0
$\begingroup$

If $C$ is a compact (semi-)topological (semi-)group, are there nonzero positive functions having zero Haar integral? In other words: is the Hermitian product associated to the Haar integral degenerate?

The motivation of my question: if $G$ is a locally-compact (semi-)topological (semi-)group, $C$ its almost periodic (AP) compactification and $\lambda$ a Haar measure on $G$, then for any AP function $f$ on $G$ and any Foelner sequence $H_n$ in $G$ one has:

$\int_{C} f \mathbb{dc} = \mathbb{lim}_{n\rightarrow\infty} \frac{\int_{H_n} f \mathbb{dg}}{\lambda (H_n)}$

But then, if $f$ is as above AND integrable (like $\frac{1}{1+x^2}$ on $R$), the previous result implies that when it is lifted to the compatification its integral becomes 0 (despite the fact that its lift is positive, since a group is dense in its compactification).

For details about the theorem see Hewitt & Ross, "Abstract Harmonic Analysis", p.252-253.

Improve this question
$\endgroup$
3
  • $\begingroup$ The group is dense in its compactification. But not open. In fact, the group $\mathbb R$ has measure zero in its Bohr compactification. Your function $1/(x^2+1)$ is identiacally zero except on that set of measure zero. $\endgroup$
    – Gerald Edgar
    Commented Jun 2, 2012 at 21:07
  • 1
    $\begingroup$ Moreover, $\frac1{x^2+2}$ is not "as above and integrable": it is not almost periodic, hence has no canonical extension (not "lift"!!) to the compactification. It goes the other way: (continuous) functions on $b\mathbf{R}$ pull back to (continuous) almost periodic functions on $\mathbf{R}$. $\endgroup$
    – Francois Ziegler
    Commented Jun 2, 2012 at 22:43
  • $\begingroup$ Gerald Edgar: The AP compactification takes CONTINUOUS bounded AP functions $f$ and gives back CONTINUOUS functions $\tilde{f}$ on the AP compactification, by the Gelfand-Naimark isomorphism. Since $\tilde{f}$ is continuous, it can't be nonzero but zero almost everywhere, as you say. Francois Ziegler: Correct me if I'm wrong: $\frac{1}{1+x^2}$ has the derivative bounded by 1 and as such $|f(x)-f(y)|\leq|x-y|$ (by Lagrange's mean value theorem), and it's now easy to see that it's AP (it's Lipschitz!). Still, my question remains (just think of a continuous, bounded, AP integrable function). $\endgroup$
    – Alex M
    Commented Jun 3, 2012 at 7:15

1 Answer 1

Reset to default
4
$\begingroup$

The answer to the question asked is no, i.e. the Haar integral $I(f)$ of a nonzero, nonnegative continuous function $f$ is always positive. See Hewitt & Ross, Theorem (15.5)(i).

The mistake in your argument is that $f(x)=1/(1+x^2)$ is not almost periodic. Indeed, recall that $f$ is almost periodic iff $\forall\varepsilon>0$ $\exists L>0$ such that every interval of length $L$ contains an $\varepsilon$-almost period of $f$, i.e. a number $P$ such that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|\leqslant\varepsilon$. But it is clear (from looking at the graph) that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|>\frac12$ for every $P$ in every interval $[10, 10+L]$ say.

Re: "just think of a continuous, bounded, AP integrable function", there is no such function $f$ except zero. Indeed if $\int_{-\infty}^\infty |f(x)|dx<\infty$ then $I(|f|)=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T |f(x)|dx=0$, whence $f=0$ by the above-quoted Theorem (15.5).

Improve this answer
$\endgroup$
2
  • $\begingroup$ Yes, you are right, thank you, I was using a wrong definition of AP. Still, the question remains if, instead of the AP compactification, one uses the uniformly continuous (UC) one (I'm working with abelian groups, so "left UC" and "right UC" are the same). In this case I have troubles reconciling the aforementioned theorem about the Haar integrals and the fact that there <b>seem</b> to exist strictly positive UC functions that, when transported to the UC compactification, have zero integral (see my example under discussion, which is not AP but is UC). Any ideas, please? $\endgroup$
    – Alex M
    Commented Jun 4, 2012 at 20:17
  • $\begingroup$ Sorry, I don't know much about the UC compactification. I suggest you make a new question about this variant, this will make things less cluttered anyway :-) $\endgroup$
    – Francois Ziegler
    Commented Jun 4, 2012 at 21:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .