TITLE: What is the role of the logarithm in Shannon's entropy?
QUESTION [4 upvotes]: I am a layman interested in understanding why the foundation of Shannon's entropy is logarithmic.
To that end I've read the answers here, at the Cross Validated Stack, but I'm not technical enough to infer the basic idea from the math.
https://stats.stackexchange.com/questions/87182/what-is-the-role-of-the-logarithm-in-shannons-entropy
But in trying to make an semi-educated guess I infer the following and am asking here if this is a reasonable explanation.
Shannon's entropy is logarithmic because the CHANCES of multiple information events occurring simultaneously are multiplied ... but should all those events occur the total VALUE of those events are summed.
For example, betting on coin flips. The chance of heads is $1/2$ while the chances for three heads in a row is $1/8 = (1/2)^3$ as each flip is independent.
Assuming the bet for each flip is also independent, the total won for winning each of the three flips is $B_1+B_2+B_3$.
This is a logarithmic relationship because we can express:

(a) An exponent as a log - the chance being 'to the power of'
(b) The result of said event as a sum.  In other words, logarithms let you express multiplications as sums and vice versa.

Thanks.

REPLY [1 votes]: Short version: What you said:

the chances of multiple information events occurring simultaneously are multiplied ... but should all those events occur the total value of those events are summed

is pretty much correct.
Suppose you are going to send a message $M_1$ and then a message $M_2$.  Let's try to invent a measure of the information content of the two messages and  write $I(M_1)$ and $I(M_2)$ for the information content of the two messages.  What properties would we like this “information content” to have?
Suppose we consider the act of sending $M_1$ and then $M_2$ as a single communication, which we could call $M_1\oplus M_2$.  Whatever $I$ means, we ought to have $$I(M_1\oplus M_2) = I(M_1) + I(M_2)$$
which says that the amount of information in the combined message is equal to the sum of the information in the two messages sent separately.
Now suppose $M_1$ could consist of any of $n_1$ distinguishable signals and similarly $M_2$ will be one of $n_2$ distinguishable signals.  It makes sense that  the information content is related to the number of possible signals: if I could send you one of a million possible messages, the message I send will carry more information than if there are only two possible messages I could have sent.  Let's see what happens if we try to express $I$ only in terms of $n_1$ and $n_2$.
Since there are $n_1$ was to send $M_1$, and $n_2$ ways to send $M_2$, there are just $n_1\cdot n_2\text{ }$ways to send $M_1\oplus M_2$.  So if $I$ can be expressed in terms of the number of possible messages that could be sent, we must have $$I(n_1\cdot n_2) = I(n_1)+I(n_2)$$ which tells us that $I$ must be logarithmic in the number of possible messages.
The choice of which logarithm we use isn't important; it is simply a change of units, like the choice of whether to measure distance in feet or meters.  One common choice is a base-2 logarithm, in which case we get a unit called a bit; sometimes we use a base-$e$ logarithm and get a unit called a nat which is 44% larger.  In the past a base-10 logarithm was sometimes used, producing a unit called a hartley which is around 3.32 bits.


TITLE: Finding the parametric equations for tangent line to an ellipse at a given point
QUESTION [0 upvotes]: The original question is:
The ellipsoid $4x^2+2y^2+z^2=16$ intersects the plane $y=2$ in an ellipse. Find parametric equations for the tangent line to this ellipse at the point $(1,2,2)$.
I know the slope is $-2$ and the equation of the tangent line at point $(1,2,2)$ is $z-2=-2(x-1), y=2$. The given solution jumped directly from here to the conclusion.
The problem is that I have no idea how to get the directional vector of this tangent line.

REPLY [1 votes]: We write down the equations of the ellipsoid and the plane:
$$
4 x^2 + 2 y^2 + z^2 = 16 \tag{1}
$$
$$
y = 2 \tag{2}
$$
When (1) and (2) intersect,  we find that
$$
4 x^2 + 2 (4) + z^2 = 16 \ \ \mbox{or} \ \
4 x^2 + 8 + z^2 = 16
$$
Thus, the intersection of (1) and (2) is an ellipse:
$$
4 x^2 + z^2 = 8 \tag{3}
$$
We know that the tangent line on the ellipsoid at the point $(1, 2, 2)$ must be an element of the plane $y = 2$.
From the equation of the ellipse, we find the slope at the point $(1, 2)$ using the equation (3):
$$
8 x + 2 z z_x = 0
$$
or
$$
z_x = - {8 x \over 2 z} = - {4 x \over z}
$$
At the point $(1, 2)$,
$$
z_{x} = - {4 \over 2} = -2
$$
Now, the equation of the tangent line to the ellipse (3) at the point (1, 2) is
$$
z - 2 = m (x - 1)   = -2 (x - 1)
$$
since $m = -2$ is the slope.
Simplifying, we get the tangent line as
$$
z = - 2 x + 4 \tag{4}
$$
This shows that the tangent line for the ellipsoid (1) at the point (1, 2, 2) is:
$$
z = - 2 x + 4, \ \ y = 2
$$
We recall that the parametric equations for a line passing through a point $(x_0, y_0, z_0)$ in the direction of the vector $(l, m, n)$ in the 3-D space is:
$$
{x - x_0 \over l} = {y - y_0 \over m} = {z - z_0 \over n}
$$
It is given that $(x_0, y_0, z_0) = (1, 2, 2)$ and $m = 0$ (Note that the line lies in the plane $y = 2$).
From the equation of the tangent line $z = - 2 x + 4$ obtained in Eq. (4), we get the equation of the tangent line as
$$
{x - 1 \over l} = {y - 2 \over 2} = {z - 2 \over 2} = t
$$
Thus,
$$
x = 1 + l t
$$
$$
y = 2
$$
$$
z = 2 + n t
$$
$$
2 + n t = - 2 ( 1 + l t) + 4
$$
Simplifying, we get
$$
2 + n t = -2 - 2 l t + 4
$$
or
$$
n t + 2 l t = (n + 2 l) t =  4 - 4 = 0
$$
This gives
$$
n = - 2 l
$$
If we take $l = 1$, then we get $n = -2$.
Thus, we write the parametric equations of the tangent line   as
$$
{x - 1 \over 1} = {y - 2 \over 0} = {z - 2 \over -2} = t
$$
or equivalently
$$
\boxed{x = t + 1, \ \ y = 2, \ \ z = - 2 t + 2}
$$


TITLE: For what range does this floor function scale to?
QUESTION [0 upvotes]: I have $\lfloor\frac{X}{(2y+1)^2}\rfloor = k$ where $X$ and $k$ are known. For what values of $y$ will this hold true?
edit: all are positive integers

REPLY [1 votes]: $\displaystyle \left\lfloor\frac{X}{(2y+1)^2}\right\rfloor = k \iff k \le \frac{X}{(2y+1)^2} < k+1$
You know that $(2y+1)^2 \ge 0$, so we may multiply it on both sides.  Moreover, since you have $k \ge 0$ and $X \ge 0$, this yields:
$\frac{X}{k+1} < (2y+1)^2 \le \frac{X}{k}$
I believe you can do the rest.


TITLE: Why does alternator become easier to turn if you disconnect a transformer from it?
QUESTION [0 upvotes]: An experiment shown in class showed an alternator connected to a transformer. When you disconnected the transformer, the alternator become easier to move.
My thoughts are that when the current induced from the generator passes through a transformer, a magnetic field is produced. This magnetic field interacts with the alternator's own magnetic field (produced when a change in magnetic flux occurs), making it harder to move.
Is this correct or is there an alternative explanation.  Please note this is a 9th grade question

REPLY [0 votes]: when the alternator is disconnected from the transformer, it "sees" an open circuit, through which no current can flow. This means when you turn the alternator shaft, no current is flowing in the coils, no power is leaving the alternator, it is performing no work and it takes very little work to turn the shaft.
When you connect the alternator to the primary coil of a transformer, the alternator "sees" a complete circuit with a small amount of resistance in it (the DC resistance of the primary coil) and so current is set into motion, current times voltage is power, power is leaving the alternator, which means it takes work to turn the shaft to deliver that power.


TITLE: Algorithm to compute maximum independent set of a chordal graph
QUESTION [0 upvotes]: If $G$ is perfect => $\omega(G) = \chi(G)$
Chordal graphs are perfect graphs

With an algorithm $A$ I can compute the perfect elimination ordering of a chordal graph $G$. I can extend $A$ in such way that it outputs the $\chi(G)$ (maximal cliques and graph coloring).
Also I know:

A graph $G$ is perfect <=> $\bar G$ is perfect

Can I conclude
If $\bar G$ is perfect => $\alpha(G) = \bar \chi(G)$
Because $\omega(G) = \chi(G) = \alpha(\bar G) = \bar \chi (\bar G)$ ?
If so than I can compute the maximal independent set $\alpha(G)$ with algorithm $A$ and $\bar G$ of the chordal graph $G$. Is this right?

