text:
Q: Find out which members verified a transaction in Hyperledger Fabric I am using Hyperledger Fabric in combination with docker. My question is, how do i find out which peers verified a transaction in Hyperledger Fabric?
I use this code to instantiate my chaincode, which works:
docker exec -e "CORE_PEER_ADDRESS=peer0.org1.example.com:7051" cli peer chaincode instantiate -o orderer.example.com:7050 -C mychannel -n mycc -l golang -v ${CC_VERSION} -c '{"Args":["entry","5"]}' -P "OutOf(2, 'Org1MSP.member', 'Org1MSP.member')"
I am using peers from the same organization to verify the transaction.
All of my code works, but after running the code (which includes invokes), can i somehow check the block to see which peers endorsed the transactions that were made?
I have tried using the configtxlator to decode and view the block, but cannot find any information about endorsements in mychannel.block.

A: Probably mychannel.block was old
You have to check the latest block, where a transaction has been committed
------ OR -----
Use query with txid

A sample snippet of NodeJS
await this.channel.queryTransaction(trxnID, peer);

Took endorser block from large payload
                    "endorser": {
                      "Mspid": "nbdMSP",
                      "IdBytes": "-----BEGIN CERTIFICATE-----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\n1O4Ak+jJMrQvOook+JRq3kHmCVKDig==\n-----END CERTIFICATE-----\n"
                    },


text:
Rumbler may refer to:

 Big Daddy (BioShock), a fictional character
 Rumbler (G.I. Joe), a fictional character
 Rumbler siren, a type of emergency vehicle siren
 Rumbler Rock, a rock in Antarctica

See also

redpajama_set_name: RedPajamaWikipedia


When asked about creativity in context with motherhood, I think, "Who has time to think about being creative"?
But if you peel the layers back of our sacred journey, you soon realize that besides love, creativity would be the next operative word.
My personal journey could be compared to Odysseus, you know, the Classic story in Greek mythology. With the last name Spiropoulos, I am Odysseus. I have fought the Cyclops and reveled in defeats against monsters, all the while being a "creative" mother of three daughters. I am not here, though, to talk about my journey, I want to tell you an inspiring tale about my sister.
My sister Stella, is the most creative person I know. She had four children in five years, the second, a special needs child. Creativity is about multitasking, thinking on your feet and seizing the moment. Stella is the master of creativity. She consciously raised her children without television. Every minute of her day was a creative challenge. Many days I would call and ask what she was doing, it ranged from a puppet show to making cookies. When she asked what I was doing, I quietly said, the kids are watching "Barney" and I am doing the wash. Whether it is reading evening stories to three different age groups or keeping each engaged in making cookies, all the while including my very special nephew, Ezra, she gracefully floats from one to another with a smile on her face and a gleam in her eye. It is one thing to be a creative mother but it is quite another to be a creative woman who is a mother.
Under the guise of motherhood lies our original creative self. So we have peeled the next layer off of my dear sister to reveal a woman who knew she had to keep her creative spirit alive, for sheer survival, if nothing else. Under this layer we find a dancer, artist, author, musician and teacher. Stella miraculously never left these parts of herself too long. She fit in violin lessons between doctor appointments, painting between the fourth grade play, and teaching dance before going to one of her children's sporting events. Now, this takes creativity to a different level; creative in the arts, creative in time management and most of all creative in motherhood.
Peel back another layer of this remarkable woman and you will find the "wife". It is a phenomenon, literally, that she can find the time to let the man that she has been with for thirty years know just how much she loves him. Now that takes creativity!
When you look at all the layers of creativity, it is difficult to tell which is more important. She makes each layer seem the most important, all the while nurturing her own creativity.
Finally, if you took every layer off, you would see the brightest light of creativity and love, her essence. Many mothers have many stories and without even realizing it, we are all the most creative beings on earth.
Debbi Baron is a special education teacher at a high school in Keller, Texas.
She has three daughters and a great sense of humor!


A study reveals that developmental differences between children and adolescents can factor in to the diagnosis of specific anxiety disorders.
Children and adolescents are often grouped together and treated similarly. Most materials used to assess mental health in minors examine a large age range from five to 18 years old. However, Polly Waite and Cathy Creswell point out that children and adolescents are different in terms of mental and physical development; such factors question a homogenized treatment approach to individuals younger than 18.
According to the latest news from the October 2014 volume of Journal of Affective Disorders, children differ from adolescents in clinical characteristics so much that separate treatment plans are needed. Waite and Creswell compared the clinical characteristics of anxiety disorder assessment between children and adolescents in attempts to redefine treatment plans for each group.
All children and adolescents were recruited from the Berkshire Healthcare NHS Foundation Trust, Child and Adolescent Mental Health Service (CAMHS) Anxiety and Depression Pathway, based at the University of Reading. Waite and Creswell chose 100 children ages six to 12 and 100 adolescents ages 13 to 18. For the purpose of this study, Waite and Creswell classified preteens as children and teenagers as adolescents. However, the researchers acknowledged that developmental changes occur at different ages and sometimes in discrete ways. All individuals had been assessed with an anxiety disorder. Patients with Obsessive-compulsive Disorder (OCD) and complex Post-traumatic Stress Disorder (PTSD) were excluded from the study because neither fall under the category of broad anxiety disorders in the DSM-5.
Anxiety Disorders Interview Schedule for Children (ADIS-C/P): A structured interview designed to assess a patient according to the DSM-IV criterion.
Spence Children's Anxiety Scale (SCAS-C/P): A 38-item survey assessing panic attacks, agoraphobia, separation anxiety, physical injury fears, social phobia, generalized anxiety, and obsessive-compulsive symptoms.
Short Mood and Feelings Questionnaire (SMFQ-C/P): A 13-item self-report measuring depression.
Strength and Difficulties Questionnaire (SDQ-P): A 5-item questionnaire assessing behavioral disturbance and externalizing symptoms.
Caregivers were also assessed in the study. Waite and Creswell used A 21-item self-report called the Depression Anxiety Stress Scales (DASS) to measure levels of emotional burdens in caregivers.
Children were significantly more likely to be diagnosed with separation anxiety disorder.
Adolescents were significantly more likely to be diagnosed with Social Anxiety Disorder (SAD).
Adolescents were commonly diagnosed with a comorbid mood disorder.
More adolescents with an anxiety disorder were not regularly attending school.
Adolescents scored higher on the SMFQ-C/P and SDQ-P.
Caregivers of both groups scored similarly.
The observations clearly show a distinction between the age groups. Waite and Creswell suggest that "adapting treatments designed for children to make the materials more 'adolescent-friendly' is unlikely to sufficiently meet the needs of adolescents."


text:
\section{Introduction}\

Let $\mathcal{A}$ be an algebra over a number field $\mathbb{K}$ and
$\mathcal{M}$ be an $\mathcal{A}$-bimodule. An additive (linear)
mapping $\delta: \mathcal{A}\rightarrow \mathcal{M}$ is called a
\textit{left (right) centralizer} if $\delta(AB)=\delta(A)B$
($\delta(AB)=A\delta(B)$) for all $A, B\in \mathcal{A}$. An additive
(linear) mapping $\delta: \mathcal{A}\rightarrow \mathcal{M}$ is
called a \textit{left (right) Jordan centralizer} if
$\delta(A^2)=\delta(A)A$ ($\delta(A^2)=A\delta(A)$) for every $A\in
\mathcal{A}$. We call $\delta$ a \textit{centralizer} if $\delta$ is
both a left centralizer and a right centralizer. Similarly, we can
define a \textit{Jordan centralizer}. It is clear that every
centralizer is a Jordan centralizer, but the converse is not true in
general. In \cite{Zalar}, Zalar proved that each left Jordan
centralizer of a semiprime ring is a left centralizer and each
Jordan centralizer of a semiprime ring is a centralizer. For some
other results, see \cite{CSR,CORA,OCS} and references therein.\

In \cite{mn}, Vukman defined a new type of Jordan centralizers, named (\textit{m,n})-\textit{Jordan centralizer}, that is
an additive mapping $\delta$ from a ring $\mathcal{R}$ into itself such that $$(m+n)\delta(x^2)=m\delta(x)x+nx\delta(x)$$ for every $x\in \mathcal{R}$,
where $m\geq0$, $n\geq0$ are fixed integers with $m+n\neq0$.
Obviously, (1,0)-Jordan centralizer is a left Jordan centralizer and (0,1)-Jordan centralizer is a right Jordan centralizer.
Moreover, each Jordan centralizer is an ($m, n$)-Jordan centralizer and
$(1, 1)$-Jordan centralizer satisfies the relation $2\delta(x^2)=\delta(x)x+x\delta(x)$ for every $x\in \mathcal{R}$.
In \cite{CSR}, Vukman showed that $(1, 1)$-Jordan centralizer of a 2-torsion free semiprime ring $\mathcal{R}$ is a centralizer.
In \cite{submit}, Guo and Li studied $(1, 1)$-Jordan centralizer of some reflexive algebras.
In \cite{mn}, Vukman investigated ($m, n$)-Jordan centralizer and proved that for $m\geq1$ and $n\geq1$, every $(m, n)$-Jordan centralizer of a prime ring $\mathcal{R}$ with $char(\mathcal{R})\neq6mn(m+n)$ is a centralizer.
Motivated by this, we define a new type of Jordan centralizer, named \textit{weak }(\textit{m,n,l})-\textit{Jordan centralizer},
that is a linear mapping $\delta$ from a unital algebra $\mathcal{A}$ into itself satisfying
$$(m+n+l)\delta(A^2)-m\delta(A)A-nA\delta(A)-lA\delta(I)A\in\mathbb{K}I$$
for every $A\in \mathcal{A}$, where $m\geq0, n\geq0, l\geq0$ are fixed integers with $m+n+l\neq 0$.
This is equivalent to say that for every $A\in \mathcal{A}$, there exists a $\lambda_A\in \mathbb{K}$ such that
\begin{eqnarray}
(m+n+l)\delta(A^2)=m\delta(A)A+nA\delta(A)+lA\delta(I)A+\lambda_AI.\label{101}
\end{eqnarray}
When $\lambda_A=0$ for every $A\in \mathcal{A}$, we call such a $\delta$ an (\textit{m,n,l})-\textit{Jordan centralizer}.
It is clear that each (\textit{m,n,l})-Jordan centralizer is a weak (\textit{m,n,l})-Jordan centralizer, each (\textit{m,n,\emph{0}})-Jordan centralizer is an (\textit{m,n})-Jordan centralizer and (0,0,1)-Jordan centralizer has
the relation $\delta(A^2)=A\delta(I)A$ for every $A\in \mathcal{A}$. In this paper, we study (weak) (\textit{m,n,l})-Jordan centralizer on some reflexive algebras and generalized matrix algebras.

