TITLE: What corresponds to the operation of taking traces in of the Fourier transformation on a finite group?
QUESTION [3 upvotes]: I have a question about the Fourier transfomation on a finite non-comutative group. I hope that it is a known fact in the Representation Theory but I cannot find it written explicitly in textbooks.
Let $G$ be a finite non-abelian group, $\hat{?}:L_2(G)\to \hat L_2(\hat G)$ be its Fourier transformation and $\check{?}:\hat L_2(\hat G)\to L_2(G)$ be the inverse Fourier transformation. The space $L_2(G)$ carries the standard norm induced by the inner product $\langle x,y\rangle=\sum_{g\in G}x(g)\cdot\overline{y(g)}$. The norm on $\hat L_2(\hat G)$ is normalized so that the Fourier transformations $\hat?$ is an isometry.
Let $P:\hat L_2(\hat G)\to \hat L_2(\hat G)$ be the function assigning to a sequence of matrices $(M_\alpha)_{\alpha\in\hat G}$ in $\hat L_2(\hat G)$ the sequence $(\frac 1{\dim(\alpha)}Tr(M_\alpha)\cdot\mathrm{Id}_{\alpha})_{\alpha\in\hat G}$ of normalized traces of those matrices multiplied by the identity matrices. So, $P$ is a linear projection of $\hat L_2(\hat G)$ onto its $|\hat G|$-dimensional subspace.

Question 1. What is the norm of a sequence $(M_\alpha)_{\alpha\in \hat G}\in P(\hat L_2(\hat G))$ in the Hilbert space $\hat L_2(\hat G)$? Is it equal to $(\frac1{|\hat G|}\sum_{\alpha\in\hat G}|\frac1{\dim(\alpha)}Tr(M_\alpha)|^2)^{1/2}$?

I am interested in describing the projector $A=\check{?}\circ P\circ \hat{?}:L_2(G)\to L_2(G)$. It should assign to each function $f\in L_2(G)$ some class function on $G$.

Question 2. Is $A$ the averaging of $f$ over conjugacy classes? If yes, where can I find a reference to this fact?

REPLY [3 votes]: Exercise 6.2 in Serre's book on representation theory of finite groups can be interpreted as saying that for functions $f,h$ on $G$ one has
$$\sum_{g\in G}\overline{f(g)}{h(g)} = \dfrac{1}{|G|}\sum_{\rho \in \hat G}\dim\rho\cdot \mathrm{Tr}(\hat f(\rho)^*\hat h(\rho))$$ and so the correct formula for the norm should be $$\|(M_{\rho})_{\rho\in \hat G}\|^2 = \dfrac{1}{|G|}\sum_{\rho \in \hat G}\dim\rho\cdot \mathrm{Tr}(M^*_{\rho}M_\rho))$$ where $*$ denotes the Hermitian adjoint. See Teras and CECCHERINI-SILBERSTEIN et al for this version.  Serre uses a slightly different bilnear form that agrees with this one on characters.
For the second question you are correct I will use the second orthogonality relations which says that $$\sum_{\rho\in \hat G}\overline{\mathrm{Tr}(\rho(g))}\mathrm{Tr}(\rho(h))$$ is $0$ unless $g,h$ are conjugate, in which case it is $\dfrac{|G|}{|\mathrm{Cl}(g)|}$ where $\mathrm{Cl}(g)$ is the conjugacy class of $g$.
Now $P\hat f$ is the sequence whose $\rho$-component is $$\sum_{h\in G}\frac{f(h)}{\dim \rho}\mathrm{Tr}(\rho(h))I$$.
The Fourier inversion theorem says that if $k$ is the function with $\hat k=P\hat f$, then $$k(g) =\frac{1}{|G|}\sum_{\rho\in \hat G}\dim \rho\cdot \mathrm{Tr}(\rho(g)^*\hat k(\rho))$$
$$=\frac{1}{|G|}\sum_{h\in G}f(h)\sum_{\rho\in \hat G}\overline{\mathrm{Tr}(\rho(g))}\mathrm{Tr}(\rho(h))=\sum_{h\sim g}\dfrac{f(h)}{|\mathrm{Cl}(g)|}$$ where $\sim$ is conjugacy by the second orthogonality relations, which is the average value of $f$ on the conjugacy class of $g$.


TITLE: Is it possible to create a new language that translates into words at least the basic operations of math, to make it easier to understand?
QUESTION [0 upvotes]: I was wondering if it's possible to make maths easier to calculate and understand, and more natural for kids and even adults. And then I realized how easy it is for human beings to manipulate words and from them make whole sentences and express complex concepts easily than what we do with maths.

So my question: Is it possible to create a new language that translates into words at least the basic operations of math? and if it is, how do we make those operations with words?

I hope you guys understand my question and that I'm not trying to reinvent math. I'm trying to make it easier for the daily routine and to kids and adults that have been struggling with math as myself.
Sorry for any mistakes I might made, as English isn't my native language.

REPLY [0 votes]: Thanks you both for take time to answer my question, and i think that what we re dealing its how Stephen Goree said disciplane litteracy. Im gonna keep studying math education because as its is said knowledge is power, thanks again.


TITLE: invariant 2-form in hyperbolic 3-space
QUESTION [2 upvotes]: Hello all
As is probably well-known to most, in the upper halfplane we have a natural action of $SL_2(\mathbb{R})$ through linear fractional transformations, and a $2$-form $\frac{dx dy}{y^2}$ which is invariant under this action.
In the case of the upper half-space $H_3 = \{ (z,t) \ | \ z\in \mathbb{C} \ , \ t>0 \ \}$ we have an analogous situation, only now we have a natural action of all of $SL_2(\mathbb{C})$. The fastest way of defining this action is by identifying a point $(z,t)$ in $H_3$ with the quaternion $q:= z+t j$, and then setting
$$   \begin{pmatrix} a&b\cr c& d\cr \end{pmatrix}.(z,t) := (aq+b)(cq+d)^{-1}$$
Now for my question; I want to try to construct an $SL_2(\mathbb{Z}[i])$-invariant two-form on $H_3$ which when restricted to the plane $Im(z)=0$ takes the form $\frac{dxdt}{t^2}$. Do such a thing exist ? How would I go about constructing it? Any thoughts/references would be greatly appreciated.

REPLY [8 votes]: In upper half space, the thing that's analogous to the invariant 2-form (which measures hyperbolic area) is the hyperbolic volume form $1/z^3 dx dy dz$.
If you want to understand 2-forms invariant by a particular discrete group, they are
the same thing as 2-forms on the quotient orbifold. In the case of $SL_2$ of the algebraic integers in a quadratic imaginary field such as this, the groups are usually called Bianchi
groups: Bianchi worked out fundamental domains for them for the first set of examples, as well as a general technique to find fundamental domains.  On Allen Hatcher's website, there's a handy list of pictures of the folded-up fundamental domains in cases where they can be
readily drawn.
In this case, the quotient orbifold has underlying space $S^3$ minus a point
(for the cusp), with singular locus in the form of the 1-skeleton of a tetrahedron. Since the underlying space is contractible, any closed 2-form is a coboundary --- i.e.
there are plenty of 2-forms, closed or otherwise, but there's no obvious reason they
should be interesting. In this particular
case, the diagonal matrix with entries $i, -i$ rotates 180 degrees about the
$z$-axis, acting as an orientation-reversing map on the hyperbolic plane, so it takes the 2-form above to its negative: it's not invariant even
before attempting to extend it.


TITLE: Cartan subalgebra of ${\frak so}(5)$
QUESTION [3 upvotes]: I would like to find effectively the Cartan subalgebra of ${\frak so}(5)$. Does anybody knows how to proceed?
Edit: I don't want to start from the simple roots and then derive it. I would like to do the pedagogical procedure, starting from the cartan subalgebra and identifying the simple roots.

REPLY [1 votes]: For the Cartan subalgebra of $B_2$ (in fact, for all classical simple Lie algebras), see this handout. Here $B_2$ is the root system of $\mathfrak{so}(5)$.


TITLE: Avoiding impossible states in an adventure game
QUESTION [2 upvotes]: I've already asked this on gamedev at https://gamedev.stackexchange.com/questions/15161/avoiding-impossible-states-in-an-adventure-game, but I think I'm after more of a mathematical answer than a programming one.  Here is the question:
I'm looking at creating a fairly complex choose-your-own adventure style of game, but I'm looking for a technique or method to help with the game design.
It will be a multi-branching story, and at times the branches will fold back in on themselves, and your major actions will be remembered and items will be collected. For example, if the player goes to the swamp and the forest before going to the castle he has picked up a dinosaur bone, killed a unicorn and grown an extra arm. If a player gets to the castle via the caves and the crypt he has picked up a bicycle and a strange smell. Now when developing puzzles for the castle I don't want to create a situation where two impossibilities are requires, for example - an extra arm and the smell are needed to kill an ogre.
Other than documenting the events and items very very carefully is there a process, a technique I can use in a spreadsheet or a piece of software that can help me?

REPLY [3 votes]: You obviously know you're writing a state engine.  In this situation, I'd probably consider the castle multiple states instead of one.
That is to say, the "swamp forest castle" and the "crypt castle" are different castles.  In your code, you may render them the same, but the states are distinct because they've resulted in arriving at a destination with different assets.
Another way to think of it is to say, "at the castle isn't really my state . . . at the castle with a bone and a smell are my state."

REPLY [1 votes]: If there are $n$ locations and $m$ items/qualities that one can either have or not have, there are (a priori) $n\cdot2^m$ possible states.  If that total isn't too large—or even if the reachable part of it isn't—it should be possible to write a program to walk through all of the possible states.
Then you can simply see if after traversing all possible states if the node in question has been reached.


TITLE: Is every basis of a finite-dimensional vector space orthonormal with respect to some inner product?
QUESTION [3 upvotes]: Given a real or complex vector space $V$ and a (finite) basis $B$ of it, does it always exist an inner product on $V$ such that $B$ is an orthonormal basis with respect to it?
The question is equivalent to asking: is there always a positive definite (symmetric) matrix $A$ such that if $B=\{v_1,...,v_n\}$, then
$$v_i^tAv_j=\delta_{ij}\;?$$

REPLY [0 votes]: Another answer with a slightly different approach: what you're asking about here is a statement on bilinear forms or sesquilinear forms, in the complex case. For convenience, I'll stick to the complex case, but all of this can, with care, be handled in the case of a real vector space.
Note that an inner product $(\cdot,\cdot):V \times V \to \Bbb C$ on a complex vector space $\Bbb C$ is simply a sesquilinear form with the additional property that:

$(x,x) \geq 0$ for all $x$
In the real case: $(x,y) = (y,x)$ (this follows from the above over $\Bbb C$)
$(x,x) = 0 \iff x = 0$

From there, we use the fact that every sesquillinear form can be written as
$$
(x,y) = x^* Ay
$$
for some complex matrix $A$, whose entries are
$$
a_{ij} = e_i^* A e_j
$$
By the above, the three requirements on our sesquilinear form are precisely the statement that $A$ is positive definite.


TITLE: Representation theory of symmetric groups over Q-algebras
QUESTION [0 upvotes]: Let $k$ be a commutative $\mathbb{Q}$-algebra, and let $\Sigma_n$ denote the symmetric group on $n$ letters. What is known about the classification of $k$-linear representations of $\Sigma_n$? (Of course, it will not be as concrete as the classification as in the case that $k$ is a field, but we may treat $k$-modules as already been classified and that these may be part of the classification data.) Can someone point me to relevant literature on this?

REPLY [1 votes]: From the standard representation theory of $\Sigma_n$, it follows that the group algebra $\mathbb{Q}\Sigma_n$ is a product of matrix rings over $\mathbb{Q}$.
So $k\Sigma_n\cong\mathbb{Q}\Sigma_n\otimes_\mathbb{Q}k$ is a corresponding product of matrix rings over $k$, which is Morita equivalent to a product of copies of $k$.
Concretely, if $V_1,\dots,V_r$ are the irreducible $\mathbb{Q}\Sigma_n$-modules, then the $k\Sigma_n$-modules are of the form
$$\bigoplus_{i=1}^r M_i\otimes_\mathbb{Q}V_i,$$
where $M_1,\dots,M_r$ are $k$-modules.


TITLE: Lipschitz continuity and sup of derivative norm
QUESTION [8 upvotes]: On this wikipedia page, it is stated:

For a differentiable Lipschitz map $f : U \rightarrow R^m$ the inequality
  $\|Df\|_{\infty,U}\le K$ holds for the best Lipschitz constant of $f$, and
  it turns out to be an equality if the domain $U$ is convex.

I have two questions about this statement.
Question 1: does a best Lipschitz constant necessarily exists? Shouldn't be $K$ the $\inf$ of all Lipschitz constants?
Question 2: do you know a proof of the wikipedia statement, or a reference where this is proved?
I tried to prove it by using Taylor expansion with integral remainder (which seems to make sense with the convex assumption for equality), but it did not work...
Thank you.

REPLY [11 votes]: Question 1:  let $K$ be the inf of all Lipschitz constants. For every $\epsilon>0$, there is $K\leq L<K+\epsilon$ with $L$ Lipschitz constant. So
$$
\|f(x)-f(y)\|\leq L\|x-y\|\leq (K+\epsilon)\|x-y\|\qquad \forall x,y\in U\quad \forall \epsilon>0.
$$
Lettting $\epsilon$ tend to $0$ shows that $K$ is itself a Lipschitz constant, hence it is a min. In other terms, the set of all Lipschitz constants is of the form $[K,+\infty)$. Of course, we could have observed from the beginning that it had to be $(K,+\infty)$ or $[K,+\infty)$ since $M$ is a Lipschitz constant for every $M\geq L$ if $L$ is a Lipschitz constant.
Question 2.1:  fix $x\in U$ and take any $h$ in the ambiant normed space containing $U$ (I guess it is $\mathbb{R}^n$ in your case, but everything works the same in the general case). Then for $t\in\mathbb{R}$ close enough to $0$, $x+th$ belongs to $U$ and
$$\|f(x+th)-f(x)\|\leq K\|x+th-x\|=K|t|\|h\|\quad\Rightarrow\quad \frac{\|f(x+th)-f(x)\|}{|t|}\leq K\|h\|.$$
Letting $t$ tend to $0$ yields
$$
\|Df_x(h)\|\leq K\|h\|\quad \forall h\quad\Rightarrow \quad \|Df_x\|\leq K \quad \forall x\in U\quad\Rightarrow\quad \|Df\|_{\infty,U}\leq K.
$$
Question 2.2:  now assume $U$ is convex. And set $L:=\|Df\|_{\infty,U}$. Fix $x,y\in U$ and note that the segment $[x,y]:=\{x_t=(1-t)x+ty\,;\, t\in[0,1]\}$ is contained in $U$ by assumption. We will show that $\|f(x)-f(y)\|\leq L\|x-y\|$, which will prove that $L$ is a Lipschitz constant, whence $K\leq L$ and finally $K=\|Df\|_{\infty,U}$ in this case.
Recall that we denote $x_t=(1-t)x+ty$. Fix $\epsilon>0$ and consider
$$
I_\epsilon:=\{t\in[0,1]\,;\,\|f(x)-f(x_s)\|\leq (L+\epsilon)\|x-x_s\|\;\forall 0\leq s\leq t\}.
$$
This is clearly a closed interval of the form $[0,t_0]$ in $[0,1]$. We will show that $t_0=1$. Since $f$ is differentiable at $x_{t_0}$ and $\|Df_{x_{t_0}}\|\leq L$, there exists $\delta>0$ such that
$$
\|f(y)-f(x_{t_0})\|\leq \|Df_{x_{t_0}}(y-x_{t_0})\|+\epsilon \|y-x_{t_0}\|\leq (L+\epsilon)\|y-x_{t_0}\|\quad\forall \|y-x_{t_0}\|<\delta.
$$
It follows that
$$
\|f(x)-f(x_s)\|\leq \|f(x)-f(x_{t_0})\|+\|f(x_{t_0})-f(x_s)\|\leq (L+\epsilon)(\|x-x_{t_0}\|+\|x_{t_0}-x_s\|)
$$
$$
=(L+\epsilon)(\|t_0(y-x)\|+\|(s-t_0)(y-x)\|)=(L+\epsilon)s\|y-x\|=(L+\epsilon)\|x-x_s\|
$$
for all $t_0\leq s\leq t_0+\frac{\delta}{\|y-x\|}$. This means that $t_0$ can not be the sup of $I_\epsilon$, unless $t_0=1$. In particular
$$
\|f(x)-f(x_1)\|\leq (L+\epsilon)\|x-x_1\| \quad \mbox{i.e.}\quad \|f(x)-f(y)\|\leq (L+\epsilon)\|x-y\| .
$$
It only remains to let $\epsilon$ tend to $0$ to deduce that $L$ is indeed a Lipschitz constant as desired.
Remark: what makes the argument lengthy in 2.2 is that there is no mean value theorem for functions with range in a space of dimension greater than one. As we have seen, we still have a "mean inequality theorem", though, when the domain is convex. But 2.2 is much easier if $Df$ is assumed to be continuous. In this case, it suffices to write
$$
f(y)-f(x)=\int_0^1Df_{x_t}(y-x)dt \Rightarrow  \|f(y)-f(x)\|\leq \int_0^1\|Df_{x_t}(y-x)\|dt\leq \|Df\|_{\infty,U}\|y-x\|.
$$


text:
\begin{document}
\title{The large diffusion limit
for the heat equation\\
with a dynamical boundary condition}

