Given a $C^2$ L-smooth function the Lipschitz condition is:
$$||\nabla f(x)-\nabla f(y)||\leq L ||x-y||$$ which give us $$\nabla^2f(x) \preceq L I.$$ Then, based on the Tayloe expansion $$f(y)=f(x)+\nabla f(x)^T(x-y) + \frac{1}{2}(x-y)^T\nabla^2 f(z)(x-y)$$ with $z$ a convex combination of $x$ and $y$ we may take $$f(y) \leq f(x) + (y-x)^T\nabla f(x) + \frac{L}{2}||y-x||^2.$$
Are these conditions only true for convex $C^2$ function? What will change If $f$ is $C^2$ and concave?
Thanks!
This condition is completely independent of convexity. As long as you can connect any two points in your domain to be able to use the argument, the two conditions are equivalent for $C^2$ functions. This is true for convex domains. No assumption on $f$ is needed, apart from regularity.