o}

\begin{proof}
Let ${\cal A}$\ be a $C^*$-algebra and $I$ be a two-sided closed ideal in
${\cal A}$, then $I$ is a $C^*$-algebra by (\cite{B-D}, Theorem~38.18),
so $I$ is weakly amenable. By Lemma~2.1, ${\cal A}$ is $I$-weakly
amenable.\hfill $\mbox{\zz{\char'245}}$
\end{proof}

\begin{coro}$\left.\right.$\vspace{.5pc}

\noindent Let $H$ be a Hilbert space{\rm,} $K(H)$ be the space of
compact operators in $B(H)$. Then $H^1(B(H),K(H)^*)=\{0\}$.
\end{coro}

\begin{coro}$\left.\right.$\vspace{.5pc}

\noindent Let $H$ be an infinite dimensional Hilbert space. Then $B(H)$
is ideally amenable{\rm,} but not amenable.
\end{coro}

Let $G$ be a locally compact topological group. Then $L^1(G)$ is an
ideal in $L^1(G)^{**}$ if and only if $G$ is compact, then by Lemma~2.1
we have the following result.

\begin{coro}$\left.\right.$\vspace{.5pc}

\noindent Let $G$ be a compact topological group. Then $L^1(G)^{**}$ is
$L^1(G)$-weakly amenable.
\end{coro}

\section{Maximal ideals}

Let $M$ be a maximal ideal in $\cal A$. We study conditions when
$\cal A$ is $M$-weakly amenable.

\setcounter{theo}{0}
\begin{theor}[\!]
Let $\cal A$ be a unital weakly amenable Banach algebra. Then for every
maximal ideal $ M$ in $\cal A${\rm,} $\cal A$ is $ M$-weakly amenable.
\end{theor}

\begin{proof}
Let $D\hbox{:}\ {\cal A}\longrightarrow { M}^*$ be a derivation. For
each $a$ in $\cal A$, $D(a)$ has an extension $\tilde{D}(a)\in \cal
A^*$ such that $\tilde{D}(1)=0$. For $ a,b \in \cal A$
there exists $\lambda_1, \lambda_2 \in \Bbb{C}$ and $m,n \in M$ such
that
\begin{align*}
\tilde{{D}}(ab) &= \tilde {D}((\lambda_1\cdot 1+m)(\lambda_2\cdot 1+n))\\[.2pc]
&= \lambda_1 {D}(n)+\lambda_2 {D}(m) +\tilde {D}(mn)\\[.2pc]
&= \lambda_1 {D}(n)+\lambda_2 {D}(m) +m\cdot {D}(n)+
{D}(m)\cdot n\\[.2pc]
&= (\lambda_1\cdot 1+m)\cdot {D}(n)+{D}(m)\cdot(\lambda_2+n)\\[.2pc]
&= a\cdot\tilde {D}(b)+\tilde {D}(a)\cdot b.
\end{align*}
Therefore ${\tilde D}\in {\cal Z}^1 ({\cal A},{\cal A}^*)$ and there
exists $a^*\in {\cal A}^*$ such that ${\tilde D}={\delta}_{a^*}$.
Obviously ${D}=\delta_{a^*\mid M}$.\hfill $\mbox{\zz{\char'245}}$
\end{proof}

\begin{theor}[\!]
Let $\cal A$ be a Banach algebra and $M$ be a closed two-sided ideal in
$\cal A$ of codimension one. If $\cal A$ is $M$-weakly amenable{\rm,}
then $M$ is weakly amenable.
\end{theor}

\begin{proof}
Let ${D}\hbox{:}\ M\longrightarrow M^*$ be a derivation. Since ${\cal
A}=M\oplus \Bbb C$, the map ${D}_1\hbox{:}\ {\cal A}\longrightarrow M^*$
defined by ${ D}_1(m+c)={ D}(m)$ for $m\in {\cal M}, \alpha \in \Bbb C$
is an inner derivation. Consequently ${D}$ is inner.\hfill $\mbox{\zz{\char'245}}$
\end{proof}

