1
$\begingroup$

Let $f$ be a real-valued $C^2$ function in $\mathbb{R}^2.$ By Second partial derivatives, we know that if the Hessian is negative definite (equivalently, has all eigenvalues negative) at a, then $f$ attains a local maximum at a.

My question is about necessary condition for local maximum.

One of necessary conditions for local maximums is that its Hessian is negative semidefinite.

I was wondering if the following is true:

If $f$ is $C^2$ and has a local maximum at $x,$ then all second partial derivatives at $x$ are nonpositive.

Please let me know if you have any comment of it. Thanks in advance!

$\endgroup$
0
$\begingroup$

No, this is not correct. If $f(x,y)=2xy-x^2-y^2$, then $f(x,y)\leq f(0,0)$ for all $x,y$, but $\frac{\partial^2f}{\partial x\partial y}(0,0)=2>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.