text:
\section{Introduction}

In 1930 Littlewood \cite{LW} has shown the following result on bilinear forms on $c_{0}\times c_{0}$, now called Littlewood's $4/3$ inequality: for any bounded bilinear form $A:c_{0}\times c_{0} \rightarrow \mathbb{C}$,
\[
\left(  \sum_{i,j=1}^{+\infty}|A(e_{i},e_{j})|^{\frac{4}{3}}\right)  ^{\frac
{3}{4}}\leq\sqrt{2}\Vert A\Vert
\]
and, moreover, the exponent $4/3$ is optimal.  From now on, $m\geq1$ is a positive
integer, $\mathbf{p}:=\left(  p_{1},\dots,p_{m}\right)  \in\left[
1,+\infty\right]  ^{m}$ and
\[
\left\vert \frac{1}{\mathbf{p}}\right\vert :=\frac{1}{p_{1}}+\dots+\frac
{1}{p_{m}}.
\]
For $1\leq p<+\infty$, let us set $X_{p}:=\ell_{p}$ and let us define
$X_{\infty}=c_{0}$. As soon as Littlewood's $4/3$ inequality appeared, it was
rapidly extended to more general frameworks. For instance:

\begin{itemize}
\item (Bohnenblust and Hille, \cite[Theorem I]{BH}, 1931 (see also \cite{dddsss})) There exists a constant $C=C(m)\geq1$ such
that
\begin{equation}
\left(  \sum\limits_{i_{1},...,i_{m}=1}^{+\infty}\left\vert A(e_{i_{^{1}}%
},...,e_{i_{m}})\right\vert ^{\frac{2m}{m+1}}\right)  ^{\frac{m+1}{2m}}\leq
C\left\Vert A\right\Vert \label{juui}%
\end{equation}
for all continuous $m$-linear forms $A:c_{0}\times\cdots\times c_{0}%
\rightarrow\mathbb{C}$ and the exponent $\frac{2m}{m+1}$ is optimal.

\item (Hardy and Littlewood, \cite{hardy.lw}, 1934 (see also \cite[page 224]{hardy})/Praciano-Pereira, \cite[Theorems A and B]{praciano}, 1981) Let $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$ with
\[
\left\vert {\frac{1}{\mathbf{p}}}\right\vert \leq\frac{1}{2},
\]
then there exists a constant $C>0$ such that, for every continuous $m$-linear
form $A:X_{p_{1}}\times\dots\times X_{p_{m}}\rightarrow\mathbb{C}$,
\begin{equation}
\left(  \sum_{i_{1},\dots,i_{m}=1}^{+\infty}|A(e_{i_{1}},\dots,e_{i_{m}%
})|^{\frac{2m}{m+1-2\left\vert {\frac{1}{\mathbf{p}}}\right\vert }}\right)
^{\frac{m+1-2\left\vert {\frac{1}{\mathbf{p}}}\right\vert }{2m}}\leq C\Vert
A\Vert.\label{juui2}%
\end{equation}

\item (Defant and Sevilla-Peris, \cite[Theorem 1]{df}, 2009) If $1\leq s\leq q\leq2,$ there exists a constant
$C>0$ such that, for every continuous $m$-linear mapping $A:c_{0}\times
\dots\times c_{0}\rightarrow\ell_{s}$, then
\[
\left(  \sum_{i_{1},\dots,i_{m}=1}^{+\infty}\Vert A(e_{i_{1}},\dots,e_{i_{m}%
})\Vert_{\ell_{q}}^{\frac{2m}{m+2\left(  \frac{1}{s}-\frac{1}{q}\right)  }}\right)
^{\frac{m+2\left(  \frac{1}{s}-\frac{1}{q}\right)  }{2m}}\leq C\Vert A\Vert.
\]

\end{itemize}

Very recently the previous results were generalized by the authors and by Dimant and Sevilla-Peris:

\begin{itemize}
\item (\cite[Corollary 1.3]{abps}, 2013) Let $1\leq s\leq q\leq2$ and $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$ such that
\begin{equation}
\frac{1}{s}-\frac{1}{q}-\left\vert{\frac{1}{\mathbf{p}}}\right\vert
\geq0.\label{assump}%
\end{equation}
Then there exists a constant $C>0$ such that, for every continuous $m$-linear mapping $A:X_{p_1} \times\dots\times X_{p_m} \rightarrow {X_{s}}$, we have
\[
\left(  \sum_{i_{1},\dots,i_{m}=1}^{+\infty}\Vert A(e_{i_{1}},\dots,e_{i_{m}%
})\Vert_{\ell_{q}}^{\frac{2m}{m+2\left(  \frac{1}{s}-\frac{1}{q}-\left\vert {\frac
{1}{\mathbf{p}}}\right\vert \right)  }}\right)  ^{\frac{m+2\left(  \frac{1}%
{s}-\frac{1}{q}-\left\vert {\frac{1}{\mathbf{p}}}\right\vert \right)  }{2m}%
}\leq C\Vert A\Vert
\]
and the exponent is optimal.
\end{itemize}

\begin{itemize}
\item (Dimant and Sevilla-Peris, \cite[Proposition 4.4]{dsp}, 2013) Let
$\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$ and $s, q \in [1,+\infty]$ be such that $s\leq q$. Then there exists a constant $C>0$ such that, for every continuous $m$-linear mapping $A : X_{p_1} \times\cdots\times X_{p_m} \rightarrow X_s$, we have
\[
\left( \sum_{i_1,\dots,i_m=1}^{+\infty}
  \left\Vert A \left( e_{i_1}, \ldots, e_{i_m}\right)  \right\Vert_{\ell_q}^\rho
\right)^\frac1\rho
\leq C\Vert A\Vert,
\]
where $\rho$ is given by

\begin{itemize}
\item[(i)] If $s \leq q \leq2$, and

\begin{itemize}
\item[(a)] if $0 \leq \left|\frac1{\mathbf p}\right| <\frac1s -\frac1q$, then $\frac{1}{\rho} = \frac12 + \frac1m \left( \frac1s - \frac1q - \left|\frac1{\mathbf p}\right| \right)$.

\item[(b)] if $\frac1s - \frac1q \leq \left|\frac1{\mathbf p}\right| < \frac12 + \frac1s - \frac1q$, then $\frac1\rho = \frac12 + \frac1s - \frac1q - \left|\frac1{\mathbf p}\right|$.
\end{itemize}

\item[(ii)] If $s \leq2 \leq q$, and

\begin{itemize}
\item[(a)] if $0 \leq \left|\frac1{\mathbf p}\right| < \frac1s - \frac12$, then $\frac1\rho = \frac12 + \frac1m \left( \frac1s - \frac12 - \left|\frac1{\mathbf p}\right|\right)$.

\item[(b)] if $\frac1s - \frac12 \leq \left|\frac1{\mathbf p}\right| < \frac1s$, then $\frac1\rho = \frac1s - \left|\frac1{\mathbf p}\right|$.
\end{itemize}

\item[(iii)] If $2 \leq s \leq q$ and $0 \leq \left|\frac1{\mathbf p}\right| < \frac1s$, then $\frac1\rho = \frac1s - \left|\frac1{\mathbf p}\right|$.
\end{itemize}

Moreover, the exponents in the cases (ia),(iib) and (iii) are optimal. Also,
the exponent in (ib) is optimal for $\frac1s-\frac1q\leq\ \left|\frac1{\mathbf p}\right| <\frac12$.
\end{itemize}

Our main intention, in this paper, is to improve the previous theorems in three directions.
\begin{enumerate}
\item We study in depth the remaining cases of the Dimant and Sevilla-Peris result. Surprisingly, we show that
in case (iia), the exponent given above is optimal whereas it is not optimal in case (ib) when $\left|\frac1{\mathbf p}\right|>\frac 12$. We give a better exponent in that case and show a necessary condition on it. These two bounds coincide when $s=1$. We can summarize this into the two following statements.

\begin{theorem}
\label{nov1} Let $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$ and let
$\rho>0$. Assume moreover that either $q\geq 2$ or $q<2$ and $\left|\frac1{\mathbf p}\right|< \frac 12$. Let

\[
\frac{1}{\lambda}
:= \frac{1}{2}+\frac{1}{s}-\frac{1}{\min\{q,2\}} - \left\vert{\frac{1}{\mathbf{p}}}\right\vert > 0.
\]

Then there exists $C>0$ such that, for every continuous $m$-linear
operator $A:X_{p_{1}}\times\dots\times X_{p_{m}}\rightarrow X_{s}$, we have%
\[
\left(  \sum_{i_{1},\dots,i_{m}=1}^{+\infty}\Vert A(e_{i_{1}},\dots,e_{i_{m}%
})\Vert_{\ell_{q}}^{\rho}\right)  ^{\frac{1}{\rho}}\leq C\Vert A\Vert
\]
if and only if
\[
\frac{m}{\rho}\leq\frac{1}{\lambda}+\frac{m-1}{\max\{\lambda,s,2\}}.
\]

\end{theorem}

The following table summarizes the optimal value of $\frac{1}{\rho}$ following
the respective values of $s,q,p_{1},...,p_{m}$:
\[%
\begin{array}
[c]{c|c}%
1\leq s\leq q\leq2,\ \lambda<2 & \displaystyle\frac{1}{2}+\frac{1}{ms}%
-\frac{1}{mq}-\frac{1}{m}\times\left\vert {\frac{1}{\mathbf{p}}}\right\vert
\\[3mm]\hline
& \\
1\leq s\leq q\leq2,\ \lambda\geq2,\ \left|\frac1{\mathbf p}\right| < \frac12 & \displaystyle\frac{1}{2}+\frac{1}{s}%
-\frac{1}{q}-\left\vert {\frac{1}{\mathbf{p}}}\right\vert \\[3mm]\hline
& \\
1\leq s\leq2\leq q,\ \lambda<2 & \displaystyle\frac{m-1}{2m}+\frac{1}%
{ms}-\frac{1}{m}\times\left\vert {\frac{1}{\mathbf{p}}}\right\vert \\[3mm]\hline
& \\
1\leq s\leq2\leq q,\ \lambda\geq2 & \displaystyle\frac{1}{s}-\left\vert
{\frac{1}{\mathbf{p}}}\right\vert \\[3mm]\hline
& \\
2\leq s\leq q & \displaystyle\frac{1}{s}-\left\vert {\frac{1}{\mathbf{p}}%
}\right\vert \\
&
\end{array}
\]

We note that (\ref{juui}) and (\ref{juui2}) are recovered by Theorem \ref{nov1} just by choosing $s=1$ and $q=2$.

When $q<2$ and $\left|\frac1{\mathbf p}\right|>\frac 12$ (observe that this automatically implies $\lambda\geq 2$), the situation
is more difficult and we get the following statement.
\begin{theorem}\label{nov2}
Let $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$, $\left|\frac1{\mathbf p}\right|>\frac 12$, $1\leq s\leq q\leq 2$ and let
$\rho>0$. Let us consider the following property.
\begin{quote}
There exists $C>0$ such that, for every continuous $m$-linear
operator $A:X_{p_{1}}\times\dots\times X_{p_{m}}\rightarrow X_{s}$, we have%
\[
\left(  \sum_{i_{1},\dots,i_{m}=1}^{+\infty}\Vert A(e_{i_{1}},\dots,e_{i_{m}%
})\Vert_{\ell_{q}}^{\rho}\right)  ^{\frac{1}{\rho}}\leq C\Vert A\Vert.
\]
\end{quote}
\begin{itemize}
\item[(A)] The property is satisfied as soon as
$$\frac 1\rho\leq \frac{\left(\frac 1s-\frac 1q\right)\left(\frac 1s-\left|\frac1{\mathbf p}\right|\right)}{\frac 12-\frac 1s}.$$
\item[(B)] If the property is satisfied, then
$$\frac 1\rho\leq 2\left(1-\left|\frac1{\mathbf p}\right|\right)\left(\frac 1s-\frac 1q\right).$$
\end{itemize}
\end{theorem}
In particular, if $s=1$, then the property is satisfied if and only if
$$\frac 1\rho\leq 2\left(1-\left|\frac1{\mathbf p}\right|\right)\left(1-\frac1q\right).$$

\item We give a simpler proof of the sufficient part of the Dimant and Sevilla-Peris theorem.  It turns out that it is easier to prove a more general result.

\begin{theorem}
\label{main2}Let $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$ and $1\leq
s\leq q\leq\infty$ be such that
\[
\left\vert \frac{1}{\mathbf{p}}\right\vert <\frac{1}{2}+\frac{1}{s}-\frac
{1}{\min\{q,2\}}.
\]

Let

\[
\frac{1}{\lambda}:=\frac{1}{2}+\frac{1}{s}-\frac{1}{\min\{q,2\}}-\left\vert
\frac{1}{\mathbf{p}}\right\vert.
\]

