User shineway - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T23:20:14Z http://mathoverflow.net/feeds/user/10331 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43754/some-questions-about-invexity Some questions about Invexity Shineway 2010-10-27T03:39:33Z 2012-03-12T10:47:05Z <p>Recently, I am looking into a non-convex optimization problem whose points satisfying KKT conditions can be obtained. Then the problem becomes how to decide whether the KKT conditions are sufficient for the global optimality of the solution. It is said on Wikipedia(<a href="http://en.wikipedia.org/wiki/Karush%25E2%2580%2593Kuhn%25E2%2580%2593Tucker_conditions#Sufficient_conditions" rel="nofollow">link text</a>) that for "invex function" the KKT is sufficient for optimality. By doing some searching, I was led to a concept called "invexity" and the book "Invexity and Optimization" ( <a href="http://www.springerlink.com/content/978-3-540-78561-3#section=155972&amp;page=1" rel="nofollow">http://www.springerlink.com/content/978-3-540-78561-3#section=155972&amp;page=1</a>).</p> <p>But, why this "invexity" research, which was first proposed in 1980's, is only confined in a small group of people. And the book is cited for only 4 times.</p> <p>In my current understanding, invexity is a generalization of convexity, and has some very good properties as in convexity, which should be very attractive and should have drawn lots of people's attention.</p> <p>Is it because this concept is not interesting, not useful? Or it is being studied under other names?</p> http://mathoverflow.net/questions/43627/can-subgradient-infer-convexity Can subgradient infer convexity? Shineway 2010-10-26T03:56:13Z 2010-10-26T04:55:46Z <p>It is known that If $f:U→ R$ is a real-valued convex function defined on a convex open set in the Euclidean space $R^n$, a vector v in that space is called a subgradient at a point $x_0$ in $U$ if for any $x$ in U one has $f(x)-f(x_0)\geq v\cdot(x-x_0)$</p> <p>What if for function $f$, at any $x_0$ I can find $v$, such that $f(x)-f(x_0)\geq v\cdot(x-x_0)$ for any $x$, does this show that $f$ is convex?</p>