Let $F_2$ be the free group on two generators. Let $\ell^0(F_2)=\{\mu\in
{\ell^1(F_2)}\hbox{:}\ \mu(F_2)=0\}$. Then by Theorem~3.1, $\ell^1(F_2)$
is $\ell^0(F_2)$-weakly amenable, and by Theorem~3.2, $\ell^0(F_2)$ is
weakly amenable.

\begin{lem}
Let $\cal A$ be a unital commutative Banach algebra and $M$ be a closed
two-sided ideal in $\cal A$ of codimension one. If $\cal A$ is
$M$\!-weakly amenable{\rm,} then $\cal A $ is weakly amenable.
\end{lem}

\begin{proof}
By the above theorem, we know that $M$ is weakly amenable. On the other
hand, we have ${\cal A}=M\oplus\Bbb C$. Therefore by (\cite{Gr1},
Proposition~2.3), $\cal A$ is weakly amenable.\hfill $\mbox{\zz{\char'245}}$
\end{proof}

Now we have the following theorem:

\begin{theor}[\!]
Let $\cal A$ be a commutative unital Banach algebra{\rm,} the following
assertions are equivalent{\rm :}
\begin{enumerate}
\renewcommand\labelenumi{\rm (\roman{enumi})}
\leftskip.35pc
\item $\cal A$ is weakly amenable{\rm,}

\item $\cal A$ is ideal weakly amenable{\rm,}

\item $\cal A$ is $M$-weakly amenable for some maximal ideal $M$
in $\cal A${\rm,}

\item $H^1({\cal A},X^*)=\{0\}$ for every commutative
Banach $\cal A$-bimodule $X$.
\end{enumerate}
\end{theor}

\begin{propo}$\left.\right.$\vspace{.5pc}

\noindent Let $\cal A$ be a commutative Banach algebra with a bounded
approximate identity and $M$ be a maximal modular ideal in $\cal A$ also
${\cal A}_z\cap M=0$ {\rm (}where ${\cal A}_z$ is the set of topological
divisor of zero elements in $\cal A${\rm )}. Let $D{\rm :}\ {\cal
A}\longrightarrow M$ be a derivation such that $D(M)=\{0\}${\rm,} then\break 
$D=0$.
\end{propo}

\begin{proof}
Let $(e_\alpha)_{\alpha\in I}$ be a bounded approximate identity for
$\cal A$. We may suppose that $e_\alpha\in {\cal A}\setminus M$ for
every $\alpha\in I$. Let $a\in\cal A$ and $D(a)\neq 0$. Then for every
$\alpha\in I$ there exists $a_\alpha\in {\cal A}\setminus M$ and
$m_\alpha\in M$ such that
\begin{equation}
e_\alpha=m_\alpha+a_\alpha a.
\end{equation}
Therefore $0=D^2(e_\alpha)=2D(a_\alpha)D(a)$. Since $D(a)\in M$ and
${\cal A}_z\cap M=0$, for every $\alpha \in I$, $D(a)=0$. On the other
hand, for every $a'\in\cal A$, we have
\begin{align*}
D(a') &= \lim_\alpha D(e_\alpha a')\\[.2pc]
&= \lim_\alpha D(e_\alpha)a' +\lim_\alpha e_\alpha D(a')\\[.2pc]
&= \lim_\alpha D(e_\alpha)a'+D(a').
\end{align*}
Consequently $\displaystyle\lim_\alpha D(e_\alpha)a'=0$ for every $a'\in
A$. Let $a'$ be a non-zero element of $M$, since ${\cal A}_z\cap M=0$,
$\displaystyle\lim_\alpha D(e_\alpha)=0$ and by (1), we have
$\displaystyle\lim_\alpha a_\alpha D(a)=0$. Therefore
$\displaystyle\lim_\alpha a_\