\section{introduction}

If $X$ is a Banach space over $\mathbb{K}$, $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, we let
$\mathcal{L}^s(^mX)$ denote the Banach space of all continuous symmetric $m$-linear forms $L:X^m\rightarrow\mathbb{K}$ with the norm
\[\|L\|=\sup\{|L(x_1,\ldots,x_m)|: \|x_1\| \leq 1,\ldots, \|x_m\| \leq 1\}\,.\]
A function $P: X\rightarrow\mathbb{K}$ is a continuous $m$-homogeneous polynomial if there is a continuous symmetric $m$-linear form $L:X^m\rightarrow\mathbb{K}$ for which $P(x)=L(x, \ldots, x)$ for all $x\in X$. In this case it is convenient to write $P=\widehat{L}$. Let $\mathcal{P}(^mX)$ denote the Banach space
of all continuous $m$-homogeneous polynomials $P: X\rightarrow\mathbb{K}$ with the norm
\[\|P\|=\sup\{|P(x)|: \|x\|\leq 1\}\,.\]
We write $L(x_1^{k_1}\ldots x_n^{k_n})$ as shorthand for $L(x_1,\ldots,x_1,\ldots,x_n,\ldots,x_n)$ where each $x_i$ appears $k_i$ times for $1\leq i\leq n$, $k_1+\ldots+k_n=m$.\\
We also need to define the norm
\[\|L\|_{(n)}=\sup_{k_1+\ldots k_n=m}\sup\{|L(x_1^{k_1}\ldots x_n^{k_n})|: \|x_1\| \leq 1,\ldots, \|x_n\| \leq 1\},\]
which for $n=2$ gives
\[\|L\|_{(2)}=\sup_{1\leq k\leq m}\sup\{|L(x_1^kx_2^{m-k})|: \|x_1\| \leq 1, \|x_2\| \leq 1\}\,.\]
Clearly, $\|\widehat{L}\|\leq\|L\|_{(n)}\leq\|L\|_{(n+1)}\leq\|L\|$.

It is known that if $L\in\mathcal{L}^s(^mX)$ and $\widehat{L}$ the associated polynomial, then
\[\|L\|\leq\frac{m^m}{m!}\|\widehat{L}\|.\]
This is the answer of Problem 73 of Mazur and Orlicz in \cite{SB}.

A natural generalization of Problem 73 is the following:\\
Let $X$ be a normed linear space, $k_1,\ldots,k_n$ be nonnegative integers whose sum is $m$ and let $c(k_1,\ldots,k_n,X)$ be the smallest number with the property that if $L$ is any symmetric $m$-linear mapping of one real normed linear space into another, then
\[|L(x_1^{k_1}\ldots x_n^{k_n})|\leq c(k_1,\ldots,k_n,X)\|\widehat{L}\|.\]
It is shown in \cite{H1} Theorem 1, that if only complex normed linear spaces and comlex scalars are considered, then
\[c(k_1,\ldots,k_n,X)=\frac{k_1!\ldots k_n!}{k_1^{k_1}\ldots k_n^{k_n}}\frac{m^m}{m!}.\]
In the next two sections we shall find bounds for the constant $c(k_1,\ldots,k_n,X)$ in the case of real normed linear spaces and specifically in the case of real $\ell_p$ spaces.


\section{Polynomials on a real normed linear space}

From \cite{H2} Corollary 7 (see also \cite{RS}, \cite{S2}), we have
\begin{equation} \label{eq Harris}
|L(x_1^{k_1}\ldots x_n^{k_n})|\leq\sqrt{\frac{m^m}{k_1^{k_1}\ldots k_n^{k_n}}}\|\widehat{L}\|
\end{equation}
for all non-negative integers $k_1,\ldots,k_n$ with $k_1+\ldots+k_n=m$.\\
For $n=2$ we get
\[|L(x_1^kx_2^{m-k})|\leq\sqrt{\frac{m^m}{k^k(m-k)^{m-k}}}\|\widehat{L}\|\]
and we can easily see that the square root takes its maximum value which is $(\sqrt{2})^m$ when $k=\frac{m}{2}$.\\
To see this, simply consider the function $f(k)=k^k(m-k)^{m-k}$.\\
Then,
\begin{eqnarray*}
f'(k) &=& (1+\ln k)k^k(m-k)^{m-k}-k^k[1+\ln(m-k)](m-k)^{m-k}\\
&=& k^k(m-k)^{m-k}\ln\frac{k}{m-k}
\end{eqnarray*}
and
\[f'(k)=0\Leftrightarrow\ln\frac{k}{m-k}=0\Leftrightarrow\frac{k}{m-k}=1\Leftrightarrow k=\frac{m}{2},\]
since $k\neq0$ and $k\neq m$.\\
Moreover, $f'(k)<0$ for $0<k<\frac{m}{2}$ (i.e. $f$ if strictly decreasing for $0<k<\frac{m}{2}$) and $f'(k)>0$ for $\frac{m}{2}<k<m$ (i.e. $f$ if strictly increasing for $\frac{m}{2}<k<m$).\\
Therefore, the minimum value of $f$ is $f\big(\frac{m}{2}\big)=\big(\frac{m}{2}\big)^m$. Thus,
\[\bigg(\frac{\|L\|_{(2)}}{\|\widehat{L}\|}\bigg)^{\frac{1}{m}}\leq\sqrt{2}.\]
If $m$ is odd, the previous inequality is strict (since $k\in\mathbb{N}$).\\
Similarly, we get the following general result:
\[\bigg(\frac{\|L\|_{(n)}}{\|\widehat{L}\|}\bigg)^{\frac{1}{m}}\leq\sqrt{n}\]