REPLY [0 votes]: The above argument fails because the complement of a chordal graph is not chordal. Hence you cannot use algorithm A to find the chromatic number of the complement graph, since algorithm A outputs the correct chromatic number only if the graph is chordal.


TITLE: Optimization: Cut wire in to two pieces one square and one circle
QUESTION [1 upvotes]: A straight piece of wire $40$ cm long is cut into two pieces. One piece is bent
into a circle and the other is bent into a square. How should wire be cut so
that the total area of both circle and square is minimized?
my work
I found the function to be $(x/4)^2 + \pi ( (40-x) / 2\pi ) ^2$. However, the answer key showed a different answer and I'm not sure how to get to that equation.
The critical point that I got was $160/(\pi+4)$, but it was $20/(\pi+4)$ for the answer key answer key

REPLY [0 votes]: You did everything correctly. You just defined $x$ to be the perimeter of the square, while the answer key defined $x$ to be the radius of the circle. Indeed, look at the last sentence: one of the pieces has a length of $\frac{160}{\pi + 4}$.


TITLE: How can I mathematically model the combinatory problem?
QUESTION [0 upvotes]: I have the following problem, and I would like to model it using a mathematical formula, for a purpose of optimization problem:
let's say that I have two tasks $[T_1, T_2]$, and $3$ resources $[R_1,R_2,R_3]$, each resource has a price $P$.
I would like to find the formula which found the best combination of [task, resource], let's say the maximum combination of the following sums:
$Max$$[(T_1/P_1 + T_2/P_1), (T_1/P_1 + T_2/P_2), (T_1/P_1 + T_2/P_3), (T_1/P_2 + T_2/P_1), (T_1/P_2 + T_2/P_2), (T_1/P_2 + T_2/P_3), (T_1/P_3 + T_2/P_1), (T_1/P_3 + T_2/P_2), (T_1/P_3 + T_2/P_3)]$
Regards.

REPLY [0 votes]: You have a classical assignment problem. The wiki entry should have enough info to get you started on modeling it.


TITLE: Prove an identity for any $a, b \in \mathbb{C} $
QUESTION [0 upvotes]: I've been trying to use the $|z|^2 = z\overline{z}$ to transform the identity and then just compare the sides but I end up having something inequal.
What's a good and efficient solution to this?
Identity to prove:
$$|1-\overline{a}b|+|a-b|^2 = (1+|ab|)^2 - (|a|^2 + |b|^2)$$

REPLY [1 votes]: After some answers/comments I've realized that the identity is wrongly written in the book. With a tiny edit in the text, it can be proven easily. Thanks for the help! (Answers given by: Lord Shark The Unknown and amsmath.


TITLE: Understanding the value of $\gamma$ parameter, so that the given process is a martingale.
QUESTION [1 upvotes]: I have a Poisson process of $X_t$ with rate $\lambda=\tilde\lambda.$ Now, given $M_t = X_t + \gamma\cdot t$ I want to uderstand, for which values of $\gamma$, $M_t$ is a martingale.
I know, that in order for $M_t$ to be a martingale, the following should hold (assuming the natural filtration, and given that $X_t$ is the only term of $M_t$ from which the variability comes):
$$
\mathbb{E}\left[M_{n+1}| \mathcal{F_n}\right] = M_n
$$
I have applied the above condition on what I have for $M_t$, but I may have done a mistake, since for some reason $\gamma$ for me does not come to be a constant, what seems quite weird..
$$
\mathbb{E}[M_{n+1}|\mathcal{F}_n] = \mathbb{E}[X_{n+1} + \gamma(n+1)|\mathcal{F}_n] = (\text{since }n \text{ is fixed}) = \gamma(n+1) + \mathbb{E}[X_{n+1}|\mathcal{F}_n] =\\= (\text{here I decided to add and subtract }X_n)=X_n + \gamma n + \gamma + \mathbb{E}[X_{n+1}|\mathcal{F}_n] - X_n = M_n +\\+\left(\gamma - X_n + \mathbb{E}[X_{n+1}|\mathcal{F}_n]\right)\implies \mathbb{E}[X_{n+1}|\mathcal{F}_n] + (\gamma - X_n) = 0~(\text{for }M_n\text{ to be a martingale}).
$$
That is, we have to have: $\mathbb{E}[X_{n+1}|\mathcal{F}_n] = X_n - \gamma \Longleftrightarrow \mathbb{E}[X_{n+1}] = X_n - \gamma \Longleftrightarrow \tilde\lambda=X_n-\gamma$. So that I finally obtain that:
$$
\gamma = X_n - \tilde\lambda.
$$
I am totally sure this result is incorrect, since $\gamma$ is definitely not supposed to involve the variable term. But how can I then obtain it, and what is my mistake?
I would appreciate any help, thank you in advance!

REPLY [3 votes]: In a martingale $(M_t)$ the mean $EM_t=EM_{t+1}$. So we must have  $E(X_t+\gamma t)=E(X_{t+1}+\gamma ((t+1)))$ which gives $t \overline{ \lambda}+\gamma t=(t+1) \overline{ \lambda}+\gamma (t+1)$. Hence, we must have $\gamma =-\overline{ \lambda}$. I will let you try to prove (using independence of increments of $(X_t)$)  that this value of $\gamma$ works. (I can post details if necessary).


TITLE: It is correct to say that a tensor is simply a multidimensional array of related quantities? But what about a tensor as a transformation?
QUESTION [0 upvotes]: It is correct to say that a tensor is simply a multidimensional array of related quantities?
More specifically a tensor is a collection or tuples of vectors where every vector in the tuple represent a different type of information but the components of the different vectors depend of each other.
I said this because of the following sentence I read in Wikipedia:
"at the end of the 19th century, the electromagnetic field was understood as a collection of two vector fields in space. nowadays, one recognizes this as a single antisymmetric 2nd-rank tensor field in spacetime"
I understand that electric and magnetic fields are grouped into a tensor because the components the magnetic field depend of some maner of the components of the electric field and viceversa.
But what about tensors as transformation objects? if you have two vectors (not tuples of vectors) then the transformation is simply a matrix (or rank 2 tensor), but what is the necessity for tensors of rank bigger than 2? Is to transform tuples of vectors?
I only have a background in physics so please try to answer in the most simple terms.

REPLY [1 votes]: A tensor is a "multidimensional array of quantities" in the same sense that a vector is an "array of quantities" -- that is, you're giving a description of the coordinate representation of said objects relative to a chosen basis.
The example of electromagnetism is just an example of a phenomenon where, when we move to a unified notion of space-time, that some classical quantities turn out to be the time and space components of some other quantity. Another example is how energy and classical 3-dimensional momentum turn out to be the time and space components of the special relativity momentum 4-vector
There is a rank 2 tensor field which breaks apart into components:

The (time, time) component is zero
The (time, space) component is the electric field vector
The (space, time) component is the negative of the electric field vector
The (space, space) component is the magnetic field pseudovector

