Inverse Function Theorem - Challenging question I found the following question on the internet, and I think it would be useful to solve it as part as studying to my midterm:
Let $f:\mathbb{R}^n \to \mathbb{R}^n $ having continuous partial derivatives up to first order , such that:
$ \| f(x)- f(y) \| > \frac{1}{10} \|x-y\| $ for every $x,y\in \mathbb{R}^n $ such that $x\neq y$.


*

*Prove that $J(f) \neq 0 $ 

*Prove that $f$ sends $\mathbb{R}^n $ onto $\mathbb{R}^n $ .
Thanks in advance
 A: Part 1:
Since $f$ is differentiable at $x$, for all $\epsilon>0$ there is some $\delta>0$ such that for all $\|h\| < \delta$, we have $\|f(x+h)-f(x) -DF(x)h\| \le \epsilon \|h\|$. Choose $\epsilon = \frac{1}{20}$.
Then
\begin{eqnarray}
\|Df(x)h\| &=& \|DF(x)h - (f(x+h)-f(x))+(f(x+h)-f(x))\| \\
&\ge& \|f(x+h)-f(x)\| - \|DF(x)h - (f(x+h)-f(x))\| \\
&\ge& \frac{1}{20} \|h\|
\end{eqnarray}
And so $DF(x)$ is invertible and $J(f(x)) \neq 0$.
Part 2:
Let $Y= f(\mathbb{R}^n)$.
Since $DF(x)$ is invertible, the implicit function theorem shows that $f$ has a local inverse, and hence $f$ maps open sets into open sets. Hence $Y$ is open.
Suppose $y_n \in Y$, and $y_n \to y$. Since $y_n \in Y$, there is some $x_n$ such that $y_n = f(x_n)$. Since $\|y_n-y_m \| = \|f(x_n)-f(x_m)\| \ge \frac{1}{10} \|x_n-x_m\|$, we see that $x_n$ is Cauchy, and hence $x_n \to x$ for some $x$. Since $f$ is continuous, we have $f(x) = y$, and so $y \in Y$. This shows that $Y$ is closed.
Since $\mathbb{R}^n$ is connected, and $Y \ne \emptyset$, we have $Y = \mathbb{R}^n$.
A: Since $f: \mathbb{R}^n \to \mathbb{R}^n$ has continuous partial derivatives, $f$ is differentiable. So we have
$$
\lim_{h \to 0} \frac{\|f(x + h) - f(x) - Df(x) \cdot h\|}{\|h\|} = 0.
$$
Triangle inequality gives
$$
\|f(x + h) - f(x)\| - \|Df(x) \cdot h\| \leq \|f(x + h) - f(x) - Df(x) \cdot h\|.
$$
Hence
$$
\lim_{h \to 0} \frac{\|f(x + h) - f(x)\|}{\|h\|} - \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} \leq 0 \\
\implies \lim_{h \to 0} \frac{1}{\|h\|}\frac{1}{10}(\|(x + h) - x\|) - \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} < 0 \\
\implies \frac{1}{10} < \lim_{h \to 0} \frac{\|Df(x) \cdot h\|}{\|h\|} \leq \lim_{h \to 0} \frac{\|Df(x)\| \cdot \|h\|}{\|h\|} \\
\implies \|Df(x)\| > \frac{1}{10}
$$
where $\|Df(x)\|$ is the operator norm. So $Df(x) \neq 0$.