If $\lambda>0$ and $t_{1},\ldots,t_{m}\in\left[  \lambda,\max\left\{  \lambda,s,2\right\}
\right]  $ are such that
\begin{equation}
\frac{1}{t_{1}}+\cdots+\frac{1}{t_{m}}\leq\frac{1}{\lambda}+\frac{m-1}%
{\max\{\lambda,s,2\}},\label{dez11}%
\end{equation}

then there exists $C>0$ satisfying, for every continuous $m$-linear map
$A:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow X_{s}$,
\begin{equation}
\left(  \sum_{i_{1}=1}^{+\infty}\left(  \ldots\left(  \sum_{i_{m}=1}^{+\infty
}\left\Vert A\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert
_{\ell_{q}}^{t_{m}}\right)  ^{\frac{t_{m-1}}{t_{m}}}\ldots\right)  ^{\frac{t_{1}%
}{t_{2}}}\right)  ^{\frac{1}{t_{1}}}\leq C\Vert A\Vert. \label{q3}%
\end{equation}
Moreover, the exponents are optimal except eventually if $q\leq 2$ and $\left|\frac1{\mathbf p}\right|>\frac 12$.
\end{theorem}

\begin{remark}
\bigskip The optimality in the above theorem shall be understood in a strong
sense: when $\lambda<2,$ we prove that if $t_{1},\ldots,t_{m}\in\left[
1,+\infty\right)  $ are so that (\ref{q3}) holds then (\ref{dez11}) is
valid. When $\lambda\geq2$, note that $\lambda=\max\left\{  \lambda
,s,2\right\}  $ and we prove that if $t=t_{1}=\cdots=t_{m}$ are in $\left[
1,+\infty\right)$ and (\ref{q3}) is valid, then we have (\ref{dez11}) and, as a direct consequence, $t\geq\lambda$.
\end{remark}

\item We prove similar results for $m$-linear mappings with arbitrary codomains which assume their cotype. For a Banach space $X$, let $q_X=\inf\{q\geq 2;\ X$ has cotype $q\}.$

 The proof that $(B)$ implies $(A)$ in the theorem below appears in \cite[Proposition 4.3]{dsp}.
\begin{theorem}\label{000aaa}
Let $\mathbf{p\in}\left[  2,{+\infty}\right]  ^{m}$, let $X$ be an infinite
dimensional Banach space with cotype $q_{X}$,  $\left\vert \frac{1}{\mathbf{p}}\right\vert <\frac{1}{q_{X}}$, and let $\rho>0.$ The following
assertions are equivalent:

(A) Every  bounded $m$-linear operator $A:X_{p_{1}}\times\dots\times X_{p_{m}%
}\rightarrow X$ is such that%
\[
\sum\limits_{i_{1},...,i_{m}=1}^{{+\infty}}\left\| A(e_{i_{^{1}}}%
,...,e_{i_{m}})\right\| ^{\rho}<{+\infty}.
\]

(B) $\frac{1}{\rho}\leq\frac{1}{q_{X}}-\left\vert \frac{1}{\mathbf{p}%
}\right\vert .$
\end{theorem}

\end{enumerate}

Finally, in the last section of the paper we obtain better estimates for the constants of vector-valued Bohnenblust--Hille inequalities.

\smallskip

We conclude this introduction by noting that our theorems can be naturally stated in the context of homogeneous
polynomials. Given an $m$-homogeneous polynomial $P:X\to Y$, we denote its coefficients $(c_\alpha(P))$.
In \cite[Lemma 5]{df}, it is shown that an inequality
$$\left(\sum_{\alpha}\|c_\alpha(P)\|^\rho\right)^{\frac 1\rho}\leq C\|P\|$$
holds for every $m$-homogeneous polynomial $P:X\to Y$ if and only if a similar inequality
$$\left(\sum_{i_1,\dots,i_m}\|T(e_{i_1},\dots,e_{i_m})\|^\rho\right)^{\frac 1\rho}\leq C'\|T\|$$
is satisfied for every $m$-linear mapping $A:X\times\dots\times X\to Y$, where $X$ is a Banach sequence space.

\medskip

\noindent\textsc{Notations.} For two positive integers $n,k$, we set
 \begin{eqnarray*}
\mkn{k}&:=&\big\{\mathbf i=(i_1,\dots,i_k); \ i_1,\dots,i_k\in\{1,\dots,n\}\big\}.\\
\end{eqnarray*}
For $q\in [1,+\infty]$, $q^*$ will denote its conjugate exponent.

\section{Proof of Theorem \ref{main2} (sufficiency)\label{88}}

\bigskip Let $1\leq q\leq{+\infty}$. We recall that a Banach space $X$ has
\emph{cotype} $q$ if there is a constant $\kappa>0$ such that, no matter how
we select finitely many vectors $x_{1},...,x_{n}\in X$,
\[
\left(  \sum_{k=1}^{n}\Vert x_{k}\Vert^{q}\right)  ^{\frac{1}{q}}\leq
\kappa\left(  \int_{I}\left\Vert \sum_{k=1}^{n}r_{k}(t)x_{k}\right\Vert
^{2}dt\right)  ^{\frac{1}{2}}
\]
where $I=[0,1]$ and $r_{k}$ denotes the $k$-th Rademacher function. To cover
the case $q={+\infty}$, the left hand side should be replaced by $\max_{1\leq
k\leq n}\Vert x_{k}\Vert$. The smallest of all these constants is denoted by
$C_{q}(X)$ and named the cotype $q$ constant of $X$.

An operator between Banach spaces $v:X\to Y$ is $(r,s)$-summing (with $s\leq
r\leq{+\infty}$) if there exists $C>0$ such that, for all $n\geq1$ and for
all vectors $x_{1},\dots,x_{n}\in X$,
\[
\left(  \sum_{k=1}^{n} \|vx_{k}\|^{r}\right)  ^{\frac1r}\leq C\sup_{x^{*}\in
B_{X^{*}}}\left(  \sum_{k=1}^{n}|x^{*}(x_{k})|^{s}\right)  ^{\frac1s}.
\]
The smallest constant in this inequality is denoted by $\pi_{r,s}(v)$.

We need a cotype $q$ version of \cite[Proposition 4.1]{abps}, whose proof can
be found in \cite[Proposition 3.1]{dsp}:

\begin{lemma}
\label{cot.q} Let $X$ be a Banach space, let $Y$ be a cotype $q$ space, let $r\in[1,q]$ and let $\mathbf p\in [1,+\infty]^m$
with
\[
\left\vert {\frac{1}{\mathbf{p}}}\right\vert <\frac{1}{r}-\frac{1}{q}.
\]
Define
\[
\frac{1}{\lambda}:=\frac{1}{r}-\left\vert {\frac{1}{\mathbf{p}}}\right\vert .
\]
Then, for every continuous $m$-linear map $A:X_{p_{1}}\times\cdots\times
X_{p_{m}}\rightarrow X$ and every $\left(  r,1\right)  $-summing operator
$v:X\rightarrow Y$, we have
\begin{equation}
\left(  \sum_{i_{k}}\left(  \sum_{\widehat{i_{k}}}\Vert vA(e_{i_{1}}%
,\cdots,e_{i_{m}})\Vert_{Y}^{q}\right)  ^{\lambda/q}\right)  ^{1/\lambda}%
\leq\left(  \sqrt{2}C_{q}(Y)\right)  ^{m-1}\pi_{r,1}(v)\Vert A\Vert\label{q2}%
\end{equation}
for all $k=1,...,m.$
\end{lemma}

The symbol $\sum_{\widehat{i_{k}}}$ means that we are fixing the $k$-th index
and that we are summing over all the remaining indices.

We shall deduce from this lemma the following theorem, which extends results
of \cite{abps} and \cite{dsp}:

\begin{theorem}
\label{thmain} Let $\mathbf{p}\in\left[  1,+\infty\right]  ^{m}$, $X$ be a
Banach space, $Y$ be a cotype $q$ space and $1\leq r\leq q$, with $\left\vert
\frac{1}{\mathbf{p}}\right\vert <\frac{1}{r}$. Define
\[
\frac{1}{\lambda}:=\frac{1}{r}-\left\vert \frac{1}{\mathbf{p}}\right\vert .
\]
If $t_{1},\ldots,t_{m}\in\left[  \lambda,\max\left\{  \lambda,q\right\}
\right]  $ are such that%
\[
\frac{1}{t_{1}}+\cdots+\frac{1}{t_{m}}\leq\frac{1}{\lambda}+\frac{m-1}%
{\max\{\lambda,q\}},
\]
then, for every continuous $m$-linear map $A:X_{p_{1}}\times\cdots\times
X_{p_{m}}\rightarrow X$ and every $\left(  r,1\right)  $-summing operator
$v:X\rightarrow Y$, we have {\small
\begin{equation}
\left(  \sum_{i_{1}=1}^{{+\infty}}\left(  \ldots\left(  \sum_{i_{m}=1}^{{+\infty}
}\left\Vert vA\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert
_{Y}^{t_{m}}\right)  ^{\frac{t_{m-1}}{t_{m}}}\ldots\right)  ^{\frac{t_{1}%
}{t_{2}}}\right)  ^{\frac{1}{t_{1}}}\leq\left(  \sqrt{2}C_{\max\left\{
\lambda,q\right\}  }(Y)\right)  ^{m-1}\pi_{r,1}(v)\left\Vert A\right\Vert .
\label{q1}%
\end{equation}
}
\end{theorem}

\begin{proof}
If $\lambda<q$, from Lemma \ref{cot.q}, we have (\ref{q1}) for
\[
\left(  t_{1},...,t_{m}\right)  =\left(  \lambda,q,...,q\right)  .
\]
Since $\lambda<q$, the mixed $\left(  \ell_{\lambda},\ell_{q}\right)  -$ norm
inequality (see \cite[Proposition 3.1]{abps}), we also have (\ref{q1}) for the
exponents
\[
\left(  t_{1},...,t_{m}\right)  =\left(  q,...,q,\lambda,q,...,q\right)
\]
with $\lambda$ in the $k$-th position, for all $k=1,...,m$. Now, using a general version of  H\"older's inequality (see \cite[Theorem 2.1]{fou}), or
interpolating
\[
\left(  \lambda,q,...,q\right)  ,\left(  q,\lambda,q,...,q\right)
,\ldots,\left(  q,...,q,\lambda\right)
\]
in the sense of \cite{abps}, we get (\ref{q1}) for all $\left(  t_{1}%
,...,t_{m}\right)  \in[\lambda,q]^{m} $ such that
\[
\frac{1}{t_{1}}+\cdots+\frac{1}{t_{m}}=\frac{1}{\lambda}+\frac{m-1}{q}%
=\frac{1}{\lambda}+\frac{m-1}{\max\{\lambda,q\}}.
\]
If $\lambda\geq q$, for any $\varepsilon>0$, let $q_{\varepsilon}%
=\lambda+\varepsilon$. So $\lambda<q_{\varepsilon}$ and this automatically implies that
$$\left|\frac1{\mathbf p}\right|<\frac 1r-\frac1{q_{\varepsilon}}.$$
Since $Y$ has cotype $q_{\varepsilon}>q$, we may apply Lemma
\ref{cot.q} to get
\[
\left(  \sum_{i_{1}=1}^{N}\left(  \sum_{i_{2},...,i_{m}=1}^{N}\left\Vert
vA\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert ^{\lambda
+\varepsilon}\right)  ^{\frac{\lambda}{^{\lambda+\varepsilon}}}\right)
^{\frac{1}{\lambda}}\leq\left(  \sqrt{2}C_{\lambda+\varepsilon}(Y)\right)
^{m-1}\pi_{r,1}(v)\Vert A\Vert
\]
for all positive integer $N$. Making $\varepsilon\rightarrow0$, we get
\[
\left(  \sum_{i_{1},...,i_{m}=1}^{N}\left\Vert vA\left(  e_{i_{1}}%
,\ldots,e_{i_{m}}\right)  \right\Vert ^{\lambda}\right)  ^{\frac{1}{\lambda}%
}\leq\left(  \sqrt{2}C_{\lambda}(Y)\right)  ^{m-1}\pi_{r,1}(v)\Vert A\Vert,
\]
for all $N$ and the proof is done.
\end{proof}

\begin{remark}
If we take $t_{1}=\dots=t_{m}$, then, upon polarization, we recover exactly
\cite[Theorem 1.2]{dsp} with a much simpler proof due to the fact that the
inequality is simpler to prove for the extremal values of $(t_{1},\dots
,t_{m})$.
\end{remark}

\medskip

We are now ready for the proof of the sufficient part of Theorem \ref{main2}.
We split the proof into three cases, and we combine Theorem \ref{thmain} with
the Bennett-Carl inequalities (\cite{ben, carl}): for $1\leq s\leq q\leq
{+\infty}$, the inclusion map $\ell_{s}\hookrightarrow\ell_{q}$ is $\left(
r,1\right)  $-summing, where the optimal $r$ is given by
\[
\frac{1}{r}:=\frac{1}{2}+\frac{1}{s}-\frac{1}{\min\{2,q\}}.
\]

\noindent\emph{(i) $s\leq q\leq2$:}  The
Bennet-Carl-inequalities ensure that the inclusion map $\ell_{s}%
\hookrightarrow\ell_{q}$ is $\left(  r,1\right)  $-summing with $\frac{1}{r} =
\frac{1}{2} + \frac{1}{s} - \frac{1}{q}$, so the results follow from Theorem
\ref{thmain}, with $t_{1},\ldots,t_{m}$ satisfying
\[
\frac{1}{t_{1}}+\cdots+\frac{1}{t_{m}}=\frac{1}{2}+\frac{1}{s}-\frac{1}%
{q}-\left\vert \frac{1}{\mathbf{p}}\right\vert +\frac{m-1}{\max\{\lambda
,2\}}.
\]