From \cite{N} Lemma 3, we have
\[|L(x_1^{k_1}\ldots x_n^{k_n})|\leq e^{\frac{m}{2}}C_m
\begin{pmatrix} m+n-1\\n-1 \end{pmatrix}
\|\widehat{L}\|.\]
The value of the constant $C_m$ needs to be determined in order to compare this result with (\ref{eq Harris}).
To do this, we need to find a bound for $\mathbb{E}(|A|^k)$ which is used in the proof of \cite{N} Lemma 3. Stirling's formula and some elementary calculations will do the job.\\
For $k\neq2$ even:
\begin{eqnarray*}
\mathbb{E}(|A|^k) &\leq& k(2p)^\frac{k}{2}\Gamma\bigg(\frac{k}{2}\bigg)=k(2p)^\frac{k}{2}\bigg(\frac{k}{2}-1\bigg)!=
k(2p)^\frac{k}{2}\bigg(\frac{\frac{k}{2}-1}{e}\bigg)^{\frac{k}{2}-1}\sqrt{\frac{k}{2}-1}\\
&=& k(2p)^\frac{k}{2}\bigg(\frac{k-2}{2e}\bigg)^\frac{k}{2}\frac{2e}{k-2}\sqrt{\frac{k-2}{2}}=
\frac{ke\sqrt{2}}{\sqrt{k-2}}\bigg[\frac{p(k-2)}{e}\bigg]^\frac{k}{2}\\
&=& \frac{ke\sqrt{2}}{\sqrt{k-2}}\bigg(\frac{k-2}{k}\bigg)^\frac{k}{2}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}=
e\sqrt{2}\sqrt{k-2}\bigg(\frac{k-2}{k}\bigg)^{\frac{k}{2}-1}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}\\
&\leq& e\sqrt{2(k-2)}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}.
\end{eqnarray*}
For $k\neq1$ odd:
\begin{eqnarray*}
\mathbb{E}(|A|^k) &\leq& k(2p)^\frac{k}{2}\Gamma\bigg(\frac{k}{2}\bigg)\leq k(2p)^\frac{k}{2}\Gamma\bigg(\frac{k+1}{2}\bigg)=k(2p)^\frac{k}{2}\bigg(\frac{k-1}{2}\bigg)!\\
&=& k(2p)^\frac{k}{2}\bigg(\frac{\frac{k-1}{2}}{e}\bigg)^{\frac{k-1}{2}}\sqrt{\frac{k-1}{2}}=
k(2p)^\frac{k}{2}\bigg(\frac{k-1}{2e}\bigg)^\frac{k}{2}\sqrt{\frac{2e}{k-1}}\sqrt{\frac{k-1}{2}}\\
&=& k\sqrt{e}\bigg[\frac{p(k-1)}{e}\bigg]^\frac{k}{2}=
k\sqrt{e}\bigg(\frac{k-1}{k}\bigg)^\frac{k}{2}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}\\
&=& (k-1)\sqrt{e}\bigg(\frac{k-1}{k}\bigg)^{\frac{k}{2}-1}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}\leq
(k-1)\sqrt{e}\bigg(\frac{pk}{e}\bigg)^\frac{k}{2}.
\end{eqnarray*}
Moreover, $\mathbb{E}(|A|)\leq\sqrt{2p}$ and $\mathbb{E}(|A|^2)\leq4p$.\\
In order to make the next calculations easier, we have to unite all the above cases. Thus, we take $\mathbb{E}(|A|^k)\leq ke\big(\frac{pk}{e}\big)^\frac{k}{2}$. The term $ke$ could be slightly better, but it would not make any essential difference.\\
At the end of the proof of \cite{N} Lemma 3, we need to calculate the $\sup_{k_i}(C_{k_1}\ldots C_{k_n})$, where $k_1+\ldots+k_n=m$. We have:
\[\sup_{k_i}(C_{k_1}\ldots C_{k_n})=\sup_{k_i}(k_1e\ldots k_ne)=e^n\sup_{k_i}(k_1\ldots k_n)\leq e^n\bigg(\frac{m}{n}\bigg)^n=\bigg(\frac{em}{n}\bigg)^n\]
Finally, we get
\begin{equation} \label{eq Nguyen}
|L(x_1^{k_1}\ldots x_n^{k_n})|\leq e^{\frac{m}{2}}\bigg(\frac{em}{n}\bigg)^n
\begin{pmatrix} m+n-1\\n-1 \end{pmatrix}
\|\widehat{L}\|.
\end{equation}
Inequality (\ref{eq Nguyen}) is worse than inequality (\ref{eq Harris}). But asymptotically it gives better estimate, so it can be useful in some cases.

