Positive function with zero Haar integral - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:18:53Z http://mathoverflow.net/feeds/question/98680 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98680/positive-function-with-zero-haar-integral Positive function with zero Haar integral Alex M 2012-06-02T20:47:20Z 2012-06-03T14:02:29Z <p>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?</p> <p>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:</p> <p><code>$\int_{C} f \mathbb{dc} = \mathbb{lim}_{n\rightarrow\infty} \frac{\int_{H_n} f \mathbb{dg}}{\lambda (H_n)}$</code></p> <p>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).</p> <p>For details about the theorem see Hewitt &amp; Ross, "Abstract Harmonic Analysis", p.252-253.</p> http://mathoverflow.net/questions/98680/positive-function-with-zero-haar-integral/98715#98715 Answer by Francois Ziegler for Positive function with zero Haar integral Francois Ziegler 2012-06-03T13:27:41Z 2012-06-03T14:02:29Z <p>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 &amp; Ross, <a href="http://books.google.com/books?id=uf11K1wXEYUC&amp;pg=PA185" rel="nofollow">Theorem (15.5)(i)</a>.</p> <p>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 <em>$\varepsilon$-almost period</em> 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.</p> <p>Re: "just think of a continuous, bounded, AP integrable function", there is no such function $f$ except zero. Indeed if $\int_{-\infty}^\infty |f(x)|dx&lt;\infty$ 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).</p>