Consider $C^1([0,1])$ the functions with continuous derivative on $[0,1]$ (one-sided derivatives at each end), and $\operatorname{Lip}([0,1])$ the Lipschitz functions on $[0,1]$. The mean value theorem of course shows that $C^1([0,1]) \subseteq \operatorname{Lip}([0,1])$.
Given $f\in \operatorname{Lip}([0,1])$ with Lipschitz constant $\leq 1$, and $\epsilon>0$, can I find $g\in C^1([0,1])$ with $\|g'\|_\infty\leq 1$ (i.e. $g$ also has Lipschitz constant $\leq 1$) and with $\|f-g\|_\infty \leq \epsilon$?
(E.g. one way to find $f$ is to let $f(x) = \int_0^x F(t) \ dt$ for some $F\in L^\infty$ with $\|F\|_\infty\leq 1$. Then to find a suitable $g$, I can set $g(x)=\int_0^x a(t) \ dt$ for some $a\in C([0,1])$ with $\|a\|_\infty\leq 1$ and $\| F-a\|_1\leq\epsilon$. I can find $a$ by convolving $F$ with a bump function. But not every Lipschitz $f$ arises in this way.)
Best Answer
Let $\varphi$ be a positive $C_c^\infty$ function so that $\int_\mathbb{R}\varphi(x)\;\mathrm{d}x=1$.
$f\ast\varphi_\epsilon$ is $C^\infty$ because $\varphi$ is, and it has Lipschitz constant $\le1$ because $f$ does. This means that $(f\ast\varphi_\epsilon)^\prime\le1$. Since $f$ is continuous on $[0,1]$, it is uniformly continuous, and therefore, $f\ast\varphi_\epsilon\to f$ uniformly. That is, $$ \|f\ast\varphi_\epsilon-f\|_\infty\to0\tag{1} $$
However, if we use the standard Lipschitz norm, $$ \|f\|_\mathrm{Lip}=\|f\|_\infty+\sup_{x\not=y}\,\left|\frac{f(x)-f(y)}{x-y}\right| $$ the same cannot be said. Take for instance $f(x)=|x|$, the absolute value function. Any $C^1$ function that matches the slope of $f$ just below $0$, will fail to match the slope of $f$ by $2$ just above $0$. Thus, the Lipschitz norm of the difference between a $C^1$ function and $f$ is at least $1$.