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?

• It's not possible for only one second-order derivative to be negative while maintaining the determinant of the Jacobian positive. Ie. $(c)$ being true implies that if $(b)$ holds, both second-order derivatives are negative. Sep 30, 2021 at 17:20
• 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.
– Lazy
Sep 30, 2021 at 17:21
• 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.
– Lazy
Sep 30, 2021 at 17:33
• I think I already proved $H$ is negative definite: "the eigenvalues of $\nabla^2 f(\theta_0)$ should be both negative", then $H$ is negative definite. Sep 30, 2021 at 17:38
• As far as I read you wrote: If it is a local maximum then the hessian is negative definite (not true by the way), then all the eigenvalues are negative, and thus the det positive. So you’d do a circular proof (where even one step is false) if you reasoned from that that the hessian was negative.
– Lazy
Sep 30, 2021 at 17:40

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).

• Can't $\left(\frac{\partial^2}{\partial\theta_1\partial\theta_2}H(\theta_1,\theta_2)\right)^2$ quantity in $\frac{\partial^2}{\partial\theta_1^2}H(\theta_1,\theta_2)\frac{\partial^2}{\partial\theta_2^2}H(\theta_1,\theta_2)-\left(\frac{\partial^2}{\partial\theta_1\partial\theta_2}H(\theta_1,\theta_2)\right)^2 >0$ be larger than $\frac{\partial^2}{\partial\theta_1^2}H(\theta_1,\theta_2)\frac{\partial^2}{\partial\theta_2^2}H(\theta_1,\theta_2)$ making the determinant less than $0$?
– time
Dec 29, 2021 at 7:20
• @time As I said, I’m elaborating on some comments. If you look at the second comment to the question I stated that these conditions are sufficient but not nescessary. I even gave a counterexample.
– Lazy
Dec 29, 2021 at 15:49