Combining inequalities (\ref{eq Harris}) and (\ref{eq Nguyen}) we obtain the following

\begin{lemma} \label{lemma asymptotic}
Let $L:X^m\rightarrow Y$ be an $m$-linear map. Then
\[
\bigg(\frac{\|L\|_{(n)}}{\|\widehat{L}\|}\bigg)^{\frac{1}{m}}\leq
\begin{cases}
\sqrt{2},\ \text{if }n=2\\
C\sqrt{e},\ \text{if }n\geq3
\end{cases}
\]
where $C=C(m,n)$ is independent of $L$, $X$, $Y$ and tends to 1 as $m\rightarrow\infty$ for fixed $n$.
\end{lemma}

The $n$th Rademacher function is defined on $[0,1]$ by $r_n=\text{sign}\sin2^n\pi t$.

\begin{lemma}({\bf{Polarization Formula}}) \label{polarization formula}
Let $X$ be a vector space and $L\in\mathcal{L}^s(^mX)$. If $x_1,\ldots,x_m\in X$, then
\begin{equation} \label{eq polarization formula}
L(x_1,\ldots,x_m) = \frac{1}{m!}\int_0^1r_1(t)\ldots r_m(t)\widehat{L}\bigg[\sum_{i=1}^mr_i(t)x_i\bigg]dt.
\end{equation}
\end{lemma}

The proof of Lemma \ref{polarization formula} is easy, so it is omitted (see \cite{S1} Lemma 2).

\begin{proposition} \label{proposition X}
Let $X$, $Y$ be real normed linear spaces and $L:X^m\rightarrow Y$ be a continuous symmetric $m$-linear mapping with associated homogeneous polynomial $\widehat{L}$. If $x_1,\ldots,x_n$ are norm-one vectors in $X$, then
\[\frac{|L(x_1^{k_1}\ldots x_n^{k_n})|}{\|\widehat{L}\|}\leq\min\bigg\{
\sqrt{\frac{m^m}{k_1^{k_1}\ldots k_n^{k_n}}}\ ,
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^m\bigg\}\]
for all non-negative integers $k_1,\ldots,k_n$ with $k_1+\ldots+k_n=m$.
\end{proposition}

\begin{proof}
Using Lemma \ref{polarization formula}, we have
\[L(x_1^{k_1}\ldots x_n^{k_n}) = \frac{1}{m!}\int_0^1r_1(t)\ldots r_m(t)\widehat{L}[(r_1(t)+\ldots+r_{k_1}(t))x_1+\ldots]dt\]

\[\Leftrightarrow L\bigg(\bigg(\frac{x_1}{k_1}\bigg)^{k_1}\ldots\bigg(\frac{x_n}{k_n}\bigg)^{k_n}\bigg) =
\frac{1}{m!}\int_0^1r_1(t)\ldots r_m(t)\widehat{L}\bigg[(r_1(t)+\ldots+r_{k_1}(t))\frac{x_1}{k_1}+\ldots\bigg]dt\]

\[\Leftrightarrow \frac{1}{k_1^{k_1}\ldots k_n^{k_n}}L(x_1^{k_1}\ldots x_n^{k_n}) =
\frac{1}{m!}\int_0^1r_1(t)\ldots r_m(t)\widehat{L}\bigg[(r_1(t)+\ldots+r_{k_1}(t))\frac{x_1}{k_1}+\ldots\bigg]dt\]
Therefore,
\begin{equation} \label{eq new}
\begin{aligned}
|L(x_1^{k_1}\ldots x_n^{k_n})| & \leq
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}\|\widehat{L}\|\int_0^1\bigg\|(r_1(t)+\ldots+r_{k_1}(t))\frac{x_1}{k_1}+\ldots\bigg\|^mdt\\
& \leq
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}\|\widehat{L}\|\int_0^1\bigg(\frac{|r_1(t)+\ldots+r_{k_1}(t)|}{k_1}+\ldots\bigg)^mdt\\
& \leq \frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^m\|\widehat{L}\|.
\end{aligned}
\end{equation}
Inequalities (\ref{eq Harris}) and (\ref{eq new}) complete the proof.
\end{proof}

Asymptotically, inequality (\ref{eq new}) gives much worse estimate than inequalities (\ref{eq Harris}) and (\ref{eq Nguyen}). It is useful though for large $n$'s and when $n$ depends on $m$, while inequality (\ref{eq Harris}) is useful only for fixed $n$'s. We now conclude that:

\begin{theorem} \label{theorem X}
Let $X$, $Y$ be real normed linear spaces and $L:X^m\rightarrow Y$ be a continuous symmetric $m$-linear mapping with associated homogeneous polynomial $\widehat{L}$. If $x_1,\ldots,x_n$ are norm-one vectors in $X$, then
\[\frac{k_1!\ldots k_n!}{k_1^{k_1}\ldots k_n^{k_n}}\frac{m^m}{m!}\leq c(k_1,\ldots,k_n,X)\leq\min\bigg\{
\sqrt{\frac{m^m}{k_1^{k_1}\ldots k_n^{k_n}}}\ ,
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^m\bigg\}\]
for all non-negative integers $k_1,\ldots,k_n$ with $k_1+\ldots+k_n=m$.
\end{theorem}

