Let $U$ be an open set $U \subset \mathbb R^n$. Let $f$ be a class-2 function $f: U → \mathbb R$.
Prove or disprove the following statement. $∇^2 f=0$ and $∇f= 0$ at $x_0 \in U$ implies $x_0$ is the saddle point.
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@JavaMan/ Yes. That notation means what you said. =ㅅ=;; – Chris kim May 07 '13 at 03:08
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1This is false. Take, for example, $f(x) = x^4$ in one dimension. – Qiaochu Yuan May 07 '13 at 03:09
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@QiaochuYuan/ Thank you. @_@ Can you give the another example in R^2 ? – Chris kim May 07 '13 at 03:12
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$f(x, y) = x^4 + y^4$. I could give you a $3$-dimensional example too... – Qiaochu Yuan May 07 '13 at 03:13
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1More general examples show that no condition on finitely many derivatives of $f$ suffices to determine exactly when a point is a saddle point. – Qiaochu Yuan May 07 '13 at 03:28
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The notation $\nabla^2 f$ is ambiguous: it can mean the Hessian matrix or its trace (the Laplacian). I will write $\Delta f$ for the Laplacian and $D^2f$ for the Hessian matrix. Here is something to fill the answer box:
If $\nabla f(x_0)=0$, $\Delta f(x_0)=0$, but $D^2f(x_0)$ is not the zero matrix, then $x_0$ is a saddle point.
Proof: Since $D^2f(x_0)$ is symmetric, it is diagonalizable and not all of the eigenvalues are zero. The sum of eigenvalues is $0$. Therefore, $D^2f(x_0)$ has at least one positive eigenvalue and at least one negative eigenvalue.
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