text:
\section{Introduction}

Let $H$ be a separable Hilbert space. A sequence $\{f_n\}\subset H$ is called a \emph{frame}
for $H$ if there exist positive constants $C_1$ and $C_2$ such that
$$C_1\|f\|^2 \le \sum_{n=1}^{\infty} |\langle f,f_n\rangle |^2 \le C_2\|f\|^2$$
for all $f\in H$. The numbers $C_1$ and $C_2$ are certainly not unique. The optimal upper
constant, $\inf C_2$, will be called the upper frame bound for $\{f_n\}$. Similarly, the
optimal lower constant, $\sup C_1$, will be called the lower frame bound for $\{f_n\}$.
A frame is called \emph{tight} if its lower and upper frame bounds are the same.
Also, a frame is called \emph{Parseval} or \emph{normalized tight} if its lower and upper
frame bounds are both $1$. See \cite{Chr} for an introduction to the theory of frames.

The \emph{singular values} or \emph{s-numbers} of a compact operator $T$ on $H$ are the
square roots of the positive eigenvalues of the operator $T^*T$, where $T^*$
denotes the adjoint of $T$. Equivalently, this is the sequence of positive eigenvalues
of $|T|=(T^*T)^{1/2}$. We always arrange the singular values of $T$, $\{\lambda_n\}$,
such that $\lambda_1\ge\lambda_2\ge\lambda_3\ge\cdots$, with each eigenvalue
of multiplicity $k$ repeated $k$ times in the sequence.

Given $0<p<\infty$, the \emph{Schatten $p$-class} of $H$, denoted $S_p(H)$ or simply $S_p$,
is defined as the space of all compact operators $T$ on $H$ with its singular value sequence $\{\lam_n\}$ belonging to $\ell^p$. It is well known that $S_p$ is a two-sided ideal in the
full algebra $\calL(H)$ of all bounded linear operators on $H$. Also, when equiped with
$$\|T\|_p=\left[\sum_{n=1}^\infty\lambda_n^p\right]^{\frac1p},$$
$S_p$ is a Banach space when $1\le p<\infty$ and a complete metric space when $0<p<1$.
Two special cases are especially interesting in operator theory: $S_1$ is called the
trace class and $S_2$ is the Hilbert-Schmidt class. See \cite{GK, Simon, Zhu} for
basic properties of Schatten classes.

Operators in Schatten classes can often be described by their action on orthonormal bases.
For example, a positive operator $T\in\calL(H)$ belongs to the trace class $S_1$ if and
only if $\sum\langle Te_n,e_n\rangle<\infty$, where $\{e_n\}$ is any given orthonormal basis
for $H$. Similarly, an operator $T\in\calL(H)$ belongs to the Hilbert-Schmidt class $S_2$ if
and only if $\sum\|Te_n\|^2<\infty$, where $\{e_n\}$ is any given orthonormal basis for $H$.
See \cite{GK, Simon, Zhu} again for these and other related results.

It is clear that any orthonormal basis is a frame, with frame bounds equal to $1$.
The purpose of this article is to study Schatten class operators in terms of frames.
We state our main results as follows.

\begin{thma}
Suppose $T$ is a compact operator on $H$ and $2\le p<\infty$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] $\{\|Te_n\|\}\in\ell^p$ for \emph{every} orthonormal basis $\{e_n\}$ in $H$.
\item[(c)] $\{\|Tf_n\|\}\in\ell^p$ for \emph{every} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, we always have
$$\|T\|_p^p=\sup\sum_{n=1}^\infty\|Te_n\|^p=\sup\sum_{n=1}^\infty\|Tf_n\|^p,$$
where the first supremum is taken over all orthonormal bases $\{e_n\}$ and the second
supremum is taken over all frames $\{f_n\}$ with upper frame bound less than or equal to $1$.
\end{thma}

\begin{thmb}
Suppose $T$ is a bounded operator on $H$ and $0<p\le2$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] $\{\|Te_n\|\}\in\ell^p$ for \emph{some} orthonormal basis $\{e_n\}$ in $H$.
\item[(c)] $\{\|Tf_n\|\}\in\ell^p$ for \emph{some} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, we always have
$$\|T\|_p^p=\inf\sum_{n=1}^\infty\|Te_n\|^p=\inf\sum_{n=1}^\infty\|f_n\|^{2-p}
\|Tf_n\|^p=\inf\sum_{n=1}^\infty\|Tf_n\|^p,$$
where the first infimum is taken over all orthonormal bases, the second infimum
is taken over all frames with lower frame bound greater than or equal to $1$, and the
third infimum is taken over all Parseval frames $\{f_n\}$.
\end{thmb}

The conditions above concerning orthonormal basis are more or less well known to
experts in the field. But the necessary \emph{and} sufficient conditions stated here do not
seem to have appeared anywhere before. Partial statements in terms of orthonormal basis can
be found in \cite{GK, Simon, Zhu} for example. We will include the treatment for orthonormal
bases in the paper for the sake of completeness.

It is interesting to observe the sharp contrast between the cases $p\ge2$ and $p\le2$:
in the first case $\sup$ is used to compute the norm $\|T\|_p$, while in the second case
$\inf$ must be used. We will also construct examples to show that the cut-off at $p=2$
is necessary, and the result at the cut-off value $p=2$ is particularly nice.

In addition to Theorems A and B, we will obtain corresponding results in terms of the
sequence $\{\langle Tf_n,f_n\rangle\}$ and the double-indexed sequence
$\{\langle Tf_n,f_k\rangle\}$. But in these cases it is sometimes \emph{necessary} to
require additional assumptions on the operator $T$, such as $T$ being positive or
self-adjoint.

The relationship between frames and operators in Schatten classes has been studied by
several authors in the past few years. See \cite{ACP,Bal,Koo} and references therein.
There is some overlap between the present paper and the papers just referenced. However,
the approach here is different, the results here are \emph{complete}, and the proofs here
are simpler and more natural.

\section{The case when $p$ is large}

The description of operators in the Schatten class $S_p$ depends on the range of $p$.
In this section we focus on the case when $p$ large. We begin with the following lemma
which is well known to experts. This is the only result from the theory of frames that
we will use in the paper, so we include a short proof here for the reader's easy reference.

\begin{lem}
Suppose $\{e_n\}$ is an othonormal basis and $\{f_n\}$ is a frame for $H$. Then the operator
$A:H\to H$ defined by
$$A\left(\sum_{k=1}^\infty c_ke_k\right)=\sum_{k=1}^\infty c_kf_k$$
is a well-defined bounded linear operator. Furthermore, $\|A\|^2$ is between the lower and
upper frame bounds of $\{f_n\}$, and $AA^*$ is invertible on $H$.
\label{1}
\end{lem}

\begin{proof}
If $f$ is any vector in $H$, then
\begin{eqnarray*}
\left|\left\langle A\left(\sum_{k=1}^N c_ke_k\right),f\right\rangle\right|^
&=& \left|\sum_{k=1}^N c_k \langle f_k ,f\rangle\right|^2 \\
&\le& \sum_{k=1}^N |c_k|^2 \sum_{k=1}^N | \langle f_k, f \rangle |^2 \\
&\le& C_2 \|f\|^2\left\|\sum_{k=1}^N c_ke_k \right\|^2,
\end{eqnarray*}
where $C_2$ is the upper frame bound for $\{f_n\}$.

Therefore, by the Hahn-Banach theorem, $A$ extends to a bounded linear operator on $H$
with $\|A\|^2\le C_2$, namely,
$$A \left(\sum_{k=1}^{\infty} c_ke_k \right) = \sum_{k=1}^{\infty} c_kf_k,$$
where $\{c_k\}\in\ell^2$. If $\langle f,f_k\rangle=0$ for all $k$, then it follows from
the definition of frame that $f=0$. Therefore, $A$ has dense range.

For any vector $f\in H$, we have
$$A^*f=\sum_{n=1}^\infty\langle A^*f,e_n\rangle\,e_n=
\sum_{n=1}^\infty\langle f,Ae_n\rangle\,e_n=\sum_{n=1}^\infty\langle f,f_n\rangle\,e_n.$$
It follows that
$$\|A^*f\|^2=\sum_{n=1}^\infty|\langle f,f_n\rangle|^2\ge C_1\|f\|^2$$
for $f\in H$, where $C_1$ is the lower frame bound for $\{f_n\}$. This shows that
$\|A\|^2=\|A^*\|^2\ge C_1$ and $A^*$ is one-to-one and has closed range. Furthermore,
for any $f\in H$, we have
$$C_1\|f\|^2\le\|A^*f\|^2=\langle AA^*f,f\rangle\le\|AA^*f\|\|f\|.$$
It follows that $\|AA^*f\|\ge C_1\|f\|$ for all $f\in H$, so that $AA^*$ is one-to-one
and has closed range. Since $\ran(AA^*)^\perp=\ker(AA^*)=(0)$, $AA^*$ must be onto.
Therefore, $AA^*$ is invertible.
\end{proof}

As a consequence of the invertibility of $AA^*$, we see that the operator $A$ above is
actually onto. Therefore, every vector $f\in H$ admits a representation of the form
$$f=\sum_{n=1}^\infty c_nf_n,$$
where $\{c_n\}\in\ell^2$. Note that $A$ is generally not one-to-one. For example, a frame
may contain a certain vector that is repeated a finite number of times. In this case, the
associated operator $A$ is obviously not one-to-one.

\begin{thm}
Suppose $T$ is a compact operator on $H$ and $2\le p<\infty$. Then
the following statements are equivalent.
\begin{enumerate}
\item[(a)] $T$ is in the Schatten class $S_p$.
\item[(b)] $\{\|Te_n\|\}\in\ell^p$ for \emph{every} orthonormal basis $\{e_n\}$ in $H$.
\item[(c)] $\{\|Tf_n\|\}\in\ell^p$ for \emph{every} frame $\{f_n\}$ in $H$.
\end{enumerate}
Moreover, we always have
$$\|T\|^p_p=\sup\sum_{n=1}^\infty\|Te_n\|^p=\sup\sum_{n=1}^\infty\|Tf_n\|^p,$$
where the first supremum is taken over all orthonormal bases $\{e_n\}$ and the second
supremum is taken over all frames $\{f_n\}$ with upper frame bound less than or
equal to $1$.
\label{2}
\end{thm}

\begin{proof}
The equivalence of conditions (a) and (b) is well known. See Theorem 1.33 of \cite{Zhu} for
example. Note that Theorem 1.33 of \cite{Zhu} was stated and proved in terms of orthonormal
sets. Since every orthonormal set can be expanded to an orthonormal basis, the result
remains true when the phrase ``orthonormal sets'' is replaced by ``orthonormal bases''.

Since every orthonormal basis is a frame, it is trivial that condition (c) implies (b).

To prove that (a) and (b) together imply (c), we fix an orthonormal basis $\{e_n\}$ and
a frame $\{f_n\}$ for $H$ and consider the operator $A$ defined in Lemma~\ref{1}. If
$T$ is in $S_p$, then so is the operator $S=TA$. Apply condition (b) to the operator $S$,
we obtain
$$\sum_{n=1}^\infty\|Tf_n\|^p=\sum_{n=1}^\infty\|TAe_n\|^p
=\sum_{n=1}^\infty\|Se_n\|^p<\infty.$$
This completes the proof of the equivalence of conditions (a), (b), and (c).