(in coordinates, the magnetic field really ought to be an antisymmetric two-dimensional array of scalars — but as history goes we first discovered how to represent such things via pseudovector, and that description has stuck in classical physics)


text:
\begin{document}
\title[On the polynomial Hardy--Littlewood inequality]{On the polynomial
Hardy--Littlewood inequality}
\author[Ara\'{u}jo]{G. Ara\'{u}jo}
\address{Departamento de Matem\'{a}tica \\
\indent Universidade Federal da Para\'{\i}ba \\
\indent 58.051-900 - Jo\~{a}o Pessoa, Brazil.}
\email{gdasaraujo@gmail.com}
\author[Jim\'{e}nez]{P. Jim\'{e}nez-Rodriguez}
\address{Departament of Mathematical Sciences\\
\indent Mathematics and computer Sciences Building\\
\indent Kent State University \\
\indent Kent, OH 44242, USA.}
\email{pjimene1@kent.edu}
\author[Mu\~{n}oz]{G. A. Mu\~{n}oz-Fern\'{a}ndez}
\address{Departamento de An\'{a}lisis Matem\'{a}tico\\
\indent Facultad de Ciencias Matem\'{a}ticas \\
\indent Plaza de Ciencias 3 \\
\indent Universidad Complutense de Madrid\\
\indent Madrid, 28040, Spain.}
\email{gustavo$\_$fernandez@mat.ucm.es}
\author[N\'{u}\~{n}ez]{D. N\'{u}\~{n}ez-Alarc\'{o}n}
\address{Departamento de Matem\'{a}tica\\
\indent  Universidade Federal de Pernambuco\\
\indent 50.740-560 - Recife, Brazil.}
\email{danielnunezal@gmail.com}
\author[Pellegrino]{D. Pellegrino}
\address{Departamento de Matem\'{a}tica \\
\indent Universidade Federal da Para\'{\i}ba \\
\indent 58.051-900 - Jo\~{a}o Pessoa, Brazil.}
\email{pellegrino@pq.cnpq.br and dmpellegrino@gmail.com}
\author[Seoane]{J.B. Seoane-Sep\'{u}lveda}
\address{Departamento de An\'{a}lisis Matem\'{a}tico\\
\indent Facultad de Ciencias Matem\'{a}ticas \\
\indent Plaza de Ciencias 3 \\
\indent Universidad Complutense de Madrid\\
\indent Madrid, 28040, Spain.}
\email{jseoane@mat.ucm.es}
\author[Serrano]{D. M. Serrano-Rodr\'iguez}
\address{Departamento de Matem\'{a}tica\\
\indent  Universidade Federal de Pernambuco\\
\indent 50.740-560 - Recife, Brazil.}
\email{dmserrano0@gmail.com}
\thanks{G. Ara\'{u}jo, D. Pellegrino and J.B. Seoane-Sep\'{u}lveda was
supported by CNPq Grant 401735/2013-3 (PVE - Linha 2). G.A. Mu\~{n}oz-Fern
\'{a}ndez and J. B. Seoane-Sep\'{u}lveda were supported by MTM2012-34341.}
\subjclass{}
\keywords{Hardy--Littlewood inequality, Bohnenblust--Hille inequality;
absolutely summing operators}

\begin{abstract}
We investigate the growth of the constants of the polynomial
Hardy-Littlewood inequality.
\end{abstract}

\maketitle

\section{Introduction}

Given $\alpha =(\alpha _{1},\ldots ,\alpha _{n})\in {\mathbb{N}}^{n}$, let $
|\alpha |:=\alpha _{1}+\cdots +\alpha _{n}$ and let $\mathbf{x}^{\alpha }$
stand for the monomial $x_{1}^{\alpha _{1}}\cdots x_{n}^{\alpha _{n}}$ for $
\mathbf{x}=(x_{1},\ldots ,x_{n})\in {\mathbb{K}}^{n}$. The polynomial
Bohnenblust--Hille inequality asserts that if $P$ is a homogeneous
polynomial of degree $m$ on $\ell _{\infty }^{n}$ given by
\begin{equation*}
P(x_{1},...,x_{n})=\sum_{|\alpha |=m}a_{\alpha }\mathbf{{x}^{\alpha },}
\end{equation*}
then
\begin{equation*}
\left( {\sum\limits_{\left\vert \alpha \right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2m}{m+1}}\right) ^{\frac{m+1}{2m}}\leq D_{m}^{\mathbb{K}
}\left\Vert P\right\Vert
\end{equation*}
for some positive constant $D_{m}^{\mathbb{K}}$ which does not depend on $n$
(the exponent $\frac{2m}{m+1}$ is optimal). Precise estimates of the growth
of the constants $D_{m}^{\mathbb{K}}$ are crucial for different
applications. The following diagram shows the evolution of the estimates of $
D_{m}^{\mathbb{C}}:$

\bigskip

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
\textbf{Authors} & \textbf{Year} & \textbf{Estimate} \\ \hline
Bohnenblust and Hille & 1931, \cite{bh} & $D_{m}^{\mathbb{C}}\leq m^{\frac{
m+1}{2m}}\left( \sqrt{2}\right) ^{m-1}$ \\ \hline
$
\begin{array}{c}
\text{Defant, Frerick, Ortega-Cerd\'{a},} \\
\text{Ouna{\"{\i}}es, and Seip}
\end{array}
$ & 2011, \cite{ann} & $D_{m}^{\mathbb{C}}\leq\left( 1+\frac{1}{m-1}\right)
^{m-1}\sqrt{m}\left( \sqrt{2}\right) ^{m-1} $ \\ \hline
Bayart, Pellegrino, and Seoane-Sep\'{u}lveda & 2013, \cite{bps} & $D_{m}^{
\mathbb{C}}\leq C(\varepsilon)\left( 1+\varepsilon\right) ^{m}$, \\ \hline
\end{tabular}
\end{center}

\bigskip

\noindent where, in the table above, $C(\varepsilon)\left( 1+\varepsilon
\right) ^{m}$ means that given $\varepsilon>0$, there is a constant $C\left(
\varepsilon\right) >0$ such that $D_{m}^{\mathbb{C}}\leq C(\varepsilon
)\left( 1+\varepsilon\right) ^{m} $ for all $m.$

For real scalars it is shown in \cite{campos}\ that
\begin{equation*}
\left( 1.1\right) ^{m}\leq D_{m}^{\mathbb{R}}\leq C(\varepsilon )\left(
2+\varepsilon \right) ^{m},
\end{equation*}
and this means that for real scalars the inequality is hypercontractive and
this result is optimal.

When replacing $\ell _{\infty }^{n}$ by $\ell _{p}^{n}$ the extension of the
Bohnenblust--Hille inequality is called Hardy--Littlewood inequality and the
optimal exponents are $\frac{2mp}{mp+p-2m}$ for $p\geq 2m$. This is a
consequence of the multilinear Hardy--Littlewood inequality (see \cite
{abps.hlw, dimant}). \ More precisely, if $P$ is a homogeneous polynomial of
degree $m$ on $\ell _{p}^{n},$ with $p\geq 2m,$ given by
\begin{equation*}
P(x_{1},\ldots ,x_{n})=\sum_{|\alpha |=m}a_{\alpha }\mathbf{{x}^{\alpha },}
\end{equation*}
then
\begin{equation*}
\left( {\sum\limits_{\left\vert \alpha \right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}}\leq
D_{m,p}^{\mathbb{K}}\left\Vert P\right\Vert ,
\end{equation*}
and $D_{m,p}^{\mathbb{K}}$ does not depend on $n$.

In this paper we look for upper and lower estimates for $D_{m,p}^{\mathbb{K}
}.$

\section{First (and probably bad) upper estimates for $D_{m,p}^{\mathbb{K}}$}

Given $\alpha =(\alpha _{1},\ldots ,\alpha _{n})\in {\mathbb{N}}^{n}$, let
us define
\begin{equation*}
\binom{m}{\alpha }:=\frac{m!}{\alpha _{1}!\cdot \ldots \cdot \alpha _{n}!},
\end{equation*}
for $|\alpha |=m\in {\mathbb{N}}^{\ast }$. A straightforward consequence of
the multinomial formula yields the following relationship between the
coefficients of a homogeneous polynomial and the polar of the polynomial
(this lemma appears in \cite{diaz} and is essentially \emph{folklore}).

\begin{lemma}
\label{lem:Multinomial} If $P$ is a homogeneous polynomial of degree $m$ on $
{\mathbb{K}}^{n}$ given by
\begin{equation*}
P(x_{1},\ldots,x_{n})=\sum_{|\alpha|=m}a_{\alpha}\mathbf{{x}^{\alpha},}
\end{equation*}
and $L$ is the polar of $P$, then
\begin{equation*}
L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha_{n}})=\frac{a_{\alpha}}{\binom {m}{
\alpha}},
\end{equation*}
where $\{e_{1},\ldots,e_{n}\}$ is the canonical basis of ${\mathbb{K}}^{n}$
and $e_{k}^{\alpha_{k}}$ stands for $e_{k}$ repeated $\alpha_{k}$ times.
\end{lemma}

\bigskip From now on, for any map $f:\mathbb{R}\rightarrow \mathbb{R}$ we
define
\begin{equation*}
f\left( \infty \right) :=\lim_{p\rightarrow \infty }f(p).
\end{equation*}

The following result is also essentially known. We present here the details
of its proof for the sake of completeness of the paper.

