Consider an indicator function $\chi_{[a,b]}$ on an interval $[a,b]$. My question - is there a decreasing sequence $f_\delta\in C^1(V)$ (where $[a,b]\subset V$ and V is open), such that $f_\delta\rightarrow \chi_{[a,b]}$ pointwisely. Further the sequence should satisfy: $\exists C>0, \, \forall \delta>0: \int_V|\partial_x f_\delta|dx<C$.
My initial approach was to do the usual smoothing: \begin{equation} f_\delta(x) = (\chi_{[a-\delta,b+\delta]}\, \ast \, \varphi_\delta)(x), \end{equation} where $\varphi(x)=\chi_{(-1,1)}\exp(-1/(1-x^2))$ is the usual mollifier and $\phi_\delta(x)=\frac{1}{\delta}\varphi(\frac{x}{\delta})$. Here, I work with $\delta< \mathrm{dist}([a,b],V)$.
Off course, $\varphi_\delta$ is a decreasing sequence that converges to $\chi_[a,b]$ pointwisely. How do I show the uniform bound on the derivative? Formally, I expect it but don't know how to show it.
Any hints are appreciated.
The functions $f_{\delta}$ are absolutely continuous. To find the constant, it suffices to note that $f_{\delta}$ are non decreasing in $(-\infty,\frac{1}{2}(a+b))$, so the integral of the absolute value of the derivative is area under the graph. Similarly for the descent.