Derivations on group algebras of a locally compact abelian group

Abstract

Let G be a locally compact abelian group. In this paper, we study derivations on the Banach algebra $L_0^\infty (G)^*$. We prove that any derivation on $L_0^\infty (G)^*$ maps it into its radical and a derivation on $L_0^\infty (G)^*$ is continuous if and only if its restriction to the right annihilator of $L_0^\infty (G)^*$ is continuous. We also show that the composition of two derivations on $L_0^\infty (G)^*$ is always a derivation on it and the zero map is the only centralizing derivation on $L_0^\infty (G)^*$. Finally, we characterize the space of inner derivations of $L_0^\infty (G)^*$ and show that G is discrete if and only if there exist $i, j, k\in {\mathbb {N}}$ such that $[d(m), n]_i^j=[m, n]_k$ for all $m, n\in L_0^\infty (G)^*$; or equivalently, any inner derivation on $L_0^\infty (G)^*$ is zero.

1 Introduction

Let G be a locally compact abelian group with a fixed left Haar measure and let $L^1(G)$ be the group algebra of G defined as in [4] equipped with the convolution product $*$ and the norm $\Vert .\Vert _1$. We denote by $L_0^\infty (G)$ the subspace of all functions $f\in L^\infty (G)$, the usual Lebesgue space as defined in [4] equipped with the essential supremum norm $\Vert .\Vert _\infty$, that for each $\varepsilon > 0$, there is a compact subset K of G for which

$$\begin{aligned} \Vert f\chi _{G\setminus K}\Vert _\infty < \varepsilon , \end{aligned}$$

where $\chi _{G\setminus K}$ denotes the characteristic function $G\setminus K$ on G. It is well-known from [6] that the subspace $L_0^\infty (G)$ is a topologically introverted subspace of $L^\infty (G)$, that is, for each $n\in L_0^\infty (G)^*$ and $f\in L_0^\infty (G)$, the function $nf\in L_0^\infty (G)$, where

$$\begin{aligned} \langle nf, \phi \rangle = \langle n, f\phi \rangle ,\quad \hbox {in which}\quad \langle f\phi , \psi \rangle = \langle f, \phi *\psi \rangle \end{aligned}$$

for all $\phi , \psi \in L^1(G)$. Hence $L_0^\infty (G)^*$ is a Banach algebra with the first Arens product “$\cdot$” defined by the formula

$$\begin{aligned} \langle m\cdot n, f\rangle =\langle m, nf\rangle \end{aligned}$$

for all $m, n\in L_0^\infty (G)^*$ and $f\in L_0^\infty (G)$. Note that $L^1(G)$ may be regarded as a subspace of $L_0^\infty (G)^*$ and then $L^1(G)$ is a closed ideal in $L_0^\infty (G)^*$ with a bounded approximate identity [6]. Let $\Lambda _0(G)$ denote the set of all weak$^*$-cluster points of an approximate identity in $L^1(G)$ bounded by one. It is easy to see that if $u\in \Lambda _0(G)$, then for every $m\in L_0^\infty (G)^*$ and $\phi \in L^1(G)$

$$\begin{aligned} m\cdot u=m\quad \hbox {and}\quad u\cdot \phi =\phi . \end{aligned}$$

Let $\pi$ denote the natural continuous operator that associates to any functional in $L_0^\infty (G)^*$ its restriction to $C_0(G)$, the space of all continuous functions on G vanishing at infinity. Then the restriction map $\pi$ from $L_0^\infty (G)^*$ into M(G), the measure algebra of G as defined in [4] endowed with the convolution product $*$ and the total variation norm, is a homomorphism and

$$\begin{aligned} \pi _u:=\pi |_{u\cdot L_0^\infty (G)^*} \end{aligned}$$

is an isomorphism for all $u\in \Lambda _0(G)$. Note that, for every $f\in L_0^\infty (G)$ and $\phi \in L^1(G)$, we have $f\phi \in C_0(G)$. Hence for every $n\in L_0^\infty (G)^*$ and $f\in L_0^\infty (G)$, we may define the function $\pi (n)f\in L^\infty (G)$ by

$$\begin{aligned} \langle \pi (n)f, \phi \rangle =\langle \pi (n), f\phi \rangle . \end{aligned}$$