\begin{proposition}
\label{pro:first_approach} If $P$ is a homogeneous polynomial of degree $m$
on $\ell _{p}^{n},$ with $p\geq 2m,$ given by
\begin{equation*}
P(x_{1},\ldots ,x_{n})=\sum_{|\alpha |=m}a_{\alpha }\mathbf{{x}^{\alpha },}
\end{equation*}
then
\begin{equation*}
\left( {\sum\limits_{\left\vert \alpha \right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}}\leq
D_{m,p}^{\mathbb{K}}\left\Vert P\right\Vert
\end{equation*}
with
\begin{equation*}
D_{m,p}^{\mathbb{K}}=C_{m,p}^{\mathbb{K}}\frac{m^{m}}{\left( m!\right) ^{
\frac{mp+p-2m}{2mp}}},
\end{equation*}
where $C_{m,p}^{\mathbb{K}}$ are the constants of the multilinear
Hardy-Littlewood inequality.
\end{proposition}

\begin{proof}
From Lemma \ref{lem:Multinomial} we have
\begin{align*}
{\sum\limits_{\left\vert \alpha\right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}} & ={\sum\limits_{\left\vert
\alpha\right\vert =m}}\left( \binom{m}{\alpha}\left\vert
L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha_{n}})\right\vert \right) ^{\frac{
2mp}{mp+p-2m}} \\
& ={\sum\limits_{\left\vert \alpha\right\vert =m}}\binom{m}{\alpha}^{\frac{
2mp}{mp+p-2m}}\left\vert L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha
_{n}})\right\vert ^{\frac{2mp}{mp+p-2m}}.
\end{align*}
However, for every choice of $\alpha$, the term
\begin{equation*}
\left\vert L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha_{n}})\right\vert ^{
\frac{2mp}{mp+p-2m}}
\end{equation*}
is repeated $\binom{m}{\alpha}$ times in the sum
\begin{equation*}
{\sum\limits_{i_{1},\ldots,i_{m}=1}^{n}}\left\vert L(e_{i_{1}},\ldots
,e_{i_{m}})\right\vert ^{\frac{2mp}{mp+p-2m}}.
\end{equation*}
Thus
\begin{equation*}
{\sum\limits_{\left\vert \alpha\right\vert =m}}\binom{m}{\alpha}^{\frac {2mp
}{mp+p-2m}}\left\vert
L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha_{n}})\right\vert ^{\frac{2mp}{
mp+p-2m}}={\sum\limits_{i_{1},\ldots,i_{m}=1}^{n}}\binom{m}{\alpha}^{\frac{
2mp}{mp+p-2m}}\frac{1}{\binom{m}{\alpha}}\left\vert
L(e_{i_{1}},\ldots,e_{i_{m}})\right\vert ^{\frac{2mp}{mp+p-2m}}
\end{equation*}
and, since
\begin{equation*}
\binom{m}{\alpha}\leq m!
\end{equation*}
we have
\begin{equation*}
{\sum\limits_{\left\vert \alpha\right\vert =m}}\binom{m}{\alpha}^{\frac {2mp
}{mp+p-2m}}\left\vert
L(e_{1}^{\alpha_{1}},\ldots,e_{n}^{\alpha_{n}})\right\vert ^{\frac{2mp}{
mp+p-2m}}\leq\left( m!\right) ^{\frac {mp-p+2m}{mp+p-2m}}{
\sum\limits_{i_{1},\ldots,i_{m}=1}^{n}}\left\vert
L(e_{i_{1}},\ldots,e_{i_{m}})\right\vert ^{\frac{2mp}{mp+p-2m}}.
\end{equation*}
We finally obtain
\begin{align*}
\left( {\sum\limits_{\left\vert \alpha\right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} &
\leq\left( \left( m!\right) ^{\frac{mp-p+2m}{mp+p-2m}}{\sum\limits_{i_{1},
\ldots,i_{m}=1}^{n}}\left\vert L(e_{i_{1}},\ldots,e_{i_{m}})\right\vert ^{
\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& =\left( m!\right) ^{\frac{mp-p+2m}{2mp}}\left( {\sum\limits_{i_{1},
\ldots,i_{m}=1}^{n}}\left\vert L(e_{i_{1}},\ldots,e_{i_{m}})\right\vert ^{
\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& \leq\left( m!\right) ^{\frac{mp-p+2m}{2mp}}C_{m,p}^{\mathbb{K}}\left\Vert
L\right\Vert .
\end{align*}
On the other hand, it is well-known that
\begin{equation*}
\left\Vert L\right\Vert \leq\frac{m^{m}}{m!}\left\Vert P\right\Vert
\end{equation*}
and, hence,
\begin{align*}
\left( {\sum\limits_{\left\vert \alpha\right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} & \leq
C_{m,p}^{\mathbb{K}}\left( m!\right) ^{\frac{mp-p+2m}{2mp}}\frac{m^{m}}{m!}
\left\Vert P\right\Vert \\
& =C_{m,p}^{\mathbb{K}}\frac{m^{m}}{\left( m!\right) ^{\frac{mp+p-2m}{2mp}}}
\left\Vert P\right\Vert .
\end{align*}
\end{proof}

\begin{remark}
Let us define the polarization constants for polynomials on $\ell_{p}$
spaces as
\begin{equation*}
{\mathbb{K}}(m,p):=\inf\{M> 0\, :\, \|L\|\leq M\|P\|\},
\end{equation*}
where the infimum is taken over all $P\in{\mathcal{P}}(^{m}\ell_{p}^{n})$
and $L$ is the polar of $P$. Notice that ${\mathbb{K}}(m,p)$ may improve the
inequality
\begin{equation*}
\left\Vert L\right\Vert \leq\frac{m^{m}}{m!}\left\Vert P\right\Vert
\end{equation*}
used in the proof of Proposition \ref{pro:first_approach}. Using ${\mathbb{K}
}(m,p)$ instead of $\frac{m^{m}}{m!}$ in the proof of Proposition \ref
{pro:first_approach} we end up with
\begin{equation*}
D_{m,p}^{\mathbb{K}}=\left( m!\right) ^{\frac{mp-p+2m}{2mp}}C_{m,p}^{\mathbb{
K}}{\mathbb{K}}(m,p).
\end{equation*}

The calculation of polarization constants has been a fruitful problem since
the 1970's and some facts are know about them. Specifically, Harris proved
in \cite{Harris} that
\begin{equation*}
{\mathbb{C}}(m,p)\leq \left(\frac{m^m}{m!}\right)^\frac{|p-2|}{p},
\end{equation*}
for all $p\geq 1$, whenever $m$ is a power of $2$. Also, Harris \cite{Harris}
proved that
\begin{equation*}
{\mathbb{C}}(m,\infty)\leq \frac{m^\frac{m}{2}(m+1)^\frac{m+1}{2}}{2^mm!}.
\end{equation*}
One more example was provided by Sarantopoulos \cite{Sarant} who proved that
\begin{equation*}
{\mathbb{K}}(m,p)=\frac{m^\frac{m}{p}}{m!},
\end{equation*}
whenever $1\leq p\leq \frac{m}{m-1}$.
\end{remark}

\section{The real polynomial Hardy--Littlewood inequality: lower bounds for
the constants}

For $m\geq 2$ and $p\geq 2m$, let us denote by $H_{\mathbb{K},m,p}^{\mathrm{
pol}}$ the optimal constants satisfying the polynomial Hardy-Littlewood
inequality with scalars in $\mathbb{K}.$ In this section we show that for
real scalars the polynomial Hardy-Littlewood inequality has at least an
hypercontractive growth.

\begin{theorem}
\label{88990}For all positive integers $m\geq 2$ and $2m\leq p<\infty $ we
have
\begin{equation*}
\left( \sqrt[16]{2}\right) ^{m}\leq 2^{\frac{mp+p-6m+4}{4p}\cdot \frac{m-1}{m
}}\leq H_{\mathbb{R},m,p}^{\mathrm{pol}}.
\end{equation*}
\end{theorem}

\bigskip

\begin{proof}
Let $m$ be an even integer. Consider the $m$-homogeneous polynomial $
P_{m}:\ell _{p}^m\rightarrow \mathbb{R}$ given by
\begin{equation*}
P_{m}(x_{1},\ldots ,x_{m})=\left( x_{1}^{2}-x_{2}^{2}\right) \left(
x_{3}^{2}-x_{4}^{2}\right) \cdots \left( x_{m-1}^{2}-x_{m}^{2}\right) .
\end{equation*}