\noindent\emph{(ii) $s\leq2\leq q$:} Also by using
Bennet-Carl inequalities, $\ell_{s}\hookrightarrow\ell_{2}$ is $\left(
s,1\right)  $-summing, thus we get (\ref{q3}) applying Theorem \ref{thmain},
with $t_{1},\ldots,t_{m}$ satisfying
\[
\frac{1}{t_{1}}+\cdots+\frac{1}{t_{m}}=\frac{1}{s}-\left\vert \frac
{1}{\mathbf{p}}\right\vert +\frac{m-1}{\max\{\lambda,2\}}.
\]

\medskip

\noindent\emph{(iii) $2\leq s\leq q$:}  Since
$\ell_{s}\hookrightarrow\ell_{s}$ is $\left(  s,1\right)  $-summing, the
result follows from Theorem \ref{thmain}, with $t_{1}=\dots=t_{m}=\lambda$ and
$\lambda\geq s$, since $r=s$ and
\[
\frac{1}{\lambda}:=\frac{1}{s}-\left\vert \frac{1}{\mathbf{p}}\right\vert
\leq\frac{1}{s}.
\]

\begin{remark}

Let us set
\[
c_{qs}:=\left\{
\begin{array}
[c]{ll}%
q\,, & \mbox{ if }s\leq q\leq2\,,\\
2\,, & \mbox{ if }s\leq2\leq q\,,\\
s\,, & \mbox{ if }2\leq s\leq q.
\end{array}
\right.
\]
\medskip

With the above notations, a careful look at the proof shows that the constant
$C$ which appears in Theorem \ref{main2} is dominated by
\[
\left(  \sqrt{2}C_{\max\left\{  \lambda,s,2\right\}  }(\ell_{c_{qs}})\right)
^{m-1}\pi_{r,1}(\ell_{s}\hookrightarrow\ell_{c_{qs}})
\]

\end{remark}

\section{Proof of Theorem \ref{main2} (optimality) \label{99}}
In this section we show that the exponents in Theorem \ref{main2} are optimal except when $q\leq 2$ and $\left|\frac1{\mathbf p}\right|>\frac 12$. More precisely, if $(t_1,\dots,t_m)\in [1,+\infty)^m$ are such that there exists $C>1$ satisfying, for any continuous multilinear map $X_{p_1}\times\dots\times X_{p_m}\to X_{s}$,
\begin{equation}
\left(  \sum_{i_{1}=1}^{{+\infty}}\left(  \ldots\left(  \sum_{i_{m}=1}^{{+\infty}
}\left\Vert A\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert
_{\ell_{q}}^{t_{m}}\right)  ^{\frac{t_{m-1}}{t_{m}}}\ldots\right)  ^{\frac{t_{1}%
}{t_{2}}}\right)  ^{\frac{1}{t_{1}}}\leq C \Vert A\Vert, \label{q4}%
\end{equation}
then we prove that (\ref{dez11}) holds.
When $\lambda\geq 2$, we will always assume that $t_1=\dots=t_m=t$, since $\lambda=\max\left\{  \lambda
,s,2\right\}  $ and our inequality holds true when all the exponents are equal. We split the proof into several cases. Most of the cases are a consequence of a random construction. The main tool is the following lemma,
from \cite[Lemma 6.2]{abps}.
\begin{lemma}\label{LEMPROBA}
Let $d,n\geq 1$, $q_1,\dots,q_{d+1}\in [1,+\infty]^{d+1}$ and let, for $q\geq 1$,
$$\alpha(q)=\left\{
\begin{array}{ll}
\frac12-\frac 1q&\textrm{if }q\geq 2\\
0&\textrm{otherwise.}
\end{array}\right.$$
Then there exists a $d$-linear mapping $A:\ell_{p_1}^n\times\dots\times\ell_{p_d}^n\to \ell_{p_{d+1}}^n$ which may be written
$$A\big(x^{(1)},\dots,x^{(d)}\big)=\sum_{i_1,\dots,i_{d+1}=1}^n \pm x_{i_1}^{(1)}\cdots x_{i_d}^{(d)}e_{i_{d+1}}$$
such that
$$\|A\|\leq C_d n^{\frac 12+\alpha(p_1)+\dots+\alpha(p_d)+\alpha(p_{d+1}^*)}.$$
\end{lemma}

\subsection{Case 1: $1\leq s\leq q\leq 2$ and $\lambda<2$}
This case has already been solved in \cite[Section 6.2]{abps}, using Lemma \ref{LEMPROBA} with $d=m$ and $(q_1,\dots,q_{m+1})=(p_1,\dots,p_m,s)$.

\subsection{Case 2: $1\leq s\leq q\leq 2$, $\lambda\geq 2$ and $\left|\frac1{\mathbf p}\right|\leq \frac 12$} This case has already been solved in \cite[Proposition 4.4(ib)]{dsp} using a Fourier matrix. We shall give an alternative probabilistic proof. Let $p\in [2,+\infty]$ be such that $\frac 1p=\left|\frac1{\mathbf p}\right|$. By Lemma \ref{LEMPROBA}, there exists a linear map $T:\ell_p^n\to\ell_s^n$ which may be written $T(x)=\sum_{i,j}\varepsilon_{i,j}x_ie_j$ with $\varepsilon_{i,j}=\pm 1$ and such that
$$\|T\|\leq Cn^{\frac 12+\frac 12-\frac 1p+\frac 12-\frac 1{s^*}}=Cn^{\frac 12+\frac 1s-\left|\frac1{\mathbf p}\right|}.$$
Let $A:\ell_{p_1}^n\times\cdots\times \ell_{p_m}^n\to\ell_{s}^n$ defined by
$$A\big(x^{(1)},\dots,x^{(m)}\big):=\sum_{i,j}\varepsilon_{i,j}x_i^{(1)}\cdots x_i^{(m)}e_j.$$
By H\"older's inequality, it is plain that $\|A\|\leq \|T\| \leq Cn^{\frac 12+\frac 1s-\left|\frac1{\mathbf p}\right|}$.
On the other hand, since $A(e_{i_{1}},\dots,e_{i_{m}})\neq 0$ if and only if $i_{1}=\ldots=i_{m}$,  and
$$\|A(e_i,\dots,e_i)\|_{\ell_{q}}=n^{1/q},$$
we have
$$\left(  \sum_{\mathbf i\in\mkn{m}}\left\Vert A\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert
_{\ell_{q}}^{t}\right)  ^{\frac{1}{t}}=n^{\frac 1q+\frac1{t}}.$$
This clearly implies
$$\frac 1t\leq \frac 12+\frac 1s-\frac 1q-\left|\frac1{\mathbf p}\right|.$$

\subsection{Case 3: $1\leq s\leq 2\leq q$ and $\lambda<2$} Let $p\in [0,+\infty]$ be defined by
$$\frac 1p=\frac 1{p_m}+\frac 1{s^*}.$$
Since $\lambda<2$, it is easy to check that $p\geq 2$ and that $p_i\geq 2$ for any $i=1,\dots,m$.
We then apply Lemma \ref{LEMPROBA} with $d=m-1$ and $(q_1,\dots,q_m)=(p_1,\dots,p_{m-1},p^*)$. We get an $(m-1)$-linear form $T:\ell_{p_1}^n\times\cdots\times \ell_{p_{m-1}}^n\to\ell_{p^*}^n$ which can be written
$$T\big(x^{(1)},\dots,x^{(m-1)}\big)=\sum_{i_1,\dots,i_m}\varepsilon_{i_1,\dots,i_m}x_{i_1}^{(1)}\cdots x_{i_{m-1}}^{(m-1)}e_{i_m}$$
and such that
$$\|T\|\leq C n^{\frac 12+\frac m2-\left|\frac1{\mathbf p}\right|-\frac 1{s^*}}=Cn^{\frac{m-1}2-\left|\frac1{\mathbf p}\right|+\frac 1s}.$$
We then define $A:\ell_{p_1}^n\times\cdots\times \ell_{p_m}^n\to\ell_{s}^n$ by
$$A\big(x^{(1)},\dots,x^{(m)}\big)=\sum_{i_1,\dots,i_m}\varepsilon_{i_1,\dots,i_m}x^{(1)}_{i_1}\cdots x_{i_m}^{(m)} e_{i_m}.$$
Then, for any $x^{(1)},\dots,x^{(m)}\in B_{\ell_{p_1}^n}\times\dots\times B_{\ell_{p_m}^n}$,
\begin{eqnarray*}
\|A\big(x^{(1)},\dots,x^{(m)}\big)\|&=&\sup_{y\in B_{\ell^n_{s^*}}}\left |\sum_{i_1,\dots,i_m}\varepsilon_{i_1,\dots,i_m}x_{i_1}^{(1)}\cdots x_{i_m}^{(m)}y_{i_m}\right|\\
&\leq&\sup_{z\in B_{\ell^n_{p}}}\left |\sum_{i_1,\dots,i_m}\varepsilon_{i_1,\dots,i_m}x_{i_1}^{(1)}\cdots x_{i_{m-1}}^{(m-1)}z_{i_m}\right|\\
&\leq&\|T\|.
\end{eqnarray*}

Moreover, given any $\mathbf i\in\mkn{m}$, $\|A(e_{i_1},\dots,e_{i_m})\|_q=\|e_{i_m}\|_q=1$, so that
$$\left(  \sum_{i_{1}=1}^{{+\infty}}\left(  \ldots\left(  \sum_{i_{m}=1}^{{+\infty}
}\left\Vert A\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  \right\Vert
_{\ell_{q}}^{t_{m}}\right)  ^{\frac{t_{m-1}}{t_{m}}}\ldots\right)  ^{\frac{t_{1}%
}{t_{2}}}\right)  ^{\frac{1}{t_{1}}}=n^{\frac {1}{t_1}+\dots+\frac{1}{t_m}}.$$
Hence, provided (\ref{q4}) is satisfied, $(t_1,\dots,t_m)$ has to satisfy
$$\frac 1{t_1}+\dots+\frac 1{t_m}\leq \frac{m-1}2+\frac 1s-\left|\frac1{\mathbf p}\right|.$$

\subsection{Case 4 and Case 5: $1\leq s\leq2\leq q$ and $\lambda\geq2$, $2\leq s\leq q$} These cases have a
deterministic proof, as noted in \cite[Proposition 4.4 (iib), (iii)]{dsp},
considering $A:\ell_{p_{1}}^n\times\cdots\times \ell_{p_{m}}^n \rightarrow \ell^n_{s}$
given by
\[
A \big(x^{(1)},\dots,x^{(m)}\big):=\sum_{i=1}^{n}x_{i}^{(1)}\cdots x_{i}^{(m)}e_{i}.
\]

\subsection{ The proof of Theorem \ref{nov1}}

From Theorem 1.3, by choosing $t_{1}=\ldots=t_{m}$ we conclude that provided
\[
\left\vert \frac{1}{\mathbf{p}}\right\vert <\frac{1}{2}+\frac{1}{s}-\frac
{1}{\min\{q,2\}},
\]
the best exponent $\rho$ in Theorem \ref{nov1} satisfies
\[
\frac{m}{\rho}=\frac{1}{\lambda}+\frac{m-1}{\max\{\lambda,s,2{\}}}.
\]
To conclude the proof, it remains to prove that, whenever
\[
\left\vert \frac{1}{\mathbf{p}}\right\vert \geq\frac{1}{2}+\frac{1}{s}%
-\frac{1}{\min\{q,2\}},
\]
we cannot find an exponent $\rho>0$ such that (\ref{nov1}) is satisfied for all
$m$-linear operators $A:X_{p_{1}}\times\cdots\times X_{p_{m}}\rightarrow
X_{s}$. In fact, everything has already been done before: if $q\leq 2$, then we have just to follow the lines of Case 2 and if $q\geq 2$, then we may consider the $m$-linear mapping of Cases 4 and 5.

\section{The case $1\leq s\leq q\leq  2$, $\lambda\geq 2$ and $\left|\frac1{\mathbf p}\right|>\frac12$}

\subsection{A reformulation of the Hardy-Littlewood type inequalities}

We shall improve in this section the bound given by Theorem \ref{nov1}. We shall proceed by interpolation.
To do this, we need a reformulation of the result of this theorem, as Villanueva and Perez-Garcia reformulated the Bohnenblust-Hille inequality in \cite{PV2}. The proof is a combination of  \cite[Corollary 3.20]{PV2} and Proposition 2.2 and will be omitted.

\begin{theorem}\label{THMREFORMULATION}
Let  $1\leq p_{1},...,p_{m}\leq+\infty$, $1\leq s\leq q\leq\infty$ and let
$\rho>0$. The following assertions are equivalent.
\begin{itemize}
\item[(A)] There exists $C>0$ such that, for every continuous $m$-linear mapping $A:X_{p_1}\times\cdots\times X_{p_m}\to X_s$, we have
$$\left(\sum_{i_1,\dots,i_m} \left\|A(e_{i_1},\dots,e_{i_m})\right\|_{\ell_q}^{\rho}\right)^{1/\rho}\leq C\|A\|.$$
\item[(B)] There exists $C>0$ such that, for any $n\geq 1$, for any Banach spaces $Y_1,\dots,Y_m$, for any
continuous $m$-linear mapping $S:Y_1\times\cdots\times Y_m\to X_s$, the induced operator
\begin{eqnarray*}
T:\ell_{p_1^*,w}^n(Y_1)\times\cdots\times \ell_{p_m^*,w}^n(Y_m)&\to&\ell_{\rho}^{n^m}(X_q)\\
\big(x^{(1)},\dots,x^{(m)}\big)&\mapsto&\big(S(x_{i_1}^{(1)},\dots,x_{i_m}^{(m)})\big)_{\mathbf i\in\mkn{m}}
\end{eqnarray*}
satisfies $\|T\|\leq C\|S\|$.
\end{itemize}
\end{theorem}
We recall that, for any $p\in[1,+\infty]$ and any Banach space $Y$,
$$\ell_{p,w}^n(Y)=\left\{(x_j)_{j=1}^n\subset Y;\ \|(x_j)\|_{w,p}:=\sup_{\varphi\in B_{Y^*}}\left(\sum_{j=1}^n |\varphi(x_j)|^p\right)^{1/p}<+\infty\right\}$$
with the appropriate modifications for $p=\infty$.

