\section{Introduction}
 For any Banach algebra $\mA$, Richard Arens (in \cite{Arens}) defined
 two products $\Box$ and $\Diamond$ on it bidual  space $\mA^{**}$ such that each product
 makes $\mA^{**}$ into a Banach algebra and the canonical isometric
 inclusion $J:\mA\to \mA^{**}$ becomes a homomorphism with respect to both
 the products. A Banach algebra $\mA$ is said to be \textit{Arens
   regular} if the two products $\Box$ and $\diamond$ agree on
 $\mA^{**}$, i.e. $f\Box g=f\diamond g$ for all $f,g\in \mA^{**}$. Even
 when a Banach algebra is not Arens regular, the size of its
 topological center carry a great deal of imformation.  Dales and Lau
 introduced the notion of strong Arens irregularity to distinguish
 between non-Arens regular Banach algebras. \smallskip

Briefly speaking, a Banach algebra $\mA$ is said to be {\em left}
(resp., {\em right}) {\em strongly Arens irregular} if its so called
{\em left topological center} $Z_t^{(l)} (\mA^{**})$ (resp., {\em
  right topological center} $Z_t^{(r)} (\mA^{**})$) equals $\mA$. And,
$\mA$ is said to be {\em strongly Arens irregular} (in short, S.A.I.)
if it is both left and right strongly Arens irregular.  Interestingly,
on the other extreme, it is known that $\mA$ is Arens regular if and
only if $Z_t^{(l)} (\mA^{**}) = \mA^{**} = Z_t^{(r)}
(\mA^{**})$. Reflexive spaces turn out to be at the junction of Arens
regularity and strong Arens irregularity because
$\mA=\mA^{**}=Z^{(l)}(\mA^{**})=Z^{(t)}(\mA^{**})$.\smallskip

 After the pioneering paper \cite{Arens} by Richard Arens, people
 started investigating the Arens regularity of various Banach algebras
 that appear in abstract Harmonic analysis. The group algebra $L^1(G)$
 has been a center of attention for quite some time. Arens himself
 proved that the semi-group algebra $\ell^1(\mathbb N)$ (with respect
 to convolution) is not Arens regular. Graham, in \cite{Graham},
 proved very elegantly that the group algebra $L^1(G)$ is not Arens
 regular for all infinite discrete amenable groups and non-discrete
 locally compact groups. Finally, Losert and Lau in \cite{Lau-losert}
 proved that $L^1(G)$ is strongly Arens irregular for all infinite
 locally compact groups.\smallskip

 It is then quite natural to analyze the strong Arens irregularity of
 generalized group algebras $L^1(G,\mA)$ for infinite locally compact
 groups $G$ and Banach algebras $A$; and, this short article is just a
 report on some progress made in this direction. A lot still remains
 to be explored. Since $L^1(G)$ is not Arens regular, and because
 $L^1(G, A) \cong L^1(G) \obp A$, we know from \cite{Ulger} that
 $L^1(G,\mA)$ is also not Arens regular. We wish to study under what
 conditions on $\mA$, the Banach algebra $L^1(G,\mA)$ is strongly
 Arens irregular. Needless to mention, if $\mA=\mathbb C^n$ with
 point-wise multiplication and any $p$-norm and $G$ is infinite, then
 $L^1(G, \mA)$ is strongly Arens irregular (due to
 \cite{Lau-losert}). Further, since there is an isometric algebra
 isomorphism from $L^1(G)\obp L^1(H)$ onto $L^1(G\times H)$
 (w.r.t. convolution on both the legs) - see \cite{Kaniuth}, it follow
 again from \cite{Lau-losert} that $L^1(G,\mA)$ is strongly Arens
 irregular for $\mA=L^1(H)$, where $H$ is an infinite group. However,
 on the other extreme, we give a class of Banach algebras $\mA$ for
 which $L^1(G,\mA)$ is not S.A.I. for any infinite discrete group $G$.
 This is interestingly in contrast to the above mentioned fact that
 the (infinite) group algebra $L^1(G)$ is always S.A.I. - see
 \cite{Lau-losert}.  Our approach relies on chasing the annihilators
 in the bidual of $L^1(G, \mA)$ using appropriate functionals
 associated to the non-principal ultra-filters on $\N$.



