Meaning of $\alpha I \preceq \nabla ^2 F$

Question

I learned that the notation $\alpha I \preceq \nabla ^2 F $ means that $\langle \alpha x, x\rangle \leq \langle \nabla^2 F x, x\rangle$.

1. What is this called? I know that $\nabla^2 F \succeq 0$ is means that $F$ is positive semidefinite. What would we call $\alpha I \preceq \nabla ^2 F $ which I guess is a weaker version?

2. Is this related to $\nabla F$ being Lipschitz continuous? It roughly looks like an upper/lower bound on $\nabla \cdot \nabla F$, but I don't understand the vector case where we work with inner products.

3. I saw in a paper that $\alpha I \preceq \nabla ^2 F $ implied that $\langle \nabla F(x) - \nabla F(y) , \; x-y \rangle \geq \alpha \|x-y\|^2_2$. I can see why $\langle \nabla^2 F(x-y) , \; x-y \rangle \geq \alpha \|x-y\|^2_2$ from the definition, but I don't know why this would apply to $\nabla F(x) - \nabla F(y)$ on the left hand side.

Answer 1

For #3:

Fix $x$ and $y$. Define $g(t) = F(y+t(x-y))$. Then $g'(t) = \langle \nabla F(y+t(x-y)), x-y\rangle$ and $g''(t) = \langle (\nabla^2 F(y+t(x-y))) x-y, x-y \rangle$.

Then $$\langle \nabla F(x) - \nabla F(y), x-y\rangle = g'(1)-g'(0).$$ By the mean value theorem, there exists some $c \in [0, 1]$ such that the above equals $$g''(c) = \langle (\nabla^2 F(y+t(x-y))) (x-y), x-y \rangle \ge \alpha \|x-y\|^2$$ where the last bound follows from the definition of $\alpha I \preceq \nabla^2 F$.