world:max
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world:max [2021/10/06 10:35] rdouc [Proof] |
world:max [2022/03/16 07:40] (current) |
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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 | ||
| $\beta^\star$. | $\beta^\star$. | ||
| - | * $\beta^\star\neq 0$, in which case $h' (\beta^\star)=0$. This | + | * **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 $2 (\beta^\star-u)+c \sgn(u)=0$ |
| - | implies $2 (\beta^\star-u)+c \sgn(\beta^\star)=0$. Therefore | + | 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, |
| - | $2u=\sgn(\beta^\star)\lr{2|\beta^\star|+c}$, 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=u \lr{1-\frac{c}{2|u|}}^+ | ||
| $$ | $$ | ||
| - | * $\beta^\star= 0$. In this case, for all $\beta \neq 0$, | + | * **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: |
| - | $h (\beta) \geq h (0)=u^2$, 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|}}^+ | ||
| - | $$ | + | $$ |
world/max.1633509341.txt.gz · Last modified: 2022/03/16 01:37 (external edit)
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