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world:max


2017/10/07 23:39 ·

$\newcommand{\sgn}{\mathrm{sgn}}$

# Statement

Let $h (\beta)= (\beta -u)^2+c |\beta|$ where $u>0$. Show that the argmin of $h$ can be written as $$\beta^\star=u \lr{1-\frac{c}{2|u|}}^+$$

# Proof

The function $h$ is strictly convex and $\lim_{\beta \to \pm \infty}|h (x)|=\infty$. This implies that $h$ admits a unique minimizer $\beta^\star$.

• Case 1: $\beta^\star\neq 0$, in which case $h' (\beta^\star)=0$. This implies $2 (\beta^\star-u)+c \sgn(\beta^\star)=0$. Therefore $2u=\sgn(\beta^\star)\lr{2|\beta^\star|+c}$, which implies $\sgn (u)=\sgn (\beta^\star)$. Therefore $2 (\beta^\star-u)+c \sgn(u)=0$

from which we deduce $\beta^\star=u \lr{1-\frac{c}{2|u|}}$. Using again $\sgn (u)=sgn (\beta^\star)$, we deduce $1-\frac{c}{2|u|}\geq 0$ and finally, $$\beta^\star=u \lr{1-\frac{c}{2|u|}}^+$$

• Case 2: $\beta^\star= 0$. In this case, for all $\beta \neq 0$, $h (\beta) \geq h (0)=u^2$, which is equivalent to $\beta^2-2 \beta u+c|\beta|\geq 0$. Dividing by $|\beta|$ and letting $\beta\to 0$, we get $-2 u\sgn(\beta)+c \geq 0$ which in turn implies $-2|u|+c\geq 0$. This shows $1-\frac{c}{2|u|}\leq 0$ and we therefore have again:

$$\beta^\star=0=u \lr{1-\frac{c}{2|u|}}^+$$