Should both of the second-order partial derivatives be negative for a two-variate function to achieve its local maximum? My textbook said:
To use two-variate calculus to verify that a non-negative function $H(\theta_1,\theta_2)$ has a local maximum at $(\hat\theta_1,\hat\theta_2)$, it must be shown that the following three conditions hold.

*

*a. The first-order partial derivatives are $0$, $$\frac{\partial}{\partial\theta_1}H(\theta_1,\theta_2)|_{\theta_1=\hat\theta_1,\theta_2=\hat\theta_2}=0$$ and $$\frac{\partial}{\partial\theta_2}H(\theta_1,\theta_2)|_{\theta_1=\hat\theta_1,\theta_2=\hat\theta_2}=0$$


*b. At least one second-order partial derivative is negative, $$\frac{\partial^2}{\partial\theta_1^2}H(\theta_1,\theta_2)|_{\theta_1=\hat\theta_1,\theta_2=\hat\theta_2}<0$$ or $$\frac{\partial^2}{\partial\theta_2^2}H(\theta_1,\theta_2)|_{\theta_1=\hat\theta_1,\theta_2=\hat\theta_2}<0$$


*c. The Jacobian of the second-order partial derivatives is positive, $$\begin{vmatrix}
\frac{\partial^2}{\partial\theta_1^2}H(\theta_1,\theta_2)&\frac{\partial^2}{\partial\theta_1\partial\theta_2}H(\theta_1,\theta_2)\\
\frac{\partial^2}{\partial\theta_1\partial\theta_2}H(\theta_1,\theta_2)&\frac{\partial^2}{\partial\theta_2^2}H(\theta_1,\theta_2)
\end{vmatrix}_{\theta_1=\hat\theta_1,\theta_2=\hat\theta_2}\\
=\frac{\partial^2}{\partial\theta_1^2}H(\theta_1,\theta_2)\frac{\partial^2}{\partial\theta_2^2}H(\theta_1,\theta_2)-\left(\frac{\partial^2}{\partial\theta_1\partial\theta_2}H(\theta_1,\theta_2)\right)^2 >0$$
I can't understand condition (b):
For (b) I think both of the second-order partial derivatives should be negative if only one second-order partial derivative is negative and the other is positive, then it should be a saddle point. And (c) also implies that both of the second-order partial derivatives should be negative.
For (c), since $f(\theta) \approx f(\theta_0) + \nabla f(\theta_0)^T (\theta - \theta_0) + \frac12 (\theta - \theta_0)^T \nabla^2 f(\theta_0) (\theta - \theta_0)$.  If $\theta_0$ is a local extremum for $f$, then $\nabla f(\theta_0) = 0$, so if $\nabla^2 f(\theta_0)$ is negative definite, then $(\theta - \theta_0)^T \nabla^2 f(\theta_0) (\theta - \theta_0) \leq 0$ for all $\theta$, so $\theta_0$ is a local maximum.  Then the eigenvalues of $\nabla^2 f(\theta_0)$ should be both negative and the determinant should be positive. Am I correct?
 A: Elaborating on the earlier comments:
For b) note that negative is $<0$. It could also be $=0$. In fact, the conditions are only sufficient, but not nescessary. Take $H(x,y)=−x^4−y^4$. Then $H$ has it’s only maximum at $(x,y)=(0,0)$, but the Hessian vanishes. Thus the direction you need to prove: If those three properties hold, then $H$ has a local maximum in that point.
For this you might want to prove: If b), c) holds then the Hessian of H is strictly negative definite. The proof for this is quite straightforward:
Suppose the Hessian is given by $h_{11},h_{22},h_{12}=h_{21}$.
From b) and c) you can easily see the in fact in b) both conditions must hold. Else: Suppose $h_{11}<0,h_{22}\geq 0$ then $h_{11}h_{22} - h_{12}^2 \leq 0$.
Then from c) we get
$$ |h_{12}|<\sqrt{h_{11}h_{22}} $$
Then if $v=(v_1,v_2)$, then:
$$ v^T\nabla^2 H v = v_1(v_1h_{11}+v_2h_{12})+v_2(v_1h_{12}+v_2h_{22})
= v_1^2 h_{11} + v_2^2h_{22} + 2v_1v_2h_{12} < v_1^2 h_{11} + v_2h_{22} \pm 2v_1v_2\sqrt{h_{11}h_{22}} $$
(where the $<$ holds as long as we do not have $v_1=v_2=0$ and the $\pm$ is so that $\pm v_1v_2\geq0$).
Now as $h_{11},h_{22}< 0$ we have $\sqrt{h_{11}}=r_1i$, $\sqrt{h_{22}}=r_2i$, $r_1,r_2>0$. Thus we have for $v\neq 0$
$$ v^T\nabla^2 H v < v_1^2 h_{11} + v_2h_{22} \pm 2v_1v_2\sqrt{h_{11}h_{22}} = (v_1r_1i)^2 + (v_2r_2i)^2 \mp 2(v_1r_1iv_2r_2i) = (v_1r_1i \mp v_2r_2i)^2 \leq 0$$
(with $=$ exactly if $v_1r_1 = \pm v_2r_2$, but that does not matter).
