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--- world:max [2021/10/06 10:36]
rdouc [Proof]
+++ world:max [2022/03/16 07:40] (current)
@@ Line 14: / Line 14: @@
 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  $2 (\beta^\star-u)+c \sgn(u)=0$
+  * **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,
 $$
 \beta^\star=u \lr{1-\frac{c}{2|u|}}^+
 $$
-  *  $\beta^\star= 0$ . 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:
+  * **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:
 $$
 \beta^\star=0=u \lr{1-\frac{c}{2|u|}}^+
 $$

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