Differences

This shows you the differences between two versions of the page.

--- world:max [2021/10/06 10:35]
rdouc [Proof]
+++ world:max [2022/03/16 07:40]
@@ Line 1: / Line 1: @@
-{{page>:defs}}
-$\newcommand{\sgn}{\mathrm{sgn}}$
-====== 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
-$\beta^\star$.
-  * $\beta^\star\neq 0$, 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,
-  $$
-\beta^\star=u \lr{1-\frac{c}{2|u|}}^+
-$$
-  * $\beta^\star= 0$. 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:
-  $$
-\beta^\star=0=u \lr{1-\frac{c}{2|u|}}^+
-  $$

Welcome to Randal Douc's wiki