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