Showing $f(x_1,...,x_n)$ = $g(x_1,...,x_{n-1})$ and $g$ is differentiable

Question

I have an open ball $E^n$ of center $(a_1, a_2, a_3,...,a_n)$ and radius $r$.

$f$ is a differentiable real-valued function on that open ball. $f_{x_n}$=$\frac{\partial f}{\partial x_n} = 0$ .

I do not understand how there is a unique real-valued function $g$ on the open ball in $E^{n-1}$ of center $(a_1,...,a_{n-1})$ and radius $r$ such that $f(x_1,...,x_n)$ = $g(x_1,...,x_{n-1})$, and this $g$ is differentiable.

Any help with showing this would be great thank you.

Answer 1

Use the mean-value theorem to show that $f(x_1,...,x_{n-1},x_n) = f(x_1,...,x_{n-1},a_n)$ for each $(x_1,...,x_n)$. This means we can just define $g(x_1,...,x_{n-1}) = f(x_1,...,x_{n-1},a_n)$ to get the desired function. (Recall $a_n$ is constant here so this is just a function of the first $n-1$ variables).