The equality
$$\|T\|_p^p=\sup\sum_{n=1}^\infty\|Te_n\|^p$$
was established in Theorem 1.33 of \cite{Zhu}. Since every orthonormal basis is a frame
with frame bounds equal to $1$, we clearly have
$$\|T\|_p^p=\sup\sum_{n=1}^\infty\|Te_n\|^p\le\sup\sum_{n=1}^\infty\|Tf_n\|^p.$$
This along with the arguments in the previous paragraph shows that
$$\sum_{n=1}^\infty\|Tf_n\|^p=\sum_{n=1}^\infty\|TAe_n\|^p\le\|TA\|_p^p.$$
It is well known (see \cite{GK,Zhu} for example) that $\|TA\|_p\le\|T\|_p\|A\|$, which
combined with the estimate for $\|A\|$ in Lemma~\ref{1} shows that $\|TA\|_p\le\|T\|_p$
whenever $\{f_n\}$ has upper frame bound less than or equal to $1$. This shows that
$$\sup\sum_{n=1}^\infty\|Tf_n\|^p_p\le\|T\|_p^p,$$
and completes the proof of the theorem.
\end{proof}

It is possible to obtain a version of Theorem~\ref{2} without the a priori assumption that
$T$ be compact. This will be done using an approximation argument based on the following lemma.

\begin{lem}
Suppose $T$ and $T_k$, $k\ge1$, are bounded linear operators on $H$. If $1<p<\infty$,
$T_k\to T$ in the weak operator topology, and $\|T_k\|_p\le C$ for some constant $C$
and all $k\ge1$, then $\|T\|_p\le C$.
\label{3}
\end{lem}

\begin{proof}
Let $S$ be a finite-rank operator and $\{e_n\}$ be an orthonormal basis of $H$ such that
$TS(e_n)=0$ for all but a finite number of $n$. Then
$$\tr(TS)=\sum_{n=1}^\infty\langle TSe_n,e_n\rangle=\lim_{k\to\infty}
\sum_{n=1}^\infty\langle T_kSe_n,e_n\rangle=\lim_{k\to\infty}\tr(T_kS).$$
Since the Banach dual of $S_p$ is $S_q$, $1/p+1/q=1$, under the pairing induced by
the trace, we have
$$|\tr(T_kS)|\le\|T_k\|_p\|S\|_q\le C\|S\|_q,\qquad k\ge1.$$
It follows that $|\tr(TS)|\le C\|S\|_q$ for all finite rank
operators $S$. Since the set of finite rank operators is dense in $S_q$, we conclude
that $T\in S_p$ and $\|T\|_p\le C$.
\end{proof}

\begin{thm}
When $2\le p<\infty$, the following conditions are equivalent for any bounded linear
operator $T$ on $H$:
\begin{enumerate}
\item[(i)] $T\in S_p$.
\item[(ii)] There exists a constant $C>0$ such that
$$\sum_{n=1}^\infty\|Te_n\|^p\le C$$
for \emph{every} orthonormal basis $\{e_n\}$.
\item[(iii)] There exists a constant $C>0$ such that
$$\sum_{n=1}^\infty\|Tf_n\|^p\le C$$
for \emph{every} frame $\{f_n\}$ with upper frame bound no greater than $1$.
\end{enumerate}
\label{4}
\end{thm}

\begin{proof}
All we have to show here is that condition (ii) implies (i) without the a priori assumption
that $T$ be compact. This can be done with the help of an approximation argument. More
specifically, we fix an increasing sequence $\{P_k\}$ of finite-rank projections such that
$\{P_k\}$ converges to the identity operator in the strong operator topology and let
$T_k=P_kT$ for $k\ge1$. Each $T_k$ is a finite-rank operator, so condition (ii) along
with Theorem~\ref{2} gives
$$\|T_k\|^p_p=\sup\sum_{n=1}^\infty\|T_ke_n\|^p\le\sup\sum_{n=1}^\infty\|Te_n\|^p\le C$$
for all $k\ge1$, where the suprema are taken over all orthonormal bases $\{e_n\}$. Since
$T_k\to T$ in the strong operator topology, it follows from Lemma~\ref{3} that
$\|T\|_p^p\le C$.
\end{proof}

It is natural to ask the following question: suppose $p\ge2$ and $\{\|Tf_n\|\}\in\ell^p$ for \emph{some} frame $\{f_n\}$, does it imply that $T\in S_p$? The answer is yes for $p=2$ but
no for $p>2$. We will get back to the case $p=2$ in Section 4 but will now settle the case
$p>2$.

\begin{prop}
Suppose $2<p<\infty$, $\varepsilon>0$ (not necessarily small), and $T$ is any operator
in $S_{p+\varepsilon}-S_p$. Then there exists a frame $\{f_n\}$ for $H$ such that
$\{\|Tf_n\|\}\in\ell^p$.
\label{5}
\end{prop}

\begin{proof}
Suppose that
$$Tx=\sum_{n=1}^\infty\lambda_n\langle x,e_n\rangle\,\sigma_n$$
is the canonical decomposition of $T$, where $\{\lambda_n\}$ is the singular value sequece
of $T$ which is arranged in nonincreasing order and repeated according to multiplicity.
Thus we have $\{\lambda_n\}\in\ell^{p+\varepsilon}$ but $\{\lambda_n\}\not\in\ell^p$.

Let $\{e_n'\}$ denote an orthonormal basis for $\ker(T)$. Then $\{e_n\}\cup\{e_n'\}$ is an
orthonormal basis for $H$. In fact, for any vector $x\in H$, we have $\langle x,e_n\rangle=0$
for every $n$ if and only if $Tx=0$. Therefore, $\{e_n\}^\perp=\{e_n'\}$.

For every $n\ge1$ choose a positive number $\delta_n$ such that $\delta_n^{p-2}
=\lambda_n^\varepsilon$. Since $p>2$, we have $\delta_n\to0$ as $n\to\infty$, so we
can choose a sequence $\{N_n\}$ of positive integers such that $N_n\delta_n^2\sim1$
as $n\to\infty$. In other words, there exist positive constants $c$
and $C$ such that $c\le N_n\delta_n^2\le C$ for all $n\ge1$. Let $\{f_n\}$ be the sequence
consisting of all vectors in $\{e_n'\}$, plus $N_1$ copies of the vector $\delta_1e_1$,
plus $N_2$ copies of the vector $\delta_2e_2$, and so on.

For any vector $f\in H$, we have
\begin{eqnarray*}
\sum_{n=1}^\infty|\langle f,f_n\rangle|^2&=&\sum_{n=1}^\infty N_n|\langle f,
\delta_ne_n\rangle|^2+\sum_{n=1}^\infty|\langle f,e_n'\rangle|^2\\
&=&\sum_{n=1}^\infty N_n\delta_n^2|\langle f,e_n\rangle|^2+\sum_{n=1}^\infty
|\langle f,e_n'\rangle|^2\\
&\sim&\sum_{n=1}^\infty|\langle f,e_n\rangle|^2+\sum_{n=1}^\infty|\langle f,e_n'\rangle|^2\\
&=&\|f\|^2.
\end{eqnarray*}
This shows that $\{f_n\}$ is a frame for $H$.

On the other hand,
\begin{eqnarray*}
\sum_{n=1}^\infty\|Tf_n\|^p&=&\sum_{n=1}^\infty N_n\|T(\delta_ne_n)\|^p
=\sum_{n=1}^\infty N_n\delta_n^p\|Te_n\|^p\\
&=&\sum_{n=1}^\infty N_n\delta_n^p\lambda_n^p
=\sum_{n=1}^\infty N_n\delta_n^2\delta_n^{p-2}\lambda_n^p\\
&\sim&\sum_{n=1}^\infty\delta_n^{p-2}\lambda_n^p
=\sum_{n=1}^\infty\lambda_n^{p+\varepsilon}<\infty.
\end{eqnarray*}
This completes the proof of the proposition.
\end{proof}

We also derive a version of the above proposition in terms of orthonormal bases.

\begin{prop}
Suppose $2<p<\infty$, $\varepsilon>0$, and $\{e_n\}$ is any orthonormal basis for $H$.
Then there exists an operator $S\in S_{p+\varepsilon}-S_p$ such that $\{\|Se_n\|\}\in\ell^p$.
\label{6}
\end{prop}

\begin{proof}
Fix any operator $T\in S_{p+\varepsilon}-S_p$ and use Proposition~\ref{5} to select a
frame $\{f_n\}$ such that $\{\|Tf_n\|\}\in\ell^p$. Let $A$ be the operator on $H$ defined
by $Ae_n=f_n$, $n\ge1$. By Lemma~\ref{1}, the operator $A$ is bounded and the operator
$AA^*$ is invertible. Let $S=TA$. Then $S\in S_{p+\varepsilon}$ because $S_{p+\varepsilon}$
is a two-sided ideal in the full operator algebra $\calL(H)$. Since $AA^*$ is invertible,
we have $S\not\in S_p$ as well. Otherwise, the operator $T(AA^*)=SA^*$ would be in
$S_p$. Multiplying from the right by $(AA^*)^{-1}$ and using the fact that
$S_p$ is a two-sided ideal in $\calL(H)$ again, we would then obtain that $T$ is in $S_p$,
a contradiction. Therefore, $S\in S_{p+\varepsilon}-S_p$ and
$$\sum_{n=1}^\infty\|Se_n\|^p=\sum_{n=1}^\infty\|TAe_n\|^p=\sum_{n=1}^\infty
\|Tf_n\|^p<\infty.$$
This completes the proof of the proposition.
\end{proof}

Characterizations of Schatten classes can also be given in terms of the
sequence $\{\langle Tf_n,f_n\rangle\}$ and the double-indexed sequence $\{\langle Tf_n,
f_k\rangle\}$. We now proceed to the characterization of $S_p$ based on the sequence
$\{\langle Tf_n,f_n\rangle\}$.

\begin{thm}
Suppose $1\le p<\infty$ and $S$ is a compact operator on $H$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $S$ belongs to $S_p$.
\item[(b)] $\{\langle Se_n,e_n\rangle\}\in\ell^p$ for \emph{every} orthonormal
basis $\{e_n\}$ in $H$.
\item[(c)] $\{\langle Sf_n,f_n\rangle\}\in\ell^p$ for \emph{every} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, if $S$ is self-adjoint, then
$$\|S\|_p^p=\sup\sum_{n=1}^\infty|\langle Se_n,e_n\rangle|^p=\sup\sum_{n=1}^\infty
|\langle Sf_n,f_n\rangle|^p,$$
where the first supremum is taken over all orthonormal bases and the second supremum is
taken over all frames with upper frame bound less than or equal to $1$.
\label{7}
\end{thm}

\begin{proof}
The equivalence of (a) and (b) follows from Theorem 1.27 in \cite{Zhu}. Note again that
Theorem 1.27 in \cite{Zhu} is stated in terms of orthonormal sets. Since every orthonormal
set can be expanded to an orthonormal basis, we see that Theorem 1.27 in \cite{Zhu} remains
valid when the phrase``orthonormal sets'' is replaced by ``orthonormal bases''.

It is trivial that (c) implies (b).

To prove that (a) implies (c), first assume that $S$ is positive. In this case, we can
write $S=T^*T$, where $T$ is the square root of $S$. Then $\langle Sf_n,f_n\rangle
=\|Tf_n\|^2$ and the desired result follows from Theorem~\ref{2} and the fact that
$S\in S_p$ if and only if $T\in S_{2p}$. When $S$ is not necessarily positive, it is
well known that we can write
$$S=(S_1-S_2)+i(S_3-S_4),$$
where each $S_k$ is positive and belongs to $S_p$. By the already proved case for positive
operators, $\{\langle S_kf_n,f_n\rangle\}\in\ell^p$ for each $1\le k\le4$. It follows that
$\{\langle Sf_n,f_n\rangle\}\in\ell^p$.

It follows from the canonical decomposition for self-adjoint compact operators and the
fact that every orthonormal basis is a frame with frame bounds $1$ that
$$\|S\|_p^p\le\sup\sum_{n=1}^\infty|\langle Se_n,e_n\rangle|^p
\le\sup\sum_{n=1}^\infty|\langle Sf_n,f_n\rangle|^p.$$
If $\{f_n\}$ is a frame with upper frame bound $1$, then by the norm estimate for
$A$ in Lemma~\ref{1} we have $\|f_n\|\le1$ for every $n$, which together with
Theorem~\ref{2} gives
$$\sum_{n=1}^\infty|\langle Sf_n,f_n\rangle|^p\le\sum_{n=1}^\infty
\|Sf_n\|^p\le\|S\|_p^p.$$
This shows that
$$\sup\sum_{n=1}^\infty|\langle Sf_n,f_n\rangle|^p\le\|S\|_p^p,$$
and completes the proof of the theorem.
\end{proof}

Note that the second assertion in Theorem~\ref{7} concerning the norm of $S$ in $S_p$ is
false for operators that are not necessarily self-adjoint. A counter-example can be found
on page 22 of \cite{Zhu}. Nevertheless, using the fact that every operator $T$ admits a
canonical decomposition $T=T_1+iT_2$ with
$$|\langle Tf,f\rangle|^2=|\langle T_1f,f\rangle^2+|\langle T_2f,f\rangle|^2,$$
where
$$T_1=\frac{T+T^*}2,\qquad T_2=\frac{T-T^*}{2i},$$
are self-adjoint, we easily show that there still exists a positive constant $C$ such that
$$C^{-1}\|T\|_p^p\le\sup\sum_{n=1}^\infty|\langle Te_n,e_n\rangle|^p\le C\|T\|_p^p$$
for all operators $T\in S_p$, where the supremum is taken over all orthonormal bases
$\{e_n\}$. See the second part of the proof of Theorem 1.27 in \cite{Zhu}. Similarly, we have
$$C^{-1}\|T\|_p^p\le\sup\sum_{n=1}^\infty|\langle Tf_n,f_n\rangle|^p\le C\|T\|_p^p$$
for all operators $T\in S_p$, where the supremum is taken over all frames $\{f_n\}$ with
upper frame bound less than or equal to $1$.