\section{Preliminaries}

\subsection{Arens regularity and strong Arens irregularity}
We quickly recall the definitions of the two products $\Box$ and
$\Diamond$.  Let $\mA$ be a Banach algebra. For $a \in \mA$,
$\omega\in \mA^*$, $f\in \mA^{**}$, consider the functionals
$\omega_a, {}_{a}\omega\in \mA^*$, $\omega_f,{}_f\omega\in \mA^{**}$
given by
\[
w_a=(L_a)^*(\omega),
{}_a\omega=(R_a)^*(\omega); \omega_f(a)=f({}_a\omega)\text{ and }
{}_f\omega(a)=f(\omega_a).
\]
Then, for $f,g\in \mA^{**}$ the
operations $\Box$ and $\Diamond$ are given by
\[
(f\Box g)
(\omega)= f({}_g\omega)\text{ and } (f\Diamond
g)(\omega)=g(\omega_f)
\]
for all $\omega \in \mA^*$. And, $\mA$ is said to be {\em Arens regular} if the
products $\Box$ and $ \Diamond$ agree.

Further, the left and the right topological centers of $\mA$ are defined,
respectively, as
\[
\begin{array}{rcl}
Z_t^{(l)} (\mA^{**}) & = & \{\varphi \in \mA^{**}: \phi\Box
\psi=\phi\Diamond\psi \text{ for all } \psi\in \mA^{**} \}; \text{ and }
\\ Z_t^{(r)} (\mA^{**}) & = & \{\varphi \in \mA^{**}:
\psi\Box\phi=\psi\Diamond \phi \text{ for all } \psi\in \mA^{**} \}.
\end{array}
\]
Identifying $\mA$ with $J(\mA)$, where $J : \mA \to \mA^{**}$ is the
canonical isometry, it is known that $\mA \subseteq Z_t^{(l)}
(\mA^{**}) \cap Z_t^{(r)} (\mA^{**})$.  It is easily seen that $\mA$ is Arens
regular if and only if $Z_t^{(l)} (\mA^{**}) = \mA^{**} = Z_t^{(r)}
(\mA^{**})$.   A Banach algebra $\mA$ is said
to be {\em left} (resp., {\em right}) {\em strongly Arens irregular}
if $Z_t^{(l)} (\mA^{**}) = \mA$ (resp., $Z_t^{(r)} (\mA^{**}) =
\mA$). And, $\mA$ is said to be {\em strongly Arens irregular} if it
is both left and right strongly Arens irregular. We refer the reader to \cite{DL} for further discussion
on topological centers.

\subsection{Projective tensor product}
Let $X$ and $Y$ be Banach spaces. Then,  there
are various ways of imposing a normed space  structure on their algebraic tensor product $X
\otimes Y$.   For instance, for each $u \in X \otimes Y$, its
projective norm is given by
\[
\Vert u \Vert_\gamma =\inf\left\{\sum_{i=1}^{n}\Vert x_i\Vert
\Vert y_i\Vert~:~u=\sum_{i=1}^{n}x_i\otimes y_i\right\}.
\]
This norm turns out to be a \textit{cross norm}, i.e., $\|x
\otimes y\|_{\gamma} = \|x\| \|y\|$ for all $(x,y) \in X \times
Y$; and, the completion of the normed space $X \otimes Y$ with
respect to this norm is  usually denoted by
$X \otimes^\gamma Y$.

Let $\mA$ and $\mB$ be Banach algebras. Then, there is a natural
multiplication structure on their algebraic tensor product $\mA
\otimes \mB$ satisfying $(a_1\otimes b_1)(a_2\otimes
b_2)=a_1a_2\otimes b_1b_2$ for all $a_i \in \mA$ and $b_i \in \mB$,
$i=1,2$. And, there are various ways of imposing a normed algebra
structure on $\mA \otimes \mB$.  For instance, completing $\mA \otimes
\mB$ with respect to the projective norm gives a Banach algebra
structure on $\mA \obp \mB$, which occupies a prominent space  in the
theory of tensor products of Banach algebras.

\subsubsection{Generalized group algebras} 
Let $G$ be a locally compact group and $A$ be a Banach algebra. Then,
recall that, $L^1(G,\mA)$ denotes the space of all Bochner integrable
functions from $G$ to $\mA$ with the norm $||f||_1:=\int_G||f||d\mu$
and forms a Banach algebra with respect to the convolution product
defined as $f\star g(x)=\int_Gf(s)g(s^{-1}x)ds$ - see \cite{Kaniuth}
for further details on generalized group algebras. When $G$ is a
discrete group, then $L^1(G,\mA)$ is nothing but the collection of all
absolutely summable $\mA$-valued functions and $||f||_1=\sum_{g\in
  G}||f(g)||$.

