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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?

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Does $\|f\|_p$ here refer to Lebesgue measure? Or $\mu$? –  user15464 Apr 5 '12 at 2:07
    
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? –  Davide Giraudo Apr 6 '12 at 8:09
    
@Davide: a straightforward computation shows that the the usual Fourier-Stieltjes transform $\hat{\mu}$ must be meant. –  t.b. Apr 6 '12 at 20:25
    
@t.b. Maybe straightforward but I should have done it. –  Davide Giraudo Apr 6 '12 at 20:32
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1 Answer

I will assume $\mu$ finite. By Hölder's inequality, $$\tag{*}\lVert f*\widehat\mu\rVert_p\lVert f\rVert_{p'}\geq \int_{\Bbb R^n}(f*\widehat \mu)(x)f(x)dx,$$ hence it's enough to show that the RHS of this inequality is the LHS in the OP. First, we can write \begin{align} (f*\widehat \mu)(x)&=\int_{\Bbb R^n}f(x-t)\widehat\mu(t)dt\\ &=\int_{\Bbb R^n}f(x-t)\int_{\Bbb R^n}e^{-2\pi its}d\mu(s)dt\\ &=\int_{\Bbb R^n\times\Bbb R^n}f(x-t)e^{-2\pi its}d\mu(s)dt, \end{align} and putting it in (*), we have, denoting $g(x):=f(-x)$, \begin{align} \lVert f*\widehat\mu\rVert_p\lVert f\rVert_{p'}&\geq \int_{(\Bbb R^n)^3}f(x-t)f(x)e^{-2\pi its}d\mu(s)dtdx\\ &=\int_{(\Bbb R^n)^3}g(t-x)f(x)e^{-2\pi its}d\mu(s)dtdx\\ &= \int_{\Bbb R^n\times\Bbb R^n}(g*f)(t)e^{-2\pi its}d\mu(s)dt\\ &=\int_{\Bbb R^n}\widehat{g*f}(s)d\mu(s)\\ &=\int_{\Bbb R^n}\widehat{g}(s)\widehat f(s)d\mu(s)\\ &=\int_{\Bbb R^n}|\widehat f|^2d\mu(s), \end{align} what is wanted.

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+1, but don't you need finiteness of $\mu$ in order for the Fourier-Stieltjes transform $\hat\mu$ to be defined e.g. in the second equality you write after $(\ast)$? –  t.b. Aug 5 '12 at 22:12
    
You are perfectly right. I worried about $\sigma$-finiteness without thinking before about well-definiteness of the Fourier transform. Thanks! –  Davide Giraudo Aug 5 '12 at 22:16
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