.0ex]
\label{FRC2}
\exists\,\delta\in\mathbb{R}_{+}: \ M_{w}(\delta)<\infty.
\end{gather}
\end{subequations}
\end{definition}

\begin{remark}
If the regularity condition for a weight functions \(w(\xi)\) is satisfied, than
its multiplicative continuity modulus \(M_{w}(\delta)\)
 is locally bounded and grow not faster then exponentially with respect to \(\delta\).
 Moreover the estimate
 \begin{equation}
 \label{MCE}
 M_{w}(\delta)\leq{}C_{w}\exp(\mu_{w}\delta), \ \ \forall\,\delta\in\mathbb{R}_{+}
 \end{equation}
 holds, where \(C_w\) and \(\mu_w\) are some constants, \(0<C_{w}<\infty,\ 0\leq\mu_{w}<\infty\).

  The estimate \eqref{MCE}
 can be easily derived from \eqref{SubSp}.

 Choosing \(\xi^{\prime}=\xi,\,\xi^{\prime\prime}=0\) in \eqref{RegVa}, we obtain the estimate
 \begin{equation}
 \label{MwE}
 w(\xi)\leq{}w(0)M_{w}(|\xi|), \ \ \forall\,\xi\in\mathbb{R}.
 \end{equation}
 Comparing \eqref{MCE} and \eqref{MwE}, we obtain the estimate of the weight function \(w(\xi)\) from above:
 \begin{equation}
 \label{McEa}
 w(\xi)\leq{}w(0)\,C_{w}\exp(\mu_{w}|\xi|), \ \ \forall\,\xi\in\mathbb{R}.
 \end{equation}
 Taking into account \textup{Remark \ref{winv}}, we obtain the estimate of the weight function
 \(w(\xi)\) from below:
 \begin{equation}
 \label{McEb}
w(0)\,C^{-1}_{w}\exp(-\mu_{w}|\xi|)\leq{}w(\xi), \ \ \forall\,\xi\in\mathbb{R}.
 \end{equation}
 The constants \(C_{w}\) and \(\mu_{w}\) in \eqref{McEa} and \eqref{McEb} are the same that in
 \eqref{MCE}.
\end{remark}
\begin{remark}
\label{IWF}
In what follows, we impose the regularity conditions on the weight functions
\(1+w_T(t)\) and \(1+w_\Omega(\omega)\), where \(w_T\) and \(w_{\Omega}\) are
the same weight functions which appear in the definition\footnote{
See Definition \ref{DMS}.
}
of the space
\(\mathcal{H}_{w_T,w_{\Omega}}\).
These regularity conditions imply the estimates
\begin{subequations}
\label{IWFE}
\begin{align}
\label{IWFEa}
w_T(t)&\leq{}C_Te^{\mu_T|t|}, \ \ t\in\mathbb{R},\\
\label{IWFEb}
w_\Omega(\omega)&\leq{}C_{\Omega}e^{\mu_\Omega|\omega|},
 \ \ \omega\in\mathbb{R},
\end{align}
where \(C_T,\infty\), \(C_\Omega,\infty\), \(\mu_T<\infty\), \(\mu_{\Omega}<\infty\)
are some constants.
\end{subequations}
\end{remark}
\subsection{The shift operator.}
\begin{definition}
 \label{DefSh}
 For a function \(x(\xi)\) defined for \(\xi\in\mathbb{R}\) and for a number \(h\in\mathbb{R}\),
 the \textsf{shifted function} \((\mathscr{T}_hx)(\xi)\) is defined as a function
 \begin{equation}
 \label{DefShf}
 (\mathscr{T}_{\eta}x)(\xi)=x(\xi-\eta), \ \ \forall\,\xi\in\mathbb{R}.
 \end{equation}
 of the variable \(\xi\).
 The number \(\eta\) is considered as a parameter, the so called \textsf{shift parameter}.
 \end{definition}

 \begin{lemma}
 \label{PrTrO}
 Let \(w(\xi)\) be a weight function which satisfy the regularity condition~
 \footnote{
 \textup{See Definition \ref{DefReCo}.}
 }.
 Then
 \begin{enumerate}
 \item[\textup{1}.]
 For each value \(\eta\in\mathbb{R}\) of the shift parameter, the shift operator \(\mathscr{T}_{\eta}\) is a bounded operator in the space \(L^2(\mathbb{R},w(\xi)d\xi)\).
 The inequality
 \begin{equation}
 \label{ENS}
 \|(\mathscr{T}_{\eta}x)(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq
 \sqrt{M_{w}(|\eta|)}\,\|x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}
 \end{equation}
 holds for every \(x(\xi)\in{}L^2(\mathbb{R},w(\xi)d\xi)\) and every \(\eta\in\mathbb{R}\), where \(M_{w}(\eta)\) is the multiplicative continuity modulus of the weight function \(w\).
 \item[\textup{2}.] The operator function \(\mathscr{T}_{\eta}:
 L^2(\mathbb{R},w(\xi)d\xi)\to{}L^2(\mathbb{R},w(\xi)d\xi)\) is strongly continuous
 with respect to \(\eta\):
 \begin{equation}
 \label{StCo}
 \lim_{|\eta|\to0}\|(\mathscr{T}_{\eta}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0,
 \ \ \forall\,x\in{}L^2(\mathbb{R},w(\xi)d\xi)\,.
 \end{equation}
 \end{enumerate}
\end{lemma}
\begin{proof} {\ }\\
 \textbf{1}. According to definition of the norm in \(L^2(\mathbb{R},w(\xi)d\xi)\),
\[\|(\mathscr{T}_{\eta}x)(\xi)\|^2_{L^2(\mathbb{R},w(\xi)d\xi)}=
\int\limits_{\mathbb{R}}|x(\xi-\eta)|^2w(\xi)d\xi.
\]
Changing variable \(\xi-\eta\to\xi\), we obtain
\[
\|(\mathscr{T}_{\eta}x)(\xi)\|^2_{L^2(\mathbb{R},w(\xi)d\xi)}=
\int\limits_{\mathbb{R}}|x(\xi)|^2w(\xi+\eta)d\xi.
\]
According to Definition \ref{DMMC}, the inequality
\[w(\xi+\eta)\leq{}w(\xi)M_{w}(|\eta|)\]
holds. Combining this inequality with the previous equality, we obtain \eqref{ENS}.\\
\textbf{2}. Given a function \(x(\xi)\in{}L^2(\mathbb{R},w(\xi)d\xi)\) and \(\varepsilon>0\),
we can split the function \(x\) into the sum
\[x(t)=\varphi(t)+r(t),\]
where \(\varphi\) is a continuous function with a compact support,
and
\[\|r\|_{L^2(\mathbb{R},w(\xi)d\xi)}<\varepsilon.\]
Hence
\begin{multline*}
\|(\mathscr{T}_{\eta}x)(\xi)-x(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}\\
\leq\|(\mathscr{T}_{\eta}\varphi)(\xi)-\varphi(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}+
\|r(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}+
\big\|(\mathscr{T}_{\eta}r)(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}.
\end{multline*}
Since the function \(\varphi\) is continuous and has compact support and the weight function \(w_{T}\)
is locally bounded,
\[\lim_{\eta\to0}\|(\mathscr{T}_{\eta}\varphi)(\xi)-\varphi(\xi)
\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0.\]
According to \eqref{ENS},
\[\varlimsup_{\eta\to0}\big\|(\mathscr{T}_{\eta}r)(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}
\leq\sqrt{M_{T}(+0)}\,\varepsilon.\]
Therefore
\[\varlimsup_{\eta\to0}\|(\mathscr{T}_{\eta}x)(\xi)-
x(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq
\Big(1+\sqrt{M_{T}(+0)}\Big)\varepsilon.\]
Since \(\varepsilon>0\) can be arbitrary small,
the limiting relation \eqref{StCo} holds.
\end{proof}
\subsection{On integration of
\(\boldsymbol{L^2\pmb{(}\pmb{\mathbb{R}},w\pmb{(}\xi\pmb{)}d\xi\pmb{)}}\)-valued functions.}

Let \(X(\xi,\eta)\) be a function of two variables \((\xi,\eta)\in\mathbb{R}^2\),
\(\mathscr{D}(X)\) is the domain of definition of \(X(\xi,\eta)\), \(X(\xi,\eta):\,\mathscr{D}(X)\to\mathbb{C}.\)
We assume that the function \(X(\xi,\eta)\) is defined for almost every \((\xi,\eta)\in\mathbb{R}^2\)            with respect to the two-dimensional Lebesgue measure \(d\xi{}d\eta\):
\begin{equation}
\label{aeR2}
\iint\limits_{\mathbb{R}^2\setminus{}\mathscr{D}(X)}d\xi{}d\eta=0.
\end{equation}
For \(\eta\in\mathbb{R}\), let \(\mathscr{D}_{\eta}(X)\) be the
 \textsf{\(\eta\)-section of the set \(\mathscr{D}(X)\)}:

 \begin{subequations}
 \label{seq}
 \begin{gather}
\label{etsec}
\mathscr{D}_{\eta}(X)=\lbrace{}\xi\in\mathbb{R}:\ (\xi,\eta)\in{}\mathscr{D}(X)\rbrace,
\ \  \eta \in \mathbb{R}:\\
\intertext{For \(\xi\in\mathbb{R}\), let \(\mathscr{D}_{\xi}(X)\) be the
 \(\xi\)-\textsf{section of the set }\(\mathscr{D}(X)\):}\notag\\
 \notag\\[-6.0ex]
 \label{xisec}
 \mathscr{D}_{\xi}(X)=\lbrace{}\eta\in\mathbb{R}:\ (\xi,\eta)\in{}\mathscr{D}(X)\rbrace,
 \ \  \xi \in \mathbb{R}.
\end{gather}
\end{subequations}
The sets \(\mathscr{D}_{\eta}(X)\) and \(\mathscr{D}_{\xi}(X)\) are domains of definition
of the functions \(X_{\eta}(\xi)\) and \(X_{\xi}(\eta)\) respectively.