Let $X$ be a Banach space over $\mathbb{K}$ and $B(X)$ be the set of all bounded operators on $X$, where $\mathbb{K}$ is the real field $\mathbb{R}$ or the complex field $\mathbb{C}$.
We use $X^*$ to denote the set of all bounded linear functionals on $X$.
For $A\in B(X)$, denote by $A^*$ the adjoint of $A$.
For any non-empty subset $L\subseteq X$, $L^\perp$ denotes its annihilator, that is,
$L^\perp=\{f\in X^*: f(x)=0~\mathrm{for}~\mathrm{all}~x\in L\}$.
By a subspace lattice on $X$, we mean a collection $\mathcal{L}$ of closed subspaces of $X$
with (0) and $X$ in $\mathcal{L}$ such that for every family $\{M_r\}$ of elements of $\mathcal{L}$,
both $\cap M_r$ and $\vee M_r$ belong to $\mathcal{L}$. For a subspace lattice $\mathcal{L}$ of $X$,
let alg$\mathcal{L}$ denote the algebra of all operators in $B(X)$ that leave members of $\mathcal{L}$ invariant;
and for a subalgebra $\mathcal{A}$ of $B(X)$, let lat$\mathcal{A}$ denote the lattice of all closed subspaces of $X$
that are invariant under all operators in $\mathcal{A}$.
An algebra $\mathcal{A}$ is called \textit{reflexive} if alglat$\mathcal{A}=\mathcal{A}$;
and dually, a subspace lattice is called \textsl{reflexive} if latalg$\mathcal{L}=\mathcal{L}$.
Every reflexive algebra is of the form alg$\mathcal{L}$ for some subspace lattice $\mathcal{L}$ and vice versa.

For a subspace lattice $\mathcal{L}$ and for $E\in \mathcal{L}$, define
$$E_-=\vee\{F\in \mathcal{L}: F\nsupseteq E\}~\mathrm{and}~E_+=\wedge\{F\in \mathcal{L}: F\nleq E\}.$$
Put
$$\mathcal{J}(\mathcal{L})=\{K\in \mathcal{L}: K\neq(0)~\mathrm{and}~K_-\neq X\}.$$
For any non-zero vectors $x\in X$ and $f\in X^*$, the rank one operator $x\otimes f$ is defined by $x\otimes f(y)=f(y)x$ for $y\in X$.
Several authors have studied the properties of the set of rank one operators in reflexive algebras (for example, see \cite{lambrou, ORO}).
It is well known (see \cite{ORO}) that $x\otimes f\in \textrm{alg}\mathcal{L}$ if and only if there exists some $K\in \mathcal{J}(\mathcal{L})$
such that $x\in K$ and $f\in K_-^\perp$.
When $X$ is a separable Hilbert space over the complex field $\mathbb{C}$, we change it to $H$. In a Hilbert space,
we disregard the distinction between a closed subspace and the orthogonal projection onto it.
A subspace lattice $\mathcal L$ on a Hilbert space $H$ is called a \textit{commutative~subspace~lattice}
(\textit{CSL}), if all projections in $\mathcal L$ commute pairwise. If $\mathcal{L}$ is a CSL, then the
corresponding algebra alg$\mathcal{L}$ is called a \textit{CSL algebra}. By \cite{CSL}, we know that if $\mathcal{L}$ is a CSL, then $\mathcal{L}$
is reflexive.
Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$ satisfying
$\vee\{L: L\in \mathcal{J}(\mathcal{L})\}=X$ or $\wedge\{L_-: L\in \mathcal{J}(\mathcal{L})\}=(0)$.
In \cite{JDRA}, Lu considered this kind of reflexive algebras which have rich rank one operators.
In Section 2, we prove that if $\delta$ is a weak (\textit{m,n,l})-Jordan centralizer from alg$\mathcal{L}$ into itself, where $\mathcal{L}$ is a CSL or satisfies $\vee\{L: L\in \mathcal{J}(\mathcal{L})\}=X$ or $\wedge\{L_-: L\in \mathcal{J}(\mathcal{L})\}=(0)$,
then $\delta$ is a centralizer.

A \textit{Morita context} is a set $(\mathcal{A}, \mathcal{B}, \mathcal{M}, \mathcal{N})$
and two mappings $\phi$ and $\varphi$, where $\mathcal{A}$ and $\mathcal{B}$ are two algebras over a number field $\mathbb{K}$, $\mathcal{M}$ is an
$(\mathcal{A}, \mathcal{B})$-bimodule and $\mathcal{N}$ is a $(\mathcal{B}, \mathcal{A})$-bimodule.
The mappings $\phi: M\otimes_\mathcal{B} N\rightarrow A$ and $\varphi: N\otimes_\mathcal{A} M\rightarrow B$ are two bimodule homomorphisms satisfying
$\phi(M\otimes N)M'=M\varphi(N\otimes M')$ and $\varphi(N\otimes M)N'=N\phi(M\otimes N')$ for any $M, M'\in \mathcal{M}$ and
$N, N'\in \mathcal{N}$. These conditions insure that the set
$$\left[
  \begin{array}{cc}
    \mathcal{A} & \mathcal{M} \\
    \mathcal{N} & \mathcal{B} \\
  \end{array}
\right]=\left\{\left[
            \begin{array}{cc}
              A& M \\
              N & B \\
            \end{array}
          \right]\mid ~A\in \mathcal{A}, M\in \mathcal{M}, N\in \mathcal{N}, B\in \mathcal{B}\right\}$$
form an algebra over $\mathbb{K}$ under usual matrix operations.
We call such an algebra a \textit{generalized matrix algebra} and denoted by
$\mathcal{U}=\left[
  \begin{array}{cc}
    \mathcal{A} & \mathcal{M} \\
    \mathcal{N} & \mathcal{B} \\
  \end{array}
\right]$, where $\mathcal{A}$ and $\mathcal{B}$ are two unital algebras and at least one of the two bimodules $\mathcal{M}$ and $\mathcal{N}$
is distinct from zero.  This kind of algebra was first introduced
by Sands in \cite{sands}.
Obviously, when $\mathcal{M}=0$ or $\mathcal{N}=0$, $\mathcal{U}$ degenerates to the triangular algebra.
In Section 3, we show that if $\delta$ is a weak (\textit{m,n,l})-Jordan centralizer
from $\mathcal{U}$ into itself, then $\delta$ is a centralizer. We also study $(m,n,l)$-Jordan centralizer on AF algebras.
Throughout the paper, we assume $m,n,l\in \mathbb{N}$ are such that $m+l\geq1$, $n+l\geq1$.

\section{Centralizers of certain reflexive algebras}
\

In order to prove our main results, we need the following  several lemmas.

\begin{lemma}\label{201}
Let $\mathcal{A}$ be a unital algebra with identity $I$ and $\delta$ be a weak \emph{(}\textit{m,n,l}\emph{)}-Jordan centralizer
from $\mathcal{A}$ into itself. Then for any $A,B\in\mathcal{A}$,
\begin{eqnarray}
&&(m+n+l)\delta(AB+BA)=m\delta(A)B+m\delta(B)A+nA\delta(B)+nB\delta(A)\nonumber\\
&&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~+lA\delta(I)B+lB\delta(I)A+(\lambda_{A+B}-\lambda_A-\lambda_B)I.\label{2001}
\end{eqnarray}
In particular, for any $A\in\mathcal{A}$,
\begin{eqnarray}
\delta(A)=\frac{m+l}{m+n+2l}\delta(I)A+\frac{n+l}{m+n+2l}A\delta(I)+\lambda(A),\label{2002}
\end{eqnarray}
where we set $\lambda(A)=\frac{1}{m+n+2l}(\lambda_{A+I}-\lambda_A)I$ for every $A \in \mathcal{A}$.
\end{lemma}

\begin{proof}
Since $\delta$ is a weak (\textit{m,n,l})-Jordan centralizer, we have
\begin{eqnarray*}
(m+n+l)\delta(A^2)=m\delta(A)A+nA\delta(A)+lA\delta(I)A+\lambda_AI
\end{eqnarray*}
for every $A\in \mathcal{A}.$
Replacing $A$ by $A+B$ in above equation, (\ref{2001}) holds. Letting $B=I$ in (\ref{2001}) gives (\ref{2002}), since $\lambda_I=0$.
\end{proof}

\begin{remark}\label{211}
\emph{For an (\textit{m,n,l})-Jordan centralizer, we could actually define it from a unital algebra $\mathcal{A}$ to an $\mathcal{A}$-bimodule. Hence when lemmas in this section are applied to an (\textit{m,n,l})-Jordan centralizer $\delta$, we will take it for granted that $\delta$ is from a unital algebra $\mathcal{A}$ to its bimodule, since all the proofs remain true if we set $\lambda_A=0$ for all $A\in\mathcal{A}$.}
\end{remark}
\begin{remark} \label{212}
\emph{Obviously, each (1,0,0)-Jordan centralizer is a left Jordan centralizer and each (0,1,0)-Jordan centralizer is a right Jordan centralizer.
So by Lemma \ref{201}, it follows that every left Jordan centralizer of unital algebras is a left centralizer
and every right Jordan centralizer of unital algebras is a right centralizer.
Therefore every Jordan centralizer of unital algebras is a centralizer.}
\end{remark}
Let $f$ be a linear mapping from an algebra $\mathcal{A}$ to its bimodule $\mathcal{M}$. Recall that $f$ is a \emph{derivation} if $f(ab)=f(a)b+af(b)$ for all $a,b\in\mathcal{A}$; it is a \emph{Jordan derivation} if $f(ab)=f(a)a+af(a)$ for every $a\in \mathcal{A}$; it is a \emph{generalized derivation} if $f(ab)=f(a)b+ad(b)$ for all $a,b\in\mathcal{A}$, where $d$ is a derivation from $\mathcal{A}$ to $\mathcal{M}$; and it is a \emph{generalized Jordan derivation} if $f(ab)=f(a)a+ad(a)$ for every $a\in\mathcal{A}$, where $d$ is a Jordan derivation from $\mathcal{A}$ to $\mathcal{M}$.
From Remarks \ref{211} and \ref{212}, we have the following corollary.
\begin{corollary}\label{202}
Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$ satisfying $\vee\{F: F\in \mathcal{J}(\mathcal{L})\}=X$ or
$\wedge\{L_-: L\in \mathcal{J}(\mathcal{L})\}=(0)$.
If $f$ is a generalized Jordan derivation from alg$\mathcal{L}$ to $B(X)$, then $f$ is a generalized derivation.
\end{corollary}
\begin{proof}
Since $f$ is a generalized Jordan derivation, we have the relation
\begin{eqnarray*}
f(A^2)=f(A)A+Ad(A)
\end{eqnarray*}
for every $A\in \textrm{alg}\mathcal{L}$, where $d$ is a Jordan derivation of alg$\mathcal{L}$.
By \cite[Theorem 2.1]{JDRA}, one can conclude that $d$ is a derivation. Let $\delta=f-d$. Then we have
$$\delta(A^2)=f(A^2)-d(A^2)=f(A)A+Ad(A)-Ad(A)-d(A)A=f(A)A-d(A)A=\delta(A)A$$
for every $A\in \textrm{alg}\mathcal{L}$. This means that $\delta$ is a left Jordan centralizer. By
Remark \ref{212}, $\delta$ is a left centralizer. Hence
$$f(AB)=d(AB)+\delta(AB)=d(A)B+Ad(B)+\delta(A)B=f(A)B+Ad(B)$$
for all $A, B\in \textrm{alg}\mathcal{L}$. In other words, $f$ is a generalized derivation.
\end{proof}