Then

$$\begin{aligned} \pi (n)f=nf\in L_0^\infty (G). \end{aligned}$$

This enable us to define the functional $m\cdot \pi (n)\in L_0^\infty (G)^*$ by

$$\begin{aligned} \langle m\cdot \pi (n), f\rangle =\langle m, \pi (n)f\rangle . \end{aligned}$$

It follows that

$$\begin{aligned} m\cdot \pi (n)=m\cdot n \end{aligned}$$

for all $m, n\in L_0^\infty (G)^*$; see [6]. Let $\hbox {Ann}_r(L_0^\infty (G)^*)$ denote the right annihilator of $L_0^\infty (G)^*$; i.e. the set of all $r\in L_0^\infty (G)^*$ such that $m\cdot r=0$ for all $m\in L_0^\infty (G)^*$. one can easily prove that

$$\begin{aligned} \hbox {Ann}_r(L_0^\infty (G)^*)= \hbox {ker}(\pi ). \end{aligned}$$

Furthermore, an easy application of the Hahn-Banach theorem shows that G is discrete if and only if

$$\begin{aligned} \hbox {Ann}_r(L_0^\infty (G)^*)=\{0\}. \end{aligned}$$

Let $\mathfrak {A}$ be a Banach algebra; a linear mapping $d: \mathfrak {A}\rightarrow \mathfrak {A}$ is called a derivation if

$$\begin{aligned} d(ab)=d(a)b+ad(b). \end{aligned}$$

A fundamental question for derivations concerns their image. Singer and Wermer [12] showed that the range of a continuous derivation on a commutative Banach algebra is contained in the radical of algebra. They conjectured that this result holds for discontinuous derivations. Thomas [13] proved this conjecture. Posner [10] gave a noncommutative version of the Singer-Wermer theorem for prime rings. He proved that the zero map is the only centralizing derivation on a noncommutative prime ring (Posner’s second theorem). These results have been extended in various directions by several authors; see for instance [1, 3, 5, 7, 8, 11, 14].

Can we apply the well-known results concerning derivations of commutative Banach algebras and derivations of prime rings to $L_0^\infty (G)^*$? This question seems natural, because $L_0^\infty (G)^*$ is neither a commutative Banach algebra nor a prime ring, when G is a non-discrete group. In this paper, we investigate the truth of these results for $L_0^\infty (G)^*$.

This paper is organized as follows: In Sect. 2, we investigate the Singer- Wermer conjecture and automatic continuity for $L_0^\infty (G)^*$. We prove that the range of a derivation on the noncommutative Banach algebra $L_0^\infty (G)^*$ is contained in the radical of $L_0^\infty (G)^*$ and a derivation on $L_0^\infty (G)^*$ is continuous if and only if its restriction to $\hbox {Ann}_r(L_0^\infty (G)^*)$ is continuous. In Sect. 3, we investigate Posner’s second theorem and show that the zero map is the only centralizing derivation on $L_0^\infty (G)^*$. In Sect. 4, we characterize the space of all inner derivations of $L_0^\infty (G)^*$ and prove that G is discrete if and only if any inner derivation on $L_0^\infty (G)^*$ is zero.

2 The Singer-Wermer conjecture for $L_0^\infty (G)^*$

We commence this section with the following result.

Theorem 1

Let G be a locally compact abelian group and d be a derivation on $L_0^\infty (G)^*$. Then d has its image in the right annihilator of $L_0^\infty (G)^*$.