\author{
Marek Fila\\
Department of Applied Mathematics and Statistics\\
Comenius University,\\
84248 Bratislava, Slovakia
\\
\\
Kazuhiro Ishige\\
Graduate School of Mathematical Sciences\\
The University of Tokyo\\
3-8-1 Komaba, Meguro-ku, Tokyo 153-8914, Japan
\\
\\
Tatsuki Kawakami\\
Department of Applied Mathematics and Informatics,\\
Ryukoku University\\
Seta Otsu 520-2194, Japan
}
\date{}
\maketitle

\begin{abstract}
We study the heat equation on a half-space with a linear dynamical boundary condition.
Our main aim is to show that, if the diffusion coefficient tends to infinity, then the solutions
converge (in a suitable sense) to solutions of the Laplace equation with the same dynamical
boundary condition.
\end{abstract}

\vspace{10pt}
Keywords: heat equation, dynamical boundary condition,
 large diffusion limit

\section{Introduction}

We consider the problem
\begin{equation}
\label{eq:1.1}
\left\{
\begin{array}{ll}
\displaystyle{\varepsilon \partial_t u_\varepsilon-\Delta u_\varepsilon=0}, & x\in{\mathbb R}^N_+,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\partial_tu_\varepsilon+\partial_\nu u_\varepsilon=0}, & x\in\partial{\mathbb R}^N_+,\,\,\, t>0,\vspace{5pt}\\
\displaystyle{u_\varepsilon(x,0)=\varphi(x)},\qquad & x\in{\mathbb R}^N_+,
\vspace{5pt}\\
\displaystyle{u_\varepsilon(x,0)=\varphi_b(x')},\qquad & x=(x',0)\in\partial{\mathbb R}^N_+,
\end{array}
\right.
\end{equation}
where $N\ge 2$, ${\mathbb R}^N_+:={\mathbb R}^{N-1}\times{\mathbb R}_+$,
$\Delta$ is the $N$-dimensional Laplacian (in $x$),
$\partial_t:=\partial/\partial t$, $\partial_\nu:=-\partial/\partial x_N$, $\varepsilon\in(0,1)$ and
$\varphi$ and $\varphi_b$ are measurable functions
in ${\mathbb R}^N_+$ and ${\mathbb R}^{N-1}$, respectively.
Our main aim is to show that, as $\varepsilon\to 0$, it holds that $u_\varepsilon\to u$ (in a suitable sense),
where $u$ is the solution of
\begin{equation}
\label{eq:1.2}
\left\{
\begin{array}{ll}
\displaystyle{\Delta u=0}, & x\in{\mathbb R}^N_+,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\partial_tu+\partial_\nu u=0}, & x\in\partial{\mathbb R}^N_+,\,\,\, t>0,\vspace{5pt}\\
\displaystyle{u(x,0)=\varphi_b(x')},\qquad & x=(x',0)\in\partial{\mathbb R}^N_+.
\end{array}
\right.
\end{equation}
This convergence does not look unexpected, see \cite{AAIY}, but we are not aware of any previous result which would
support this natural conjecture. In particular, convergence of this type means that the influence of the initial function
$\varphi$ is lost in the limit, and we shall describe this phenomenon in more detail.

A result in a similar spirit was established in \cite{AAIY} for the eikonal equation with the same dynamical boundary condition
as in \eqref{eq:1.1}. More precisely, the following problem was considered in \cite{AAIY}:
$$
\left\{
\begin{array}{ll}
\displaystyle{\varepsilon \partial_t u_\varepsilon+ |\nabla_x u_\varepsilon|=1}, & x\in\Omega,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\partial_tu_\varepsilon+\partial_\nu u_\varepsilon=0}, & x\in\partial\Omega,\,\,\, t>0,\vspace{5pt}\\
\displaystyle{u_\varepsilon(x,0)=\varphi(x)},\qquad & x\in\overline\Omega,
\end{array}
\right.
$$
where $\varepsilon\in(0,1)$, $\Omega\subset {\mathbb R}^N$ is a bounded domain with $C^1$-boundary,
and $\nu$ is the outer normal of $\partial\Omega$. It was shown in \cite{AAIY} that $u_\varepsilon\to u$
as $\varepsilon\to 0$, where $u$ is the solution of
\[
\left\{
\begin{array}{ll}
\displaystyle{|\nabla_x u|=1}, & x\in\Omega,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\partial_tu+\partial_\nu u=0}, & x\in\partial\Omega,\,\,\, t>0,
\end{array}
\right.
\]
with an appropriate initial condition.

In the context of diffusion, the boundary condition from
\eqref{eq:1.1} describes thermal contact with a perfect conductor or
diffusion of solute from a well-stirred fluid or vapour (see e.g. \cite{C}). Various aspects of analysis of parabolic
and elliptic equations with dynamical boundary conditions have been treated by many authors, see for example
\cite{AQR}--\cite{BPR}, \cite{EMR, E2, E3, FQ2, FV, GH, H, KI}, \cite{VV1}--\cite{V1} for the parabolic case
and \cite{E1}, \cite{E4}--\cite{FQ1}, \cite{GM}, \cite{KI}--\cite{L}, \cite{V2, Y} for the elliptic one. Here we
demonstrate on the simplest possible linear example how are these two classes of problems linked.

Throughout this paper we often identify ${\mathbb R}^{N-1}$ with $\partial{\mathbb R}^N_+$.
We introduce some notation.
Let $\Gamma_D=\Gamma_D(x,y,t)$ be the Dirichlet heat kernel on ${\mathbb R}^N_+$, that is,
\begin{equation}
\label{eq:1.3}
\Gamma_D(x,y,t):=(4\pi t)^{-\frac{N}{2}}
\left[\exp\biggr(-\frac{|x-y|^2}{4t}\biggr)-\exp\biggr(-\frac{|x-y_*|^2}{4t}\biggr)\right]
\end{equation}
for $(x,y,t)\in\overline{{\mathbb R}^{N}_+}\times{\mathbb R}^{N}_+\times(0,\infty)$,
where $y_*=(y',-y_N)$ for $y=(y',y_N)\in{\mathbb R}^N_+$.
Define
\begin{equation}
\label{eq:1.4}
[S_1(t)\phi](x):=\int_{{\mathbb R}^N_+}\Gamma_D(x,y,t)\phi(y)\, dy,
\quad (x,t)\in\overline{{\mathbb R}^N_+}\times(0,\infty),
\end{equation}
for any measurable function $\phi$ in ${\mathbb R}^N_+$.
For $x=(x',x_N)\in\overline{{\mathbb R}^N_+}$ and $t>0$,
set
$$
P(x',x_N,t):=c_N(x_N+t)^{1-N}\left(1+\left|\frac{x'}{x_N+t}\right|^2\right)^{-\frac{N}{2}},
$$
where $c_N$ is a constant chosen so that
$$
\int_{{\mathbb R}^{N-1}}P(x',x_N,t)\,dx'=1\quad\mbox{for all $x_N\ge 0$ and $t>0$}.
$$
Then $P=P(x',x_N,t)$ is the fundamental solution of
the Laplace equation in ${\mathbb R}^N_+$
with the homogeneous dynamical boundary condition, that is,
$P$ satisfies
$$
\left\{
\begin{array}{ll}
\displaystyle{-\Delta P=0}, & x\in{\mathbb R}^N_+,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\partial_tP+\partial_\nu P=0}, & x\in\partial{\mathbb R}^N_+,\,\,\, t>0,\vspace{5pt}\\
\displaystyle{P(x,0)=\delta(x')},\qquad & x=(x',0)\in\partial{\mathbb R}^N_+,
\end{array}
\right.
$$
where $\delta=\delta(\cdot)$ is the Dirac delta function on $\partial{\mathbb R}^N_+={\mathbb R}^{N-1}$.
Define
\begin{equation}
\label{eq:1.5}
[S_2(t)\psi](x):=\int_{{\mathbb R}^{N-1}}P(x'-y',x_N,t)\psi(y')\, dy',
\quad (x,t)\in\overline{{\mathbb R}^N_+}\times(0,\infty),
\end{equation}
for any measurable function $\psi$ in ${\mathbb R}^{N-1}$.
\vspace{5pt}

We formulate the definition of a solution of \eqref{eq:1.1}
by the use of the two integral kernels $\Gamma_D$ and $P$.
For simplicity, let $\varphi_b=\varphi_b(x')$ and $g=g(x',t)$ be continuous functions
in ${\mathbb R}^{N-1}$ and ${\mathbb R}^{N-1}\times(0,\infty)$, respectively, such that
$\varphi_b(x')$ and $g(x',t)$ decay rapidly as $|x'|\to\infty$.
Then the function
\begin{equation}
\begin{split}
\label{eq:1.6}
 & w(x,t)=w(x',x_N,t)

 :=[S_2(t)\varphi_b](x)+\int_0^t[S_2(t-s)g(s)](x)\,ds

\end{split}
\end{equation}
can be defined for $x=(x',x_N)\in {\mathbb R}^N_+$ and $t>0$
and it is a classical solution of the Cauchy problem for the Laplace equation with a nonhomogeneous dynamical boundary condition
\begin{equation}
\label{eq:1.7}
\left\{
\begin{array}{ll}
\displaystyle{-\Delta w=0},
& x\in{\mathbb R}^N_+,\quad t>0,\vspace{5pt}\\
\displaystyle{\partial_tw+\partial_\nu w=g},
& x\in\partial{\mathbb R}^N_+,\quad t>0,\vspace{5pt}\\
\displaystyle{w(x,0)=\varphi_b(x')},\quad
&x=(x',0)\in\partial{\mathbb R}^N_+.
\end{array}
\right.
\end{equation}
It follows from \eqref{eq:1.6} that
\begin{equation}
\label{eq:1.8}
\begin{split}
\partial_tw(x,t):= & \,\,\int_{{\mathbb R}^{N-1}}\partial_tP(x'-y',x_N,t)\varphi_b(y')\,dy'
\\
&\,\,\,\,
+\int_{{\mathbb R}^{N-1}}P(x'-y',x_N,0)g(y',t)\,dy'\\
 & \,\,\,\,\,\,
+\int_0^t\int_{{\mathbb R}^{N-1}}\partial_tP(x'-y',x_N,t-s)g(y',s)\,dy'\,ds
\end{split}
\end{equation}
for $x=(x',x_N)\in{\mathbb R}^N_+$ and $t\in(0,T)$.

Set
\begin{equation}
\label{eq:1.9}
\Phi(x):=\varphi(x)-[S_2(0)\varphi_b](x).
\end{equation}
Then the function
$$
v_\varepsilon(x,t)
:=[S_1(\varepsilon^{-1}t)\Phi](x)
-\int_0^t[S_1(\varepsilon^{-1}(t-s))]\partial_t w(s)](x)\, ds

$$
satisfies
\begin{equation}
\label{eq:1.10}
\left\{
\begin{array}{ll}
\displaystyle{\varepsilon \partial_tv_\varepsilon=\Delta v_\varepsilon -\varepsilon\partial_t w

},\quad
& x\in{\mathbb R}^N_+,\quad t>0,\vspace{5pt}\\
\displaystyle{v_\varepsilon=0},
& x\in\partial{\mathbb R}^N_+,\quad t>0,\vspace{5pt}\\
\displaystyle{v_\varepsilon(x,0)=\Phi(x)},
&x\in{\mathbb R}^N_+.
\end{array}
\right.
\end{equation}
Let $\partial_{x_N}:=\partial/\partial x_N$.
If $g_\varepsilon(x',t):=\partial_{x_N} v_\varepsilon(x',0,t)$ for $x'\in{\mathbb R}^{N-1}$, $t>0$, and
$w_\varepsilon$ is defined as in \eqref{eq:1.6} with $g_\varepsilon$ instead of $g$,
then it follows from \eqref{eq:1.7}, \eqref{eq:1.8} and \eqref{eq:1.10} that $v_\varepsilon$ and $w_\varepsilon$
satisfy
\begin{equation}
\label{eq:1.11}
\left\{
\begin{array}{ll}
\displaystyle{\varepsilon \partial_t v_\varepsilon=\Delta v_\varepsilon-\varepsilon F_1[\varphi_b]-\varepsilon F_2[v_\varepsilon]},
\qquad & x\in{\mathbb R}^N_+,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{\Delta w_\varepsilon=0}, & x\in{\mathbb R}^N_+,\,\,\,t>0,\vspace{5pt}\\
\displaystyle{v_\varepsilon=0},\quad
\displaystyle{\partial_tw_\varepsilon-\partial_{x_N} w_\varepsilon=\partial_{x_N}v_\varepsilon}, & x\in\partial{\mathbb R}^N_+,\,\,\, t>0,\vspace{5pt}\\
\displaystyle{v_\varepsilon(x,0)=\Phi(x)},\qquad & x\in{\mathbb R}^N_+,\vspace{5pt}\\
\displaystyle{w_\varepsilon(x,0)=\varphi_b(x')}, & x=(x',0)\in\partial{\mathbb R}^N_+,
\end{array}
\right.
\end{equation}
where
\begin{align}
\label{eq:1.12}
F_1[\varphi_b](x,t) &
:=\int_{{\mathbb R}^{N-1}}\partial_t P(x'-y',x_N,t)\varphi_b(y')\,dy',\\
\begin{split}
\label{eq:1.13}
F_2[v](x,t) &
:=\int_{{\mathbb R}^{N-1}}P(x'-y',x_N,0)\partial_{x_N}v(y',0,t)\,dy'\\
 &
\quad
+\int_0^t\int_{{\mathbb R}^{N-1}}\partial_t P(x'-y',x_N,t-s)\partial_{x_N}v(y',0,s)\,dy'\,ds.
\end{split}
\end{align}
Furthermore, the function $u_\varepsilon:=v_\varepsilon+w_\varepsilon$ is a classical solution of \eqref{eq:1.1}.
Motivated by this observation,
we formulate the definition of a solution of \eqref{eq:1.1} via problem~\eqref{eq:1.11}.

\begin{definition}
\label{Definition:1.1}
Let $\varphi$ and $\varphi_b$ be measurable functions
in ${\mathbb R}^N_+$ and ${\mathbb R}^{N-1}$, respectively.
Let $0<T\le\infty$ and
$$
v_\varepsilon,\,\,\partial_{x_N}v_\varepsilon,\,\, w_\varepsilon\in C(\overline{{\mathbb R}^N_+}\times(0,T)).
$$
We call $(v_\varepsilon,w_\varepsilon)$ a solution of \eqref{eq:1.11} in ${\mathbb R}^N_+\times(0,T)$
if $v_\varepsilon$ and $w_\varepsilon$ satisfy
\begin{align*}
\begin{split}
v_\varepsilon(x,t)
 & =[S_1(\varepsilon^{-1}t)\Phi](x)
-\int_0^t[S_1(\varepsilon^{-1}(t-s))F_1[\varphi_b](s)](x)\,ds\\
 & \qquad\qquad\qquad\qquad\quad\,\,\,
-\int_0^t[S_1(\varepsilon^{-1}(t-s))F_2[v_\varepsilon](s)](x)\,ds,
 \end{split}
 \notag
\\
w_\varepsilon(x,t)
&
=[S_2(t)\varphi_b](x)+\int_0^t[S_2(t-s)\partial_{x_N}v_\varepsilon(s)](x)\, ds,
\end{align*}
for $x\in\overline{{\mathbb R}^N_+}$ and $t\in(0,T)$.