Since every Jordan derivation of CSL algebras is a derivation \cite{JDCSL}, we also have the following corollary.

\begin{corollary}\label{203}
Let $\mathcal{L}$ be a \textit{CSL} on a Hilbert space $H$.
If $f$ is a generalized Jordan derivation from alg$\mathcal{L}$ into itself, then $f$ is a generalized derivation.
\end{corollary}

\begin{lemma}\label{204}
Let $\mathcal{A}$ be a unital algebra
and $\delta$ be a weak \emph{(}\textit{m,n,l}\emph{)}-Jordan centralizer from $\mathcal{A}$ into itself.
Then for every idempotent  $P \in \mathcal{A}$ and every $A\in\mathcal{A}$, \\
\emph{(i)}~~$\delta(P)=P\delta(I)=\delta(I)P$;\\
\emph{(ii)}~~$\delta(AP)=\delta(A)P+\lambda(AP)-\lambda(A)P$;\\
\emph{(iii)}~~$\delta(PA)=P\delta(A)+\lambda(PA)-\lambda(A)P$.
\end{lemma}

\begin{proof}
(i) Suppose $P$ is an idempotent in $\mathcal{A}$. It follows from Lemma \ref{201} that
\begin{eqnarray}
(m+n+2l)\delta(P)=(m+l)\delta(I)P+(n+l)P\delta(I)+(\lambda_{P+I}-\lambda_P)I.\label{2003}
\end{eqnarray}
Right and left multiplication of (\ref{2003}) by $P$ gives $$P\delta(P)P=P\delta(I)P+\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_P)P.$$
Since $(m+n+l)\delta(P)=m\delta(P)P+nP\delta(P)+lP\delta(I)P+\lambda_PI$,
multiplying $P$ from the right leads to
\begin{eqnarray*}
(n+l)\delta(P)P&=&n(P\delta(I)P+\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_P)P)+lP\delta(I)P+\lambda_PP\\
&=&(n+l)P\delta(I)P+(\frac{n}{m+n+2l}(\lambda_{P+I}-\lambda_P)+\lambda_P)P,
\end{eqnarray*}
whence
\begin{eqnarray}
\delta(P)P=P\delta(I)P+\varepsilon_PP\label{2007}
 \end{eqnarray}
for some $\varepsilon_P\in \mathbb{C}$.
Similarly, $P\delta(P)=P\delta(I)P+\varepsilon_P'P$ for some $\varepsilon_P'\in \mathbb{C}$. \\
Hence $\delta(P)P-\varepsilon_PP=P\delta(P)-\varepsilon_P'P$. Right and left multiplication of $P$ gives $\varepsilon_P=\varepsilon_P'$, which implies \begin{eqnarray}
\delta(P)P=P\delta(P).\label{2012}
 \end{eqnarray}

Replacing $P$ by $I-P$ in the above equation gives $\delta(I)P=P\delta(I)$.

Now, we have from (\ref{2003})
\begin{eqnarray}
\delta(P)=\delta(I)P+\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_P)I.\label{2008}
\end{eqnarray}
On the other hand, (\ref{2007}) and (\ref{2012}) yields
\begin{eqnarray*}
(m+n+l)\delta(P)&=&m\delta(P)P+nP\delta(P)+lP\delta(I)P+\lambda_PI\\
&=&(m+n+l)\delta(P)P+\lambda_PI-l\varepsilon_PP,
\end{eqnarray*}
right multiplication of which by $P$ gives $\lambda_P=l\varepsilon_P$. Hence
\begin{eqnarray}
\delta(P)=\delta(P)P+\frac{1}{m+n+l}\lambda_P(I-P)\label{2006}.
\end{eqnarray}
We then have from (\ref{2008}) that
\begin{eqnarray}
\delta(P)P=\delta(I)P+\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_{P})P \label{2014}
\end{eqnarray}
 and hence
\begin{eqnarray*}
\delta(P)=\delta(I)P+\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_{P})P+\frac{1}{m+n+l}\lambda_P(I-P),
\end{eqnarray*}
which together with (\ref{2008}) implies
$$\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_{P})=\frac{1}{m+n+l}\lambda_{P}.$$
Thus we have
\begin{eqnarray}
\delta(P)=\delta(I)P+\frac{1}{m+n+l}\lambda_{P}I,\label{2009}
\end{eqnarray}
while
\begin{eqnarray}
\delta(P)=\frac{m+n}{m+n+l}\delta(P)P+\frac{l}{m+n+l}\delta(I)P+\frac{1}{m+n+l}\lambda_{P}I. \label{2010}
\end{eqnarray}
Comparing (\ref{2009}) and (\ref{2010}) gives $$\delta(I)P=\delta(P)P.$$
This together with (\ref{2014}) gives
$$\lambda(P)=\frac{1}{m+n+2l}(\lambda_{P+I}-\lambda_{P})I=\frac{1}{m+n+l}\lambda_{P}I=0,$$ whence
$$\delta(P)=\delta(I)P=P\delta(I).$$
(ii) By Lemma \ref{201} and (i), we have
\begin{eqnarray*}
\delta(AP)&=&\frac{m+l}{m+n+2l}\delta(I)AP+\frac{n+l}{m+n+2l}AP\delta(I)+\lambda(AP)\\
&=&(\frac{m+l}{m+n+2l}\delta(I)A+\frac{n+l}{m+n+2l}A\delta(I))P+\lambda(AP)\\
&=&(\delta(A)-\lambda(A))P+\lambda(AP)\\
&=&\delta(A)P+\lambda(AP)-\lambda(A)P.
\end{eqnarray*}

(iii) The proof is analogous to the proof of (ii).
\end{proof}

An subset $\mathcal{I}$ of an algebra $\mathcal{A}$ is called a \emph{left separating set} of $\mathcal{A}$ if for every $A\in \mathcal{A}$, $A\mathcal{I}={0}$ implies $A=0$. We have the following simple but noteworthy result.

\begin{corollary}\label{205}
Suppose $\mathcal{I}$ is a left separating left ideal of a unital algebra $\mathcal{A}$ and is contained in the algebra generated by all idempotents in $\mathcal{A}$. Then each weak \emph{(}\textit{m,n,l}\emph{)}-Jordan centralizer $\delta$ from $\mathcal{A}$ into itself is a centralizer.
\end{corollary}
\begin{proof}
Since $\mathcal{I}$ is contained in the algebra generated by all idempotents in $\mathcal{A}$ and by (i) of Lemma \ref{204}, we have that $\delta(I)\in \mathcal{I}'$, where $\mathcal{I}'$ denotes the commutant of $\mathcal{I}$. Hence $\delta(A)=\delta(I)A+\lambda(A)=A\delta(I)+\lambda(A)$ for every $A\in \mathcal{I}$ according to (\ref{2002}). For any $A(\neq \mathbb{K}I)\in\mathcal{I}$, we have
\begin{eqnarray*}
&&(m+n+l)(\delta(I)A^2+\lambda(A^2))\\
&=&(m+n+l)\delta(A^2)\\
&=&m\delta(A)A+nA\delta(A)+lA\delta(I)A+\lambda_AI\\
&=&m(\delta(I)A^2+\lambda(A)A)+n(A^2\delta(I)+A\lambda(A))+lA^2\delta(I)+\lambda_AI,
\end{eqnarray*}
which implies $\lambda(A)A=kI$ for some $k\in\mathbb{K}$. Hence $\lambda(A)=0$ and $\delta(A)=\delta(I)A=A\delta(I)$ for every $A\in \mathcal{I}$.
Then Lemma \ref{204} yields $A\delta(I)\mathcal{I}=A\mathcal{I}\delta(I)=\delta(A\mathcal{I})=\delta(I)A\mathcal{I}$, and since $\mathcal{I}$
is a separating left ideal, we have $A\delta(I)=\delta(I)A$ for every $A\in \mathcal{A}$.
Therefore,
$\delta(A)=\delta(I)A+\lambda(A)=A\delta(I)+\lambda(A)$ for every $A\in \mathcal{A}$. Now by the same argument as above, we have that $\delta(A)=\delta(I)A=A\delta(I)$ for every $A\in \mathcal{A}$ and this completes the proof.
\end{proof}

\begin{remark}
\emph{By \cite[Proposition 2.2]{Hadwin}, \cite[Example 6.2]{Samei}, we see that the class of algebras we discussed in Corollary \ref{205} contains a lot of algebras and is therefore very large.}

\end{remark}

The proof of the following lemma is similar to the proof of \cite[Proposition 1.1]{LDRA} and we omit it.

\begin{lemma}\label{206}
Let $E$ and $F$ be non-zero subspaces of $X$ and $X^*$ respectively.
Let $\phi: E\times F\rightarrow B(X)$ be a bilinear mapping such that
$\phi(x,f)X\subseteq \mathbb{K}x$ for all $x\in E$ and $f\in F$.
Then there exists a linear mapping $S: F\rightarrow X^*$ such that
$\phi(x,f)=x\otimes Sf$ for all $x\in E$ and $f\in F$.
\end{lemma}

\begin{lemma}\label{207}
Let $\mathcal{L}$ be a subspace lattice on a Banach space X
and $\delta$ be a weak \emph{(}\textit{m,n,l}\emph{)}-Jordan centralizer from $alg\mathcal{L}$ into itself.
Suppose that $E$ and $L$ are in $\mathcal{J}(\mathcal{L})$ such that $E_{-}\ngeq L$.
Let $x$ be in E and $f$ be in $L_-^\perp$. Then $(\delta(x\otimes f)-\lambda(x\otimes f))X\subseteq \mathbb{K}x$.
\end{lemma}

\begin{proof}
Since $E_{-}\ngeq L$, we have that $E\leq L$. So $x\otimes f\in \textrm{alg}\mathcal{L}$. Suppose $f(x)\neq0$,
it follows from Lemmas \ref{201} and \ref{204} that $\lambda(x\otimes f)=0$ and $\delta(x\otimes f)=x\otimes f\delta(I)$. Thus $\delta(x\otimes f)X\subseteq \mathbb{K}x$.
\

Now we assume $f(x)=0$. Choose $z$ from $L$ and $g$ from $E_-^\perp$ such that $g(z)=1$. Then
\begin{eqnarray*}
&&(m+n+2l)(m+n+l)\delta(x\otimes f)\\
&&~~~~~~~=(m+n+2l)(m+n+l)\delta((x\otimes g)(z\otimes f)+(z\otimes f)(x\otimes g))\\
&&~~~~~~~=(m+n+2l)(m\delta(x\otimes g)(z\otimes f)+n(x\otimes g)\delta(z\otimes f)+l(x\otimes g)\delta(I)(z\otimes f))\\
&&~~~~~~~~~~+(m+n+2l)(m\delta(z\otimes f)(x\otimes g)+n(z\otimes f)\delta(x\otimes g)+l(z\otimes f)\delta(I)(x\otimes g))\\
&&~~~~~~~~~~+(m+n+2l)(\lambda_{x\otimes g+z\otimes f}-\lambda_{x\otimes g}-\lambda_{z\otimes f})I\\
&&~~~~~~~=(m^2+ml)\delta(I)x\otimes f+(n^2+nl)x\otimes f\delta(I)\\
&&~~~~~~~~~~+2(mn+ml+nl+l^2)(x\otimes g\delta(I)z\otimes f+z\otimes f\delta(I)x\otimes g)+\lambda_1I
\end{eqnarray*}
for some $\lambda_1\in \mathbb{K}$.