\noindent
For \(\eta\in\mathbb{R}\), the \(\eta\)-section \(X_{\eta}(\xi)\) of the function \(X(\xi,\eta)\)
is defind as
\begin{gather}
X_{\eta}(\xi)=X(\xi,\eta), \ \ \xi\in{}\mathscr{D}_{\eta}, \ \ \eta\in\mathbb{R}.
\intertext{For \(\xi\in\mathbb{R}\), the \(\xi\)-section \(X_{\xi}(\eta)\) of the function \(X(\xi,\eta)\) is defind as}\notag\\
\notag\\[-6.0ex]
X_{\xi}(\eta)=X(\xi,\eta), \ \ \eta\in{}\mathscr{D}_{\xi}, \ \ \xi\in\mathbb{R}.
\end{gather}
For any \(\eta\in\mathbb{R}\), the set \(\mathscr{D}_{\eta}(X)\) is the domain of definition
of the function \(X_{\eta}(\xi)\).\\
From \eqref{aeR2} and Fubini's Theorem it follows that
\begin{equation}
\label{aeae}
\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}\setminus{}\mathscr{D}_{\eta}}d\xi\bigg)d\eta=0\,,
\qquad
\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}\setminus{}\mathscr{D}_{\xi}}d\eta\bigg)d\xi=0\,.
\end{equation}
First of the equalities \eqref{aeae} means that for \(d\eta\)-almost every \(\eta\in\mathbb{R}\),
the \(\eta\)-section \(X_{\eta}(\xi)\) of the function \(X(\xi,\eta)\) is defined
for \(d\xi\)-almost every \(\xi\in\mathbb{R}\). The second one means that for \(d\xi\)-almost every \(\xi\in\mathbb{R}\), the \(\xi\)-section \(X_{\xi}(\eta)\) of the function \(X(\xi,\eta)\) is defined for \(d\eta\)-almost every \(\eta\in\mathbb{R}\).

In  particular, the value \(\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\) is well defined
for \(d\eta\)-almost every \(\eta\in\mathbb{R}\):
\begin{equation}
\label{wdes}
0\leq\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq\infty, \ \ \textup{for}
\ d\xi-\textup{almost every} \ \xi\in\mathbb{R}.
\end{equation}

 The following Lemma is related to integration of functions
 which values are elements of the inner space product \(L^2(\mathbb{R},w(\xi)d\xi)\).
 \begin{lemma}
\label{NIIN}
Let \(w(\xi)\) be a weight function satisfying the condition \eqref{StriPos}.

Let \(X(\xi,\eta)\) be a function of two variables which is defined for
almost every \((\xi,\eta)\in\mathbb{R}^2\) with respect to two-dimensional Lebesgue
measure \(d\xi{}d\eta\), that is the condition \eqref{aeR2} holds for the domain of definition \(\mathscr{D}(X)\)
of the function \(X\).\\
We assume that\footnote{In particular, \(\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}<\infty\)
 for \(d\eta\)-almost every \(\eta\in\mathbb{R}\).}
\begin{equation}
\label{FIVF}
\int\limits_{\mathbb{R}}\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}d\eta<\infty
\end{equation}

Then the function \(Y(\xi)\) which is defined by the integral
\begin{equation}
\label{FDBI}
Y(\xi)=\int\limits_{\mathbb{R}}X_{\xi}(\eta)\,d\eta
\end{equation}
exists for \(d\xi\)-almost every \(\xi\in\mathbb{R}\). Moreover,
\(Y(\xi)\in{}L^2(\mathbb{R},d\xi)\) and the inequality holds
\begin{equation}
\label{NIlIN}
\|Y(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq
\int\limits_{\mathbb{R}}\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}d\eta\,.
\end{equation}
\end{lemma}
\begin{proof} Let
 \(\displaystyle{}m(\xi)=\int\limits_{\mathbb{R}}|X_{\xi}(\eta)|\,d\eta, \ \
0\leq{}m(\xi\leq\infty\).
The integral \eqref{FDBI} exists for those \(\xi\) for which \(m(\xi)<\infty\).
Let
\begin{equation}
\label{Conr}
 \|r(\xi)\|_{L^2(\mathbb{R},w(\xi)\,d\xi)}\leq1, \ \ r(\xi)\geq0.
\end{equation}
By Fubini's Theorem,
\[\int\limits_{\mathbb{R}}m(\xi)r(\xi)w(\xi)\,d\xi=
\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}}|X_{\eta}(\xi)|r(\xi)w(\xi)\,d\xi\bigg)\,d\eta.\]
Substituting the inequality
\[\int\limits_{\mathbb{R}}|X_{\eta}(\xi,\eta)|r(\xi)w(\xi)\,d\xi\leq
\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)\,d\xi)},
\ \ \textup{\(\forall\,r\) \ satisfying \eqref{Conr},}
 \]
into the previous inequality, we obtain the inequality
\[\int\limits_{\mathbb{R}}m(\xi)r(\xi)w(\xi)\,d\xi\leq
\int\limits_{\mathbb{R}}\,
\|(X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\,d\eta\]
which holds for each \(r\) satisfying \eqref{Conr}. Taking \(\sup\) over all such \(r\),
we come to the inequality
\[\|m(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq\int\limits_{\mathbb{R}}
\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\,d\eta.\]
If \eqref{FIVF} holds, then also \(\|m(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}<\infty\).
In particular, under the conditions \eqref{StriPos}, \eqref{FIVF}, the inequality \(m(\xi)<\infty\) holds
for \(d\xi\)-almost every \(\xi\in\mathbb{R}\). So the integral \eqref{FDBI} exists for
\(d\xi\)-almost every \(\xi\in\mathbb{R}\).
Since \(|Y(\xi)|\leq{}m(\xi)\), the inequality \eqref{NIlIN} holds.
\end{proof}
\subsection{Approximative Identities.}
Our proof of the completeness of Gaussians in the space \(\mathcal{H}_{w_T,w_{\Omega}}\)
is based on some approximation procedure. In this approximation procedure two families
of operators play role: the family \(\lbrace{}\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
of integral operators and the family \(\lbrace{}\mathscr{M}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\) of multiplication operators.

The kernels of integral operators \(\lbrace{}\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
involves the functions \(\lbrace{}G_{\alpha}(\eta)\rbrace_{\alpha\in\mathbb{R}_{+}}\):
\begin{equation}
\label{MAKe}
G_{\alpha}(\eta)=\sqrt{\alpha}\exp(-\pi\alpha\eta^2),\ \ \eta\in\mathbb{R}.
\end{equation}
The functions \(G_{\alpha}(\eta)\) posses the properties:
\begin{subequations}
\label{PropG}
\begin{enumerate}
\item{\ }\\[-6.0ex]
\begin{equation}
\label{Pos}
G_{\alpha}(\eta)>0, \ \ \forall\,\eta\in\mathbb{R},\,\forall\,\alpha\in\mathbb{R}_{+}.
\end{equation}
\item {\ }\\[-6.0ex]
\begin{equation}
\label{Norm}
\int\limits_{\mathbb{R}}G_{\alpha}(\eta)\,d\eta=1, \\ \ \forall\,\alpha\in\mathbb{R}_{+}.
\end{equation}
\item For each \(\mu\in\mathbb{R}_{+}\) and \(\delta\in\mathbb{R}_{+}\),
\begin{equation}
\label{conc}
\lim_{\alpha\to\infty}\int\limits_{\mathbb{R}\setminus(-\delta,\delta)}
G_{\alpha}(\eta)\exp(\mu|\eta|)d\eta=0.
\end{equation}
\end{enumerate}
\end{subequations}