In the case when $T=\infty$, we call $(v_\varepsilon,w_\varepsilon)$ a global-in-time solution of \eqref{eq:1.11}
and $u_\varepsilon$ a global-in-time solution of \eqref{eq:1.1}.
\end{definition}

We are ready to state the main results of this paper.
For $1\le r\le\infty$, we write
$|\cdot|_{L^r}:=\|\cdot\|_{L^r(\partial{\mathbb R}^N_+)}$ and $\|\cdot\|_{L^r}:=\|\cdot\|_{L^r({{\mathbb R}^N_+})}$
for simplicity.

\begin{theorem}
\label{Theorem:1.1}
Let $N\ge 2$, $\varepsilon\in(0,1)$, $\varphi\in L^\infty({\mathbb R}^N_+)$ and $\varphi_b\in L^\infty({\mathbb R}^{N-1})$.
Then problem~\eqref{eq:1.11} possesses a unique global-in-time solution
$(v_\varepsilon,w_\varepsilon)$
satisfying
\begin{equation}
\label{eq:1.14}
\sup_{0<t<T}\,\left[\|v_\varepsilon(t)\|_{L^\infty}
+(\varepsilon^{-1}t)^{\frac{1}{2}}\|\partial_{x_N}v_\varepsilon(t)\|_{L^\infty}
+\|w_\varepsilon(t)\|_{L^\infty}\right]<\infty
\end{equation}
for any $T>0$.
Furthermore,
$v_\varepsilon$ and $w_\varepsilon$ are bounded and smooth in $\overline{{\mathbb R}^N_+}\times I$
for any bounded interval $I\subset(0,\infty)$ and have the following properties for any $\tau>0$:
\begin{itemize}
  \item[{\rm (a)}]
  There exists $C(\tau)>0$ such that
  \begin{equation*}
  \begin{split}
   &
  \sup_{0<t<\tau}\,\left[\|v_\varepsilon(t)\|_{L^\infty}
  +(\varepsilon^{-1}t)^{\frac{1}{2}}\|\partial_{x_N}v_\varepsilon(t)\|_{L^\infty}
  +\|w_\varepsilon(t)\|_{L^\infty}\right]
  \\
  & \qquad\qquad\qquad\qquad\qquad\qquad\qquad
  \le C_\tau\bigg(\|\varphi\|_{L^\infty}+|\varphi_b|_{L^\infty}\bigg);
  \end{split}
  \end{equation*}
  \item[{\rm (b)}]
  $\displaystyle{\lim_{\varepsilon\to0}\sup_{0<t<\tau}\, t^{\frac{1}{2}}\|v_\varepsilon(t)\|_{L^\infty({\mathbb R}^{N-1}\times(0,L))}=0}$
  for any $L>0$;
  \item[{\rm (c)}]
$\displaystyle{\lim_{\varepsilon\to0}\sup_{0<t<\tau}\,\|w_\varepsilon(t)-S_2(t)\varphi_b\|_{L^\infty}=0}$.
\end{itemize}
\end{theorem}

As a corollary of Theorem~\ref{Theorem:1.1},
we see that the solution $u_\varepsilon=v_\varepsilon+w_\varepsilon$ of \eqref{eq:1.1}
converges to the solution $S_2(t)\varphi_b$ of \eqref{eq:1.2}.

\begin{corollary}
\label{Corollary:1.1}
Assume the same conditions as in Theorem~{\rm\ref{Theorem:1.1}}.
Let

$(v_\varepsilon,w_\varepsilon)$ be the solution given in Theorem~{\rm\ref{Theorem:1.1}}.
Then $u_\varepsilon=v_\varepsilon+w_\varepsilon$ is a classical solution of \eqref{eq:1.1} and it satisfies
$$
\lim_{\varepsilon\to 0}\sup_{\tau_1<t<\tau_2}\,\|u_\varepsilon(t)-S_2(t)\varphi_b\|_{L^\infty(K)}=0
$$
for any compact set $K$ in $\overline{{\mathbb R}^N_+}$ and $0<\tau_1<\tau_2<\infty$.
\end{corollary}

We prepare some useful lemmata in Section~2 and then we give a proof of Theorem~\ref{Theorem:1.1} in Section~3.

\section{Preliminaries}

In this section
we prove several lemmata on $S_1(t)\phi$, $F_1[\varphi_b]$ and $F_2[v]$.
In what follows, by the letter $C$
we denote generic positive constants (independent of $x$ and $t$)
and they may have different values also within the same line.
\vspace{3pt}

We first recall the following properties of $S_1(t)\phi$
(see e.g. \cite{GGS}).

\begin{itemize}
\item[(${\rm G_1}$)]
For any $1\le q\le r\le\infty$,
$$
\|S_1(t)\phi\|_{L^r}\le C t^{-\frac{N}{2}(\frac{1}{q}-\frac{1}{r})}\|\phi\|_{L^q},\qquad t>0,
$$
for all $\phi\in L^q({\mathbb R}^N_+)$.
In particular, if $q=r$, then
\begin{equation}
\label{eq:2.1}
\sup_{t>0}\,\|S_1(t)\phi\|_{L^r}\le \|\phi\|_{L^r}.
\end{equation}
\item[(${\rm G_2}$)]
Let $\phi\in L^q({\mathbb R}^N_+)$ with $1\le q\le\infty$.
Then, for any $T>0$,
$S_1(t)\phi$ is bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T,\infty)$.
\end{itemize}

Furthermore, we have:

\begin{lemma}
\label{Lemma:2.1}
Let $\phi\in L^\infty({\mathbb R}^N_+)$.
Then
\begin{equation}
\label{eq:2.2}
\sup_{t>0}\,t^{\frac{1}{2}}\|\partial_{x_N}[S_1(t)\phi]\|_{L^\infty}\le \|\phi\|_{L^\infty}.
\end{equation}
Furthermore,
\begin{equation}
\label{eq:2.3}
\lim_{\varepsilon\to0}\sup_{t>0}\,\,t^{\frac{1}{2}}
\|S_1(\varepsilon^{-1}t)\phi\|_{L^\infty({\mathbb R}^{N-1}\times(0,L))}=0
\quad\mbox{for any $L>0$}.
\end{equation}
\end{lemma}

\noindent
{\bf Proof.}

It follows from \eqref{eq:1.3} that
\begin{equation}
\label{eq:2.4}
\begin{split}
 & K(x,y,t):=\partial_{x_N}\Gamma_D(x,y,t)\\
 & =\Gamma_{N-1}(x'-y',t)\times
 \\
 &\qquad
 \times
\left(-\frac{x_N-y_N}{2t}\Gamma_1(x_N-y_N,t)+\frac{x_N+y_N}{2t}\Gamma_1(x_N+y_N,t)\right)
\end{split}
\end{equation}
for $(x,y,t)\in\overline{{\mathbb R}^N_+}\times{\mathbb R}^N_+\times(0,\infty)$,
where $\Gamma_d$ $(d=1,2,\dots)$ is the Gauss kernel in $\mathbb R^d$.
Then
\begin{equation}
\label{eq:2.5}
\begin{split}
&
\int_{{\mathbb R}^N_+}|K(x,y,t)|\,dy
\\
&
\le\int_0^\infty \left(\frac{|x_N-y_N|}{2t}\Gamma_1(x_N-y_N,t)+\frac{x_N+y_N}{2t}\Gamma_1(x_N+y_N,t)\right)\,dy_N\\
 & =(\pi t)^{-\frac{1}{2}}\int_0^\infty 2\eta e^{-\eta^2}\,d\eta=(\pi t)^{-\frac{1}{2}}
\end{split}
\end{equation}
for $x\in\overline{{\mathbb R}^N_+}$ and $t>0$.
By \eqref{eq:1.4} and \eqref{eq:2.5} we have
$$
|\partial_{x_N}[S_1(t)\phi](x)|
\le\int_{{\mathbb R}^N_+}|K(x,y,t)||\phi(y)|\,dy
\le t^{-\frac{1}{2}}\|\phi\|_{L^\infty}
$$
for $x\in\overline{{\mathbb R}^N_+}$ and $t>0$.
This implies \eqref{eq:2.2}.

On the other hand,
for any $L>0$, it follows from \eqref{eq:1.3} that
\begin{equation}
\label{eq:2.6}
\begin{split}
& \int_{{\mathbb R}^N_+}\Gamma_D(x,y,\varepsilon^{-1}t)\,dy\\
&
=\int_0^\infty\bigg(\Gamma_1(x_N-y_N,\varepsilon^{-1}t)-\Gamma_1(x_N+y_N,\varepsilon^{-1}t)\bigg)\,dy_N\\
& =2(4\pi\varepsilon^{-1}t)^{-\frac{1}{2}}\int_0^{x_N}\exp\left({-\frac{\varepsilon\eta^2}{4t}}\right)\,d\eta
\le 2(4\pi\varepsilon^{-1}t)^{-\frac{1}{2}}L\le C(\varepsilon^{-1} t)^{-\frac{1}{2}}
\end{split}
\end{equation}
for $x\in {\mathbb R}^{N-1}\times(0,L)$, $t>0$ and $\varepsilon>0$.
For any $\phi\in L^\infty({\mathbb R}^N_+)$,
by \eqref{eq:1.4} and \eqref{eq:2.6} we have
$$
|[S_1(\varepsilon^{-1}t)\phi](x)|
\le
\int_{{\mathbb R}^N_+}\Gamma_D(x,y,\varepsilon^{-1}t)|\phi(y)|\,dy
\le C(\varepsilon^{-1} t)^{-\frac{1}{2}}\|\phi\|_{L^\infty}
$$
for $x\in {\mathbb R}^{N-1}\times(0,L)$, $t>0$ and $\varepsilon>0$.
This implies \eqref{eq:2.3},
and the proof of Lemma~\ref{Lemma:2.1} is complete.
$\Box$
\vspace{5pt}

Next we recall some properties of $S_2(t)\psi$.

\begin{itemize}
  \item[(${\rm P}_1$)]
  Let $\psi\in L^r({\mathbb R}^{N-1})$ for some $r\in[1,\infty]$
  and $t$, $t'>0$. Then
  \begin{align*}
    & [S_2(t)\psi](x',x_N)=[S_2(t+x_N)\psi](x',0),\vspace{3pt}
    \\
    & [S_2(t+t')\psi](x)=[S_2(t)(S_2(t')\psi)](x),
  \end{align*}
  for $x=(x',x_N)\in\overline{{\mathbb R}^N_+}$.
  Furthermore,
  $$
  \lim_{t\to 0}|S_2(t)\psi-\psi|_r=0\quad\mbox{if $1\le r<\infty$};
  $$
  \item[(${\rm P}_2$)]
  For any $1\le q\le r\le\infty$,
 $$
  |S_2(t)\psi|_{L^r}\le Ct^{-(N-1)(\frac{1}{q}-\frac{1}{r})}|\psi|_{L^q},\qquad t>0,
 $$
  for all $\psi\in L^q({\mathbb R}^{N-1})$.
  In particular, if $q=r$, then
  \begin{equation}
  \label{eq:2.7}
  \sup_{t>0}\,|S_2(t)\psi|_{L^r}\le |\psi|_{L^r}.
  \end{equation}
 \end{itemize}
Properties~(${\rm P}_1$) and (${\rm P}_2$) easily follow from \eqref{eq:1.5} (see e.g. \cite{FIK02}) and imply that
\begin{align}
\label{eq:2.8}
\sup_{t>0}\,\|S_2(t)\psi\|_{L^\infty}\le |\psi|_{L^\infty}
\end{align}
for all $\psi\in L^\infty({\mathbb R}^{N-1})$.
Furthermore,
by a similar argument as in the proof of property~($G_2$)
we have:
\begin{itemize}
  \item[(${\rm P}_3$)]
  Let $\psi\in L^q({\mathbb R}^{N-1})$ with $1\le q\le\infty$.
  Then, for any $T>0$, $S_2(t)\psi$ is bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T,\infty)$.
\end{itemize}

\begin{lemma}
\label{Lemma:2.2}
Let $\psi\in L^\infty({\mathbb R}^{N-1})$.
Set
\begin{equation}
\label{eq:2.9}
D_\varepsilon[\psi](x,t)
:=\int_0^t[S_1(\varepsilon^{-1}(t-s))F_1[\psi](s)](x)\,ds
\end{equation}
for $x\in\overline{{\mathbb R}^N_+}$, $t>0$ and $\varepsilon>0$.
Then $D_\varepsilon[\psi]$ and $\partial_{x_N}D_\varepsilon[\psi]$ are bounded and smooth
in $\overline{{\mathbb R}^N_+}\times(T,\infty)$ for any $T>0$.
Furthermore, there exists $C>0$ such that
\begin{equation}
\label{eq:2.10}
\|D_\varepsilon[\psi](t)\|_{L^\infty}\le
Ct^{\frac{1}{4}}(\varepsilon^{\frac{1}{2}}+t^{\frac{3}{4}})|\psi|_{L^\infty}
\end{equation}
for $t>0$ and $\varepsilon>0$.
Moreover,
\begin{equation}
\label{eq:2.11}
\lim_{\varepsilon\to0}\sup_{t\in(0,T_1)}\,
\|D_\varepsilon[\psi](t)\|_{L^\infty({\mathbb R}^{N-1}\times(0,L))}=0
\end{equation}
for $T_1>0$ and $L>0$.
\end{lemma}