On the other hand,
\begin{eqnarray*}
&&(m+2n+l)(m+n+l)\delta(x\otimes f)\\
&&~~~=(m+n+l)((m+l)\delta(I)x\otimes f+(n+l)x\otimes f\delta(I)+(\lambda_{x \otimes f+I}-\lambda_{x\otimes f})I)\\
&&~~~=(m^2+2ml+l^2+mn+nl)\delta(I)x\otimes f+(ml+mn+l^2+2nl+n^2)x\otimes f\delta(I)+\lambda_2I
\end{eqnarray*}
for some $\lambda_2 \in \mathbb{K}$.

So
\begin{eqnarray}
\delta(I)x\otimes f+x\otimes f\delta(I)=2x\otimes g\delta(I)z\otimes f+2z\otimes f\delta(I)x\otimes g+ \lambda I \label{2004}
\end{eqnarray}
for some $\lambda\in \mathbb{K}$.

Notice that this equation is valid for all $z$ in $L$ satisfying $g(z)=1$. Applying (\ref{2004}) to $x$, we have
\begin{eqnarray}
f(\delta(I)x)x=2g(x)f(\delta(I)x)z+\lambda x.\label{2005}
\end{eqnarray}

If $g(x)=0$ and $f(z)=0$, then $f(\delta(I)x)=\lambda$. Substituting $z+x$ for $z$ in (\ref{2004}) gives
\begin{eqnarray}
\delta(I)x\otimes f+x\otimes f\delta(I)=2x\otimes g\delta(I)(z+x)\otimes f+2\lambda(z+x)\otimes g+\lambda I\label{2011}.
\end{eqnarray}
Comparing (\ref{2004}) with (\ref{2011}) yields
\begin{eqnarray}
g(\delta(I)x)x\otimes f+\lambda x\otimes g=0.
\end{eqnarray}
Applying this equation to $z$ yields $\lambda x=0,$ which means $f(\delta(I)x)=\lambda=0$.

If $g(x)=0$ and $f(z)\neq0$, we also have $f(\delta(I)x)=\lambda$, and it follows from Lemma \ref{204} that
\begin{eqnarray*}
\delta(I)x\otimes f+x\otimes f\delta(I)&=&2x\otimes g\delta(I)z\otimes f+2z\otimes f\delta(I)x\otimes g+\lambda I\\
&=&2(x\otimes g)(z\otimes f)\delta(I)+2\delta(I)(z\otimes f)(x\otimes g)+\lambda I\\
&=&2x\otimes f\delta(I)+\lambda I,
\end{eqnarray*}
whence $$\delta(I)x\otimes f=x\otimes f\delta(I)+\lambda I.$$
Applying the above equation to $x$ leads to $f(\delta(I)x)=-\lambda$. Hence $f(\delta(I)x)=\lambda=0$.

If $g(x)\neq 0$, replacing $z$ by $\frac{1}{g(x)}x$ in (\ref{2005}) gives $f(\delta(I)x)=-\lambda$,
while
\begin{eqnarray}
\delta(I)x\otimes f+x\otimes f\delta(I)&=&2x\otimes g\delta(I)z\otimes f+2 z\otimes f\delta(I)x\otimes g+\lambda I\nonumber\\
&=&2\delta(I)(x\otimes g) (z\otimes f)+2 (z\otimes f) (x\otimes g)\delta(I)+\lambda I\nonumber\\
&=&2\delta(I)(x\otimes f)+\lambda I.\nonumber
\end{eqnarray}
Hence
\begin{eqnarray}
x\otimes f\delta(I)=\delta(I)x\otimes f+\lambda I.\label{2013}
\end{eqnarray}
Applying (\ref{2013}) to $x$ yields $f(\delta(I)x)=\lambda$. Therefore, $f(\delta(I)x)=\lambda=0$.

So by (\ref{2004}), we obtain $\delta(I)x\otimes f=2g(\delta(I)z)x\otimes f-x\otimes f\delta(I)$.
It follows from Lemma \ref{201} that
\begin{eqnarray*}
\delta(x\otimes f)&=&\frac{m+l}{m+n+2l}\delta(I)(x\otimes f)+\frac{n+l}{m+n+2l}(x\otimes f)\delta(I)+\lambda(x\otimes f)\\
&=&\frac{m+l}{m+n+2l}(2g(\delta(I)z)x\otimes f-x\otimes f\delta(I))+\frac{n+l}{m+n+2l}(x\otimes f)\delta(I)+\lambda(x\otimes f)\\
&=&\frac{2(m+l)}{m+n+2l}g(\delta(I)z)x\otimes f+\frac{n-m}{m+n+2l}(x\otimes f)\delta(I)+\lambda(x\otimes f).
\end{eqnarray*}
Hence $(\delta(x\otimes f)-\lambda(x\otimes f))X\subseteq \mathbb{K}x$.
\end{proof}

\begin{theorem}\label{208}
Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$ satisfying $\vee\{F: F\in \mathcal{J}(\mathcal{L})\}=X$.
If $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $alg\mathcal{L}$ into itself, then $\delta$ is a centralizer.
In particular, the conclusion holds if $\mathcal{L}$ has the property $X_-\neq X$.
\end{theorem}

\begin{proof}
Let $E$ be in $\mathcal{J}(\mathcal{L})$. By $\vee\{F: F\in \mathcal{J}(\mathcal{L})\}=X$,
there is an element $L$ in $\mathcal{J}(\mathcal{L})$ such that $E_-\ngeq L$.
Let $x$ be in $E$ and $f$ be in $(L_-)^\bot$. Let $\overline{\delta}=\delta-\lambda$. Then $\overline{\delta}(I)=\delta(I)$, and it follows from Lemmas \ref{206} and \ref{207} that there exists a linear mapping
$S: (L_-)^\bot\rightarrow X^*$ such that
$$\overline{\delta}(x\otimes f)=x\otimes Sf.$$
This together with $$\frac{m+l}{m+n+2l}\overline{\delta}(I)x\otimes f+\frac{n+l}{m+n+2l}x\otimes f\overline{\delta}(I)=\overline{\delta}(x\otimes f)$$ leads to
$$x\otimes (Sf-\frac{n+l}{m+n+2l}\overline{\delta}(I)^*f)=\frac{m+l}{m+n+2l}\overline{\delta}(I)x\otimes f.$$
Thus there exists a constant $\lambda_E$ in $\mathbb{K}$ such that $\overline{\delta}(I)x=\lambda_Ex$ for every $x\in E$.
Similarly, for every $y\in L$, we have $\overline{\delta}(I)y=\lambda_Ly$.

If $f(x)\neq 0$, then $\overline{\delta}(x\otimes f)=\overline{\delta}(I)x\otimes f=x\otimes f\overline{\delta}(I)$.

If $f(x)=0$, according to the proof of Lemma \ref{207}, we can choose $z$ from $L$ and $g$ from $E_-^\perp$ such that $g(z)=1$ and
$\overline{\delta}(I)x\otimes f=2g(\overline{\delta}(I)z)x\otimes f-x\otimes f\overline{\delta}(I)$.
Since $x\in E\leq L$, we have $\overline{\delta}(I)x=\lambda_Lx$. Thus
$$\overline{\delta}(I)x\otimes f=2\lambda_Lx\otimes f-x\otimes f\overline{\delta}(I)=2\overline{\delta}(I)x\otimes f-x\otimes f\overline{\delta}(I).$$
Hence $\overline{\delta}(x\otimes f)=\overline{\delta}(I)x\otimes f=x\otimes f\overline{\delta}(I)$.\

Then for any $x\in E$, $f\in (L_-)^\bot$ and $A\in \textrm{alg}\mathcal{L}$, we have
\begin{eqnarray*}
A\overline{\delta}(I)x\otimes f=Ax\otimes f\overline{\delta}(I)=\overline{\delta}(I)Ax\otimes f.
\end{eqnarray*}
So we have $A\overline{\delta}(I)x=\overline{\delta}(I)Ax$ for any $x\in E$. By $\vee\{F: F\in \mathcal{J}(\mathcal{L})\}=X$,
we have $\overline{\delta}(A)=A\overline{\delta}(I)=\overline{\delta}(I)A$ for any $A\in \textrm{alg}\mathcal{L}$, this means $\delta(A)=A\delta(I)+\lambda(A)=\delta(I)A+\lambda(A)$. The remaining part goes along the same line as the proof of Corollary \ref{205} and this completes the proof.
\end{proof}

\begin{remark}
\emph{By \cite{SRL}, a subspace lattice $\mathcal{L}$ is said to be \textit{completely~distributive}
if $L=\vee\{E\in \mathcal{L}: E_-\ngeq L\}$ or $L=\wedge\{E_-: E\in \mathcal{L}~\mathrm{and}~E\nleq L\}$ for all $L\in \mathcal{L}$.
It follows that completely distributive subspace lattices satisfy the condition
$\vee \{E: E\in \mathcal{J}(\mathcal{L}) \}=X$. Thus Theorem \ref{208} applies to completely distributive subspace lattice algebras.
A subspace lattice $\mathcal{L}$ is called a \textit{$\mathcal{J}$-subspace lattice} on $X$ if
$\vee\{K: K\in \mathcal{J}(\mathcal{L})\}=X$,
$\wedge\{K_-: K\in \mathcal{J}(\mathcal{L})\}=(0)$,
$K\vee K_-=X$ and
$K\wedge K_-=(0)~\mathrm{for~any}~K\in \mathcal{J}(\mathcal{L})$. Note also that the condition
$\vee \{K: K\in \mathcal{J}(\mathcal{L}) \}=X$ is part of the definition of $\mathcal{J}$-subspace lattices, thus Theorem \ref{208} also
applies to $\mathcal{J}$-subspace lattice algebras.}
\end{remark}

With a proof similar to the proof of Theorem \ref{208}, we have the following theorem.
\begin{theorem}\label{209}
Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$
satisfying $\wedge\{L_-: L\in \mathcal{J}(\mathcal{L})\}=(0)$.
If $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $alg\mathcal{L}$ into itself, then $\delta$ is a centralizer.
In particular, the conclusion holds if $\mathcal{L}$ has the property $(0)_+\neq (0)$.
\end{theorem}
As for the cases of $(m,n,l)$-Jordan centralizers, we have from Remark \ref{211} , Theorem \ref{208} and Theorem \ref{209} the following theorem.
\begin{theorem}
Let $\mathcal{L}$ be a subspace lattice on a Banach space $X$
satisfying $\vee\{F: F\in \mathcal{J}(\mathcal{L})\}=X$ or $\wedge\{L_-: L\in \mathcal{J}(\mathcal{L})\}=(0)$. If $\delta$ is an $(m,n,l)$-Jordan centralizer from $alg\mathcal{L}$ to $B(X)$, then $\delta$ is a centralizer.
\end{theorem}

