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For a positive function $f$ and positive measures $\mu, \nu$, does $$\mu\ast f\leq \nu\ast f \Rightarrow \|\mu\|\leq \|\nu\|?$$

More details: Let $G$ be a locally compact group, $C(G)$ be the space of continuous functions on $G$, $M(G)$ be the finite, Borel measures on $G$. The convolution of a measure and function is defined by $$\mu\ast f(x) = \int f(y^{-1}x)d\mu (y)$$.

If $0\neq f\in C(G)$ is positive and compactly supported, and $\mu, \nu\in M(G)^+$ then the above implication holds. Simply take the Haar integral and evaluate the first inequality to find the second.

A similar approach works if $G$ is amenable and $f$ has nonzero mean value.

My questions: Does the implication above hold for all (edit: bounded) $f\in C(G)^+$ if $G$ is amenable? If $G$ is not amenable, what are the functions for which it is true?

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It seems that Justin Moore has some results on single functions witnessing non-amenability in his answer to his question mathoverflow.net/questions/60247/… –  Ben Willson Aug 27 at 1:38
    
In the non-topological setting, I think this question is very closely related to supramenability. –  Ben Willson Nov 12 at 2:36

1 Answer 1

Do you want to require $f$ to be bounded? Otherwise there are easy counterexamples: take $G = \mathbb{Z}$, $\mu(1) = 2$, $\nu(0) = 1$, and $\mu$ and $\nu$ zero at all other points, and $f(n) = 2^n$. Then $\mu * f = \nu*f$ but $\|\mu\| = 2 > 1 = \|\nu\|$.

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Thanks. I am mostly interested in bounded $f$, but it is good to be reminded of unbounded examples. –  Ben Willson Aug 26 at 7:41

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