Let $f$ be a square integrable function, strictly positive almost everywhere. Consider the family of functions $f_a=f(x+a)$, where $a$ is any real number. I want to prove that if a function is orthogonal to all this family then it must be zero. The hint I was given is to use Fourier transform, that is to show that $(g,e^{ika}\hat f)=0$ implies that $g=0$. But from here I am not sure how to go on... I see that $(g,e^{ika}\hat f)$ is proportional to the inverse Fourier transform of $g^{*}\hat f$ and therefore $g^{*}\hat f$ is zero almost everywhere, but can I say that $\hat f$ is also positive, since it is the Fourier transform of a positive function?

  • $\begingroup$ The exercise seems to be from a book, but as Julian Aguirre showed, it's not true. I guess there is a missing hypothesis. $\endgroup$ – Davide Giraudo Jul 20 '12 at 12:54
  • $\begingroup$ I would imagine this to be true if, in addition, $\hat{f}(\omega) \neq 0$ a.e.? $\endgroup$ – copper.hat Jul 20 '12 at 21:39
  • $\begingroup$ I am sorry - I have just discovered there was a misprint in the problem. The hypothesis is that $\hat f>0$ a.e., not $f$. This makes the proof straightforward. $\endgroup$ – quark1245 Jul 20 '12 at 22:28

Your claim is not true in general. Let $$ f(x)=\Bigl(\frac{\sin x}{x}\Bigr)^2,\quad x\in\mathbb{R}. $$ $f$ is square integrable and $\{x\in\mathbb{R}:f(x)\le0\}=\{k\,\pi:k\in\mathbb{Z},k\ne0\}$ has measure zero. Moreover $\hat f$ is supported on $[-2,2]$.

Let $g\in L^2(\mathbb{R})$ be such that $\hat g$ is supported on $\mathbb{R}\setminus[-2,2]$. Then $\widehat{f\ast g}=\hat f\cdot \hat g\equiv0$, so that $f\ast g(a)=0$ for all $a\in\mathbb{R}$, that is $$ \int_{\mathbb{R}}g(x)f(x-a)dx=0\quad\forall a\in\mathbb{R}. $$

  • $\begingroup$ +1 I presume the $()^2$ on $f$'s definition is a mistake? $\endgroup$ – copper.hat Jul 19 '12 at 23:24
  • $\begingroup$ @copper.hat Why do you presume so? $()^2$ is there to make $f$ positive. The Fourier transform of $\sin x/x$ is (up to a multiplicative constant) $\chi_{[-1,1]}$, the characteristic function of $[-1,1]$. The Fourier transform of $f$ is then a multiple of $\chi_{[-1,1]}\ast\chi_{[-1,1]}$, a tent function supported on $[-2,2]$. $\endgroup$ – Julián Aguirre Jul 20 '12 at 9:00
  • $\begingroup$ My mistake. Nice answer. $\endgroup$ – copper.hat Jul 20 '12 at 14:22
  • $\begingroup$ @copper.hat No problem. We all make mistakes. $\endgroup$ – Julián Aguirre Jul 20 '12 at 21:32

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