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The question I want to pose is the following

assume $f:\mathbb R^p\times \mathbb R\times \mathbb R\to \mathbb R$, $(w,x,y)\mapsto f(w,x,y)$ is such that in a neighborhood $U$ of the origin it holds that both

$xf\in C^\infty(U)$,

$yf\in C^\infty(U)$.

Prove then that $f\in C^\infty (U)$.

The fact is that I encountered this problem as a technical lemma in a paper to a subsequent theorem, and it is stated that the proof is trivial and therefore left to the reader. I was unsuccesful in proving this so I am asking for an help.

Bonus Question... The previous statement remains true if we substitute smooth with analytic.

Thanks for the help.

guido giuliani
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1 Answers1

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I think this goes along the lines of Quotient of two smooth functions is smooth.

First, the assumptions imply that $f$ is $C^\infty$ on the set $\{x\ne 0\}\cup \{y\ne 0\}$. Therefore, $xf$ is equal to zero on $\{x=0\}\cap \{y\ne 0\}$, which by continuity implies that $xf$ is equal to zero on $\{x=0\}$. Let $g = xf$; then $$ g(x,y,z) = x\int_0^1 D_1g(sx,y,z)\,ds \tag1 $$ where $D_1g$ is the derivative of $g$ in the first-placed argument, and therefore a smooth function. Thus,
$$ f(x,y,z) = \int_0^1 D_1g(sx,y,z)\,ds \tag2 $$ is also smooth.

If $g$ is analytic, integration in (2) preserves that too.

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