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Assume one has real valued functions $f(x)$ and $g(x)$ that belong to the Schwartz space. I know that the Fourier transforms of $f^3(x)$ and $f^2(x)g(x)$ can be expressed straightforward in terms of some complicated convolutions as

\begin{equation} \cal{F}\{ f^3(x) \} \propto \intop_{\mathbb{R}}\intop_{\mathbb{R}}\widehat{f}(k-p-q)\widehat{f}(q)\widehat{f}(p)dpdq \end{equation} \begin{equation} \cal{F}\{ f^3(x) \} \propto \intop_{\mathbb{R}}\intop_{\mathbb{R}}\widehat{f}(k-p-q)\widehat{f}(q)\widehat{g}(p)dpdq \mbox{,} \end{equation}

where the 'hat' denotes the Fourier transform of the function. However, considering that $f(x)$ is a real function, i.e., $f^2(x)$ is non-negative, is there some simplification of the above convolution integrals?

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No, $f$ being real does not simplify the formulas. If it did, then by writing an arbitrary complex-valued function as $f+ig$, we'd have a simpler formula for the transform of its square, $$(f+ig)^2=(f^2-g^2)+\frac{i}{2}( (f+g)^2-(f-g)^2)$$ But it's just that a convolution of Fourier transforms.

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