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Anton Petrunin
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$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ There are no such examples.

Suppose $ $$f$ is such a function. Choose a sequence of points $ $$p_n$ such that $ $$|\nabla_{p_n}f|\to\infty$. Let $ $$f_n$ be a function with $ $$p_n$ shifted to a fixed point $ $$p$. So $ $$f_n(x)=f\circ\iota_n(x)$ where $ $$\iota_n$ is a motion such that $ $$\iota(p)=p_n$. Pass to a converging subsequence of the functions $$\phi_n=\frac{f_n-f_n(p)}{|\nabla_pf_n|}$$ denote its limit by $ $$\phi_\infty$.

Note that $ $$\phi_\infty$ has vanishing Hessian and nonvanishing gradinet $ $--- a contradiction.

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There are no such examples.

Suppose $f$ is such a function. Choose a sequence of points $p_n$ such that $|\nabla_{p_n}f|\to\infty$. Let $f_n$ be a function with $p_n$ shifted to a fixed point $p$. So $f_n(x)=f\circ\iota_n(x)$ where $\iota_n$ is a motion such that $\iota(p)=p_n$. Pass to a converging subsequence of the functions $$\phi_n=\frac{f_n-f_n(p)}{|\nabla_pf_n|}$$ denote its limit by $\phi_\infty$.

Note that $\phi_\infty$ has vanishing Hessian and nonvanishing gradinet --- a contradiction.

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Anton Petrunin
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The hyperbolic plane can be described by a Riemannian metric $ (\begin{smallmatrix} 1&0 \\ 0&e^x \end{smallmatrix}) $ in on the$ $ $(x,y)$-plane.

Consider a function$ $ $f\colon(x,y)\mapsto x^2$; its gradient is not bounded while Hessian is.$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $

The hyperbolic plane can be described by a Riemannian metric $ (\begin{smallmatrix} 1&0 \\ 0&e^x \end{smallmatrix}) $ in on the $(x,y)$-plane.

Consider a function $f\colon(x,y)\mapsto x^2$; its gradient is not bounded while Hessian is.

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Post Deleted by Anton Petrunin
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Anton Petrunin
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The hyperbolic plane can be described by a Riemannian metric $ (\begin{smallmatrix} 1&0 \\ 0&e^x \end{smallmatrix}) $ in on the $(x,y)$-plane.

Consider a function $f\colon(x,y)\mapsto x^2$; its gradient is not bounded while Hessian is.