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Let $\hat{f}$ be the fourier transform of $f$.

By Parseval-Plancherel identity, for suitable $f,g$, we have $$\left\|\hat{f}*\hat{h}\right\|_{L^2_{\xi}}^2=\left\|f\cdot h\right\|_{L^2_{x}}^2.$$

Let $f,g$ be good enough functions. I want to know if for some universal constant $C>0$, $$\left\||\hat{f}|*|\hat{h}| \right\|_{L^2_{\xi}}^2\leq C\left\||f|\cdot|h|\right\|_{L^2_{x}}^2.$$

Or is there a counterexample?

Thanks a lot!

Let $\hat{f}$ be the fourier transform of $f$.

By Parseval-Plancherel identity, for suitable $f,g$, we have $$\left\|\hat{f}*\hat{h}\right\|_{L^2_{\xi}}^2=\left\|f\cdot h\right\|_{L^2_{x}}^2.$$

I want to know if for some universal constant $C>0$, $$\left\||\hat{f}|*|\hat{h}| \right\|_{L^2_{\xi}}^2\leq C\left\||f|\cdot|h|\right\|_{L^2_{x}}^2.$$

Or is there a counterexample?

Thanks a lot!

Let $\hat{f}$ be the fourier transform of $f$.

By Parseval-Plancherel identity, for suitable $f,g$, we have $$\left\|\hat{f}*\hat{h}\right\|_{L^2_{\xi}}^2=\left\|f\cdot h\right\|_{L^2_{x}}^2.$$

Let $f,g$ be good enough functions. I want to know if for some universal constant $C>0$, $$\left\||\hat{f}|*|\hat{h}| \right\|_{L^2_{\xi}}^2\leq C\left\||f|\cdot|h|\right\|_{L^2_{x}}^2.$$

Or is there a counterexample?

Thanks a lot!

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Parseval-Plancherel identity involving absolute value

Let $\hat{f}$ be the fourier transform of $f$.

By Parseval-Plancherel identity, for suitable $f,g$, we have $$\left\|\hat{f}*\hat{h}\right\|_{L^2_{\xi}}^2=\left\|f\cdot h\right\|_{L^2_{x}}^2.$$

I want to know if for some universal constant $C>0$, $$\left\||\hat{f}|*|\hat{h}| \right\|_{L^2_{\xi}}^2\leq C\left\||f|\cdot|h|\right\|_{L^2_{x}}^2.$$

Or is there a counterexample?

Thanks a lot!