I want to compute the Moreau envelope of the function
$$h:\mathbb{R} \to \mathbb{R}, h(x) = \begin{cases} \frac{1}{2}x & x\leq 0 \\ 2x & x>0. \end{cases}$$
I know the the Moreau envelope will be given by:
$$M_{\lambda h}(x) = \inf_{u}\space\{h(u)+\frac{1}{2\lambda}(u-x)^2\}.$$ I am thinking of considering cases based on the value of $x$, but I get confused when trying to see when a certain infimum occurs.
Let us denote $f(x) = \inf\limits_{u \le 0} \left\{ h(u) + \frac{1}{2\lambda} (u-x)^2 \right\}$, $g(x) = \inf\limits_{u > 0} \left\{ h(u) + \frac{1}{2\lambda} (u-x)^2 \right\} $. Then $$f(x) = \inf\limits_{u \le 0} \left\{ \frac{1}{2}u + \frac{1}{2\lambda} (u-x)^2 \right\} = \inf\limits_{u \le 0} \left\{\frac{1}{2\lambda}u^2+\left(\frac{1}{2} - \frac{x}{\lambda}\right)u +\frac{1}{2\lambda}x^2 \right\} = $$ $$= \begin{cases} \frac{1}{2\lambda}(x -\lambda/2)^2+\left(\frac{1}{2} - \frac{x}{\lambda} \right)(x - \lambda/2) + \frac{1}{2\lambda}x^2 , \quad x \le \lambda/2, \\ \frac{1}{2\lambda}x^2 , \quad x > \lambda/2\end{cases} =$$ $$ = \begin{cases} \frac{x}{2} - \frac{\lambda}{8}, \quad x \le \lambda/2, \\ \frac{1}{2\lambda}x^2 , \quad x > \lambda/2 .\end{cases}$$ $$g(x) = \inf\limits_{u > 0} \left\{ 2u + \frac{1}{2\lambda} (u-x)^2 \right\} = \inf\limits_{u > 0} \left\{\frac{1}{2\lambda}u^2+\left(2 - \frac{x}{\lambda}\right)u +\frac{1}{2\lambda}x^2 \right\} =$$ $$= \begin{cases} \frac{1}{2\lambda}(x -2\lambda)^2+\left(2 - \frac{x}{\lambda} \right)(x - 2\lambda) + \frac{1}{2\lambda}x^2 , \quad x > 2\lambda, \\ \frac{1}{2\lambda}x^2 , \quad x \le 2\lambda \end{cases} =$$ $$= \begin{cases} 2x- 2\lambda , \quad x > 2\lambda, \\ \frac{1}{2\lambda}x^2 , \quad x \le 2\lambda. \end{cases} $$ So, $M_{\lambda h}(x) = \min\limits_{x}(f(x), g(x))$ (we use fact $\inf\limits_{A \cup B} (\cdot) = \min(\inf\limits_{A} (\cdot), \inf\limits_{B}(\cdot))$).
If $x<\lambda/2$ then $\frac{x}{2} - \frac{\lambda}{8} < \frac{1}{2\lambda}x^2 $, because $$\frac{x}{2} - \frac{\lambda}{8} < \frac{1}{2\lambda}x^2 \, \Longleftrightarrow \, 4x\lambda -\lambda^2<4x^2 \, \Longleftrightarrow \, (2x-\lambda)^2>0.$$ So, $M_{\lambda h}(x) = \frac{x}{2} - \frac{\lambda}{8}$ in this case.
If $\lambda/2 \le x \le 2\lambda$ then $M_{\lambda h}(x) = \frac{1}{2\lambda}x^2$.
If $x>2\lambda$ then $\frac{1}{2\lambda}x^2 > 2x-2\lambda$, because $$\frac{1}{2\lambda}x^2 > 2x-2\lambda \, \Longleftrightarrow \, x^2> 4x\lambda - 4\lambda^2 \, \Longleftrightarrow \, (x-2\lambda)^2>0$$ So, $M_{\lambda h}(x) = 2x - 2\lambda$ in this case.
Finally, $$M_{\lambda h}(x) = \begin{cases} \frac{x}{2} - \frac{\lambda}{8}, \quad x<\lambda/2, \\ \frac{1}{2\lambda}x^2, \quad \lambda/2 \le x \le 2\lambda, \\ 2x - 2\lambda, \quad x>2\lambda. \end{cases}$$ For clarity, I have included a picture when $\lambda = 1$ (blue curve is graph of function $h(x)$, red curve is graph of function $M_{\lambda h}(x)$)
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