I would like to produce, for every $n \in \mathbb{N}$, a function $f_n: \mathbb{R}^d \to \mathbb{R}^d$ with the following properties:
- $f_n \in C^\infty(\mathbb{R}^d; \mathbb{R}^d)$,
- $f_n(x)=x$ for every $x \in \mathbb{R}^d$ with $|x| \le n$,
- $|f_n(x)| \le C_n$ for every $x \in \mathbb{R}^d$ and some constant $C_n>0$,
- the operator norm of the Jacobian matrix of $f$ at $x$, $\|D f(x)\|$, is uniformly bounded by $1$ (even $1+\epsilon_n$ with $\epsilon_n \downarrow 0$ would be ok).
This is easy to be done in dimension $d=1$ (it is enough to smoothly join the identity function with the constant $2n$ from $x=n$ to $x=2n$ keeping the absolute value of the derivative bounded by $1$), but even in dimension $d=2$, I cannot come up with an example.
Thank you in advance for your help!
EDIT: maybe I have a solution: consider the function $g_n:\mathbb{R}^d \to \mathbb{R}^d$ defined as $$ g_n(x):= \begin{cases} x \quad &\text{ if } |x|\le n+1, \\ x/|x|(n+1) \quad &\text{ if } |x| \ge n+1. \\ \end{cases} $$ Then $g_n$ is $1$-Lipschitz and $(n+1)$-boudned. Consider a standard mollifier $\varrho$ and then define $f_n:= g_n \ast \varrho$. This is again $1$-Lipschtiz and smooth hence the norm of its derivative is bounded above by $1$, it also coincides with $x$ if $|x| \le n$ and it is bounded.