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Let $\Phi:\mathbb{R^{n-1}\to R}$ be a vector function. Suppose that exists a set $S_1\subset \mathbb R^{n-1}$ on which $G=\nabla\Phi$, the gradient of the function is unbounded.

Does that imply necessarily that exist a set $S_2$ on which the Hessian determinant $H=\det (\frac{\partial^2\Phi}{\partial x_i\partial x_j})$ or at least one of the eigenvalues is non-vanishing?

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  • $\begingroup$ @user1952009 I meant gradient. Thanks. $\endgroup$ – Differential Jun 13 '16 at 3:56
  • $\begingroup$ if the gradient is unbounded on the neighborhood of $x=a $ there exists $k$ such that $ \frac{\partial \phi}{x_k}$ is unbounded on the neighborhood of $x=a$ so that it reduces to the 1 dimensional case $f(x_k) = \phi(x_k,a)$, and if $f$ is twice differentiable then $f'$ unbounded $\implies f''$ unbounded $\endgroup$ – reuns Jun 13 '16 at 3:58
  • $\begingroup$ and finally, since $H$ is symmetric, by the spectral theorem its eigenvalues are $\ge 0$, and $tr(H) = \sum \lambda(H)$ is unbounded hence some of its eigenvalues are unbounded $\endgroup$ – reuns Jun 13 '16 at 4:03
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    $\begingroup$ @user1952009 how the positive definiteness of the Hessian follows from the spectral theorem? $\endgroup$ – Differential Jun 13 '16 at 4:22
  • $\begingroup$ @user1952009 sorry about the late response but I wonder now also about $f$: if $f''$ is unbounded why does it imply that the Hessian det is unbounded? $\endgroup$ – Differential Jun 13 '16 at 4:49
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Suppose $f(x,y) = x^2.$ Then $\nabla f(x,y) = (2x,0)$ is unbounded. But the determinant of the Hessian matrix of $f$ is $0$ everywhere.

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