{\bf Proof.}

We prove \eqref{eq:2.10} first.
Since
$$
\partial_tP(x',x_N,t)
=\frac{1}{x_N+t}\frac{|x'|^2-(N-1)(x_N+t)^2}{|x'|^2+(x_N+t)^2}P(x',x_N,t),
$$
it follows that
\begin{equation}
\label{eq:2.12}
|\partial_tP(x',x_N,t)|
\le C(x_N+t)^{-1}P(x',x_N,t).
\end{equation}
By \eqref{eq:1.12}, \eqref{eq:2.7} and \eqref{eq:2.12} we have
\begin{equation}
\label{eq:2.13}
\begin{split}
\|F_1[\psi](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}
&
\le C(y_N+s)^{-1}\|S_2(s+y_N)\psi\|_{L^\infty({\mathbb R}^{N-1})}
\\
&
\le C(y_N+s)^{-1}|\psi|_{L^\infty}
\end{split}
\end{equation}
for $y_N\in[0,\infty)$ and $s>0$.
Since
\begin{equation}
\label{eq:2.14}
(y_N+s)^{-1}\le
\left\{
\begin{array}{ll}
y_N^{-\frac{3}{4}}s^{-\frac{1}{4}}
&
\quad\mbox{for}\quad 0\le y_N\le 1,
\vspace{5pt}
\\
1
&
\quad\mbox{for}\quad y_N>1,
\end{array}
\right.
\end{equation}
by \eqref{eq:1.3}, \eqref{eq:2.9} and \eqref{eq:2.13}
we see that
\begin{equation}
\label{eq:2.15}
\begin{split}
 & |D_\varepsilon[\psi](x,t)|
 \le\int_0^t\int_{{\mathbb R}^N_+}\Gamma_D(x,y,\tau_\varepsilon)|F_1[\psi](y,s)|\,dy\,ds\\
 &
 \le
 C \int_0^t\int_0^\infty\Gamma_1(x_N-y_N,\tau_\varepsilon)\|F_1[\psi](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\,dy_N\,ds\\
 &
 \le C|\psi|_{L^\infty}\int_0^t\int_0^\infty\tau_\varepsilon^{-\frac{1}{2}}
 \exp\left(-\frac{(x_N-y_N)^2}{4\tau_\varepsilon}\right)
 \left(y_N+s\right)^{-1}\,dy_N\,ds\\
 &
 \le C|\psi|_{L^\infty}\left\{
 \int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{-\frac{1}{4}}\int_0^1y_N^{-\frac{3}{4}}\,dy_N\,ds+\int_0^t\,ds\right\}\\
 &
 \le C|\psi|_{L^\infty}\left\{\varepsilon^{\frac{1}{2}}
 \int_0^t(t-s)^{-\frac{1}{2}}s^{-\frac{1}{4}}\,ds+t\right\}\\
 &
 \le C|\psi|_{L^\infty}(\varepsilon^{\frac{1}{2}}t^{\frac{1}{4}}+t)
 = C|\psi|_{L^\infty}t^{\frac{1}{4}}(\varepsilon^{\frac{1}{2}}+t^{\frac{3}{4}})
\end{split}
\end{equation}
for $x\in\overline{{\mathbb R}^N_+}$, $t>0$ and $\varepsilon>0$,
where $\tau_\varepsilon:=\varepsilon^{-1}(t-s)$.
Here $\Gamma_1$ is the one-dimensional Gauss kernel.
This implies \eqref{eq:2.10}.

We prove \eqref{eq:2.11}.
Let $L>0$.
Similarly to \eqref{eq:2.15},
we obtain
\begin{equation}
\label{eq:2.16}
\begin{split}
 & |D_\varepsilon[\psi](x,t)|
 \le\int_0^t\int_{{\mathbb R}^N_+}\Gamma_D(x,y,\tau_\varepsilon)|F_1[\psi](y,s)|\,dy\,ds\\
 & \le C
 \int_0^t\int_0^\infty\bigg(\Gamma_1(x_N-y_N,\tau_\varepsilon)-\Gamma_1(x_N+y_N,\varepsilon^{-1}(t-s))\bigg)\\
 & \qquad\qquad\qquad\qquad\qquad\qquad
 \times
 \|F_1[\psi](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\,dy_N\,ds\\
 & \le C|\psi|_{L^\infty}
 \int_0^t\tau_\varepsilon^{-\frac{1}{2}}
 \int_0^\infty
 (y_N+s)^{-1}
 \\
 &\hspace{1cm}
 \times
 \left[\exp\left({-\frac{(x_N-y_N)^2}{4\tau_\varepsilon}}\right)-\exp\left({-\frac{(x_N+y_N)^2}{4\tau_\varepsilon}}\right)\right]
\,dy_N\,ds
 \end{split}
\end{equation}
for $x\in\overline{{\mathbb R}^N_+}$, $t>0$ and $\varepsilon>0$.
This together with \eqref{eq:2.6}, \eqref{eq:2.13} and \eqref{eq:2.14}
implies that
\begin{equation*}
\begin{split}
 & |D_\varepsilon[\psi](x,t)|
 \\
 & \le C|\psi|_{L^\infty}\bigg\{\int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{-\frac{1}{4}}\int_0^1y_N^{-\frac{3}{4}}\,dy_N\,ds\\
 & \,\,\,
 +\int_0^t\tau_\varepsilon^{-\frac{1}{2}}
 \int_1^\infty
 \left[\exp\left({-\frac{(x_N-y_N)^2}{4\tau_\varepsilon}}\right)-\exp\left({-\frac{(x_N+y_N)^2}{4\tau_\varepsilon}}\right)\right]
 \, dy_N \,ds\bigg\}\\
 & \le C|\psi|_{L^\infty}\left\{\int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{-\frac{1}{4}}\,ds
 +\int_0^t\tau_\varepsilon^{-\frac{1}{2}}\,ds\right\}\\
 & \le C|\psi|_{L^\infty}\varepsilon^{\frac{1}{2}}\left\{\int_0^t(t-s)^{-\frac{1}{2}}s^{-\frac{1}{4}}\,ds
 +\int_0^t(t-s)^{-\frac{1}{2}}\,ds\right\}\\
 & \le C|\psi|_{L^\infty}(\varepsilon^{\frac{1}{2}}t^{\frac{1}{4}}+(\varepsilon t)^\frac{1}{2})
\end{split}
\end{equation*}
for $x\in {\mathbb R}^{N-1}\times(0,L)$, $t>0$ and $\varepsilon>0$.
Thus \eqref{eq:2.11} holds.

On the other hand,
it follows from the semigroup property of $S_1(t)$ that
\begin{equation*}
\begin{split}
 & D_\varepsilon[\psi](x,t)
 =\int_0^t[S_1(\varepsilon^{-1}(t-s))F_1[\psi](s)](x)\,ds\\
 & =S_1(\varepsilon^{-1}(t-T/2))D_\varepsilon[\psi](x,T/2)
+\int_{T/2}^t[S_1(\varepsilon^{-1}(t-s))F_1[\psi](s)](x)\,ds
\end{split}
\end{equation*}
for $x\in\overline{{\mathbb R}^N_+}$ and $0<T<t<\infty$.
Then, by \eqref{eq:2.9} and (${\rm G_2}$)
we see that
$$
S_1(\varepsilon^{-1}(t-T/2))D_\varepsilon[\psi](x,T/2)
$$
is bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T,\infty)$.
Furthermore, by \eqref{eq:2.13}
we apply the same argument as in \cite[Section~3, Chapter 1]{F}
to see that
$$
\int_{T/2}^t[S_1(\varepsilon^{-1}(t-s))F_1[\psi](s)](x)\,ds
$$
is also bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T,\infty)$.
(See also \cite[Proposition~5.2]{FIK01} and \cite[Lemma~2.1]{IKK01}.)
Therefore we deduce that
$D_\varepsilon[\psi]$ and $\partial_{x_N}D_\varepsilon[\psi]$ are bounded and smooth
in $\overline{{\mathbb R}^N_+}\times(T,\infty)$.
Thus Lemma~\ref{Lemma:2.2} follows.
$\Box$\vspace{5pt}

\begin{lemma}
\label{Lemma:2.3}
Let $0\le \alpha<1$. Then there exists $C>0$ such that
\begin{equation}
\label{eq:2.17}
\int_0^\infty
\frac{|x\pm y|}{t}\Gamma_1(x\pm y,t)y^{-\alpha}\,dy
\le Ct^{-\frac{\alpha+1}{2}}
\end{equation}
for $x\ge 0$ and $t>0$.
Here $\Gamma_1$ is the one-dimensional Gauss kernel.
\end{lemma}

{\bf Proof.}
Let $x\ge 0$ and $t>0$.
It follows that
\begin{equation*}
\begin{split}
 & \int_0^\infty
\frac{|x-y|}{t}\Gamma_1(x-y,t)y^{-\alpha}\,dy\\
 & =(4\pi t)^{-\frac{1}{2}}\biggr(\int_0^{x_2}+\int_{x/2}^\infty\biggr)
\frac{|x-y|}{t}\exp\left(-\frac{|x-y|^2}{4t}\right)y^{-\alpha}\,dy.
\end{split}
\end{equation*}
Since $y^{-1}\le |x-y|^{-1}$ for $0\le x\le 2y$,
we have
\begin{equation}
\label{eq:2.18}
\begin{split}
 & \int_0^\infty
\frac{|x-y|}{t}\Gamma_1(x-y,t)y^{-\alpha}\,dy\\
 & \le Ct^{-\frac{1}{2}}\frac{x}{t}
 \exp\left(-\frac{x^2}{16t}\right)
 \int_0^{x/2}y^{-\alpha}\,dy_N\\
 & \qquad\quad
 +Ct^{-\frac{3}{2}+\frac{1-\alpha}{2}}\int_{x/2}^\infty
 \left(\frac{|x-y|}{t^{1/2}}\right)^{1-\alpha}
 \exp\left(-\frac{|x-y|^2}{4t}\right)\,dy\\
 & \le Ct^{-\frac{3}{2}}x^{2-\alpha}\exp\left(-\frac{x^2}{16t}\right)
 +Ct^{-1+\frac{1-\alpha}{2}}\le Ct^{-\frac{\alpha+1}{2}}.
\end{split}
\end{equation}
Since $y^{-1}\le 2(x+y)^{-1}$ for $0\le x\le y$,
similarly to \eqref{eq:2.18}, we obtain
$$
\int_0^\infty
\frac{x+y}{t}\Gamma_1(x+y,t)y^{-\alpha}\,dy
\le Ct^{-\frac{\alpha+1}{2}}.
$$
Thus \eqref{eq:2.17} holds and Lemma~\ref{Lemma:2.3} follows.
$\Box$\vspace{5pt}

\begin{lemma}
\label{Lemma:2.4}
Let $\psi\in L^\infty({\mathbb R}^{N-1})$.
Then there exists $C>0$ such that
\begin{equation}
\label{eq:2.19}
\|\partial_{x_N}D_\varepsilon[\psi](t)\|_{L^\infty}
\le
C\varepsilon^{\frac{3}{4}}t^{-\frac{1}{4}}|\psi|_{L^\infty}
\end{equation}
for $t>0$ and $\varepsilon>0$.
\end{lemma}

{\bf Proof.}

By \eqref{eq:2.4}, \eqref{eq:2.9} and \eqref{eq:2.13}
we see that
\begin{equation}
\label{eq:2.20}
\begin{split}
& |\partial_{x_N} D_\varepsilon[\psi](x,t)|\\
& \le\int_0^t\int_{{\mathbb R}^N_+}|K(x,y,\tau_\varepsilon)||F_1[\psi](y,s)|\,dy\,ds\\
& \le C
\int_0^t\int_0^\infty \tilde{K}(x_N,y_N,\tau_\varepsilon)
\|F_1[\psi](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\, dy_N\,ds\\
&
\le C|\psi|_{L^\infty}
\int_0^ts^{-\frac{1}{2}}\int_0^\infty
\tilde{K}(x_N,y_N,\tau_\varepsilon)
y_N^{-\frac{1}{2}}\, dy_N\,ds
\end{split}
\end{equation}
for $x\in\overline{{\mathbb R}^N_+}$, $t>0$ and $\varepsilon>0$,
where $\tau_\varepsilon:=\varepsilon^{-1}(t-s)$ and
\begin{equation}
\label{eq:2.21}
\tilde{K}(x_N,y_N,t)=\frac{|x_N-y_N|}{t}\Gamma_1(x_N-y_N,t)
+\frac{x_N+y_N}{t}\Gamma_1(x_N+y_N,t)
\end{equation}
for $x_N\ge0$, $y_N>0$ and $t>0$.
By \eqref{eq:2.17} with $\alpha=1/2$ and \eqref{eq:2.20}
we deduce that
\begin{equation*}
\begin{split}
|\partial_{x_N} D_\varepsilon[\psi](x,t)|
&
\le C|\psi|_{L^\infty}
\int_0^ts^{-\frac{1}{2}}\tau_\varepsilon^{-\frac{3}{4}}\,ds
\\
&
= C|\psi|_{L^\infty}
\int_0^ts^{-\frac{1}{2}}(\varepsilon^{-1}(t-s))^{-\frac{3}{4}}\,ds
\le C\varepsilon^{\frac{3}{4}}t^{-\frac{1}{4}}|\psi|_{L^\infty}
\end{split}
\end{equation*}
for $x\in\overline{{\mathbb R}^N_+}$, $t>0$ and $\varepsilon>0$.
Thus \eqref{eq:2.19} follows.
$\Box$

\section{Proof of Theorem~\ref{Theorem:1.1}}

We introduce some notation.
Let $T>0$ and $\varepsilon\in(0,1)$.
Set
$$
X_T:=\bigg\{v, \partial_{x_N}v\in C(\overline{{\mathbb R}^N_+}\times(0,T))\,:\,\|v\|_{X_T}<\infty\bigg\},
\quad
\|v\|_{X_T}:=\sup_{0<t<T}E_\varepsilon[v](t),
$$
where
$$
E_\varepsilon[v](t):=\|v(t)\|_{L^\infty}
+(\varepsilon^{-1}t)^{\frac{1}{2}}\|\partial_{x_N}v(t)\|_{L^\infty}.
$$
Then $X_T$ is a Banach space equipped with the norm $\|\cdot\|_{X_T}$.
For the proof of Theorem~\ref{Theorem:1.1}
we apply the Banach contraction mapping principle in $X_T$
to find a fixed point of
\begin{equation}
\label{eq:3.1}
Q_\varepsilon[v](t)
:=S_1(\varepsilon^{-1}t)\Phi
-D_\varepsilon[\varphi_b](t)-\int_0^tS_1(\varepsilon^{-1}(t-s))F_2[v](s)\,ds,
\end{equation}
where $\Phi$, $F_2[v]$ and $D_\varepsilon[\varphi_b]$ are as in \eqref{eq:1.9}, \eqref{eq:1.13} and \eqref{eq:2.9}, respectively.

\begin{lemma}
\label{Lemma:3.1}
There exists $C>0$ such that
\begin{equation}
\label{eq:3.2}
F_2[v](x,t)\le C\varepsilon^\frac{1}{2}\left(t^{-\frac{1}{2}}+h(x_N,t)\right)\|v\|_{X_T}
\end{equation}
for $x\in{\mathbb R}^N_+$, $0<t<T$, $\varepsilon\in(0,1)$ and $v\in X_T$.
Here
$$
h(x_N,t):=x_N^{-\frac{3}{4}}t^{\frac{1}{4}}\quad\mbox{if}\quad 0<x_N\le1
\quad\mbox{and}\quad
h(x_N,t):=x_N^{-\frac{1}{2}}\quad\mbox{if}\quad x_N>1.
$$
\end{lemma}