\begin{definition}
\label{DApU}
Let \(\alpha\in\mathbb{R}_{+}\).
\begin{enumerate}
\item[\textup{1}.]
The operator \(\mathscr{I}_{\alpha}:
L^2(\mathbb{R},w(\xi)d\xi)\to{}L^2(\mathbb{R},w(\xi)d\xi)\) is defined as
\begin{equation}
\label{AppUnO}
(\mathscr{I}_{\alpha}x)(\xi)=
\int\limits_{\mathbb{R}}G_{\alpha}(\eta)(\mathscr{T}_{\eta}x)(\xi)d\xi,
\end{equation}
where \(\mathscr{T}_{\eta}\) is the shift operator defined by \eqref{DefShf}.
\item[\textup{2}.]
The operator \(\mathscr{M}_{\alpha}:
L^2(\mathbb{R},w(\xi)d\xi)\to{}L^2(\mathbb{R},w(\xi)d\xi)\) is defined as
\begin{equation}
\label{AppMuO}
(\mathscr{M}_{\alpha}x)(\xi)=\exp\big(-\tfrac{\pi}{\alpha}\,{\textstyle\xi^2}\big)\,x(\xi).
\end{equation}
\end{enumerate}
\end{definition}
\begin{lemma}
\label{AppIdL}
Let \(w(\xi)\) be a weight function which satisfy the regularity condition\footnote{
See Definition \ref{DefReCo}.
}.\\
Then
\begin{enumerate}
\item[\textup{1.}]
For each \(\alpha>0\), the operator \(\mathscr{I}_{\alpha}\) is \, a bounded operator \,
in the space \(L^2(\mathbb{R},w(\xi)d\xi)\),
and the estimate
\begin{equation}
\label{IFN}
\|\mathscr{I}_{\alpha}\|_{L^2(\mathbb{R},w(\xi)d\xi)
\to{L^2(\mathbb{R},w(\xi)d\xi)}}\leq
\int\limits_{\mathbb{R}}G_{\alpha}(\eta)\sqrt{M_{w}(|\eta|)}\,d\eta, \ \ \forall\,\alpha\in\mathbb{R}^{+},
\end{equation}
holds, where \(M_{w}\) is the multiplicative modulus of continuity\footnote{
See Definition \ref{DMMC}.
}
of the weight function \(w\). In particular,
\begin{equation}
\label{IFNb}
\|\mathscr{I}_{\alpha}\|_{L^2(\mathbb{R},w(\xi)d\xi)
\to{L^2(\mathbb{R},w(\xi)d\xi)}}\leq
\sqrt{C_{w}}\int\limits_{\mathbb{R}}G_{\alpha}(\eta)\exp(\mu_{w}|\eta|/2)\,d\eta, \ \ \forall\,\alpha\in\mathbb{R}^{+},
\end{equation}
where \(C_{w}\) and \(\mu_{w}\) are the same that in \eqref{MCE}.
\item[\textup{2}.] The family \(\lbrace\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
is an approximative identity in the space \(L^2(\mathbb{R},w(\xi)d\xi)\):
\begin{equation}
\label{ApprUn}
\lim_{\alpha\to\infty}
\big\|(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0,
\ \ \textup{for each} \ x\in{}L^2(\mathbb{R},w(\xi)d\xi).
\end{equation}
\end{enumerate}
\end{lemma}
\begin{proof} {\ }\\
\textsf{1.} We apply Lemma \ref{NIIN} to the function
\(X(\xi,\eta)=G_{\alpha}(\eta)(\mathscr{T}_{\eta}x)(\xi)\). According to Lemma
\ref{PrTrO}, the inequality \eqref{ENS} holds. This inequality implies the estimate for
\(L^2(\mathbb{R},w(\xi)d\xi)\)-norm of \(\eta\)-sections \(X_{\eta}(\xi)\) of the function \(X\):
\[\|X_{\eta}(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq{}G_{\alpha}(\eta)\sqrt{M_{w}(\eta)}
\|x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}.\]
The inequality \eqref{IFN} is a consequence of the last inequality. Using the estimate
\eqref{MCE} for \(M_{w}\), we come to the inequality \eqref{IFNb}.\\
\textsf{2.}  In view of \eqref{Norm},
\[(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)=
\int\limits_{\mathbb{R}}G_{\alpha}(\eta)\big((\mathscr{T}_{\eta}x)(\xi)-x(\xi)\big)d\xi.\]
Let us fix  \(x(\xi)\in{}L^2(\mathbb{R},w(\xi)d\xi)\).
According to Lemma \ref{NIIN},
\begin{equation}
\label{PreEst}
\|(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq
\int\limits_{\mathbb{R}}G_{\alpha}(\eta)
\|(\mathscr{T}_{\eta}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}d\eta.
\end{equation}
By statement 2 of Lemma \ref{PrTrO}, for any \(\varepsilon>0\) there exists \(\delta>0\),
\(\delta=\delta(\varepsilon,x)\), such that
\begin{equation}
\label{ShEs}
\|(\mathscr{T}_{\eta}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}<\varepsilon
\ \ \textup{if} \ \ |\eta|<\delta.
\end{equation}
Splitting the integral in the right hand side of \eqref{PreEst} into the sum of integrals
taken over \((-\delta,\delta)\) and \(\mathbb{R}\setminus(-\delta,\delta)\), we obtain
\begin{multline}
\label{abc}
\|(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq
\!\int\limits_{(-\delta,\delta)}G_{\alpha}(\eta)
\|(\mathscr{T}_{\eta}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}d\eta\\
\hfill+
\int\limits_{\mathbb{R}\setminus(-\delta,\delta)}G_{\alpha}(\eta)
\Big(\|(\mathscr{T}_{\eta}x)(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}
+\|x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\Big)d\eta.
\end{multline}
From \eqref{abc} and from the inequalities \eqref{ShEs}, \eqref{ENS}, \eqref{MCE}
we obtain that
\begin{multline}
\label{ABC}
\|(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq\\
\varepsilon\int\limits_{(-\delta,\delta)}G_{\alpha}(\eta)d\eta+\big(\sqrt{C_{w}}+1\big)\!\!\!\!
\int\limits_{\mathbb{R}\setminus(-\delta,\delta)}G_{\alpha}(\eta)
\exp\big(\tfrac{\mu_{w}}{2}|\eta|\big)d\eta
\,\,\,\,
\|x\|_{L^2(\mathbb{R},w(\xi)d\xi)}.
\end{multline}
From \eqref{ABC} and from the properties \eqref{PropG} of the functional family
\(\lbrace{}G_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\) it follows that
\[\varlimsup_{\alpha\to\infty}
\|(\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\|_{L^2(\mathbb{R},w(\xi)d\xi)}\leq\varepsilon.\]
Since \(\varepsilon>0\) is arbitrary, \eqref{ApprUn} holds.
\end{proof}
\begin{lemma}
\label{ApIdWN}
Let \(w(\xi)\) be a weight function which satisfy the regularity condition.

Then each of the two families
\(\lbrace\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\) and
\(\lbrace\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
of operators
is an approximative identity in the space \(L^2(\mathbb{R},w(\xi)d\xi)\):
\begin{subequations}
\label{ApIdwn}
\begin{gather}
\label{ApIdwn1}
\lim_{\alpha\to\infty}
\big\|(\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}x)
(\xi)-x(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0,
\ \ \textup{for each} \ x\in{}L^2(\mathbb{R},w(\xi)d\xi),\\
\label{ApIdwn2}
\lim_{\alpha\to\infty}
\big\|(\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}x)(\xi)-x(\xi)\big\|_{L^2(\mathbb{R},w(\xi)d\xi)}=0,
\ \ \textup{for each} \ x\in{}L^2(\mathbb{R},w(\xi)d\xi).
\end{gather}
\end{subequations}
\end{lemma}
\begin{proof}
From \eqref{IFNb} it follows that the family of operators
\(\lbrace\mathscr{I}_{\alpha}\rbrace_{\alpha\in[1,\infty)}\) is uniformly bounded:
\begin{equation}
\label{UniBou}
\sup_{\alpha\in[1,\infty)}
\|\mathscr{I}_{\alpha}\|_{L^2(\mathbb{R},w(\xi)d\xi)\to{}L^2(\mathbb{R},w(\xi)d\xi)}<\infty.
\end{equation}
The operator \(\mathscr{M}_{\alpha}\) is contractive for any \(\alpha\in\mathbb{R}_{+}\):
\[\|\mathscr{M}_{\alpha}\|_{L^2(\mathbb{R},w(\xi)d\xi\to{}L^2(\mathbb{R},w(\xi)d\xi}=1,
\ \ \forall\,\alpha>0.
\]
In particular, the family \(\lbrace\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
is uniformly bounded.

In Lemma \ref{AppIdL} we established that
the family \(\lbrace\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\)
is an approximative identity in the space \(L^2(\mathbb{R},w(\xi)d\xi)\):
\[\lim_{\alpha\to\infty}\mathscr{I}_{\alpha}=\mathscr{I},\]
where \(\mathscr{I}\) is the identity operator in the space \(L^2(\mathbb{R},w(\xi)d\xi)\)
and  convergence is the strong convergence in this space. It is clear that
the family \(\lbrace\mathscr{M}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\) also is
an approximative identity in the space \(L^2(\mathbb{R},w(\xi)d\xi)\):
\[\lim_{\alpha\to\infty}\mathscr{M}_{\alpha}=\mathscr{I},\]
where convergence is the strong convergence in \(L^2(\mathbb{R},w(\xi)d\xi)\).
The assertion of Lemma \ref{ApIdWN} follows now from the equalities
\begin{align*}
\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}-\mathscr{I}&=
\mathscr{I}_{\alpha}(\mathscr{M}_{\alpha}-\mathscr{I})+(\mathscr{I}_{\alpha}-\mathscr{I}),\\
\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}-\mathscr{I}&=
\mathscr{M}_{\alpha}(\mathscr{I}_{\alpha}-\mathscr{I})+(\mathscr{M}_{\alpha}-\mathscr{I}).
\end{align*}
\end{proof}

\section{The completeness of a system of Gaussians\\ in the space \(\boldsymbol{\mathcal{H}_{w_T,w_{\Omega}}}\).}
\begin{definition}
\label{DeGa}
For each \(\alpha\in\mathbb{R}_{+}\) and \(\tau\in\mathbb{R}\), \textsf{the Gaussian} \(\boldsymbol{g_{\alpha,\tau}}\)
 is a function
\begin{equation}
\label{Ga}
g_{\alpha,\tau}(t)=\sqrt{\alpha}\exp\big(-\pi\alpha(t-\tau)^2)\big)
\end{equation}
 of the variable \(t\in\mathbb{R}\). The Gaussians form a two-parametric family
 \(\lbrace{}g_{\alpha,\tau}\rbrace_{\alpha,\tau}\) of functions on \(\mathbb{R}\) which is parametrized
 by the parameters \(\alpha\in\mathbb{R}_{+}\) and \(\tau\in\mathbb{R}\).
\end{definition}

