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Let $\langle\xi\rangle=(1+|\xi|^2)^{\frac 12}$. Is $\mathcal{F}^{-1}(\frac{\langle\xi\rangle^{-n}}{1+\log{\langle\xi\rangle}})$ a bounded function? $\mathcal{F}^{-1}$ denotes the inverse Fourier transform.

My idea was to construct some case of the Cauchy-Schwarz inequality, but what I managed is only the upper bound. Any idea or hint would be appreciated.

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  • $\begingroup$ I changed $<\xi>$ to $\langle\xi\rangle$. That is standard usage. $\endgroup$ – Michael Hardy Dec 3 '12 at 1:06

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