\noindent
{\bf Proof.}
Let $T>0$, $\varepsilon\in(0,1)$ and $v\in X_T$.
It follows from \eqref{eq:1.13} that
\begin{equation}
\label{eq:3.3}
F_2[v](x,t)=F_2'[v](x,t)+F_2''[v](x,t)
\end{equation}
for $x\in{\mathbb R}^N_+$ and $t>0$,
where
\begin{align*}
 & F_2'[v](x,t):=\int_{{\mathbb R}^{N-1}}P(x'-y',x_N,0)\partial_{x_N}v(y',0,t)\, dy',\\
 & F_2''[v](x,t):=\int_0^t\int_{{\mathbb R}^{N-1}}\partial_t P(x'-y',x_N,t-s)\partial_{x_N}v(y',0,s)\, dy'\,ds.
\end{align*}
Since $v\in X_T$,
by \eqref{eq:1.5} and \eqref{eq:2.7} we see that
\begin{equation}
\label{eq:3.4}
\begin{split}
|F_2'[v](x,t)|
&
\le \|S_2(x_N)\partial_{x_N}v(\cdot,0,t)\|_{L^\infty({\mathbb R}^{N-1})}
\\
&
\le |\partial_{x_N}v(t)|_{L^\infty}
\le \varepsilon^{\frac{1}{2}}t^{-\frac{1}{2}}\|v\|_{X_T}
\end{split}
\end{equation}
for $x\in{\mathbb R}^N_+$ and $t>0$.
On the other hand,
it follows from \eqref{eq:2.12} that
$$
|\partial_t P(x',x_N,t)|
\le\left\{
\begin{array}{ll}
CP(x',x_N,t)x_N^{-\frac{3}{4}}t^{-\frac{1}{4}} \quad & \mbox{if}\quad x_N\le1,
\vspace{5pt}\\
CP(x',x_N,t)x_N^{-\frac{1}{2}}t^{-\frac{1}{2}} & \mbox{if}\quad x_N>1,
\end{array}
\right.
$$
for $x\in{\mathbb R}^N_{+}$ and $t>0$.
Then we obtain
\begin{equation}
\label{eq:3.5}
\begin{split}
&
|F_2''[v](x,t)|
\\
&
\le Cx_N^{-\frac{3}{4}}\int_0^t(t-s)^{-\frac{1}{4}}
\int_{{\mathbb R}^{N-1}}P(x'-y',x_N,t-s)|\partial_{x_N}v(y',0,s)|\,dy'\,ds
\\
&
\le C\varepsilon^{\frac{1}{2}}x_N^{-\frac{3}{4}}\bigg(\sup_{0<s<t}\,(\varepsilon^{-1}s)^{\frac{1}{2}}|\partial_{x_N}v(s)|_{L^\infty}\bigg)
\int_0^t(t-s)^{-\frac{1}{4}}s^{-\frac{1}{2}}\,ds
\\
&
\le C\varepsilon^{\frac{1}{2}}x_N^{-\frac{3}{4}}t^{\frac{1}{4}}\|v\|_{X_T}
=C\varepsilon^{\frac{1}{2}}h(x_N,t)\|v\|_{X_T}
\end{split}
\end{equation}
for $x'\in{\mathbb R}^{N-1}$, $0<x_N\le 1$ and $0<t<T$.
Similarly, we deduce that
\begin{equation}
\label{eq:3.6}
\begin{split}
&
|F_2''[v](x,t)|
\\
&
\le C\int_0^t(t-s)^{-\frac{1}{2}}\int_{{\mathbb R}^N_+}P(x'-y',x_N,t-s)|\partial_{x_N}v(y',0,s)|\,dy'\,ds\\
 &
\le C\varepsilon^{\frac{1}{2}}x_N^{-\frac{1}{2}}
\bigg(\sup_{0<s<t}\,(\varepsilon^{-1}s)^{\frac{1}{2}}|\partial_{x_N}v(s)|_{L^\infty}\bigg)
\int_0^t(t-s)^{-\frac{1}{2}}s^{-\frac{1}{2}}\,ds\\
 &
\le C\varepsilon^{\frac{1}{2}}x_N^{-\frac{1}{2}}\|v\|_{X_T}
=C\varepsilon^{\frac{1}{2}}h(x_N,t)\|v\|_{X_T}
\end{split}
\end{equation}
for $x'\in{\mathbb R}^{N-1}$, $x_N>1$ and $0<t<T$.
Therefore, by \eqref{eq:3.3}, \eqref{eq:3.4}, \eqref{eq:3.5} and \eqref{eq:3.6} we obtain \eqref{eq:3.2}.
Thus Lemma~\ref{Lemma:3.1} follows.
$\Box$

\begin{lemma}
\label{Lemma:3.2}
For any $v\in X_T$ and $\varepsilon\in(0,1)$,
set
\begin{equation}
\label{eq:3.7}
\tilde{D}_\varepsilon[v](t):=
\int_0^tS_1(\varepsilon^{-1}(t-s))F_2[v](s)\,ds.
\end{equation}
Then there exists $T_*=T_*(N)>0$ such that
\begin{equation}
\label{eq:3.8}
\|\tilde{D}_\varepsilon[v]\|_{X_{T_*}}\le\frac{1}{4}\|v\|_{X_{T_*}}
\end{equation}
for $v\in X_{T_*}$ and $\varepsilon\in(0,1)$.
Furthermore,
$\tilde{D}_\varepsilon[v]$ and $\partial_{x_N}\tilde{D}_\varepsilon[v]$ are bounded and smooth in $\overline{{\mathbb R}^N_+}\times(\tau,T_*)$ for any $0<\tau<T_*$.
\end{lemma}

\noindent
{\bf Proof.}
Let $T>0$.
By \eqref{eq:2.9} and \eqref{eq:3.7} we see that
$\tilde{D}_\varepsilon$ is defined analogously as  $D_\varepsilon$ with $F_1$ replaced by $F_2$.
Then it follows from \eqref{eq:2.15} and \eqref{eq:3.2} that
\begin{equation*}
\begin{split}
& |\tilde{D}_\varepsilon[v](x,t)|\\
& \le C\int_0^t\int_0^\infty\Gamma_1(x_N-y_N,\tau_\varepsilon)\|F_2[v](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\,dy_N\,ds\\
& \le C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\int_0^t\int_0^\infty\tau_\varepsilon^{-\frac{1}{2}}\exp\left(-\frac{(x_N-y_N)^2}{4\tau_\varepsilon}\right)\left(s^{-\frac{1}{2}}+h(y_N,s)\right)\,dy_N\,ds\\
& \le C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\left\{\int_0^ts^{-\frac{1}{2}}\,ds
+\int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{\frac{1}{4}}\int_0^1y_N^{-\frac{3}{4}}\,dy_N\,ds
\right.\\
& \qquad\qquad\qquad\qquad
\left.
+\int_0^t\int_1^\infty\tau_\varepsilon^{-\frac{1}{2}}\exp\left(-\frac{(x_N-y_N)^2}{4\tau_\varepsilon}\right)\,dy_N\,ds
\right\}\\
&
\le C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\left\{t^{\frac{1}{2}}
+\varepsilon^{\frac{1}{2}}\int_0^t(t-s)^{-\frac{1}{2}}s^{\frac{1}{4}}\,ds+t\right\}
\le C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}
(t^{\frac{1}{2}}+\varepsilon^{\frac{1}{2}} t^{\frac{3}{4}}+t)
\end{split}
\end{equation*}
for $x\in\overline{{\mathbb R}^N_+}$ and $0<t<T$,
where $\tau_\varepsilon=\varepsilon^{-1}(t-s)$.
Then, taking a sufficiently small $T>0$ if necessary, we obtain
\begin{equation}
\label{eq:3.9}
\sup_{0<t<T}\|\tilde{D}_\varepsilon[v](t)\|_{L^\infty}
\le\frac{1}{8}\|v\|_{X_T}.
\end{equation}
On the other hand,
similarly to \eqref{eq:2.20},
by \eqref{eq:3.2}
we see that
\begin{equation*}
\begin{split}
& \left|\partial_{x_N}\tilde{D}_\varepsilon[v](x,t)\right|\\
& \le
C\int_0^t\int_0^\infty
\tilde{K}(x_N,y_N,\tau_\varepsilon)
\|F_2[v](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\,dy_N\,ds\\
& \le
C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\int_0^t\int_0^\infty
\tilde{K}(x_N,y_N,\tau_\varepsilon)
\left(s^{-\frac{1}{2}}+h(y_N,s)\right)\,dy_N\,ds\\
& \le
C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\int_0^t\int_0^\infty
\tilde{K}(x_N,y_N,\tau_\varepsilon)
\left(s^{-\frac{1}{2}}+y_N^{-\frac{3}{4}}s^{\frac{1}{4}}+y_N^{-\frac{1}{2}}\right)\,dy_N\,ds
\end{split}
\end{equation*}
for $x\in\overline{{\mathbb R}^N_+}$ and $0<t<T$,
where $\tilde{K}$ is the function given by \eqref{eq:2.21}.
By \eqref{eq:2.17} we have
\begin{equation*}
\begin{split}
& \left|\partial_{x_N}\tilde{D}_\varepsilon[v](x,t)\right|
\le
C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\bigg(\int_0^ts^{-\frac{1}{2}}\tau_\varepsilon^{-\frac{1}{2}}\,ds
+\int_0^ts^{\frac{1}{4}}\tau_\varepsilon^{-\frac{7}{8}}\,ds
+\int_0^t\tau_\varepsilon^{-\frac{3}{4}}\,ds\bigg)
\\
& =
C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\bigg(\int_0^ts^{-\frac{1}{2}}(\varepsilon^{-1}(t-s))^{-\frac{1}{2}}\,ds
\\
&\qquad\qquad\qquad
+\int_0^ts^{\frac{1}{4}}(\varepsilon^{-1}(t-s))^{-\frac{7}{8}}\,ds
+\int_0^t(\varepsilon^{-1}(t-s))^{-\frac{3}{4}}\,ds\bigg)
\\
& \le
C\varepsilon^{\frac{1}{2}}\|v\|_{X_T}\bigg(\varepsilon^{\frac{1}{2}}+\varepsilon^{\frac{7}{8}}t^{\frac{3}{8}}+\varepsilon^{\frac{3}{4}}t^{\frac{1}{4}}\bigg)
\\
&
\le
C(\varepsilon^{-1}t)^{-\frac{1}{2}}\|v\|_{X_T}\bigg((\varepsilon t)^{\frac{1}{2}}+(\varepsilon t)^{\frac{7}{8}}+(\varepsilon t)^{\frac{3}{4}}\bigg)
\end{split}
\end{equation*}
for $x\in\overline{{\mathbb R}^N_+}$ and $0<t<T$.
Taking a sufficiently small $T>0$ if necessary, we see that
\begin{equation}
\label{eq:3.10}
\sup_{0<t<T}\,(\varepsilon^{-1}t)^{\frac{1}{2}}\|\partial_{x_N}\tilde{D}_\varepsilon[v](t)\|_{L^\infty}
\le\frac{1}{8}\|v\|_{X_T}.
\end{equation}
Therefore, by \eqref{eq:3.9} and \eqref{eq:3.10}
we have \eqref{eq:3.8}.
Furthermore,
by \eqref{eq:3.2} we apply a similar argument as in the proof of Lemma~\ref{Lemma:2.2}
and deduce that
$\tilde{D}_\varepsilon[v]$ and $\partial_{x_N}\tilde{D}_\varepsilon[v]$ are bounded and smooth in $\overline{{\mathbb R}^N_+}\times(\tau,T)$ for any $0<\tau<T$.
Thus Lemma~\ref{Lemma:3.2} follows.
$\Box$ \vspace{5pt}

Now we are ready to complete the proof of Theorem~\ref{Theorem:1.1}.
\vspace{5pt}
\newline

{\bf Proof of Theorem~\ref{Theorem:1.1}.}
Let
\begin{equation}
\label{eq:3.11}
m:=16\max\{\|\varphi\|_{L^\infty}, |\varphi_b|_{L^\infty}\}.
\end{equation}
Let $T_*>0$ be as in Lemma~\ref{Lemma:3.2}
and $v\in X_{T_*}$ with $\|v\|_{X_{T_*}}\le m$.
Then, by property~($G_2$), Lemmata~\ref{Lemma:2.2} and~\ref{Lemma:3.2}
we see that $Q_\varepsilon[v]\in X_{T_*}$.
Since it follows from \eqref{eq:1.9} and \eqref{eq:2.8} that
$$
\|\Phi\|_{L^\infty}\le\|\varphi\|_{L^\infty}+|\varphi_b|_{L^\infty},
$$
by \eqref{eq:2.1}, \eqref{eq:2.2} and \eqref{eq:3.11} we have
\begin{equation}
\label{eq:3.12}
\|S_1(\varepsilon^{-1}t)\Phi\|_{L^\infty}
+(\varepsilon^{-1} t)^{\frac{1}{2}}\|\partial_{x_N}[S_1(\varepsilon^{-1}t)\Phi]\|_{L^\infty}
\le2\|\Phi\|_{L^\infty}\le\frac{m}{4}
\end{equation}
for $0<t<T_*$.
Furthermore, by \eqref{eq:2.10} and \eqref{eq:2.19},
taking a sufficiently small $T_*>0$ if necessary, we see that
\begin{equation}
\label{eq:3.13}
\|D_\varepsilon[\varphi_b](t)\|_{L^\infty}
+(\varepsilon^{-1} t)^{\frac{1}{2}}\|\partial_{x_N}D_\varepsilon[\varphi_b](t)\|_{L^\infty}
\le CT_*^{\frac{1}{4}}(1+T_*^{\frac{3}{4}})|\varphi_b|_\infty
\le\frac{m}{4}
\end{equation}
for $0<t<T_*$.
Lemma~\ref{Lemma:3.2} together with \eqref{eq:3.1}, \eqref{eq:3.12} and \eqref{eq:3.13} implies that
$$
\left\|Q_\varepsilon[v]\right\|_{X_{T_*}}\le
\|S_1(\varepsilon^{-1}\cdot)\varphi\|_{X_{T_*}}+\|D_\varepsilon[\varphi_b]\|_{X_{T_*}}+\|\tilde{D}_\varepsilon[v]\|_{X_{T_*}}
\le\frac{3m}{4}<m.
$$
Similarly, we obtain
$$
\left\|Q_\varepsilon[v_1]-Q_\varepsilon[v_2]\right\|_{X_{T_*}}
=\|\tilde{D}_\varepsilon[v_1]-\tilde{D}_\varepsilon[v_2]\|_{X_{T_*}}\le\frac{1}{4}\|v_1-v_2\|_{X_{T_*}}
$$
for $v_i\in X_{T_*}$ with $\|v_i\|_{X_{T_*}}\le m$ $(i=1,2)$.
Then, the contraction mapping theorem ensures that
there exists a unique $v_\varepsilon\in X_{T_*}$ with $\|v_\varepsilon\|_{X_{T_*}}\le m$ and
\begin{equation}
\label{eq:3.14}
v_\varepsilon=Q_\varepsilon[v_\varepsilon]
=S_1(\varepsilon^{-1}t)\Phi
-D_\varepsilon[\varphi_b](t)
-\tilde{D}_\varepsilon[v_\varepsilon](t)
\quad\mbox{in}\quad X_{T_*}.
\end{equation}
In particular, we see that
\begin{equation}
\label{eq:3.15}
\|v_\varepsilon\|_{X_{T_*}}\le C(\|\varphi\|_{L^\infty}+|\varphi_b|_{L^\infty}).
\end{equation}
Furthermore, by $(G_2)$ and Lemmata~\ref{Lemma:2.2},~\ref{Lemma:3.2}, we see that $v_\varepsilon$ is bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T_1,T_*)$
for any $0<T_1<T_*$.
As before, set
$$
w_\varepsilon(x,t)=[S_2(t)\varphi_b](x)+\int_0^t[S_2(t-s)\partial_{x_N}v_\varepsilon(s)](x)\, ds
$$
for $x\in\overline{{\mathbb R}^N_+}$ and $t\in(0,{T_*})$.
By \eqref{eq:2.8} and \eqref{eq:3.11} we obtain
\begin{equation}
\label{eq:3.16}
\begin{split}
&
\|w_\varepsilon(t)\|_{L^\infty}
\le \|S_2(t)\varphi_b\|_{L^\infty}+\int_0^t\|S_2(t-s)\partial_{x_N}v_\varepsilon(s)\|_{L^\infty}\,ds\\
&\quad
  \le |\varphi_b|_{L^\infty}+\int_0^t|\partial_{x_N}v_\varepsilon(s)|_{L^\infty}\,ds
  \\
  &\quad
 \le \frac{m}{16}+\varepsilon^{\frac{1}{2}}m\int_0^ts^{-\frac{1}{2}}\, ds
 \\
  &\quad
 \le C(1+T_*^{\frac{1}{2}})m
  \le C(1+T_*^{\frac{1}{2}})(\|\varphi\|_{L^\infty}+|\varphi_b|_{L^\infty})<\infty
\end{split}
\end{equation}
for all $0<t<T_*$.
Furthermore, by (${\rm P_3}$)
we apply a similar argument as in Lemma~\ref{Lemma:2.2}
and see that $w_\varepsilon$ is bounded and smooth in $\overline{{\mathbb R}^N_+}\times(T_1,T_*)$ for any $0<T_1<T_*$.
Therefore we deduce that
$(v_\varepsilon,w_\varepsilon)$ is a solution of \eqref{eq:1.11} in ${\mathbb R}^N_+\times(0,T_*)$.
In addition, by \eqref{eq:3.15} and \eqref{eq:3.16}
we have assertion~(a) for any $\tau\in(0,T_*)$.
Since $T_*$ is independent of $m$,
due to the semigroup properties of $S_1(t)$ and $S_2(t)$,
we see that $(v_\varepsilon,w_\varepsilon)$ is a global-in-time solution of \eqref{eq:1.11}
and it satisfies assertion~(a) for any $\tau>0$.

