**Motivation/Hand-Wavy Question:**
In [this post](https://mathoverflow.net/questions/335736/low-degree-polynomial-approximation-of-the-piecewise-linear-function-x-mapsto), it was asked what the best local approximation of $f(x):=\max\{0,x\}$ is by a *polynomial* of a given degree; with the answer provided by Chebyshev's [Equioscillation theorem](https://en.wikipedia.org/wiki/Equioscillation_theorem). Instead, I'm looking for a smooth (not necessarily analytic) global approximation of $f$ (which should be possible since I can now leverage smooth bump functions) such that $f$'s Lipschitz constant is also approximated by the smooth function's $C^k$-norms:


$$
\|f\|\_k\,:=\,\max\_{0\le j \le k} \sup\_{x\in\mathbb{R}} \left |\partial^{j} f (x) \right |.
$$




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**Rigerous Question:**
For every $\delta\_1,\, \delta\_2$ and each $k\in \mathbb{N}$, I'm looking for a smooth (not necessarily analytic) approximation $f\_{k,\delta}\in C^{k,1}(\mathbb{R})$ satisfying


* $\sup\_{x\in \mathbb{R}}\, \big|f(x)-f\_{k,\delta}(x)\big|<\delta\_1$,
* $\big|\|f\_{k,\delta}\|\_{C^{k}} - \operatorname{Lip}(f)\big| = \big|\|f\_{k,\delta}\|\_{C^{k}} - 1\big| <\delta\_2$.


Is such an approximation possible? Atleast can I get $\delta\_1$ arbitrarily small even if I cannot get $\delta\_2$ to approach 1?




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If no, what what is the largest $k$ for which this is possible. If yes, can we also require that $f\_{k,\delta}$ belong to the Sobolev spaces $W^{\tilde{k},2}(\mathbb{R})$ for appropriately large $\tilde{k}$ (given by the Sobolev embedding theorem)?



Smooth approximation of the $\max\{0,x\}$ function with controlled derivatives
Let us consider $[-1,1]$ instead.


**Lemma 1**. For any $\epsilon>0$, there exists $f\_\epsilon\in C^{\infty}\left(\left[0,1\right]\right)$,
a bijection from $\left[0,1\right]$ onto itself, such that
\begin{align\*}
f\_\epsilon(0) & =0,\\
f\_\epsilon (1) & =1,\\
\forall k\in\mathbb{N},k\geq1,f\_\epsilon^{(k)}\left(0\right)=f\_\epsilon^{(k)}\left(1\right) & =0,\\
\sup\_{\left[0,1\right]}\left|f\_\epsilon^{\prime}\left(x\right)\right| & <1+\epsilon.
\end{align\*}


A constructive proof is [here](https://math.stackexchange.com/questions/1736090/cut-off-function-construction/4365567#4365567). With this lemma, you see that for any $\epsilon$, define $f\_{1,\delta} = \begin{cases} 0 &\textrm{if } x<0 \\ f\_\delta &\text{otherwise } \end{cases}$ then $f\_{1,\delta}\in C^\infty[-1,1]$ and fits the bill for $k=1$ (and $\delta\_1=\delta\_2=\delta$ wlog).


For $k\geq2$ that isn't possible. Pietro Majer in the comments to your question has given a nice counter example, and Giorgio Metafune has explained why this cannot be for the same reason even if you only ask the second derivative to be bounded by an arbitrary constant ($\delta\_1\to0$, $\delta\_2=10$ for example). By Arzela-Ascoli, the sequence of first derivative would be equicontinuous and (up to extraction of a subsequence) converge uniformly in $C^1$ to $x^+$ which is not in $C^1$.


