This is probably a silly question with which I am stuck however. Let $G$ be a locally compact group. It seems to me that there is a canonical map of $L\_\infty(G)$ into $M=C\_0(G)^{\*\*}$ (the latter is the enveloping von Neumann algebra of $C\_0(G)$). I would reason as follows:


1. Let $I$ be the annihilator of $L\_1(G)$ in $M$. Then a) $I$ is weakly closed (obvious); b) $I$ is an ideal. To prove the second thing, consider the quotient map $Q: M\to L\_1(G)^\*$. If we identify $L\_1(G)^\*$ with $L\_\infty(G)$, then $Q$ becomes a homomorphism (it is a homomorphism on $C\_0(G)$, and is weakly continuous). So the kernel of $Q$ is an ideal.
2. For every weakly closed ideal in a von Neumann algebra, there is a complementing ideal $J$ such that $M=I\oplus J$. (it is known that there is a central projection $p\in M$ such that $I=Mp=pM$, then put $J=(1-p)M$).
3. Now $M/I\simeq J\simeq L\_\infty(G)$, so this isomorphism gives an inclusion in the title.


Why this worries me: since $C\_0(G)^{\*\*}$ is the dual space of $M(G)$, the space of Radon measures, then such an inclusion $T$ would mean that we can associate to every $f\in L\_\infty(G)$ its value at every point of $G$, by $f(t)=Tf(\delta\_t)$, where $\delta\_t$ is the probability measure concentrated at $t$.


Am I doing a silly mistake? Thank you in advance!



Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$?
Since you are in the commutative setting, you can present the construction more simply. $M(G)=L\_1(G)\oplus\_1 S(G)$, where $S(G)$ is the space of complex measures that are singular with respect to Haar measure, and hence $M(G)^∗=L\_\infty (G)\oplus\_\infty S(G)^∗$.