\subsection{Proof of the sufficient condition}
We now prove our better upper bound in the case $1\leq s\leq q\leq 2$, $\left|\frac1{\mathbf p}\right|>\frac 12$ (namely we prove the first part of Theorem \ref{nov2}). Let $n\geq 1$, let $Y_1,\dots,Y_m$ be Banach spaces and let $S:Y_1\times\dots\times Y_m\to X_s$ be bounded. Let $n\geq 1$ and let $T$ be the operator induced by $S$ on
$\mathcal Y=\ell_{p_1^*,w}^n(Y_1)\times\cdots\times \ell_{p_m^*,w}(Y_m)$, defined by
$$T\big(x^{(1)},\dots,x^{(m)}\big)=\big(S(x_{i_1}^{(1)},\dots,x_{i_m}^{(m)}\big)).$$
Then $T$ is bounded as an operator from $\mathcal Y$ into $\ell_{\infty}^{n^m}(X_s)$ (this is trivial). $T$ is also bounded as an operator from $\mathcal Y$ into $\ell_\rho^{n^m}(X_s)$ with $\frac 1\rho=\frac 1s-\left|\frac1{\mathbf p}\right|$
(this is Theorem \ref{nov1} for $1\leq s\leq 2$ and $q\geq 2$). We can interpolate between these two extreme situations. Hence, let $q\in [s,2]$ and let $\theta\in [0,1]$ be such that
$$\frac 1q=\frac{1-\theta}s+\frac \theta2\iff \theta=\frac{\frac 1s-\frac 1q}{\frac 1s-\frac 12}.$$
By \cite[Theorem 4.4.1]{berg.lofst}, $T$ is bounded as an operator from $\mathcal Y$ into $\ell_t^{n^m}(X_q)$
where
$$\frac 1t=\frac{1-\theta}{\infty}+\frac{\theta}{\rho}=\frac{\left(\frac 1s-\frac 1q\right)\left(\frac 1s-\left|\frac1{\mathbf p}\right|\right)}{\frac 1s-\frac 12}.$$
\begin{remark}
It is easy to check that, for $1\leq s\leq q\leq 2$ and $\left|\frac1{\mathbf p}\right|\geq\frac 12$, then the bound $\frac{\left(\frac 1s-\frac 1q\right)\left(\frac 1s-\left|\frac1{\mathbf p}\right|\right)}{\frac 1s-\frac 12}$ is always better (namely larger) than the bound
$\frac 12+\frac 1s-\frac 1q-\left|\frac1{\mathbf p}\right|$ obtained in Theorem \ref{nov1}.
\end{remark}

\subsection{The necessary condition}
We now prove the second part of Theorem \ref{nov2}. It also uses a probabilistic device  for linear maps when the two spaces do not need to have the same dimension. The forthcoming lemma can be found in \cite[Proposition 3.2]{ben}.
\begin{lemma}\label{LEMBENNETT}
Let $n,d\geq 1$, $1\leq p,s\leq 2$. There exists $T:\ell_p^d\to\ell_s^n$, $T(x)=\sum_{i,j}\pm x_je_i$ such that
$$\|T\|\leq C_{p,s}\max \left(d^{1/s},n^{1-\frac 1p}d^{\frac 1s-\frac 12}\right).$$
\end{lemma}
Coming back to the proof of Theorem \ref{nov2}, we first observe that we may always assume that $\left|\frac1{\mathbf p}\right|<1$. Otherwise, we can consider
the $m$-linear map $A:X_{p_1}\times\cdots\times X_{p_m}\to X_s$ defined by
\[
A\big(x^{(1)},\dots,x^{(m)}\big)=\sum_{i\geq 1}x_i^{(1)}\dots x_i^{(m)}e_0
\]
and observe that it is bounded whereas it has infinitely many coefficients equal to 1. We then define $p\in [1,2]$ by $\frac 1p=\left|\frac1{\mathbf p}\right|$ and we consider $T:\ell_p^d\to\ell_s^n$, $T(x)=\sum_{i,j}\varepsilon_{i,j}x_je_i$ the map given by Lemma \ref{LEMBENNETT}.
We then define
\begin{eqnarray*}
A:\ell_{p_1}^d\times\cdots\times\ell_{p_m}^d&\to&\ell_s^n\\
\big(x^{(1)},\dots,x^{(m)}\big)&\mapsto&\sum_{i,j}\varepsilon_{i,j}x_j^{(1)}\cdots x_j^{(m)}e_i
\end{eqnarray*}
and we observe that, by H\"older's inequality, $\|A\|\leq \|T\|$. Furthermore,
$$\left(\sum_{i_1,\dots,i_m}\|A(e_{i_1},\dots,e_{i_m})\|_{\ell_q}^t\right)^{1/t}=n^{1/t}d^{1/q}.$$
Taking $d^{1/2}=n^{1-\frac 1p}$ (this is the optimal relation between $d$ and $n$), we get that if
$$\left(\sum_{i_1,\dots,i_m}\|A(e_{i_1},\dots,e_{i_m})\|_{\ell_q}^t\right)^{1/t}\leq C\|A\|,$$
then it is necessary that
$$\frac 1t\leq 2\left(1-\frac 1p\right)\left(\frac 1s-\frac 1q\right).$$
\begin{remark}
This last condition is optimal when $s=1$ or when $\left|\frac1{\mathbf p}\right|=\frac 12$ (with, in fact, the same proof as in Case 2 above). When $1<s<2$, another necessary condition is
$$\frac 1t\leq\frac 1s-\left|\frac1{\mathbf p}\right|$$
(see Case 4 or Case 5 above).
\end{remark}

\section{Optimal estimates under cotype assumptions}\label{s8}

For a Banach space $X$, let $q_X:=\inf\{q\geq 2;\ X \textrm{ has cotype } q\}$. For scalar-valued multilinear operators it is easy to observe that summability
in multiple indexes behaves in a quite different way than summability in just one
index. For instance, for any bounded bilinear form $A:c_{0}\times
c_{0}\rightarrow\mathbb{C}$,
\[
\left(  \sum_{i,j=1}^{{+\infty}}|A(e_{i},e_{j})|^{\frac{4}{3}}\right)  ^{\frac
{3}{4}}\leq\sqrt{2}\Vert A\Vert
\]
and the exponent $4/3$ is optimal. But, if we sum diagonally $\left(
i=j\right)  $ the exponent $4/3$ can be reduced to $1$ since%
\[
\sum_{i=1}^{{+\infty}}|A(e_{i},e_{i})|\leq\Vert A\Vert
\]
for any bounded bilinear form $A:c_{0}\times c_{0}\rightarrow\mathbb{C}$. Now we prove Theorem \ref{000aaa} which shows  that when replacing the scalar field by infinite-dimensional
spaces the situation is quite different.

\begin{proof}

$(A)\Rightarrow(B).$ From a deep result of Maurey and Pisier (\cite{pisier} and \cite[Section 14]{djt}), $\ell_{q_{X}}$ is finitely
representable in $X$, which means that, for any $n\geq1$, one may find unit
vectors $z_{1},\dots,z_{n}\in X$ such that, for any $a_{1},\dots,a_{n}%
\in\mathbb{C}$,
\[
\sum_{i=1}^{n}\Vert a_{i}z_{i}\Vert_{X}\leq2\left(  \sum_{i=1}^{n}%
|a_{i}|^{q_{X}}\right)  ^{1/q_{X}}.
\]
We then consider the $m$-linear map $A:\ell_{p_{1}}^{n}\times\cdots
\times\ell_{p_{m}}^{n}\rightarrow X$ defined by
\[
A\left(  x^{(1)},\cdots,x^{(m)}\right)  :=\sum_{i=1}^{n}x_{i}^{(1)}\cdots
x_{i}^{(m)}z_{i}.
\]
Then, for any $(x^{(1)},\dots,x^{(m)})$ belonging to $B_{\ell_{p_{1}}^{n}%
}\times\dots\times B_{\ell_{p_{m}}^{n}}$,
\begin{align*}
\left\Vert A\left(  x^{(1)},\cdots,x^{(m)}\right)  \right\Vert  & \leq2\left(
\sum_{i=1}^{n}|x_{i}^{(1)}|^{q_{X}}\cdots|x_{i}^{(m)}|^{q_{X}}\right)
^{1/{q_{X}}}\\
& \leq2n^{\frac{1}{q_{X}}-\left\vert \frac{1}{\mathbf{p}}\right\vert }%
\end{align*}
where the last inequality follows from H\"{o}lder's inequality applied to the exponents $$\frac{p_{1}}{q_{X}},...,\frac{p_{m}}{q_{X}},\left(
1-q_{X}\left\vert \frac{1}{\mathbf{p}}\right\vert \right)  ^{-1}.$$

 On the other hand,
\[
\left(  \sum_{i=1}^{n}\Vert A\left(  e_{i},\dots,e_{i}\right)  \Vert^{\rho
}\right)  ^{1/\rho}=n^{\frac{1}{\rho}}%
\]
and we obtain $\left(  3\right)  $.

$(B)\Rightarrow(A).$ This implication is proved in \cite[Proposition 4.3]{dsp}.
\end{proof}

If $X$ does not have cotype $q_X$, the condition remains necessary. But now we just have the following sufficient condition:
\[
\frac{m}{\rho}<\frac{1}{q_X}-\left|\frac{1}{\mathbf p}\right|.
\]

Of course, it would be nice to determine what happens in this case. A look at \cite[page 304]{djt} shows that the situation does not look simple.

\bigskip As a consequence of the previous result we conclude that under certain circumstances the
concepts of absolutely summing multilinear operator and multiple summing
multilinear operator (see \cite{ir, matos, PV}) are precisely the same.

\begin{corollary}
\bigskip Let $p\in[  2,{+\infty}]  $, let $X$ be an infinite
dimensional Banach space with cotype $q_{X}<\frac{p}{m}$ and let $\rho>0$. The following
assertions are equivalent:

(A) Every bounded $m$-linear operator $A:X_{p}\times\dots\times X_{p}%
\rightarrow X$ is absolutely $\left(  \rho;p^{\ast}\right)  $-summing.

(B)  Every bounded $m$-linear operator $A:X_{p}\times\dots\times
X_{p}\rightarrow X$ is  multiple $\left(  \rho;p^{\ast}\right)  $-summing.

(C) $\frac{1}{\rho}\leq\frac{1}{q_{X}}- \frac{m}{p}.$
\end{corollary}

\bigskip We stress the equivalence between $(A)$ and $(B)$ is not true, in general. For instance, every bounded bilinear
operator $A:\ell_{2}\times\ell_{2}\rightarrow\ell_{2}$ is absolutely $\left(
1;1\right)  $-summing but this is no longer true for multiple summability.