Notice that
\begin{equation*}
\left\Vert P_{m}\right\Vert =P_{m}\left( \frac{1}{\sqrt[p]{m/2}},0,\frac{1}{
\sqrt[p]{m/2}},...,\frac{1}{\sqrt[p]{m/2}},0\right) =\left( \frac{1}{\sqrt[p]
{m/2}}\right) ^{m}.
\end{equation*}
From the Hardy-Littlewood inequality for $P_{m}$ we have
\begin{equation*}
\left( {\sum\limits_{\left\vert \alpha \right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}}\leq H_{
\mathbb{R},m,p}^{\mathrm{pol}}\left\Vert P_{m}\right\Vert ,
\end{equation*}
i.e.,
\begin{equation*}
H_{\mathbb{R},m,p}^{\mathrm{pol}}\geq \frac{\left( 2^{\frac{m}{2}}\right) ^{
\frac{mp+p-2m}{2mp}}}{\left( \frac{1}{\sqrt[p]{m/2}}\right) ^{m}}=2^{\frac{
mp+p-2m}{4p}}\left( \frac{m}{2}\right) ^{\frac{m}{p}}=2^{\frac{mp+p-6m}{4p}
}m^{\frac{m}{p}}\geq 2^{\frac{mp+p-6m}{4p}}.
\end{equation*}

If $m$ is odd, define
\begin{equation*}
Q_{m}(x_{1},\ldots,x_{m})=\left( x_{1}^{2}-x_{2}^{2}\right) \left(
x_{3}^{2}-x_{4}^{2}\right) \cdots \left( x_{m-2}^{2}-x_{m-1}^{2}\right)
x_{m}.
\end{equation*}
Then
\begin{equation*}
\left\Vert Q_{m}\right\Vert \leq\left\Vert P_{m-1}\right\Vert =\left( \frac{1
}{\sqrt[p]{\left( m-1\right) /2}}\right) ^{m-1}.
\end{equation*}
From the Hardy-Littlewood inequality for $Q_{m}$ we have
\begin{equation*}
\left( {\sum\limits_{\left\vert \alpha\right\vert =m}}\left\vert a_{\alpha
}\right\vert ^{\frac{2mp}{mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}}\leq H_{
\mathbb{R},m,p}^{\mathrm{pol}}\left\Vert Q_{m}\right\Vert ,
\end{equation*}
i.e.,
\begin{eqnarray*}
H_{\mathbb{R},m,p}^{\mathrm{pol}} & \geq & \frac{\left( 2^{\frac{m-1}{2}
}\right) ^{\frac{mp+p-2m}{2mp}}}{\left( \frac{1}{\sqrt[p]{\left( m-1\right)
/2}}\right) ^{m-1}} \\
& = & 2^{\frac{\left( mp+p-2m\right) \left( m-1\right) }{4mp}}\left( \frac{
m-1}{2}\right) ^{\frac{m-1}{p}} \\
& = & 2^{\frac{mp+p-6m+4}{4p}\cdot \frac{m-1}{m} }\left( m-1\right) ^{\frac{
m-1}{p}} \\
& \geq & 2^{\frac{mp+p-6m+4}{4p}\cdot\frac{m-1}{m}}.
\end{eqnarray*}
\end{proof}

\begin{remark}
From the estimates of the last proof note that if $m$ is even, and $m\geq 4$
, then
\begin{equation}
\begin{array}{rcl}
H_{\mathbb{R},m,p}^{\mathrm{pol}} & \geq  & 2^{\frac{mp+p-6m}{4p}}m^{\frac{m
}{p}}\vspace{0.2cm} \\
& \geq  & 2^{\frac{mp+p-6m}{4p}}\left( 2^{\frac{3}{2}}\right) ^{\frac{m}{p}}
\vspace{0.2cm} \\
& = & \left( \sqrt[4]{2}\right) ^{m+1}.
\end{array}
\label{par}
\end{equation}
If $m$ is odd, and $m\geq 5$, then
\begin{equation}
\begin{array}{rcl}
H_{\mathbb{R},m,p}^{\mathrm{pol}} & \geq  & 2^{\frac{mp+p-6m+4}{4p}\cdot
\frac{m-1}{m}}\left( m-1\right) ^{\frac{m-1}{p}}\vspace{0.2cm} \\
& \geq  & 2^{\frac{mp+p-6m+4}{4p}\cdot \frac{m-1}{m}}\left( 2^{\frac{3}{2}
}\right) ^{\frac{m-1}{p}}\vspace{0.2cm} \\
& = & \allowbreak 2^{\frac{pm^{2}+4m-p-4}{4mp}}\vspace{0.2cm} \\
& \geq  & \left( \sqrt[4]{2}\right) ^{m-\frac{1}{m}}.
\end{array}
\label{impar}
\end{equation}
Thus, by (\ref{par}) and (\ref{impar}), if $m\geq 4$
\begin{equation*}
H_{\mathbb{R},m,p}^{\mathrm{pol}}\geq \left( \sqrt[4]{2}\right) ^{m-\frac{1}{
m}}.
\end{equation*}
\end{remark}

\section{Real versus complex estimates}

As it happens with the constants of the Bohnenblust-Hille inequality, we
observe that
\begin{equation}
H_{\mathbb{R},m,p}^{\mathrm{pol}}\leq 2^{m-1}H_{\mathbb{C},m,p}^{\mathrm{pol}
}.  \label{989}
\end{equation}
In fact, from \cite{muno} we know that if $P:\ell _{p}\rightarrow \mathbb{R}$
is an $m$-homogeneous polynomial and $P_{\mathbb{C}}:\ell _{p}\rightarrow
\mathbb{C}$ is the same polynomial, then
\begin{equation*}
\left\Vert P_{\mathbb{C}}\right\Vert \leq 2^{m-1}\left\Vert P\right\Vert .
\end{equation*}
We thus obtain (\ref{989}). So if one succeeds in proving that the constants
of the complex Hardy-Littlewood polynomial inequality have a subexponential
growth (as it happens with the constants of the complex polynomial
Bohnenblust-Hille inequality) then we immediately conclude that the
constants of the real polynomial Hardy littlewood inequality have an
exponential growth and this result is optimal (due to Theorem \ref{88990}).

\section{The complex polynomial Hardy-Littlewood inequality: trying to find
upper estimates}

\bigskip In this section we try to improve the estimates for the upper
bounds of the complex polynomial Hardy-Littlewwod inequality from Section 2.
However, for $p<\infty $ the validity of our estimates depend on a
conjecture (see (\ref{888})). For the case $p=\infty $ our results are
exactly those from (\cite[Remark 2.2]{bps}).

The following multi-index notation will come in handy for us: for positive
integers $m,n$, we set
\begin{align*}
\mathcal{M}(m,n)& :=\left\{ \mathbf{i}=(i_{1},\dots ,i_{m});\,i_{1},\dots
,i_{m}\in \{1,\dots ,n\}\right\} , \\
\mathcal{J}(m,n)& :=\left\{ \mathbf{i}\in \mathcal{M}(m,n);\,i_{1}\leq
i_{2}\leq \dots \leq i_{m}\right\} ,
\end{align*}
and for $k=1,\dots ,m$, $\mathcal{P}_{k}(m)$ denotes the set of the subsets
of $\{1,\dots ,m\}$ with cardinality $k$. For $S=\{s_{1},\dots ,s_{k}\}\in
\mathcal{P}_{k}(m)$, its complement will be $\widehat{S}:=\{1,\dots
,m\}\setminus S$, and $\mathbf{i}_{S}$ shall mean $(i_{s_{1}},\dots
,i_{s_{k}})\in \mathcal{M}(k,n)$. For a multi-index $\mathbf{i}\in \mathcal{M
}(m,n)$, we denote by $|\mathbf{i}|$ the cardinality of the set of
multi-indexes $\mathbf{j}\in \mathcal{M}(m,n)$ such that there is a
permutation $\sigma $ of $\{1,\dots ,m\}$ with $i_{\sigma (k)}=j_{k}$, for
every $k=1,\dots ,m$. The equivalence class of $\mathbf{i}$ is denoted by $[
\mathbf{i].}$ When we write $c_{[\mathbf{i]}}$ for $\mathbf{i\in }\mathcal{M}
(m,n)$ we mean $c_{\mathbf{j}}$ for $\mathbf{j}\in \mathcal{J}(m,n)$ and $
\mathbf{j}$ equivalent to $\mathbf{i}.$

The following very recent generalization of the famous Blei inequality will
be crucial for our estimates (see \cite[Remark 2.2]{bps}).