The following natural identifications are well known  and will be used ahead - see, for
instance, \cite{Kaniuth, Ryan} .
\begin{theorem}\label{obp-facts}
  \begin{enumerate}
  \item For any two Banach spaces $X$ and $Y$, there is a canonical
    surjective isometry $\theta: B(X, Y^*) \to (X \obp Y)^* $
    satisfying
    \[
\theta(f)(x\obp y) = f(x)(y) \text{ for all } f \in B(X, Y^*), x \in X, y \in Y.
    \]
\item For any locally compact group $G$ and Banach algebra $A$, there
  exists an isometric algebra isomorphism $T:L^1(G)\obp A\to L^1(G,A)$
  satisfying
  \[
  T(f\otimes a)(g)=f(g)a \text{ for all } f \in f \in L^1(G), a \in A, g \in G.
  \]
  \end{enumerate}
\end{theorem}
The following elementary observations  will be used
ahead. We include their proofs just for the sake of convenience.


\begin{lemma}\label{countable-support}
	Let $G$ be an infinite discrete group and $\mA$ be a Banach
        algebra. Then,
        \begin{enumerate}
\item if $x= \sum_{n\in F} \delta_{g_n} \ot a_n \in L^1(G) \obp \mA$
  for some $F \subseteq \N$, a  collection $\{g_n: n \in F\} \subseteq G$ with no repetitions and a collection $\{a_n: n
  \in F\} \subset \mA$,  then $\|x\|_\gamma= \sum_{n}\|a_n\|$; and,
\item for each $x\in L^1(G) \obp \mA$, there exists
        an $F \subset \N$, a  collection $\{h_n: n \in F\}\subset G$ with no repetitions and
a collection        $\{b_n: n \in F\} \subset \mA\setminus \{0\}$ such that $x = \sum_{n\in F}
        \delta_{h_n} \ot b_n $. 
\end{enumerate}
\end{lemma}
\begin{proof}
  By \Cref{obp-facts}(2), we know that $L^1(G) \obp \mA \cong L^1(G,
  \mA)$ isometrically algebraically.

  (1): Identifying $x$ with its corresponding element in $L^1(G,
  \mA)$, we see that $x(g) = a_n$ if $g=g_n$ for some $n$ and $0$ otherwise; so,
  \[
\|x\|_{\gamma}=  \|x\|_1 = \sum_{g\in G} \|x(g)\| = \sum_{n\in F} \|x(g_n)\| = \sum_{n\in F} \|a_n\|.
  \]

  (2): Again, identifying $x$ with its
  corresponding element in $L^1(G, \mA)$, it is a standard fact that $x: G \to \mA$
  has a (possibly finite) countable support, say, $\{ h_n : n \in F\}$ for some
     $F \subset \N$. Let $b_n:=x(h_n)\in \mA$, $n \in F$.     Next,  note that, $\|x\|_\gamma =\|x\|_1= \sum_{g\in G} \|x(g)\| =
    \sum_{n\in F} \|x(h_n)\|$. Hence, $\{b_n: n \in F\}$ converges
    absolutely in $\mA$ and we also have $ x = \sum_{n\in F}
    \delta_{h_n}\ot b_n$. 
    \end{proof}


\section{Topological center of Generalized group algebras}
Throughout this section, $G$ will denote an infinite discrete
group.

Recall that, for a Banach space $Z$ with a normalized Schauder basis
$\{z_n\}$, the dual functionals $\{z_n^*\}$ are given by $z_n^*(\sum_m
\alpha_m z_m)= \alpha_m$ and it is easily seen that each $z_n^*$ is a
bounded linear functional on $Z$. In fact, the collection $\{z_n^*\}$ is uniformly
bounded - see \cite[Chapter 4]{Fabian}.

The following observation is the crux of this article.
\begin{prop}\label{phi-U} 
	Let $\mA$ be a Banach algebra with a (normalized)
	Schauder basis and let $X:=L^1(G,\mA)$. Then, for each non-principial ultrafilter
	$\mU$ on $\N$, there exists a  $\varphi_{\mU} \in X^{**}\setminus
	J(X)$.
\end{prop}

\begin{proof}
	Let  $\mU$ be a non-principal ultrafilter on $
	\mathbb N$.  Fix a countable set $S=\{h_n\}_{n=1}^\infty$ consisting of  distinct elements of $G$. 
	