\noindent
The Fourier transform \(\widehat{g_{\alpha,\tau}}(\omega)\) of the Gaussian \(g_{\alpha,\tau}(t)\) is
\begin{equation}
\label{FTGa}
\widehat{g_{\alpha,\tau}}(\omega)=
\exp\big(-\tfrac{\pi}{\alpha}\omega^2-2\pi{}i\tau\omega\big).
\end{equation}
\begin{lemma}[I.Schur]
\label{ScTe}
For  a function \(K(t,\tau),\,t\in\mathbb{R},\,\tau\in\mathbb{R}\) of two variables, let as define
the values
\begin{equation}
\label{SchVal}
N_1(K)=\sup_{\tau\in\mathbb{R}}\int\limits_{\mathbb{R}}|K(t,\tau)|dt,\quad
N_\infty(K)=\sup_{t\in\mathbb{R}}\int\limits_{\mathbb{R}}|K(t,\tau)|d\tau.
\end{equation}
If the condition
\begin{gather}
\label{SchCon}
N_{1}(K)<\infty,\quad{}N_{\infty}(K)<\infty
\end{gather}
are fulfilled, then for each \(f(t)\in{}L^2(\mathbb{R},dt)\), \(g(\tau)\in{}L^2(\mathbb{R},d\tau)\)
the double integral
\begin{equation}
\label{DoIn}
\iint\limits_{\mathbb{R}\times\mathbb{R}}f(t)K(t,\tau)\overline{g(\tau)}\,dtd\tau
\end{equation}
exists and admits the estimate
\begin{equation}
\label{DoInEs}
\bigg|\iint\limits_{\mathbb{R}\times\mathbb{R}}f(t)K(t,\tau)\overline{g(\tau)}\,dtd\tau\bigg|
\leq\sqrt{N_1(K)N_\infty(K)}\,\,\big\|f\big\|_{L^2(\mathbb{R},dt)}\,\big\|g\big\|_{L^2(\mathbb{R},d\tau)}.
\end{equation}
\end{lemma}
\begin{proof}
The integral \eqref{DoIn} exists if \
\(\displaystyle\iint\limits_{\mathbb{R}\times\mathbb{R}}|f(t)|\,|K(t,\tau)|\,|g(\tau)|\,dtd\tau<\infty.\) \
Applying Cauchy-Schwarz inequality for double integral and Fubini Theorem, we obtain:
\begin{gather*}
\iint\limits_{\mathbb{R}\times\mathbb{R}}|f(t)|K(t,\tau)|g(\tau)|\,dtd\tau=
\iint\limits_{\mathbb{R}\times\mathbb{R}}\Big(|f(t)|K(t,\tau)^{1/2}\Big)
\Big(K(t,\tau)^{1/2}|g(\tau)|\Big)\,dtd\tau\\
\leq\Big(\iint\limits_{\mathbb{R}\times\mathbb{R}}\Big(|f(t)|^2K(t,\tau)dt\,d\tau\Big)^{\frac{1}{2}}
\cdot\Big(\iint\limits_{\mathbb{R}\times\mathbb{R}}\Big(|g(\tau)|^2K(t,\tau)dt\,d\tau\Big)^{\frac{1}{2}}\\
=\Bigg(\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}}K(t,\tau)\,d\tau\bigg)|f(t)|^2dt\Bigg)^{\frac{1}{2}}
\cdot\Bigg(\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}}K(t,\tau)\,dt\bigg)|g(\tau)|^2d\tau\Bigg)^{\frac{1}{2}}\\
\hspace*{0.6\linewidth}\leq\sqrt{N_\infty(K)N_1(K)}\,\big\|f\big\|^2_{L^2(\mathbb{R},dt)},
\end{gather*}
\end{proof}
\begin{theorem}
\label{GaIn}
We assume that:
\begin{enumerate}
\item[\textup{1.}]
 The weight functions \(w_{T}\) and \(w_{\Omega}\) satisfy
the non-degeneracy conditions,
\textup{(Definition \ref{DeCo})};
\item[\textup{2.}]
The weight functions \(1+w_T\) and \(1+w_{\Omega}\) satisfy the regularity conditions, \textup{(Definition \ref{DefReCo})}.\\
\end{enumerate}
Then
\begin{enumerate}
\item[\textup{1}.]
Each Gaussian \(g_{\alpha,\tau}\) belongs to the space \(\mathcal{H}_{w_T,w_{\Omega}}\).
\item[\textup{2}.]
The functional family \(\lbrace{}g_{\alpha,\tau}\rbrace_{\alpha,\tau}\), where \(\alpha\) runs
over \(\mathbb{R}_{+}\) and \(\tau\) runs over \(\mathbb{R}\), is complete\footnote{
That is the linear hall of this family is dense in \(\mathcal{H}_{w_T,w_{\Omega}}\).
}
in the space
\(\mathcal{H}_{w_{T},w_{\Omega}}\).
\end{enumerate}
\end{theorem}
\begin{proof} {\ }\\
\textbf{1}.
From  \eqref{Ga}, \eqref{FTGa} and \eqref{IWFE} it is evident that
\begin{equation*}
\int\limits_{\mathbb{R}}\big|g_{\alpha,\tau}(t)\big|^2w_{T}(t)dt+
\int\limits_{\mathbb{R}}\big|\widehat{g_{\alpha,\tau}}(\omega)\big|^2w_{\Omega}(\omega)d\omega<\infty
\ \ \forall\,\alpha\in\mathbb{R}_{+},\,\ \forall\,\tau\in\mathbb{R}.
\end{equation*}
\textbf{2}. \textsf{Step 1}. According to Theorem \ref{Com}, the inner product space \(\mathcal{H}_{w_{T},w_{\Omega}}\)
is compete. Given \(f\in\mathcal{H}_{w_{T},w_{\Omega}}\),  we have to prove that from the orthogonality condition
\begin{equation}
\label{OrCo}
\langle{}f,g_{\alpha,\tau}\rangle_{\mathcal{H}_{w_{T},w_{\Omega}}}=0, \ \ \forall\,\alpha\in\mathbb{R}_{+}, \ \forall\,\tau\in\mathbb{R}
\end{equation}
it follows that \(\langle{}f,f\rangle_{\mathcal{H}_{w_{T},w_{\Omega}}}=0\). According to \eqref{InPr}, \eqref{Ga}, \eqref{FTGa}, the condition \eqref{OrCo} can be presented in the form
\begin{multline}
\label{OrCo1}
\hfil
\int\limits_{\mathbb{R}}f(t)\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt+
\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega=0, \\
\forall\,\alpha\in\mathbb{R}_{+}, \ \forall\,\tau\in\mathbb{R}.
\end{multline}
Is appropriate to recall\footnote{
See Corollary \ref{Cor}.
}
 that since \(f\in\mathcal{H}_{w_T,w_{\Omega}}\),
\begin{subequations}
\label{ATR}
\begin{gather}
\label{ATRa}
f(t)\in{}L^2(\mathbb{R},(1+w_T(t))dt),\\
\label{ATRb}
\hat{f}(\omega)\in{}L^2(\mathbb{R},(1+w_{\Omega}(\omega))d\omega).
\end{gather}
\end{subequations}
 We multiply the equality \eqref{OrCo1} with \(e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}\)
and integrate by the measure \(d\tau\) over \(\mathbb{R}\).
\begin{gather}
\int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}f(t)\sqrt{\alpha}
e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\notag\\[-1.0ex]
\label{MulIn}\\[-1.0ex]
+\int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}
\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau=0.
\notag
\end{gather}
\textbf{Step 2}. Let us proof the existence of the iterated integral
 \begin{equation}
 \label{ItInt1}
 \int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}f(t)\sqrt{\alpha}
e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau
\end{equation}
which appears as the first summand of \eqref{MulIn}.
Using the inequality
\[w_{T}(t)\leq{}C_{T}e^{\mu_{T}|t-\tau|}e^{\mu_{T}|\tau|}, \ \forall\,t\in\mathbb{R},\,\tau\in\mathbb{R},\]
which follows from \eqref{IWFEa}, we obtain the inequality
\begin{equation}
\label{Ker}
\iint\limits_{\mathbb{R}\times\mathbb{R}}|f(t)|
\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}w_{T}(t)e^{-\frac{\pi}{\alpha}\tau^2}|f(\tau)|\,dtd\tau
\leq\iint\limits_{\mathbb{R}\times\mathbb{R}}|f(t)|K_{T}(t,\tau)|f(\tau)|\,dtd\tau,
\end{equation}
where the kernel \(K_{T}\) is of the form
\begin{equation}
\label{K1}
K_{T}(t,\tau)=C_{T}\sqrt{\alpha}\exp\big(-\pi\alpha(t-\tau)^2+\mu_{T}|t-\tau|\big)
\exp\big(-\tfrac{\pi}{\alpha}\tau^2+\mu_{T}|\tau|\big).
\end{equation}
(The constants \(C_{T}\) and \(\mu_{T}\) are the same that in \eqref{IWFEa}.)
By direct calculation,
\begin{equation*}
\max_{\tau\in\mathbb{R}}\exp\big(-\tfrac{\pi}{\alpha}\tau^2+\mu_{T}|\tau|\big)=
\exp\big(\tfrac{\mu_{T}^2\alpha}{4\pi}\big).
\end{equation*}
So, the kernel \(K_{T}(t,\tau)\) admits estimate
\[0\leq
K_{T}(t,\tau)\leq{}C_T\sqrt{\alpha}\exp\big(\tfrac{\mu_{T}^2\alpha}{4\pi}\big)
\exp\big(-\pi\alpha(t-\tau)^2+\mu_{T}|t-\tau|\big), \ \ \forall\,t\in\mathbb{R},\,\tau\in\mathbb{R}.
\]
From the last inequality, the estimates
\[\sup_{\tau\in\mathbb{R}}\int\limits_{\mathbb{R}}K_{T}(t,\tau)dt\leq{}I(\alpha), \quad \sup_{t\in\mathbb{R}}\int\limits_{\mathbb{R}}K_{T}(t,\tau)dt\leq{}I(\alpha)\]
hold, where
\[I(\alpha)=C_T\sqrt{\alpha}\exp\big(\tfrac{\mu_{T}^2\alpha}{4\pi}\big)
\int\limits_{\mathbb{R}}\exp\big(-\pi\alpha\xi^2+\mu_{T}|\xi|\big)\,d\xi,
\ \ I(\alpha)<\infty \ \forall\,\alpha\in\mathbb{R}_{+}.\]

Applying Lemma \ref{ScTe} to the kernel \(K_{T}(t,\tau)\), \eqref{K1}, we come
to the inequality
\[\iint\limits_{\mathbb{R}\times\mathbb{R}}|f(t)|K_{T}(t,\tau)|f(\tau)|\,dtd\tau
\leq I(\alpha)\big\|f\big\|^2_{L^2(\mathbb(R),dt)}<\infty.\]

Therefore the double integral in the left hand side of \eqref{Ker} is finite for every \(\alpha\in\mathbb{R}_{+}\)
and for every \(f\in{}L^2(\mathbb{R},dt)\), in particular, for every \(f\in\mathcal{H}_{w_{T},w_{\Omega}}\).

