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Let $g(x)=f(x)|f(x)|^{q-2}$ where $q>2$ and $f\in \mathcal C^\infty _0(\mathbb R^d)$. First of all, is $g\in \mathcal C_0^\infty (\mathbb R^d)$ ?

Then I was wondering if $$|\hat g(\xi)|^2=|\widehat{f^{q-1}}(\xi)|^2$$

was true. The only thing I get is $$|\hat g(\xi)|^2=\hat g(\xi)\overline{\hat g(\xi)}=\int_{\mathbb R^d}\int_{\mathbb R^d}f(x)|f(x)|^{q-2}f(y)|f(y)|^{q-2}e^{-2i\pi x\cdot \xi}e^{2i\pi y\cdot \xi}dxdy, $$ but I can't get $$\int_{\mathbb R^d}\int_{\mathbb R^d}f(x)^{q-1}f(y)^{q-1}e^{-2i\pi x\cdot \xi}e^{2i\pi y\cdot \xi}dxdy.$$

Any idea ?

user349449
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  • The title does not match the question. In general, the absolute value of the fourier transform does not equal the fourier transform of the absolute value. – robjohn Jan 11 '18 at 17:07

1 Answers1

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In general, the absolute value of the fourier transform does not equal the fourier transform of the absolute value. Consider, for example, this answer.

robjohn
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  • Thanks a lot for your answer. In fact my question refer to my other question here. If it was true, then I could understand the inequality, but if it's not true in general, how can I conclude in my question in the post ? – user349449 Jan 11 '18 at 17:39