	Notice that, for each $\tau \in X^*$, via the identifications illustrated in \Cref{obp-facts}, we have
	\[
	|\tau
	(\delta_{h_n}\otimes a_n)| \leq \|\tau\| \text{ for all } n \in \N
	.
	\]
	Thus, since limit along an ultrafilter in a compact set always
        exists, we have a map $\varphi_\mU: X^* \to \C$ given by
	\begin{equation}\label{phi-U-defn}
	\varphi_\mU(\tau)=\lim_{\mU}\tau (\delta_{h_n}\otimes a_n)   \text{ for all }
	\tau \in X^*.
	\end{equation}
	Clearly, $\varphi_\mU$ is a bounded linear functional on
        $X^*$. Also, $\varphi_\mU= \lim^{w^*}_\mU J(\delta_{h_n}
        \otimes a_n)$.
	
	We first assert that $\varphi_\mU \neq 0$. Consider $\eta:L^1(G)\to \mA^*$ defined
	as
	\[
	\eta(\sum_{g\in
		G}\alpha_g\delta_g)=\sum_{i=1}^\infty\alpha_{h_i}a_i.
	\]
	One can
	easily verify $\eta \in B(L^1(G), \mA^*)$ and that $||\eta||\leq 1
	$. Clearly, $\varphi_{\mU}(\eta)=1$. Hence, $\varphi_{\mU}\neq 0$.
	\medskip
	
	
	Next, consider the subset $\Delta:=\operatorname{span}\{\mT_{g,i}~\vert ~g\in
	G, i\in \mathbb N\}$ of $X^*$, where $\mT_{g,i}: L^1(G) \to \mA^*$ is
	defined as
	\[
	\mT_{g,i}(\sum_{h\in G}c_h\delta_h)=c_ga_i^*.
	\]
	(One can easily see that $\mT_{g,i}\in B(L^1(G), \mA^*)$ and are uniformly bounded and, hence, in
	view of \Cref{obp-facts}, $\Delta$ is a well defined non-trivial set
	in $X^*$.)
	
	We now claim that $J_x$ does not vanish on $\overline{\Delta}$ for any
	$0 \neq x\in X$, while $\varphi_{\mU}$ vanishes (on $\Delta$ and
	hence) on $\overline{\Delta}$. Since $\varphi_\mU \neq 0$, this will then imply that $\varphi_\mU
	\notin J(X)$.
	
	The latter claim is evident from the fact that
	$\varphi_{\mU}(\mT_{g,i})=0$ for each $g\in G$ and $i\in \mathbb
	N$.
	
	To prove the first part of the claim, suppose $x \in
        X\setminus \{0\}$. Then,  as observed in \Cref{countable-support},
        there exist countable sets $F\subset \N$, $\{u_j : j\in F\}
        \subset G$ and $\{b_j: j \in F\}\subset \mA\setminus\{0\}$ such that
        $x=\sum_{j=1}^\infty\delta_{u_j}\otimes b_j$.

        Note that, each $b_j$ can be written as $b_j=\sum_{n=1}^\infty
        c^{(j)}_na_n$. So, for each $j\in \N$, we can choose an $n_j$
        such that $c^{(j)}_{n_j}\neq 0$.  Consider $\mT\in X^*$
        given by $\mT=\sum_{j=1}^\infty\frac{1}{j^2}\beta_j
        \mT_{u_j,n_j}$, where
        $\beta_j=\frac{\overline{c^{(j)}_{n_j}}}{|c^{(j)}_{n_j}|}$. (Note that 
        $\mT\in X^*$ because
        its defining series is absolutely convergent in the norm of
        $X^*$.) Clearly, $\mT\in \overline{\Delta}$. 