Proof

Let $u\in \Lambda _0(G)$. Define the function $D: M(G)\rightarrow M(G)$ by

$$\begin{aligned} D(\mu )=\pi \circ \tilde{d}\circ \pi ^{-1}_u(\mu ), \end{aligned}$$

where $\tilde{d}=d|_{u\cdot L_0^\infty (G)^*}$. It is routine to check that D is derivation on the commutative semisimple Banach algebra M(G). Hence D is zero. It follows that

$$\begin{aligned} \tilde{d}\circ \pi ^{-1}_u(M(G))\subseteq \hbox {ker}(\pi )=\hbox {Ann}_r(L_0^\infty (G)^*). \end{aligned}$$

Since $\pi _u$ maps $u\cdot L_0^\infty (G)^*$ onto M(G), we have

$$\begin{aligned} d(u\cdot L_0^\infty (G)^*)\subseteq \hbox {Ann}_r(L_0^\infty (G)^*). \end{aligned}$$

On the one hand,

$$\begin{aligned} m\cdot d(r)=d(m\cdot r)-d(m)\cdot r=0 \end{aligned}$$

for all $m\in L_0^\infty (G)^*$ and $r\in \hbox {Ann}_r(L_0^\infty (G)^*)$. So

$$\begin{aligned} d(\hbox {Ann}_r(L_0^\infty (G)^*))\subseteq \hbox {Ann}_r(L_0^\infty (G)^*). \end{aligned}$$

Now, we only need to recall that $L_0^\infty (G)^*$ is the Banach space direct sum of $u\cdot L_0^\infty (G)^*$ and $\hbox {Ann}_r(L_0^\infty (G)^*)$. $\square$

Before we give the following consequence of Theorem 1, let us recall that a linear mapping T on $L_0^\infty (G)^*$ is called spectrally bounded if there is a non-negative number $\alpha$ such that $r(T(m))\le \alpha r(m)$ for all $m\in L_0^\infty (G)^*$, where $r(\cdot )$ stands for the spectral radius.

Corollary 1

Let G be a locally compact abelian group. Then the following statements hold.

• (i) Every derivation on $L_0^\infty (G)^*$ maps it into its radical.

• (ii) Primitive ideals of $L_0^\infty (G)^*$ are invariant under derivations on $L_0^\infty (G)^*$.

• (iii) Every derivation on $L_0^\infty (G)^*$ is spectrally bounded.

• (iv) The composition of two derivations on $L_0^\infty (G)^*$ is always a derivation on $L_0^\infty (G)^*$.

Proof

The statement (i) follows from Theorem 1 together with the fact that the set of nilpotent elements is contained in the radical of the algebra. The statement (ii) follows immediately from (i). For (iii), note that if d is a derivation on $L_0^\infty (G)^*$, then

$$\begin{aligned} d(m)^i=0 \end{aligned}$$

for all $m\in L_0^\infty (G)^*$ and $i\ge 2$. Finally, the statement (iv) follows from Theorem 1. $\square$

As an another consequence of Theorem 1, we have the following result.

Corollary 2

Let G be a locally compact abelian group. Then the following statements hold.

• (i) If d is a derivation on $L_0^\infty (G)^*$, then $d|_{L^1(G)}$ is zero.

• (ii) The zero map is the only $\hbox {weak}^{*}-\hbox {weak}^{*}$ continuous derivation on $L_0^\infty (G)^*$.

Proof

First note that

$$\begin{aligned} r\cdot \phi =\phi \cdot r=0 \end{aligned}$$

for all $r\in \hbox {Ann}_r(L_0^\infty (G)^*)$ and $\phi \in L^1(G)$. So if d is a derivation on $L_0^\infty (G)^*$, then

$$\begin{aligned} d(\phi _1*\phi _2)=d(\phi _1)\cdot \phi _2+\phi _1\cdot d(\phi _2)=0 \end{aligned}$$

for all $\phi _1,\phi _2\in L^1(G)$. In view of Cohen’s factorization theorem, $d=0$ on $L^1(G)$. So (i) holds. The statement (ii) follows from Goldstein’s theorem (see e.g. [2, chapter 5, Proposition 4.1]) and (i). $\square$

Theorem 2

Let G be a locally compact abelian group and d be a derivation on $L_0^\infty (G)^*$. Then the following statements hold.