If we remove the a priori assumption that $T$ be compact, we obtain the following slightly
different version of Theorem~\ref{7}.

\begin{thm}
If $1\le p<\infty$ and $S$ is a bounded linear operator on $H$, then the following
conditions are equivalent:
\begin{enumerate}
\item[(i)] $S\in S_p$.
\item[(ii)] There exists a positive constant $C$ such that
$$\sum_{n=1}^\infty|\langle Se_n,e_n\rangle|^p\le C$$
for \emph{every} orthonormal basis $\{e_n\}$.
\item[(iii)] There exists a positive constant $C$ such that
$$\sum_{n=1}^\infty|\langle Sf_n,f_n\rangle|^p\le C$$
for \emph{every} frame $\{f_n\}$ with upper frame bound less than or equal to $1$.
\end{enumerate}
\label{8}
\end{thm}

\begin{proof}
This follows from Theorem~\ref{7}, the remarks immediately following Theorem~\ref{7},
and the same approximation argument used in the proof of Theorem~\ref{4}.
\end{proof}

\begin{prop}
If $1<p<\infty$, $\varepsilon>0$, and $S\in S_{p+\varepsilon}-S_p$ is positive, then
there exists \emph{some} frame $\{f_n\}$ such that $\{\langle Sf_n,f_n\rangle\}\in\ell^p$.
\label{9}
\end{prop}

\begin{proof}
Write $S=T^*T$, where $T=\sqrt S$. Then $T\in S_{2p+2\varepsilon}-S_{2p}$ and
$\|Tf_n\|^{2p}=\langle Sf_n,f_n\rangle^p$. The desired result then follows from
Proposition~\ref{5}.
\end{proof}

\begin{prop}
If $1<p<\infty$, $\varepsilon>0$, and $\{e_n\}$ is an orthonormal basis for $H$, then
there exists a positive operator $S\in S_{p+\varepsilon}-S_p$ such that
$\{\langle Se_n,e_n\rangle\}\in\ell^p$.
\label{10}
\end{prop}

\begin{proof}
By Proposition \ref{6}, there exists an operator $T\in S_{2p+2\varepsilon}-S_{2p}$
such that $\{\|Te_n\|\}\in\ell^{2p}$. Let $S=T^*T$. Then $S\in S_{p+\varepsilon}-S_p$
and the sequence $\langle Se_n,e_n\rangle=\|Te_n\|^2$ belongs to $\ell^p$.
\end{proof}

Next we proceed to the characterization of operators in Schatten classes in terms of
the double-indexed sequence $\{\langle Tf_n,f_k\rangle\}$. We need the following lemma.

\begin{lem}
For any frame $\{f_n\}$ in $H$ there exist positive constants $c$ and $C$ with the
following properties.
\begin{enumerate}
\item[(a)] If $2\le p<\infty$, then
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\le C\sum_{n=1}^\infty
\|Tf_n\|^p$$
for all bounded linear operators $T$ on $H$.
\item[(b)] If $0<p\le2$, then
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\ge c\sum_{n=1}^\infty
\|Tf_n\|^p$$
for all bounded linear operators on $H$.
\end{enumerate}
\label{11}
\end{lem}

\begin{proof}
The desired estimates follow from H\"older's inequality and the definition of frames.

For $2\le p<\infty$, we have $0<2/p\le1$, so
$$\left[\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\right]^{\frac2p}
\le\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^2\le C_1\|Tf_n\|^2,$$
where $C_1$ is upper frame bound for $\{f_n\}$. It follows that for $C=C_1^{p/2}$ we have
$$\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\le C\|Tf_n\|^p,\qquad n\ge1,$$
so that
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p
\le C\sum_{n=1}^\infty\|Tf_n\|^p.$$

Similarly, for $0<p\le 2$, we have $0<p/2\le1$, so
\begin{eqnarray*}
\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p&=&
\sum_{n=1}^\infty\sum_{k=1}^\infty\left(|\langle Tf_n,f_k\rangle|^2\right)^{\frac p2}\\
&\ge&\sum_{n=1}^\infty\left[\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^2\right]^{\frac p2}\\
&\ge&C_2^{p/2}\sum_{n=1}^\infty\left[\|Tf_n\|^2\right]^{\frac p2}\\
&=&c\sum_{n=1}^\infty\|Tf_n\|^p,
\end{eqnarray*}
where $C_2$ is the lower frame bound for $\{f_n\}$ and $c=C_2^{p/2}$.
\end{proof}

It is clear that if $\{f_n\}$ happens to be an orthonormal basis, then both $C$ and $c$ can
be taken to be $1$ in Lemma~\ref{11}.

\begin{thm}
Suppose $T$ is a compact operator on $H$ and $2\le p<\infty$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] The condition
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p<\infty$$
for \emph{every} orthonormal basis $\{e_n\}$ in $H$.
\item[(c)] The condition
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p<\infty$$
holds for \emph{every} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, there exists a positive constant $c$ such that
$$c\|T\|_p^p\le\sup\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p
\le\sup\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\le\|T\|_p^p,$$
where the first supremum is taken over all orthonormal bases $\{e_n\}$ and the second
supremum is taken over all frames $\{f_n\}$ with upper frame bound less than or equal
to $1$.
\label{12}
\end{thm}

\begin{proof}
That condition (a) implies (c) follows from Theorem~\ref{2} and part (a)
of Lemma~\ref{11}. Since every orthonormal basis is a frame, it is trivial that
condition (c) implies (b).

It remains to show that condition (b) implies (a). So we assume that condition (b) holds for
an operator $T$. It is clear that condition (b) also holds for $T^*$, which implies that
condition (b) holds for $T+T^*$ and $T-T^*$ as well. Write
$T=T_1+iT_2$, where $T_1=(T+T^*)/2$ and $T_2=(T-T^*)/(2i)$, and apply condition (b) to the
self-adjoint operators $T_1$ and $T_2$, we may as well assume that $T$ is already self-adjoint.

But if $T$ is self-adjoint, its canonical decomposition takes the form
$$Tx=\sum_{n=1}^\infty\lambda_n\langle x,e_n\rangle\,e_n,$$
where $\{\lambda_n\}$ is the singular value sequence of $T$ and $\{e_n\}$ is an orthonormal
set. If $\{\sigma_n\}$ is an orthonormal basis for $\ker(T)$, then $\{e_n'\}=\{e_n\}\cup
\{\sigma_n\}$ is an orthonormal basis for $H$. Therefore, it follows from condition (b)
and the relation $\ran(T)^\perp=\ran(T^*)^\perp=\ker(T)$ that
$$\sum_{n=1}^\infty\lambda_n^p=\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p
=\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n',e_k'\rangle|^p<\infty.$$
The first norm estimate follows from the decomposition $T=T_1+iT_2$ of $T$ into a linear
combination of self-adjoint operators and the canonical decomposition of self-adjoint
compact operators. The second norm estimate is trivial. The third norm estimate follows
from Theorem~\ref{2} and part (a) of Lemma~\ref{11}.
\end{proof}

Note that if $T$ is self-adjoint, then the proof above actually shows
$$\|T\|_p^p=\sup\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p
=\sup\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p.$$
We are not sure if this holds for general operators as well.

Once again, if we do not make the a priori assumption that $T$ be compact, then
Theorem~\ref{12} should be modified as follows.

\begin{thm}
For $2\le p<\infty$ and any bounded linear operator $T$ the following conditions
are equivalent:
\begin{enumerate}
\item[(i)] $T\in S_p$.
\item[(ii)] There exists a positive constant $C$ such that
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p\le C$$
for \emph{every} orthonormal basis $\{e_n\}$.
\item[(iii)] There exists a positive constant $C$ such that
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\le C$$
for \emph{every} frame $\{f_n\}$ with upper frame bound less than or equal to $1$.
\end{enumerate}
\label{13}
\end{thm}

\begin{proof}
This follows from Theorem~\ref{12} and the approximation argument used in the proof
of Theorems~\ref{4} and \ref{8}.
\end{proof}

\begin{prop}
Let $2<p<\infty$, $\varepsilon>0$, and $T\in S_{p+\varepsilon}-S_p$. There exists a
frame $\{f_n\}$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Tf_n,f_k\rangle|^p<\infty.$$
\label{14}
\end{prop}

\begin{proof}
This follows from Proposition~\ref{5} and part (a) of Lemma~\ref{11}.
\end{proof}

\begin{prop}
Suppose $2<p<\infty$, $\varepsilon>0$, and $\{e_n\}$ is an orthonormal basis for $H$.
Then there exists an operator $T\in S_{p+\varepsilon}-S_p$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Te_n,e_k\rangle|^p<\infty.$$
\label{15}
\end{prop}

\begin{proof}
This follows from Proposition~\ref{6} and part (a) of Lemma~\ref{11}.
\end{proof}

\section{The case when $p$ is small}

We begin this section with a simple example to show that the characterizations for
operators in Schatten classes $S_p$ obtained in the previous section for $2\le p<\infty$
are not true for the range $0<p<2$.

Fix any orthonormal basis $\{e_n\}$ and consider the vector
$$h=\sum_{n=1}^\infty\frac{e_n}{\sqrt n\,\log(n+1)}$$
in $H$. Define a rank one operator $T$ on $H$ by $Tx=\langle x,h\rangle\,h$. We have
$$Te_n=\langle e_n,h\rangle\,h=\frac h{\sqrt n\,\log(n+1)},\qquad n\ge1.$$
It follows that
$$\sum_{n=1}^\infty\|Te_n\|^p=\|h\|^p\sum_{n=1}^\infty\frac1{[\sqrt n\,\log(n+1)]^p}
=\infty$$
for any $0<p<2$. This shows that the characterizations obtained in Theorem~\ref{2}
are no longer true for \emph{any} $0<p<2$.

Later in this section we will actually show that for \emph{any} operator $T\in S_p$,
$0<p<2$, there exists a frame $\{f_n\}$ such that $\{\|Tf_n\|\}\not\in\ell^p$.
Nevertheless, there is still a nice characterization for operators in $S_p$, $0<p\le2$,
in terms of orthonormal bases and frames.

\begin{thm}
Suppose $T$ is a positive operator on $H$ and $0<p\le1$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] $\{\langle Te_n,e_n\rangle\}\in\ell^p$ for \emph{some} orthonormal
basis $\{e_n\}$ in $H$.
\item[(c)]$\{\langle Tf_n,f_n\rangle\}\in\ell^p$ for \emph{some} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, we have
\begin{eqnarray*}
\|T\|^p_p&=&\inf\sum_{n=1}^\infty|\langle Te_n,e_n\rangle|^p\\
&=&\inf\sum_{n=1}^\infty\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p\\
&=&\inf\sum_{n=1}^\infty\langle Tf_n,f_n\rangle^p,
\end{eqnarray*}
where the first infimum is taken over all orthonormal bases, the second
infimum is taken over all frames with lower frame bound greater than
or equal to $1$, and the third infimum is taken over all Parseval frames.
\label{16}
\end{thm}

\begin{proof}
If $T$ is positive and in $S_p$, then its canonical decomposition takes the form
$$Tx=\sum_{n=1}^\infty\lambda_n\langle x,\sigma_n\rangle\,\sigma_n,$$
where $\{\lambda_n\}\in\ell^p$ is the singular value sequence of $T$ and
$\{\sigma_n\}$ is an orthonormal set in $H$. Since each $\lambda_n$ is positive, we
have $Tx=0$ if and only if $\langle x,\sigma_n\rangle=0$ for every $n$. Therefore,
$\ker(T)=\{\sigma_n\}^\perp$. If $\{\sigma_n'\}$ is an orthonormal basis for $\ker(T)$,
then $\{e_n\}=\{\sigma_n\}\cup\{\sigma_n'\}$ is an orthonormal basis for $H$. Since
$T(\sigma_n)=\lambda_n \sigma_n$ for every $n$ and $\{\lambda_n\}\in\ell^p$, we have
$$\sum_{n=1}^\infty|\langle Te_n,e_n\rangle|^p=\sum_{n=1}^\infty|\langle T\sigma_n,
\sigma_n\rangle|^p=\sum_{n=1}^\infty\lambda_n^p=\|T\|_p^p<\infty.$$
This shows that condition (a) implies (b).