In the rest of this section we will investigate weak $(m,n,l)$-Jordan centralizer on CSL algebras.
Let $H$ be a complex separable Hilbert space and $\mathcal{L}$ be a CSL on $H$.
Let $\mathcal{L^\perp}$ be the lattice \{$I-E: E\in \mathcal{L}$\} and $\mathcal{L^\prime}$ be the commutant of $\mathcal{L}$.
It is easy to verify that $(\textrm{alg}\mathcal{L})^*=\textrm{alg}\mathcal{L^\perp}$ for any lattice $\mathcal{L}$ on $H$ and the diagonal
$(\textrm{alg}\mathcal{L})\cap(\textrm{alg}\mathcal{L})^*=\mathcal{L^\prime}$ is a von
Neumann algebra. Given a CSL $\mathcal{L}$ on a Hilbert space $H$,
we define $G_1(\mathcal{L})$ and $G_2(\mathcal{L})$ to be the
projections onto the closures of the linear spans of $\{EA(I-E)x:
E\in \mathcal{L}, A\in alg\mathcal{L}, x\in H\}$ and $\{(I-E)A^*Ex:
E\in \mathcal{L}, A\in alg\mathcal{L}, x\in H\}$, respectively. For simplicity, we
write $G_1$ and $G_2$ for $G_1(\mathcal{L})$ and $G_2(\mathcal{L})$.
Since CSL is reflexive, it is easy to verify that $G_1\in
\mathcal{L}$ and $G_2\in \mathcal{L}^\bot$. In \cite{JDCSL}, Lu
showed that $G_1\vee G_2\in \mathcal{L}\cap\mathcal{L}^\bot$ and
$\textrm{alg}\mathcal{L}(I-G_1\vee G_2)\subseteq \mathcal{L}^\prime$.

\begin{theorem}\label{210}
Let $\mathcal{L}$ be a CSL on $H$. If $\delta$ is a bounded weak $(m,n,l)$-Jordan centralizer from $alg\mathcal{L}$ into itself,
then $\delta$ is a centralizer.
\end{theorem}
\begin{proof}
We divide the proof into two cases.\\
Case 1: Suppose $G_1\vee G_2=I$.\

Let $A\in \textrm{alg}\mathcal{L}$. For any $T\in \textrm{alg}\mathcal{L}$ and $P\in\mathcal{L}$,
since $PT(I-P)=P-(P-PT(I-P))$ is a difference of two idempotents, it follows from Lemma \ref{204} that
\begin{eqnarray*}
\delta(I)APT(I-P)=A\delta(I)PT(I-P)=\delta(APT(I-P))=\delta(A)PT(I-P)-\lambda(A)PT(I-P).
\end{eqnarray*}

By arbitrariness of $P$ and $T$, we have $A\delta(I)G_1=\delta(I)AG_1=(\delta(A)-\lambda(A))G_1$. That is,
\begin{eqnarray*}
\delta(A)G_1=(A\delta(I)+\lambda(A))G_1=(\delta(I)A+\lambda(A))G_1,
\end{eqnarray*}
whence
\begin{eqnarray}
\delta(AG_1)=\delta(A)G_1+\lambda(AG_1)-\lambda(A)G_1=\delta(I)AG_1+\lambda(AG_1)=A\delta(I)G_1+\lambda(AG_1)\label{3001}.
\end{eqnarray}

Define $\delta^*(A^*)=\delta(A)^*$ for every $A^*\in \textrm{alg}\mathcal{L}^\bot$. So
\begin{eqnarray*}
(m+n+l)\delta^*((A^*)^2)&=&((m+n+l)\delta(A^2))^*\\
&=&(m\delta(A)A+nA\delta(A)+lA\delta(I)A+\lambda_AI)^*\\
&=&mA^*\delta^*(A^*)+n\delta^*(A^*)A^*+lA^*\delta^*(I)A^*+\lambda_{A^*},
\end{eqnarray*}
where $\lambda_{A^*}=\overline{\lambda_A}$.

With the proof similar to the proof of (\ref{3001}), we have $$G_2\delta(I)A=G_2A\delta(I)=G_2(\delta(A)-\lambda(A)).$$
So by $G_1\vee G_2=I$,
\begin{eqnarray*}
(I-G_1)\delta(I)A=(I-G_1)A\delta(I)=(I-G_1)(\delta(A)-\lambda(A)),
\end{eqnarray*}
whence
\begin{eqnarray}
\delta((I-G_1)A)&=&(1-G_1)\delta(A)+\lambda((I-G_1)A)-\lambda(A)(I-G_1)\nonumber\\
&=&(1-G_1)(\delta(A)-\lambda(A))+\lambda((I-G_1)A)\nonumber\\
&=&(1-G_1)\delta(I)A+\lambda((I-G_1)A)\nonumber\\
&=&(I-G_1)A\delta(I)+\lambda((I-G_1)A)\label{3004}.
\end{eqnarray}

Hence by (\ref{3001}) and (\ref{3004}),
\begin{eqnarray*}
\delta(A)&=&\delta(AG_1+G_1A(I-G_1)+(I-G_1)A)\\
&=&A\delta(I)G_1+\lambda(AG_1)+G_1A(I-G_1)\delta(I)+(I-G_1)A\delta(I)+\lambda((1-G_1)A)\\
&=&G_1A\delta(I)G_1+G_1A\delta(I)(I-G_1)+(I-G_1)A\delta(I)\\
&&+\lambda(AG_1)+\lambda((1-G_1)A)+\lambda(G_1A(1-G_1))\\
&=&A\delta(I)+\lambda(A),
\end{eqnarray*}
and similarly, $\delta(A)=\delta(I)A+\lambda(A)$. The remaining part goes along the same line as the proof of Corollary \ref{205} and we conclude that $\delta$ is a centralizer in this case.\\
Case 2: Suppose $G_1\vee G_2<I$.

Let $G=G_1\vee G_2$. Since $G\in \mathcal{L}\cap\mathcal{L}^\bot$ and $\textrm{alg}\mathcal{L}(I-G)\subseteq\mathcal{L}^\prime$,
so $(I-G)\textrm{alg}\mathcal{L}(I-G)$ is a von Neumann algebra. The algebra $\textrm{alg}\mathcal{L}$ can be written as the direct sum
$$\textrm{alg}\mathcal{L}=\textrm{alg}(G\mathcal{L}G)\oplus \textrm{alg}((I-G)\mathcal{L}(I-G)).$$
By Lemma \ref{204} we have that $$\delta(GAG)=G\delta(A)G~\mathrm{and}~\delta((I-G)A(I-G))=(I-G)\delta(A)(I-G)$$
for every $A\in \textrm{alg}\mathcal{L}$. Therefore $\delta$ can be written as $\delta^{(1)}\oplus\delta^{(2)}$,
where $\delta^{(1)}$ is a weak (\textit{m,n,l})-Jordan centralizer from $\textrm{alg}(G\mathcal{L}G)$ into itself
and $\delta^{(2)}$ is a weak (\textit{m,n,l})-Jordan centralizer from $\textrm{alg}((I-G)\mathcal{L}(I-G))$ into itself.
It is easy to show that $G_1(G\mathcal{L}G)\vee G_2(G\mathcal{L}G)=G$.
So it follows from Case 1 that $\delta^{(1)}$ is a centralizer on $\textrm{alg}(G\mathcal{L}G)$.
$(I-G)\textrm{alg}\mathcal{L}(I-G)$ is a von Neumann algebra and $\delta^{(2)}$ is continuous, so by Corollary \ref{205},
$\delta^{(2)}$ is a centralizer on $\textrm{alg}((I-G)\mathcal{L}(I-G))$. Consequently, $\delta$ is a centralizer on $\textrm{alg}\mathcal{L}$.
\end{proof}

\section{Centralizers of generalized matrix algebras}

\

We call $\mathcal M$ a \textsl{unital $\mathcal A$-bimodule} if $\mathcal M$ is an
$\mathcal A$-bimodule and satisfies $I_{\mathcal A}M=M I_{\mathcal A}=M$ for every $M\in{\mathcal M}$.
We call $\mathcal{M}$ a \textit{faithful left $\mathcal{A}$-module} if for any $A\in \mathcal{A}$,
$A\mathcal{M}=0$ implies $A=0$. Similarly, we can define a \textit{faithful right $\mathcal{B}$-module}. \

Throughout this section, we denote the generalized matrix algebra originated from the Morita context
$(\mathcal{A}, \mathcal{B}, \mathcal{M}, \mathcal{N}, \phi_{\mathcal{M}\mathcal{N}}, \varphi_{\mathcal{N}\mathcal{M}})$
by
$\mathcal{U}=\left[\begin{array}{cc}\mathcal{A} & \mathcal{M} \\ \mathcal{N} & \mathcal{B} \\ \end{array} \right]$
,
where $\mathcal{A}, \mathcal{B}$ are two unital algebras over a number field $\mathbb{K}$ and $\mathcal{M}, \mathcal{N}$ are two unital bimodules, and
at least one of $\mathcal{M}$ and $\mathcal{N}$ is distinct from zero.
We use the symbols $I_{\mathcal A}$ and $I_\mathcal{B}$ to denote the unit element in $\mathcal A$ and $\mathcal{B}$, respectively.
Moreover, we make no difference between $\lambda(A)=\frac{1}{m+n+2l}(\lambda_{A+I}-\lambda_A)I$ and $\frac{1}{m+n+2l}(\lambda_{A+I}-\lambda_A)\in\mathbb{K}$.
\begin{lemma}\label{30101}
Let $\delta$ be a weak $(m,n,l)$-Jordan centralizer from $\mathcal{U}$ into itself. Then $\delta$ is of the form
$$\delta\left(\left[\begin{array}{cc}  A   & M \\  N  & B  \\ \end{array}\right]\right)=
\left[\begin{array}{cc}  a_{11}(A)+\lambda\left(\left[
                                            \begin{array}{cc}
                                              0 & M \\
                                              N & B \\
                                            \end{array}
                                          \right]\right)I_\mathcal{A}
 &c_{12}(M) \\ d_{21}(N) &  b_{22}(B)+\lambda\left(\left[
                                               \begin{array}{cc}
                                                 A & M \\
                                                 N & 0 \\
                                               \end{array}
                                             \right]\right)I_\mathcal{B}
   \\   \end{array}\right],$$
for any $A\in \mathcal{A}$, $M\in \mathcal{M}$, $N\in \mathcal{N}$, $B\in \mathcal{B}$,
where $a_{11}: \mathcal{A}\rightarrow \mathcal{A}$, $c_{12}: \mathcal{M}\rightarrow \mathcal{M}$,
$d_{21}: \mathcal{N}\rightarrow \mathcal{N}$, $b_{22}: \mathcal{B}\rightarrow \mathcal{B}$
are all linear mappings satisfying
$$c_{12}(M)=a_{11}(I_\mathcal{A})M=Mb_{22}(I_\mathcal{B})~~~and~~~d_{21}(N)=Na_{11}(I_\mathcal{A})=b_{22}(I_\mathcal{B})N.$$
\end{lemma}

\begin{proof}
Assume that $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $\mathcal{U}$ into itself.
Because $\delta$ is linear, for any $A\in \mathcal{A}$, $M\in \mathcal{M}$, $N\in \mathcal{N}$, $B\in \mathcal{B}$,
we can write
 \begin{eqnarray*}
&&\delta\left(\left[\begin{array}{cc}A & M \\N & B \\\end{array}\right]\right)\\
&=&\left[\begin{array}{cc}a_{11}(A)+b_{11}(B)+c_{11}(M)+d_{11}(N) & a_{12}(A)+b_{12}(B)+c_{12}(M)+d_{12}(N) \\a_{21}(A)+b_{21}(B)+c_{21}(M)+d_{21}(N) & a_{22}(A)+b_{22}(B)+c_{22}(M)+d_{22}(N) \\\end{array}\right],
\end{eqnarray*}
where
$a_{ij}$, $b_{ij}$, $c_{ij}$, $d_{ij}$ are linear mappings, $i, j\in\{1, 2\}$.