• (i) For every $u\in \Lambda _0(G)$, $d|_{u\cdot L_0^\infty (G)^*}$ is always continuous.

• (ii) d is continuous if and only if $d|_{\hbox {Ann}_r(L_0^\infty (G)^*)}$ is continuous.

Proof

(i) Let $u\in \Lambda _0(G)$ and $(u\cdot m_\alpha )_{\alpha \in A}$ be a net in $L_0^\infty (G)^*$ such that $u\cdot m_\alpha \rightarrow 0$. It follows from Theorem 1 that

$$\begin{aligned} \Vert d(u\cdot m_\alpha )\Vert= & {} \Vert d(u\cdot u\cdot m_\alpha )\Vert \\= & {} \Vert d(u)\cdot u\cdot m_\alpha \Vert \le \Vert d(u)\Vert \;\Vert u\cdot m_\alpha \Vert \end{aligned}$$

for all $\alpha \in A$. Hence $d(u\cdot m_\alpha )\rightarrow 0$. This shows that $d|_{u\cdot L_0^\infty (G)^*}$ is continuous. (ii) Let $m\in L_0^\infty (G)^*$ and $u\in \Lambda _0(G)$. Then $m-u\cdot m$ is an element of $\hbox {Ann}_r(L_0^\infty (G)^*)$. If $d|_{\hbox {Ann}_r(L_0^\infty (G)^*)}$ is continuous, then for some $\alpha >0$

$$\begin{aligned} \Vert d(m-u\cdot m)\Vert \le \alpha \Vert m-u\cdot m\Vert \le 2\;\alpha \Vert m\Vert . \end{aligned}$$

By (i) there exists $\beta >0$ such that

$$\begin{aligned} \Vert d(u\cdot m)\Vert \le \beta \;\Vert u\cdot m\Vert \le \beta \;\Vert m\Vert . \end{aligned}$$

Thus

$$\begin{aligned} \Vert d(m)\Vert =\Vert d(u\cdot m)+d(m-u\cdot m)\Vert \le (2\alpha +\beta )\Vert m\Vert . \end{aligned}$$

It follows that d is continuous. $\square$

Our last result of this section is an immediate consequence of Theorem 2(ii).

Corollary 3

Let G be a discrete abelian locally compact group.Then every derivation on $L_0^\infty (G)^*$ is continuous.

3 Posner’s second theorem for $L_0^\infty (G)^*$

Let $\hbox {Z} (L_0^\infty (G)^*)$ denote the center of $L_0^\infty (G)^*$; that is, the set of all $m\in L_0^\infty (G)^*$ such that $m\cdot n=n\cdot m$ for all $n\in L_0^\infty (G)^*$.

Proposition 1

Let G be a locally compact abelian group. Then

$$\begin{aligned} Z (L_0^\infty (G)^*)=L^1(G). \end{aligned}$$

Proof

Let $u\in \Lambda _0(G)$. Since $L^1(G)$ is an ideal in $L_0^\infty (G)^*$ and $\pi$ is identity on $L^1(G)$, we have

$$\begin{aligned} \phi \cdot m= & {} \pi (\phi \cdot m)=\pi (\phi )*\pi (m)\\= & {} \pi (m)*\pi (\phi )=\pi (m\cdot \phi )=m\cdot \phi \end{aligned}$$

for all $\phi \in L^1(G)$ and $m\in L_0^\infty (G)^*$. So $L^1(G)$ is contained in $\hbox {Z}(L_0^\infty (G)^*)$. For $m\in \hbox {Z}(L_0^\infty (G)^*)$, we have

$$\begin{aligned} m=m\cdot u=u\cdot m. \end{aligned}$$

This shows that

$$\begin{aligned} m\in \cap _{u\in \Lambda _0(G)}u\cdot L_0^\infty (G)^*. \end{aligned}$$

Hence $\hbox {Z} (L_0^\infty (G)^*)$ is contained in $L^1(G)$; see Theorem 2.11 of [6]. $\square$

For any positive integer k, a mapping $T: L_0^\infty (G)^*\rightarrow L_0^\infty (G)^*$ is called k -centralizing if

$$\begin{aligned}{}[T(m), m^k]\in Z(L_0^\infty (G)^*) \end{aligned}$$

for all $m\in L_0^\infty (G)^*$; in a special case when $[T(m),m^k]=0$ for all $m\in L_0^\infty (G)^*$, T is called k-commuting, where $[m,n]:=m\cdot n-n\cdot m$ for all $m,n\in L_0^\infty (G)^*$.