Let $(\tilde{v}_\varepsilon,\tilde{w}_\varepsilon)$ be a global-in-time solution of \eqref{eq:1.11}
satisfying \eqref{eq:1.14}.
Since
$$
v_\varepsilon-\tilde{v}_\varepsilon
=Q_\varepsilon[v_\varepsilon]-Q_\varepsilon[\tilde{v}_\varepsilon]
=\tilde{D}_\varepsilon[v_\varepsilon-\tilde{v}_\varepsilon]\quad\mbox{in}\quad X_{T_*},
$$
by \eqref{eq:3.8} we have
$$
\|v_\varepsilon-\tilde{v}_\varepsilon\|_{X_{T_*}}\le\frac{1}{4}\|v_\varepsilon-\tilde{v}_\varepsilon\|_{X_{T_*}}.
$$
This implies that $v_\varepsilon=\tilde{v}_\varepsilon$ in $X_{T_*}$.
Repeating this argument, we see that $v_\varepsilon=\tilde{v}_\varepsilon$ in $X_T$ for any $T>0$.
Therefore we deduce that $(v_\varepsilon,w_\varepsilon)$ is a unique global-in-time solution of \eqref{eq:1.11}
satisfying \eqref{eq:1.14}.

It remains to prove assertions~(b) and (c).
Let $T'>0$ and $L>0$.
By \eqref{eq:1.14} and \eqref{eq:2.8}
we have
\begin{equation*}
\begin{split}
&
\|w_\varepsilon(t)-S_2(t)\varphi_b\|_{L^\infty}
\le
\int_0^t\|S_2(t-s)\partial_{x_N}v_\varepsilon(s)\|_{L^\infty}\, ds
\\
&\quad
\le
\int_0^t|\partial_{x_N}v_\varepsilon(s)|_{L^\infty}\,ds
\le C\|v_\varepsilon\|_{X_{T'}}\varepsilon^{\frac{1}{2}}\int_0^ts^{-\frac{1}{2}}\, ds
\le C\|v_\varepsilon\|_{X_{T'}}\varepsilon^{\frac{1}{2}}T'^{\frac{1}{2}}
\end{split}
\end{equation*}
for all $t\in(0,T')$.
This implies assertion~(c).
On the other hand,
since $\tilde{D}_\varepsilon[v_\varepsilon]$ is given with $F_1[\psi]$ replaced by $F_2[v_\varepsilon]$,
by \eqref{eq:3.2} we apply a similar argument as in \eqref{eq:2.16}
to obtain
\begin{equation*}
\begin{split}
&
\left|\tilde{D}_\varepsilon[v_\varepsilon](x,t)\right|
\le
C
 \int_0^t\int_0^\infty\bigg(\Gamma_1(x_N-y_N,\tau_\varepsilon)-\Gamma_1(x_N+y_N,\tau_\varepsilon)\bigg)\\
 & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
 \times
 \|F_2[v_\varepsilon](\cdot,y_N,s)\|_{L^\infty({\mathbb R}^{N-1})}\,dy_N\,ds\\
&
\le C\varepsilon^{\frac{1}{2}}\|v_\varepsilon\|_{X_{T'}}
\int_0^t\int_0^\infty
\bigg(\Gamma_1(x_N-y_N,\tau_\varepsilon)-\Gamma_1(x_N+y_N,\tau_\varepsilon)\bigg)\\
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
 \times
\left(s^{-\frac{1}{2}}+h(y_N,s)\right)\,dy_N\,ds
\\
&
\le C\varepsilon^{\frac{1}{2}}\|v_\varepsilon\|_{X_{T'}}
\int_0^t\int_0^\infty
\bigg(\Gamma_1(x_N-y_N,\tau_\varepsilon)-\Gamma_1(x_N+y_N,\tau_\varepsilon)\bigg)\\
& \qquad\qquad\qquad\qquad\qquad\qquad
 \times
\left(s^{-\frac{1}{2}}+\chi_{y_n\ge1}+y_N^{-\frac{3}{4}}s^{\frac{1}{4}}\chi_{0\le y_N<1}\right)\,dy_N\,ds
\end{split}
\end{equation*}
for $x \in {\mathbb R}^{N-1}\times(0,L)$, $0<t<T'$ and $\varepsilon>0$,
where $\tau_\varepsilon=\varepsilon^{-1}(t-s)$.
By \eqref{eq:2.6} we have
\begin{equation*}
\begin{split}
&
\left|\tilde{D}_\varepsilon[v_\varepsilon](x,t)\right|
\\
&
\le C\varepsilon^{\frac{1}{2}}\|v_\varepsilon\|_{X_{T'}}\left\{\int_0^t(1+s^{-\frac{1}{2}})\tau_\varepsilon^{-\frac{1}{2}}\,ds
+\int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{\frac{1}{4}}\int_0^1y_N^{-\frac{3}{4}}\,dy_N\,ds
\right\}
\\
&
\le
C\varepsilon^{\frac{1}{2}}\|v_\varepsilon\|_{X_{T'}}\left\{\int_0^t(1+s^{-\frac{1}{2}})\tau_\varepsilon^{-\frac{1}{2}}\,ds
+\int_0^t\tau_\varepsilon^{-\frac{1}{2}}s^{\frac{1}{4}}\,ds
\right\}
\\
&
\le
C\varepsilon\|v_\varepsilon\|_{X_{T'}}\left\{\int_0^t(1+s^{-\frac{1}{2}})(t-s)^{-\frac{1}{2}}\,ds
+\int_0^t(t-s)^{-\frac{1}{2}}s^{\frac{1}{4}}\,ds
\right\}
\\
&
\le
C\varepsilon\|v_\varepsilon\|_{X_{T'}}(1+t^{\frac{1}{2}}+t^{\frac{3}{4}})
\end{split}
\end{equation*}
for $x \in {\mathbb R}^{N-1}\times(0,L)$, $0<t<T'$ and $\varepsilon>0$.
This implies that
\begin{equation}
\label{eq:3.17}
\lim_{\varepsilon\to0}\sup_{t\in(0,T')}\,\|\tilde{D}_\varepsilon[v_\varepsilon](t)\|_{L^\infty({\mathbb R}^{N-1}\times(0,L))}=0.
\end{equation}
Therefore,
applying \eqref{eq:2.3}, \eqref{eq:2.11} and \eqref{eq:3.17} to \eqref{eq:3.14}, we obtain assertion~(b).
Thus the proof of Theorem~\ref{Theorem:1.1} is complete.
$\Box$\vspace{5pt}

\noindent
{\bf Proof of Corollary~\ref{Corollary:1.1}.}
Corollary~\ref{Corollary:1.1} immediately follows from Theorem~\ref{Theorem:1.1} and Definition~\ref{Definition:1.1}.
$\Box$
\vspace{8pt}

\noindent
{\bf Acknowledgment.} The first author was supported in part by the Slovak
Research and Development Agency under the contract No. APVV-14-0378 and by the VEGA grant
1/0347/18. Part of this work was carried out while the first author visited
the Research Alliance Center for Mathematical Sciences, Tohoku University.
The second author was supported in part by the Grant-in-Aid for Scientific Research (A)(No.~15H02058)
from Japan Society for the Promotion of Science.
The third author was supported by the Grant-in-Aid for Young Scientists (B)
(No.~16K17629)
from Japan Society for the Promotion of Science.

config: arxiv


TITLE: Random reflections unexpectedly produce banded distributions
QUESTION [8 upvotes]: Start with $p_1$ a random point on the origin-centered unit circle $C$.
At step $i$, select a random point $q_i$ on $C$, and a random mirror line
$M_i$ through $q_i$, and reflect $p_i$ in $M_i$ to point $p_{i+1}$.
(Note that if every $M_i$ passed through the origin rather than $q_i \in C$,
all $p_i$ would lie on $C$.)
In the example below, $p_1$ on $C$ reflects to $p_2$ near the origin,
$p_2$ reflects out to $p_3$ near $(1,1.5)$, etc.

Green connects $p_1,\ldots,p_5$. Dark lines are mirrors $M_i$.

I expected that repeating this process would produce some smooth
distribution around the origin, as the points reflect further and further out,
when the mirrors effectively pass through the origin at that expanded scale.
But I find not infrequently bands of density, as depicted below.

Each distribution includes $n=5000$ reflected points. The unit circle is red.
Approx. radii: $107,67,47,148$.

Extending $n$ to larger values seems to continue banding effects.
Here is the 3rd example above, the most uniform of those four, extended.
Its radius increases from $47$ to $225$:

Continuation of 3rd example above, to $n=25000$. Approx. radius $225$.

Q. What explains this radial clustering/banding behavior?

REPLY [2 votes]: As said by Sangchui Lee, we are interested in $r_i = |p_i|$ and we assume that the reflections induce enougth random rotation such that the "bands" of radius $r$ just reveal the number of time $r_i$ was closed to $r$. And we have
$$| p_{i+1}| =|p_i -2\langle q_i, n_i \rangle n_i|$$
We define then a random march $X_i$ defined by $X_0=0$ and $$X_{i+1}=X_i-2\langle q_i, n_i \rangle n_i$$
We immediately have that $|p_i|$ has the same law as $|X_i|$ and that for large $i$, $X_i$ behave like the brownian motion on $\mathbb{R}^2$.
One can calculate that the variance of $2\langle q_i, n_i \rangle n_i$ is $\sigma^2 = 2$, the points should be then in a disk of randius $\sim \sqrt{n \sigma^2 }=100$ for $n=5000$ and $224$ for $n=25000$ .
Conclusion for large $i$, $|p_i|$ is given by the Bessel process. And as Sangchui Lee said we are interested in the local time.
https://en.wikipedia.org/wiki/Local_time_(mathematics).
The Ray-Knigh theorem should be usefull here even if it is state for the Brownian motion and not for the Bessel process.
(original paper of Knigh https://www.jstor.org/stable/pdf/1993647.pdf)


TITLE: How to systematically simplify deduction in formal first order logic into daily English
QUESTION [3 upvotes]: Suppose the function
\begin{equation}
h(x,y) = \frac{\sin\sqrt{1-x^2-2y^2}}{\sqrt{1-x^2-2y^2}}
\end{equation}
is defined on $\left\{(x,y)\vert x^2 + 2y^2 < 1\right\}$. We need to prove that
\begin{equation}
\forall x_0 \in \mathbb{R}, \forall y_0 \in \mathbb{R}, {x_0}^2 + 2{y_{0}}^{2} = 1 \longrightarrow \lim_{(x,y)\to(x_0,y_0)}h(x,y) = 1
\end{equation}
In my proof, I assumed the existence of the following conclusions:
\begin{equation}
\lim_{u\to 0}\frac{\sin u}{u} = 1,
\end{equation}
\begin{equation}
\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0.
\end{equation}
To strictly follow the rules of first order logic, I applied the fitch style. The proof is divided into two parts. In the first part, I deduced the following proposition:
\begin{equation}
\begin{aligned}
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0\right) \longrightarrow\\
&\left(\forall u, \forall v, u^2 + 2v^2 = 1 \longrightarrow
\forall \varepsilon > 0, \exists \delta > 0, \forall x, \forall y, \left(0 < \sqrt{(x-u)^2+(y-v)^2} < \delta \wedge x^2+2y^2 < 1\right) \longrightarrow \sqrt{1-x^2-2y^2} < \varepsilon\right)
\end{aligned}
\end{equation}
The formal proof is as follows:

In the second proof, the final conclusion is deduced. The formal proof is given below:

The formal proof of this seemingly simple conclusion is very tedious. However, without proper use of first order logic, there is no systematic way to examine a deduction. Recalling my first classes in undergraduate, I feel very upset about the type of proof I learned at that time. Every deduction was based on intuitive use of math rules, especially for quantification. I am wondering how to train myself to prove in English while have a picture of first order logic in mind.

REPLY [1 votes]: Actually the proof you wrote is structurally very close to what a mathematician would write, the trick is to make the right omissions, and to be conservative when expanding definitions.
For example, by making the right omissions, translating a hypothesis by "suppose", universal elimination by "let", a conclusion by "then", and existential elimination by "there exists ... such that", then your first proof almost literally translates to:

Suppose $$\forall u. \forall v. (u^2 + 2v^2 = 1 \longrightarrow\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0),$$ let $u_0,v_0$ be arbitrary, then suppose  $$u_0^2 + 2v_0^2 = 1,$$ then $$\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0,$$ let $\epsilon_0$ be arbitrary, suppose $\epsilon_0
> 0$, then there exists $\delta_0$ such that $$\delta_0>0,$$ $$\forall x.\forall y. 0 < \sqrt{(x - u_0)^2 + (y - v_0)^2} < \delta_0\longrightarrow|1x^2-2y^2|<\epsilon_0^2,$$
let $x_0,y_0$ be arbitrary, then suppose $$0 <\sqrt{(x_0-u_0)^2+(y_0-v_0)^2}<\delta_0,$$ $$x_0^2 + 2y_0^2 < 1,$$ then $$1 - x_0^2 - 2y_0^2 > 0,$$ $$|1 - x_0^2 - 2y_0^2| < \epsilon_0^2,$$ $$\sqrt{1 - x_0^2 - 2y_0^2} < \epsilon_0,$$ thus $$\forall \varepsilon > 0. \exists \delta > 0. \forall x. \forall y. \left(0 <\sqrt{(x-u)^2+(y-v)^2} < \delta \wedge x^2+2y^2 < 1 \\\longrightarrow\sqrt{1-x^2-2y^2} < \varepsilon\right).$$

You could write this in even more natural language as

Suppose that $$\forall u,v. (u^2 + 2v^2 = 1 \longrightarrow \lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0).$$ Now let $u_0,v_0$ be arbitrary, and suppose
that $u_0^2 + 2v_0^2 = 1,$ then $\lim_{(x,y)\to (u,v)} (1-x^2-2y^2) = 0.$
So for $\epsilon_0 > 0$,
there exists $\delta_0>0$ such that for all $x$ and $y$,
$0 < \sqrt{(x - u_0)^2 + (y - v_0)^2} < \delta_0$ implies $|1x^2-2y^2|<\epsilon_0^2.$
Now let $x_0,y_0$ be arbitrary such that
$0 <\sqrt{(x_0-u_0)^2+(y_0-v_0)^2}<\delta_0$ and
$x_0^2 + 2y_0^2 < 1,$
then since $1 - x_0^2 - 2y_0^2 > 0,$
we know
$$1 - x_0^2 - 2y_0^2 = |1 - x_0^2 - 2y_0^2| < \epsilon_0^2,$$
which implies that $\sqrt{1 - x_0^2 - 2y_0^2} < \epsilon_0$, concluding the proof.