By Fubini's Theorem, the iterated integral \eqref{ItInt1} exists. Moreover we can
 interchange the order of
integration in this integral:
\begin{multline*}
\int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}
f(t)\,\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\\=
\int\limits_{\mathbb{R}}f(t)\bigg(\int\limits_{\mathbb{R}}
\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\bigg)w_{T}(t)dt.
\end{multline*}
Changing the variable \(\tau\to{}t-\tau\) in the inner integral of the right hand side
of the previous equality, we come to the equality
\begin{multline*}
\int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}
f(t)\,\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\\=
\int\limits_{\mathbb{R}}f(t)\bigg(\int\limits_{\mathbb{R}}
\sqrt{\alpha}e^{-\pi\alpha\tau^2}
e^{-\frac{\pi}{\alpha}(t-\tau)^2}\overline{f(t-\tau)}d\tau\bigg)w_{T}(t)dt.
\end{multline*}
In other words,
\begin{equation}
\label{ChOrd}
\int\limits_{\mathbb{R}}\bigg(\int\limits_{\mathbb{R}}
f(t)\sqrt{\alpha}e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau=
\langle{}f,\mathfrak{I}_{\alpha}\mathscr{M}_{\alpha}f\rangle_{L^2(\mathbb{R},w_{T}(t)dt)},
\end{equation}
where \(\mathscr{I}_{\alpha}\) and \(\mathscr{M}_{\alpha}\) are the operators which were
introduced in Definition \ref{DApU}.

Since \(f\in\mathcal{H}_{w_T,w_{\omega}}\), the inclusion \eqref{ATRa} holds.
According to Lemma \ref{ApIdWN}, the family
\(\lbrace\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}\rbrace_{\alpha\to\infty}\)
is an approximative identity
in \(L^2(\mathbb{R},(1+w_T(t))dt)\):
\[
\lim_{\alpha\to\infty}\big\|(\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}x)(t)
-x(t)
\big\|_{L^2(\mathbb{R},(1+w_{T}(t))dt)}=0, \ \ \forall\,x\in{}L^2(\mathbb{R},(1+w_{T}(t))dt).
\]

All the more,
\[
\lim_{\alpha\to\infty}\big\|(\mathscr{I}_{\alpha}\mathscr{M}_{\alpha}f)(t)
-f(t)
\big\|_{L^2(\mathbb{R},w_{T}(t)dt)}=0, \ \ \textup{for} \ f \ \textup{which appears in} \
\eqref{ChOrd}.
\]
Comparing the last equality with \eqref{ChOrd}, we conclude that
\begin{equation}
\label{LS1}
\lim_{\alpha\to\infty}\int\limits_{\mathbb{R}}
\bigg(\int\limits_{\mathbb{R}}f(t)\tfrac{1}{\sqrt{\alpha}}
e^{-\pi\alpha(t-\tau)^2}w_{T}(t)dt\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau=
\langle{}f,f\rangle_{L^2(\mathbb{R},w_{T}(t)dt)}.
\end{equation}
\textsf{Step3}. Let us elaborate the second summand in \eqref{MulIn}, i.e. the expression
\begin{equation}
\label{ItInt2}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau.
\end{equation}
Using the inequality
\[w_{\Omega}(\omega)\leq{}C_{\Omega}e^{\mu_{\Omega}|\omega|}, \ \forall\,\omega\in\mathbb{R},\]
which is the inequality \eqref{IWFEa}, we obtain
\begin{equation}
\label{Ker2}
\iint\limits_{\mathbb{R}\times\mathbb{R}}|\hat{f}(\omega)|
e^{-\frac{\pi}{\alpha}\omega^2}w_{\Omega}(\omega)
e^{-\frac{\pi}{\alpha}\tau^2}|f(\tau)|\,d\omega{}d\tau
\leq\iint\limits_{\mathbb{R}\times\mathbb{R}}
|\hat{f}(\omega)|K_{\Omega}(\omega,\tau)|f(\tau)|\,d\omega{}d\tau,
\end{equation}
where the kernel \(K_{\Omega}\) is of the form
\begin{equation}
\label{K2}
K_{\Omega}(\omega,\tau)=C_{\Omega}\exp\big(-\tfrac{\pi}{\alpha}\omega^2+\mu_{\Omega}|\omega|\big)
\exp\big(-\tfrac{\pi}{\alpha}\tau^2\big).
\end{equation}
Since
\[\max_{\omega\in\mathbb{R}}\exp\big(-\tfrac{\pi}{\alpha}\omega^2+\mu_{\Omega}|\omega|\big)
=\exp\big(\tfrac{\mu_{\Omega}^2\alpha}{4\pi}\big), \qquad\max_{\tau\in\mathbb{R}}
\exp\big(-\tfrac{\pi}{\alpha}\tau^2\big)=1,\]
the kernel \(K_{\Omega}(\omega,\tau)\), \eqref{K2}, admits two estimates:
\begin{subequations}
\label{EstK2}
\begin{align}
\label{EstK21}
K_{\Omega}(\omega,\tau)\leq{}C_{\Omega}\exp\big(-\tfrac{\pi}{\alpha}\omega^2+\mu_{\Omega}|\omega|\big),
\ \ &\forall\,\omega\in\mathbb{R},\,\forall\,\tau\in\mathbb{R},\\[-3.5ex]
\intertext{and}
\notag\\[-7.5ex]
\label{EstK22}
K_{\Omega}(\omega,\tau)\leq{}C_{\Omega}\exp\big(\tfrac{\mu_{\Omega}^2\alpha}{4\pi}\big)
\exp\big(-\tfrac{\pi}{\alpha}\tau^2\big),
\ \ &\forall\,\omega\in\mathbb{R},\,\forall\,\tau\in\mathbb{R}.
\end{align}
\end{subequations}
From \eqref{EstK21} it follows that
\begin{equation*}

\int\limits_{\mathbb{R}}K_{\Omega}(\omega,\tau)\,d\omega\leq{}I_1(\alpha), \ \
\forall\,\tau\in\mathbb{R},
\end{equation*}
where
\[
I_1(\alpha)=C_{\Omega}\int\limits_{\mathbb{R}}
\exp\big(-\tfrac{\pi}{\alpha}\omega^2+\mu_{\Omega}|\omega|\big)\,d\omega<\infty, \ \
\forall\,\alpha\in\mathbb{R}_{+}.
\]
From \eqref{EstK22} it follows that
\begin{equation*}

\int\limits_{\mathbb{R}}K_{\Omega}(\omega,\tau)\,d\tau\leq{}I_\infty(\alpha), \ \
\forall\,\omega\in\mathbb{R},
\end{equation*}
where
\[
I_\infty(\alpha)=C_{\Omega}\exp\big(\tfrac{\mu_{\Omega}^2\alpha}{4\pi}\big)\int\limits_{\mathbb{R}}
\exp\big(-\tfrac{\pi}{\alpha}\tau^2\big)\,d\tau<\infty, \ \
\forall\,\alpha\in\mathbb{R}_{+}.
\]
Applying Lemma \ref{ScTe} to the kernel \(K_{\Omega}(\omega,\tau)\), \eqref{K2}, we come
to the inequality
\[\iint\limits_{\mathbb{R}\times\mathbb{R}}|\hat{f}(\omega)|K_{\Omega}(\omega,\tau)|f(\tau)|\,dtd\tau
\leq \sqrt{I_1(\alpha)I_{\infty}(\alpha)}
\big\|\hat{f}\big\|_{L^2(\mathbb{R},d\omega)}
\big\|f\big\|_{L^2(\mathbb{R},d\tau)}<\infty.\]
Therefore the double integral in the left hand side of \eqref{Ker2} is finite for every \(\alpha\in\mathbb{R}_{+}\)
and for every \(\hat{f}\in{}L^2(\mathbb{R},d\omega)\), i.e. \(f\in{}L^2(\mathbb{R},d\tau)\), in particular, for every \(f\in\mathcal{H}_{w_{T},w_{\Omega}}\).

