# Local maximum or minimum of a smooth function.

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.

• The answer should be $A$. – Jacky Chong Oct 12 '16 at 5:14
• But sir how $A$ is answer....please give some explanation..... – neelkanth Oct 12 '16 at 5:15
• My intuition tells me the function has to be convex (concave up). – Jacky Chong Oct 12 '16 at 5:17
• 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. – Jacky Chong Oct 12 '16 at 5:19
• But in our question at every point hassian matrix is positive definite... – neelkanth Oct 12 '16 at 5:20

If the Hessian is positive semi definite then $f$ is convex and any stationary point is a global minimum.
• 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?) – copper.hat Oct 12 '16 at 5:31
• Can i say that if Hessian matrix negative semi definite then $f$ is concave?? – neelkanth Oct 12 '16 at 5:48
• Yes, because $f$ is concave iff $-f$ is convex. – copper.hat Oct 12 '16 at 5:49