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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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    @t.b. Maybe straightforward but I should have do$n$e it.2012-04-06

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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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    You are perfectly right. I worried about $\sigma$-finiteness without thinking before about well-definiteness of the Fourier transform. Thanks!2012-08-05