So even though the natural language proof looks informal and contains omissions, the idea is that it should contain enough information to be able to recover the fully formal proof you have written. However different people write proofs with varying degree of details, depending on how much they expect the reader to be familiar with the topic.


TITLE: The proof of Cantor's Intersection Theorem on nested compact sets
QUESTION [1 upvotes]: The book "Metric Spaces" by Babu Ram says this about the proof of Cantor's Intersection Theorem:

Create nested intervals $F_{n+1}\subset F_n$ such that $\lim_{n\to\infty}\text{diameter}(F_n)=0$. Choose points $x_i\in F_i$. $\{x_i\}$ is a cauchy sequence. Hence, by the completeness property of the metric space, it has a limit point in the space. Let it be $a$. Our claim is that $a\in\bigcap_{i=1}^{\infty} F_i$.
Let us assume to the contrary. Then there is an $F_k$ such that $a\notin F_k$, and hence $a\notin F_j$ for all $j\geq k$. Let $r=d(a,F_k)$. Then $r>0$. Hence, $B(a,\frac{r}{2})$ does not contain any of the points $x_e$, $e\geq k$. Therefore, it can't be the limit point of $\{x_i\}$.

My question
Why can't $r=0$? For example, if $F_k=(1,2)$, and $a=1$, then $$d(a,F_k)=\inf\{ d(a,y): y\in F_k\}=0$$ If $r=0$, then doesn't this whole proof fall flat? Clearly checking whether $B(a,0)$ contains any of the points of $\{x_i\}$, and hence checking whether it is a convergent point of $\{x_i\}$ makes no sense.

REPLY [1 votes]: The intervals $F_n$ are closed intervals, and hence, are a positive distance from any point outside of them. More generally, in any metric space, closed sets are a positive distance from any point outside them, since the closed set's complement is an open neighborhood of the point.


TITLE: Confusion about the indicator function in probability.
QUESTION [3 upvotes]: For a random variable $X$, we consider $X\textbf{1}_A$, where $A$ is some event. For some $x>0$, it seems that \begin{align}P(X\textbf{1}_A<x)&=P(X\textbf{1}_A<x, A\text{ is true})+P\left(X\mathbf{1}_A<x, A\text{ is not true}\right)\\[0.2cm]&=P(X<x)+P(0<x)\\[0.2cm]&=P(X<x)+1\end{align}
It is obviously wrong since any probability can not greater than 1. However, I don't know where I am wrong.

REPLY [4 votes]: $X\cdot 1_A$ is a random variable which is equal to $X$ when the event $A$ occurs, and is equal to zero otherwise. Therefore if $x>0$ then
$$ \mathbb{P}(X\cdot 1_A<x)=\mathbb{P}(\{X<x\}\cap A)+\mathbb{P}(A^c) $$
This probability is at most $\mathbb{P}(A)+\mathbb{P}(A^c)=1$, as expected.


TITLE: How can black hole increase its mass?
QUESTION [8 upvotes]: From observer point of view an object, which falls into black hole never crosses its horizon. Then how does black hole appears and grows its mass?
Or does any black hole looks (and feels by all other information sources) for us like a void sphere with all mass around its horizon?

REPLY [5 votes]: The Schwarzschild coordinates (which seem to suggest that no object ever crosses the event horizon when viewed from far outside) were derived for stationary case: no matter flows onto the black hole, the black hole has constant mass. In fact, Schwarzschild was assuming zero stress-energy tensor (vacuum solution).
However if you start adding a lot of mass to the black hole, the situation changes. Imagine you throw a little object towards the event horizon. It "seems" to freeze on the surface of the horizon (it actually visually disappears due to the red shift). Later on, there is a huge amount of material streaming to the black hole. It is thousands of times more mass than the original mass of the black hole. At this point the conditions under which Schwarzschild found his solution no longer stand, because the stress-energy tensor is far from being zero. The event horizon will grow, since it forms wherever the gravitational potential reaches certain value. By adding more mass you unavoidably enlarge the volume where the potential has the required value to form the event horizon.
The case of non-constant mass is described by the Vaidya metric. Mathematically this is described on pages 133-134 of this book.


TITLE: Notation with random variable $\overline{X}_{n^2}$ in Strong Law of Large Numbers proof.
QUESTION [2 upvotes]: I'm reading the proof for the strong law of large numbers. It says:

Let $X_1,X_2,\ldots$ be a sequence of independent and i.i.d. random variables with finite mean $\mu$ and finite variance $\sigma^2$. Then:
  $$\overline{X}_n=\frac{X_1+x_2+\cdots+x_n}{n}\underset{n\to\infty}{\stackrel{\text{a.s.}}{\longrightarrow}} \mu. $$

The proof reads:

Without loss of generality we can assume that $\mu =0$. Now
  $$\mathbb E\left(\left[\overline{X_n}\right]^2\right)=\frac{1}{n^2}\sum\limits_{1\leq j,k\leq n}\mathbb E\left(X_jX_k\right) = \frac{1}{n^2}\sum\limits_{j=1}^n\mathbb E\left(X_j^2\right) = \frac{\sigma ^2}{n}.$$
  Therefore:
  $$\lim\limits_{n\to\infty}\mathbb E\left(\left[\overline{X_n}\right]^2\right)=0.$$

Here comes the notation I don't understand and have never seen before:

Moreover $\mathbb E\left(\left[\overline{X}_n\right]^2\right)=\dfrac{\sigma^2}{n}$ implies that:
  $$\sum\limits_{n=1}^\infty\mathbb E\left(\left[\overline{X}_{n^2}\right]^2\right) = \sum\limits_{n=1}^\infty\frac{\sigma^2}{n^2}<\infty.$$

I know that
$$\sum\limits_{n=1}^\infty\frac{\sigma^2}{n^2}=\sum\limits_{n=1}^\infty \frac{1}{n}\mathbb E\left(\left[\overline{X}_n\right]^2\right),$$
So this $_{n^2}$ would be adding that factor $\dfrac{1}{n}$ into the summation.
Any idea of what $\overline{X}_{n^2}$ could mean? If you have seen this before I appreciate you share your knowledge about this notation.

REPLY [2 votes]: $$\overline{X}_{n^2}=\frac{1}{n^2}\sum_{i=1}^{n^2}X_i$$

Indeed since you have already proved that $$\mathbb E\left(\left[\overline{X_n}\right]^2\right)= \frac{1}{n^2}\sum\limits_{j=1}^n\mathbb E\left(X_j^2\right) = \frac{\sigma ^2}{n}$$ then you have that
$$\mathbb E\left(\left[\overline{X}_{n^2}\right]^2\right)= \frac{1}{n^4}\sum\limits_{j=1}^{n^2}\mathbb E\left(X_j^2\right) = \frac{1}{n^4}\cdot n^2σ^2=\frac{\sigma ^2}{n^2}$$ as the author of proof claims.


TITLE: About Mrowka spaces.
QUESTION [1 upvotes]: I'm studying some properties of Mrowka spaces and I'm stuck proving the next things. First, consider $\mathcal{M}$ a maximal almost disjoint family such that $|\mathcal{M}|>\aleph_0$. Let be $\psi=\omega\cup\mathcal{M}$ and let $$B=\{\{n\}\mid n\in\omega \}\cup\{\{M\}\cup S\mid M\in\mathcal{M} \ \text{and $S$ is a cofinite subset of $M$ } \}$$an open basis for $\psi$. I want to prove

$\mathcal{M}$ is a zero set of $\psi$, i.e., there exist $f\in C(\psi)$ such that $f^{-1}[\{0\}]=\mathcal{M}$.
Each subset of $\psi$ is $G_\delta$.

For the first one consider the next function.
\begin{equation}
f(x)=\left\{
\begin{array}{ccc}
\frac{1}{x} & \text{if} & x\in\omega \\\\
0 & \text{if} & x\in\mathcal{M}
\end{array}
\right.
\end{equation}
We need to prove that $f\in C(\psi)$. Let $U$ be an open set of $\mathbb{R}$. If $0\in U$ then $U$ contains infinitely many points of the form $1/n$ and only finitely many are outside $U$. Thus, $f^{-1}[U]=\mathcal{M}\cup\{n\in\omega\mid 1/n\in U \}$. This set is open because if $x\in f^{-1}[U]$ then we have two cases: if $1/x\in U$ then $\{ x\}$ is an open neighborhood of $x$ contained in $f^{-1}[U]$. If $x\in \mathcal{M}$ then, by the fact that $\omega\setminus \{n\in\omega\mid 1/n\in U \}$ is finite, there exist $S\subseteq \{n\in\omega\mid 1/n\in U \}$ such that $\{ x\}\cup S$ is an open neighborhood of $x$ contained in $f^{-1}[U]$.
If $0\notin U$ then $f^{-1}[U]\subseteq\omega$ and by the fact that every element of $\omega$ is open in $\psi$ then $f^{-1}[U]$ is open. It is correct?
For the second, I don't know how to proceed. Any hint? I really appreciate any help you can provide me.

REPLY [2 votes]: Your $f$ is fine (maybe better use $\frac{1}{x+1}$, as $0 \in \omega$ too..).
And if $O \subseteq \Bbb R$ does not contain $0$, $f^{-1}[O] \subseteq \omega$ so is open, and if $0 \in O$, cofinally many $f(n)$ are contained in $O$, say we miss only $f(n), n \in F$, where $F \subseteq \omega$ finite. Then for every $A \in \mathcal{M}$, $\{A\} \cup (A\setminus F) \subseteq f^{-1}[O]$ so all points of $\psi$ are interior points of $f^{-1}[O]$, and $f^{-1}[O]$ is open.
Any zero-set like $\mathcal{M}$ is a closed $G_\delta$. And $\mathcal{M}$ is a discrete subspace, so any $B \subseteq \mathcal{M}$ is open in $\mathcal{M}$, so also a $G_\delta$ in $\psi$ (as $(\psi\setminus B) \cap \mathcal{M}$, the intersection of a $G_\delta$ and an open set). It follows that any subset of $\psi$ is $G_\delta$: $A = (A \cap \mathcal{M}) \cup (A \cap \omega)$ is a union of a $G_\delta$ and an open set, so $G_\delta$. Or note that all subsets of $\mathcal{M}$ are closed in $\psi$ (closed in closed is closed) and adding countably many points from $\omega$ keeps it an $F_\sigma$. So all subsets are $F_\sigma$ too.
Note that if $\mathcal{M}$ has size $\kappa$ such that $2^\kappa > 2^\omega$ then Jones' lemma implies $\psi$ is not normal. It is Tychonoff, being Hausdorff and locally compact, or because it's zero-dimensional (it has a clopen base).


TITLE: proving Cartesian product of two metric space is a metric space
QUESTION [1 upvotes]: I saw somewhere that Cartesian product $X = X_1 \times  X_2$ of two metric spaces $(X_1,d_1)$ and $(X_2,d_2)$ can be made into a metric space $(X,d)$ like following:
$d(x,y) = (d_1(x_1,y_1)^p + d_2(x_2,y_2)^p)^{1/p}$
How do I prove $(X,d)$ is metric space? More specifically, how to prove triangular inequality?
[Note: $p\ge1$, $x=(x_1,x_2)$ and $y= (y_1,y_2)$]

REPLY [1 votes]: Hint: Suppose $X_1 = X_2 = \mathbb{R}$. You know how to make a metric on $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ out of the metric on $\mathbb{R}$ in a way that looks like the one above (think: Pythagorean theorem). Try and generalize this.


TITLE: Equivalence definition of rapidly decreasing in $\mathbb R^n$
QUESTION [1 upvotes]: I am trying to prove:

a continuous function $f:\mathbb R^n\to \mathbb R$ is rapidly decreasing iff $\sup_{x\in \mathbb R^n}|x|^k|f(x)| <\infty $ for all $k$

where a conitnuous $f$ is defined to be rapidly decreasing iff $\sup_{x\in\mathbb R^n}|x^\alpha f|<\infty$ for all multi-index $\alpha\in\mathbb N^n$. Here $x^\alpha:=x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$.
I was trying to show some inequality relations between $|x^\alpha|$ and $|x|^k$. I found that if $1\le |x_i|$ for all $i$, then $|x^\alpha|\le|x|^{2k||\alpha||_{\infty}}$, as by expanding $|x|^{2k}$, we have $x_1^2...x_n^2\le |x|^{2k}$, and therefore $|x^\alpha|\le x_1^{2||\alpha||_\infty}...x_n^{2||\alpha||_\infty}$ as $1\le |x_i|$.
I am stuck here. Any help or comment will be appreciated.

REPLY [0 votes]: For your first inequality, you actually don't need any case distinction, because for any $i\in\{1,\dots, n\}$ we have $|x_i|\leq \|x\|_2$, so
\begin{align}
|x^{\alpha}|&:=|x_1^{\alpha_1}\cdots x_n^{\alpha_n}|\leq \|x\|_2^{\alpha_1+\dots +\alpha_n}.
\end{align}
For the other, note that clearly, $|x_i|\leq |x_1|+\dots +|x_n|$ for all $i$, so
\begin{align}
\|x\|_2&:=\sqrt{x_1^2+\dots +x_n^2} \leq \sqrt{n}(|x_1|+\dots + |x_n|)
\end{align}
So, if you raise both sides to the power $k$, then on the RHS, by expanding everything out, you get a finite sum of terms of the form $|x^{\alpha}|$; i.e there is a $C>0$ (which is like the maximum of all the combinatorical factors obtained when expanding the RHS) and a finite set $A\subset \Bbb{N}_0^n$ of multindices such that
\begin{align}
\|x\|_2\leq C\sum_{\alpha\in A}|x^{\alpha}|.
\end{align}


TITLE: Fine tuning: why is multiverse with anthropic principle considered a better explanation than anthropic principle alone?
QUESTION [1 upvotes]: I got the impression from several public lectures that, when it comes to the fine tuning problem, physicists tend to give multiverse together with anthropic principle as a plausible solution, but seldom anthropic principle alone.
However it seems the core of this kind of solution is really anthropic principle, that is, once we decide to invoke anthropic principle, the concept of multiverse becomes irrelevant(if fine tuning problem is our only concern), adding multiverse to the explanation does not seem to strengthen the solution by much. Is there anything I missed or misunderstood?
Of course it is possible that I overestimate the popularity of multiverse explanation in academia, after all I got this impression from public lectures and I never talked to a cosmologist in person, in this case I would be happy if anyone better informed can confirm this.

REPLY [2 votes]: It seems plausible that the probability of the universe being just right for human life by pure chance is pretty low, and possibly absurdly low. That means if the universe only had one chance at existing the probability we'd be around to see it is vanishingly small.
The various multiverse theories suggest that huge numbers of disconnected universes get created, so even though the probability of human life in any single universe is small there are enough universes that it becomes almost inevitable that at least one of them contains humans. This is why the anthropic principle only makes sense if multiple universes exist.
Response to comment:
Ideally physicists test their theories with experiment. Sadly we can't create universes so we can't easily test our theories of how the universe came to exist. It seems likely we'll never be able to prove beyond reasonable doubt whether our ideas are correct.
Given this situation we look at the probability that any particular explanation for the universe could be true. If we accept the idea that we are here by chance then it is much much more probable that the universe had multiple goes than that it had only one go. When I say much much more probable no-one can put a figure on this but if we believe ideas about the string landscape the probability that a single universe got it right is around $1$ in $10^{500}$. This number is so small as to be indistinuishable from zero. The number of atoms in the observable universe is only around $10^{80}$.
Similar arguments have been made before. In 1802 William Paley observed that the probability of a living being appearing by chance was so low that there had to have been a divine watchmaker to create it. Scientists agreed this was improbable, and in 1859 Darwin explained that living objects didn't appear by chance but were the result of evolution having lots of goes at it.
The same applies to the origin of the universe. Asking us to believe a single universe got it right first time is at the six impossible things before breakfast level. It seems vastly more likely a multiverse is responsible.


