Is there an inclusion of $L_\infty(G)$ into $C_0(G)^{**}$?

Question

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!

Answer 1

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 $M(G)^∗=L_\infty (G)\oplus_\infty S(G)^∗$.

Answer 2

Undeleted: This is perhaps a little more tangential to the original question than I'd hoped. But maybe it gives some hints as to why the conclusions aren't that "worrying"...

I don't think this is silly. For example, if $G=\mathbb R$ then for each $t\in\mathbb R$ consider $$ f_{t,n} = \frac{n}{2} \chi_{[t-1/n,t+1/n]} \in L^1(\mathbb R). $$ Then each $f_{t,n}$ is a unit vector in $L^1(\mathbb R)$. Then given $F\in L^\infty(G)$, we define $$\tilde F(t) = \lim_n \langle F,f_{t,n} \rangle = \lim_n \frac{n}{2} \int_{t-1/n}^{t+1/n} F(s) \ ds$$ Maybe this limit doesn't actually exist-- but the sequence is bounded, so just force it to converge via an ultra-filter limit, or similar. I think you have just described an abstract version of this construction. The point is that $\tilde F$ is little more than a function $\mathbb R\rightarrow\mathbb C$ which is bounded; I don't see why it need have any continuity or measurability properties...

Edit: Actually, maybe a better argument is the following. If you convolve an $L^1(G)$ function by an $L^\infty(G)$ function, then you get a (left or right, depending on taste) uniformly continuous function, which you can then integrate against a bounded measure. So if $(e_\alpha)$ is a bai for $L^1(G)$, then define $T:L^\infty(G)\rightarrow M(G)^*$ by taking an ultrafilter limit: $\langle T(F),\mu\rangle = \lim_\alpha \int_G e_\alpha * F \ d\mu$. This gives the identity on $C_0(G)$ (indeed, on left/right uniformly continuous functions). I think this construction would be well-known to Banach algebraists...