\begin{lemma}[Bayart, Pellegrino, Seoane, \protect\cite{bps}]
\label{blei.interp} Let $m,n$ positive integers, $1\leq k\leq m$ and $1\leq
s\leq q$, satisfying $\frac{m}{\rho }=\frac{k}{s}+\frac{m-k}{q}$. Then for
all scalar matrix $\left( a_{\mathbf{i}}\right) _{\mathbf{i}\in\mathcal{M}
(m,n)}$,

\begin{equation*}
\left( \sum_{\mathbf{i}\in\mathcal{M}(m,n)}\left\vert a_{\mathbf{i}
}\right\vert ^{\rho}\right) ^{\frac{1}{\rho}}\leq\prod_{S\in\mathcal{P}
_{k}(m)}\left( \sum_{\mathbf{i}_{S}}\left( \sum_{\mathbf{i}_{\widehat{S}
}}\left\vert a_{\mathbf{i}}\right\vert ^{q}\right) ^{\frac{s}{q}}\right) ^{
\frac{1}{s}\cdot\frac{1}{\binom{m}{k}}}.
\end{equation*}
\end{lemma}

Let us use the following notation: $S_{\ell _{p}^{n}}$ denotes the unit
sphere on $\ell _{p}^{n}$ if $p<\infty $, and $S_{\ell _{\infty }^{n}}$
denotes the $n$-dimensional torus. More precisely: for $p\in \left( 0,\infty
\right) $
\begin{equation*}
S_{\ell _{p}^{n}}:=\left\{ \mathbf{z}=\left( z_{1},...,z_{n}\right) \in
\mathbb{C}^{n}:\left\Vert \mathbf{z}\right\Vert _{\ell _{p}^{n}}=1\right\} ,
\end{equation*}
and
\begin{equation*}
S_{\ell _{\infty }^{n}}:=\mathbb{T}^{n}=\left\{ \mathbf{z}=\left(
z_{1},...,z_{n}\right) \in \mathbb{C}^{n}:\left\vert z_{i}\right\vert
=1\right\} .
\end{equation*}
For $p\in \left[ 1,\infty \right] $, let $\mu _{p}^{n}$ be the normalized
Lebesgue measure on $S_{\ell _{p}^{n}}$. For the sake of simplicity, $\mu
_{p}^{n}$ will be simply denoted by $\mu ^{n}.$ The following inequality is
a variant of a result due to Bayart \cite[Theorem 9]{bayart} (see, for
instance, \cite[Lemma 5.1]{bps}).

\begin{lemma}[Bayart]
\label{bay_in} Let $1\leq s\leq 2$. For every $m$-homogeneous polynomial $P(
\mathbf{z})=\sum_{|\alpha |=m}a_{\alpha }\mathbf{z}^{\alpha }$ on $\mathbb{C}
^{n}$ with values in $\mathbb{C}$, we have
\begin{equation*}
\left( \sum_{|\alpha |=m}\left\vert a_{\alpha }\right\vert ^{2}\right) ^{
\frac{1}{2}}\leq \left( \frac{2}{s}\right) ^{\frac{m}{2}}\left( \int_{
\mathbb{T}^{n}}\left\vert P(\mathbf{z})\right\vert ^{s}d\mu ^{n}(\mathbf{z}
)\right) ^{\frac{1}{s}}.
\end{equation*}
\end{lemma}

For $m\in \lbrack 2,\infty ]$ let us define $p_{0}(m)$ as the infimum of the
values of $p\in \left[ 2m,\infty \right] $ such that for all $1\leq s\leq
\frac{2p}{p-2}$ there is a $K_{s,p}>0$ such that
\begin{equation}
\left( \sum_{|\alpha |=m}\left\vert a_{\alpha }\right\vert ^{\frac{2p}{p-2}
}\right) ^{\frac{p-2}{2p}}\leq K_{s,p}^{m}\left( \int_{S_{\ell
_{p}^{n}}}\left\vert P(\mathbf{z})\right\vert ^{s}d\mu ^{n}(\mathbf{z}
)\right) ^{\frac{1}{s}}  \label{888}
\end{equation}
for all positive integers $n$ and all $m$-homogeneous polynomials $P:\mathbb{
C}^{n}\rightarrow \mathbb{C}$. For the sake of simplicity $p_{0}(m)$ will be
simply denoted by $p_{0}.$ From Bayart's lemma we know that this definition
makes sense, since from Bayart's lemma we know that (\ref{888}) is valid for
$p=\infty .$ We conjecture that $p_{0}\leq m^{2}.$

Now, let us state and prove the main result of this section. The argument of
the proof follows the lines of that in \cite{ann, bps}. We will use the
following result due L. Harris (see \cite[Theorem 1]{Harris}):

\begin{lemma}[Harris]
\label{haharr} Let $X$ be a complex normed linear space. If $P$ is a
homogeneous polynomial of degree $m$ on $X$ and $L$ is the polar of $P$,
then, for any nonnegative integers $m_1,...,m_k$ with $m_1+\cdots+m_k=m$ and
for any $x^{(1)},...,x^{(k)}$ unit vectors in $X$,
\begin{equation*}
|L( \underbrace{x^{(1)},...,x^{(1)}}_{m_1 \text{ times}},...,\underbrace{
x^{(k)},...,x^{(k)}}_{m_k \text{ times}})|\leq \frac{m_1 ! \cdots m_k !\cdot
m^m}{m_1^{m_1}\cdots m_k^{m_k}\cdot m!}\|P\|.
\end{equation*}
\end{lemma}

\begin{theorem}
\label{hardy}Let $m\in \lbrack 2,\infty ]$ and $1\leq k\leq m-1.$ If $
p_{0}(m-k)<p\leq \infty $ (and $p=\infty $ if $p_{0}(m-k)=\infty $) then,
for every $m$-homogeneous polynomial $P:\ell _{p}^{n}\rightarrow \mathbb{C}$
, defined by $P(\mathbf{z})=\sum_{|\alpha |=m}a_{\alpha }\mathbf{z}^{\alpha
},$ we have
\begin{align*}
& \left( \sum_{|\alpha |=m}\left\vert a_{\alpha }\right\vert ^{\frac{2mp}{
mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& \leq K_{\frac{2kp}{kp+p-2k},p}^{m-k}\cdot \frac{m^{m}}{(m-k)^{m-k}}\cdot
\left( \frac{(m-k)!}{m!}\right) ^{\frac{p-2}{2p}}\left( \frac{2}{\sqrt{\pi }}
\right) ^{\frac{2k\left( k-1\right) }{p}}\cdot \left( B_{\mathbb{C},k}^{
\mathrm{mult}}\right) ^{\frac{p-2k}{p}}\Vert P\Vert ,
\end{align*}
where $B_{\mathbb{C},k}^{\mathrm{mult}}$ is any constant satisfying the $k$
-linear Bohnenblust-Hille inequality.
\end{theorem}

\begin{proof}
We can also write
\begin{equation*}
P\left( \mathbf{z}\right) =\sum_{\mathbf{i\in}\mathcal{J}(m,n)}c_{\mathbf{i}
}z_{i_{1}}....z_{i_{m}}.
\end{equation*}
Consider
\begin{align*}
\rho & =\frac{2mp}{mp+p-2m}, \\
s_{k} & =\frac{2kp}{kp+p-2k}, \\
q & =\frac{2p}{p-2}.
\end{align*}
Note that
\begin{equation*}
s_{k}\leq2<q,
\end{equation*}
\begin{equation*}
\frac{m}{\rho}=\frac{mp+p-2m}{2p},
\end{equation*}
and
\begin{align*}
\frac{k}{s_{k}}+\frac{m-k}{q} & =\frac{kp+p-2k}{2p}+\frac{\left( m-k\right)
\left( p-2\right) }{2p} \\
& =\frac{kp+p-2k}{2p}+\frac{mp-kp-2m+2k}{2p} \\
& =\frac{mp+p-2m}{2p}.
\end{align*}
and thus
\begin{equation*}
\frac{m}{\rho}=\frac{k}{s_{k}}+\frac{m-k}{q}
\end{equation*}
and we can use Lemma \ref{blei.interp}.

