For $f$ geodesically convex with $L$-Lipschitz-gradient on hyperbolic space, is $f(x)-f(x^*)\leq(\mathrm{const}) \cdot L r$ for all $x \in B(x^*, r)$?

Question

$\DeclareMathOperator\dist{dist}$Setting: Let $M$ be a hyperbolic space of sectional curvature $-1$, and let $f \colon M\to \mathbb{R}$ be a $C^2$, geodesically convex function which has $L$-Lipschitz-gradient. In other words, $0 \preceq \nabla^2 f(x) \preceq L I$ for all $x \in M$. Further assume $f$ admits a minimizer $x^*$ with $\nabla f(x^*) = 0$. ($\nabla f(x)$, $\nabla^2 f(x)$ are the (Riemannian) gradient and Hessian of $f$ at $x$, respectively.) Lastly let $B(x^*, r)$ denote the closed geodesic ball of radius $r$ centered at $x^*$. It is easy to show that $f(x) - f(x^*) \leq \frac{1}{2} L r^2$ for all $x \in B(x^*, r)$.

Question: Is the following true: for $r$ large enough, there exists an absolute constant $c$, such that $f(x) - f(x^*) \leq c \cdot L r$ for all $x \in B(x^*, r)$.

My attemps:

1. I have not been able to find a counter example. For example, the Riemannian squared distance function $x \mapsto \frac{1}{2} \dist(x, x^*)^2$ ($L = \frac{r}{\tanh(r)}$), the Riemannian distance function $x \mapsto \dist(x, x^*)$ (sufficiently smoothed near $x^*$), and the squared distance function $x \mapsto \frac{1}{2} \dist(x, S)^2$ to a totally geodesic submanifold $S$ of $M$ all satisfy $f(x) - f(x^*) \leq c \cdot L r$ for some $c$ big enough (say $c=10$ and $r$ bigger than some constant).
2. One can in fact show (using hyperbolic trigonometry) that for any such $f$, there must be a point $y \in \partial B(x^*, r)$ such that $f(y) - f(x^*) \leq c \cdot L r$ (again, for say $c=10$ and $r$ bigger than some constant). This implies that any radial function must satisfy $f(x) - f(x^*) \leq c \cdot L r$ for all $x \in B(x^*, r)$. One can then use this to show that for any such $f$ (not necessarily radial) at least half of the points $x$ in $B(x^*, r)$ must satisfy $f(x) - f(x^*) \leq c \cdot L r$. However, I am unable to show that this is true for all points $x \in B(x^*, r)$, or exhibit a counterexample.

Answer 1

The function $f$ has to be Lipschitz.

Assume contrary, choose a sequence of points $p_n$ such that $\lambda_n=|\nabla_{p_n}f|\to \infty$. Let $$f_n(x)=\tfrac1{\lambda_n}\cdot f\circ m_n(x)-f(p_n)$$ where $m_n$ is a motion of the plane that sends $p\mapsto p_n$.

The Hession of $f_n$ converges to zero. So, after passing to a subsequence we can assume that $f_n$ converges to an affine function, say $f$ (it is convex and concave at the same time). Note that $|\nabla_p f|=1$. So, there is no such function on the hyperbolic plane --- a contradiction.

Мore direct proof. Suppose $f$ has bounded Hessian. Observe that the gradient $v=\nabla f$ is almost parallel; that is, there is a constant $C$ such that $|\nabla_u v|\le C$ for any unit tangent vector $u$. In particular, the parallel translation of $v(p)$ around a circle has to be close to $v(p)$. If $|v(p)|$ is large, the latter contradicts the Gauss--Bonnet formula.

Comment. It is a nice problem --- I plan to include it in my PIGTIKAL.