\bigskip

\section{Constants of vector-valued Bohnenblust--Hille inequalities}

A particular case of our main result is the following vector-valued
Bohnenblust--Hille inequality (see \cite[Lemma 3]{df} and also \cite[Section 2.2]{ursula}):

\begin{theorem}
\label{61} Let $X$ be a Banach space, $Y$ a cotype $q$ space and
$v:X\rightarrow Y$ an $(r,1)$-summing operator with $1\leq r\leq q$. Then, for
all $m$-linear operators $T:c_{0}\times\cdots\times c_{0}\rightarrow X$,
\[
\left(  \sum_{i_{1},...,i_{m}=1}^{+\infty}\Vert vT\left(  e_{i_{1}}%
,\ldots,e_{i_{m}}\right)  \Vert_{Y}^{\frac{qrm}{q+(m-1)r}}\right)
^{\frac{q+\left(  m-1\right)  r}{qrm}}\leq C_{Y,m}\pi_{r,1}(v)\Vert T\Vert
\]
with $C_{Y,m}=\left(  \sqrt{2}C_{q}\left(  Y\right)  \right)  ^{m-1}.$
\end{theorem}

In this section, in Theorem \ref{ttrr}, we improve the above estimate for $C_{Y,m}$. The proof of Theorem \ref{ttrr} follows almost word by word the proof of \cite[Proposition 3.1]{bps} using \cite[Lemma 2.2]{defant33} and Kahane's inequality instead of the Khinchine inequality. We present the proof for the sake of completeness. We need the following inequality due to Kahane:

\begin{kahane}
Let $0<p,q<+\infty$. Then there is a constant $K_{p,q}>0$ for which
\[
\left(  \int_{I}\Vert\sum_{k=1}^{n}r_{k}(t)x_{k}\,dt\Vert^{q}\right)
^{\frac{1}{q}}\leq K_{p,q}\left(  \int_{I}\Vert\sum_{k=1}^{n}r_{k}%
(t)x_{k}\,dt\Vert^{p}\right)  ^{\frac{1}{p}},
\]
regardless of the choice of a Banach space $X$ and of finitely many vectors
$x_{1},...,x_{n}\in X$.
\end{kahane}

\bigskip

\begin{theorem}\label{ttrr}
\label{62}For all $m$ and all $1\leq k<m,$%
\[
C_{Y,m}\leq\left(  C_{q}(Y)K_{\frac{qrk}{q+(k-1)r},2}\right)  ^{m-k}C_{Y,k}.
\]

\end{theorem}

\begin{proof}
Let $\rho:=\frac{qrm}{q+(m-1)r}$ and to simplify notation let us write%
\[
vTe_{\mathbf{i}}=vT\left(  e_{i_{1}},\ldots,e_{i_{m}}\right)  .
\]
Let us make use of \cite[Remark 2.2]{bps} with $m\geq2,\ 1\leq k\leq m-1$ and
$s=\frac{qrk}{q+(k-1)r}$. So we have
\begin{equation}
\left(  \sum_{\mathbf{i}}\Vert vTe_{\mathbf{i}}\Vert_{Y}^{\rho}\right)
^{\frac{1}{\rho}}\leq\prod_{S\in P_{k}(m)}\left(  \sum_{\mathbf{i}_{S}}\left(
\sum_{\mathbf{i}_{\hat{S}}}\Vert vT\left(  e_{\mathbf{i}_{S}},e_{\mathbf{i}%
_{\hat{S}}}\right)  \Vert_{Y}^{q}\right)  ^{\frac{s}{q}}\right)  ^{\frac
{1}{s\binom{m}{k}}},\label{55f}%
\end{equation}
where $P_{k}(m)$ denotes the set of all subsets of $\{1,...,m\}$ with
cardinality $k.$ For sake of clarity, we shall assume that $S=\{1,\dots,k\}$.
By the multilinear cotype inequality (see \cite[Lemma 2.2]{defant33}) and the
Kahane inequality, we have {\small
\begin{align*}
&  \left(  \sum_{\mathbf{i}_{\hat{S}}}\Vert vT\left(  e_{\mathbf{i}_{S}%
},e_{\mathbf{i}_{\hat{S}}}\right)  \Vert_{Y}^{q}\right)  ^{\frac{s}{q}}\\
&  \leq\left(  C_{q}(Y)K_{s,2}\right)  ^{s(m-k)}\int_{I^{m-k}}\left\Vert
\sum_{\mathbf{i}_{\hat{S}}}r_{\mathbf{i}_{\hat{S}}}(t_{\hat{S}})vT\left(
e_{\mathbf{i}_{S}},e_{\mathbf{i}_{\hat{S}}}\right)  \right\Vert _{Y}%
^{s}\,dt_{\hat{S}}\\
&  =\left(  C_{q}(Y)K_{s,2}\right)  ^{s(m-k)}\int_{I^{m-k}}\left\Vert
vT\left(  e_{\mathbf{i}_{S}},\sum_{\mathbf{i}_{\hat{S}}}r_{\mathbf{i}_{\hat
{S}}}(t_{\hat{S}})e_{\mathbf{i}_{\hat{S}}}\right)  \right\Vert _{Y}%
^{s}\,dt_{\hat{S}}\\
&  =\left(  C_{q}(Y)K_{s,2}\right)  ^{s(m-k)}\int_{I^{m-k}}\left\Vert v\left(
T\left(  e_{i_{1}},\dots,e_{i_{k}},\sum_{i_{k+1}}r_{k+1}(t_{k+1})e_{k+1}%
,\dots,\sum_{i_{m}}r_{m}(t_{m})e_{m}\right)  \right)  \right\Vert _{Y}%
^{s}\,dt_{k+1}\dots dt_{m}\\
&
\end{align*}
} But for a fixed choice of $\left(  t_{k+1},\dots,t_{m}\right)  \in
I^{m-k}=[0,1]^{m-k}$, we know, by Theorem \ref{61}, that
\[
\sum_{i_{1},\dots,i_{k}}\left\Vert v\left(  T\left(  e_{i_{1}},\dots,e_{i_{k}%
},\sum_{i_{k+1}}r_{k+1}(t_{k+1})e_{k+1},\dots,\sum_{i_{m}}r_{m}(t_{m}%
)e_{m}\right)  \right)  \right\Vert _{Y}^{s}\newline\leq\left(  C_{Y,k}%
\pi_{r,1}(v)\Vert T\Vert\right)  ^{s}.
\]
Thus,
\begin{align}
&  \sum_{\mathbf{i}_{S}}\left(  \sum_{\mathbf{i}_{\hat{S}}}\Vert vT\left(
e_{\mathbf{i}_{S}},e_{\mathbf{i}_{\hat{S}}}\right)  \Vert_{Y}^{q}\right)
^{\frac{s}{q}}\nonumber\\
&  \leq\left(  C_{q}(Y)K_{s,2}\right)  ^{s(m-k)}\cdot\sum_{i_{1},\dots,i_{k}%
}\left\Vert v\left(  T\left(  e_{i_{1}},\dots,e_{i_{k}},\sum_{i_{k+1}}%
r_{k+1}(t_{k+1})e_{k+1},\dots,\sum_{i_{m}}r_{m}(t_{m})e_{m}\right)  \right)
\right\Vert _{Y}^{s}\nonumber\\
&  \leq\left(  \left(  C_{q}(Y)K_{s,2}\right)  ^{m-k}\pi_{r,1}(v)C_{Y,k}\Vert
T\Vert\right)  ^{s},\label{estrela}%
\end{align}
namely
\[
\left(  \sum_{\mathbf{i}_{S}}\left(  \sum_{\mathbf{i}_{\hat{S}}}\Vert
vT\left(  e_{\mathbf{i}_{S}},e_{\mathbf{i}_{\hat{S}}}\right)  \Vert_{Y}%
^{q}\right)  ^{\frac{s}{q}}\right)  ^{\frac{1}{s}}\leq\left(  C_{q}%
(Y)K_{s,2}\right)  ^{m-k}\pi_{r,1}(v)C_{Y,k}\Vert T\Vert.
\]
From (\ref{55f}) we conclude that
\[
\left(  \sum_{\mathbf{i}}\Vert vTe_{\mathbf{i}}\Vert_{Y}^{\rho}\right)
^{\frac{1}{\rho}}\leq\left(  C_{q}(Y)K_{s,2}\right)  ^{m-k}C_{Y,k}\pi
_{r,1}(v)\Vert T\Vert.
\]

\end{proof}

\bigskip When $m$ is even, the case $k=\frac{m}{2}$ recovers the constants
from \cite{ursula}.

\begin{corollary}
For all $m,$%
\[
C_{Y,m}\leq C_{q}(Y)^{m-1}%
{\textstyle\prod\limits_{k=1}^{m-1}}
K_{\frac{qrk}{q+(k-1)r},2}.
\]

\end{corollary}

\section{Other exponents}

From now on $1\leq r\leq q$ and $\left(  q_{1},\ldots,q_{m}\right)  \in\lbrack
r,q]^{m}$ so that
\[
\frac{1}{q_{1}}+\cdots+\frac{1}{q_{m}}=\frac{q+(m-1)r}{qr}=\frac{1}{r}%
+\frac{m-1}{q}%
\]
are called vector-valued Bohnenblust--Hille exponents. From Theorem
\ref{thmain} we have:

\begin{theorem}
[Multiple exponent vector-valued Bohnenblust--Hille inequality]Let $X$ be a
Banach space and $Y$ a cotype $q$ space with $1\leq r\leq q$. If $\left(
q_{1},\ldots,q_{m}\right)  \in\lbrack r,q]^{m}$ are vector-valued
Bohnenblust--Hille exponents, then there exists $C_{Y,q_{1},\dots,q_{m}}\geq1$
such that, for all $m$-linear operators $T:c_{0}\times\cdots\times
c_{0}\rightarrow X$ and every $(r,1)$-summing operator $v:X\rightarrow Y$, we
have
\begin{equation}
\left(  \sum_{i_{1}=1}^{+\infty}\dots\left(  \sum_{i_{m}=1}^{+\infty}\Vert
vTe_{\mathbf{i}}\Vert_{Y}^{q_{m}}\right)  ^{\frac{q_{m-1}}{q_{m}}}%
\dots\right)  ^{\frac{1}{q_{1}}}\leq C_{Y,q_{1},\ldots,q_{m}}\pi_{r,1}(v)\Vert
T\Vert,\label{eq}%
\end{equation}
with $C_{Y,q_{1},\ldots,q_{m}}=\left(  \sqrt{2}C_{q}\left(  Y\right)  \right)
^{m-1}.$
\end{theorem}

Our final result gives better estimates for the constants $C_{Y,q_{1}%
,\ldots,q_{m}}:$

\begin{theorem}
If $\left(  q_{1},\dots,q_{m}\right)  $ is a vector-valued Bohnenblust--Hille
exponent, then%
\[
C_{Y,q_{1}\dots,q_{m}}\leq\prod_{k=1}^{m}\left(  \left(  C_{q}(Y)K_{\frac
{kqr}{q+(k-1)r},2}\right)  ^{m-k}C_{Y,k}\right)  ^{\theta_{k}}%
\]
with
\begin{equation}
\theta_{m}=m\left(  \frac{1}{r}-\frac{1}{q}\right)  ^{-1}\left(  \frac
{1}{q_{m}}-\frac{1}{q}\right)  \label{222}%
\end{equation}
and
\begin{equation}
\theta_{k}=k\left(  \frac{1}{r}-\frac{1}{q}\right)  ^{-1}\left(  \frac
{1}{q_{k}}-\frac{1}{q_{k+1}}\right)  ,\ \text{ for }k=1,\dots,m-1. \label{333}%
\end{equation}
\bigskip
\end{theorem}

\begin{proof}
It suffices to consider $q_{i}\leq q_{j}$ whenever $i<j$. For each
$k=1,\dots,m$, define
\[
s_{k}=\frac{kqr}{q+(k-1)r}.
\]
From the proof of Theorem \ref{62} we have (\ref{eq}) for each exponent
$\left(  s_{k},\overset{k\text{ times}}{\dots},s_{k},q\dots,q\right)  .$ More
precisely, from (\ref{estrela}) we have
\[
\left(  \sum_{i_{1},\dots,i_{k}}\left(  \sum_{i_{k+1},\dots,i_{m}}\Vert
vTe_{\mathbf{i}}\Vert_{Y}^{q}\right)  ^{\frac{s_{k}}{q}}\right)  ^{\frac
{1}{s_{k}}}\leq\left(  C_{q}(Y)K_{s_{k},2}\right)  ^{m-k}C_{Y,k}\pi
_{r,1}(v)\Vert T\Vert.
\]
Consequently, for each $k=1,\dots,m$ we have
\[
C_{Y,s_{k},\overset{k\text{ times}}{\dots},s_{k},q\dots,q}\leq\left(
C_{q}(Y)K_{s_{k},2}\right)  ^{m-k}C_{Y,k}.
\]
Since every vector-valued Bohnenblust--Hille exponent $\left(  q_{1}%
,\ldots,q_{m}\right)  $ with $q_{1}\leq\cdots\leq q_{m}$ is obtained by
interpolation of $\alpha_{1},...,\alpha_{m}$ with $\alpha_{k}=\left(
s_{k},\overset{k\text{ times}}{\dots},s_{k},q\dots,q\right)  $, and
$\theta_{1},...,\theta_{m}$ as in (\ref{222}) and (\ref{333}), we conclude
that
\[
C_{Y,q_{1},\dots,q_{m}}\leq\prod_{k=1}^{m}\left(  C_{Y,s_{k},\overset{k\text{
times}}{\dots},s_{k},q,\dots,q}\right)  ^{\theta_{k}}\leq\prod_{k=1}%
^{m}\left(  \left(  C_{q}(Y)K_{s_{k},2}\right)  ^{m-k}C_{Y,k}\right)
^{\theta_{k}}%
\]