Let $L:\ell_{p}^{n}\times\dots\times\ell_{p}^{n}\rightarrow\mathbb{C}$ be
the unique symmetric $m$-linear map associated to $P$. Note that
\begin{equation*}
L\left( \mathbf{z}^{(1)},...,\mathbf{z}^{(m)}\right) =\sum_{\mathbf{i\in}
\mathcal{M}(m,n)}\frac{c_{[\mathbf{i}]}}{\left\vert \mathbf{i}\right\vert }
z_{i_{1}}^{(1)}...z_{i_{m}}^{(m)}.
\end{equation*}
Thus
\begin{align*}
\sum_{|\alpha|=m}\left\vert a_{\alpha}\right\vert ^{\frac{2mp}{mp+p-2m}} &
=\sum_{\mathbf{i\in}\mathcal{J}(m,n)}\left\vert c_{\mathbf{i}}\right\vert ^{
\frac{2mp}{mp+p-2m}} \\
& =\sum_{\mathbf{i\in}\mathcal{M}(m,n)}\left\vert \mathbf{i}\right\vert ^{
\frac{-p}{mp+p-2m}}\left( \frac{\left\vert c_{[\mathbf{i}]}\right\vert }{
\left\vert \mathbf{i}\right\vert ^{\frac{1}{q}}}\right) ^{\frac {2mp}{mp+p-2m
}} \\
& \leq\sum_{\mathbf{i\in}\mathcal{M}(m,n)}\left( \frac{\left\vert c_{\left[
\mathbf{i}\right] }\right\vert }{\left\vert \mathbf{i}\right\vert ^{\frac {1
}{q}}}\right) ^{\frac{2mp}{mp+p-2m}}
\end{align*}
Using Lemma \ref{blei.interp} with $s_{k}=\frac{2kp}{kp+p-2k}$ and $q=\frac{
2p}{p-2}$, we get
\begin{align*}
& \left( \sum_{|\alpha|=m}\left\vert a_{\alpha}\right\vert ^{\frac {2mp}{
mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& \leq\left[ \prod_{S\in\mathcal{P}_{k}}\left( \sum_{\mathbf{i}_{S}\in
\mathcal{M}(k,n)}\left( \sum_{\mathbf{i}_{\hat{S}}\in\mathcal{M}
(m-k,n)}\left( \frac{\left\vert c_{\left[ \mathbf{i}\right] }\right\vert }{
\left\vert \mathbf{i}\right\vert ^{\frac{1}{q}}}\right) ^{q}\right) ^{\frac{
s_{k}}{q}}\right) ^{\frac{1}{s_{k}}}\right] ^{\frac{1}{\binom{m}{k}}}.
\end{align*}

Note that $|\mathbf{i}|\leq |\mathbf{i}_{\hat{S}}|\frac{m!}{(m-k)!}$, and
thus
\begin{align*}
& \left( \sum_{|\alpha |=m}\left\vert a_{\alpha }\right\vert ^{\frac{2mp}{
mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& \leq \left( \frac{m!}{(m-k)!}\right) ^{\frac{q-1}{q}}\left[ \prod_{S\in
\mathcal{P}_{k}}\left( \sum_{\mathbf{i}_{S}\in \mathcal{M}(k,n)}\left( \sum_{
\mathbf{i}_{\hat{S}}\in \mathcal{M}(m-k,n)}\frac{\left\vert c_{\left[
\mathbf{i}\right] }\right\vert ^{q}}{\left\vert \mathbf{i}\right\vert }
\left( \frac{\left\vert \mathbf{i}_{\hat{S}}\right\vert }{\left\vert \mathbf{
i}\right\vert }\right) ^{q-1}\right) ^{\frac{s_{k}}{q}}\right) ^{\frac{1}{
s_{k}}}\right] ^{\frac{1}{\binom{m}{k}}} \\
& =\left( \frac{m!}{(m-k)!}\right) ^{\frac{q-1}{q}}\left[ \prod_{S\in
\mathcal{P}_{k}}\left( \sum_{\mathbf{i}_{S}\in \mathcal{M}(k,n)}\left( \sum_{
\mathbf{i}_{\hat{S}}\in \mathcal{M}(m-k,n)}\frac{\left\vert c_{\left[
\mathbf{i}\right] }\right\vert ^{q}}{\left\vert \mathbf{i}\right\vert ^{q}}
\left\vert \mathbf{i}_{\hat{S}}\right\vert ^{q-1}\right) ^{\frac{s_{k}}{q}
}\right) ^{\frac{1}{s_{k}}}\right] ^{\frac{1}{\binom{m}{k}}}.
\end{align*}
Let us fix $S\in \mathcal{P}_{k}(m)$. There is no loss of generality in
supposing $S=\left\{ 1,...,k\right\} $. We then fix some $\mathbf{i}_{S}\in
\mathcal{M}(k,n)$ and we introduce the following $\left( m-k\right) $
-homogeneous polynomial on $\ell _{p}^{n}$:
\begin{equation*}
P_{\mathbf{i}_{S}}(\mathbf{z})=L\left( e_{i_{1}},...,e_{i_{k}},\mathbf{z}
,...,\mathbf{z}\right) .
\end{equation*}
Observe that
\begin{align*}
P_{\mathbf{i}_{S}}(\mathbf{z})& =\sum_{\mathbf{i}_{\hat{S}}\in \mathcal{M}
(m-k,n)}\frac{c_{\left[ \mathbf{i}\right] }}{\left\vert \mathbf{i}
\right\vert }z_{\mathbf{i}_{\hat{S}}} \\
& =\sum_{\mathbf{i}_{\hat{S}}\in \mathcal{J}(m-k,n)}\frac{c_{\left[ \mathbf{i
}\right] }}{\left\vert \mathbf{i}\right\vert }\left\vert \mathbf{i}_{\hat{S}
}\right\vert z_{\mathbf{i}_{\hat{S}}}
\end{align*}
so that
\begin{equation*}
\left\Vert P_{\mathbf{i}_{S}}(\mathbf{z})\right\Vert _{q}=\left( \sum_{
\mathbf{i}_{\hat{S}}\in \mathcal{J}(m-k,n)}\frac{\left\vert c_{\left[
\mathbf{i}\right] }\right\vert ^{q}}{\left\vert \mathbf{i}\right\vert ^{q}}
\left\vert \mathbf{i}_{\hat{S}}\right\vert ^{q}\right) ^{\frac{1}{q}}=\left(
\sum_{\mathbf{i}_{\hat{S}}\in \mathcal{M}(m-k,n)}\frac{\left\vert c_{\left[
\mathbf{i}\right] }\right\vert ^{q}}{\left\vert \mathbf{i}\right\vert ^{q}}
\left\vert \mathbf{i}_{\hat{S}}\right\vert ^{q-1}\right) ^{\frac{1}{q}}.
\end{equation*}
By the definition of $p_{0}$ we have
\begin{equation*}
\left\Vert P_{\mathbf{i}_{S}}(\mathbf{z})\right\Vert _{q}^{s_{k}}\leq
K_{s_{k},p}^{(m-k)s_{k}}\int_{S_{\ell _{p}^{n}}}\left\vert L\left(
e_{i_{1}},...,e_{i_{k}},\mathbf{z},...,\mathbf{z}\right) \right\vert
^{s_{k}}d\mu ^{n}(\mathbf{z}).
\end{equation*}
Thus,
\begin{align*}
& \sum_{\mathbf{i}_{S}}\left( \sum_{\mathbf{i}_{\hat{S}}}\frac{\left\vert c_{
\left[ \mathbf{i}\right] }\right\vert ^{q}}{\left\vert \mathbf{i}\right\vert
^{q}}\left\vert \mathbf{i}_{\hat{S}}\right\vert ^{q-1}\right) ^{\frac{1}{q}
\times s_{k}} \\
& \leq K_{s_{k},p}^{(m-k)s_{k}}\int_{S_{\ell _{p}^{n}}}\sum_{\mathbf{i}
_{S}}\left\vert L\left( e_{i_{1}},...,e_{i_{k}},\mathbf{z},...,\mathbf{z}
\right) \right\vert ^{s_{k}}d\mu ^{n}(\mathbf{z}).
\end{align*}
Now fixing $\mathbf{z}\in S_{\ell _{p}^{n}}$ we apply the multilinear
Hardy-Littlewood inequality to the $k-$linear form
\begin{equation*}
\left( \mathbf{z}^{(1)},...,\mathbf{z}^{(k)}\right) \mapsto L\left( \mathbf{z
}^{(1)},...,\mathbf{z}^{(k)},\mathbf{z},...,\mathbf{z}\right)
\end{equation*}
and we obtain, from \cite[Theorem 1.1]{apd} and Lemma \ref{haharr},
\begin{align*}
& \sum_{\mathbf{i}_{S}}\left\vert L\left( e_{i_{1}},...,e_{i_{k}},\mathbf{z}
,...,\mathbf{z}\right) \right\vert ^{s_{k}} \\
& \leq \left( \left( \frac{2}{\sqrt{\pi }}\right) ^{\frac{2k\left(
k-1\right) }{p}}\cdot \left( B_{\mathbb{C},k}^{\mathrm{mult}}\right) ^{\frac{
p-2k}{p}}\cdot \sup_{\mathbf{z}^{(1)},...,\mathbf{z}^{(k)}\in S_{\ell
_{p}^{n}}}\left\vert L\left( \mathbf{z}^{(1)},...,\mathbf{z}^{(k)},\mathbf{z}
,...,\mathbf{z}\right) \right\vert \right) ^{s_{k}} \\
& \leq \left( \left( \frac{2}{\sqrt{\pi }}\right) ^{\frac{2k\left(
k-1\right) }{p}}\cdot \left( B_{\mathbb{C},k}^{\mathrm{mult}}\right) ^{\frac{
p-2k}{p}}\cdot \frac{(m-k)!\cdot m^{m}}{(m-k)^{m-k}\cdot m!}\left\Vert
P\right\Vert \right) ^{s_{k}}\text{.}
\end{align*}
Thus
\begin{align*}
& \left( \sum_{|\alpha |=m}\left\vert a_{\alpha }\right\vert ^{\frac{2mp}{
mp+p-2m}}\right) ^{\frac{mp+p-2m}{2mp}} \\
& \leq \left( \frac{m!}{(m-k)!}\right) ^{\frac{q-1}{q}}\cdot
K_{s_{k},p}^{m-k}\cdot \frac{(m-k)!\cdot m^{m}}{(m-k)^{m-k}\cdot m!}\cdot
\left( \frac{2}{\sqrt{\pi }}\right) ^{\frac{2k\left( k-1\right) }{p}}\cdot
\left( B_{\mathbb{C},k}^{\mathrm{mult}}\right) ^{\frac{p-2k}{p}}\Vert P\Vert
.
\end{align*}
\end{proof}

