4
$\begingroup$

I'm reading a paper where the following inequality appears. $$ \| \widehat{f} \|^2_{L^2(d\mu)} \leq \| f \ast \widehat{\mu} \|_p \| f \|_{p^\prime} $$ where $f$ is a real-valued measurable function on $\mathbb{R}^n$, $\mu$ is a positive measure on $\mathbb{R}^n$, and $\frac{1}{p} + \frac{1}{p^{\prime}} = 1$. I think $\| \cdot \|_p$ and $\| \cdot \|_{p^{\prime}}$ are with respect to Lebesgue measure.

$$ \widehat{\mu}(\xi) = \int e^{-2 \pi i x \xi} d\mu(x) $$

I feel like this should be a consequence of Hölder's inequality and some identities relating convolution and the Fourier transform, but I can't figure it out.

Can someone please help?

  • 0
    Does $\|f\|_p$ here refer to Lebesgue measure? Or $\mu$?2012-04-05
  • 0
    What is $\widehat{d\mu}$. I understand the notation when we have only $\mu$, but does $d\mu$ mean the Radon-Nikodym derivative, or what?2012-04-06
  • 0
    @Davide: a straightforward computation shows that the the usual Fourier-Stieltjes transform $\hat{\mu}$ must be meant.2012-04-06
  • 0
    @t.b. Maybe straightforward but I should have done it.2012-04-06

1 Answers 1