\begin{proof}
The right hand side inequality is an immediate consequence of Proposition \ref{proposition X}.\\
For the left hand side, let $x^i=(x_n^i)_{n=1}^\infty\in X$, $i=1,\ldots,m$ and $l\in\mathcal{L}^s(^mX)$ be defined by
\[L(x^1,\ldots,x^m)=\frac{1}{m!}\sum_{\sigma\in S_m}x_{\sigma(1)}^1\ldots x_{\sigma(m)}^m,\]
where $S_m$ is the set of permutations of the first $m$ natural numbers.
Then
\[L\big((x^1)^{k_1}\ldots(x^n)^{k_n}\big)=\frac{1}{m!}\sum_{\sigma\in S_m}x_{\sigma(1)}^1\ldots x_{\sigma(k_1)}^1\ldots x_{\sigma(k_1+\ldots k_{n-1}+1)}^n\ldots x_{\sigma(k_1+\ldots k_n)}^n.\]
Take $e_i$ to be the $i$th coordinate vector of $X$ and define
\[y^1=\frac{1}{k_1}(e^1+\ldots+e^{k_1})\]
\[y^2=\frac{1}{k_2}(e^{k_1+1}+\ldots+e^{k_1+k_2})\]
\[\vdots\]
\[y^n=\frac{1}{k_n}(e^{k_1+\ldots+k_{n-1}+1}+\ldots+e^{k_1+\ldots+k_n}).\]
Then an easy calculation shows that $y^1,\ldots,y^n$ are unit vectors in $X$ and
\[L\big((y^1)^{k_1}\ldots(y^n)^{k_n}\big)=\frac{1}{m!}\frac{k_1!\ldots k_n!}{k_1^{k_1}\ldots k_n^{k_n}}.\]
On the other hand, $\|\widehat{L}\|\leq\frac{1}{m^m}$, since
\[|\widehat{L}(x)|=|x_1\ldots x_m|=\big[(|x_1|\ldots|x_m|)^\frac{1}{m}\big]^m\leq\bigg(\frac{|x_1|+\ldots+|x_m|}{m}\bigg)^m\]
by the arithmetic-geometric mean inequality. Thus
\[\big|L\big((y^1)^{k_1}\ldots(y^n)^{k_n}\big)\big|\geq\frac{k_1!\ldots k_n!}{k_1^{k_1}\ldots k_n^{k_n}}\frac{m^m}{m!}\|\widehat{L}\|.\]
\end{proof}

\begin{remark}
For $k_1=\ldots=k_n=1$, the upper and lower bound of $c(k_1,\ldots,k_n,X)$ in Theorem \ref{theorem X} give the same estimate which is $\frac{m^m}{m!}$.
\end{remark}


\section{polynomials on a real $\ell_p$ space}

In \cite{S1} Theorem 2, Y. Sarantopoulos proved that in the case of $L_p(\mu)$, for $1\leq p\leq m'$, $\frac{1}{m}+\frac{1}{m'}=1$ holds that
\[\|L\|\leq\frac{m^\frac{m}{p}}{m!}\|\widehat{L}\|.\]
This is an improved estimate of the one in \cite{H1} by L. A. Harris.
Here we shall give some estimates of the constant $c(k_1,\ldots,k_n,\ell_p)$ for which $|L(x_1^{k_1}\ldots x_n^{k_n})|\leq c(k_1,\ldots,k_n,\ell_p)\|\widehat{L}\|$.

If $f$ is a measurable function on $(X,\mathcal{A},\mu)$, we define its distribution function $\lambda_f:(0,+\infty)\rightarrow[0,+\infty]$ by
\[\lambda_f(a)=\mu(\{x:|f(x)|>a\}).\]
From \cite{F}, we have the following

\begin{proposition}
If $\lambda_f(a)<\infty$ for every $a>0$ and $\phi$ is a nonnegative Borel function on $(0,\infty)$, then
\[\int_X\phi\circ|f|d\mu=-\int_0^\infty\phi(a)d\lambda_f(a).\]
\end{proposition}

The case of this result we are interested in, is $\phi(a)=a^p$, which gives
\[\int|f|^pd\mu=-\int_0^{\infty}a^pd\lambda_f(a).\]
Integrating the right side by parts, we obtain
\[\int|f|^pd\mu=p\int_0^{\infty}a^{p-1}\lambda_f(a)da.\]
The validity of this calculation becomes clear if we consider that $a^p\lambda_f(a)\rightarrow0$ as $a\rightarrow0$ and $a\rightarrow\infty$ (since $\lambda_f$ is strictly decreasing). In the following Proposition the function $f$ will be of the form $f(t)=r_1(t)+\ldots+r_k(t)$, $k\in\mathbb{N}$. Therefore, using Hoeffding's inequality (see \cite{H} Theorem 2), we get that
\[\lambda_f(x):=\lambda_k(x)=P(|r_1(t)+\ldots+r_k(t)|\geq x)\leq2e^{-\frac{x^2}{2k}}.\]

\begin{proposition} \label{proposition l_p}
Let $1\leq p\leq\infty$ and $L:(\ell_p)^m\rightarrow\mathbb{R}$ be a continuous symmetric $m$-linear mapping with associated homogeneous polynomial $\widehat{L}$. If $x_1,\ldots,x_n$ are norm-one vectors in $\ell_p$ with disjoint supports, then
\[\frac{|L(x_1^{k_1}\ldots x_n^{k_n})|}{\|\widehat{L}\|}\leq
\begin{cases}
\min\bigg\{\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^\frac{m}{p},
\frac{p2^\frac{p}{2}\Gamma\big(\frac{p}{2}\big)m^\frac{p}{2}}{m!}\bigg\},\text{ if }p\geq m\\
\min\bigg\{\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^\frac{m}{p},
\frac{n^\frac{m-p}{p}m2^\frac{m}{2}\Gamma\big(\frac{m}{2}\big)m^\frac{m}{2}}{m!}\bigg\},\text{ if } p<m
\end{cases}\]
for all non-negative integers $k_1,\ldots,k_n$ with $k_1+\ldots+k_n=m$.
\end{proposition}

\begin{proof}
Working as in the proof of Proposition \ref{proposition X}, we get
\begin{equation} \label{eq l_p}
\begin{aligned}
|L(x_1^{k_1}\ldots x_n^{k_n})| & \leq
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}\|\widehat{L}\|\int_0^1\bigg\|(r_1(t)+\ldots+r_{k_1}(t))\frac{x_1}{k_1}+\ldots\bigg\|_p^mdt\\
& \leq
\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}\|\widehat{L}\|\int_0^1\bigg[\bigg(\frac{|r_1(t)+\ldots+r_{k_1}(t)|}{k_1}\bigg)^p+\ldots\bigg]^\frac{m}{p}dt\\
& \leq \frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^\frac{m}{p}\|\widehat{L}\|.
\end{aligned}
\end{equation}

We shall now use a different technique and we need to distinguish two cases.

