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Let $G$ be a locally compact group. Let $M(G)$ denote the measure algebra and $L^1(G)$ denote the group algebra on $G$. Then $M(G)$ acts on $L^1(G)$ by convolution. So by duality $M(G)$ acts on $L^1(G)^{*}$ via $$\langle f\cdot\mu,\phi\rangle=\langle f,\mu\star \phi\rangle,$$

where $\mu\in M(G)$, $f\in L^{\infty}(G)$ and $\phi\in L^1(G)$.

Again via duality we get an action of $M(G)$ on $L^1(G)^{**}$ $$\langle \mu \cdot n,f\rangle=\langle n,f\cdot\mu\rangle,$$ where $\mu\in M(G)$, $f\in L^{\infty}(G)$ and $n\in L^1(G)^{**}$.

Let $C_0(G)$ denote the continuous functions on $G$ vanishing at infinity.

I'm working on a problem and I can boil it down to the following question:

Question. Suppose that $n\in L^1(G)^{**}$ such that the mapping $$\psi_{n,f}:G\to\mathbb{C},\ \ x\mapsto \langle\delta_x \cdot n,f\rangle$$ belongs to $C_0(G)$, for all $f\in L^{\infty}(G)$. Can I infer any information regarding the element $n\in L^1(G)^{**}$? For what $n\in L^1(G)^{**}$ this can be true?

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Non-mathematical suggestion: use \langle and \rangle instead of < and > and use \cdot instead of the usual "full stop". –  Yemon Choi Apr 7 at 3:19
    
One easy observation is that if $G$ is non-compact, $n$ must annihilate constant functions (take $f$ to be a constant function). My suggestion would be to approach this problem by considering the cases $G={\bf T}$ and $G={\bf Z}$ in more detail, and then looking at what happens for $G={\bf R}$, before seeing what the general picture might be. –  Yemon Choi Apr 7 at 4:03
    
To expand on my previous comment: note that if $G$ is discrete and infinite then you can take $n=\delta_e$ and then you are asking if $x\mapsto f(x)$ belongs to $c_0(G)$ for every $f\in \ell^\infty(G)$, to which the answer is obviously "no". This leads me to suspect that when $G$ is non-compact, the only $n$ which works is $n=0$ -- but I haven't tried to write down a proof. –  Yemon Choi Apr 7 at 4:08
    
In case $G$ is a discrete group, taking the constant function $1$ will tell you that for the points in the spectrum of $=l^{\infty}(G)^*=luc(G)^*$, $\psi_{\delta_x,1}$ doesn't belong to $C_0(G)$. Then you can take into account that $luc(G)^*=M(c)\oplus c_0(G)^{\perp}$ and that $G^{luc}$ is weak-star dense in $luc(G)^*$ and try to build some element that satisfy this condition. At least that's the approach I have in mind. –  Bob Apr 7 at 4:15
    
In the case where $G$ is discrete there is no point in identifying $\ell^\infty(G)$ with ${\rm luc}(G)$ unless you just want to quote theorems rather than think about the problem from first principles. And this is the kind of problem where I think one should try to think from first principles rather than pull results off the shelf. –  Yemon Choi Apr 7 at 4:26

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