# Geodesic convexity and the Geometric Hessian

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.

• The Riemannian metric can be used to turn one Hessian into the other. So the assumptions are the same in both versions. – Sebastian Goette Mar 28 at 17:16
• @SebastianGoette Thank you for the comment. I have posted a proof. – user1952770 Mar 31 at 21:38

Let $$\mathfrak{X}(\mathcal{M})$$ and $$\mathfrak{F}(\mathcal{M})$$ be the set of smooth vector fields and scalar functions on $$\mathcal{M}$$, respectively. By Theorem 6.2 in [1], $$f$$ is geodesically convex in $$\mathcal{X}$$ if and only if its second covariant derivative $$\nabla^2 f: \mathfrak{X}(\mathcal{M}) \times \mathfrak{X}(\mathcal{M}) \to \mathfrak{F}(\mathcal{M})$$ is positive semidefinite on $$\mathcal{X}$$, i.e., for all vector fields $$\eta \in \mathfrak{X}(\mathcal{M})$$, the corresponding function $$\nabla^2 f[\eta, \eta]$$ is non-negative on $$\mathcal{X}$$. Let $$\nabla$$ be the Riemannian connection. At a particular point $$x \in \mathcal{X}$$, $$$$\nabla^2 f(x)[\eta, \eta] = \langle \nabla_{\eta_x} \text{grad} f, \eta_x \rangle_x = \langle \text{Hess} f(x) [\eta_x], \eta_x \rangle_x \geq 0,$$$$ where the second equality holds by definition of the geometric Hessian (Definition 5.5.1 in [2]).