To prove that condition (b) implies (a), we use Theorem 1.26 and part (b) of
Proposition 1.31 in \cite{Zhu}. More specifically,
$$\|T\|_p^p=\|T^p\|_1^p=\sum_{n=1}^\infty\langle T^pe_n,e_n\rangle
\le\sum_{n=1}^\infty\langle Te_n,e_n\rangle^p<\infty$$
whenever $\{\langle Te_n,e_n\rangle\}\in\ell^p$. This shows that condition (b) implies (a),
so conditions (a) and (b) are equivalent for any positive operator on $H$ and
$$\|T\|_p^p=\inf\sum_{n=1}^\infty\langle Te_n,e_n\rangle^p,$$
where the infimum is taken over all orthonormal bases $\{e_n\}$.

Since every orthonormal basis is a frame with both upper and lower frame bounds equal
to $1$, it is trivial that condition (b) implies (c), and
$$\inf\sum_{n=1}^\infty\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p\le\inf\sum_{n=1}^\infty
\langle Te_n,e_n\rangle^p,$$
where the first infimum is taken over all frames $\{f_n\}$ with lower frame bound at
least $1$ and the second infimum is taken over all orthonormal bases $\{e_n\}$.

Finally, we assume that $\{\langle Tf_n,f_n\rangle\}\in\ell^p$ for some frame $\{f_n\}$.
Fix any orthonormal basis $\{e_n\}$, let $A$ be the operator defined in Lemma~\ref{1}, and
set $S=A^*TA$. Then $S$ is positive again and
$$\langle Se_n,e_n\rangle=\langle A^*TAe_n,e_n\rangle=\langle TAe_n,Ae_n\rangle
=\langle Tf_n,f_n\rangle.$$
Thus $\{\langle Se_n,e_n\rangle\}\in\ell^p$. By the equivalence of (a) and (b),
$S$ is in $S_p$. Since $S_p$ is a two-sided ideal in $\calL(H)$, the operator
$(AA^*)T(AA^*)=ASA^*$ also belongs to $S_p$. Multiply from both sides by $(AA^*)^{-1}$ (see
Lemma~\ref{1}), we conclude that $T$ is in $S_p$ as well. This shows that condition (c)
implies (a), and completes the proof of the equivalence of (a), (b), and (c).

We now proceed to prove that
$$\|T\|_p^p\le\sum_{n=1}^\infty\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p$$
whenever $\{f_n\}$ is a frame with lower frame bound greater than or equal to $1$. To
this end, we fix such a frame and, without loss of generality, assume that the right-hand
side above is finite (otherwise, the desired inequality is trivial) and $f_n\not=0$
for each $n$. By part (b) of Proposition 1.31 in \cite{Zhu}, we have
\begin{eqnarray*}
+\infty&>&\sum_{n=1}^\infty\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p\\
&=&\sum_{n=1}^\infty\|f_n\|^2\left\langle T\frac{f_n}{\|f_n\|},
\frac{f_n}{\|f_n\|}\right\rangle^p\\
&\ge&\sum_{n=1}^\infty\|f_n\|^2\left\langle T^p\frac{f_n}{\|f_n\|},
\frac{f_n}{\|f_n\|}\right\rangle\\
&=&\sum_{n=1}^\infty\langle T^pf_n,f_n\rangle.
\end{eqnarray*}
By the equivalence of (a) and (c), the operator $T^p$ is in the trace class. If
$$T^px=\sum_{k=1}^\infty\mu_k\langle x,\sigma_k\rangle\,\sigma_k$$
is the canonical decomposition for $T^p$, then
$$\langle T^pf_n,f_n\rangle=\sum_{k=1}^\infty\mu_k|\langle f_n,\sigma_k\rangle|^2$$
for every $n$. Since the lower frame bound of $\{f_n\}$ is greater than or equal to $1$,
it follows from Fubini's theorem and Theorem 1.26 in \cite{Zhu} that
\begin{eqnarray*}
\sum_{n=1}^\infty\langle T^pf_n,f_n\rangle^p&=&\sum_{k=1}^\infty\mu_k
\sum_{n=1}^\infty|\langle f_n,\sigma_k\rangle|^2\\
&\ge&\sum_{k=1}^\infty\mu_k\|\sigma_k\|^2=\sum_{k=1}^\infty\mu_k\\
&=&\|T^p\|_1=\|T\|_p^p.
\end{eqnarray*}
This completes the proof that
$$\|T\|_p^p=\inf\sum_{n=1}^\infty\langle Te_n,e_n\rangle^p=\inf\sum_{n=1}^\infty
\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p,$$
where the first infimum is taken over all orthonormal bases and the second infimum is
taken over all frames with lower frame bound at least $1$.

Finally, if $\{f_n\}$ is a Parseval frame, then by the norm estimate for $A$ in Lemma~\ref{1},
we have $\|f_n\|\le1$ for every $n$. It follows that
$$\sum_{n=1}^\infty\|f_n\|^{2(1-p)}\langle Tf_n,f_n\rangle^p\le\sum_{n=1}^\infty
\langle Tf_n,f_n\rangle^p.$$
Combining this with the fact that every orthonormal basis is a Parseval frame, we obtain
$$\|T\|_p^p=\inf\sum_{n=1}^\infty\langle Tf_n,f_n\rangle^p,$$
where the infimum is taken over all Parseval frames $\{f_n\}$.
\end{proof}

Note that, unlike Theorem \ref{7}, we need to make the additional assumption that $T$ be
positive here. Without any extra assumption, Theorem~\ref{16} will be false. For
example, if $\{e_n\}$ is any fixed orthonormal basis for $H$ and $T$ is the unilateral shift
operator defined by $T(e_n)=e_{n+1}$, $n\ge1$. Then it is clear that
$\{\langle Te_n,e_n\rangle\}\in\ell^p$ for any $p>0$, but $T$ is not even compact.

As a consequence of Theorem \ref{16} we obtain the following.

\begin{thm}
Suppose $T$ is a bounded linear operator on $H$ and $0<p\le2$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] $\{\|Te_n\|\}\in\ell^p$ for \emph{some} orthonormal basis $\{e_n\}$ in $H$.
\item[(c)] $\{\|Tf_n\|\}\in\ell^p$ for \emph{some} frame $\{f_n\}$ in $H$.
\end{enumerate}
Furthermore, we have
$$\|T\|^p_p=\inf\sum_{n=1}^\infty\|Te_n\|^p=\inf\sum_{n=1}^\infty\|f_n\|^{2-p}\|Tf_n\|^p
=\inf\sum_{n=1}^\infty\|Tf_n\|^p,$$
where the first infimum is taken over all orthonormal bases, the second infimum is taken
over all frames with lower frame bound greater than or equal to $1$, and the third infimum
is taken over all Parseval frames.
\label{17}
\end{thm}

\begin{proof}
Consider $S=T^*T$ and apply Theorem \ref{16} to the operator $S$. The desired result then
follows from the identity $\langle Sf_n,f_n\rangle=\|Tf_n\|^2$ and the fact that
$T\in S_p$ if and only if $S\in S_{p/2}$.
\end{proof}

\begin{thm}
Suppose $0<p\le2$ and $T$ is a self-adjoint operator on $H$. Then the following
conditions are equivalent.
\begin{enumerate}
\item[(a)] $T\in S_p$.
\item[(b)] There exists \emph{some} orthonormal basis $\{e_n\}$ in $H$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Te_n,e_k\rangle|^p<\infty.$$
\item[(c)] There exists \emph{some} frame $\{f_n\}$ in $H$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Tf_n,f_k\rangle|^p<\infty.$$
\end{enumerate}
Moreover, we have
\begin{eqnarray*}
\|T\|_p^p&=&\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p\\
&=&\inf\sum_{n=1}^\infty\sum_{k=1}^\infty\|f_n\|^{2-p}|\langle Tf_n,f_k\rangle|^p\\
&=&\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p,
\end{eqnarray*}
where the first infimum is taken over all orthonormal bases, the second infimum is taken
over all frames with lower frame bound at least $1$, and the third infimum is taken over
all Parseval frames.
\label{18}
\end{thm}

\begin{proof}
If $T\in S_p$ is self-adjoint, then there exists an orthonormal set $\{\sigma_n\}$
such that
$$Tx=\sum_{n=1}^\infty\lambda_n\langle x,\sigma_n\rangle\,\sigma_n$$
for all $x\in H$, where $\{\lambda_n\}\in\ell^p$ is the nonzero eigenvalue sequence of
$T$. Since each $\lambda_n$ is nonzero, we see that $Tx=0$ if and only if $x\perp\sigma_n$
for every $n$. Therefore, if $\{\sigma_n'\}$ is an orthonormal basis for $\ker(T)$, then
$\{e_n\}=:\{\sigma_n\}\cup\{\sigma_n'\}$ is an orthonormal basis for $H$. Moreover,
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p=
\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle T\sigma_n,\sigma_k\rangle|^p=
\sum_{n=1}^\infty|\lambda_n|^p<\infty.$$
This shows that condition (a) implies (b).

Since every orthonormal basis is a frame, it is trivial that condition (b) implies (c).
That (c) implies (a) follows from Theorem \ref{17} and part (b) of Lemma~\ref{11}.

If $\{f_n\}$ is a Parseval frame, then by Theorem~\ref{17} and the proof for part (b)
of Lemma~\ref{11},
$$\|T\|_p^p\le\sum_{n=1}^\infty\|Tf_n\|^p\le\sum_{n=1}^\infty\sum_{k=1}^\infty
|\langle Tf_n,f_k\rangle|^p.$$
It follows that
$$\|T\|_p^p\le\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p.$$
Since every orthonormal basis is a Parseval frame, we clearly have
$$\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p\le
\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p.$$
The inequality
$$\inf\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p\le\|T\|_p^p$$
follows from the first paragraph of this proof.

It follows from the proof of Lemma~\ref{11} that we always have
$$\sum_{n=1}^\infty\|f_n\|^{2-p}\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^p
\ge\sum_{n=1}^\infty\|f_n\|^{2-p}\|Tf_n\|^p,$$
where $0<p\le2$ and $\{f_n\}$ is any frame with lower frame bound at least $1$.
Combining this with Theorem~\ref{17}, we see that
$$\|T\|_p^p=\inf\sum_{n=1}^\infty\sum_{k=1}^\infty\|f_n\|^{2-p}
|\langle Tf_n,f_k\rangle|^p,$$
where the infimum is taken over all frames with lower frame bound greater than or
equal to $1$.
\end{proof}

We are not sure if the additional assumption that $T$ be self-adjoint is necessary in
Theorem~\ref{18} above. But we can show by an example that its proof will definitely not
work if no additional assumption is placed on $T$. To see this, fix any orthonormal
basis $\{e_n\}$ and set
$$h_1=\sum_{n=1}^\infty\frac c{\sqrt n\,\log(n+1)}\,e_n,$$
where $c$ is a normalizing constant such that $\|h_1\|=1$. Expand $h_1$ to an orthonormal
basis $\{h_n\}$. Now define an operator $T$ on $H$ by
$$Tx=\sum_{n=1}^\infty\frac1{2^n}\langle x,h_n\rangle\,e_n,\qquad
T^*x=\sum_{n=1}^\infty\frac1{2^n}\langle x,e_n\rangle\,h_n.$$
It is easy to show that $T\in S_p$ for every $p>0$, but for $0<p<2$ we have
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Te_n,e_k\rangle|^p=
\sum_{n=1}^\infty\frac1{2^{np}}\sum_{k=1}^\infty|\langle h_n,e_k\rangle|^p=\infty,$$
because in this case we have
$$\sum_{k=1}^\infty|\langle h_1,e_k\rangle|^p=c^p\sum_{k=1}^\infty
\frac1{[\sqrt k\,\log(k+1)]^p}=\infty.$$

It is now natural for us to ask whether the ``converse'' of the theorems above is true.
We show that the answer is no when $p$ is not the upper end-point. The end-point case
will be discussed in the next section. The next three propositions only require the
operator $T$ to be bounded, not necessarily in $S_p$.

\begin{prop}
Let $0<p<1$ and let $T$ be any nonzero operator on $H$. Then there exists a frame
$\{f_n\}$ such that $\{\langle Tf_n,f_n\rangle\}\not\in\ell^p$.
\label{19}
\end{prop}

\begin{proof}
Since $T\not=0$, there exists a unit vector $h$ such that
$\langle Th,h\rangle\not=0$. Fix such a vector $h$ and set
$$e_n'=\frac{h}{\sqrt n\log(n+1)},\qquad n\ge1.$$
For any $f\in H$ we have
$$\sum_{n=1}^\infty|\langle f,e_n'\rangle|^2=\sum_{n=1}^\infty\frac{|\langle f,h\rangle|^2}
{n[\log(n+1)]^2}\le\|f\|^2\sum_{n=1}^\infty\frac1{n[\log(n+1)]^2}.$$
Let $\{e_n\}$ be any orthonormal basis for $H$ and let $\{f_n\}=\{e_n\}\cup\{e_n'\}$.
Since the last series above converges and
$$\sum_{n=1}^\infty|\langle f,f_n\rangle|^2=\sum_{n=1}^\infty|\langle f,e_n\rangle|^2+
\sum_{n=1}^\infty|\langle f,e_n'\rangle|^2=
\|f\|^2+\sum_{n=1}^\infty|\langle f,e_n'\rangle|^2,$$
we conclude that $\{f_n\}$ is a frame for $H$.