For $p\geq m$:
\begin{eqnarray*}
|L(x_1^{k_1}\ldots x_n^{k_n})| &\leq&
\frac{\|\widehat{L}\|}{m!}\int_0^1\|(r_1(t)+\ldots+r_{k_1}(t))x_1+\ldots\|_p^mdt\\
&\leq& \frac{\|\widehat{L}\|}{m!}\int_0^1(|r_1(t)+\ldots+r_{k_1}(t)|^p+\ldots)^{\frac{m}{p}}dt\\
&\leq& \frac{\|\widehat{L}\|}{m!}\bigg(p\int_0^{\infty}x^{p-1}\lambda_{k_1}(x)dx+\ldots+p\int_0^{\infty}x^{p-1}\lambda_{k_n}(x)dx\bigg)\\
&\leq& \frac{\|\widehat{L}\|}{m!}\bigg(p\int_0^{\infty}x^{p-1}2e^{-\frac{x^2}{2k_1}}dx+\ldots+p\int_0^{\infty}x^{p-1}2e^{-\frac{x^2}{2k_n}}dx\bigg)\\
&=& \frac{\|\widehat{L}\|}{m!}\bigg(p(2k_1)^{\frac{p}{2}}\Gamma\big(\frac{p}{2}\big)+\ldots+p(2k_n)^{\frac{p}{2}}\Gamma\big(\frac{p}{2}\big)\bigg)\\
&=& \frac{p2^\frac{p}{2}\Gamma\big(\frac{p}{2}\big)}{m!}\sum_{i=1}^nk_i^\frac{p}{2}\|\widehat{L}\|\leq
\frac{p2^\frac{p}{2}\Gamma\big(\frac{p}{2}\big)}{m!}\bigg(\sum_{i=1}^nk_i\bigg)^\frac{p}{2}\|\widehat{L}\|\\
&=& \frac{p2^\frac{p}{2}\Gamma\big(\frac{p}{2}\big)m^\frac{p}{2}}{m!}\|\widehat{L}\|
\end{eqnarray*}

For $p<m$:
\begin{eqnarray*}
|L(x_1^{k_1}\ldots x_n^{k_n})| &\leq&
\frac{\|\widehat{L}\|}{m!}\int_0^1\|(r_1(t)+\ldots+r_{k_1}(t))x_1+\ldots\|_p^mdt\\
&\leq& \frac{\|\widehat{L}\|}{m!}\int_0^1(|r_1(t)+\ldots+r_{k_1}(t)|^p+\ldots)^{\frac{m}{p}}dt\\
&=& \frac{n^\frac{m}{p}\|\widehat{L}\|}{m!}\int_0^1\bigg(\frac{|r_1(t)+\ldots+r_{k_1}(t)|^p+\ldots}{n}\bigg)^{\frac{m}{p}}dt\\
&\leq& \frac{n^\frac{m}{p}\|\widehat{L}\|}{m!}\int_0^1\frac{|r_1(t)+\ldots+r_{k_1}(t)|^m+\ldots}{n}dt\\
&\leq& \frac{n^\frac{m-p}{p}\|\widehat{L}\|}{m!}\bigg(m\int_0^{\infty}x^{m-1}\lambda_{k_1}(x)dx+\ldots+m\int_0^{\infty}x^{m-1}\lambda_{k_n}(x)dx\bigg)\\
&\leq& \frac{n^\frac{m-p}{p}\|\widehat{L}\|}{m!}\bigg(m\int_0^{\infty}x^{m-1}2e^{-\frac{x^2}{2k_1}}dx+\ldots+m\int_0^{\infty}x^{m-1}2e^{-\frac{x^2}{2k_n}}dx\bigg)\\ &=& \frac{n^\frac{m-p}{p}\|\widehat{L}\|}{m!}\bigg(m(2k_1)^{\frac{m}{2}}\Gamma\big(\frac{m}{2}\big)+\ldots+m(2k_n)^{\frac{m}{2}}\Gamma\big(\frac{m}{2}\big)\bigg)\\
&=& \frac{n^\frac{m-p}{p}m2^\frac{m}{2}\Gamma\big(\frac{m}{2}\big)}{m!}\sum_{i=1}^nk_i^\frac{m}{2}\|\widehat{L}\|\leq
\frac{n^\frac{m-p}{p}m2^\frac{m}{2}\Gamma\big(\frac{m}{2}\big)}{m!}\bigg(\sum_{i=1}^nk_i\bigg)^\frac{m}{2}\|\widehat{L}\|\\
&=& \frac{n^\frac{m-p}{p}m2^\frac{m}{2}\Gamma\big(\frac{m}{2}\big)m^\frac{m}{2}}{m!}\|\widehat{L}\|
\end{eqnarray*}
\end{proof}