Let $P=\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]$ and
for any $A\in \mathcal{A}$, $S=\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]$.
By Lemma \ref{204}, $\delta(PS)=P\delta(S)+\lambda(PS)-\lambda(S)P$ and $\delta(SP)=\delta(S)P+\lambda(SP)-\lambda(S)P$, so we have
\begin{eqnarray*}
&&\left[\begin{array}{cc}a_{11}(A) & a_{12}(A) \\a_{21}(A) & a_{22}(A)\\\end{array}\right]
=\delta\left(\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]\right)
=\delta\left(\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]
\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]\right)\\
&=&\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]
\delta\left(\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]\right)+\left[
                                                                                \begin{array}{cc}
                                                                                \lambda(PS)I_\mathcal{A} & 0 \\
                                                                                0 & \lambda(PS)I_\mathcal{B} \\
                                                                                \end{array}
                                                                               \right]-\left[
                                                                                \begin{array}{cc}
                                                                                \lambda(S)I_\mathcal{A}& 0 \\
                                                                                0 & 0 \\
                                                                                \end{array}
                                                                                \right]\\
&=&\left[\begin{array}{cc}a_{11}(A) & a_{12}(A) \\0 & \lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  A & 0 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}
\\\end{array}\right]
\end{eqnarray*}
and
\begin{eqnarray*}
&&\left[\begin{array}{cc}a_{11}(A) & a_{12}(A) \\a_{21}(A) & a_{22}(A)\\\end{array}\right]
=\delta\left(\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]\right)
=\delta\left(\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]
\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]\right)\\
&=&\delta\left(\left[\begin{array}{cc} A & 0 \\  0 & 0  \end{array}\right]\right)
\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]+\left[
                                                                                \begin{array}{cc}
                                                                                \lambda(SP)I_\mathcal{A} & 0 \\
                                                                                0 & \lambda(SP)I_\mathcal{B} \\
                                                                                \end{array}
                                                                               \right]-\left[
                                                                                \begin{array}{cc}
                                                                                \lambda(S)I_\mathcal{A}& 0 \\
                                                                                0 & 0 \\
                                                                                \end{array}
                                                                                \right]\\
&=&\left[\begin{array}{cc}a_{11}(A) &  0\\a_{21}(A) & \lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  A & 0 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}\\\end{array}\right].
\end{eqnarray*}
So we have $a_{12}(A)=0$, $a_{21}(A)=0$ and $a_{22}(A)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  A & 0 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}$.
\

Similarly, by considering $S=\left[\begin{array}{cc}0 & M \\0 & 0 \\\end{array}\right]$ and
$P=\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]$, we obtain that
$c_{11}(M)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A}$, $c_{21}(M)=0$ and $c_{22}(M)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}$ for every $M\in \mathcal{M}$.\

By considering $S=\left[\begin{array}{cc}0 & 0 \\N & 0 \\\end{array}\right]$ and
$P=\left[\begin{array}{cc}I_\mathcal{A} & 0 \\0 & 0 \\\end{array}\right]$, we obtain
$d_{11}(N)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & 0 \\
                                                                  N & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A}$, $d_{12}(N)=0$ and $d_{22}(N)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & 0 \\
                                                                  N & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}$ for every $N\in \mathcal{N}$.\

By considering $S=\left[\begin{array}{cc}0 & 0 \\0 & B \\\end{array}\right]$ and
$Q=\left[\begin{array}{cc}0 & 0 \\0 & I_\mathcal{B} \\\end{array}\right]$, we obtain
$b_{11}(B)=\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & 0 \\
                                                                  0 & B \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A}$, $b_{12}(B)=0$ and $b_{21}(B)=0$ for every $B\in \mathcal{B}$.\

For any $A\in \mathcal{A}$, $M_1\in \mathcal{M}$, $M_2\in \mathcal{M}$ and $B\in \mathcal{B}$,
let $S=\left[\begin{array}{cc}A & M_1 \\0 & 0 \\\end{array}\right]$ and $T=\left[\begin{array}{cc}0 & M_2 \\0 & B \\\end{array}\right]$.

Then
\begin{eqnarray*}
&&(m+n+l)\left[\begin{array}{cc}0 & c_{12}(AM_2+M_1B) \\0 & 0 \\\end{array}\right]\\
&=&(m+n+l)\delta(ST)=(m+n+l)\delta(ST+TS)\\
&=&m\left[\begin{array}{cc}a_{11}(A)+\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_1 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A} & c_{12}(M_1) \\0 & \lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  A & M_1 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B} \\\end{array}\right] \left[\begin{array}{cc}0 & M_2 \\0 & B \\\end{array}\right]\\
&&+ m\left[\begin{array}{cc}\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_2 \\
                                                                  0 & B \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A} & c_{12}(M_2) \\0 & b_{22}(B)+\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_2 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B} \\\end{array}\right] \left[\begin{array}{cc}A & M_1 \\0 & 0 \\\end{array}\right]\\
&&+n\left[\begin{array}{cc}A & M_1 \\0 & 0 \\\end{array}\right]
\left[\begin{array}{cc}\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_2 \\
                                                                  0 & B \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A} & c_{12}(M_2) \\0 & \lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_2 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{B}+b_{22}(B) \\\end{array}\right]
\end{eqnarray*}
\begin{eqnarray*}
&&+n\left[\begin{array}{cc}0 & M_2 \\0 & B \\\end{array}\right]
\left[\begin{array}{cc}a_{11}(A)+\lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  0 & M_1 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                              \right]\right)I_\mathcal{A} & c_{12}(M_1) \\0 & \lambda\left(\left[
                                                                \begin{array}{cc}
                                                                  A & M_1 \\
                                                                  0 & 0 \\
                                                                \end{array}
                                                             \right]\right)I_\mathcal{B}\\\end{array}\right]\\
&&+ l\left[\begin{array}{cc}A & M_1 \\0 & 0 \\\end{array}\right] \left[\begin{array}{cc}a_{11}(I_\mathcal{A}) & 0 \\0 & b_{22}(I_\mathcal{B}) \\\end{array}\right]
\left[\begin{array}{cc}0 & M_2 \\0 & B \\\end{array}\right]\\
&&+ \left[\begin{array}{cc}(\lambda_{S+T}-\lambda_S-\lambda_T) I_\mathcal{A} & 0 \\0 & (\lambda_{S+T}-\lambda_S-\lambda_T) I_\mathcal{B} \\\end{array}\right].
\end{eqnarray*}
The above matrix equation implies
\begin{eqnarray}
&&(m+n+l)c_{12}(AM_2+M_1B)\nonumber\\
&=&ma_{11}(A)M_2+m\lambda\left(\left[
                                       \begin{array}{cc}
                                         0 & M_1 \\
                                         0 & 0 \\
                                       \end{array}
                                     \right]\right)M_2
+ mc_{12}(M_1)B+m\lambda\left(\left[
                   \begin{array}{cc}
                     0 & M_2 \\
                     0 & B \\
                   \end{array}
                 \right]\right)M_1\nonumber\\
&&+nAc_{12}(M_2)+n\lambda\left(\left[
                        \begin{array}{cc}
                          0 & M_2 \\
                          0 & 0 \\
                        \end{array}
                      \right]\right)M_1
+nM_1b_{22}(B)+n\lambda\left(\left[
                         \begin{array}{cc}
                           A & M_1 \\
                           0 & 0 \\
                         \end{array}
                       \right]\right)M_2\nonumber\\
&&+lAa_{11}(I_\mathcal{A})M_2+lM_1b_{22}(I_\mathcal{B})B.\label{3000001}
\end{eqnarray}

Taking $B=0$ , $A=I_\mathcal{A}$ and $M_1=0$ in (\ref{3000001}),
we have $c_{12}(M)=a_{11}(I_\mathcal{A})M$ for every $M\in \mathcal{M}$.
Taking $A=0$, $B=I_\mathcal{B}$ and $M_2=0$ in (\ref{3000001}), we have $c_{12}(M)=Mb_{22}(I_\mathcal{B})$ for every $M\in \mathcal{M}$.

Symmetrically, $d_{21}(N)=b_{22}(I_\mathcal{B})N=Na_{11}(I_\mathcal{A})$ for every $N\in \mathcal{N}$.
\end{proof}

\begin{theorem}\label{30102}
Let $\delta$ be a weak $(m,n,l)$-Jordan centralizer from $\mathcal{U}$ into itself.
If one of the following conditions holds:\\
\emph{(1)}~~${\mathcal M}$ is a faithful left $\mathcal{A}$-module and a faithful right ${\mathcal B}$-module;\\
\emph{(2)}~~${\mathcal M}$ is a faithful left $\mathcal{A}$-module and ${\mathcal N}$ is a faithful left ${\mathcal B}$-module;\\
\emph{(3)}~~${\mathcal N}$ is a faithful right ${\mathcal A}$-module and a faithful left ${\mathcal B}$-module;\\
\emph{(4)}~~${\mathcal N}$ is a faithful right ${\mathcal A}$-module and ${\mathcal M}$ is a faithful right ${\mathcal B}$-module.\\
Then $\delta$ is a centralizer.
\end{theorem}

\begin{proof}
Let $\delta$ be an $(m,n,l)$-Jordan centralizer from $\mathcal{U}$ into itself. By Lemma \ref{30101}, we have
\begin{eqnarray}
c_{12}(M)=a_{11}(I_\mathcal{A})M=Mb_{22}(I_\mathcal{B})\label{30102}
\end{eqnarray}
for every $M\in \mathcal{M}$,
\begin{eqnarray}
d_{21}(N)=Na_{11}(I_\mathcal{A})=b_{22}(I_\mathcal{B})N\label{30103}
\end{eqnarray}
for every $N\in \mathcal{N}$.
\

