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Let $G$ be a locally compact Hausdorff group. In the proof of theorem 4.32 of Folland's book "A course in abstract harmonic analysis", the following result is used:

If $U$ is an open neighborhood of a point $x \in G$, then there exists $f \in \mathcal{B}^1$ with $f(x)\ne 0$ and $\operatorname{supp}(f)\subseteq U$. This is supposed to be a consequence of the following theorem:

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Recall that $\mathcal{B}^1:= [\operatorname{span}\mathcal{P}(G)]\cap L^1(G)$, where $\mathcal{P}(G)$ are the continuous functions of positive type on $G$.

How does this Urysohn-like result follow from the above result?

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    $\begingroup$ According to my copy of Folland's book (2nd ed), just before Lemma 4.20 ${\mathcal B}^1$ is defined to be the intersection of ${\mathcal B}(G)$ with $L^1(G)$. Perhaps the notation in the book is not globally consistent? $\endgroup$
    – Yemon Choi
    May 15, 2022 at 16:25
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    $\begingroup$ Also (for others reading this question) Theorem 4.20 is the Pontrjagin duality theorem for locally compact abelian groups, although I think that the hypothesis of being abelian is not needed for the step which the OP is asking about. $\endgroup$
    – Yemon Choi
    May 15, 2022 at 16:27
  • $\begingroup$ @YemonChoi Yes thanks, I will edit. $\endgroup$
    – Andromeda
    May 15, 2022 at 16:34

1 Answer 1

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[Nitpick: I think Urysohn's lemma concerns existence of functions which are 1 and 0 on two given closed subsets; the property here, of being supported inside an open neighbourhood and non-zero, is noticeably weaker. The property described/needed here is more like being a "regular" function algebra. Compare the difference between a normal topological space and a completely regular topological space]


I will formulate everything for general locally compact groups, even though at the place where Folland needs this result, he is assuming all groups are locally compact abelian.

Let me first address the case where $x=e$ and $U$ is an open neighbourhood of $e$. By general topology, since $G$ is locally compact we may assume that there is an open neighbourhood $V\ni e$ whose closure is compact and contained in $U$. Moreover, we can find a compact neighbourhood $K\ni e$ in $G$ such that $KK^{-1}:=\{ tu^{-1} \mid t,u\in K\} \subseteq V$.

Since $K$ has non-empty interior it has strictly positive Haar measure. Writing $|K|$ for the (Haar) measure of $K$, we now define $\xi = |K|^{-1/2} 1_K$ and form the coefficient function

$$ f(s) = \langle \lambda_s \xi, \xi\rangle = |K|^{-1} \int 1_K(s^{-1}t) 1_K(t)\,dt =|K|^{-1} |K\cap sK| $$

  • $f$ is a positive definite function (in the language used in Folland's book, it is a function of positive type). In particular $f\in {\mathcal B}(G)$.

  • Moreover, $f$ is supported inside $V$: for if $K\cap sK\neq\emptyset $, let $t\in K\cap sK$ and put $y=s^{-1}t\in K$, yielding $s=ty^{-1}\in K K^{-1} \subseteq V$.

  • Since $V$ has compact closure, $f\in C_c(G)\subseteq L^1(G)$; and since $f$ vanishes outside $V$ it certainly vanishes outside $U$, as originally required.


The general case, as stated in the OP's question, now follows from the case $x=e$ that was done above, and the fact that ${\mathcal B}^1(G)$ is invariant under left translations. Alternatively one can run the argument above but with an appropriate left translation inserted everywhere: the resulting $f$ will still be a coefficient function of the left regular representation, but it won't usually be a function of positive type.

Remarks.

  1. If you haven't seen this construction of small "triangular" bump functions it is instructive to do this for, e.g. $G={\mathbb T}$ and $x=1$, $U= \{ e^{2\pi it} \mid -\delta< t <\delta\}$ for some small positive $\delta$.
  2. I am not quite sure why Folland refers to Proposition 3.33, except that it says ${\mathcal B}(G)$ does contain $f*g$ for any $f,g\in C_c(G)$. Initially I thought he might be using one of the density statements that are mentioned in the statement of the proposition, but this doesn't seem to play a role.
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  • $\begingroup$ Thanks for your answer! $\endgroup$
    – Andromeda
    May 15, 2022 at 17:15
  • $\begingroup$ If instead of taking $\xi=\eta=|K|^{-1/2} 1_K$, we instead take $\eta = |L|^{-1/2} 1_L$ for a really "small" compact $L$, then the resulting coefficient functional will be identically $1$ on a "large" subset of $K$, and zero off $KL^{-1}$ (or similar). In this way, I think one can prove a result which is close to Urysohn's lemma (which of a locally compact space would normally ask for a function identically 1 on a compact subset, and $0$ on some specified disjoint closed subset). $\endgroup$ May 15, 2022 at 18:35
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    $\begingroup$ @Andromeda It depends slightly which definition one is using of ${\mathcal B}(G)$. I tend to define it as the space of all possible coefficient functions of all possible unitary representations, in which case translating such a coefficient function just corresponds to shifting one of the defining vectors. In the part of Folland's book you're referring to, G is LCA, and so you can use Bochner's theorem (together with the fact that translation on $G$ corresponds by Fourier transform to a phase shift on the dual group $\widehat{G}$) $\endgroup$
    – Yemon Choi
    May 15, 2022 at 19:15
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    $\begingroup$ Ah, yes, you are correct about the normalisation: so you can build suitable functions for Urysohn, but the natural norms have little control. $\endgroup$ May 15, 2022 at 20:17
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    $\begingroup$ @Andromeda Yes, that's what I meant with the second approach. $\endgroup$
    – Yemon Choi
    May 15, 2022 at 23:28

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