Note that the second technique used in the proof of Proposition \ref{proposition l_p} gives better estimates than (\ref{eq l_p}) for large $m$'s and small $n$'s and its special case where $p\geq m$ holds asymptotically only for $\ell_\infty$. Moreover, observe that Proposition \ref{proposition l_p} gives better estimates than Proposition \ref{proposition X}.

\begin{remark}
The term $\sum_{i=1}^nk_i^\frac{p}{2}$ \big(resp. $\sum_{i=1}^nk_i^\frac{m}{2}$\big) in the proof of Proposition \ref{proposition l_p} can be bounded by $(m-n+1)^\frac{p}{2}+n-1$ \big(resp. $(m-n+1)^\frac{m}{2}+n-1$\big) instead of $m^\frac{p}{2}$ \big(resp. $m^\frac{m}{2}$\big) since it takes its maximum value when all but one $k_i$'s equal to 1 and the last one equals to $m-n+1$.
\end{remark}

We now conclude that:

\begin{theorem} \label{theorem l_p}
Let $1\leq p\leq\infty$ and $L:(\ell_p)^m\rightarrow\mathbb{R}$ be a continuous symmetric $m$-linear mapping with associated homogeneous polynomial $\widehat{L}$. If $x_1,\ldots,x_n$ are norm-one vectors in $\ell_p$ with disjoint supports, then:\\
if $p\geq m$
\[\frac{k_1!\ldots k_n!}{k_1^\frac{k_1}{p}\ldots k_n^\frac{k_n}{p}}\frac{m^\frac{m}{p}}{m!}\leq c(k_1,\ldots,k_n,\ell_p)\leq\min\bigg\{\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^\frac{m}{p},
\frac{p2^\frac{p}{2}\Gamma\big(\frac{p}{2}\big)m^\frac{p}{2}}{m!}\bigg\}\]
and if $p<m$
\[\frac{k_1!\ldots k_n!}{k_1^\frac{k_1}{p}\ldots k_n^\frac{k_n}{p}}\frac{m^\frac{m}{p}}{m!}\leq c(k_1,\ldots,k_n,\ell_p)\leq\min\bigg\{\frac{k_1^{k_1}\ldots k_n^{k_n}}{m!}n^\frac{m}{p},
\frac{n^\frac{m-p}{p}m2^\frac{m}{2}\Gamma\big(\frac{m}{2}\big)m^\frac{m}{2}}{m!}\bigg\}\]
for all non-negative integers $k_1,\ldots,k_n$ with $k_1+\ldots+k_n=m$.
\end{theorem}

\begin{proof}
The right hand side inequality is an immediate consequence of Proposition \ref{proposition l_p}.\\
For the left hand side, we just need some simple adjustments to the proof of Proposition \ref{proposition X}.
\end{proof}

\begin{remark}
For $k_1=\ldots=k_n=1$, the upper and lower bound of $c(k_1,\ldots,k_n,\ell_p)$ in Theorem \ref{theorem l_p} give the same estimate which is $\frac{m^\frac{m}{p}}{m!}$.
\end{remark}


\section{Radius of analyticity of a power series on a real Banach space}

A power series centered at $a\in X$ is a formal sum
\begin{equation} \label{eq power series}
\sum_{m=0}^\infty P_m(x-a)
\end{equation}
where for each $m$, $P_m:X\rightarrow Y$ is a continuous $m$-homogeneous polynomial.

The radius of uniform convergence of this power series is defined to be
\[\rho:=\{r:(\ref{eq power series})\text{ converges uniformly on }|x-a|\leq r\},\]
which is given by the following standard formula:
\[\rho=\frac{1}{\limsup_{m\rightarrow\infty}\|P_m\|^\frac{1}{m}}.\]
If $\rho>0$, then for every $0<r<\rho$, (\ref{eq power series}) is a uniformly and absolutely convergent series for every $x\in B_r(a)$. hence in this case, (\ref{eq power series}) defines a function on $B_{\rho}(a)$ taking values in $Y$.

The Taylor series of an infinitely differentiable function $F$ defined in a neighborhood of $a$ is the power series defined by
\[T_aF(x)=\sum_{m=0}^\infty\frac{1}{m!}D^mF(a)((x-a)^m),\]
where $D^mF(a):X^m\rightarrow Y$ is the symmetric $m$-linear map given by taking the Fr\'{e}chet derivative of $F$ $m$ times.

$F$ is called analytic at $a$ if $T_aF(x)$ has a positive radius of uniform convergence and equals $F(x)$ within the domain of uniform convergence. If $U\subset X$ is open, we say that $F$ is analytic in $U$ if it is analytic at every $a\in U$. If furthermore, $T_aF(x)$ converges uniformly in every closed ball centered at $a$ contained in $U$, for each $a\in U$, $F$ is called fully analytic in $U$.

Let $F(x)$ be a power series centered at $a$ with radius of uniform convergence $\rho>0$. The radius of analyticity $\rho_{\substack{\\ A}}=\rho_{\substack{\\ A}}(F)$ of $F(x)$ at $a$ is the largest $r>0$ such that $F(x)$ is fully analytic in $B_r(a)$.