On the other hand, the sequence $\{\langle Tf_n,f_n\rangle\}$ contains the subsequence
$\{\langle Te_n',e_n'\rangle\}$, which is not in $\ell^p$ for $0<p<1$. In fact,
$$\langle Te_n',e_n'\rangle=\frac{\langle Th,h\rangle}{n[\log(n+1)]^2}$$
for all $n\ge1$, which clearly shows that $\{\langle Te_n',e_n'\rangle\}\not\in\ell^p$
for $0<p<1$. This shows that $\{\langle Tf_n,f_n\rangle\}$ is not in $\ell^p$ and
completes the proof of the proposition.
\end{proof}

\begin{prop}
Let $0<p<2$ and let $T$ be any nonzero operator on $H$. There exists a frame
$\{g_n\}$ such that $\{\|Tg_n\|\}\notin\ell^p$.
\label{20}
\end{prop}

\begin{proof}
Consider the operator $S=T^*T$ and use Proposition~\ref{19} to find a frame $\{f_n\}$
such that $\{\langle Sf_n,f_n\rangle\}$ is not in $\ell^{p/2}$. This is clearly the
same as $\{\|Tf_n\|\}\not\in\ell^p$.
\end{proof}

\begin{prop}
For any $0<p<2$ and any nonzero operator $T$ on $H$ there exists a frame $\{f_n\}$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Tf_n,f_k\rangle|^p=\infty.$$
\label{21}
\end{prop}

\begin{proof}
This follows from Proposition~\ref{20} and part (b) of Lemma~\ref{11}.
\end{proof}

We can also obtain versions of these propositions in terms of orthonormal bases. Note that
the operators $T$ obtained in the next three propositions are in $S_p$, not just bounded.

\begin{prop}
Suppose $0<p<1$ and $\{e_n\}$ is any orthonormal basis for $H$. Then there exists a positive
operator $S\in S_p$ such that $\{\langle Se_n,e_n\rangle\}\not\in\ell^p$.
\label{22}
\end{prop}

\begin{proof}
Fix a nonzero operator $T\in S_p$ and use Proposition~\ref{19} to find a frame $\{f_n\}$
such that $\{\langle Tf_n,f_n\rangle\}\not\in\ell^p$. Let $A$ denote the operator from
Lemma~\ref{1} and consider the operator $S=A^*TA$. Then $S$ is a positive operator in
$S_p$ and $\{\langle Se_n,e_n\rangle\}=\{\langle Tf_n,f_n\rangle\}\not\in\ell^p$.
\end{proof}

\begin{prop}
Suppose $0<p<2$ and $\{e_n\}$ is any orthonormal basis for $H$. Then there exists a positive
operator $S\in S_p$ such that $\{\|Se_n\|\}\not\in\ell^p$.
\label{23}
\end{prop}

\begin{proof}
By Proposition~\ref{22}, there exists a positive operator $T\in S_{p/2}$ such that
$\{\langle Te_n,e_n\rangle\}\not\in\ell^{p/2}$. Let $S=\sqrt T$. Then $S$ is a positive
operator in $S_p$, $\|Se_n\|^2=\langle Te_n,e_n\rangle$, and $\{\|Se_n\|\}\not\in\ell^p$.
\end{proof}

\begin{prop}
For any $0<p<2$ and any orthonormal basis $\{e_n\}$ there exists a positive operator
$S\in S_p$ such that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty|\langle Se_n,e_k\rangle|^p=\infty.$$
\label{24}
\end{prop}

\begin{proof}
This follows from Proposition~\ref{23} and part (b) of Lemma~\ref{11}. Note that the
operator constructed right before Proposition~\ref{19} is not positive.
\end{proof}

\section{The trace class and Hilbert-Schmidt class}

In this section we focus on two special classes of operators: the trace class $S_1$ and
the Hilbert-Schmidt class $S_2$. It is well known that a bounded linear operator $T$ on $H$
is in $S_2$ if and only if $\sum\|Te_n\|^2<\infty$, where $\{e_n\}$ is any given
orthonormal basis for $H$. Also, for $T\ge0$, $T\in S_1$ if and only if $\sum\langle
Te_n,e_n\rangle<\infty$. We show that these results remain true when the orthonormal
basis $\{e_n\}$ is replaced by a frame.

\begin{thm}
Suppose $T$ is a positive operator on $H$. Then the following conditions are equivalent.
\begin{enumerate}
\item[(a)] $T$ is in the trace class $S_1$.
\item[(b)] $\{\langle Tf_n,f_n\rangle\}\in\ell^1$ for \emph{every} frame $\{f_n\}$.
\item[(c)] $\{\langle Tf_n,f_n\rangle\}\in\ell^1$ for \emph{some} frame $\{f_n\}$.
\end{enumerate}
\label{25}
\end{thm}

\begin{proof}
This follows from Theorems \ref{7} and \ref{16}.
\end{proof}

An equivalent version of Theorem \ref{25} above is the following.

\begin{thm}
Let $T$ be a bounded linear operator on $H$. Then the following conditions are equivalent.
\begin{enumerate}
\item[(a)] $T$ is in the Hilbert-Schmidt class $S_2$.
\item[(b)] $\{\|Tf_n\|\}\in\ell^2$ for \emph{every} frame $\{f_n\}$.
\item[(c)] $\{\|Tf_n\|\}\in\ell^2$ for \emph{some} frame $\{f_n\}$.
\end{enumerate}
\label{26}
\end{thm}

\begin{proof}
Note that $T$ is Hilbert-Schmidt if and only if $T^*T$ is trace class. Since
$\langle T^*Tf_n,f_n\rangle=\|Tf_n\|^2$, the desired result follows from Theorem~\ref{25}.
Alternatively, the desired result follows from Theorems~\ref{2} and \ref{17}.
\end{proof}

When $\{f_n\}$ is a frame, it is clear tha the condition $\{\|Tf_n\|\}\in\ell^2$ is
equivalent to the condition
$$\sum_{n=1}^\infty\sum_{k=1}^\infty|\langle Tf_n,f_k\rangle|^2<\infty.$$
Therefore, conditions (b) and (c) in Theorem~\ref{26} above can also be stated in terms of
the double-indexed sequence $\{\langle Tf_n,f_k\rangle\}$.

\section{An application}

In this section, we consider a special class of frames in the Bergman space of the unit
disk, namely, normalized reproducing kernels induced by sampling sequences. We use this
to obtain an integral condition for a bounded linear operator on the Bergman space to
belong to the Schatten class $S_p$.

Thus we let $A^2$ denote the space of analytic functions $f$ in the unit disk $\DD$ such that
$$\|f\|^2=\ind|f(z)|^2\,dA(z)<\infty,$$
where $dA$ is area measure on $\DD$ normalized so that $\DD$ has area $1$. As a closed subspace
of $L^2(\DD,dA)$, $A^2$ is a Hilbert space. In fact, $A^2$ is a reproducing Hilbert space whose
reproducing kernel is the well-known Bergman kernel
$$K_w(z)=K(z,w)=\frac1{(1-z\overline w)^2}.$$
For any $w\in\DD$ let $k_w$ denote the function in $A^2$ defined by
$$k_w(z)=\frac{K(z,w)}{\sqrt{K(w,w)}}=\frac{1-|w|^2}{(1-z\overline w)^2}.$$
Each $k_w$ is a unit vector in $A^2$, called the normalized reproducing kernel at $w$.

A sequence $\{w_n\}$ in $\DD$ is called a sampling sequence for the Bergman space $A^2$ if
there exists a positive constant $C$ such that
$$C^{-1}\|f\|^2\le\sum_{n=1}^\infty(1-|w_n|^2)^2|f(w_n)|^2\le C\|f\|^2$$
for all $f\in A^2$. This condition can be written as
$$C^{-1}\|f\|^2\le\sum_{n=1}^\infty|\langle f,k_{w_n}\rangle|^2\le C\|f\|^2.$$
Therefore, $\{w_n\}$ is a sampling sequence for the Bergman space if and only if the
sequence $\{k_{w_n}\}$ is a frame in $A^2$. See \cite{HKZ} for the theory
of Bergman spaces, including the notions of normalized reproducing kernels and
sampling sequences. Sampling sequences for the Bergman space are characterized in \cite{Seip}.

Some results obtained in the paper can be stated in terms of sampling sequences. As one
particular example, we infer from Theorem~\ref{26} that if $\{w_n\}$ is a sampling
sequence for the Bergman space, then a bounded linear operator $T$ on $A^2$ belongs to
the Hilbert-Schmidt class $S_2$ if and only if it satisfies the condition
$$\sum_{n=1}^\infty\|Tk_{w_n}\|^2<\infty.$$
Equivalently, a positive operator $T$ on $A^2$ belongs to the trace class if and only if
$$\sum_{n=1}^\infty\langle Tk_{w_n},k_{w_n}\rangle<\infty.$$

\begin{lem}
Suppose $T$ is a bounded linear operator on $A^2$ and $0<p<\infty$. Then the function
$F(w)=\|TK_w\|^p$ is subharmonic in $\DD$.
\label{27}
\end{lem}

\begin{proof}
Without loss of generality we assume that $T\not=0$. It is then clear that the function
$F$ has isolated zeros in $\DD$. Furthermore, away from the zeros of $F$, it follows from
$$F(w)=\langle TK_w,TK_w\rangle^{\frac p2}$$
that
$$\frac{\partial F}{\partial\overline w}=\frac p2\langle TK_w,TK_w\rangle^{\frac p2-1}
\langle TK_w^*,K_w\rangle,$$
where
$$K_w^*(z)=\frac{\partial}{\partial\overline w}\frac1{(1-z\overline w)^2}=
\frac{2z}{(1-z\overline w)^3}.$$
Differentiating one more time, we obtain
\begin{eqnarray*}
\frac{\partial^2F}{\partial w\partial\overline w}&=&\frac p2\left(\frac p2-1\right)
\langle TK_w,TK_w\rangle^{\frac p2-2}\langle TK_w^*,K_w\rangle\langle TK_w,TK_w^*\rangle\\
&&\ +\frac p2\langle TK_w,T_w\rangle^{\frac p2-1}\langle TK_w^*,TK_w^*\rangle\\
&=&\frac p2\!\left[\frac p2-1\right]\!\|TK_w\|^{p-4}|\langle TK_w^*,TK_w\rangle|^2
+\frac p2\|TK_w\|^{p-2}\|TK_w^*\|^2\\
&=&\frac p2\|TK_w\|^{p-4}\left[\left(\frac p2-1\right)|\langle TK_w^*,TK_w\rangle|^2
+\|TK_w\|^2\|TK_w^*\|^2\right]\\
&\ge&\frac p2\|TK_w\|^{p-4}\left[\|TK_w\|^2\|TK_w^*\|^2-|\langle TK_w^*,TK_w\rangle|^2\right]\\
&=&\frac p2\|TK_w\|^{p-4}\left[\|TK_w\|\|TK_w^*\|+|\langle TK_w^*,TK_w\rangle|\right]\cdot\\
&&\ \cdot\left[\|TK_w\|\|TK_w^*\|-|\langle TK_w^*,TK_w\rangle|\right]\\
&\ge&0.
\end{eqnarray*}
The last inequality above is a consequence of the Cauchy-Schwarz inequality.
\end{proof}

\begin{thm}
Suppose $T$ is a bounded linear operator on $A^2$ and
$$d\lambda(w)=\frac1{(1-|w|^2)^2}\,dA(w)$$
is the M\"obius invariant area measure on $\DD$. Then the condition
$$\ind\|Tk_w\|^p\,d\lambda(w)<\infty$$
is sufficient for $T\in S_p$ when $0<p\le2$ and it is necessary for $T\in S_p$
when $2\le p<\infty$. Consequently, $T$ is Hilbert-Schmidt if and only if
$$\ind\|Tk_w\|^2\,d\lambda(w)=\ind\|TK_w\|^2\,dA(w)<\infty.$$
\label{28}
\end{thm}

\begin{proof}
The result follows from Theorem 6.6 in \cite{Zhu}, because $T\in S_p$ if and only if
the positive operator $S=T^*T$ is in $S_{p/2}$,
$$\widetilde S(w)=\langle Sk_w,k_w\rangle=\|Tk_w\|^2$$
for all $w\in\DD$, and $0<p\le2$ if and only if $0<p/2\le1$.