        Now, notice that \begin{align*}
          J_x(\mT)&=\mT(x)\\&=\sum_{j=1}^\infty\sum_{i=1}^\infty\frac{1}{j^2}\beta_j\mT_{u_j,n_j}(\delta_{u_i})(b_i)\\&=\sum_{j=1}^\infty\frac{1}{j^2}\beta_j
          a_{n_j}^*(b_j)\\&=\sum_{j=1}^\infty\frac{1}{j^2}\beta_jc^{(j)}_{n_j}\\&=\sum_{j=}^\infty
          \frac{1}{j^2}|c_{n_j}^{(j)}|\neq 0.
	\end{align*}
	Thus, $\varphi_\mU\in X^{**}\setminus J(X)$.
\end{proof}

The following definition is well known to the experts - see, for instance, \cite[Chapter 4]{Fabian}.
\begin{definition}\label{eqdef}
	A (normalized) Schauder basis $\{z_n\}$ of a Banach space $Z$
        is said to be equivalent to the canonical Schauder basis of
        $\ell^1(\mathbb N)$ if there exists a positive scalar $C$ such
        that
        \[
        \LL\sum_{i=1}^N\lambda_iz_i\RR\leq\sum_{i=1}^N|\lambda_i|\leq
        C\LL\sum_{i=1}^N\lambda_iz_i\RR
        \]
        for any set of  scalars $\{\lambda_i\}_{i=1}^N$ and $N\in \mathbb N$.  
\end{definition}
Notice that the first inequality in the preceding definition is
naturally true for any normalized Schauder basis in any Banach space. It is
worth noticing that $C\geq 1$ always.\\

\begin{lemma}\label{equiv}
	Let $G$ be an infinite discrete group and $Z$ be a Banach
        space with a normalied Schauder basis $\{z_n\}_{n=1}^\infty$ which is
        equivalent to the canonical Schauder basis of $\ell^1(\mathbb
        N)$ with equivalence constant $C$ (as in
        \Cref{eqdef}). Then,
        \[
        \frac{1}{C}\sum_{i=1}^N|\lambda_i|\leq
        \LL\sum_{i=1}^N\lambda_i\delta_{g_i}\otimes
        z_{n_i}\RR_\gamma
        \]
        for any $\{g_1,g_2,\ldots
        ,g_N\}\subset G$, $\{n_1,n_2, \ldots ,n_N\}\subset \mathbb N$
        and $N \in \N$.
\end{lemma}

\begin{proof}
We prove this by induction on $N$.  Notice that the statement is true for $N=2$ because
  \[
  ||\lambda_1\delta_{g_1}\otimes z_{n_1}+\lambda_2\delta_{g_2}\otimes z_{n_2}||_\gamma=\begin{cases}
 |\lambda_1|+|\lambda_2| & \textrm{if $g_1\neq g_2$}; \text{ and}\\
||\lambda_1z_{n_1}+\lambda_2z_{n_2}|| & \textrm{if $g_1=g_2$},
  \end{cases}
  \]
where the first equality follows from \Cref{countable-support}(1).
Thus, since $C \leq 1$, we have
\[
        \frac{1}{C}\sum_{i=1}^2|\lambda_i|\leq
        \LL\sum_{i=1}^2\lambda_i\delta_{g_i}\otimes
        z_{n_i}\RR_\gamma        .
        \]

        Now, let the statement be true for $N=r>2$. Then, for
any given collections $\{g_1, \ldots, g_{r+1}\} \subset G$ and $\{n_1, \ldots, n_{r+1}
\}\subset \N$, consider
\[
f:=\sum_{i=1}^{r+1}\lambda_i
\delta_{g_i}\otimes z_{n_i}\in L^1(G) \obp Z.
\]
If all the $g_i$'s are same then 
$||f||_\gamma=||\sum_{i=1}^{r+1}\lambda_iz_{n_i}||\geq
\frac{1}{C}\sum_{i=1}^{r+1}|\lambda_i|$, because $\|\cdot\|_\gamma$ is a cross norm and  the Schauder basis
$\{z_n\}$ is normalized and equivalent with the canonical Schauder basis of $\ell^1(\mathbb N)$.


On the other hand, if some of the $g_i$'s are distinct,  we
can choose a $g_j$ such that it has less than $r+1$ (say $k$)
repetitions (we may even have $k=1$); and, suppose that $\{n_t: 1 \leq t \leq k\} = \{m: g_j = g_m\}$. Then, 
\begin{align*}
  ||f||_\gamma&=||\delta_{g_j}\otimes
  (\sum_{t=1}^k\lambda_{j_t}z_{n_{j_t}})||_\gamma+||\sum_{i\neq
    j_t~\text{for any $t$}}\lambda_i\delta_{g_i}\otimes
  z_{n_i}||_\gamma\\ &\geq\LL\sum_{t=1}^k\lambda_{j_t}z_{n_{j_t}}\RR+
  \frac{1}{C} \sum_{i\neq j_t~\text{for any $t$}}|\lambda_i| &&
  (\text{by induction hypothesis}) \\ &\geq
  \frac{1}{C}\sum_{t=1}^k|\lambda_{j_t}|+ \frac{1}{C} \sum_{i\neq
    j_t~\text{for any $t$}}|\lambda_i| && (\text{by equivalence of
    bases})
  \\&=\frac{1}{C}\sum_{i=1}^{r+1}|\lambda_i|.
\end{align*}
Thus, by the principle of induction, the statement is true for all $N\in \mathbb N$.
\end{proof}