Theorem 3

Let G be a locally compact abelian group, d be a derivation on $L_0^\infty (G)^*$ and k be a positive integer. Then the following assertions are equivalent.

• (a) $d=0$.

• (b) d is k-centralizing.

• (c) d is k-commuting.

Proof

It is clear that (a) implies (b). If (b) holds, then by Theorem 1 and Proposition 1, we obtain

$$\begin{aligned}{}[d(m),m^k]= & {} d(m)\cdot m^k\\= & {} d(m^{k+1})\in \hbox {Ann}_r( L_0^\infty (G)^*)\cap L^1(G)=\{0\} \end{aligned}$$

for all $m\in L_0^\infty (G)^*$. Thus (c) holds. Now, let d be k-commuting. Choose $u\in \Lambda _0(G)$. Then

$$\begin{aligned} d(u)=[d(u),u]=[d(u),u^k]=0. \end{aligned}$$

(1)

For every $r\in \hbox {Ann}_r(L_0^\infty (G)^*)$, we have $(r+u)=(r+u)^k.$ Hence

$$\begin{aligned} d(r)=[d(r+u),(r+u)]=[d(r+u),(r+u)^k]=0. \end{aligned}$$

(2)

From (1) and (2) we infer that

$$\begin{aligned} d(m)= & {} d(u\cdot m)+d(m-u\cdot m)\\= & {} d(u)\cdot m+d(m-u\cdot m)\\= & {} 0 \end{aligned}$$

for all $m\in L_0^\infty (G)^*$. Thus (c) implies that (a). $\square$

As an immediate consequence from Theorem 3, we have the following result.

Corollary 4

Let G be a locally compact abelian group. Then the zero map is the only centralizing derivation on $L_0^\infty (G)^*$.

Let $[m,n]_1=[m, n]$ and $[m, n]_k=[[m, n]_{k-1}, n]$ for all $m,n\in L_0^\infty (G)^*$ and all positive integers $k>1$.

Corollary 5

Let G be a locally compact abelian group and d be a derivation on $L_0^\infty (G)^*$. Then the following assertions are equivalent.

• (a) $d=0$.

• (b) d is centralizing.

• (c) For every $k\in {\mathbb {N}}$, d is k-centralizing.

• (d) There exists $k\in {\mathbb {N}}$ such that d is k-centralizing.

• (e) There exist positive integers k, l such that $l\ge 2$ and $[d(m), n]_k=[m, n]^l$ for all $m, n\in L_0^\infty (G)^*$

Proof

This follows from Theorem 3 with the observation that for every $m, n\in L_0^\infty (G)^*$, we have $[m, n]\in \hbox {Ann}_r(L_0^\infty (G)^*)$ and so $[m, n]^l=0$ for all $l\ge 2$.$\square$

We conclude the section with the following result.

Theorem 4

Let G be a locally compact abelian group and d be a derivation on $L_0^\infty (G)^*$. Then the following assertions are equivalent.

• (a) G is discrete.

• (b) $L_0^\infty (G)^*$ is commutative.

• (c) There exist $i, j, k\in {\mathbb {N}}$ such that $[d(m), n]_i^j=[m, n]_k$ for all $m, n\in L_0^\infty (G)^*$.

In this case, $d=0$.

