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I want to know suppose that i prove that $$ f_x^{''}(x,y)<0, \forall y, $$ and $$f_y^{''}(x,y)<0, \forall x. $$ If i can say that f(x,y) is concave? because, $$ f(x+\delta x,y+\delta y)<f(x+\delta x,y)<f(x,y). $$

The test by hessain matrix is just suitable for certain point. I can't find any law for my problem. many thanks!!

Jianzheng Wang
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  • I mean $$f^{''}{xx}$$, and $$f^{''}{yy}$$. It's a typo. –  Jan 26 '17 at 04:52
  • I downvoted because this question is more suitable for math.se. Anyway, consider $f(x) = x^T \begin{bmatrix} -1 & 10\ 10 & -1\end{bmatrix}x$, the hessian of $f(x)$ is indefinite. This problem is non concave even though $f_{xx} = f_{yy} > 0$ – Pushpendre Jan 26 '17 at 05:35
  • thank you for the explaination. but as for your function, $$f=-x^2_1+20x_1x_2-x^2_2$$. it is also concave according to matlab simulation. i don't know how to add the picture here. –  Jan 26 '17 at 06:36
  • The restriction of $f$ to the diagonal $x_1=x_2$ is $18 x_1^2$... –  Jan 26 '17 at 10:44
  • Please accept my answer or indicate what is missing. – LinAlg Mar 12 '18 at 13:33

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The answer is negative, consider $$f(x) = x^T \begin{pmatrix} -10 & 1 \\ 1 & -10 \end{pmatrix}x,$$ whose hessian is indefinite.

LinAlg
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