Recall that a Schauder basis $\{a_n\}$ of a Banach algebra $\mA$ is
said to be orthogonal if $a_n\cdot a_m=\delta_{mn}a_n$ for all $m,n\in
\mathbb N$.  We suggest the reader to see \cite{Watson} for more on
orthogonal Schauder bases of Banach algebras.

\begin{theorem} \label{mainth}
	Let $G$ be an infinite discrete group, $\mA$ be a Banach
        algebra with an orthonormal Schauder basis equivalent to the
        canonical basis of $\ell^1(\mathbb N)$ and $\mU$ be a
        non-principal ultrafilter on $ \mathbb N$.  Then,
        $\varphi_{\mU}\in Z^{(l)}_t((L^1(G, \mA))^{**})$.

 In particular, the generalized group algebra $L^1(G,\mA)$ is not strongly
        Arens irregular.
\end{theorem}
\begin{proof}
  Let $ X:=L^1(G,\mA)$ and $\{a_n\}$ denote an orthonormal Schauder
  basis of $\mA$ which is equivalent to the standard Schauder basis of
  $\ell^1(\N)$. Cosider $S :=\{h_n\} \subseteq G$ as in
  \Cref{phi-U}. Then, for any $\psi\in X^{**}$ and $\mT\in X^*$, we
  have
  \begin{align*}
  ( \varphi_\mU\square \psi)(\mT)&=\lim_{\mU}\ {_\psi}\mT(\delta_{h_n}\otimes a_n)\\
  &=\lim_{\mathcal U}\psi(\mathcal T_{\delta_{h_n}\otimes a_n}).
  \end{align*}
	We claim that $\lim_{n}\psi(\mT_{\delta_{h_n}\otimes
          a_n})=0$.

        To prove this we assume the contrary. Then, there exists an
        $\epsilon >0$ and an increasing sequence $\{n_t\}_{t\in
          \mathbb N}\subset \N$ such that
        $|\psi(\mT_{\delta_{h_{n_t}}\otimes a_{n_t}})|\geq \epsilon $
        for all $t\in \mathbb N$.

        Consider the functionals
        $h_N:=\sum_{t=1}^Nc_{n_t}\mT_{\delta_{h_{n_t}}\otimes
          a_{n_t}}$ for $N \in \N$, where
        $c_{n_t}:=\frac{\overline{\psi(\mT_{\delta_{h_{n_t}}\otimes
              a_{n_t}})}}{|\psi(\mT_{\delta_{h_{n_t}}\otimes
            a_{n_t}})|}$. Then, for any two finite sets $\{g_i : 1
        \leq i \leq k\}\subset G$ and $\{s_i : 1 \leq i \leq
        k\}\subset\N$ (with possible repetitions), if
          $x:=\sum_{i=1}^k\lambda_i\delta_{g_i}\otimes a_{s_i}\in
          L^1(G)\otimes \mA$, then we have
       	\begin{align*}
	||h_N(x)||&=\left|\mT(\sum_{t=1}^N\sum_{i=1}^k\lambda_i c_{n_t}\delta_{h_{n_t}}\star\delta_{g_i}\otimes a_{n_t}\cdot a_{s_i})\right|\\
	&=\left|\mT(\sum_{i=1}^k\lambda_ic_{s_i}\delta_{h_{s_i}+g_i}\otimes a_{s_i})\right|& & \textrm{($\because$~ $a_i\cdot a_j=\delta_{i,j}a_i$ )}\\
	&\leq||\mT||\cdot  \left|\left| \sum_{i=1}^k\lambda_ic_{s_i}\delta_{h_{s_i}+g_i}\otimes a_{s_i}\right|\right|_\gamma \\
	&\leq||\mT||\cdot \sum_{i=1}^k|\lambda_i|& & \textrm{($\because |c_{s_t}|=1$)}\\
        & \leq C \|\mT\| \|x\|_\gamma. & & \textrm{(due to \Cref{equiv})}
	\end{align*}
Since such $x$'s form a  dense subspace of  $L^1(G) \obp \mA$,  it follows
that $||h_N||\leq C||\mT||$ for all $N$. Now notice that
\[
\psi(h_N)=\sum_{t=1}^N\left|\psi(\mT_{\delta_{h_n}\otimes
  a_{n_t}})\right|\geq N\epsilon
\]
for all $N \in \N$. But this is
absurd because $\psi$ is a bounded linear functional and $\{h_N\}$ is a
bounded sequence. Thus, our assumption is wrong, and we must have
        $$
        \lim_{n}\psi(\mT_{\delta_{h_n}\otimes a_n})=0.
        $$ Thus, $(\varphi_\mU\square \psi)(\mT)=0$ (because the limit
        along an ultrafilter in a compact Hausdorff space is unique and
        every non-principal ultrafilter contains the cofinite
        filter). Since $\psi$ and $\mT$ were arbitrary, we conclude
        that $\varphi_\mU\square \psi=0$ for all $\psi\in X^{**}$.

