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world:max [2021/10/06 10:37] rdouc [Proof] |
world:max [2022/03/16 07:40] (current) |
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\beta^\star=u \lr{1-\frac{c}{2|u|}}^+ | \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: | + | * **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|}}^+ | \beta^\star=0=u \lr{1-\frac{c}{2|u|}}^+ | ||
$$ | $$ | ||