The proof of Theorem 6.6 in \cite{Zhu} depends on the notion of the Berezin transform and
the spectral decomposition for positive operators. Here we give an independent proof in
the case $0<p\le2$ that is based on sampling sequences and subharmonicity.

Fix a sampling sequence $\{w_n\}\subset\DD$ for the Bergman space $A^2$ such that $\{w_n\}$ is
separated in the Bergman metric $\beta$, say, $\beta(w_i,w_j)>2r$ for some positive number $r$
and all $i\not=j$. See \cite{HKZ} for the existence of such a sequence. Let $D(w_n,r)=\{z\in\DD:
\beta(z,w_n)<r\}$ denote the Bergman metric ball centered at $w_n$ with radius $r$.

By Lemma \ref{27}, the function $w\mapsto\|TK_w\|^p$ is subharmonic. It follows from the
proof of Proposition 4.13 in \cite{Zhu} that there exists a positive constant $C$,
independent of $n$ and $T$, such that
$$\|TK_{w_n}\|^p\le\frac C{|D(w_n,r)|}\int_{D(w_n,r)}\|TK_w\|^p\,dA(w)$$
for all $n\ge1$, where $|D(w_n,r)|$ is the area of $D(w_n,r)$. By Proposition 4.5 in
\cite{Zhu} and the remarks following it, we have
$$|D(w_n,r)|\sim(1-|w_n|^2)^2\sim(1-|w|^2)^2$$
for $w\in D(w_n,r)$. It follows that there exists another positive constant $C$, independent
of $n$ and $T$, such that
$$\|Tk_{w_n}\|^p\le C\int_{D(w_n,r)}\|Tk_w\|^p\,d\lambda(w)$$
for all $n\ge1$. Since the Bergman metric balls $D(w_n,r)$ are mutually disjoint, we have
$$\sum_{n=1}^\infty\|Tk_{w_n}\|^p\le C\ind\|Tk_w\|^p\,d\lambda(w).$$
The desired result now follows from Theorem \ref{17}.
\end{proof}

The ideas and results of this section clearly generalize to many other reproducing
kernel Hilbert spaces, including weighted Bergman spaces on various domains and Fock spaces
on $\CC^n$.

redpajama_set_name: RedPajamaArXiv


text:
Oral-B Pro 2 2000 vs Pro 2 2500
Last Updated: March 10, 2021 by Jon Love
Our choice: Oral-B Pro 2 2000 (view on Amazon)
2nd choice: Oral-B Pro 2 2500 (view on Amazon)
The Pro 2 2000 is the winner here.
This is because it is usually the cheaper of the 2 models.
The Pro 2 2500 does however become our choice, providing it is no more than S$10-15 more than the 2000.
The 2500 comes with a travel case which the 2000 does not. This is worth a small premium.
But, more often than not it is the 2000 that is cheaper, making that the more sensible purchase.
Given the limited differences, it is more about what is right for you and what the price is at the time of purchase.
Oral-B Pro 2 2000
99.00S$
154.00S$
The following table lists the key differences between each brush.
A more detailed list and explanation of the differences can be found beneath the comparison table.
Pro 2 2000
Recommended Retail Price (RRP) S$170 S$170
Cleaning Action 3D Oscillating Rotating & Pulsating 3D Oscillating Rotating & Pulsating
Number of Cleaning Modes 2 2
Timer & Pacer Yes Yes
Pressure Sensor Yes
(including visible) Yes
(including visible)
Battery Life (Days – Manufacturer claimed) 14+ 14+
Battery Type Li-Ion Li-Ion
Number of Brush Heads Included 1 1
Travel Case – Yes
Bluetooth – –
Position Detection Technology – –
Handle Colour(s) Dark Blue
Other Key Information – –
The Pro 2 2500 comes with a travel case in the box whereas the Pro 2 2000 does not.
The Pro 2 2500 comes in 4 colour options (Black/White, White/White, Pink/White & Black Edition colour options) compared to the 2 options of the Pro 2 2000 (Dark Blue/White & Pink/White)
Brush heads provided (subject to model).
Oral-B Pro 2 2000 Review
Please note.
There is another model called Pro 2 2900 which many have asked about how it compares to the Pro 2 2000 & 2500. CLICK HERE to see our summary and verdict.
Detailed comparison: what's the difference between the Oral-B Pro 2 2000 and Pro 2 2500?
OK, so let's be clear from the start here the differences between these two models are minimal.
It is very easy to get confused with the array of different options available from Oral-B, so you need to pay attention to ensure you are getting the right brush for you.
Technically, the Pro 2 2500 sits above the Pro 2 2000 in the range as it offers more in the box.
With the 2 2500 you get a travel case included.
A fairly basic but functional affair, the case is made from plastic.
Hinged on the left side 2 halves of the case come together to lock tight and provide a relatively strong travel case, despite the relatively limited thickness to the plastic.
Subject to the variant, the exact design of the case and colour can differ. It is either clear, white or black in colour.
4 sides of the case are usually flat, with the right and left sides curved.
The clear case is a bit more bulbous with limited flat edges.
The case holds the brush handle and up to 2 brush heads.
Functional for the backpacker to the businessperson, it does not have the level of quality and premium construction some might desire but it does the job of avoiding accidental activation and keeping the brush out of contact with other items in your bag.
Aside from the travel case, the other key difference is the handle colour.
The physical size and features of the brushes are identical, but the colours are different.
Made from a gloss white plastic, the Pro 2 2000 comes with either a blue or pink rubber grip that runs down the front of the handle, contrasting against the white plastic, The Pro 2 2500 comes with either black, white or pink rubber down the handle.
Hopefully, this all makes sense so far.
Another slight difference is the brush head that comes provided. Both models will come supplied with 1 brush head as standard.
However, there are different package options where the brush head provided differs.
As standard, it is a CrossAction head that is included, but UltraThin and 3D white heads can be included too.
The differences between the brush heads are explained in our ultimate Oral-B brush head guide, if you would like to know more.
As a result of the case being included, it is generally the Pro 2 2500 is more expensive than the Pro 2 2000.
The Pro 2 2500 has a recommended retail price of S$170, just like the Pro 2 2000.
Both brushes typically sell with a 25% discount at around S$130.
Prices do vary, but at times the difference in price can be minimal at other times much more significant.
Either are good brushes and it's personal opinion which is right or best for you.
S$10-15 extra is worth it for the case.
However, more often than not the Pro 2 2000 is going to be the cheaper option.
You can purchase different brush heads and even the case separately if you require, so you are not limited should you decide in the future you want a slightly different configuration.
With the key differences covered I want to reiterate how the brushes are so very similar.
Both have 2 cleaning modes, Daily Clean and Sensitive.
A timer, quadpacer and pressure sensor are built into both handles.
2 weeks battery life is possible from the built in lithium-ion battery.
The handle is water resistant and comes with a 1 year warranty.
No there is no difference. Technically they are the same from the perspective of cleaning abilities so to claim one cleans better than the other is unjustified.
With the right technique and regular cleaning, you will leave your teeth feeling and looking healthier, but if you are using the Pro 2 2000 or 2 2500 on the daily clean or sensitive mode, there is no known difference in their actual results.
Despite the same retail price and both being subject to a discount of around 25%, because the Pro 2 2000 is more widely stocked, it is normally the cheaper of the 2 models.
At times the price of the 2500 is lower, but this is not always true.
The 2500 is worth more because of the travel case included.
You can purchase the travel case separately, this would normally be anywhere between S$15-20 so spending $10-15 extra to have this included in the box I think is worth it.
Thus if you think you would like or benefit from a travel case it might be worth paying a few dollars more over and above the Pro 2 2000 to benefit from the case.
However, given that the 2000 is sold more widely the prices tend to be more competitive and more significantly discounted, making it the more sensible purchase.
I generally recommend the Pro 2 2000 as the one to go for as for most this is the better buy.
As I have pointed out though with so little between them the Pro 2 2500 sometimes will be the better buy. Particularly if you want the travel case.
Ultimately, choose the brush you are most happy with and pay a price you are content with.
Oral-B Pro 2 2900 vs Pro 2 2000/2500
As the article above has highlighted, the fundamental difference between the Pro 2 2000 and the Pro 2 2500 is the inclusion of the travel case with the 2500.
However, to complicate things further there is also a Pro 2 2900 within the Oral-B range. You may have come across it or been offered it as an option for you. No doubt you are wondering how it compares. Let me explain.
There is not easy answer here really.
If the decision was on price alone, then the Pro 2 2900 would be our overall choice, because although it is more expensive to buy, when you factor in that for the higher price you get 2 brush handles, it actually works out better value per handle.
However, it is not necessarily as black and white as this and depends on your personal circumstances to some extent.
If you need or want dual handled package then the 2900 can be a really good value purchase.
But for most it will be the Pro 2 2000 as for most, this is the better buy.
This is because unless you need 2 brush handle, there is little point buying the 2900.
You might think you can give the other handle to a friend or relative, you certainly could. However, they would then need to buy a charging stand for it, if they did not have one.
Even if you do need 2 brush handles, the 2000 can be a more attractive option, because if you buy 2 of the 2000 models you will get 2 charging stands compared to only a single charging stand with the 2900.
Lets assume you are a couple living together and you buy the 2900. It would be cheaper to buy outright, but if you went on holiday, neither of you have a travel case to store the toothbrush handle and head in.
If you needed to charge the toothbrush, you have to share a charging stand, whereas the Pro 2000 comes with 1 with each handle and that useful travel case.
The typical price for the 2900 appears to be around S$199 compared to the $130 for the 2000. That would make 2 of the 2000 S$60, $15 more expensive than the 2900. However, factor in the cost for an extra charging stand, should you need it and the difference is now not so great.
To conclude, both brushes are technically the same. The cleaning performance is identical, its what you get in the box and the value that each offers you. You need to decide what is important.
Pro 2 2000 vs Pro 2 2900
The Pro 2 2900 has 2 brush handles in the box whereas the Pro 2 2000 has 1.
The Pro 2 2900 comes with either 2 x Black/White handles or 1 x Pink/White and 1 x Black/White. The Pro 2 2000 comes with 1 x Dark Blue/White or 1 x Pink/White handle.
Pro 2 2900 comes with 2 x CrossAction brush heads. Pro 2 2000 comes with 1 x CrossAction or 1 x UltraThin head, subject to variant.
The Pro 2 2900 comes with 1 charging stand to shared between the 2 brush handles.
The Pro 2 2000 has a retail price of S$170 compared to $240 of the 2 2900. Both discounted by around 25%+. Price per handle generally cheaper with Pro 2 2900.
The Pro 2 2900 comes with either 2 x Black/White handles or 1 x Pink/White and 1 x Black/White. The Pro 2 2500 comes with 1 x Black/White, 1 x White/White or 1 x Pink/White handle.
Pro 2 2900 comes with 2 x CrossAction brush heads. Pro 2 2500 comes with 1 x CrossAction or 1 x 3D White or 1 x UltraThin head, subject to variant.
The Pro 2 2500 comes with a travel case.
Oral-B Genius X Review
Philips Sonicare 9900 Prestige Review
Sonicare vs Oral-B: 2023 Comparison
Oral-B Pro 1000 Review

redpajama_set_name: RedPajamaCommonCrawl


text:
Nobody can go scot-free as far as I'm concerned
President Buhari said no person accused of corruption can go scot-free even if they join his party, APC. The President stated this during a recent interview with NTA .
When asked about the prospect of suspected corrupt politicians going scot-free by simply defecting to the governing APC, President Buhari responded "They cannot"
He added
"And nobody can go scot-free as far as I'm concerned. A lot of confusion could have been caused, but luckily, I think we try to occupy offices with integrity and that is the best thing for Nigerian leadership at all levels.
If that is done, we would achieve a better result and if that is not done, a lot of time would be wasted talking and quarrelling and a lot of noise."
He said the APC tries to insist on accountability and, that the party is very serious about people who aspire to lead at whatever level to be people of integrity.
The President said further;
"The problem is that, we have to work within the system. Even if we suspect individuals, unless there is evidence, there isn't much you can do. So it is the system that permits people to be corrupt."
Many Nigerians believe that opposition politicians accused of corruption or on trial for corruption join the APC to avoid being prosecuted.
Movie Review: Rancor; A definition of everything wrong in a film
MC Oluomo's ex-lover Ehi Ogbebor knocks troll after saying her 'juju' will soon expire on her new man
Manchester City star, Riyad Mahrez 'secretly marries Taylor Ward in a traditional Muslim ceremony


