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Let $\mu$ be a finite measure supported by $\Gamma $ (a smoth finite curve) and absolutely continuous with respect to the length measure on $\Gamma$ such that $\Gamma \cap (\Gamma+x)$ is a finite set for any non-zero $x\in \mathbb{R^2}.$

$\int_{R^2}\varphi(x)d\mu(x)=0$ ,for every $2\pi$-periodic function $\varphi$.

that means that the locally finite measure defined by the sum $\sum_{n\in \mathbb Z^2} d\mu(x-2\pi n)$.

is in fact the zero measure ???

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  • $\begingroup$ Please make clear what question you are asking. $\endgroup$
    – Chris Ramsey
    Commented Jan 3, 2016 at 17:50
  • $\begingroup$ please why $\int_{R^2}\varphi(x)d\mu(x)=0$ implies $\sum_{n\in{\mathbb{Z}}^2} d\mu(x-2\pi n)=0$?thank you . $\endgroup$
    – Akram Akram
    Commented Jan 3, 2016 at 18:55
  • $\begingroup$ When you integrate against the measure defined by the sum, you're integrating the periodization against $d\mu$. In symbols, $\langle \phi, \chi_{\mathbb{Z}^2}\ast d\mu\rangle = \langle \chi_{\mathbb{Z}^2}\ast\phi, d\mu\rangle = 0$. $\endgroup$
    – Brendan Murphy
    Commented Jan 3, 2016 at 20:08
  • $\begingroup$ Thank you sir ,but what's this '∗' ,$\chi_{{\mathbb{Z}}^2}$? ,i need some explanations thank you so much $\endgroup$
    – Akram Akram
    Commented Jan 3, 2016 at 20:19
  • $\begingroup$ For other readers of this question: from comments on the OP's other question it seems that these questions relate to sciencedirect.com/science/article/pii/S0007449710000953 $\endgroup$
    – Yemon Choi
    Commented Jan 5, 2016 at 16:32

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I don't think the definition of $\mu$ is relevant to your problem, you only need that $\int\phi d\mu =0 $ if $\phi$ is $2\pi$-periodic.

The reason is that the measure $$ d\nu(x) = \sum_{n\in\mathbb{Z}^2}d\mu(x-2\pi n) $$ (which is the convolution of the indicator function of the lattice $\chi_{2\pi\mathbb{Z}^2}$ with $d\mu$) is defined to make integration of $\phi$ against $d\nu$ the same as integration of the periodization of the $\phi$ against $d\mu$: $$ \int \phi(x)d\nu(x) =\int\phi(x)\left(\sum_{n\in\mathbb{Z}^2}d\mu(x-2\pi n)\right) = \int\left(\sum_{n\in\mathbb{Z}^2}\phi(y+2\pi n)\right)d\mu(y) $$ The periodization $$ \sum_{n\in\mathbb{Z}^2}\phi(y+2\pi n) $$ is $2\pi$-periodic, so the second integral is always zero.

(Throughout, let's assume that $\phi$ has enough decay to make all the integrals converge. Also, we need to exchange the order of the integral and the sum to make the change of variables $x\to y+2\pi n$, then change the order back.)

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  • $\begingroup$ Thank you for your help and your patience but ,what is $d\mu(x)$ ? $\endgroup$
    – Akram Akram
    Commented Jan 20, 2016 at 23:53
  • $\begingroup$ It's any measure such that $\int\phi d\mu = 0$ if $\phi$ is $2\pi$-periodic, so in particular, your measure. $\endgroup$
    – Brendan Murphy
    Commented Jan 21, 2016 at 18:20

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