        Next, we turn to the second Arens product. We have 
        $(\varphi_\mU\Diamond
        \psi)(\mT)=\psi(\mT_{\varphi_{_{\mU}}})$ for all $\psi \in X^{**}$, $\mT \in X^*$. Notice that for any $u\in
        L^1(G)\obp \mA$, we have
         \begin{align*}
         \mT_{\varphi_{_{\mU}}}(u)&=\varphi_{\mU}({_u}{\mT})\\&=\lim_{\mU}{_u\mT}(\delta_{h_n}\otimes a_n)\\&=\lim_{\mU}\mT_{\delta_{h_n}\otimes a_n}(u).
        \end{align*}
        Thus,
        $\mT_{\varphi_{_\mU}}=\overset{w^*}{\lim_{\mU}}\mT_{\delta_{h_n}\otimes
          a_n}$ and hence
        $\varphi_{_\mU}\Diamond\psi(\mT)=\psi(\overset{w^*}{\lim_{\mU}}\mT_{\delta_{h_n}\otimes
          a_n})$ for all $\psi \in X^{**}$, $\mT \in X^*$. Further,
        for an arbitrary $u=\sum_{i=1}^\infty \lambda_if_i\otimes
        b_i\in L^1(G,\mA)$, we have \begin{align*} \lim_{n}\left|
          \mT_{\delta_{h_{n}}\otimes a_n}(u)\right|&= \lim_n
          \left|\mT(\sum_{i=1}^\infty\lambda_i\delta_{h_n}\star
          f_i\otimes a_n^*(b_i))\right|&& (\because
          a_n\cdot b_i=a_n^*(b_i)~ \text{due to orthonormality})\\ &=0.
          &&(\because \lim_na_n^*(b_i)=0~\forall i)
      \end{align*} Thus,
        $\overset{w^*}{\lim_{\mU}}\mT_{\delta_{g_n}\otimes
          e_n}=0$. Hence, $\varphi_\mU\Diamond\psi(\mT)=0$ for all
        $\psi\in X^{**}$ and all $\mT\in X^*$. 
	
	Thus, we have shown that $\varphi_\mU\square\psi= 0 =\varphi_{\mU}\Diamond \psi$ for all $\psi\in X^{**}$. In particular,
	$ \varphi_{\mU}\in Z^{(l)}_t(X^{**})$.

        By \Cref{phi-U}, we know that $\varphi_\mU \notin  J(L^1(G,\mA))$. Hence, $L^1(G,\mA)$ is not strongly Arens irregular.
\end{proof}


\begin{remark}
A Banach algebra $\mA$ satisfying the hypothesis of \Cref{mainth} is
in fact forced to be algebraically isomorphic (not necessarily
isometric) to $\ell^1(\N)$ (with respect to pointwise
multiplication). Indeed, the map
$\theta:\ell^1(\mathbb N)\to \mA$ defined as $\theta(\sum_{n=1}^\infty
c_ne_n)=\sum_{n=1}^\infty c_na_n$ defines a algebra isomorphism. Thus, in other words, such an $\mA$ can always be thought
of as $\ell^1(\N)$ with some equivalent Banach algebra norm. 


\end{remark}
\color{black}
