# Inverse of a composite function from $\mathbb R$ to $\mathbb{R}^p$ to $\mathbb R$ again with a non-zero continuous gradient in a point

Let $D \subset \mathbb{R}^p$ and $f: D \to \mathbb{R}$ a continious function. Let $a$ be a point in $D$. Let $\nabla f(a) \neq 0$ and let the gradient be continious at $a$.

To proof: there exists a function $c: [a, b] \to \mathbb{R}^p$ such that $f(c(t)) = t$ for all $t \in [a, b]$.

I know that an inverse function exists if the function is injective, and what partial derivatives are. I don't know how to do the chain rule in multiple variables however.

• What is your question ? – Red shoes Jun 16 '17 at 20:02
• Why there exists a function $c: [a, b] \to \mathbb{R}^p$ such that $f(c(t)) = t$ for all $t \in [a, b]$ – Pel de Pinda Jun 16 '17 at 20:06
• Can you clarify the relation between $a \in D$ and function $c$? – Red shoes Jun 16 '17 at 20:12
• There isn't really a relation between them. The gradient of f is non zero and continious at a, and that should guarentee the existance of a function c as described in the post, but I have no idea why that is the case. – Pel de Pinda Jun 16 '17 at 20:14

So Let assume $\nabla f$ exists in a neighborhood of $a$!

$\nabla f(a) \neq 0$ so WLOG let assume $\frac{\partial f }{\partial x_1} (a) \neq 0$ and $f(a)=0$.

This means the function $g(t) = f(a+ t e_1)$ is continuously differentiable at $t=0$ and $g' (0) = \frac{\partial f }{\partial x_1} (a) \neq 0$, so according to the inverse mapping theorem (one dimension case) $g: R \to R$ has an inverse locally around $t=0$ i.e., $g^{-1} : (- \delta , \delta ) \to (- \epsilon , + \epsilon)$ is continuous. Now define $c :(- \delta , \delta ) \to R^p$ with $c(t) = a + g^{-1} (t) e_1$.

P.S : $e_1 = (1,0,0,...,0)$

If it is high price to use invers function theorem , you can argue like as follow WLOG assume $g' (0)= \frac{\partial f }{\partial x_1} (a) > 0$ this guarantees that for small enough $\delta >0,$ $g'(x) > 0$ for all $x \in (-\delta ,+ \delta )$ because $g'$ is continuous at $t=0$! So $g$ is strictly increasing, thus it has an inverse.

• How do you know that $g'(x)$ exists for all $x\in (-\delta,\delta)$? Isn't it possible that $\text{Dom}(g')=\{0\}$? – Tyron Jun 19 '17 at 9:11
• By continuouty of $\nabla f$ at $a$ I assumed $\nabla f$ exists in a neighborhood of $f$, other wise there is counter example for problem ! – Red shoes Jun 19 '17 at 16:21
• @Tyron See an example I constructed above ! – Red shoes Jun 19 '17 at 16:40

Question's claim as stated is not true:

For $p=1$, let $\phi : R \to R$ be the well-known continuous function, but differentiable no where.

Now Take $f(x):=x + x^2 \phi(x)$ then $f$ is continuous every where $Dom (f')= \{0\}$ and $f' (0) =1 >0$ so $f'$ is continuous. But there is no sub interval $(a,b)$ on which $f$ be invertible on. !

• But if it is differentiable no where, does the gradient exist? Because the gradient has to be non-zero in a point a. – Pel de Pinda Jun 25 '17 at 8:14
• @PeldePinda $\phi$ is differentiable no where not $f$ ! $f$ is only differentiable at $x=0$ – Red shoes Jun 25 '17 at 8:16