Smoothing a piecewise-linear real-valued function relative to a closed set

The function $f$ is continuous on $\mathbb{R}$ and smooth on the open neighbourhood $(-\mathrm{\infty },-1\mathrm{/}2)$ of $A$.

Let $\epsilon :\mathbb{R}\to (0,\mathrm{\infty })$ be any continuous error function. By continuity at $0$, choose $0<\delta <\min \{1\mathrm{/}2,\epsilon (0)\mathrm{/}2\}$ such that $\epsilon (x)>\epsilon (0)\mathrm{/}2$ whenever $\mathrm{\mid }x\mathrm{\mid }\le \delta$.

Let $\sigma$ be the standard smooth step from The standard smooth step function and set $\tilde{f}(x)=x\sigma ((x+\delta )\mathrm{/}(2\delta ))$. This is smooth, equals $0=f(x)$ for $x\le -\delta$, and equals $x=f(x)$ for $x\ge \delta$. In particular it agrees with $f$ on the open neighbourhood $(-\mathrm{\infty },-\delta )$ of $A$.

For $\mathrm{\mid }x\mathrm{\mid }\le \delta$, the bound $0\le \sigma \le 1$ gives $\mathrm{\mid }\tilde{f}(x)-f(x)\mathrm{\mid }\le \mathrm{\mid }x\mathrm{\mid }\le \delta <\epsilon (0)\mathrm{/}2<\epsilon (x)$; outside this interval the error is $0<\epsilon (x)$. Thus this explicit smoothing has the stated relative agreement and arbitrary positive continuous error control.