# Does an unbounded gradient implies a non-vanishing hessian determinant

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?

• @user1952009 I meant gradient. Thanks. – Differential Jun 13 '16 at 3:56
• 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 – reuns Jun 13 '16 at 3:58
• 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 – reuns Jun 13 '16 at 4:03
• @user1952009 how the positive definiteness of the Hessian follows from the spectral theorem? – Differential Jun 13 '16 at 4:22
• @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? – Differential Jun 13 '16 at 4:49

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.