config: arxiv


TITLE: What is the correct way to calculate flux across given surface?
QUESTION [0 upvotes]: Suppose I need to find the flux of vector Field $\vec{F} = <y^2, xz , -1>,$ across the cone  $z = 2\sqrt{x^2 + y^2}$ between the planes $z = 0$ and $z = 2$,
As we can see, there will be two surfaces  $S_1$ the conical part and $S_2$ the circular part on top of the cone of radius  $1$ .
My question is : Do I need to find the flux  through each of the surfaces or do I need to just find it acroos $S_1$ ?,
I have solved a number of questions like these , sometimes the problem is solved just by considering it across $S_1$(ie the conical part) , while some time I also have to take care of $S_2$ the circular part.
Can anyone please tell me in detail when Do I have to include both the surfaces and when do I have to exclude one ?
(The given problem is from Thomas Calculus, and the flux is calculated simply by considering $S_1$)
Thank you

REPLY [0 votes]: If the problem asks for the flux through the surface, I cant imagine why you would calculate it through only part of the surface!  Could you give an example in which you "I have solved a number of questions like these , sometimes the problem is solved just by considering it across S1 (ie the conical part)"?


TITLE: Determine the probability that two random vectors over a finite field are orthogonal
QUESTION [0 upvotes]: Hi all,
Suppose that $\mathbf{f}=[f_1, f_2,\ldots,f_m]$ and $\mathbf{g}=[g_1,g_2,\ldots,g_m]$ are two $m$-dimensional vectors. All $f_i$'s are chosen uniformly randomly from a finite field $\mathbb{F}_q$, where $q$ is the finite field size. For convenience, we denote the elements of $\mathbb{F}_q$ as $\{0, 1, 2,\ldots,q-1\}$. Each $g_i$ is distributed as $\mathrm{Pr}(g_i=0)=1-p_i$ and $\mathrm{Pr}(g_i=r)=p_i/(q-1),r=1,2,\ldots,q-1$. Note that $p_i$ might not be equal for each $i\in\{1,2,\ldots,m\}$.
My question is, is that possible to determine the probability that $\mathbf{f}$ and $\mathbf{g}$ are orthogonal, i.e., $\sum_{i=1}^mf_ig_i=0$? If a closed form is not possible, is that possible to bound the probability? What I want to know is, whether such a probability would approach $0$ when I use a sufficiently large $q$. Any suggestions or references are appreciated.
Thanks sincerely.

REPLY [3 votes]: If $\vec g = \vec 0$, then all vectors $\vec f$ are orthogonal to $\vec g$.
If $\vec g \ne \vec 0$, then $1/q$ of the vectors $\vec f$ are orthogonal to $\vec g$. Given the previous coordinates of $\vec f$, there is a unique choice for the last coordinate of $\vec f$ paired with a nonzero coordinate of $\vec g$ so that $\vec f \cdot \vec g =  0$.
So, $Prob(\vec f \perp \vec g) = \prod (1-p_i) + \frac{1}{q}\bigg( 1 - \prod (1-p_i)\bigg).$ If you fix the set $\lbrace p_i \rbrace$, the limit as $q \to \infty$ is $\prod(1-p_i)$, not $0$.


TITLE: Are sums of the inverses of prime siblings finite?
QUESTION [4 upvotes]: PART I (Initial version)
Let   $P$   be the set of all primes   $2\ 3\ \ldots$.   Let
$$P_d\ \ :=\ \ \{\ p\in P\ :\ \exists_{q\in P}\ \ 0 < |p-q|\le d\ \}$$
and
$$S_d\ :=\ \sum_{p\in P_d}\ \frac 1p$$
for every real   $d>0$.   Thus   $d\mapsto S_d$   is non-decreasing,   $S_2 < \infty$,   and   $\lim_{d\rightarrow\infty} S_d = \infty$.   What else is known about   $S_d$ ?   For which values of   $d$   the sum   $S_d$   is finite?
PART II (additional)
Let   $d\ m\ n$   be positive integers. Is it true that
$$ (m < n)\quad\Rightarrow\quad \left( S_{m+d}-S_m\ \ge\ S_{n+d}-S_n \right)$$
? - It seems (to me) that it should be true.

REPLY [14 votes]: It turns out that $S_d$ is finite for every $d$. See my answer to this very similar question.

REPLY [3 votes]: The following heuristic study suggests that for any gap $d \ge 2$ the sum of the reciprocal of primes having a gap of $d$ should be convergent.
Let $L(r,x)$ be the number of conjectured occurrences of gaps of size $2r$ between successive prime $\le x$. We have
$$
L(r,x) = \int_{2}^{x} \sum_{}{}\frac{(-1)^k A(r,k)}{(\log t)^{k+1}} dt
$$
The coefficients $A(r,k)$ as well the details of the above formula are explained in the following link:
http://mac6.ma.psu.edu/primes/
The above study shows that till the point $1.7427435732 * 10^{35}$ the gap of 6 is most frequent. Since we sum of the reciprocal of twin primes are convergent and they are asymptotically more denser than primes with any other gaps, we expect that the sum of the reciprocal of primes with a gap $d > 2$ to be convergent as well.


TITLE: "Almost-ideals" in the (simple) Lie algebra of an algebraic group?
QUESTION [6 upvotes]: Let $G<\mathrm{GL_n}$ be a simple linear algebraic group defined over a finite field $K$. Let $\mathfrak{g}$ be its Lie algebra. Assume $\mathfrak{g}$ is simple.
Is it necessarily the case that there is no subspace $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}_g$ for every $g\in G(K)$?
Note: it is clear that there is no $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}$ for every $g\in G(\overline{K})$. It is also clear that the answer to the question above is "yes" when the number of elements of $K$ is larger than a constant depending only on $n$: since $\mathfrak{v}$ is not an ideal, there is a $v\in\mathfrak{v}$ such that all $g\in G(\overline{K})$ such that $\mathrm{Ad}_g(v)\in \mathfrak{v}$ lie in a proper subvariety of $G$.
Note 2: A friend has just proposed over the breakfast table that there are linear algebraic groups with no non-trivial rational points over $K$. That would obviously imply an answer of "no" to my question. I am not convinced that such a thing is really possible, at least not when we are talking about the group $G(K)$, $G$ simple (as opposed to more exotic groups of Lie type).
EDIT: as per Venkataramana's comment below, this situation cannot, in fact, occur for $G$ simple.

REPLY [2 votes]: You are looking for the following theorem of Steinberg:
Let q=|K|. An irreducible algebraic representation of $G(\overline{K})$ of highest weight λ remains irreducible when restricted to G(K) if $\langle \lambda,a^\vee\rangle <q$ for all simple coroots $\alpha^\vee$.
The reference is here. When G is simply connected, Steinberg proves more, namely that all irreducible modules for G(K) arise from this construction.
In the situation at hand, we need to apply Steinberg's theorem to the adjoint representation. There are only a small number of cases where the condition $\langle \lambda,a^\vee\rangle <q$ does not apply, easily classified on a case by case basis. I haven't checked, but I would suspect that in each of these cases, $\mathfrak{g}$ is not simple.