Let $[a,b]$ be a closed and bounded interval and let $f:[a,b]\to\mathbb{R}$ be a continuously differentiable function $f\in C^1([a,b])$. I am trying to bound from above the following expression $$\frac{1}{\varepsilon}\int_{\{|f|<\varepsilon\}}|f'(x)|dx$$ uniformly as $\varepsilon\downarrow 0$.
I would like to prove $$\sup\left\{\frac{1}{\varepsilon}\int_{\{|f|<\varepsilon\}}|f'(x)|dx:\varepsilon\in(0,1]\right\}$$ is finite but I have no idea of how to do it. Any suggestions?
Olivier made me realise I should probably explain how this came up, so I will explain the original question. Consider the step function approximation of Dirac's delta $\delta^\varepsilon(t)=\frac{1}{2\varepsilon}\textbf{1}_{\{|t|<\varepsilon\}}$. I would like to justify for $f\in C^1(\mathbb{R})$ the equation $$\lim_{\varepsilon\downarrow 0}\int_0^T\delta^\varepsilon(f(t))|f'(t)|dt=\int_0^T\lim_{\varepsilon\downarrow0}\delta^\varepsilon(f(t))|f'(t)|dt$$
Any hints on this would be equally helpful.
This is not true.
Suppose $f$ has an infinite number of zeros $z$ for which $f'(x)$ is a constant $M_z\not = 0$ in a small neighborhood of $z$. For any $\varepsilon > 0$, let $z_1, z_2, \dots, z_n$ be zeros of $f$ and $E_i \ni z_i$ be connected components of $\{x : |f(x)| < \varepsilon\}$ such that $f'|_{E_i} = M_i \not = 0$. Denote by $l(E_i)$ the measure of $E_i$. Then $l(E_i) |M_i| = 2\varepsilon$ and $$ \frac{1}{\varepsilon} \int_{|f| < \varepsilon} |f'(x)| dx \geq \frac{1}{\varepsilon}\sum_{i=1}^n l(E_i)|M_i| = 2n. $$ Taking $\varepsilon > 0$ sufficiently small, we can find an arbitrarily large number $n$ of such zeros $z_i$ and components $E_i$. The supremum is therefore infinite.
Note: The argument could be refined and applied to $f(x) = x^3 \sin (1/x)$, defined on $[0,1]$, for instance.