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Let $f\in C^{\infty}(\mathbb{R}^n)$ be such that $f(x_1,\dots,x_n)=0$ whenever $c_1x_1+\dots+c_nx_n+q=0$, for some $c_1,\dots,c_n,q\in\mathbb{R}$. Can the function $$g(x) = \frac{f(x)}{c_1x_1+\dots+c_nx_n+q}$$ be extended as to define a smooth function in $\mathbb{R}^n$? If so, can $g$ be bounded by $f$ and it's derivatives?

Following the ideas in Quotient of two smooth functions is smooth, I believe I've managed to prove that is the case when $n=1$, by writing $g$ as: $$ g(x) = \frac{1}{c_1}\int_0^1f'\left(tx+(1-t)\frac{-q}{c_1}\right)dt$$ However I am having some difficulty generalizing this idea to the case $n>1$. Is there an easier way to prove this? Or is this not true in general? I couldn't think of a counter-example.

MathNewbie
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I believe I managed to prove it, though it's "asymmetry" worries me a bit. I'll prove the case $n=2$ for simplicity, but the general case is analogous. First note that we may assume $c_1$ or $c_2$ are not $0$, otherwise we are dealing with $f=0$ or there are no restrictions. Without loss of generality, suppose $c_2\neq0$. Then note that $$ c_1x_1+c_2\left(\frac{-q-c_1x_1}){c_2}\right)+q=0$$ so that $$f(x_1,x_2) = \int_0^1\frac{\partial f}{\partial x_2}\left(x_1,tx_2+(1-t)\left(\frac{-q-c_1x_1}{c_2}\right)\right)dt\left(x_2+\frac{q+c_1x_1}{c_2}\right)$$ therefore we may write: $$\frac{f(x_1,x_2)}{c_1x_1+c_2x_2+q} = \frac{1}{c_2} \int_0^1\frac{\partial f}{\partial x_2}\left(x_1,tx_2+(1-t)\left(\frac{-q-c_1x_1}{c_2}\right)\right)dt$$ so it is smooth and bounded by $\partial_{x_2}f$. Is this ok?

MathNewbie
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