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world:kkt
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world:kkt [2025/10/08 02:09] rdouc [Weak duality] |
world:kkt [2025/10/08 02:30] (current) rdouc |
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| <hidden Click here to see the proof> | <hidden Click here to see the proof> | ||
| $\bproof$ | $\bproof$ | ||
| - | Assume now that $f(x^*)=\inf_{x\in \mcD} f(x)$ for some $x^* \in\mcD$. Then, $\{f-f(x^*)<0\} \cap \mcD =\emptyset$ so that there exists $\lambda^* \geq 0$ satisfying for all $x\in \mcD$, $f(x)-f(x^*)+\sum_{i=1}^n \lambda^*_i h_i(x) \geq 0$. And this implies that $(x^*,\lambda^*)$ is a saddle point since: for all $\lambda \geq 0$ and $x \in \Xset$, | + | Assume now that $f(x^*)=\inf_{x\in \mcD} f(x)$ for some $x^* \in\mcD$. Then, $\{f-f(x^*)<0\} \cap \mcD =\emptyset$ so that there exists $\lambda^* \geq 0$ satisfying for all $x\in \Xset$, $f(x)-f(x^*)+\sum_{i=1}^n \lambda^*_i h_i(x) \geq 0$. And this implies that $(x^*,\lambda^*)$ is a saddle point since: for all $\lambda \geq 0$ and $x \in \Xset$, |
| $$ | $$ | ||
| f(x^*)+\sum_{i=1}^n \lambda_i h_i(x^*) \leq f(x^*) \leq f(x)+\sum_{i=1}^n \lambda^*_i h_i(x) | f(x^*)+\sum_{i=1}^n \lambda_i h_i(x^*) \leq f(x^*) \leq f(x)+\sum_{i=1}^n \lambda^*_i h_i(x) | ||
world/kkt.1759882195.txt.gz · Last modified: 2025/10/08 02:09 by rdouc
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