Timeline for Optimality condition for non-differentiable constrained convex optimization problem
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 21, 2014 at 13:11 | comment | added | Yury | If $f$ is defined on $D$, then $0\in\partial f(x)$. | |
| May 21, 2014 at 7:46 | comment | added | martin | @Yury Found proof along those lines (edited main post). | |
| May 21, 2014 at 7:45 | history | edited | martin | CC BY-SA 3.0 |
found proof
|
| May 21, 2014 at 7:19 | comment | added | martin | @Yury That's the case for unconstrained problems, but in a constrained problem a point may be optimal without $0\in\partial f(x)$, in particular when the constraint is tight. I'm trying to avoid having to compute the subdifferential of $f(x) + \mathbf{I}_D(x)$, where $\mathbf{I}_D(x)\in\{0,\infty\}$ is the usual indicator function. | |
| May 21, 2014 at 3:39 | comment | added | Yury | If $f$ attains its minimum at point $x$, then $0\in \partial f(x)$. Condition $v^T (y-x) \geq 0$ holds for $v=0\in \partial f(x)$, as required. | |
| May 21, 2014 at 2:42 | history | asked | martin | CC BY-SA 3.0 |