Can we approximate $f(x) = \chi_{(0,\infty)}(x)$ by smooth monotone functions?

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Given $f(x) = \text{sign}^+(x) = \chi_{(0,\infty)}(x)$, is it possible to a sequence $f_n$ such that $f_n$ is smooth, $f_n' \geq 0$ and $$f_n \to f?$$ In some sense of convergence?

Preferably $f_n''$ should be bounded but this is not necessary.

How to do these approximations? Does it have something to do with Yosida approximation?

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There are already some examples on how to construct such an approximation in the comments above. So I'll just say something about the limitations:

It is easy to get pointwise convergence. However you will not get uniform convergence (uniformly converging $f_n$ would imply that the limit $f$ is continuous, which it isn't). Also you cannot get any uniform bounds on the derivative, since $\int_{-\epsilon}^\epsilon f_n'(x) dx = f_n(\epsilon) - f_n(-\epsilon) \rightarrow 1$ for any $\epsilon > 0$, which also implies that none of the higher derivatives are bounded.

The one additional thing you can get are integral bounds, so $$\|f-f_n\|_{L^p} = \sqrt[p]{ \int_{\mathbb{R}} |f(x)-f_n(x)|^p dx} \rightarrow 0$$ is possible for any $p\in [1,\infty)$, but again not uniformly for all $p$ (this would imply uniform convergence). However there are again no such bounds on the derivative, so $\|f'-f_n'\|_{L^p}$ does not converge to 0. (This would imply that $f$ has a weak derivative, which it hasn't. If you want more information about that, you will have to look into Sobolev spaces.)