By Fubini's Theorem, the iterated integral \eqref{ItInt2} exists. Moreover we can
interchange the order of integration in this integral:
\begin{multline}
\label{ChOrd2}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\\=
\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
\bigg(\int\limits_{\mathbb{R}}
e^{-\frac{\pi}{\alpha}\tau^2+2\pi{}i\omega\tau}\overline{f(\tau)}d\tau\bigg)w_{\Omega}(\omega)d\omega.
\end{multline}
The Fourier Transform of the exponential \(e^{-\frac{\pi}{\alpha}\tau^2+2\pi{}i\omega\tau}\),
considered as a function of the variable \(\tau\), is
\[\int\limits_{\mathbb{R}}e^{-\frac{\pi}{\alpha}\tau^2+2\pi{}i\omega\tau}
e^{-2\pi{}i\lambda\tau}d\tau=\sqrt{\alpha}e^{-\pi\alpha(\omega-\lambda)^2}.\]
In view of the Parseval equality,
\[\int\limits_{\mathbb{R}}e^{-\frac{\pi}{\alpha}\tau^2+2\pi{}i\omega\tau}
\overline{f(\tau)}d\tau=
\int\limits_{\mathbb{R}}\sqrt{\alpha}e^{-\pi\alpha(\omega-\lambda)^2}
\overline{\hat{f}(\lambda)}\,d\lambda.
\]
Thus the iterated integral in the right hand side of \eqref{ChOrd2} takes the form
\begin{multline*}

\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
\bigg(\int\limits_{\mathbb{R}}
e^{-\frac{\pi}{\alpha}\tau^2+2\pi{}i\omega\tau}\overline{f(\tau)}d\tau\bigg)
w_{\Omega}(\omega)d\omega\\
=\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
\bigg(\int\limits_{\mathbb{R}}\sqrt{\alpha}e^{-\pi\alpha(\omega-\lambda)^2}
\overline{\hat{f}(\lambda)}\,d\lambda\bigg)w_{\Omega}(\omega)\,d\omega.
\end{multline*}
Comparing the last equality with \eqref{ChOrd2}, we see that
\begin{multline*}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\\
=\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
\bigg(\int\limits_{\mathbb{R}}\sqrt{\alpha}e^{-\pi\alpha(\omega-\lambda)^2}
\overline{\hat{f}(\lambda)}\,d\lambda\bigg)w_{\Omega}(\omega)\,d\omega.
\end{multline*}
Changing the variable \(\lambda\to\omega-\lambda\) in the inner integral of the right hand side
of the previous equality, we come to the equality
\begin{multline*}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau\\
=\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
\bigg(\int\limits_{\mathbb{R}}\sqrt{\alpha}e^{-\pi\alpha\lambda^2}
\overline{\hat{f}(\omega-\lambda)}\,d\lambda\bigg)w_{\Omega}(\omega)\,d\omega.
\end{multline*}

In other words,
\begin{equation}
\label{ToApId2}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau=
\langle{}\hat{f},\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}
\hat{f}\rangle_{L^2(\mathbb{R},w_{\Omega}(\omega)\,d\omega)},
\end{equation}
where \(\mathscr{I}_{\alpha}\) and \(\mathscr{M}_{\alpha}\) are the operators which were
introduced in Definition \ref{DApU}.
Since \(f\in\mathcal{H}_{w_T,w_{\omega}}\), the inclusion \eqref{ATRb} holds.
According to Lemma \ref{ApIdWN}, the family
\(\lbrace\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}\rbrace_{\alpha\to\infty}\) is an approximative identity
in \(L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)\):
\[
\lim_{\alpha\to\infty}\big\|(\mathscr{M}_{\alpha}\mathscr{I}_{\alpha}y)(\omega)
-y(\omega)
\big\|_{L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)}=0, \ \
\forall\,y\in{}L^2(\mathbb{R},w_{\Omega}(\omega)d\omega).
\]
Taking \(\hat{f}(\omega)\) as \(y(\omega)\) in the last equality and comparing
with the equality \eqref{ToApId2}, we conclude that
\begin{equation}
\label{LS2}
\lim_{\alpha\to\infty}
\int\limits_{\mathbb{R}} \bigg(\int\limits_{\mathbb{R}}\hat{f}(\omega)e^{-\frac{\pi}{\alpha}\omega^2}
e^{2\pi{}i\tau\omega}w_{\Omega}(\omega)d\omega\bigg)
e^{-\frac{\pi}{\alpha}\tau^2}\overline{f(\tau)}d\tau=
\langle{}\hat{f},\hat{f}\rangle_{L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)}.
\end{equation}

\textsf{Step 4}. Taking into account the limiting relations \eqref{LS1} and \eqref{LS2}, we pass to
the limit in the equality \eqref{MulIn} as \(\alpha\to\infty\). We obtain that
\begin{equation}
\label{Fin}
\langle{}f,f\rangle_{L^2(\mathbb{R},w_{T}(t)dt)}+
\langle{}\hat{f},\hat{f}\rangle_{L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)}=0\,,
\ \ \textup{ i.e } \ \ \langle{}f,f\rangle_{\mathcal{H}_{w_{T},w_{\Omega}}}=0\,.
\end{equation}
\end{proof}
\section{A generalization.}
 The system of Gaussians is of the form
\begin{equation}
\label{nong}
g_{\alpha,\tau}(t)=\alpha\,g(\alpha(t-\tau)),
\end{equation}
where
\begin{equation}
\label{gau}
g(t)=e^{-\pi{}t^2}.
\end{equation}
We established that under certain non-degeneracy and regularity conditions imposed on
the weight functions \(w_T\) and \(w_\Omega\) the  system of Gaussians
\(g_{\alpha,\tau}\), where \(\alpha\) runs over \(\mathbb{R}_{+}\) and \(\tau\) runs over
\(\mathbb{R}\), is a complete system in the space \(\mathcal{H}_{w_T,w_{\Omega}}\).
However our reasoning remains true for more general functions \(g(t)\) than the function~
\eqref{gau}. Let us formulate the appropriate generalization.

As before, we assume that the weight functions \(w_T\) and \(w_{\Omega}\) satisfy the non-degeneracy
condition (Definition \ref{DeCo}). We assume  also that the functions \(1+w_{T}(t)\),
\(1~+~w_{\Omega}(\omega)\) satisfy the regularity conditions (Definition \ref{DefReCo}).
In our formulation, the modules of continuity \(M_{1+w_{T}}\), \(M_{1+w_{\Omega}}\)
 (Definition \ref{DMMC}) corresponding to to the weight functions \(1+w_{T}\), \(1+w_{\Omega}\)
appear.
\begin{theorem} {\ }\\
We assume that a function \(g:\,\mathbb{R}\to\mathbb{C}\), \(g\in{}L^{1}(\mathbb{R},dt)\), is given which satisfy the conditions
\begin{subequations}
\label{ApIde}
\begin{align}
\label{ApIde1}
\int\limits_{\mathbb{R}}g(t)\,dt=1;\\
\label{ApIde2}
\lim_{\alpha\to\infty}\int\limits_{\mathbb{R}\setminus(-\delta,\delta)}
\alpha\,|g(\alpha\eta)|\sqrt{M_{1+w_T}(|\eta|)}\,d\eta=0& \ \ \ \textup{for each} \ \ \delta>0;\\
\label{ApIde3}
\lim_{\alpha\to\infty}\int\limits_{\mathbb{R}\setminus(-\delta,\delta)}
\alpha\,|g(\alpha\eta)|\sqrt{M_{1+w_{\Omega}}(|\eta|)}\,d\eta=0& \ \ \ \textup{for each} \ \ \delta>0.
\end{align}
\end{subequations}
Moreover we assume that the following conditions are fulfilled.
\begin{subequations}
\label{KT}
\begin{gather}
\label{KT1}
\int\limits_{\mathbb{R}}|g(\alpha\eta)|\,M_{1+w_{T}}(|\eta|)\,d\eta<\infty \ \ \
\textup{for each} \ \ \alpha>0;\\
\label{KT2}
\sup_{\eta\in\mathbb{R}}|\hat{g}(\alpha\eta)|\,M_{1+w_{T}}(|\eta|)<\infty \ \ \
\textup{for each} \ \ \alpha>0;
\end{gather}
\end{subequations}
and
\begin{subequations}
\label{KO}
\begin{gather}
\label{KO1}
\int\limits_{\mathbb{R}}|\hat{g}(\alpha\eta)|\,M_{1+w_{\Omega}}(|\eta|)\,d\eta<\infty \ \ \
\textup{for each} \ \ \alpha>0;\\
\label{KO2}
\sup_{\eta\in\mathbb{R}}|\hat{g}(\alpha\eta)|\,M_{1+w_{\Omega}}(|\eta|)<\infty \ \ \
\textup{for each} \ \ \alpha>0;
\end{gather}
\end{subequations}
Then
\begin{enumerate}
\item[\textup{1}.]
For each \(\alpha\in\mathbb{R}_{+}\), \(\tau\in\mathbb{R}\), the function \(g(\alpha(t-\tau))\)
belongs to the space \(\mathcal{H}_{w_T,w_{\Omega}}\).
\item[\textup{2}.]
The system of functions \(\lbrace{}g(\alpha(t-\tau))\rbrace_{\alpha,\tau}\), where \(\alpha\) runs over
\(\mathbb{R}_{+}\), \(\tau\) runs over \(\mathbb{R}\), is a complete system in the space
\(\mathcal{H}_{w_T,w_{\Omega}}\).
\end{enumerate}
\end{theorem}
\vspace{2.0ex}
\noindent
{\large\textbf{Comments}.}\\
1. The conditions \eqref{ApIde} ensure that the operator family \(\lbrace\mathscr{I}_{\alpha}\rbrace_{\alpha\in\mathbb{R}_{+}}\) (Definition \ref{DApU})
is an approximative identity
in each of the spaces \(L^2(\mathbb{R},w_{T}(t)dt\) and \(L^2(\mathbb{R},w_{\Omega}(\omega)d\omega\).\\
2. Since \(M_w(\eta)\geq1\) for any weight function \(w\), the condition \eqref{KO1} implies
that \(\int\limits_{\mathbb{R}}|\hat{g}(\eta)|\,d\eta<\infty\). \ \
Hence \(\sup_{\eta\in\mathbb{R}}|g(\eta)|<\infty\).
 From \eqref{KT1} it follows now that \(g(\alpha(t-\tau))\in{}L^2(\mathbb{R},w_{T}(t)dt)\)
 for each \(\alpha\in\mathbb{R}_{+},\tau\in\mathbb{R}\). From \eqref{KO} it follows that
\(\hat{g}(\alpha^{-1}\omega)\in{}L^2(\mathbb{R},w_{\Omega}(\omega)d\omega)\)
for each \(\alpha\in\mathbb{R}_{+}\). Thus  the function \(g(\alpha(t-\tau))\)
belongs to the space \(\mathcal{H}_{w_T,w_{\Omega}}\) for each \(\alpha\in\mathbb{R}_{+}\), \(\tau\in\mathbb{R}\).\\
3. The conditions \eqref{KT} are used to prove the convergence of the double integral analogous to the integral in \eqref{Ker}.\\
4. The congitions \eqref{KO} are used to prove the convergence of the double integral analogous to the integral in \eqref{Ker2}.

