Implicit function theorem: Let $E$ be an open subset of $\mathbb{R}^n$, let $f : E \to \mathbb{R}$ be continuously differentiable, and let $y = ( y_1, … , y_n)$ be a point in $E$ such that $f(y) = 0$ and $\frac{\partial{f}}{\partial{x_n}}(y) \not= 0$. Then there exists an open subset $U$ of $\mathbb{R}^{n-1}$ containing $(y_1, … , y_{n-1})$, an open subset $V$ of $E$ containing $y$, and a function $g : U \to \mathbb{R}$ such that $g(y_1, … ,y_{n-1}) = y_n$, and
$$\{(x_1, … ,x_n) \in V : f(x_1, … ,x_n) = 0\} = \{(x_1, … ,x_{n-1}, g(x_1, … ,x_{n-1}) : (x_1, … ,x_n) \in U\}.$$
Moreover, $g$ is differentiable at $(y_1, … ,y_{n-1})$, and we have
$$\frac{\partial{g}}{\partial{x_j}}(y_1, … , y_{n-1}) = \frac{\partial{f}}{\partial{x_j}}(y)/ \frac{\partial{f}}{\partial{x_n}}(y)$$
for all $1 \le j \le n-1$.
I don't understand some steps in the proof. I highlight with purple lines the step I am struggling with.
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The author let $F^{-1} (x) = (h_1(x), …, h_n(x))$. Since $F$ is invertible, $x = F(h_1(x), … ,h_n(x))$, but why $h_j(x) = x_j$?
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We know that $W$ is open in $\mathbb{R}^n$, but how do we know that $U$ is open in $\mathbb{R}^{n-1}$?
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Similarly, we know that $h_n$ is differentiable at $(y_1, … ,y_{n-1}, 0)$, but how can we prove that $g$ is differentiable at $(y_1, … ,y_{n-1})$?
I really appreciate if you give some help!
Best Answer
$$(x_1, \ldots, x_j, \ldots, x_n) = x = F(F^{-1}(x)) = F((h_1(x), \ldots, h_j(x), \ldots, h_n(x))) = (h_1(x), \ldots, h_j(x), \ldots, f(h_1(x), \ldots, h_n(x)))$$ Therefore $x_j = h_j(x)$.
Theorem 2.1.5 (c) states that if $f$ is a continuous function then the inverse image/preimage of an open set is also open. We can make a continuous function $f: \mathbb{R}^{n-1} \rightarrow \mathbb{R}^{n}$ that maps $(x_1, \ldots, x_{n-1})$ to $(x_1, \ldots, x_{n-1}, 0)$. By that theorem, $$f^{-1}(W) = \{(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1} \vert f((x_1, \ldots, x_{n-1})) = (x_1, \ldots, x_{n-1}, 0) \in W\} = U$$ is open as $W$ is open.
The derivative of $g$ evaluated at $(y_1, \ldots, y_{n-1})$ is the derivative of $h_n$ evaluated at $(y_1, \ldots, y_{n-1}, 0)$ which exists.