Does there exist a real-valued function on the hyperbolic plane which has bounded hessian norm and unbounded gradient norm?

Specifically, consider the poincare half-plane model of the 2d hyperbolic manifold, given by $\mathbb{H} = \{(x,y):y>0\}$ with metric $(dx^2 + dy^2)/y^2$. Let $grad f(x,y)$ and $Hess f(x,y)$ denote the Riemannian gradient and hessian, respectively. Does there exists a function $f : \mathbb{H} \rightarrow \mathbb{R}$ for which $||grad f(x,y)||$ is unbounded but $||Hess f(x,y)||_{op}$ is bounded, say by $1$?

My attempts:

We will use $f^{(i,j)}(x,y)$ to denote $\frac{\partial^{i+j} f}{\partial x^i \partial y^j}$.

It is straightforward to show that $$grad f(x,y) = y \left( \begin{array}{c} f^{(1,0)}(x,y) \\ f^{(0,1)}(x,y) \\ \end{array} \right)$$ and $$Hess f(x,y) = y^2 \left( \begin{array}{cc} f^{(2,0)}(x,y) & f^{(1,1)}(x,y) \\ f^{(1,1)}(x,y) & f^{(0,2)}(x,y) \\ \end{array} \right)+y \left( \begin{array}{cc} -f^{(0,1)}(x,y) & f^{(1,0)}(x,y) \\ f^{(1,0)}(x,y) & f^{(0,1)}(x,y) \\ \end{array} \right).$$

It is also straightforward to show that $$||grad f(x,y)|| = \sqrt{\left(y f^{(0,1)}(x,y)\right)^2+\left(y f^{(1,0)}(x,y)\right)^2}$$ and $$||Hess f(x,y)|| = \left| \frac{1}{2} \left(f^{(0,2)}(x,y) y^2+f^{(2,0)}(x,y) y^2\right)\right| +\frac{1}{2} \sqrt{\left(2 y^2 f^{(1,1)}(x,y)+2 y f^{(1,0)}(x,y)\right)^2+\left(y^2 \left(-f^{(0,2)}(x,y)\right)+y^2 f^{(2,0)}(x,y)-2 y f^{(0,1)}(x,y)\right)^2}.$$

Using these formulas, it is straightforward to show that any function independent of $x$ or $y$ (i.e., $f(x,y) = h(y)$ or $f(x,y) = h(x)$) will not work. You can also show that any function of the form $f(x,y) = h(dist((x,y),p)^2)$ for smooth $h$ and fixed $p \in \mathbb{H}$ will also not work.

After many more attempts, which I will not detail here, I still haven't found such a function or a proof that none exists. My intuition says that such a function does exist. I was hoping anyone was familiar with this type of problem and could help. Thanks!

For Euclidean space, an obvious example of such a function is a quadratic.

(Crossposted on mathstackexchange, currently no responses)


1 Answer 1


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.

  • $\begingroup$ Thanks for your response. How do we know the sequence of functions $\phi_n$ has a pointwise convergent subsequence? $\endgroup$ May 29, 2020 at 2:25
  • $\begingroup$ @Doggyy Since Hessian is bounded, $|\nabla f|$ is Lipschitz. Therefore $f$ is Lipschitz in a ball $B(p,R)$ with constant $L= L(|\nabla_pf|, R)$. $\endgroup$ May 29, 2020 at 2:56

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