Space of All Smooth Lipschitz Functions

Question

This is a followup question to this post. Let $F$ be the collection of all Lipschitz functions from $\mathbb{R}^d$ to itself, which admit $k>0$ derivatives everywhere and such that the $k^{th}$ differential is itself Lipschitz. Is this space a complete metric space, under the metric $$ d(f,g) \triangleq \sum_{i=0}^k \|f^{(i)}(x)-g^{(i)}(x)\| + Lip(f^{(k)}-g^{(k)})? $$ Here $Lip(h)$ is the minimal Lipschitz constant of a Lipschitz function $h$.

Answer 1

Note the following:

1. If $f_n$ is a sequence of bounded continuous functions that are Cauchy wrt the supremum norm, then a bounded continuous function $f$ exists so that $\|f_n -f\|\to0$.
2. If $f_n$ is as above and additionally differentiable with $df_n$ being bounded continuous and Cauchy wrt sup-norm, then $f$ is differentiable and $df=\lim_n df_n$.
3. If $f:\Bbb R^d\to \Bbb R^d$ has bounded differential, then $f$ is Lipschitz (with Lipschitz constant $\|df\|$).
4. If $f_n$ is a sequence like in 1. and additionally $f_n$ are Lipschitz with the Lipschitz constants being $≤$ a common bound, then $f$ is Lipschitz.

Now combine this stuff to see:

If you have a sequence of $k$-times differentiable functions $f_n$ so that all $k$-derivatives are bounded and your sequence is Cauchy wrt the $k$ semi-norms $\|d^i f_n\|$, $i\in\{0,...,k-1\}$, then a limit $f$ exists and is also $k$-times differentiable (apply 1. and 2. $k$-times). It follows that the first $k-1$ differentials are Lipschitz (from 3.).

If you also ask that $f_n$ is Cauchy wrt the semi-norm $\mathrm{Lip}(d^k f)$, then 4. tells you that the limit $f$ also has the property that $d^k$ is Lipschitz.

Now note that a sequence is Cauchy wrt a finite amount of semi-norms $\|\cdot\|_{i}$ iff it is Cauchy wrt $\sum_{i} \|\cdot\|_i$. Thus if you have a sequence that is Cauchy wrt your norm, then it admits a limit that also has the properties of being $k$-times differentiable with the $k$-th derivative being Lipschitz. This means your space is complete.