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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.

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  • $\begingroup$ The Riemannian metric can be used to turn one Hessian into the other. So the assumptions are the same in both versions. $\endgroup$ Mar 28, 2020 at 17:16
  • $\begingroup$ @SebastianGoette Thank you for the comment. I have posted a proof. $\endgroup$ Mar 31, 2020 at 21:38

1 Answer 1

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Thanks to Prof. Absil for verifying the following proof.

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 $x \in \mathcal{X}$, \begin{equation} \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, \end{equation} where the second equality holds by definition of the geometric Hessian (Definition 5.5.1 in [2]).

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.

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