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Let $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ be a smooth function with positive definite Hassian at every . Let $(a,b)\in\mathbb{R}^{2}$ be a critical point of $f$ then

$A.$ $f$ has global minimum at $(a,b).$

$B.$ $f$ has a local , but not global minimum at $(a,b).$

$C.$ $f$ has a local, but not global maximum at $(a,b).$

$D.$ $f$ has a global maximum at $(a,b).$

I only know that if Hassian matrix is positive definite definite at $(a,b)$ then it is a point of local minimum. Now what about global minimum? Please help me. Thanks a lot.

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  • $\begingroup$ The answer should be $A$. $\endgroup$ – Jacky Chong Oct 12 '16 at 5:14
  • $\begingroup$ But sir how $A$ is answer....please give some explanation..... $\endgroup$ – neelkanth Oct 12 '16 at 5:15
  • $\begingroup$ My intuition tells me the function has to be convex (concave up). $\endgroup$ – Jacky Chong Oct 12 '16 at 5:17
  • $\begingroup$ Here's an example that comes to my mind: consider the function $f(x, y) = x^2+y^2$ then $\nabla^2 f = 2I$ at every point and $(0, 0)$ is a critical point. $\endgroup$ – Jacky Chong Oct 12 '16 at 5:19
  • $\begingroup$ But in our question at every point hassian matrix is positive definite... $\endgroup$ – neelkanth Oct 12 '16 at 5:20
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If the Hessian is positive semi definite then $f$ is convex and any stationary point is a global minimum.

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  • $\begingroup$ Yes as i thought in my last comment...but how? $\endgroup$ – neelkanth Oct 12 '16 at 5:30
  • $\begingroup$ It is not hard to show that a differentiable convex function satisfies $f(y)-f(x) \ge Df(x)(y-x)$. Hence if $Df(x) = 0$, we have $f(y) \ge f(x)$ for all $y$. (Is that what you were asking?) $\endgroup$ – copper.hat Oct 12 '16 at 5:31
  • $\begingroup$ Can i say that if Hessian matrix negative semi definite then $f$ is concave?? $\endgroup$ – neelkanth Oct 12 '16 at 5:48
  • $\begingroup$ Yes, because $f$ is concave iff $-f$ is convex. $\endgroup$ – copper.hat Oct 12 '16 at 5:49
  • $\begingroup$ Ok sir thanks.... $\endgroup$ – neelkanth Oct 12 '16 at 5:50

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