The norm we need to control when we expand a power series at a new point is $\|L\|_{(2)}$. To see this, consider the power series (\ref{eq power series}) centered at $a=0$, which we may rewrite as
\begin{equation} \label{eq series at 0}
F(x)=\sum_{m=0}^\infty L_m(x^m).
\end{equation}
Observe that by the binomial formula, given any $y\in X$
\[L_m(x^m)=L_m((y+x-y)^m)=\sum_{k=0}^\infty
\begin{pmatrix} m\\k \end{pmatrix}
L_m(y^{m-k},(x-y)^k).\]
Then, if in
\begin{equation} \label{eq double series}
\sum_{m=0}^\infty L_m(x^m)=\sum_{m=0}^\infty\sum_{k=0}^m
\begin{pmatrix} m\\k \end{pmatrix}
L_m(y^{m-k},(x-y)^k)
\end{equation}
the double series on the right converges absolutely, we can interchange summations and obtain
\[F(x)=\sum_{m=0}^\infty L_m(x^m)=\sum_{k=0}^\infty\sum_{m=k}^\infty
\begin{pmatrix} m\\k \end{pmatrix}
L_m(y^{m-k},(x-y)^k)\]
Thus, if we can perform this change of summation for all $x\in B_r(y)$, for some $r>0$, then we will have expressed $F(x)$ as a power series centered at $y$ whose $k$-homogeneous polynomial coefficients are given by
\begin{equation} \label{eq coefficients}
A_k(z):=\sum_{m=k}^\infty
\begin{pmatrix} m\\k \end{pmatrix}
L_m(y^{m-k},z^k).
\end{equation}
Observe that the absolute convergence of the double sum (\ref{eq double series}) for $x\in B_r(y)$ implies the absolute convergence of the $A_k$ in $B_r(0)$ and hence on all $X$ by homogeneity.

Absolute convergence of (\ref{eq double series}) holds if
\[\sum_{m=0}^\infty\sum_{k=0}^m
\begin{pmatrix} m\\k \end{pmatrix}
\|L_m\|_{(2)}|y|^{m-k}|x-y|^k=\sum_{m=0}^\infty\|L_m\|_{(2)}(|y|+|x-y|)^m<\infty.\]
This holds when
\begin{equation} \label{eq convergence1}
|y|+|x-y|<\frac{1}{\limsup\|L_m\|_{(2)}^\frac{1}{m}}.
\end{equation}
Choose a subsequence $m_i$ such that
\[\lim_{i\rightarrow\infty}\|L_{m_i}\|_{(2)}^\frac{1}{m_i}=\limsup\|L_m\|_{(2)}^\frac{1}{m}.\]
Let $\rho=\frac{1}{\limsup\|\widehat{L}_m\|_{(2)}^\frac{1}{m}}>0$ be the radius of uniform convergence of (\ref{eq series at 0}) and suppose $\rho<\infty$. Then (\ref{eq convergence1}) is satisfied if
\begin{equation}
\begin{aligned} \label{eq convergence2}
|y|+|x-y| & < \rho\cdot\frac{\limsup_{i\rightarrow\infty}\|\widehat{L}_{m_i}\|^\frac{1}{m_i}}{\lim_{i\rightarrow\infty}\|L_{m_i}\|_{(2)}^\frac{1}{m_i}}\\
& = \rho\cdot\limsup_{i\rightarrow\infty}\bigg(\frac{\|\widehat{L}_{m_i}\|}{\|L_{m_i}\|_{(2)}}\bigg)^\frac{1}{m_i}\\
& := \bar{\rho}.
\end{aligned}
\end{equation}
Thus, for $|y|<\bar{\rho}$ and $|x-y|<\bar{\rho}-|y|$, the series (\ref{eq double series}) converges absolutely.

Altogether then, we have shown that for any fixed $|y|<\bar{\rho}$, we have
\begin{equation} \label{eq series}
F(x)= \sum_{k=0}^\infty A_k(x-y)
\end{equation}
for $|x-y|<\bar{\rho}-|y|$.

From \cite{C} Corollary 1, p. 165, we have the following

\begin{lemma} \label{lemma Chae}
For any power series $F(x)= \sum_{k=0}^\infty A_k(x-y)$ centered at $y$ with positive radius of uniform convergence, $A_k=\frac{1}{k!}D^kF(y)$ as $k$-homogeneous polynomials.
\end{lemma}

\begin{theorem} \label{theorem radius}
Let $F(x)$ be a power series in a Banach space $X$, which we may take to be centered at the origin. Let $\rho>0$ denote its radius of uniform convergence and $\rho_{\substack{\\ A}}$ its radius of analyticity. Then
\begin{enumerate}
\item [(i)] $\rho_{\substack{\\ A}}\geq\frac{\rho}{\sqrt{2}}$,
\item [(ii)] for every $n$, the $n$th Fr\'{e}chet derivative $D^nF:X\rightarrow\mathcal{L}_n(X,Y)$ of $F(x)$, viewed as a map from $X$ to the Banach space $\mathcal{L}_n(X,Y)$ of continuous $n$-linear maps from $X$ to $Y$, has a Taylor series centered at the origin with radius of uniform convergence at least $\frac{\rho}{\sqrt{2}}$ for $n=2$ and $\frac{\rho}{\sqrt{e}}$ for $n\geq3$,
\item [(iii)] the radius of analyticity of the power series $D^nF(x)$ is at least $\frac{\rho}{\sqrt{2}}$.
\end{enumerate}
\end{theorem}

