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Let $\phi\in L^2(\mathbb{R}^3)$. Since $|\phi|^2\in L^1$, it has a distributional derivative. At least formally, $$ \nabla |\phi|^2 = \phi \nabla\overline{\phi} + \overline{\phi}\nabla\phi. $$

Is $\overline{\phi}\nabla\phi$ a well-defined distribution in itself?

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up vote 1 down vote accepted

No, it is not. Write $\phi=f+ig$ where $f$ and $g$ are real valued. The following expressions can be interpreted as distributions: $$\begin{split}f\nabla f&=\frac12 \nabla |f|^2 \\ g\nabla g&=\frac12 \nabla |g|^2 \\ f\nabla g + g\nabla f&= \nabla (fg) \end{split}$$ If we could also interpret $\bar \phi \nabla \phi $ as a distribution, using $$\bar \phi \nabla \phi = (f-ig)(\nabla f+i \nabla g) = f\nabla f+g\nabla g+i(f\nabla g-f\nabla f)$$ together with the above formulas we would be able to make sense out of $f\nabla g$ for arbitrary $f,g\in L^2$.

But $f\nabla g$ does not make distributional sense in general. For example, in one dimension take $f(x)=|x|^{-2/5}$ and $g(x)=x^{1/5}$ near the origin (you can smoothly cut them off at infinity). Then $\nabla g(x)=\frac15 |x|^{-4/5}$ as a distribution, but multiplying $\nabla g$ by $f$ just doesn't work.

You can make this example 3-dimensional, e.g., by letting $f(x)=|x_1|^{-2/5}\chi(x)$ and $g(x)=x_1^{1/5}\chi(x)$ where $\chi$ is a smooth cutoff function that is identically $1$ in a neighborhood of the origin.

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