Proof

If G is discrete, then by Proposition 3.1 of [9], we have $L_0^\infty (G)^*=L^1(G)$. Since G is an abelian, $L_0^\infty (G)^*$ is commutative. Thus (a) implies (b). It is clear that (b) implies (c) and (d). Now, let $i, j, k\in {\mathbb {N}}$ and

$$\begin{aligned} d(m\cdot n^i)^j=[m, n]\cdot n^k. \end{aligned}$$

Then for every $u\in \Lambda _0(G)$, we have

$$\begin{aligned} d(u)^j=d(u\cdot u^i)^j=[u, u]\cdot u^k=0. \end{aligned}$$

On the one hand, for every $r\in \hbox {Ann}_r(L_0^\infty (G)^*)$, we get

$$\begin{aligned} d(u)^j= & {} d(u\cdot (u+r))^j=d(u\cdot (u+r)^i)^j\\= & {} [u, u+r]\cdot (u+r)^k=[u, u+r]\cdot (u+r)=-r. \end{aligned}$$

Hence

$$\begin{aligned} \hbox {Ann}_r(L_0^\infty (G)^*)=\{0\}, \end{aligned}$$

which implies that G is discrete. To complete the proof, it suffices to notice that the assertion (c) implies that $d(m\cdot n^i)^j=[m, n]\cdot n^{k-1}$. $\square$

4 Inner derivations of $L_0^\infty (G)^*$

A derivation d on $L_0^\infty (G)^*$ is said to be inner if there exists $n_0\in L_0^\infty (G)^*$ such that $d(m)=[m,n_0]$ for all $m\in L_0^\infty (G)^*$.

Proposition 2

Let G be a locally compact abelian group and d be a derivation on $L_0^\infty (G)^*$. Then the following assertions are equivalent.

• (a) d is inner.

• (b) There exists $n_0\in L_0^\infty (G)^*$ such that for each $k\in {\mathbb {N}}$ the mapping $m\mapsto d(m)+n_0\cdot m$ is k-commuting.

• (c) There exists $n_0\in L_0^\infty (G)^*$ and $k\in {\mathbb {N}}$ such that the mapping $m\mapsto d(m)+n_0\cdot m$ is k-commuting.

• (d) There exists $n_0\in L_0^\infty (G)^*$ and $k\in {\mathbb {N}}$ such that the mapping $m\mapsto d(m)+n_0\cdot m$ is k-centralizing.

Proof

Let there exist $n_0\in L_0^\infty (G)^*$ such that $d(m)=[m,n_0]$ for all $m\in L_0^\infty (G)^*$. For $k\in {\mathbb {N}}$ and $m\in L_0^\infty (G)^*$, we obtain

$$\begin{aligned} m^k\cdot n_0\cdot m= & {} m^k\cdot \pi (n_0\cdot m)=m^k\cdot \pi (n_0)*\pi (m)\\= & {} m^k\cdot \pi (m)*\pi (n_0)=m^{k+1}\cdot n_0. \end{aligned}$$

It follows that

$$\begin{aligned}{}[d(m)+n_0\cdot m, m^k]= & {} d(m)\cdot m^k+n_0\cdot m^{k+1}-m^k\cdot n_0\cdot m\\= & {} d(m^{k+1})+n_0\cdot m^{k+1}-m^{k+1}\cdot n_0\\= & {} 0. \end{aligned}$$

Hence (a) implies (b). It is obvious that (b)$\Rightarrow$(c)$\Rightarrow$ (d). To complete the proof, let (d) hold. Define the function $D:L_0^\infty (G)^*\rightarrow L_0^\infty (G)^*$ by

$$\begin{aligned} D(m)=d(m)-[m,n_0]. \end{aligned}$$

It is clear that D is a derivation on $L_0^\infty (G)^*$. So

$$\begin{aligned}{}[D(m), m^k]=D(m)\cdot m^k=[d(m)+n_0 m, m^k]\in \hbox {Z} (L_0^\infty (G)^*). \end{aligned}$$

We now invoke Corollary 4 to conclude that $D=0$. So, we obtain (a). $\square$

In the sequel, let $\hbox {InnD}(L_0^\infty (G)^*)$ be the space of all inner derivations on $L_0^\infty (G)^*$.