Practices of experimentation lie at the heart of creative research and teaching in higher education in arts. The Department of Art & Media at Zurich University of the Arts offers a unique teaching and research environment as a laboratory of converging and diverging practices of experimentation. Its Bachelor and Master's programs are supported by two research institutes within the department, the Institute for Contemporary Art Research (IFCAR) and the Institute for Critical Theory (ith). Practices of Experimentation investigates how the different fields of fine arts, photography, media arts and theory interlace with each other, inspire and differentiate one and another. The book presents 15 positions in text, image, video and sound by theorists and artists. They enquire how practices of experimentation constitute one of the most advanced approaches to research and teaching in arts worldwide.
They ask how practices of experimentation are able to unfold, take position and enquire current discourses on artistic creation, the relation between art schools and society, the specific production of knowledge in the arts and the particularities of inter- and trans-disciplinary teaching and research in the arts. Contains essays by Essays by Ute Meta Bauer, Maria Eichhorn, Knowbotic Research, Jorg Huber, Marianne Muller, Gerald Raunig, Nils Roller, Richard Wentworth. With a foreword by Giaco Schiesser and Christoph Brunner.
Buy Practices of Experimentation by Dep of Art & Media, Zurich University from Australia's Online Independent Bookstore, Boomerang Books.
Giaco Schiesser read philosophy and German literature at the Freie Universitat Berlin. He conceived and established as its head the new department New Media at the School of Art and Design Zurich in 1997-2002. He is professor for the theory and history of media and culture and director of the Department of Art & Media at the Zurich University of the Arts. Christoph Brunner is artistic and research assistant to the head of the Department in Art & Media at Zurich University of the Arts and doing his PhD on research-creation at Concordia University in Montreal. He is also a member of SenseLab (www.senselab.ca), co-editor of Inflexions (www.inflexions.org) and researcher at the Institute for Critical Theory (www.ith-z.ch).


But sharks are in a global decline. Over fishing has reduced many shark populations around the world to levels that threaten their continued existence. Shark numbers have fallen by more than 80% in many cases, and the continued existence of some species is at immediate risk in some regions.
Certified as a (Junior) Open Water Diver or have a qualifying certification from another training organization.


Locations wanted to make spanking films in.
One of the perennial problems that all film producers have is finding good locations to make our spanking movies in.
We want to make realistic depictions of domestic and schoolgirl spankings but finding good and varied locations takes up a lot of our time.
If you think you have a suitable location, either your home or access to a business or other establishment that you think would make a good spanking film venue then please let us have details and we shall get back to you.
We are looking for good quality, clean, locations with ideally several usable rooms which are large enough for us to install our temporary flash and film lighting into. Ideally rooms should be at least 16ft x 16ft or larger, the bigger the better. We need nearby safe parking for at least 2-3 cars and lift access if above one floor up from the entrance – the kit we use is heavy and we are not getting any younger!
We do not expect you to do this out of the goodness of your heart so we offer competitive location hire fees and you also get to see the films being made ….. and you never know, you may even get to star in a film!


text:
Aleksandr Iúrievitx "Xaixa" Kaleri (; nascut a Jūrmala, Letònia, 13 de maig de 1956) és un cosmonauta veterà rus d'estades prolongades a l'estació espacial Mir i l'Estació Espacial Internacional (ISS). Kaleri recentment ha estat a l'espai a bord de la ISS servint com a enginyer de vol de les missions de llarga durada Expedició 25/26. És la segona persona que ha passat més temps a l'espai que qualsevol altra persona no nascuda en el que ara és Rússia.

Referències

Enllaços externs

 Spacefacts biography of Aleksandr Kaleri

Cavallers de la Legió d'Honor
Persones de Jūrmala
Tripulants de l'Estació Espacial Internacional
Astronautes soviètics

redpajama_set_name: RedPajamaWikipedia


text:
Q: How to Solve Button Alignment Issue? Recently I updated Ultimate Blocks plugin on one of my WordPress website and right after the update, all the buttons that I used aligned left automatically.
I want to align all buttons to center, but the problem is that I have to edit all posts manually which would take a lot of time. Is there any way to do it in once?
Problem screenshot:

Issue page: https://icemakercage.com/best-ice-makers/
Almost all the posts published before the update has the same issue. New ones are fine.

A: The class of the container of your button is "align-button-left"
You can change this class to "align-button-center" and the modification will be effective on all of your button
In jQuery you can do this :
$('.ub-buttons').removeClass('align-button-left')
$('.ub-buttons').addClass('align-button-center')

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Home Football Some Division Two clubs have better visions than many Premier League clubs,...
Some Division Two clubs have better visions than many Premier League clubs, says Frimpong Manso
The Porcupine Warriors former boss has reiterated that he is currently better off coaching in the lower tier than taking up a role at a top club which lacks vision
Former Asante Kotoko coach Frimpong Manso has insisted that he is currently satisfied with his role as the head coach of Division One League side Nkoranza Warriors and not itching to take up a challenge in the top-flight.
The experience gaffer insists that several top-flight clubs lack proper structure and vision which will help an ambitious coach to achieve his aim.
"I don't always consider the league a club I'm coaching plays in because many times I consider the vision the club that is seeking my help has before I join," Manso told Otec FM.
"Many at times, some of the Division Two clubs [playing in the third tier] here in Ghana have better vision and are important than some of the clubs in the Premier League," he added.
Manso would be hoping to do the unthinkable in Zone One by guiding Nkoranza Warriors to qualify to the Premier League ahead of Tamale City FC, B.A United, Bofoakwa Tano, Berekum Arsenals and even Techiman City FC.
Frimpong Manso
Nkoranza Warriors
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New 'Bold And The Beautiful' Spoilers For April 26, 2019 Episode Revealed
Happy Thursday, "Bold And The Beautiful" fans. Today's April 25,2019 episode featured some pretty interesting scenes. We saw a new Bill and Katie scene today, and man did it end with a shocker! Bill did a lot of sweet-talking and charming with Katie. He talked to her about what young Will wants the most, which is a fully intact family and so on and so forth.
Katie told Bill, "Will has that. So, he should be very happy." Then Bill talked about how he's on a couple of the most eligible Bachelor's lists, but he doesn't feel like he deserves it or something like that. He said a lot of people don't like him once they really get to know him.
From there, Katie gave Bill encouraging words saying, "I've known you for a while, and I like you." Shortly after that, Bill pulled out an engagement ring and tried to propose to Katie! I was like, "What? Did I miss something? I know they have deep history, but this is still too soon!"
Anyways, unfortunately for Bill, Katie turned him down. She basically told Bill that he's hurt her too much in the past. At one point, she told Bill, "Do you know how much that hurt me when you left me to be with my sister? I'm sorry I can't give you the answer you wanted."
Later on, Bill was seen mingling with Florence aka Flo's mother Shauna at the Il Giardino restaurant. They seem to get along quite well too. At one point, Shauna told Bill,"Now that I've gotten to talk to you, you don't seem so bad." It definitely seems like the producers could be plotting for Shauna to be Bill's next hook up. We'll see.
Elsewhere, Flo and Shauna had new scenes. Flo brought up that she should tell Hope about her stolen baby, but Shauna wasn't having it. Shauna kept reminding Flo how good of a future she may have since she's a Logan. Shauna also told Flo more about her father Storm. Shauna couldn't stop bragging about how great a man she thought he was.
Brooke had a new scene with Zoe. During it, Brooke revealed to Zoe that Flo was Storm's daughter, and she's a Logan. Zoe tried to take the angle that Flo might try to take advantage of their family, but Brooke expressed that she didn't think so. She thinks Flo and Shauna are just wonderful people.
Katie told Brooke about Bill's proposal. Brooke was all for it, hoping Katie said yes. Katie quickly told her she said no.
Today's episode capped off with Zoe confronting Flo about what she plans to do next. Flo told her she still wants to tell Hope about her stolen baby! Zoe said, "They're going to hate you if you do." Flo said, "They might forgive me for Storm's sake." The scene closed out with Zoe looking like she was about to go off, and we're about to get more into that.
Tomorrow's new week-ending, Friday, April 26, 2019 episode sounds like it will feature some very dramatic and interesting scenes as Brooke and Donna talk to Katie about Bill. Flo and Zoe's intense scene continues, and more!
We'll go ahead and start off today's spoiler session with the Brooke, Donna and Katie situation. It turns out that both Brooke and Donna will have a sitdown chat with Katie to work her into giving Bill yet another shot at marriage!
CBS' official description tells us, "Mortified that Katie said "no", Brooke and Donna convince her to give Bill a second chance." An alternate description reads like this, "Brooke and Donna learn that Katie told Bill 'no' and urge her to give him another chance."
Will Katie heed their words and give Bill another shot? That's the huge question for this particular situation. we do know that Katie is going to see Bill chatting it up with Shauna at the Il Giardino restaurant at some point in a future episode. So, I can't imagine that would exactly help things. We'll have to wait and see.
The second spoiler teaser reveals that Flo and Zoe's very intense scene will get even more intense! These two are going to end up physically brawling at some point because Flo makes it perfectly clear to Zoe that she wants to tell Hope the truth about her stolen baby!
CBS' description tells us, "Flo and Zoe wrestle with disastrous results as Flo prepares to tell Hope the truth about Beth." An alternate description reads like this, "Flo and Zoe clash as Flo prepares to tell Hope the truth about her baby."
We previously reported about this. At some point, Shauna will barge in on this fight and pull out her gun! So, that will be extremely wild and crazy! We recently got a new, official spoiler teaser for the aftermath of this scene.
It turns out that Flo,Zoe and Shauna will make a pact of some sort. So, it sounds like this crazy situation will calm down and no one will get shot. However, the whole pact thing disturbs me because it sounds like that could mean they're going still keep Hope's stolen baby a secret and never end this freaking storyline. We'll have to wait and see.
Alright guys. That's all the spoiler scoops we were able to track down for tomorrow's new April 26, 2019 episode, but definitely expect to see a lot of other scenes play out that were not mentioned in this spoiler session, especially since this is the week-ending episode.
The "Bold And The Beautiful" airs every Monday through Friday at 12:30 pm central standard time on CBS.
Be sure to like or follow us on our "Bold And The Beautiful" Facebook page for more Exclusive "Bold And The Beautiful" news and updates by Clicking Here.
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News / TV News & Spoilers
Tags: Bold And The Beautiful TV Show News
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This is a bicameral parliament. Switch to theSenate
At the first sitting of the chamber
When the member is sworn in
On the last day of the chamber's term (including in the case of early dissolution)
The Speaker is designated from outside parliament and becomes a member of parliament ex officio
Salaries and allowances
Basic salary The figures are the gross amount before tax in national currency.
Basic salary in PPP (Purchasing Power Parity) dollars The figures were provided by parliaments in national currency and converted to PPP dollars using World Bank conversion tables.
24215.25Compare data of this field.
Names of the basic allowances
travel, duty, entertainment, consituency
Members pay income tax on salaries Salaries and/or allowances may be subject to income tax or equivalent.
Members pay income tax on allowances Salaries and/or allowances may be subject to income tax or equivalent.
Members can have remunerated employment outside parliament Certain parliaments allow MPs to exercise other remunerated employment outside parliament.
Members are required to declare income from other employment to parliament In certain countries, MPs exercising other employment may be required to declare their income.
Facilities, services and allowances for constituency work
Personal staff to work directly for the parliamentarian The personal staff to work directly for the parliamentarian is different from parliamentary staff work for the secretariat of Parliament.
Office at parliament
Constituency office Certain countries provide for an office at constituency if it's different from a seat of parliament, which is often in the capital. The office at constituency is applicable mainly to the countries using the Plurality/Majority system of elections.
Other cash or in-kind allowances for constituency work The common in-kind allowances for constituency work include those for travel, overnight stay, etc.
Additional salaries and allowances for leadership positions
List of salaries and allowances for leadership positions
Speaker yes no
Committee Chair no no
Other leadership positions that receive additional salaries or allowances Some parliaments provides for additional salaries or allowances to the Deputy Speaker/Vice President, heads of parliamentary groups, etc.
Parliament determines the amount of salaries Some parliaments determine the amount of salaries of MPs. The amount may be set by other institution than parliament.
Name of the institution outside parliament which determines the amount This question is only relevant if the amount is set by other institution than parliament.
Parliament determines the amount of allowances Some parliaments determine the amount of allowances of MPs. The amount may be set by other institution than parliament.
The parliamentary salary is determined in reference to another salary scale In some countries, the amount of parliamentary salary is determined in reference to another salary scale, such as civil service salary scale, ministerial salary or the minimum wage of the country.
The data valid for the year
Can be a member of parliament
Number of members of the parliament/chamber who are also members of the government
Historical data for Number of members of the parliament/chamber who are also members of the government
Year Number of members of the parliament/chamber who are also members of the government


This is a 43 years old African American mother of four from Los Angeles, whose tummy tuck before and after pictures are shown here. According to her, multiple pregnancies left her with unsightly stretch marks and a hanging pannus. She research tummy tuck surgery on Google and looked for board certified plastic surgeons in Los Angeles who had experience with doing cosmetic surgery for African American women. She found Dr. Younai's before and after photos very impressive and liked the fact that she was able to see other women with similar body built on his web site. On her cosmetic surgery consultation, she asked many questions about different types of tummy tuck techniques and which on would be the best for her. Dr. Younai and her staff spent time to instruct her about her pre-operative, as well as, her post-tummy tuck recovery tips. She was able to recover with two weeks of her abdominoplasty and return to her work as a teacher. Looking back, she is very happy about her decision to have body-contouring surgery, and is grateful with care that she received at the California Center for Plastic Surgery.