config: arxiv


TITLE: How to show $X=Spec(A)$ has maximal components $V(p)$?
QUESTION [3 upvotes]: This is from Atiyah Macdonald, Chapter 1, Execrise 20.
If $A$ is a ring and $X=Spec(A)$, then the irreducible components of $X$ are the closed sets $V(p)$, where $p$ is a minimal prime ideal of $A$.
It is not difficult to prove the closed sets $V(p)$ are irreducible. However it isn't clear to me how to prove the second part. It is clear that when closed sets of the form $V(q),q\in X$ then $V(p)$ did provide a maximal irreducible closed set. Neverthelss it is not clear to me why every closed set in $X$ must be of this form(they only provide a basis of closed sets). So theoretically there is the possiblity of some closed set $U$ that contains $V(p)$, and in particular contains some prime ideal $k$ such that $p\not\subset k$. Then what can we do to prove this cannot hold and $k$ must contain $p$?
One way to address this is to take the closure of $U\cap X_{p}=K$. Since the $V(q),q\in X$ form a basis there is some $V(l)\subset \overline K$. Then $V_{l}\cup V_{p}\subset U$, while $V_{l}\cup V_{p}=V_{l\cap p}$. But by definition $l\cap p=p$; so such $V_{l}$ could not exist, and $K$ must be empty.
I am wondering if this approach is long-winded, as I found almost every other proof used the fact every irreducible component of $X$ must be of the form $V_{r(p)}$, where $r(p)$ denotes the radical of $p$. Why this is true, and how it might help to simplify the above proof?

REPLY [4 votes]: Let $I$ be an ideal of $A$.
We denote by $rad(I)$ the set $= \{x \in A\colon x^n \in I$ for some integer $n > 0$, where $n$ depends on $x\}$.
It is easy to see that $rad(I)$ is an ideal and $V(I) = V(rad(I))$.
Let $E$ be a subset of $X = Spec(A)$.
We denote by $\mathfrak{I}(E)$ the ideal $\bigcap_{p \in E} p$.
Lemma 0
Let $I$ be an ideal of $A$.
Then $\mathfrak{I}(V(I)) = rad(I)$.
Proof:
This is an immediate consequence of proposition 1.8 of Atiyah-MacDonald(it states that  $\mathfrak{I}(X) = rad(0)$.
Lemma 1
Let $E$ be a subset of $X$.
Then $V(\mathfrak{I}(E))$ is the closure of $E$.
Proof:
Let $\bar E$ be the closure of $E$.
Since $E \subset V(\mathfrak{I}(E))$, $\bar E \subset V(\mathfrak{I}(E))$.
Conversely suppose $E \subset V(I)$ for some ideal $I$.
Then $I \subset \mathfrak{I}(E)$.
Hence $V(\mathfrak{I}(E)) \subset V(I)$.
Hence $V(\mathfrak{I}(E)) \subset \bar E$.
QED
Lemma 2
$X = Spec(A)$ is irreducible if and only if $A/rad(0)$ is an integral domain.
Proof:
Since $X = Spec(A/rad(0))$, we can assume $rad(0) = 0$.
Suppose $X$ is reducible.
There exist proper closed subsets $F_1, F_2$ of $X$ such that $X = F_1 \cup F_2$.
Hence $0 = \mathfrak{I}(X) = \mathfrak{I}(F_1) \cap \mathfrak{I}(F_2)$.
By Lemma 1, $V(\mathfrak{I}(F_1)) = F_1$.
Since $F_1 \neq X$, $\mathfrak{I}(F_1) \neq 0$.
Similarly $\mathfrak{I}(F_2) \neq 0$.
Hence $A$ is not an integral domain.
Conversely suppose $A$ is not an integral domain.
There exist elements $f, g$ of $A$ such that $f \neq 0, g \neq 0$, $fg = 0$.
Then, by Lemma 0, $V(f) \neq X$, $V(g) \neq X$ and $X = V(f) \cup V(g)$.
Hence $X$ is reducible.
QED
Lemma 3
Let $I$ be an ideal of $A$.
Then $V(I)$ is irreducible if and only if $rad(I)$ is a prime ideal.
Proof:
This follows immediately from Lemma 2.
By Lemma 3, an irreducible closed subset of $X$ is of the form $V(q)$, where $q$ is a prime ideal of $A$.
Hence $V(p)$ is a maximal irreducible closed subset of $X$, if $p$ is a minimal prime ideal of $A$.


TITLE: How does spacetime metric become dynamical (gravity quantized) in string theory?
QUESTION [3 upvotes]: I asked a similar question in What do we mean by worldsheet metric fluctuating in string theory, when we have a "target manifold"?, but the question had my misunderstanding that in Polyakov action, worldsheet metric is an independent variable, when it should just be understood as being redundant. This should have been obvious, but I was confused for the reasons below.
It is said that string theories allow metric to be "dynamic" instead of "fixed." Or said differently, correspondence like AdS/CFT allows one to study conformal field theories (on a fixed background manifold/metric) to get to a string theory indirectly, but without correspondence we study string theories that allow metric tensor to be dynamical. Or that has been my elementary understanding.
I can think of intuitive picture that "gravity" is supposed to be quantized, so I guess this metric tensor being dynamical should be the right understanding.
But again, thinking of Polyakov action as an example, if equation of motion removes degree of freedom (worldsheet metric comes out of target manifold metric tensor by equation of motion), I cannot see how one can make spacetime metric tensor dynamical, and by this one means that spacetime manifold is changing for sure. Target spacetime manifold/metric is already fixed, so how can "gravity" ever be quantized in Polyakov interpretation of string theories?

REPLY [4 votes]: In any string theory, the graviton arises in the spectrum of the string, that is, in the ordinary way in canonical quantisation. For the bosonic string, $\tilde\alpha^i_{-1}\alpha^j_{-1}|0\rangle$ transforms in a representation reducible to a symmetric traceless part along with others, which we identify as the graviton, $G_{\mu\nu}(X)$.
$G_{\mu\nu}(X)$ is a field which carries target manifold indices, and is a function of the embedding functions which can be viewed as coordinates, $X^\mu(\tau,\sigma)$.
Now, we can consider a non-linear sigma model, of the form,
$$S \sim \int d^2\sigma \sqrt{h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X^\nu G_{\mu\nu}(X).$$
Now comes the question: if string theory has a graviton in the spectrum, then shouldn't $G_{\mu\nu}$ be built up from these gravitons, in the same way light is comprised of photons?
The answer is yes. If we expand $G_{\mu\nu} = \eta_{\mu\nu}+f_{\mu\nu}$ then the partition function for the theory is related to the partition function of a string in flat space by,
$$Z=\int DX Dg \, e^{-S_{\mathrm{poly}}-V}$$
where
$$V\sim \int d^2\sigma \sqrt{h}h^{\alpha\beta}\partial_\alpha X^\mu \partial_\beta X^\nu f_{\mu\nu}.$$
If we now choose $f_{\mu\nu} \sim c_{\mu\nu}e^{ip\cdot X}$ then the expression agrees with the vertex operator of a graviton in string theory, and adding $e^V$ to the path integral shifts the metric by $f_{\mu\nu}$.
Thus, to recap: a massless spin 2 gauge boson arises in the spectrum of the string upon quantisation, and the background curved metric is comprised of such gravitons, in the sense described above.
Caveat: We are not directly trying to quantise the metric in string theory, or in the description above. That approach is canonically quantised gravity. However, we can from string theory derive effective actions and these include the graviton as the metric. To obtain amplitudes from these effective actions we would quantise in the usual way.


TITLE: Equivalent Affirmations Regarding a Closed $1$-Form combined with a function of class $C^1$
QUESTION [0 upvotes]: I am studying analysis by using Elon Lages Lima's book "Curso de Análise (vol. 2)" (the book is written in portuguese). I am trying to solve the following problem about path integrals but I got stuck in some items.

Let $\omega$ be a closed $1$-form, different than zero in all points of the open set $U \subset \mathbb{R^m}$, and $f: U \rightarrow \mathbb{R}$ a function of class $C^1$ with no critical points. Prove that the following affirmations are equivalent:
(a) - the form $f \cdot \omega$ is closed;
(b) - $df$ is a multiple of $\omega$;
(c) - for every $x \in U$, and every $v \in \mathbb{R^m}$, $\omega(x) \cdot v = 0 \iff \langle \text{grad } f(x), v \rangle = 0$.