\end{proof}

A particular case of Kahane's inequality is Khintchine's inequality: if
$(\varepsilon_{i})$ is a sequence of independent Rademacher variables, then,
for any $p\in\lbrack1,2]$, there exists a constant $\mathrm{A}_{\mathbb{R},p}$
such that, for any $n\geq1$ and any $a_{1}\dots,a_{n}\in\mathbb{R}$,
\[
\left(  \sum_{i=1}^{n}|a_{i}|^{2}\right)  ^{\frac{1}{2}}\leq\mathrm{A}%
_{\mathbb{R},p}\left(  \int_{\Omega}\left\vert \sum_{i=1}^{n}a_{i}%
\varepsilon(\omega)\right\vert ^{p}d\omega\right)  ^{\frac{1}{p}}.
\]
It has a complex counterpart: for any $p\in\lbrack1,2]$, there exists a
constant $\mathrm{A}_{\mathbb{C},p}$ such that, for any $n\geq1$ and any
$a_{1}\dots,a_{n}\in\mathbb{C}$,
\[
\left(  \sum_{i=1}^{n}|a_{i}|^{2}\right)  ^{\frac{1}{2}}\leq\mathrm{A}%
_{\mathbb{C},p}\left(  \int_{\mathbb{T}^{n}}\left\vert \sum_{i=1}^{n}%
a_{i}z_{i}\right\vert ^{p}dz\right)  ^{\frac{1}{p}}.
\]
The best constants $\mathrm{A}_{\mathbb{R},p}$ and $\mathrm{A}_{\mathbb{C},p}$
are known (see \cite{haagerup} and \cite{kk}):

\begin{itemize}
\item $\mathrm{A}_{\mathbb{R},p} =
\begin{cases}
2^{\frac{1}{p}-\frac{1}{2}}, \text{ if } 0<p \leq p_{0} \approx1.847\\
\frac{1}{\sqrt{2}} \left(  \frac{\Gamma\left(  \frac{1+p}{2}\right)  }%
{\sqrt{\pi}} \right)  ^{-\frac{1}{p}}, \text{ if } p > p_{0} ;
\end{cases}
$

\item $\mathrm{A}_{\mathbb{C},p} = \Gamma\left(  \frac{1+p}{2}\right)
^{-\frac{1}{p}}$, if $1 <p \leq2$.
\end{itemize}

\bigskip

Taking $X=Y=\mathbb{K}$ and $r=1$ we obtain estimates for the constants of the
scalar-valued Bohnenblust--Hille inequality with multiple exponents:

\begin{corollary}
If $\left(  q_{1},\ldots,q_{m}\right)  \in\lbrack1,2]^{m}$ are so that%
\[
\frac{1}{q_{1}}+\cdots+\frac{1}{q_{m}}=\frac{m+1}{2},
\]
then%
\[
\left(  \sum_{i_{1}=1}^{+\infty}\dots\left(  \sum_{i_{m}=1}^{+\infty}\left\vert
T\left(  e_{i_{1}},...,e_{i_{m}}\right)  \right\vert ^{q_{m}}\right)
^{\frac{q_{m-1}}{q_{m}}}\dots\right)  ^{\frac{1}{q_{1}}}\leq C_{\mathbb{K}%
,m}^{2m\left(  \frac{1}{q_{m}}-\frac{1}{2}\right)  }\left(  \prod_{k=1}%
^{m-1}\left(  \mathrm{A}_{\mathbb{K},\frac{2k}{k+1}}^{m-k}C_{\mathbb{K}%
,k}\right)  ^{2k\left(  \frac{1}{q_{k}}-\frac{1}{q_{k+1}}\right)  }\right)
\Vert T\Vert
\]
for all $m$-linear operators $T:c_{0}\times\cdots\times c_{0}\rightarrow
\mathbb{K}.$ In particular, for complex scalars, the left hand side of the
above inequality can be replaced by
\[
\left(
{\displaystyle\prod\limits_{j=1}^{m}}
\Gamma\left(  2-\frac{1}{j}\right)  ^{\frac{j}{2-2j}}\right)  ^{2m\left(
\frac{1}{q_{m}}-\frac{1}{2}\right)  }\left(  \prod_{k=1}^{m-1}\left(
\Gamma\left(  \frac{3k+1}{2k+2}\right)  ^{\left(  \frac{-k-1}{2k}\right)
\left(  m-k\right)  }%
{\displaystyle\prod\limits_{j=1}^{k}}
\Gamma\left(  2-\frac{1}{j}\right)  ^{\frac{j}{2-2j}}\right)  ^{2k\left(
\frac{1}{q_{k}}-\frac{1}{q_{k+1}}\right)  }\right)  \Vert T\Vert.
\]

\end{corollary}

\bigskip

Acknowledgement. The authors thank the referee for important comments and suggestions.

redpajama_set_name: RedPajamaArXiv


text:
Home News Local News Six arrested in home invasion and shooting
Six arrested in home invasion and shooting
Belleview, Florida — On January 19, at approximately 9:45 p.m., Marion County Sheriff's Office deputies, along with Belleview Police Department officers, were dispatched to 5469 Southeast 104th Place, Belleview, in reference to a home invasion and shooting.
When officers arrived on scene, they found victims Dominique J. Johnson, 19, Emanuel Estada-Portalatin, 17 and Christopher Hillard, who were inside the house at the time the home invasion.
According to repots, Estada-Portalatin was shot during the home invasion and transported to a local hospital to be treated for his injuries.
Johnson told MCSO deputies that two white males had broken into the residence and that one of them hit Hillard in the back of the head with a metal bar. The other suspect pointed a gun at Johnson and shot Estada-Portalatin in the arm.
When deputies attempted a traffic stop on a silver Navigator, which was reported leaving the scene of the shooting, the driver fled at a high rate of speed and later crashed into a ditch.
The occupants of the vehicle, the driver, Cedric Mee, 23, Gregory Nelson, 24, Rebecca 18, and Jaclyn Ericson, 22, were arrested at the scene of the crash.
During questioning, Mee stated that prior to the home invasion shooting, he, Nelson, Polley, and Ericson met at Ericson's house. They discussed Mee's plan to retaliate against Estada-Portalatin and Johnson for an armed home invasion robbery they committed on Ericson and another man at the Vacation Host Inn on November 1, 2015.
According to a press release, Mee stated that all four drove to the incident location where he committed the armed home invasion with Nelson and shot Estada-Portalatin.
Victims Estada-Portalatin and Johnson were also arrested and charged with the armed home invasion robbery they committed in November.
Cedric Mee- Aggravated Battery, Aggravated Assault with a Deadly Weapon, Use/Display of a Firearm while Committing a Felony, Occupied Burglary with Assault or Battery, Possession of Weapon by a Convicted Felon, Resist Officer without Violence, Resist Officer/Flee Elude Law Enforcement Officer with Lights and Sirens Active. His bond was set at $49,500.
Gregory Nelson- Aggravated Battery, Aggravated Assault with a Deadly Weapon, Occupied Burglary with Assault or Battery. His bond was set at $37,000.
Rebecca Polley- Occupied Burglary with Assault or Battery. She is being held on a $25,000 bond.
Jaclyn Ericson- Occupied Burglary with Assault or Battery. She is being held on a $25,000 bond.
Dominique J. Johnson- Home Invasion Robbery with a Firearm. Bond was set at $25,000.
Emanuel Estada-Portalatin- Home Invasion Robbery with a Firearm, Possession of Weapon or Ammo by Delinquent, Display of Firearm during a Felony, Violation of Probation. He is being held without bond at this time.
Jaclyn Ericson also made headlines in January 2014, after she got high on cocaine and killed her 5-month-old baby. Read that full story here.
Previous articleFog advisory for January 16, 2016
Next articleSecurity guard killed at work by drunk driver
Three arrested in home invasion
Firefighters battle blaze at Market of Marion
Showering and urinating outside gets Belleview man arrested
Thanksgiving Day murder suspect located in Daytona
Information sought in downtown Ocala beating, robbery

redpajama_set_name: RedPajamaCommonCrawl


We serve a dozen meat- and vegetable-filled pies, called Xian Bing, as well as savory pancakes, noodle soups and other Northern Chinese specialties began here in Monterey Park, early 2011. To ensure that every customer receives the best quality food and experience. Beijing Pie House make sure everything meets our highest standards before serving it to our customers. We believe make to order is the best way to serve. It is a guarantee that we do not add MSG to our pies. Our great pies are only made possible with freshest ingredients. Comments, Questions and More? Visit our Yelp Page.
According to legends, niu rou bing originated from the Tang Dynastyduring Xuan Zhong's reign. The story relates to his discovery of shao bing when he was fleeing with his imperial concubine Yang Gui Fei and tried one on the street. One day he disguised himself and went to the streets of Xi'an. He saw a small shop lined with hundreds of people waiting with cake in their hands. The emperor was lured into line by the delicious smells, and found out that everyone was waiting in line for niu rou bing. After waiting in line and ordering two, he ate the first one so fast that he didn't even know what it looked like. He saw the second one and realized that it looked like the moon. After, he couldn't forget the store or the memory of the food, so he often went back to buy and eat it.


Galaxy Shuttle from the Japan-only Victory cartoon series is a Holy Grail to many Victory series fans. It was released only in Japan and is pretty hard to find in good, unyellowed condition.
Most Korean-made toys for the local market had to include the manufacturing details. If only these guys were still around, I would have tried to track them down for more info on the moulds!
V is for Victory, maybe?
Instructions ripped straight from the original.
Let's take a quick look at the real deal, to remind us what this figure is REALLY supposed to look like. (Photo from wilson-ng.blogspot.com) Pretty cool figure. I'm not a huge Victory fan, but I can see why so many collectors consider this their Holy Grail, and not just for its rarity. Who doesn't love space shuttles?!
And now, we have our KO version. Very interesting color scheme and wait a min…WTF is that star flower sticker where the Autobot insignia should be? Is it some reference to the North Korean flag? NO. It isn't. After Korea was split in half at the 38th Parallel by the DMZ, South Korean companies and people were very careful to not appear as North Korean (and therefore Communist) sympathizers. Besides, much of what has been associated with North Korean design is a relic of the previously united culture, repurposed for its political and sociocultural devices. It is simply an element of Korean traditional design, dating back hundreds of years to the culture's Buddhist roots.
Red, white and blue works… sometimes. Add gold for some bling?
The star symbol is most likely clipart of a flower seen from the top down, such as that of a lotus flower, a motif commonly used throughout Korea and other parts of Asia touched by Buddism. I couldn't track down the exact same symbol, but here are some other pretty similar ones that are commonly seen in Korea.
It could very well be the topmost flower design, minus the circular elements. These kinds of symmetrical patterns are used everywhere in Korean design, from furniture to clothing.
Drink enough milk and you'll get to 10 inches someday, buddy!
The most unfortunate thing about this KO, like very many other KOs, is the brittle plastic used in making this reproduction. There is a ton of mould degradation as well, but that pales in comparison to the shoddy plastic they used for the colored parts. It almost feels like cheap toy model plastic or the kind of plastic used for those toys you get for Christmas from somebody outside of the family who picked up something from a mini-mart along the way to your house. I'm sure many kids broke this guy right out of the box after the first transformation. Well, not surprised as Haesung is a manufacturer of toy models and ripoffs of Japanese toys, sold for cheap prices. The perfect gift for your high school buddy's kid!
Hope you enjoyed this little post – I'll keep hunting for more of these interesting KO's from Korea and elsewhere as I find them.
Hello, my name is Mike and I really enjoyed this article and all the great pics! I love the mold, especially this KO, and I'm curious to know if it is available for sale or a tip as to where I might be able to acquire one. Please email me at: chezmyke1@gmail.com if you can help in any way. Thanks for your time and keep up the great work.
Hey Mike! Thanks for the kind words. Unfortunately it has already been sold, but I occasionally find items like these on my trips to Seoul. I'll keep an eye out for you.
Yes, please do my friend. Even if it's years from now please let me know if you ever find one. I have wanted one ever since I saw your page over a year ago and haven't had any luck. Do you have any other KO transformers for sale that are like this?
Yes, I do. Give me till next week and I'll rustle up some photos for you when I get back from my biz trip. Thanks!


