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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and assume that locally $\partial_1f(x)\ge C > 0$ for some constant $C>0.$

Then I saw an application of the implicit function theorem that claimed that this would imply locally

$f(x)=g(x)(x_1-h(x_2,...,x_n))$ for some functions $g,h$ however I do not see how this follows from the implicit function theorem. Can anybody shed some light on how this follows from the theorem?

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To get an explicit representation of this special form you probably need to assume that $f(a)= 0 $ for some $a$.

In this case the implicit function theorem tells you that, in a neighbourhood $U$ of $a$, the zero set $ L_0:=\{x\in U:f(x)= 0\}$ can be written as

$$L_0 = \{(x_1,\ldots, x_n):x_1 - h(x_2,\ldots, x_n) = 0\}$$ You can then define $$g(x) = \frac{f(x)}{x_1 - h(x_2,\ldots, x_n)}$$ whenever $x\in U$ and $f(x)\neq 0$ which is equivalent to the fact that the denominator in that expression is not $=0$ (so it's well defined).

The point is that it does not matter how $g$ is defined on $L_0$, since $ x_1 - h(x_2,\ldots, x_n)= 0$ on that set. This says nothing about regularity of $g$, though.

If there is no such $a$ then you can repeat the same with $f(x)-f(a)$ but with a slightly different result.

Also note that this reasoning only works locally.

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