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Let $G$ be a locally compact group. Suppose $1<p<\infty$. We denote by $CV_p(G)$ the space of operators $T$ on $L^p(G)$ such that $T(f*g)=(Tf)*g$.

1) In the book "Amenable locally compact groups" of Pier, Proposition 9.2 (page 83), the author asserts that each operator of $CV_p(G)$ is a convolution operator with a measure on $G$. This proposition seems false to me. Does is it true?

2) (If the proposition 9.2 of Pier is false) Suppose that $G$ is a discrete group. Let $T\in CV_p(G)$. Does there exist a measure $\mu$ on $G$ such that $T(g)=\mu*g$ for any continuous function $g$ with compact support?

3) Let $G$ be an abelian locally compact group. We denote $\varepsilon_a$ the Dirac measure in the point $a\in G$. If $|\lambda|=1$, the operator $\lambda(\varepsilon_a*\cdot)$ is a extremal point of the closed unit ball of $CV_p(G)$.

Does there exist other known exemples of extremal points of this ball? Does there exist references on this subject?

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up vote 2 down vote accepted

This is a slightly rambling answer. If you have stated this right, then the book is false.

Take $p=2$. Then $CV_2(G)$ is simply the group von Neumann algebra $VN(G)$-- it's certainly not true that every member of $VN(G)$ is a measure, unless $G$ is finite. (I think this is not trivial to show. If $VN(G)=M(G)$ then as the map $M(G)\rightarrow VN(G)$ is always weak$^*$-continuous, we'd have that $A(G)=C_0(G)$, where $A(G)$ is the Fourier algebra. This is utterly false if, say, $G$ is abelian and infinite. I'm slightly embarrassed to say that I'll have to resort to Arens regularity-- from work of Young we know that $L^1(G)$ is not Arens regular when $G$ is infinite, and so neither is $M(G)$, but $VN(G)$ being a C$^*$-algebra is. Does anyone else know a nicer proof?)

When $p\not=2$, the same idea works-- $M(G)$ is not Arens regular unless $G$ is finite, but by construction $CV_p(G)$ is a closed subalgebra of $B(L^p(G))$, and by a result of mine, this algebra is Arens regular.

For (2), just argue by continuity-- both $T$ and left convolution by $\mu$ are continuous, so if $T(g)=\mu*g$ for all $g\in C_{00}(G)$, then the same will be true for all $g\in L^p(G)$. That is, in general, "no".

I haven't thought about (3).

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Just seen this, Matt. What Pier says (going from memory, I lent the book to a student) is that an operator in CV_p can be represented as convolution with some possibly infinite Radon measure on the group. Like you, I had an initial reaction of "that can't be right!" before eventually decoding terminology/notation. – user16299 Jun 15 '12 at 9:37
Ah, that sounds plausible. Indeed, if in (2) we're allow infinite measure, then the answer is easily seen to be "yes"-- evaluate $T$ at the point mass at the identity to get an $\ell^p$ function-- then convolution by this induces $T$. – Matthew Daws Jun 15 '12 at 9:41
BTW, I think M(X) can't be w*-isomorphic as a Banach space to any infinite-dimensional vN alg, because then this would give rise to an isomorphism of Banach spaces C_0(X) to predual of vN, hence by Akemann et al we have a weakly compact isomorphism, hence the identity on C_0 is weakly compact, hence C_0 is reflexive, which is only possible if X is finite. – user16299 Jun 15 '12 at 9:46

(This should perhaps be merely a comment, but I am leaving it as an answer in case people read Matthew's answer without reading the comments.)

When Pier says, in his book, that every convolution operator on $L^p(G)$ can be regarded as convolution with some (Radon) measure, he is not demanding that this measure be finite. So the answer to your question 1) is "the proposition is correct".

The answer to your question 2) is "yes", as observed in Matthew's comment above.

I have not thought too deeply about your question 3), but I believe the answer should be "yes, there are other extreme points of the unit ball of $CV_p(G)$". For $p=2$ then you are essentially asking if every extreme point of $L^\infty(H)$, $H$ a LCA group, is the Fourier transform of a point mass, and I think that is highly unlikely. Probably Rudin's book on Fourier Analysis on Groups would have details that help you build a counter-example.

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