Riemmanian Hessian and convexity

[Petrichor]

Can we conclude that a function is convex if its Riemannian Hessian is positive?

In Euclidean space, my function is convex. Referring to your earlier response: "On compact manifolds (such as the circle), geodesically convex functions are necessarily constant (this is standard; see for example Corollary 11.10 in my book)." However, when I calculate the Riemannian Hessian, I find it to be positive.

Is there something I'm overlooking regarding the connection between the Riemannian Hessian and convexity in this context?

Thank you for your help

[Nicolas Boumal]

> Can we conclude that a function is convex if its Riemannian Hessian is positive?

Yes (if by "positive" you mean "positive definite"). For example, Theorem 11.23 in my book, but this is standard.

> In Euclidean space, my function is convex.

This, unfortunately, is not enough. In the sentence above, it is important that the Riemannian Hessian is positive definite; the Euclidean Hessian is not the right object to study in order to make this determination.

[Petrichor]

Thank you. My function is convex (I attached it to the message. A is a positive scalar and y and I are complex vectors), and when I calculate the Riemannian Hessian, I find it to be positive definite. According to Corollary 11.10 in your book, a function that is both convex and g-convex must be constant. However, my function is not constant, even though it is both g-convex and convex.  

[Nicolas Boumal]

That is not what Corollary 11.10 says.

[Petrichor]

my bad. The manifold I am working on is not compact!