Let $\rho(x)$ be a function such that its derivative is bounded and its limit at $x \to -\infty$ exists. We can rewrite $\rho$ like so and define $\rho_{\alpha}$:
$$ \rho(x)=\rho(-\infty)+\int_{-\infty}^x\rho'(y) dy=\rho(-\infty)+\int_{\mathbb{R}}\rho'(y)1_{x>0}(x-y) dy $$ $$ \rho_{\alpha}(x)=\rho(-\infty)+\int_{\mathbb{R}}\rho'(y)h_{\alpha}(x-y)dy $$ where we define $h_{\alpha}$ as $$ h_{\alpha}(x)=\frac{1}{2}(1+\tanh(\alpha x)) $$
How do I show that the following holds? $$ \sup_x |\rho(x)-\rho_{\alpha}(x)| \leq |\rho'|_{L^{\infty}(\mathbb{R})} |1_{x>0}-h_{\alpha}|_{L^1} $$
We have:
\begin{align*}|\rho(x)-\rho_{\alpha}(x)| &= \left|\int_{\mathbb{R}}\rho'(y)(1_{x>0}-h_\alpha)(x-y) dy\right| \\ &\le \int_{\mathbb{R}}\left|\rho'(y)| \cdot |(1_{x>0}-h_\alpha)(x-y)\right| dy \\ &\le \int_{\mathbb{R}}\left|\rho'|_{L^\infty}|(1_{x>0}-h_\alpha)(x-y)\right| dy \\ &= |\rho'|_{L^\infty} \int_{\mathbb{R}}|(1_{x>0}-h_\alpha)(x-y)dy \\ &= |\rho'|_{L^{\infty}(\mathbb{R})} |1_{x>0}-h_{\alpha}|_{L^1} \end{align*}
The results follows by taking $\sup_x$ on both sides