# questions on convolutors on $L^p(G)$

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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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".

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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

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".
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.