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Suppose for some function $\Phi$ we have:

$$ \nabla^2 \Phi(\mathbf{r})=\phi(\mathbf{r}) $$

where $\phi(\mathbf{r})$ is some well-behaved smooth function, which is finite everywhere.

Does this mean that $\Phi(\mathbf{r})$ itself doesn't have any singularities?

Could you please point me out any useful theorems?

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What do you mean by the Laplacian of a singular function? –  Qiaochu Yuan Nov 29 '12 at 22:46
    
I was just wondering whether $\Phi$ may have discontinuities or not, if $\phi$ is continous –  molkee Nov 29 '12 at 23:45

1 Answer 1

up vote 2 down vote accepted

The Laplace operator is hypoelliptic (since it's elliptic with smooth coefficients) and any hypoelliptic operator $L$ has the property that if $Lf \in C^\infty$, then $f \in C^\infty$.

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