We assume that (1) holds. The proofs for the other cases are analogous.\

For any $A\in \mathcal{A}$ and $M\in \mathcal{M}$,
$a_{11}(I_\mathcal{A})AM=AMb_{22}(I_\mathcal{B})=Aa_{11}(I_\mathcal{A})M$.
Since ${\mathcal M}$ is a faithful left $\mathcal{A}$-module, we have
\begin{eqnarray*}
a_{11}(I_\mathcal{A})A=Aa_{11}(I_\mathcal{A})
\end{eqnarray*}
whence
\begin{eqnarray}
a_{11}(A)=Aa_{11}(I_\mathcal{A})+\lambda\left(\left[
                                     \begin{array}{cc}
                                       A & 0 \\
                                       0 & 0 \\
                                     \end{array}
                                   \right]\right)I_\mathcal{A}=a_{11}(I_\mathcal{A})A+\lambda\left(\left[
                                     \begin{array}{cc}
                                       A & 0 \\
                                       0 & 0 \\
                                     \end{array}
                                   \right]\right)I_\mathcal{A}.\label{30104}
\end{eqnarray}
For any $B\in \mathcal{B}$ and $M\in \mathcal{M}$,
$MBb_{22}(I_\mathcal{B})=a_{11}(I_\mathcal{A})MB=Mb_{22}(I_\mathcal{B})B$.
Since ${\mathcal M}$ is a faithful right $\mathcal{B}$-module, we have
\begin{eqnarray}
b_{22}(B)=b_{22}(I_\mathcal{B})B+\lambda\left(\left[
                                     \begin{array}{cc}
                                       0 & 0 \\
                                       0 & B \\
                                     \end{array}
                                   \right]\right)I_\mathcal{B}=Bb_{22}(I_\mathcal{B})+\lambda\left(\left[
                                     \begin{array}{cc}
                                       0 & 0 \\
                                       0 & B \\
                                     \end{array}
                                   \right]\right)I_\mathcal{B}.\label{30105}
\end{eqnarray}
\

For any $A\in \mathcal{A}$, $M\in \mathcal{M}$, $N\in \mathcal{N}$ and $B\in \mathcal{B}$,\\
$$\delta\left(\left[\begin{array}{cc}      A        &       M          \\       N        &      B      \\         \end{array}\right]\right)=
\left[\begin{array}{cc}  a_{11}(A)+\lambda\left(\left[
                                     \begin{array}{cc}
                                       0 & M \\
                                       N & B \\
                                     \end{array}
                                   \right]\right)I_\mathcal{A}     &     c_{12}(M)    \\   d_{21}(N)    &  b_{22}(B)+\lambda\left(\left[
                                     \begin{array}{cc}
                                       A & M \\
                                       N & 0 \\
                                     \end{array}
                                   \right]\right)I_\mathcal{B}  \\         \end{array}\right],$$
$$\delta(I)\left[\begin{array}{cc}      A        &       M          \\       N        &      B      \\         \end{array}\right]
=\left[\begin{array}{cc}  a_{11}(I_\mathcal{A})A  &  a_{11}(I_\mathcal{A})M \\   b_{22}(I_\mathcal{B})N &  b_{22}(I_\mathcal{B})B \\ \end{array}\right]$$
and
$$\left[\begin{array}{cc}      A        &       M          \\       N        &      B      \\         \end{array}\right]\delta(I)=\left[\begin{array}{cc}   Aa_{11}(I_\mathcal{A})  &  Mb_{22}(I_\mathcal{B}) \\   Na_{11}(I_\mathcal{A}) & Bb_{22}(I_\mathcal{B}) \\ \end{array}\right].$$
So by (\ref{30102})-(\ref{30105}), we have for every $S\in \mathcal{U}$,
$$\delta(S)=\delta(I)S+\lambda(S)=S\delta(I)+\lambda(S).$$
The remaining part goes along the same line as the proof of Corollary \ref{205} and this completes the proof.
\end{proof}

Note that a unital prime ring $\mathcal{A}$ with a non-trivial idempotent $P$ can be written as the matrix form
$\left[
  \begin{array}{cc}
    P\mathcal{A}P & P\mathcal{A}P^\perp \\
    P^\perp\mathcal{A}P & P^\perp\mathcal{A}P^\perp \\
  \end{array}
\right]$.
Moreover, for any $A\in \mathcal{A}$, $PAP\mathcal{A}(I-P)=0$ and $P\mathcal{A}(I-P)A(I-P)$ imply $PAP=0$ and $(I-P)A(I-P)=0$, respectively.

\begin{corollary}\label{303}
Let $\mathcal{A}$ be a unital prime ring with a non-trivial idempotent $P$.
If $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $\mathcal{A}$ into itself, then $\delta$ is a centralizer.
\end{corollary}

As von Neumann algebras have rich idempotent elements and factor von Neumann algebras are prime, the following corollary is obvious.
\begin{corollary}\label{304}
Let $\mathcal{A}$ be a factor von Neumann algebra and $\delta$ be a weak $(m,n,l)$-Jordan centralizer from $\mathcal{A}$ into itself,
then $\delta$ is a centralizer.
\end{corollary}

Obviously, when $\mathcal{N}=0$, $\mathcal{U}$ degenerates to an upper triangular algebra. Thus we have the following corollary.
\begin{corollary}\label{305}
Let $\mathcal{U}=Tri(\mathcal{A},\mathcal{M},\mathcal{B})$ be an upper triangular algebra
such that $\mathcal{M}$ is a faithful ($\mathcal{A}$,~$\mathcal{B}$)-bimodule.
If $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $\mathcal{A}$ into itself, then $\delta$ is a centralizer.
\end{corollary}
Let $\mathcal{N}$ be a nest on a Hilbert space $H$ and alg$\hspace{1pt}$$\mathcal{N}$ be the associated algebra.
If $\mathcal{N}$ is trivial, then alg$\hspace{1pt}$$\mathcal{N}$ is $B(H)$. If $\mathcal{N}$ is nontrivial, take a nontrivial projection $P\in \mathcal{N}$.
Let $\mathcal{A}=P\mathrm{alg}\hspace{1pt}\mathcal{N}P$, $\mathcal{M}=P\mathrm{alg}\hspace{1pt}\mathcal{N}(I-P)$ and $\mathcal{B}=(I-P)\mathrm{alg}\hspace{1pt}\mathcal{N}(I-P)$.
Then $\mathcal{M}$ is a faithful ($\mathcal{A}$, $\mathcal{B}$)-bimodule, and alg$\hspace{1pt}$$\mathcal{N}$=Tri($\mathcal{A}$, $\mathcal{M}$, $\mathcal{B}$)
is an upper triangular algebra.
Thus as an application of Corollaries \ref{304} and \ref{305}, we have the following corollary.
\begin{corollary}
Let $\mathcal{N}$ be a nest on a Hilbert space $H$ and $\mathrm{alg}\hspace{1pt}\mathcal{N}$ be the associated algebra.
If $\delta$ is a weak $(m,n,l)$-Jordan centralizer from $\mathrm{alg}\hspace{1pt}\mathcal{N}$ into itself, then $\delta$ is a centralizer.
\end{corollary}

In the following, we study $(m,n,l)$-Jordan centralizer on AF algebras. A unital $C^*$-algebra $\mathcal{B}$ is called \emph{approximately finite} (AF) if $\mathcal{B}$ contains an increasing chain $\mathcal{B}_n\subseteq \mathcal{B}_{n+1}$ of finite-dimensional $C^*$-subalgebra, all containing the unit $I$ of $\mathcal{B}$, such that $\bigcup_{n=1}^\infty\mathcal{B}_n$ is dense in $\mathcal{B}$. For more details and related terms, we refer the readers to \cite{Derivable}.

\begin{lemma}\label{306}
Let $\mathcal{M}_n(\mathbb{C})$ be the set of all $n \times n$ complex matrices,
$\mathcal{A}$ be a CSL subalgebra of
$\mathcal{M}_{n_1}(\mathbb{C}) \oplus \cdots \oplus \mathcal{M}_{n_k}(\mathbb{C})$, and
$\mathcal{B}$ be an algebra such that $\mathcal{M}_{n_1}(\mathbb{C})
\oplus \cdots \oplus \mathcal{M}_{n_k}(\mathbb{C}) \subseteq \mathcal{B}$ as
an embedding. If $\delta$ is an $(m,n,l)$-Jordan centralizer from $\mathcal{A}$
into $\mathcal{B}$, then $\delta$ is a centralizer.
\end{lemma}

\begin{proof}
Let $\mathcal{A}$ be the linear span of its matrix units $\{E_{ij}\}$, and since $\delta$ is linear, we only need to show that for any $i$, $j$,
\begin{eqnarray}
\delta(E_{ij})=E_{ij}\delta(I)=\delta(I)E_{ij}.\label{300003}
\end{eqnarray}

If $i=j$, by Lemma \ref{202}, (\ref{300003}) is clear.\

Next, we will prove (\ref{300003}) for $i\neq j$. By Lemma \ref{201} and Remark \ref{211}, we have
\begin{eqnarray*}
(m+n+l)\delta(E_{ij})&=&(m+n+l)\delta(E_{ii}E_{ij}+E_{ij}E_{ii})\\
&=&m\delta(E_{ii})E_{ij}+nE_{ii}\delta(I)E_{ij}+lE_{ii}\delta(I)E_{ij}\\
&=&(m+n+l)\delta(E_{ii})E_{ij},
\end{eqnarray*}
Hence $\delta(E_{ij})=\delta(E_{ii})E_{ij}$ for any $i$, $j$.\

Similarly, we have $\delta(E_{ij})=E_{ij}\delta(E_{jj})$ for any $i$, $j$.\

Hence for any $i$, $j$,
\begin{eqnarray*}
E_{ij}\delta(I)=E_{ij}\sum_{k=1}^n \delta(E_{kk})=E_{ij}\sum_{k=1} ^n E_{kk}\delta(E_{kk})=E_{ij}\delta(E_{jj})=\delta(E_{ij}).
\end{eqnarray*}

Similarly, we have for any $i$, $j$, $\delta(I)E_{ij}=\delta(E_{ij})$ and the proof is complete.
\end{proof}

\begin{theorem}
Let $\mathcal{A}$ be a canonical subalgebra of an AF $C^*$-algebra $\mathcal{B}$. If $\delta$ is a bounded $(m,n,l)$-Jordan centralizer from $\mathcal{A}$ into $\mathcal{B}$, then $\delta$ is a centralizer.
\end{theorem}

$Proof$. Suppose $\delta$ is a bounded $(m,n,l)$-Jordan centralizer from $\mathcal{A}$ into $\mathcal{B}$.
Since $\mathcal{A}_n$ is a CSL algebra, $\delta |_{\mathcal{A}_n}$ is a centralizer by Lemma \ref{306}; that is,
for any $S$ in $\mathcal{A}_n$, $$\delta(S)= \delta(I)S=S\delta(I).$$
Since $\delta$ is norm continuous and $\cup _{i=1} ^\infty A_n$ is
dense in $A$, it follows that $\delta$ is a
centralizer.

