Which is the simplest one? For example, we smooth $f(x)=|x|$ to $$f(x)=\begin{cases} \frac{x^2}{\epsilon}+\frac{\epsilon}{2} & |x| \le \epsilon\\ |x| & |x|\ge epsilon \end{cases}$$
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What do you mean by smoothing? – mStudent Jun 15 '13 at 15:47
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like $|x|$, We could change its value around 0, make it is smooth. – Vivian Jun 15 '13 at 18:31
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The simplest (and only) method I know is mollification.
Daniel Robert-Nicoud
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2+1. And though it is not compactly supported, the heat kernel can yield similar results by playing the rolw of a mollifier. In the case of the absolute value function this leads to smoothings like $$\frac{e^{-25 x^2}}{5 \sqrt{\pi}}+x \operatorname{erf}(5x)$$ or simply $$x \operatorname{erf}(5x)$$ if you prefer. – Antonio Vargas Jun 15 '13 at 21:26
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In convex analysis we also have the Moreau-Yosida regularization of $f$, which is the "infimal convolution" of $f$ with the convex function $\frac{1}{2\mu} |x|_2^2$. In other words, $f^{(\mu)}(x) = \inf_u f(u) + \frac{1}{2\mu} |u-x|_2^2$. – littleO Sep 15 '13 at 19:16
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In convex analysis, maybe the most natural or best way to smooth a convex function $f$ is to use the Moreau-Yosida regularization of $f$:
\begin{equation} f^{(\mu)}(x) = \inf_u f(u) + \frac{1}{2\mu} \|u - x\|_2^2. \end{equation}
This is discussed in lecture 15 ("Multiplier methods") of Vandenberghe's 236c notes.
If $f(x) = |x|$ for all $x \in \mathbb R$, and $\mu > 0$, then \begin{equation} f^{(\mu)}(x) = \begin{cases} |x| - \frac{\mu}{2} \quad \text{if } |x| > \mu \\ \frac{x^2}{2\mu} \quad \text{otherwise}. \end{cases} \end{equation}
littleO
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