If $N(x) = x-[f^{\prime}(x)]^{-1}f(x)$ and $f(x^\ast)=0$, how can I prove that $N^\prime(x^\ast) = 0$?

Question

Let $\Omega\subseteq \mathbb{R}^n$ open in $\mathbb{R}^n$ and $f\in C^{2}(\Omega,\mathbb{R}^n)$. Let $N(x) = x-[f^{\prime}(x)]^{-1}f(x)$, supose that:

• there exist $x^\ast\in \Omega$ such that $f(x^\ast)=0$,
• for all $x\in\Omega$, we have $N(x)\in\Omega$.
• $f^\prime(x^\ast)$ is not singular,

How can I prove that $N^\prime(x^\ast) = 0$?

My attempt. We have $N(x^\ast)=x^\ast$ if, only if, $f(x^\ast)=0$. My strategy is to make Taylor's development of $$ N(x^\ast+v)=N(x^\ast)+N^\prime(x^\ast)v+O(v^2) $$ and show that the term $N^\prime(x^\ast)=0$ equals zero. Note that \begin{align} N(x^\ast+v) =& (x^\ast+v)-[f^\prime(x^\ast+v)]^{-1}f(x^\ast+v) \\ =& N(x^\ast)+v-[f^\prime(x^\ast+v)]^{-1}f(x^\ast+v) \\ =& N(x^\ast)+v-[f^\prime(x^\ast)]^{-1}f(x^\ast+v) \\ &\qquad+ \Big([f^\prime(x^\ast)]^{-1}-[f^\prime(x^\ast+v)]^{-1}\Big)f(x^\ast+v) \\ &= N(x^\ast)+v-[f^\prime(x^\ast)]^{-1}\Big(f(x^\ast)+f^\prime(x^\ast)\cdot v+O(v^2)\Big) \\ &\qquad+ \Big([f^\prime(x^\ast)]^{-1}-[f^\prime(x^\ast+v)]^{-1}\Big)\Big(f(x^\ast)+f^\prime(x^\ast)\cdot v+O(v^2)\Big) \\ &= N(x^\ast)+v-[f^\prime(x^\ast)]^{-1}\Big(f^\prime(x^\ast)\cdot v+O(v^2)\Big) \\ &\qquad+ \Big([f^\prime(x^\ast)]^{-1}-[f^\prime(x^\ast+v)]^{-1}\Big)\Big(f^\prime(x^\ast)\cdot v+O(v^2)\Big) \\ &= N(x^\ast)+O(v^2) \\ &\qquad+ \Big([f^\prime(x^\ast)]^{-1}-[f^\prime(x^\ast+v)]^{-1}\Big)\Big(f^\prime(x^\ast)\cdot v+O(v^2)\Big) \end{align} But I can not eliminate the term $ v $ from the above expression.

Answer 1

Why not simply evaluating $N'(x)$ from its definition?

Assuming there exists $g(x) = f^{-1}(x)$ in a neighborhood around $x^*$, as your problem satisfies the conditions of the Inverse Function Theorem.

Then let $J[N](x)$ denote the Jacobian matrix of $N$ (which is preferable as notation when not working on a single dimension).

Rewriting: $$ N(x) = x - (J[f](x))^{-1}f(x) $$

By inverse function theorem:

$$ N(x) = x - J[g](x)f(x) $$

Now, taking the Jacobian of the above expression:

$$ N(x) = x - J[g](x)f(x) $$

Deriving: $$ J[N](x) = J[x](x) -\left(J\left[J[g](x)\right](x)\right)f(x)-J[g](x)J[f](x) $$ The first term is trivial: $$ J[N](x) = I -\left(J\left[J[g](x)\right](x)\right)f(x)-J[g](x)J[f](x) $$ The last term is the identity due to the aforementioned theorem: $$ J[N](x) = I -\left(J\left[J[g](x)\right](x)\right)f(x)-I $$ $$ J[N](x) = -\left(J\left[J[g](x)\right](x)\right)f(x) $$

Now, it might be difficult to compute $J\left[J[g](x)\right](x)$, and it is also a 3 dimensional tensor, and this notation may be crappy. But we have that $f(x^*)=0$, hence that term does not matter. So evaluating at $x^*$, we get a null matrix.

Answer 2

Since \begin{eqnarray*} N(x^*)v&=&\lim_{t\to0}\frac{N(x^*+tv)-N(x^*)}{t} \\ &=&\lim_{t\to0}\bigg\{ v-\frac{[f^{\prime}(x^*+tv)]^{-1}f(x^*+tv)-[f^{\prime}(x^*)]^{-1}f(x^*)}{t}\bigg\}\\ &=&v-\lim_{t\to0}\frac{[f^{\prime}(x^*+tv)]^{-1}[f(x^*+tv)-f(x^*)]}{t}-\lim_{t\to0}\frac{\{[f^{\prime}(x^*+tv)]^{-1}-[f^{\prime}(x^*)]^{-1}\}f(x^*)}{t}\\ &=&v-v=0, \end{eqnarray*} one has $N(x^*)v=0$ and hence $N(x^*)=0$.