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

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)

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