TITLE: How do we know $\pi$ is a constant?
QUESTION [2 upvotes]: How did the ancient Greeks discover that the ratio of a circle's circumference to its diameter is constant? It does not seem so intuitive. Thanks!

REPLY [1 votes]: Probably they used an intuitive (or maybe exact) idea of geometric similarity.


TITLE: Finite difference method for a PDE
QUESTION [1 upvotes]: I am trying to solve the following PDE using FDM:
$\frac{\partial^2 u}{\partial y^2} = x \frac{\partial u}{\partial x}$.
I discretized the independent variables as $x=x_0+i \Delta x$ and $y=y0+j \Delta y$. In my case,
$x_0=0=y_0$. I also descritized the derivaties and obtained the FDM as
$\frac{u_{i,j+1} -2 u_{i,j}+ u_{i,j-1} }{{\Delta y}^2} = i\Delta x \frac{u_{i,j}-u_{i-1,j}}{\Delta x}$.
Now, my question is, isn't it strange that the $\Delta x$ cancels? It apparently implies that the
accuracy of the solution is independent of the choice of $\Delta x$. However, I know a large $\Delta x$ introduces error into the solution.
The $x$-domain is semi-infinite in my problem.
Thanks.

REPLY [1 votes]: The equation you give corresponds to a system of linear equations, whose solution is your estimation for u. However note that for any scalar $\alpha$ the equations $Ax = b$ and $\alpha A x = \alpha b$ have the same solution for $x$. So, you can always express this system of linear equations that only $\Delta x$ or $\Delta y$ show up, by applying some scalar first. In fact in might be beneficial to apply such a scalar for some numerical methods, although good solvers will be able to handle that themselves.
You should see both $\Delta x$ and $\Delta y$ appear in you error estimate! Usually you can find an error estimate by expanding the "real" solution as a Taylor series, see for example here.


TITLE: Line Integral Work Done
QUESTION [0 upvotes]: I have a problem when comes to question 2, I don't know how to put this into parametric form. And I am not sure if this is a parabola. Thanks in advance.

REPLY [1 votes]: If $y=4x^2$ then the $y$-coordinate of 4 times the square of the $x$-coordinate.
Let the $x$-coordinate be $t$, then the $y$-coordinate of $4t^2$.
Your path is given by $\gamma(t) = (t,4t^2)$, where $0 \le t \le 1$.
The curve $y=4x^2$ is indeed a parabola, with focus $(x,y)=(0,\frac{1}{16})$ and directrix $y=-\frac{1}{16}$.


TITLE: compute the integral $\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$
QUESTION [3 upvotes]: Compute this integral:
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
my solution is I used derivative of Cauchy integral formula, which is
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\int \frac{f(z_0)}{(z-z_0)^{n+1}}$$
Then, I got
$$\int_{|z|=1}\left[\frac{z-2}{2z-1}\right]^3dz$$
$$ = \int_{|z|=1} \frac{(z-2)^3}{(2z-1)^3}dz$$
$$ = \frac{2 \pi i}{2!}\left(\frac{1}{2} - 2\right)^3 = \frac{27\pi i}{8}$$
That's what I get from the Cauchy integral formula. So, am I correct for this approach. If not, can someone please show me how to get the right one? Thank you

REPLY [1 votes]: You were close
$$f^{(2)}(1/2)=\frac{2!}{2\pi i}\oint_{|z|=1}\frac{(z-2)^3}{(2z-1)^3}dz=\frac{2!}{2\pi i}\oint_{|z|=1}\frac{\frac18(z-2)^3}{(z-1/2)^3}dz$$
where $f(z)=\frac18(z-2)^3$ and thus $f^{(2)}(1/2)=\frac18 (6)(-3/2)=-\frac98$
Finally, we have
$$\bbox[5px,border:2px solid #C0A000]{\oint_{|z|=1}\frac{(z-2)^3}{(2z-1)^3}dz=-\frac98 \pi i}$$


TITLE: wanted: a lucid demonstration that these $m+n$ bilinear equations have a solution
QUESTION [3 upvotes]: I shall first state the problem, and if it arouses any interest, I may gradually add a few notes, to provide a little context for those who, like myself, appreciate such background information. let me say once that all quantities are positive reals.
DATA an $m \times n$ matrix $X=\{x_{ij}\}$. an $m$-vector $P=\{p_{i}\}$ and an $n$-vector $Q=\{q_{j}\}$.
CONDITION $$\sum_{i=1}^m p_i = T = \sum_{j=1}^n q_j $$
PROBLEM
(a) find $m+n$ numbers $\lambda_i$ and $\rho_j$ satisfying:
$$ \lambda_i\sum_{j=1}^n  \rho_j x_{ij} = p_i \\
 \rho_j \sum_{i=1}^m \lambda_ix_{ij} = q_j \\ $$
(b) show, in a lucid and clear manner, that under the stated conditions  a solution always exists, and that for a given $X,P,Q$ it is unique apart from a multiplicative constant - obviously if ($\lambda, \rho$) is a solution, so is ($\alpha^{-1}\lambda, \alpha \rho$).
NOTES
this problem is of considerable practical importance - for example in certain areas of study connected with spatial diffusion in social systems.
it has a simple and elegant algorithmic solution which was perhaps discovered more by accident than design.
the convergence of the algorithm and the essential uniqueness of the solution have been demonstrated, but the only such proof I have seen runs to several pages of rather formidable algebraic reasoning. whilst I greatly admire those with the fortitude to engage successfully in such a hammer-and-chisel approach, the convergence of the algorithm is so impressive that one feels there may be some possibility of a more elegant and lucid treatment.
some years ago I heard a rumour that the result has been treated as a problem in pure mathematics, perhaps in the early 1960's. I have not been able to chase up a reference, however, and would be very grateful if anyone can provide a link to such work.
the desire to provide a proper setting for this question has been one of the chief motives for my wishing to study mathematics more seriously. MSE is a great learning environment and I am grateful to many people who have set me right about certain false assumptions. it is a steep learning curve, but I have already made considerable progress with this formal side of things, which, even in the absence of the required solution, is aesthetically satisfying. I have been led to the study of certain finite-dimensional Banach algebras which do seem the natural setting for what, in the end, is a fixed-point theorem.
there are obvious possibilities to look towards generalizations.
for example one could look at the infinite-dimensional case - say in a sequence space. it is possible that the same problem has been treated in the context of integral equations, and that some more general theorem of this type exists, so that the finite-dimensional case would be a simple corollary.
I noticed many years ago, when I was exploring the issue via a computer implementation, that convergence could be obtained in some cases when the initial matrix X had complex entries. however, without an appropriate theoretical framework, I thought it better to leave such exploration aside until I had a better grasp of the basic situation.
perhaps the most interesting generalization is to the case where we have three or more marginal constraints.
in the hope of possibly finding an applicable general theorem which can be used to obtain the desired result as a corollary, I have tentatively reformulated the problem as a differential equation  here, but so far have had no feedback at all on this from the MSE community.
I have described in brief my evolving terminology  here, in the context of a serious question about the norm restriction on multiplication in a banach algebra. again, no useful feedback has as yet been forthcoming.

REPLY [2 votes]: Since $p$ and $\rho$ are so similar typographically, I took the liberty of using $\mu_j$ instead of $\rho_j$.
For brevity, let $C^n$ denote the (closed) non-negative orthant in $\mathbf{R}^n$ with the origin removed, $C_+^n$ the (open) positive orthant (a.k.a. the interior of $C^n$), $\Delta^n$ the (closed) $(n - 1)$-simplex
$$
\{(x_j) \text{ in }\mathbf{R}^n: \text{$x_j \geq 0$ for all $i$, and $\sum_j x_j = T$}\},
$$
and $\Delta_+^n = \Delta^n \cap C_+^n$ the interior.
For each $\mu$ in $C^n$, define $\lambda$ in $C_+^m$ by
$$
\lambda_i = \frac{p_i}{\sum_{k} x_{ik} \mu_k}.
$$
Since the $x_{ik}$ are all positive, and $\mu_k \geq 0$ for all $k$, and at least one $\mu_k$ is is non-zero, the denominator is positive for each $i$. Since $p_i > 0$ for all $i$, $\lambda_i > 0$ for all $i$. In words, for each $\mu$ in $C^n$, there exists a unique $\lambda$ in $C_+^m$ satisfying the first set of equations,
$$
\lambda_i \sum_{j=1}^n x_{ij} \mu_j = p_i,
$$
and the dependence is continuous in $\mu$.
Define a rational map $F:C^n \to C^n$ by taking the $j$th component of $F(\mu)$ to be
$$
F(\mu)_j = \mu_j \sum_{i=1}^m \frac{x_{ij} p_i}{\sum_k x_{ik} \mu_k}.
$$
By the preceding observations, $F$ is continuous. Moreover, (i) $F$ is constant along rays through the origin (i.e., $F(t\mu) = F(\mu)$ for all $t > 0$); (ii) $F$ maps each face of $C^n$ into itself (i.e., the hyperplane $\{\mu_j = 0\}$ maps into itself) and similarly for intersections of faces; and (iii) $F$ maps $C_+^n$ to itself (if $\mu_j > 0$ for all $j$, then $F(\mu)_j > 0$ for all $j$).
Both sets of equations
$$
\lambda_i \sum_{j=1}^n x_{ij} \mu_j = p_i,\qquad
\mu_j \sum_{i=1}^m x_{ij} \lambda_i = q_j,
$$
are solved if there exists a $\mu$ in $C_+^n$ such that $q = F(\mu)$.
Existence: By following $F$ with a radial scaling, we obtain a continuous map $f:\Delta^n \to \Delta^n$. Explicitly, if $F(\mu) = y = (y_k)$, then
$$
f(\mu) = \frac{T}{\sum_{k} y_k}\, y.
$$
In particular, $f(\mu) = F(\mu)$ if and only if $\sum_k y_k = T$, if and only if $y \in \Delta^n$.
Since $F$ maps each intersection of faces of $C^n$ to itself, $f$ maps each sub-simplex of $\Delta^n$ to itself. By this MO post, $f$ is surjective. But $q \in \Delta^n$, so $f(\mu) = q$ for some $\mu$ in $\Delta^n$. The preceding argument shows $F(\mu) = q$ as well.
Uniqueness (up to scaling): For each $j$, the $j$th component function of $F$ may be written
$$
F(\mu)_j = \sum_{i=1}^m \frac{x_{ij} p_i}{\sum_k x_{ik} (\mu_k/\mu_j)}.
$$
The right-hand side is strictly decreasing as a function of the $(n - 1)$ ratios $\mu_k/\mu_j$ with $k \neq j$. Consequently, if $F(\mu) = F(\mu')$, then for each $j$, we have $\mu_k/\mu_j = \mu'_k/\mu'_j$ for all $k$. It follows that $\mu$ and $\mu'$ are proportional as vectors in $C^n$.


TITLE: Finding parabola tangents equations given parabola equation and a point
QUESTION [0 upvotes]: I have parabola equation (notice it's "sideways", given x =)
$\displaystyle x = \frac{y^2}{2}-4y+3$ and a point $(2, -1)$.
Find parabola tangents (equations) that go through point $(2,-1)$.
Ive watched several videos and looked similar solved problems, however I'm getting it wrong somewhere.
Maybe it's because I cant take into account that parabola is given through $x$ and not $y$. Im not sure.
But here is what I tried to do.
I take derivative of parabola equation : $x'= y-4$
so $x'$ is the slope of tangents
I can take some point on parabola $x,y$ and replace $x$ with equation so point is $(\frac{y^2}{2}-4y+3, y)$
now I put that point into slope forumla with given point
$\displaystyle \frac{y-(-1)} {\frac{y^2}{2}-4y+3} - 2 = y - 4$
if I slove this I should get y coordinates of points where tangents touch parabola ?
However the results do not make sense and the resulting formulas I make for tangents are not correct.
There is correct answer given to this problem, which is tangents equations are either
y = −0,1202...x − .0, 7595..
or
y = −0,5940...x + 0,188

I cant reach them, what am I missing here?

REPLY [0 votes]: Your derivative is incorrect and hence you are getting wrong answer.
Parabola equation is $x = \frac{y^2}{2}-4y+3$
you should differentiate wrt. $x$
$1 = y \frac{dy}{dx} - 4 \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{y-4}$
So equation of tangent line is
$y + 1 = \frac{1}{y_0-4} (x -2)$ ...(i)
(if the tangent goes through point $(x_0, y_0)$ on the parabola).
As the point $(x_0, y_0)$ is both on the line and the parabola,
$y_0 + 1 = \frac{1}{y_0-4} (x_0 -2) \implies y_0^2 - 3y_0 -2 = x_0$
and also $x_0 = \frac{y_0^2}{2}-4y_0+3$
Solving both,
$y_0 = \pm \sqrt 11 - 1$ gives you slopes $(\frac{1}{y_0-4}) \approx -0.59404, -0.12024$
Substituting both values in equation $(i)$, you get two tangents to parabola passing through point $(2, -1)$.


TITLE: From $L^p$ to $L^q$ Norm for finite trigonometric polynomial
QUESTION [1 upvotes]: Let $1\leq p\leq q \leq \infty$ and let $f\in L^p([0,2\pi])$ such that $\hat{f}(k)=0$ for all $\vert k\vert > N$. I would like to show that
$$\Vert f \Vert_p \leq (2N-1)^{1/p-1/q} \Vert f \Vert_q.$$
I managed to show the case $p=\infty$, $q=2$ using Young's Inequality for $f= f * g$, where $g$ is a function such that $\hat{g}(k)=1$ for all $k= -N,\ldots, N$ and zero otherwise, and Parseval Identity. How can I obtain the general case?
Thank you very much for any help.

REPLY [1 votes]: Your proof generalises with little effort to a large number of pairs $(p,q)$ but not all of them. Young's inequality gives
$$ \| f * g \|_{L^p} \le \|f\|_{L^q} \|g\|_{L^r} $$
where $1+\frac1p = \frac1q +\frac1r$. The natural replacement for Parseval is (the dual inequality of) Hausdorff-Young, which works for $r\ge 2$ i.e. $\frac12 + \frac1p \le \frac1q$, giving
$$ \|g\|_{L^r} = \|\mathcal F^{-1} \hat g\|_{L^r} \le \|
\hat g\|_{\ell^{r'}}, \quad \frac1r + \frac1{r'}=1 \iff \frac1{r'} = \frac1q - \frac1p$$
which gives $$\|g\|_{L^r} \le (2N+1)^{1/r'} = (2N+1)^{1/q - 1/p} $$
since $\hat g$ is supported on the $2N+1$ integers $|n|\le N$ where it takes the value $1$. (PS I think your inequality is not quite right)
This proof doesn't work when $\frac1q-\frac1p < \frac12$, and I suspect the result isn't true with constant 1 in this case since all sources I look at do not phrase it (Bernstein's inequality on the torus, see this answer and the referenced "EBP lecture notes on singular stochastic PDEs" by Gubinelli and Perkowski, pages 37 and onwards) with a constant 1. If one is prepared to lose a constant, then  the computation here also finishes the proof, since your $g$ is nothing but the Dirichlet kernel $g=\sum_{|k|\le N} e^{ikx}$.