This is an elementary question in differential geometry. We know that for a smooth real-valued function $f$ defined on an open geodesically convex set of a Riemannian manifold $ \mathcal{X} \subset \mathcal{M}$, $f$ is geodesically convex if and only if its Riemannian Hessian is positive semidefinite on $\mathcal{X}$. Here, the Riemannian Hessian is defined as the second covariant derivative of $f$, hence a tensor of type (0,2). This result can be found in [1].

My question is whether the above result applies to the alternative definition of Riemannian Hessian as a linear mapping on the tangent space at a point, hence a tensor of type (1,1). This alternative definition is also called the ''geometric Hessian'' in [2]. Specifically, let $T_x{\mathcal{M}}$ be the tangent space at $x \in \mathcal{X}$. The geometric Hessian at $x$, denoted by $\text{Hess} f(x): T_x{\mathcal{M}} \to T_x{\mathcal{M}}$, is defined as,

$$\text{Hess} f(x) [\eta_x] = \nabla_{\eta_x} \text{grad} f, \;\forall \eta_x \in T_x{\mathcal{M}},$$

where $\nabla$ is the Riemannian connection and $\text{grad} f$ is the gradient vector field.

I am wondering if the following statement is true:

For a smooth function $f: \mathcal{X} \to \mathbb{R}$, $f$ is geodesically convex if and only if $\text{Hess} f(x) \succeq 0$ for all $x \in \mathcal{X}$.

References:

[1] C. Udriste (1994), *Convex Functions and Optimization Methods on Riemannian manifolds*, Mathematics and its Applications. Dordrecht: Kluwer Academic Publishers. ISBN 0-7923-3002-1.

[2] P-A. Absil , R. Mahony, and R. Sepulchre (2008), *Optimization Algorithms on Matrix Manifolds*, Princeton University Press.