First post on math.stackexchange! I have a question which probably requires a bit of Differential Geometry knowledge which I'm lacking.
By definition a convex function over a vector space $V$ is such that $f(\lambda x+(1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y)$ for $\lambda \in [0,1]$ and $x, y \in V$. Two ways to check convexity are:
- Compute the Hessian and verify that it is positive semidefinite everywhere
- Check that $\frac{d^2 f(x+ty)}{dt^2}\mid_{t=0} \geq 0$ for all $x, y \in V$
Even though these two are equivalent, sometimes I feel it is a bit easier to use the second condition.
Now, if we define convexity over a manifold similarly to $(1)$, i.e. the Hessian is positive semi-definite everywhere, then what a natural equivalent of $(2)$ would be? I guess some additional term will pop-up because of the curvature of the manifold? Or if we differentiate similarly along a geodesic would be fine?
Just a reference would be enough, thanks.
We say that $f : M\to \mathbb R$ is a geodesically convex function if
$$f\circ \gamma$$
is convex (in the usual sense) for all geodesic $\gamma$. This is the same as
$$\frac{d^2}{dt^2} (f\circ \gamma) (t) \ge 0$$
for all $t$. Now we compute
$$\frac{d^2}{dt^2} (f\circ \gamma) (t) = \nabla_{\dot\gamma} \langle \nabla f ,\dot\gamma\rangle = \langle\nabla_{\dot\gamma} \nabla f, \dot\gamma\rangle= Hf(\dot\gamma, \dot\gamma),$$
where $Hf$ is the hessian of $f$ given by
$$Hf(X, Y) = \langle \nabla_X \nabla f, Y\rangle.$$
Thus in this sense (1) and (2) are also equivalent: $f$ is geodesically convex if and only if $Hf$ is semi-positive definite.