text:
Everything About Veronica Gutierrez (Devin Booker's Mother) Wiki, Bio, Age, and More
By admin August 14, 2022 16 Mins Read
Veronica Gutierrez, a model, and actress from Chicago, Illinois, was born in the United States. Her acting career has made her a household name. She began her modeling profession as a means of showcasing her natural ability in the field. Alma, Chicago Fire, and P.D. are just a few of the shows she has been in since 2012. (2015). Veronica Gutierrez Wiki, Height, Weight, and Family Life: Here, you'll learn all there is to know about the actress. She is a model and actress from Mexico City. Throughout high school, she appeared in several productions, including Little Shop of Horrors as well as Joseph and the Colorful Dreamcoat. When she went to a modeling training program as a child, she developed a greater interest in the arts. As a model and actress, she made a name for herself in the entertainment business at an early age.
Veronica's acting and modeling careers brought her enormous recognition and fortune. She's appeared in a number of indie films. During the 2016 Chicago Latino Film Festival, she made her film debut as Jessica in the film Alma, which had its world premiere there. In addition, she has visited several film festivals in the U.S. At the Mexico International Film Festival, she won the Silver Palm Award. Additionally, she recently starred in the short film Genesis as the main actress. Her personal life is a mystery at this time. She does a lot of boxing-style physical training. She enjoys reading, photography, studying, traveling, perusing the internet, and concocting smoothie bowls. Alfonso Herrera, as well as Kate Del Castillo, are said to be possible co-stars. You can find out all we know about her in the information provided below. See Veronica Gutierrez's age, height, weight, and other personal information.
Veronica Gutierrez Wiki
A lot of people know who Veronica Gutierrez is. Because she hasn't made her birthday public yet, we don't know how old she really is. She has no zodiac sign since her date of birth is unknown. Veronica was born and raised in Mexico, where she has Mexican citizenship. She hails from a Hispanic family and is a practicing Christian. Veronica, on the other hand, keeps her upbringing a secret from the media and her family. Veronica's siblings and parents remain a mystery to the media. Her educational background, as well as that of her husband, has not been made public in any of the media reports on the case.
Full Name: Veronica Gutierrez
Popular For: Mother of Devin Booker
Profession: Cosmetologist
Nationality: Mexican
Birth Place: Chicago, Illinois, USA
Children: Devin Booker
Who is Veronica Gutierrez?
Famous Mexican-American cosmetologist Veronica Gutierrez is well-known for being the mother of NBA star Melvin Booker. The "Phoenix Suns" of the National Basketball Association are most known for their star player, Melvin Booker, an American. Veronica Gutierrez has had a successful career, despite being a famous mother. We're going to tell you a little bit about Veronica's life, including how old she is, how many children she has, who her husband is, and how much money she makes. Let's have a peek at her profile before we go any further.
How Old Is Veronica Gutierrez?
Veronica Gutierrez looks to be in her late forties, according to her appearance. Her actual date of birth is unknown to us. Because of her reserved demeanor, we don't know much about her upbringing or family. When her kid became famous, his proud mother was unknown. As a result, her life before Devin's fame is uncertain.
As a well-known public figure, Veronica Gutierrez is a well-liked mother. However, since she has yet to publicly disclose her birth date, there is currently no way of knowing her precise age. In the same way, her horoscope is a mystery due to the fact that her birth date has not been released. Veronica, on the other hand, is a Mexican citizen by birth. As for her race and religious affiliation, she is Hispanic and a practicing Christian. Veronica, on the other hand, has not disclosed much about her upbringing or family members in the media. Similar to her siblings and parents, there is no media coverage of her family members. In addition, it's worth mentioning that she has a college degree. Her educational history has been kept a secret by the media.
Family of Veronica Gutierrez
In addition to her boyfriend Melvin Booker, Gutierrez accepted Devin into her home. In spite of this, the two never wed. Melvin, a former NBA player, is also an author. While his wife Veronica looked after their kid, he went on tours around Europe and Asia. As a matter of fact, Gutierrez and Booker first met in the CBA when Gutierrez was a member of the Grand Rapids Mackers. They quickly developed a strong friendship.
Melvin began teaching Devin the art of fencing at a young age. His kid attended Moss Point High School, where he worked as an assistant coach. Veronica also has two children from a previous relationship, so there's that. Mya Powell and Davon Wade are her children. However, the identity of her children's father remains a mystery. Located in Pinnacle Peak, Davon is a real estate agent. Mya, on the other hand, has a genetic disease called microdeletion syndrome.
In reference to her notoriety, Veronica is well known since she is the mother of the well-known professional basketball star Devin Booker. As a matter of fact. As a cosmetologist, Veronica works with people's hair. Similarly, Booker, her eldest son, is a paid professional basketball player in the United States. He is now a member of the Phoenix Suns of the National Basketball Association. In addition, Devin cites his father, Melvin Booker, as an influence on his decision to pursue a career in the NBA. His father Melvin's way of life had a significant impact on him. Furthermore, he was inspired by his father and had been playing basketball since he was in elementary school. Additionally, he would use e-mail to stay in touch when on business trips to other nations and play games with them.
What is Her Nationality?
There are not many people who are aware that Veronica, who works in the beauty industry, is of Mexican-American background and is derived from Puerto Rican forebears. Veronica is also a cosmetologist. Although she was supposedly born in Puerto Rico, she's most likely the daughter of a Mexican immigrant who came to the United States from Los Nogales, Mexico. While she is Mexican, Veronica unintentionally blocked her children, especially Devin, from learning about their family's heritage. In fact, her son Devin once said that he didn't feel a connection to his Latino heritage until he relocated to Phoenix in 2015, soon after earning his degree from the University of Kentucky.
A Mexican model and actress, Veronica Gutierrez is a household name. Veronica began mimicking Gloria Trevi as a child and dreamed of appearing in telenovelas at a young age. Because of this, she developed an interest in the arts when she was only a little girl. She started her adventure by participating in practically all of her high school plays and musicals, including productions of Little Shop of Horrors & Joseph as well as the Technicolor Dreamcoat, after she had completed a modeling training program. This was the beginning of her career.
As well as student and indie films, Veronica has appeared a number of them. Alma, which premiered at the Chicago Latino Film Festival in 2016, featured her in the character of Jessica. She has shown at several film festivals around the United States and was honored with the Silver Palm Award there at Mexico International Film Festival. Her most recent performance was as the lead in the film Genesis. Gutierrez has also worked as a voiceover actress, demonstrating her adaptability. As a result, her voice may be heard in numerous huge lots and Kmart advertising on television and radio.
Veronica is a cosmetologist when it comes to her net worth and earnings. She, too, is generating a big salary in her work. According to several online estimates, she has a net worth of between $1.6 million as well as $2.5 million, which she earned as a cosmetologist.
Is Veronica Gutierrez Currently Dating or Single?
Since divorcing Melvin, Veronica has a new boyfriend. She kept the man's identity a secret. It's unclear whether she'll be dating or not in 2021. Veronica Gutierrez is married or involved in a relationship, according to the available information. Devin Booker's mother does not seem to be married or in a relationship at this moment. Veronica is Mexican-American as well as a licensed cosmetologist, but this is a fact that only a few people are aware of, despite her dual citizenship and Puerto Rican roots. Even though her father was born in the Mexican town of Los Nogales, she is said to have been born on the Caribbean Island of Puerto Rico.
The reality is that, in spite of Veronica's Mexican heritage, she has unintentionally cut her children, particularly Devin, apart from their past. Following his short stint playing college football at the University of Kentucky, her son Devin said that he felt like he was getting near to having Latino ancestry after coming to Phoenix in 2015. Veronica is not a member of any of the several social media networks now in use by the general public.
Let's take a look at Veronica's love life. So, does she have a significant other? She is now single and unmarried as of 2021. Gutierrez, on the other hand, had a long-term relationship with Melvin Booker, with whom she was much in love. When he was a student at the University of Missouri, Melvin was a professional basketball player. The pair met in Grand Rapids, the hometown of Gutierrez, where Melvin was a member of the Grand Rapids Hoops. Devin Booker was the outcome of their relationship, which started shortly afterward. A son, Devin, was born to them on October 30, 1996, in Grand Rapids, Michigan, USA. Afterward, the unmarried couple split up, but they continued to work together as co-parents for Devin.
Gutierrez also has three children of her own. She also has two more children from a previous relationship, one of them is Devin. Davon Wade and Mya Powell were born on February 17, 1993. The father of the children, on the other hand, is still a mystery. As a child, Devin grew raised in Grandville, Michigan, and his mother inter-cropped to see Melvin in Europe. Moss Point High School at Moss Point, Mississippi, was Melvin's first stop as an assistant coach. As a member of the McDonald's All-American Game, Booker participated in 2014.
Veronica Gutierrez Husband
However, as of the time that this item was published on our website, Thrill NG, Veronica has not yet tied the knot. However, we learned that Melvin Booker, a former professional basketball player from the United States, was her ex-boyfriend. During the course of his career, Booker appeared in over 30 games. On a related topic, Melvin Booker used to play for the Houston Rockets back in 1995-1996. In a similar vein, he used to play for the highly regarded University of Missouri. As it turned out, his professional basketball career as an American ended in 2008 when he opted to hang up his cleats.
Veronica and Melvin Booker first met in 1995, while Veronica was still married to Melvin. The day they met, Melvin was playing for the Grand Rapids Mockers there in CBA. Devin Booker, their sole biological son, was born as a result of the couple's personal relationship. Despite the fact that they parted ways after a few years owing to a lack of communication and, as a result, conflicting schedules, the pair remains friendly. They split up because they were both preoccupied with their professional lives. They were, and still are, excellent parents to their children, as they have always been.
Her Son Devin Plays in the NBA
Devin Booker, Veronica's second child, is an American basketball player who now competes for the NBA's Phoenix Suns. The Phoenix Suns selected him 13th overall there in the 2015 NBA Draft. As one of the league's top players, Devin has been selected to the NBA All-Star team three times. In addition, he holds the record for the most points scored by a player under the age of 24 in a game. In 2018, he was the winner of the NBA's three-point contest. Later in 2021, Booker led the Suns to the NBA Finals. Kendall Jenner, an American model, is now dating Devin.
There is no activity on Veronica Gutierrez's Facebook or Twitter accounts. When she split from her lover, she chose to keep her personal life out of the public spotlight. Veronica, on the other hand, often attended with her kid. Her son has a sizable fan base on social media, thanks to his prolific postings. His Instagram account now has over 4.6 million followers.
What Presently Is Veronica Gutierrez Doing?
We may never know. Gutierrez makes a living as a cosmetologist. We don't know where she started or where she is now in her career. Prior to becoming the NBA star's mother, she wished to keep her life as secret as it was before. Veronica Gutierrez's secluded lifestyle makes it difficult to know what she is doing in both her personal and professional lives. It is unknown if she is presently married or involved with anybody else in a romantic relationship at this time. Additionally, there is no word on how successful she is as a cosmetologist. According to rumors, she's doing well and enjoying the life of her dreams.
Veronica Gutierrez: 5 Quick Facts
In 2008, Devin Booker's mother, Veronica Gutierrez, gave her son permission to fly to Mississippi to focus on his basketball career with his father. Booker is a professional basketball player. Devin's choice surprised his father since he hadn't spent much time with him since returning from Europe and Asia. It was also Veronica who had raised her child the whole time. In the end, it was shown that she had made the right choice for herself and her child. Devin is already a household name in the NBA, and he played a key role in the United States, taking home the gold medal in the Olympics in Tokyo in 2020.
The Spouse Of Veronica Gutierrez Was Once A Basketball Player
Veronica and Devin's father, as well as Veronica's baby daddy, is Melvin Jermaine Booker. Ex-basketball star from the United States of America There was no marriage between Melvin and Gutierrez, even though many people assume that they are married.
During his time at the University of Missouri, Melvin Booker played the position of a point guard. Booker was a part of the Houston Rockets during the 1995-1996 NBA championship season. From 1996-1997, he played for the Nuggets and Warriors. Booker played 32 games for the NBA during his time as a professional basketball player.
In 2008, he made the decision to end his career as a professional basketball player. After that, he traveled to Moss Point, Mississippi, to practice with his son Devin Booker, who was staying with his mother there. To assist his son in succeeding in his chosen area, he even went so far as to hire a career coach in 2011.
Gutierrez Has Three Children
Devin Booker's mom has three kids, all of whom have different fathers. Her son, Devin Booker, is a famous athlete. As a professional basketball player for the Phoenix Suns of the NBA, he is a well-known American athlete. Devin Gutierrez was born in Grand Rapids, Michigan, on October 30, 1996, at the age of 25. Devin's father, Melvin Booker, is a prosperous businessman.
Before becoming a mother to Devin Booker, Veronica Gutierrez already had a child called Davon Wade. There is a considerable age gap between Davon and Devin. As a Detroit-area real estate agent, he has a bachelor's degree in business administration from Western Michigan University.
Mya Renee Powell, Veronica's third child, had a chromosomal genetic condition called microdeletion syndrome when she was born in 2002, although she has now outgrown it. In May of 2021, she will get her high school graduation. Devin has always looked up to Mya as a source of motivation and strength.
Gutierrez Was The Only Parent For Devin Booker
Melvin Booker, as well as Veronica Gutierrez Booker, met for the first time in 1995. At the time, Melvin was playing for the Grand Rapids Mackers in the CBA. After that, the two started dating and became close. Their first child, Devin Booker, was born in 1996. Melvin did not spend his whole basketball career in the U.S.
Instead, he took his games on the road and visited Europe and Asia. They were unable to consummate their relationship as a consequence. Because of this, Devin's upbringing would be solely her responsibility as a single mother. She did not leave the state of Michigan with all of her three children, all of whom she had raised in Midland, which is a suburb of Grand Rapids, the city where she was born. As far as Melvin went, he wasn't entirely absent.
The father helped raise his son by going to his sports games during the summer. During Devin's 12th year of life, Melvin, his father, announced his retirement from the NBA in 2008. On his return to his native state of Mississippi, Veronica convinced him to bring his son, who resides in Michigan, to Mississippi and help their kid with his basketball career. For as long as Veronica and Melvin were together, they've always supported their son's basketball career, even if they were never married.
She Does Not Have A Husband
Veronica Gutierrez and Melvin Booker were never married, despite the widespread belief that she is the latter's wife. It was in the 1990s, when they started dating that they discovered that she already had one kid to deal with. On the other hand, the intricacies of that connection remain a mystery. In the early 2000s, after a second marriage ended in divorce, Powell gave birth to her daughter, Mya, with her second husband. Furthermore, Mya Powell's father's identity remains a mystery. Veronica Gutierrez hasn't been in a relationship since at least 2022. By 2021, she expects to be alone and unmarried, and thus she has decided to keep her personal life a secret.
The Highest Contract Ever Signed by A Phoenix Suns Player Belongs To Veronica Gutierrez's Son
The Phoenix Suns signed Devin Booker to a five-year, $158 million deal in 2018, making it the most expensive contract in the team's history and elevating him to the position of league MVP. His salary for the NBA regular season of 2023–2024 is set at $36 million according to the terms of the deal. However, by the time his professional career comes to a close in 2021, he will have amassed a total of $65 million in earnings. For the year 2020, Devin earned $6 million in sponsorships alone, according to Forbes, putting his net worth at $50,000,000.
Veronica Gutierrez
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Cases & Deals
Targa Resources Partners forms two joint ventures with subsidiary of Sanchez Energy Corporation
Clients Targa Resources Partners LP
Jones Day advised Targa Resources Partners LP ("TRP") on the formation of two joint ventures with a subsidiary of Sanchez Energy Corporation ("SN") to construct a new 200 MMcf/d cryogenic natural gas processing plant in La Salle County, Texas and approximately 45 miles of associated high pressure gathering pipelines that will connect SN's Catarina Ranch acreage in the Eagle Ford shale play to the processing plant, costing approximately $240 million. TRP and SN will each have a 50 percent ownership interest in the joint ventures. TRP will operate both the processing plant and the gathering pipelines and will hold all transportation capacity on the gathering pipelines.
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Experimental data is a dynamic network of data points much like a tree, but common tools only allow two dimensional relationships to be defined. This limitation of our tools impedes scientific discovery.
R&D organizations want global real-time access to their data in a format that is structured for statistical analysis.
Riffyn SDE was designed as a platform for global R&D organizations to capture, structure and integrate their data automatically. It allows scientists to visually design their experimental processes. Riffyn SDE then automatically generates a flexible data model to represent complex relationships between experimental data points.
The last twenty years in R&D has seen a proliferation of digital, automated, high-throughput devices that generate mountains of data. There are hundreds of such devices which allow R&D teams to conduct more experiments and generate more data points every year. However, this expansion of experimentation has raised the question of what to do with all that data. More data often doesn't translate into more insights and faster results. Data are only useful if they are clearly annotated, linked to all relevant data upstream and downstream, and most importantly, accessible to globally distributed teams.
Experimental data is often kept in siloed spreadsheets and databases.
Data analytics software and data storage solutions for life science R&D have evolved in parallel the past 20 years, but not necessarily in harmony. On one hand, currently available data analytics tools are fantastic, including JMP, Spotfire, Tableau, R, Python…the list of outstanding analytical tools is really extensive and growing. On the other hand, data storage solutions, such as Spreadsheets (Excel), Lab Information Management Systems (LIMS), Electronic Lab Notebooks (ELNs), Relational Databases, have been a bit of a mixed bag. They're well understood, and they're good for their primary purpose—storing data. However, R&D organizations are not merely interested in storing data. What they actually need is a better way to use data to extract scientific insights. Today's data storage solutions are ill-designed to solve the data-use challenges of data annotation, joining and extraction (steps necessary for meaningful data analysis).
Why are today's data storage solutions inadequate?
The problem is 2-fold: First, they're rigid in their design and aren't easily adapted to change. LIMS still require months of planning and a support team of a dozen people to implement. The result is that they are designed to solve yesterday's scientific questions and need to be constantly re-configured to support the future. That becomes a huge barrier to scientific progress, which demands constant change and continuous improvement in process and experiment designs.
Second, they're not very interoperable with each other, which makes the extraction, cleaning and formatting of data sets a huge chore. For example, joining procedural data (usually stored in an ELN) with results data (usually stored in spreadsheets or LIMS systems) is no easy task; annotating sample analytical data with all relevant procedural information used to generate the samples usually involves heroic effort, and extracting data from multiple linked unit operations in an analysis-ready format is often impossible.
So where have we gone astray?
The ramifications of these limitations are significant — they can stunt scientific discovery. New drugs are slowly discovered or go undiscovered. Novel products go undeveloped.
One oft-unaddressed challenge faced by legacy architectures is that lab data looks more like a tree with a network of relationships between data points, not like the square we want it to be for machine learning and statistics. Typically data and analytical systems need data in defined rows and columns.
Let's consider, for example, a standard bacterial fermentation. A 100 L fermentor may require sampling every 15 minutes to check for pH, optical density, product concentration (yield) and a number of additional tests. Inlet and outlet gas compositions also need to be collected and stored (often in real-time). Each fermentor is fed a prepared media to sustain the bacteria. Further, the 100 L fermentor may have been inoculated from a cell culture generated from a series of other fermentations starting from a 200 μL vial - each having their own set of media, samples, and gas compositions.
In order to conduct a meaningful analysis of this fermentation (that can control for things like random variances), these sheets need to be joined together into a statistical data format. This means all time points need to be aligned; sample data need to be connected to their sources; different units must be normalized (Celsius vs Fahrenheit); data needs to be annotated with consistent terms; sample IDs need to be rectified. Perhaps the most difficult part is simply finding the data. They reside in LIMS systems, spreadsheets, SOPs, batch sheets, and data historians. This process can take weeks to complete, even for common procedures such as fermentation.
All this is considered normal! We've lived in this world for decades so nothing feels strange about it. The problem underlying this issue is that scientific data is multi-dimensional with many complex relationships between samples, devices, and systems. It is more akin to a network or a social graph. Complex relationships aren't fully captured within stand-alone legacy databases and spreadsheets. This makes it incredibly difficult to restructure the data stored within them.
Scientific experiments are designed specifically to test the effect of varying conditions chosen by the experimenter. As such, the structure of a "master" data set (all the design, contextual and results data for an experiment) is also determined by the experimenter. So for a data system to deliver the annotated,"flattened", and joined data sets necessary for analysis, it actually needs a clear and continuously updated set of instructions from the scientist regarding the experiment context. But this heart of the problem, current data systems don't do this.
In typical labs today, data scientists are empowered with the skills to reshape and analyze data, they don't have the necessary context. They have to chase down and interview many busy R&D scientists and staff in order to understand how data should be annotated, cleaned and joined for analysis. This is a challenging, never-ending undertaking that takes a lot of time and a strong understanding of the underlying scientific design. Thus scientific activity and analyses tend to get bottlenecked on the few people with the needed computational skills. It also tends to limit access of lab scientists to the cleaned-up data and analysis tools.
Moreover science is also incredibly dynamic. Experiments you run today may be drastically different than those you recorded 3 months ago (or 3 months from now). Methods, processes, and analyses are always changing. Maybe you change an assay procedure or use very different operating conditions. Our data systems must adapt to these changes without sacrificing your ability to work with historical data.
To solve these issues, the next generation of data systems need to be more adaptive to the constantly changing designs of scientific experiments, and they need to put the power of data analytics more directly into the hands of the scientists that generate it.
Every life science company from biopharma, to chemicals, to food producers all want to do more with their data and are making significant investments in it. Merck is partnering with Accenture to develop a Research Life Science Cloud to improve cost and efficiency across the entire industry. GSK is making investments to deploy AI to revolutionize its drug discovery business. BASF has made significant investments into its digital transformation initiative within their research organization. There's real money behind this transformation and everyone is looking for the best platform on which to base their business.
Riffyn was created to serve as such a platform. We've developed our hypergraph data model to accurately reflect the complex structure of lab data. We've also incorporated a data versioning functionality to allow for new data to be easily combined with older versions.
Data across multiple related experiments can all be easily pulled together in a structured format, ready for statistical analysis, in as fast as 30 seconds. This fully annotated and cleaned data set can then be used to ask questions such as "Which operating conditions lead to optimal results?", "What's the effect of material lot numbers on my yield?", or "How do my Day 14 data compare to my Day 3 data?" These types of questions can all be discerned through data analysis without conducting additional time consuming experiments.
Further, Riffyn SDE puts the power of data directly in the hands of the scientists that generate it. The process design environment allows users to design their experiment structure and material flow, while Riffyn SDE automatically generates the data model to match this structure and flow without any coding or UI design needed. Teams can duplicate each other's experimental processes to leverage each other's best practices. They can collaborate amongst each other to design processes, execute experiments, and determine how one group's work impacts other teams.
With Riffyn SDE, R&D organizations have an opportunity to adopt a new approach to scientific discovery that delivers flexibility, context, structure, and analysis all at the same time. Data can inform next steps. Processes can be continuously, iteratively improved. Organizational turnover doesn't have to represent a loss of scientific experience and insight. Lab data can finally start serving its true purpose in driving organizational effectiveness.
In short, Riffyn delivers data you can use and reuse to drive faster discovery.
Want to better understand how to structure your data for analysis? Check out this related article by Loren Perelman: Data is Like Ikea Furniture, It's Best Shipped Flat.