\section*{Acknowledgement}
This work is supported by NSF of China.

redpajama_set_name: RedPajamaArXiv


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Psychologists Public Relations Managers Public Relations Specialists Purchasing Managers Radiation Therapists Radio and Telecommunications Equipment Installers and Repairers Radio Operators Radiologic Technicians Rail Transportation Occupations Real Estate Agents Real Estate Appraisers and Assessors Receptionists and Information Clerks Recreation Workers Recreational Therapists Registered Nurses Reporters Reservation and Transportation Ticket Agents Respiratory Therapists Retail Salespersons Roofers Sales Engineers Sales Managers Sales Representatives Sales Worker Supervisors Science Technicians Secretaries Security Guards and Gaming Surveillance Officers Semiconductor Processors Sheet Metal Workers Shipping, Receiving, and Traffic Clerks Small Engine Mechanics Social and Human Service Assistants Social Scientists Social Workers Sound Engineering Technicians Speech-language Pathologists Stationary Engineers and Boiler Operators Statisticians Stock Brokers Stock Clerks and Order Fillers Structural and Reinforcing Iron and Metal Workers Surgeons Surgical Technologists Surveyors Systems Administrators Tax Examiners, Collectors, and Revenue Agents Taxi Drivers Teacher Assistants Teachers—adult Literacy and Remedial Education Teachers—postsecondary Teachers—preschool, Kindergarten, Elementary, Middle, and Secondary Teachers—self-enrichment Education Teachers—special Education Television, Video, and Motion Picture Camera Operators and Editors Tellers Textile, Apparel, and Furnishings Occupations Tool and Die Makers Top Executives Travel Agents Truckdrivers and Driver / Sales Workers Urban and Regional Planners Veterinarians Veterinary Technicians Water and Liquid Waste Treatment Plant and System Operators Water Transportation Occupations Weighers, Measurers, Checkers, and Samplers, Recordkeeping Welding, Soldering, and Brazing Workers Woodworkers Writers
Desktop Publishers
Office and administrative support occupations
Significant Points
About 35 percent work for newspaper, periodical, book, and directory publishers, while almost 25 percent work in the printing industry.
Overall employment is expected to experience little or no change over the 2006-2016 decade.
Most employers prefer to hire experienced desktop publishers; among persons without experience, opportunities should be best for those with certificates or degrees in desktop publishing or graphic design.
Nature of Work
Desktop publishers use computer software to format and combine text, data, photographs, charts, and other graphic art or illustrations into prototypes of pages and other documents that are to be printed. They then may print the document using a high resolution printer or they may send the materials, either in print form or electronically, to a commercial printer. Examples of materials produced by desktop publishers include books, brochures, calendars, magazines, newsletters and newspapers, packaging, and forms.
Desktop publishers typically design and create the graphics that accompany text, convert photographs and illustrations into digital images, and manipulate the text and images to display information in an attractive and readable format. They design page layouts, develop presentations and advertising campaigns, and do color separation of pictures and graphics material. Some desktop publishers may write some of the text or headlines used in newsletters or brochures. They also may translate electronic information onto film or other traditional media if the final product will be sent to an off-set printer. As companies bring the production of marketing, promotional, and other kinds of materials in-house, they increasingly employ desktop publishers to produce such materials in house.
Desktop publishers use a computer and appropriate software to enter and select formatting properties, such as the size and style of type, column width, and spacing. Print formats are stored in the computer and displayed on a computer monitor. Images and text can be rearranged, column widths altered, or material enlarged or reduced. New information, such as charts, pictures, or additional text can be added. Scanners are used to capture photographs, images, or art as digital data that can be either incorporated directly into electronic page layouts or further manipulated with the use of computer software. The desktop publisher can make adjustments or compensate for deficiencies in the original color print or transparency. An entire newspaper, catalog, or book page, complete with artwork and graphics, can be created on the screen exactly as it will appear in print. Digital files are then used to produce printing plates. Like photographers and multimedia artists and animators, desktop publishers also can create special effects or other visual images using film, video, computers, or other electronic media. (Separate statements on photographers and on artists and related workers appear elsewhere in the Handbook.)
Desktop publishing encompasses a number of different kinds of jobs. Personal computers enable desktop publishers to more easily perform many of the design and layout tasks that would otherwise require large and complicated equipment and extensive human effort. Advances in computer software and printing technology continue to enhance desktop publishing work, making desktop publishing more economical and efficient than before. For example, desktop publishers get the material as computer files delivered over the Internet or on a portable disk drive instead of receiving simple typed text and instructions from customers. Other innovations in the occupation include digital color page makeup systems, electronic page layout systems, and off-press color proofing systems. In addition, most materials are reproduced on the Internet as well as printed; therefore, desktop publishers may need to know electronic publishing software, such as Hypertext Markup Language (HTML) and may be responsible for converting text and graphics to an Internet-ready format.
Some desktop publishers may write and edit as well as lay out and design pages. For example, in addition to laying out articles for a newsletter, desktop publishers may be responsible for copyediting content or for writing original content themselves. Desktop publisher's writing and editing responsibilities may vary widely from project to project and employer to employer. Smaller firms typically use desktop publishers to perform a wide range of tasks, while desktop publishers at larger firms may specialize in a certain part of the publishing process. (writers and editors are discussed elsewhere in the Handbook.)
Desktop publishers also may be called publications specialists, electronic publishers, DTP operators, desktop publishing editors, electronic prepress technicians, electronic publishing specialists, image designers, typographers, compositors, layout artists, and Web publications designers. The exact name may vary by the specific tasks performed or simply by personal preference.
Work environment. Desktop publishers usually work in clean, air-conditioned office areas with little noise. They generally work a standard workweek; however, some may work night shifts, weekends, or holidays depending upon the production schedule for the project or to meet deadlines.
These workers often are subject to stress and the pressures of short deadlines and tight work schedules. Like other workers who spend long hours working in front of a computer monitor, desktop publishers may be susceptible to eyestrain, back discomfort, and hand and wrist problems.
Most desktop publishers learn their skills by taking classes, completing a certificate program offered on line or through an accredited academic program, or through experience on the job. Experience is the best training and many desktop publishers get started just by experimenting with the software and developing a knack for designing and laying out material for publication.
Education and training. There is generally no educational requirement for the job of desktop publisher. Most people learn on the job or by taking classes on line or through local learning centers that teach the latest software. For those who are interested in pursuing a career in desktop publishing, an associate degree or a bachelor's degree in graphic arts, graphic communications, or graphic design is preferred. Graphic arts programs are a good way to learn about the desktop publishing software used to format pages; assign type characteristics; and import text and graphics into electronic page layouts. The programs teach print and graphic design fundamentals and provide an extensive background in imaging, prepress operations, print reproduction, and emerging media. Courses in other aspects of printing also are available at vocational-technical institutes, industry-sponsored update and retraining programs, and private trade and technical schools.
Other qualifications. Although formal training is not always required, those with certificates or degrees will have the best job opportunities. Most employers prefer to hire people who have at least a high school diploma and who possess good communication skills, basic computer skills, and a strong work ethic. Desktop publishers should be able to deal courteously with people, because they have to interact with customers and clients and be able to express design concepts and layout options with them. They also may have to do simple math calculations and compute ratios to scale graphics and artwork and estimate job costs. A basic understanding and facility with computers, printers, scanners, and other office equipment and technologies also is needed to work as a desktop publisher.
Desktop publishers need good manual dexterity, and they must be able to pay attention to detail and work independently. Good eyesight, including visual acuity, depth perception, a wide field of view, color vision, and the ability to focus quickly also are assets. Artistic ability often is a plus. Employers also seek persons who are even tempered and adaptable—important qualities for workers who often must meet deadlines and learn how to operate new equipment.
Advancement. Workers with limited training and experience assist more experienced staff on projects while they learn the software and gain practical experience. They advance on the basis of their demonstrated mastery of skills. Desktop publishing software continues to evolve and gain in technological sophistication. As a result, desktop publishers need to keep abreast of the latest developments and how to use new software and equipment. As they gain experience, they may advance to positions with greater responsibility. Some may move into supervisory or management positions. Other desktop publishers may start their own companies or work as independent consultants, while those with more artistic talent and further education may find job opportunities in graphic design or commercial art positions.
Desktop publishers held about 32,000 jobs in 2006. About 35 percent worked for newspaper, periodical, book, and directory publishers, while 24 percent worked in the printing and related support activities industry. Other desktop publishers work for professional, scientific, and technical services firms and in many other industries that produce printed or published materials.
The printing and publishing industries are two of the most geographically dispersed industries in the United States, and desktop publishing jobs are found throughout the country. Although most jobs are in large metropolitan cities, electronic communication networks and the Internet allow some desktop publishers to work from other locations.
Employment of desktop publishers is expected to experience little or no change over the 2006-2016 decade because more people are learning basic desktop publishing skills as a part of their regular job functions in other occupations and because more organizations are formatting materials for display on the web rather than designing pages for print publication.
Employment change. Employment of desktop publishers is expected to grow 1 percent between 2006 and 2016, which is considered little or no change in employment. Desktop publishing has become a frequently used and common tool for designing and laying out printed matter, such as advertisements, brochures, newsletters, and forms. However, increased computer processing capacity and widespread availability of more elaborate desktop publishing software will make it easier and more affordable to use for people who are not printing professionals. As a result, the need for people who specialize in desktop publishing will slow, as more people are able to do this work.
In addition, organizations are increasingly moving their published material to the Internet to save the cost of printing and distributing materials. This change will slow the growth of desktop publishers, especially in smaller membership and trade organizations, which publish newsletters and small reports. Companies that produce large reports and rely on high quality and high resolution color and graphics within their publications, however, will continue to use desktop publishers to lay out publications for offset printing.
Job prospects. Despite the little to no change in projected employment, job opportunities for desktop publishers are expected to be good because of the need to replace workers who move into managerial positions, transfer to other occupations, or leave the labor force. However, job prospects will be better for those with experience as many employers prefer to hire experienced desktop publishers because of the long time it takes to become good at this work. Among individuals with little or no experience, opportunities should be best for those with computer backgrounds, certification in desktop publishing, or who have completed a postsecondary program in desktop publishing, graphic design, or web design.
Projections Data
Projections data from the National Employment Matrix
Occupational title
SOC Code
Employment, 2006
Change, 2006-16
43-9031 32,000 32,000 300 1 PDF zipped XLS
NOTE: Data in this table are rounded. See the discussion of the employment projections table in the Handbook introductory chapter on Occupational Information Included in the Handbook.
Earnings for desktop publishers vary according to level of experience, training, geographic location, and company size. Median annual earnings of desktop publishers were $34,130 in May 2006. The middle 50 percent earned between $26,270 and $44,360. The lowest 10 percent earned less than $20,550, and the highest 10 percent earned more than $55,040 a year. Median annual earnings of desktop publishers in May 2006 were $36,460 in printing and related support services and $31,450 in newspaper, periodical, book, and directory publishers.
Related Occupations
Desktop publishers use artistic and editorial skills in their work. These skills also are essential for artists and related workers; commercial and industrial designers; prepress technicians and workers; public relations specialists; and writers and editors.
Sources of Additional Information
Details about training programs may be obtained from local employers such as newspapers and printing shops or from local offices of the State employment service.
For information on careers and training in printing, desktop publishing, and graphic arts, write to:
Graphic Arts Education and Research Foundation, 1899 Preston White Dr., Reston, VA 20191-4367. Internet: http://www.gaerf.org
Graphic Arts Information Network, 200 Deer Run Rd., Sewickley, PA 15143. Internet: http://www.gain.net
Source: Bureau of Labor Statistics Occupational Outlook Handbook

redpajama_set_name: RedPajamaCommonCrawl


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The local newspaper has a "Community Voices" section and I've been published! It was one of my short stories, a bit different from what I usually write, but it's nice to see myself in print regardless. I'm still waiting on a hard copy of "Der Reizen", but it could still be in the mail today. I haven't checked the mailbox. It's too cold and snowy to be walking down to the mailbox today. Maybe tomorrow.