Here's my attempt to solve the problem:
$(a) \implies (b)$:

Let $\omega = \sum{a_i dx_i}$. Then, $f \cdot w = \sum f a_i dx_i$. We have:
$$ \text{$f \cdot w$ is closed } \implies \frac{\partial (f a_i)}{\partial x_j} = \frac{\partial (f a_j)}{\partial x_i} \implies
 \frac{\partial f}{\partial x_j} a_i + \frac{\partial a_i}{\partial x_j} f = \frac{\partial f}{\partial x_i} a_j + \frac{\partial a_j}{\partial x_i} f $$
Since $\omega$ is a closed form we have $\frac{\partial a_i}{\partial x_j} = \frac{\partial a_j}{\partial x_i}$ and the expression above is simplified to:
$$ \frac{\partial f}{\partial x_j} a_i = \frac{\partial f}{\partial x_i} a_j$$
and therefore:
$$ a_i(x) [\frac{\partial f}{\partial x_i}(x)]^{-1} = a_j(x) [\frac{\partial f}{\partial x_j}(x)]^{-1} = k(x) $$
where $k(x)$ does not depend on the choice of the index $i$. I think I should have obtained a $k$ that also does not depend on the point $x$ to conclude that $df$ is a multiple of $\omega$, right? Or the dependency means that at every point $x$ we have $df(x) \cdot v = k(x) (\omega(x) \cdot v)$ and in this case the calculations above are enough?

$(b) \implies (c)$ - I think I was able to solve this one:

Indeed for every $x \in U$ and every $v \in \mathbb{R^m}$ we have
  $\omega(x) \cdot v = 0 \iff df(x) \cdot v = 0$ (because $df$ is a multiple of $\omega$) $\iff \langle \text{grad} f(x), v \rangle = 0$.

$ (c) \implies (a)$:

I couldn't go very far here. I notice that by the calculations done in item $(a) \implies (b)$ it suffices to prove:
$$ \frac{\partial f}{\partial x_j} a_i = \frac{\partial f}{\partial x_i} a_j$$
and I also noticed that by hypothesis, picking $v = e_i$ we have that $a_i(x) = 0 \iff \frac{\partial f}{\partial x_i}(x) = 0$. I was not able to proceed any further.

Can anyone help me in cases $(a) \implies (b)$ and $(c) \implies (a)$? Thanks in advance.

REPLY [1 votes]: HINTS: Don't write coordinates. Just write the product rule for $d(f\omega)$ and use the fact that $\omega$ is closed.
I honestly think that (c) is an inconvenient detour. I would just check (b)$\implies$ (a) (immediately) and (c)$\implies$ (b) (immediately).


TITLE: Asymptotics of $\int_1^{\infty}\frac{f(x)}{x-1-iy}\,\mathrm{d}x$ as $y\to0$
QUESTION [0 upvotes]: Consider a real analytic, positive-valued function $f:(1,\infty)\rightarrow\mathbb{R}$ satisfying
\begin{equation}
f(x)\sim(x-1)^{-\alpha}\qquad(x\to1)
\end{equation}
for some $\alpha\in(0,1)$. For all $y>0$, consider the integral
\begin{equation}
I(y)=\int_1^{\infty}\frac{f(x)}{x-1-iy}\,\mathrm{d}x,
\end{equation}
supposing that $f(x)$ vanishes at $x\to\infty$ fast enough to ensure $I(y)$ to converge. As $y\to0$, numerical evidence suggests me that $I(y)$ diverges like this:
\begin{equation}
I(y)\sim y^{-\alpha}\qquad(y\to0),
\end{equation}
i.e.
\begin{equation}
\lim_{y\to0}y^\alpha I(y)\text{  is finite, }
\end{equation}
but I am currently unable to prove it, nor to find sufficient additional conditions for this to hold.
My attempt: for every $\epsilon>0$, we can split the integral as such:
\begin{equation}
y^\alpha I(y)=\int_1^{1+\epsilon}\frac{y^\alpha f(x)}{x-1-iy}\,\mathrm{d}x+\int_{1+\epsilon}^{\infty}\frac{y^\alpha f(x)}{x-1-iy}\,\mathrm{d}x.
\end{equation}
When taking the limit, the second term in the right-hand side should converge to zero; as for the first term, heuristically speaking, when taking $\epsilon$ "sufficiently small" we should be able to "approximate" it by substituting $f(x)$ with $(x-1)^{-\alpha}$, and thus its asymptotic behaviour should be close to the one of the integral
\begin{equation}
\int_1^{1+\epsilon}\frac{y^\alpha (x-1)^{-\alpha}}{x-1-iy}\,\mathrm{d}x,
\end{equation}
which can be computed exactly and does converge to a finite value when $y\to0$. However, this is far from being an actual proof.

REPLY [0 votes]: OK, that was easier than I thought. I will write down here my procedure for completeness.
Without loss of generality, set $f(x)=g(x)/(x-1)^\alpha$ for some real analytic function $g$ in $(1,\infty)$ such that $\lim_{\lambda\to1^+}g(x)\equiv g(1)$ is finite. Then
\begin{equation}
y^\alpha I(y)=\int_1^\infty\frac{g(x)}{(x-1)^\alpha}\frac{y^\alpha}{x-1+iy}\,\mathrm{d}x.
\end{equation}
Now, performing the change of variable $x'=(x-1)/y$, we obtain
\begin{equation}
y^\alpha I(y)=\int_0^\infty\frac{g(xy+1)}{x^\alpha}\frac{1}{x+i}\,\mathrm{d}x.
\end{equation}
On the other hand, we have
\begin{equation}
\int_0^\infty\frac{x^{-\alpha}}{x+i}\,\mathrm{d}x=\pi(-i)^\alpha\csc\pi\alpha,
\end{equation}
thus, by dominated convergence,
\begin{equation}
\lim_{y\to0}y^\alpha I(y)=g(1)\pi(-i)^\alpha\csc\pi\alpha.
\end{equation}


TITLE: Limit of n!^{1/n} two ways that give distinct results. Contradiction?
QUESTION [1 upvotes]: I tried two methods to compute the limit, $\lim \limits_{n \rightarrow \infty} \sqrt[n]{n!}$ and i got distinct results.
Method 1: Observe that $(2n)! \ge \prod \limits_{k=n}^{2n} k \ge n^{n+1}$. Thus, $\sqrt[2n]{(2n)!} \ge \sqrt[2n]{n^{n+1}} \ge \sqrt{n}$. Same case for $k=2n+1$.So $\lim \sqrt[n]{n!} = +\infty$.
Method 2: We use algebra of limits, giving
$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \ldots (\lim \sqrt[n]{1}) = 1$. Since $\lim \sqrt[n]{n} = 1$.
What mistake did I make in these proofs? Thank you!

REPLY [2 votes]: You can't break apart the product like this$$\lim \sqrt[n] {n!} = (\lim \sqrt[n]{n})(\lim \sqrt[n]{n-1}) \cdots (\lim \sqrt[n]{1})$$simply because the number of terms also goes toward infinity. To illustrate this, take the related example $\lim_{n \to \infty}\sqrt[n]{n^n}$, or written out:
$$
1,\quad \sqrt2\cdot \sqrt2,\quad \sqrt[3]3\cdot \sqrt[3]3 \cdot \sqrt[3]3,\quad \sqrt[4]4\cdot \sqrt[4]4\cdot \sqrt[4]4\cdot \sqrt[4]4,\ldots
$$
which can be described as
$$
\lim_{n \to \infty}\prod_{i = 1}^n a_{i,n} = \lim_{n \to \infty}a_{1, n}a_{2, n}\cdots a_{n,n}\tag{*}
$$
where $a_{i,n} = \sqrt[n]{n}$. This product clearly has no limit, but using your second technique you still get $1$ as the answer.
Your sequence can also be described using $\text{(*)}$, but with $a_{i,n} = \sqrt[n]{i}$ instead. The argument is equally false.

REPLY [1 votes]: Your first proof is correct. The second one is not, because you cannot use "algebra of limits" as you did: in the product you write the number of terms is not fixed but is becoming larger and larger (is equal to $n$); so you cannot say say that the limit of the product is equal to the product of the limits. The same kind of things happen with infinite series: you cannot always pass to the limit under the $\sum$ sign.


TITLE: Finding a Suitable Telescoping Sum to Use in Problem
QUESTION [0 upvotes]: I am currently lost on the following problem:
Use a telescoping sum to give a proof without induction that for each $n \in \mathbb{N},$
$$1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1)^2}{4}$$
I have followed other examples where they show equalities like this, but I don't understand how they seem to come up with the telescoping series they use to solve them.
Thank you!

REPLY [0 votes]: Using the identity
$$ k^3=\frac14\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg] $$
it is easy to see
$$ \sum_{k=1}^n k^3=\frac14\sum_{k=1}^n\bigg[k^2(k+1)^2-(k-1)^2k^2\bigg]=\frac14n^2(n+1)^2. $$


TITLE: Angle between vector and one vector of general orthonormal basis
QUESTION [0 upvotes]: My question asks to consider a vector a=2e+3f+6g where {e,f,g} is an orthonormal basis. I am then asked to show that the angle between a and g belongs to the interval (π/6, π/4). I can show this easily using scalar product if the basis is the standard orthonormal basis, however i am unsure how to prove this for all orthonormal bases. Any guidance would be appreciated.

REPLY [0 votes]: Hint: the $\cos$ of the angle between $a$ and $g$ is $\frac{\langle a,g\rangle}{\|a\|\cdot\|g\|}$. Now why is the numerator $6$ and the denominator $7$?