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coming from the dangerous and innovative to your lighthearted and entertaining. By way of example, "What Would You Master at school Today" which mocks the manner where by classes young ones continually discovers about nationalism and subservience to expertise. The melody "Jimmy Newman" narrates the plot in regards to a perishing soldier and "My Son John" analyzes a soldier who earnings home and can not even learn to share what he has experienced. "Beau John" may be a structure from the civil protection under the law motion about having a stand alongside racial injustice. Paxton has done a great deal to keep up with the American layout and then he may be a popular music artists and teacher within the model (Holk, n.p).
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Pensions and animal rights top ballot sheets
The three-tier Swiss pension scheme is considered unique Keystone
Voters have the final say on Sunday on a reform of the occupational pension scheme opposed by a coalition of the centre-left parties, unions and consumer groups.
This content was published on March 7, 2010 - 10:06 March 7, 2010 - 10:06
Urs Geiser
Worked in radio and newspaper journalism, as well as teaching and tourism before joining swissinfo.ch's predecessor Swiss Radio International in the 1980s. He reports from parliament, and focuses on direct democracy issues.
More from this author | English Department
In two other ballots, the Swiss will decide on the creation of special attorneys to defend animal rights in legal cases and on a constitutional amendment on human research.
The proposed cuts in occupational benefits, agreed by a majority in parliament last year, were challenged to a referendum by the collection of more than 200,000 signatures.
The centre-left says the fight against lower benefits is a key aspect of their efforts to halt attacks on the country's welfare system.
It says the plan is tantamount to "benefits theft" by the business community and the insurance industry in particular.
But supporters, including the government and most centre-right and rightwing parties, have argued that the reform is necessary to ensure the future of the pension scheme. Rejection would undermine the country's unique social security scheme and threaten economic prosperity, they argue.
Under the proposal, annual occupational pension benefits, also known as the second pillar, will be reduced to make up for an ageing population and for lower revenue from investment.
High profile campaigns
Voters were asked to decide on reducing the so called minimum conversion rate by about 0.6 per cent to 6.4 per cent by 2016.
This technical ratio is used to calculate the annual pension based on accumulated capital. In practice new pensions of low income earners would be cut by several hundred francs.
The occupational pension, which is part of a three-tier scheme including the old age pension and individual savings plans, was made mandatory in 1985 and was subject to a revamp seven years ago.
The run-up to Sunday's vote was marked by an unusually early campaign launch by supporters of the pension reform. It is estimated that they spent about SFr10 million ($9.3 million) on posters, advertisements and other political marketing elements.
For their part, unions and the centre-left Social Democrats ran a high-profile campaign with leaflets, online gadgets, brochures and public events organised by the grassroots.
Pensions (Three Pillar system)
This content was published on Sep 18, 2006 Sep 18, 2006 The old-age pension system consists of three levels (pillars). The first is the state pension plan. Every resident must pay contributions to a...
A constitutional amendment to come to a vote on Sunday seeks to boost the legal standing of animals. The initiative was launched by a leading animal rights organisation and is aimed at institutionalising the position of special attorney in the country's 26 cantons.
Supporters say the current laws are not strong enough to crack down on animal cruelty - allegations rejected by opponents.
However, opponents argue adoption of the proposal would lead to a wave of legal cases and infringe the autonomy of the cantonal authorities.
The centre-left Social Democrats and the Greens have recommended a yes vote, while the government and a majority of the centre-right and rightwing parties have come out against it – albeit with a fair number of dissenting voices.
Switzerland would be the first country in the world to introduce mandatory animal lawyers. So far Zurich is the only canton with such an attorney.
A separate proposal by the government and parliament on the legal principles of human research has prompted little public controversy ahead of the ballot.
The constitutional amendment aims at preserving human dignity in biomedical experiments while ensuring freedom of research.
The amendment is to set the basis for more detailed cantonal legislation on transplantations, gene technology and stem cell research at a later stage.
Supporters stress the need to streamline regulations and to consult experts on bioethics to define the scope of the regulations.
Opponents, notably from the centre-left, have criticised the perceived lack of protection for people who cannot give informed consent to experiments, including children and the mentally handicapped.
The rightwing Swiss People's Party has warned of interference by the federal authorities in cantonal matters.
Urs Geiser, swissinfo.ch
March 7 votes
Some five million Swiss, including registered expatriates, are eligible to cast ballots in the nationwide votes on pension reform, animal rights and human research.
Votes on a wide range of issues, as well as elections, are taking place in more than half of the country's 26 cantons and in numerous cities and communes.
As a rule ballots take place four times a year.
Committee in favour of pension cuts – Pension fund association (German, French)
Federal Statistics Office – Results vote March 7 (German, French)
Swiss Animal Protection (SAP)
Committee against pension cuts – Unia trade union
Human research - Federal Health Office
Federal Veterinary office
Occupational benefits - Federal Social Security Office, Interior ministry
gfs berne polling institute - vote March 7 (German)
swissworld - politics
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