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world:max [2021/10/06 10:37] rdouc [Proof] |
world:max [2021/10/06 10:38] rdouc [Proof] |
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The function $h$ is strictly convex and $\lim_{\beta \to \pm \infty}|h (x)|=\infty$. This implies that $h$ admits a unique minimizer | The function $h$ is strictly convex and $\lim_{\beta \to \pm \infty}|h (x)|=\infty$. This implies that $h$ admits a unique minimizer | ||
$\beta^\star$. | $\beta^\star$. | ||
- | 1. **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 $2 (\beta^\star-u)+c \sgn(u)=0$ | + | * **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, | 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=u \lr{1-\frac{c}{2|u|}}^+ | ||
$$ | $$ | ||
- | 2. **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|}}^+ | ||
$$ | $$ | ||