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Indigenous foodways
"When you are wanting something to grow, you need to start with the children." – Learning Circle Participant In 2019 F2CC, in partnership with the Social Planning & Research Council of BC, provided a Local Food to School Learning Circle Grant to Q'wemtsín Health Society (QHS) to support Sk'elep School of Excellence, from Tk'emlups te...
Learning from the Land – A written compilation of resources and stories
November 18, 2021F2CC adminEnglish, Farm to School, Featured Resources, Home Page Blog Feed
"Learning from the Land: Resources and stories from K-12 schools engaging with Indigenous plants and pedagogy", published in November 2021 by our BC project partner Public Health Agency of BC (PHABC), is a written compilation of resources and stories highlighting programs related to Indigenous plants and pedagogy within schools located in what is colonially known...
Food Forests — edible landscapes that do more than feed us
In celebration of this year's Farm to School Month, which is focusing on land-based learning, we wanted to dive into the benefits of food forests, and why schools and communities across the country are putting them in place. In this CBC article, for instance, Port Elgin Regional School in Ontario highlighted how they "worked with...
F2S Month 2021: Celebrating Land-Based Learning
October 1, 2021F2CC adminHome Page Blog Feed
Farm to school, also known as local food to school, is grounded in re-learning, re-establishing and acknowledging and supporting existing place-based food systems. As F2CC works to advance food systems rooted in place, and as we work to better honour Indigenous ways of knowing and being, we were pleased to share that during Farm to School...
Traditional Foods and Indigenous Recipes in B.C.'s Public Institutions
June 9, 2021F2CC adminEnglish, Home Page Blog Feed
This report, prepared by Queenswood Consulting Group for the Government of British Columbia, was developed to support the government's objective to increase the use of B.C. foods, improve the patient food experience in health care, and efforts to increase access to traditional foods and Indigenous recipes in public institutions. Traditional Foods and Indigenous Recipes in...
Webinar Recap – The Three sisters: history, culture and ways of growing
June 8, 2021F2CC adminEnglish, Home Page Blog Feed, Videos, Webinar
Please note: a recording of this webinar is not available. The following provides a recap of what was shared. In this webinar, hosted in collaboration with our partners at the Public Health Association of BC, Elder Lix Lopez explained the ways of growing squash, beans, and corn, along with his story of coming to Vancouver,...
Celebrating National Indigenous History Month
June 1, 2021F2CC adminEnglish, Farm to School, Home Page Blog Feed, Indigenous
Photo: Elsipogtog School, Elsipogtog First Nation, NB The month of June is National Indigenous History Month, a time to honour the history, heritage and diversity of Indigenous peoples in Canada. We'd like to start this month by sharing some inspiring stories of school communities who are embracing Indigenous food ways, as well as some resources...
Seeking Indigenous Lead: Local Food to School
May 18, 2021F2CC adminEnglish, F2CC News, Farm to School, Home Page Blog Feed
Location: Open to any location in Canada/Indigenous Territories (or Nations) Job Terms: 28-35 hours / week; 1-year contract from July 1 2021 – June 31, 2022 (with potential to extend) Salary: $50,000-55,000 Application Deadline: Wednesday June 23rd, 2021 5pm PST Position Summary: Indigenous Nations, communities and their schools, as well as communities and schools that...
The Three Sisters: history, culture and ways of growing
April 12, 2021F2CC adminEnglish, Farm to School, Home Page Blog Feed, Webinar
In this webinar, Elder Lix Lopez, member of the Maya in Exile Garden, will highlight the importance of the Three Sisters – corn, beans and squash – for Mayan diet and culture. Through storytelling, he will describe the origin of this popular plant sisterhood, its relevance for the Mayan people and culture, and other Indigenous...
Cultivating Change: Antiracist principles in school food & Indigenous-led school food programs
December 10, 2020F2CC adminHome Page Blog Feed
From November 12-18th, Food Secure Canada's online gathering, Cultivating Change, presented a rich program of events that looked at the intersections between food systems, anti-racism, decolonization and environmental justice movements. As part of the event the Coalition for Healthy School Food hosted two important conversations: one on showcasing Indigenous-led school food programs, and one on...


The elderly woman from Fakenham was contacted by someone pretending to be from Her Majesty's Revenue Customs telling her she owed a substantial amount in taxes going back a number of years.
She was told to go and buy Google vouchers amounting to £1900 from a local supermarket, however staff recognised that it could have been a scam and stopped her from losing her money.
Rich Dawson, police beat manager in Fakenham, said: "This is a very common scam right now. HMRC will never contact you and ask you to pay back taxes by way of vouchers or gift cards.
"The supermarkets and banks have become more aware of this because of the publicity that has already gone out but the message still needs to be pushed as the issue is still on the rise.
To report a scam, visit actionfraud.police.uk.


Facebook is the most widely used social media platform on the planet. With over 1 billion active users, it is a great tool for digital marketing if you know what you're doing. A lot of businesspeople have heard about Facebook Ads as a way to promote their website and the products or services it sells. However, you don't need to use Facebook Ads to take advantage of the traffic on Facebook for this purpose. There are a lot of free ways to promote your website by simply utilizing the communication services of the platform.
Let's start with the basic way to advertise on Facebook. If you have a personal Facebook profile with at least a few hundred friends and family connected to your account, then it is easy to share text, pictures, and video content with them. You can make one post and it will automatically be shared with everyone on your friend's list. Once your friends see the post, they can add reply comments or "like" it. Plus, they can share the post with their own friend's list too. This is called word-of-mouth marketing which can generate a lot of additional free traffic from people with whom you don't even know.
Of course, if you're a businessperson, then you probably don't want to turn your friends and family into customers. That is why you should create a Facebook Fan page for your business. While ordinary profiles are supposed to be for individual people, a Facebook Fan Page can be made for companies, charities, or any other third-party entity that you control. A Fan page is a great way to gain followers to your company through means of social media. These followers can "like" or comment on posts just like on the regular profile pages.
There is also something called Facebook Groups. This is perhaps the best free way to advertise on Facebook. These groups are created by other businesspeople and users of Facebook who have similar interests in a particular niche. For example, there may be a Facebook group devoted to website design services. If you are a web designer looking to educate themselves in the business or possibly find new customers, you would join a Facebook group like this to promote yourself. Again, it is a free way to gain a new following from people who have an interest in the types of things you sell.


Want to book a private, still have questions, and would like to learn more? Please reach out, and I'd be happy to connect with you. Many thanks for your interest.
Thank you! I will respond to your inquiries shortly.


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Motorola confirms Moto X; will be built in Texas
Tom Krazit May 29, 2013 - 4:15 PM CDT
Credit: Asa Mathat/D: All Things Digital
Motorola CEO Dennis Woodside confirmed Wednesday at D11 that Motorola will introduce the Moto X smartphone later this year, calling it a "hero" device that the Google-owned company hopes will make Motorola relevant once again. The device will also be built in the U.S. just outside of Fort Worth, Tex., in what Woodside called "a first."
The Moto X is "more contextually aware of what's going on around it," Woodside said, teasing attendees by saying he's carrying the phone in his pocket but declining to show it off. Other products are coming later, but expect this device to be the centerpiece of Motorola's strategy this fall. Reports have been circulating for some time about the "X" project inside Motorola but this is the first time the company has confirmed it.
The smartphone industry is famously concentrated in China, where low-cost manufacturing and a lot of specialized factories and suppliers have set up shop. Motorola will assemble 70 percent of the Moto X in Texas, with other component work having to be sourced and manufactured elsewhere. It will employ 2,000 people at the plant.
Woodside reiterated that Motorola, despite being owned by the creator of Android, has "no access to Android code. … There is no advantage that has been conferred to us." He defended Google's decision to invest $12.5 billion in Motorola by saying "we don't need necessarily to see the exact strategic fit with our current business" in order to pull the trigger on a deal; it's more about investing in technologies that will help shape the future, he said.
Woodside was joined onstage by Regina Dugan, formerly of DARPA and now currently senior vice president for Advanced Technology & Projects at Motorola. She showed off an "electronic tattoo" that could be used for authentication as well as a concept called "vitamin authentication," producing a tiny pill with a tiny chip in it that apparently can be swallowed and kept inside the body to unlock something. It appears to be a take on the "ingestible sensor" developed and marketed by Proteus Digital Health.
Here's a look at the electronic tattoo courtesy of AllThingsD's Asa Mathat:
Regina Dugan of Motorola shows off an "electronic tattoo" at D11. Credit: Asa Mathat/D: All Things Digital
Dennis Woodside
2 Responses to "Motorola confirms Moto X; will be built in Texas"
Mike C May 30, 2013
If I purchase this phone will it's software upgrades be abandoned like they were on my Moto Photon? (Still stuck on Gingerbread with Sprint!)
Moto X Forum May 29, 2013
The Moto X Forum is now open at http://MotoXforum.net
The App Store saved us from carriers. But who will save us now?
By Katie Jansen, CMO of AppLovin Jul 18, 2018 - 1:55 PM CDT
Gigaom #MobilityTalk Tweetchat on 12/7: Innovating the Enterprise through Mobile, Security & Productivity
Bob Egan Nov 30, 2017 - 10:53 AM CST
Sector Roadmap
Sector Roadmap: Enterprise Mobility Management
Cormac Foster


Save 54% on Cloudbuilt this Weekend – New Content!
If you've been waiting to jump into the fastest 3D platformer in the world it's about time you put your rocket-powered skates on and head down to your favourite virtual storefront, Steam. Cel-shaded, freerunner Cloudbuilt is on special promotion for a limited time only for less than half price and there's never been a better time give Cloudbuilt a blast! Need another reason? Here's three.
Since release Cloudbuilt has benefited from a number of updates, both fine tuning the game and adding new features. Here are the most recent additions that are freely available to new and existing Cloudbuilt players.
We've all had that feeling where we are stuck and would love to put down a checkpoint somewhere that is convenient to us – now you can. Coilworks have developed a system that makes Cloudbuilt more accessible to the casual player by allowing them to drop their own checkpoint. It's a great feature that helps newcomers complete levels for the first time and the more experienced players by providing an efficient way to test alternative paths at any point in the level.
Everyone loves customization, right?! Unlockable skins are now available for Demi which are awarded for completing specific achievements which can be viewed on the selection screen. With 21 new skins, there's plenty to choose from and a lot to play for. Many of the skins will take skill and determination to unlock which makes it all the sweeter to wear them.
In addition to the 22 original stages, Coilworks have added 2 shiny new remix levels which contain a selection of the most interesting sequences presented in a unique theme exclusive to these stages. Of course, competition will be fierce to crown the king of the leaderboards so keep an eye out. For the rest us mere mortals, it's awesome to have an even bigger selection of intricate levels to explore and conquer.
There you have it, three more reasons to take advantage of Cloudbuilt's half price sale this weekend! A big thank you to Coilworks for continuing to bring awesome content to Cloudbuilt and we can't to see what's next.
Flash new suit, same beautiful world.
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