Suppose we have 4 functions $f,g,h,j$ all of them are functions from $R^n$ to $R$, in particular, $f,g$ are smooth. I’m curious, when given following condition: $$fh+gj=0$$can we claim that $h$ and $j$ are also smooth?
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What is *? Multiplication? – Mnifldz Apr 16 '20 at 21:23
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Yes, correct. @Mnifldz – GK1202 Apr 16 '20 at 21:28
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I edited your post slightly. It can get confusing because $*$ is used for other operations besides multiplication (i.e. convolution and other products). – Mnifldz Apr 16 '20 at 21:30
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Assuming that $g$ does not cancel,
$$j=-\frac fgh$$
solves the equation, so $h,j$ have no reason to be smooth. Not even continuous.
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Counterexample: Let $f(x) = x$ and $g(x) = x^2$. Then we could choose $h(x) = \frac{1}{x}$ and $j(x) = -\frac{1}{x^2}$ neither of which are smooth (or even defined) at $x=0$. To be extra clear, we can extend the definition of either $h$ or $j$ to make $h(0) = a$ and $j(0) = b$ for some points $a$ and $b$, but no extension of these functions make them continuous.
Mnifldz
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$h, j$ are not functions from $\mathbb{R}$ then, but from $\mathbb{R}\setminus{0}$ instead (although this is easily solved by defining them to be 0 in 0). – Dasherman Apr 16 '20 at 21:47
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I suppose to be fair, neither one of them has a continuous extension. Any point would suffice, however you do raise a good point here. – Mnifldz Apr 16 '20 at 21:49