\begin{proof}
\begin{enumerate}
\item [(i)] Suppose $\rho<\infty$. Lemma \ref{lemma asymptotic} implies
    \begin{equation} \label{eq limsup}
    \limsup_{m\rightarrow\infty}\bigg(\frac{\|\widehat{L}_m\|}{\|L_m\|_{(2)}}\bigg)^\frac{1}{m}\geq\frac{1}{\sqrt{2},}
    \end{equation}
    hence $\bar{\rho}\geq\frac{\rho}{\sqrt{2}}$ by (\ref{eq convergence2}). Thus the preceding analysis shows that for $|y|<\rho$, the Taylor series (\ref{eq series}) is absolutely convergent in $\{x:|x-y|<\frac{\rho}{\sqrt{2}}-|y|\}$. From this, we get uniform convergence of (\ref{eq series}) in $\{x:|x-y|<r\}$ for every $r<\frac{\rho}{\sqrt{2}}-|y|$, since one can bound the tail of (\ref{eq series}) as follows:
    \begin{equation} \label{eq bound}
    \begin{aligned}
    \bigg|\sum_{k=N}^\infty A_k(x-y)\bigg| & \leq \sum_{k=N}^\infty\sum_{m=k}^\infty
    \begin{pmatrix} m\\k \end{pmatrix}
    |L_m(y^{m-k},(x-y)^k)|\\
    & \leq \sum_{m=N}^\infty\|L_m\|_{(2)}(|y|+|x-y|)^m
    \end{aligned}
    \end{equation}
    where (\ref{eq bound}) tends to zero uniformly in $x$ as $N\rightarrow0$ so long as $|y|+|x-y|\leq|y|+r$ is bounded away from $\bar{\rho}\geq\frac{\rho}{\sqrt{2}}$. This follows because we have shown that (\ref{eq bound}) viewed as power series in a single real variable has radius of uniform convergence at least $\bar{\rho}$. Finally, Lemma \ref{lemma Chae} implies that the power series (\ref{eq series}) is a Taylor series of $F(x)$ centered at $y$. So for $\rho<\infty$, this proves (i) since we have shown $\rho_{\substack{\\ A}}\geq\frac{\rho}{\sqrt{2}}$.

    For the remaining case $\rho=\infty$, i.e. $\limsup_{m\rightarrow\infty}\|\widehat{L}_m\|^\frac{1}{m}=0$, (\ref{eq limsup}) implies also that $\limsup_{m\rightarrow\infty}\|L_m\|_{(2)}^\frac{1}{m}=0$. From (\ref{eq convergence1}), we can apply the previous analysis for every $\bar{\rho}>0$, hence $\rho_{\substack{\\ A}}=\rho=\infty$.
\item [(ii)] By Chae, given any power series $F(x)=\sum_{m=0}^\infty L_m(x^m)$ with radius of uniform convergence $\rho>0$, for every $n$, $D^nF(x)$ has a Taylor series centered at the origin given by
    \begin{equation} \label{eq Taylor series}
    T_0D^nF(x)=n!\sum_{m=0}^\infty
    \begin{pmatrix} m+n\\n \end{pmatrix}
    L_{m+n}(x^m).
    \end{equation}
    In (\ref{eq Taylor series}), the linear maps $L_{m+n}$ are only evaluated on at most $n+1$ distinct arguments ($D^nF(x)$ takes values in $n$-linear maps). So by Lemma \ref{lemma asymptotic}, for $n\geq3$, the lower bound for the radius of uniform convergence of (\ref{eq Taylor series}) is $\frac{\rho}{\sqrt{e}}$ since
    \begin{eqnarray*}
    \limsup_{m\rightarrow\infty}\bigg\|
    \begin{pmatrix} m+n\\n \end{pmatrix}
    L_{m+n}\bigg\|_{(n+1)}^\frac{1}{m} &\leq& \bigg(\limsup_{m\rightarrow\infty}C(m,n+1)\sqrt{e}\bigg)\\
    & & \times\bigg(\limsup_{m\rightarrow\infty}\|\widehat{L}_{m+n}\|^\frac{1}{m}\bigg)\\
    &\leq& \frac{\sqrt{e}}{\rho}.
    \end{eqnarray*}
    Similarly, the lower bound for the radius of uniform convergence of $T_0DF(x)=\sum_{m=0}^\infty(m+1)L_{m+1}(x^m)$ is $\frac{\rho}{\sqrt{2}}$ since
    $\limsup_{m\rightarrow\infty}\|(m+1)L_{m+1}\|_{(2)}^\frac{1}{m}\leq\frac{\sqrt{2}}{\rho}$.\\
    By Lemma \ref{lemma Chae} and the formula (\ref{eq coefficients}) for the $A_k$, it follows that for $n\geq3$, $T_0D^nF(y)=D^nF(y)$ for all $y$ such that $|y|<\frac{\rho}{\sqrt{e}}$ and $T_0DF(y)=DF(y)$ for all $y$ such that $|y|<\frac{\rho}{\sqrt{2}}$.
\item [(iii)] We need to show that the radius of analyticity of (\ref{eq Taylor series}) is at least $\frac{\rho}{\sqrt{2}}$. For this, we need to control $\|L_{m+n}\|_{(n+2)}$, but this is precisely the $\|\cdot\|_{(2)}$-norm of $L_{m+n}$ viewed as a map from $X$ to $\mathcal{L}_n(X,Y)$. Applying Lemma \ref{lemma asymptotic} the proof is completed.
\end{enumerate}
\end{proof}