Theorem 5

Let G be a locally compact abelian group. Then $InnD (L_0^\infty (G)^*)$ is continuously linearly isomorphic to $L_0^\infty (G)^*/L^1(G)$.

Proof

We define the mapping $\mathfrak {I}$ from $L_0^\infty (G)^*/L^1(G)$ into $\hbox {InnD}(L_0^\infty (G)^*)$ by

$$\begin{aligned} \mathfrak {I}(m+L^1(G))=\mathfrak {I}_m, \end{aligned}$$

where $\mathfrak {I}_m(n)=[n, m]$ for all $n\in L_0^\infty (G)^*$. By Proposition 1, the mapping $\mathfrak {I}$ is well defined. Obviously, $\mathfrak {I}$ is a linear map from $L_0^\infty (G)^*/L^1(G)$ onto $\hbox {InnD}(L_0^\infty (G)^*)$. To see that $\mathfrak {I}$ is injective, let $m\in L_0^\infty (G)^*$ and

$$\begin{aligned} \mathfrak {I} (m+L^1(G))=0. \end{aligned}$$

Then

$$\begin{aligned} \mathfrak {I}_m(n)=n\cdot m-m\cdot n=0 \end{aligned}$$

for all $n\in L_0^\infty (G)^*$. It follows that

$$\begin{aligned} m\in \hbox {Z}(L_0^\infty (G)^*)=L^1(G). \end{aligned}$$

Hence $m+L^1(G)=L^1(G)$. Consequently, $\mathfrak {I}$ is an isomorphism. Now, let $n\in L_0^\infty (G)^*$ and $\phi \in L^1(G)$. Then

$$\begin{aligned} \Vert \mathfrak {I}_m(n)\Vert= & {} \Vert n\cdot m-m\cdot n\Vert \\\le & {} \Vert n\cdot m-\phi \cdot n\Vert +\Vert \phi \cdot n-m\cdot n\Vert \\\le & {} \Vert n\Vert \;\Vert m-\phi \Vert +\Vert \phi -m\Vert \;\Vert n\Vert \\= & {} 2\Vert n\Vert \;\Vert m-\phi \Vert \end{aligned}$$

for all $m\in L_0^\infty (G)^*$. This implies that

$$\begin{aligned} \Vert \mathfrak {I} (m+L^1(G))\Vert =\Vert \mathfrak {I}_m\Vert \le 2\Vert m-\phi \Vert \end{aligned}$$

for all $m\in L_0^\infty (G)^*$ and $\phi \in L^1(G)$. Hence

$$\begin{aligned} \Vert \mathfrak {I} (m+L^1(G))\Vert\le & {} 2\inf \{\Vert m-\phi \Vert : \phi \in L^1(G)\}\\= & {} 2\inf \{\Vert m+\phi \Vert : \phi \in L^1(G)\}=2\;\Vert m+L^1(G)\Vert . \end{aligned}$$

Therefore, $\mathfrak {I}$ is continuous. $\square$

We finish the paper with following result.

Theorem 6

Let G be a locally compact abelian group. Then the following assertions are equivalent.

• (a) G is discrete.

• (b) Any derivation on $L_0^\infty (G)^*$ is zero.

• (c) Any inner derivation on $L_0^\infty (G)^*$ is zero.

Proof

If G is discrete, then $\hbox {Ann}_r(L_0^\infty (G)^*)=\{ 0\}$. By Theorem 1,

$$\begin{aligned} d(L_0^\infty (G)^*)\subseteq \hbox {Ann}_r(L_0^\infty (G)^*)=\{ 0\}. \end{aligned}$$

Hence (a) implies (b). It is plain that (b) implies (c). Finally, if (c) holds, then $[m,n]=0$ for all $m,n\in L_0^\infty (G)^*$. This implies that

$$\begin{aligned} \hbox {Z} (L_0^\infty (G)^*)=L_0^\infty (G)^*. \end{aligned}$$

So $L^1(G)=L_0^\infty (G)^*$. This shows that G is discrete; see Proposition 3.1 of [9]. $\square$