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Assume that $\lambda_1$ is the smallest eigenvalue of the Dirichlet Laplacean for the domain $\Omega\subset \mathbf{R}^n$ and let $0<\alpha\le 1$. Is the following statement well-known?

Let $f\in L^2(\Omega)$, $0< \alpha\le 1$ and let $$\hat f(x)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbf{R}^n} f(y) e^{-i \left<x,y\right>} dy.$$ Then $$\int_{\mathbf{R}^n}\frac{|\hat f(x)|^2}{|x|^{2\alpha}}dx \le \lambda_1^{-2\alpha}\int_{\Omega}|f(x)|^2 dx .$$

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  • $\begingroup$ How do you define the Fourier transform if your $f$ is only defined in $\Omega$? Do you consider some kind of extension? $\endgroup$
    – András Bátkai
    Commented Feb 24, 2014 at 20:53
  • $\begingroup$ $f(x)=0$, for $x\in D^C$. $\endgroup$
    – user47005
    Commented Feb 24, 2014 at 21:04
  • $\begingroup$ If $f$ is very close to the delta function, the Fourier transform will be very close to constant, and the LHS might not be bounded (say for $\alpha>0.5$). $\endgroup$
    – Asaf
    Commented Feb 24, 2014 at 21:30
  • $\begingroup$ I decided to roll it back to preserve here. It is not obviuos at the first sight that it is not true. $\endgroup$
    – András Bátkai
    Commented Mar 6, 2014 at 20:50
  • $\begingroup$ It seems to me that the inequality follows from Plancherel inequality and the identity $\widehat{(-\Delta)^{-\alpha}f}=|x|^{-2\alpha}\widehat{f}$. $\endgroup$
    – Gian Maria Dall'Ara
    Commented Mar 6, 2014 at 22:31

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