Consider a Lipschitz function $f:\mathbb{R}\to\mathbb{R}$ satisfying: $ |f(x)-f(y)|\leq L_f|x-y|. $ For any $\epsilon>0$, a piecewise approximation $f_\epsilon$ can be constructed as $$ f_\epsilon(x)= \sum_{k\in \mathbb{Z}} f(k\delta)\mathbf{1}_{(k\delta,(k+1)\delta]}(x), $$ with $\delta=\epsilon/L_f$, such that $|f(x)-f_\epsilon(x)|<\epsilon$ for any $x\in \mathbb{R}$. My question is, if we consider $f$ to be a locally Lipschitz function: $$ |f(x)-f(y)|\leq L_f(1+|x|^m+|y|^m)|x-y|, $$ where $m$ is a fixed positive integer, can we still construct a piecewise approximation $f_\epsilon$ such that $|f(x)-f_\epsilon(x)|<\epsilon$ for any $x\in \mathbb{R}$?
I am not an expert in the theory of function approximation. Please let me know if there are any parts not clear in my question. It will be better to have an explicit construction of this approximation but an abstract proof of existence is also useful. Even an approximation that is not piecewise is also okay here. More related references are also appreciated.
So far I think the difficulty here is since we no longer have the global Lipshitz continuity, the original piecewise constant approximation does not work without squeezing the value of $\delta$ locally (depending on $x$). Note that the piecewise constant approximation can be treated as using a piecewise polynomial with order $0$ to approximate the $f$ with linear growth ) because a Lipschitz $f$ satisfies $|f(x)|\leq C(1+|x|)$). For a locally Lipschitz $f$, we can show it has polynomial growth $|f(x)|\leq C(1+|x|^m)$, my guess is there might be an explicit piecewise polynomial approximation (with degree $m-1$). I really appreciate it if anyone has any thoughts on this question. Thanks again.