[T_D_K]

Ok. I stared at it long enough, and I think I understand. Being Lipschitz-continuous means (in a non-rigorous way?) that a the gradient / slope of a function has an upper bound. And Hessian-Lipschitz means the same, but for the second derivative / hessian.

So, f(x) = x^2 is not Lipschitz-continuous (because the slope gets arbitrarily large), but something like f(x) = sin(x) is Lipschitz-continuous because the slope never exceeds some upper bound.

Funny how trying to write down the question gives the brain the kick it needs sometimes :)

[Choco31415]

That's correct.

In some situations, it's enough to prove an equation is Lipschitz-continuous on a range. Example, y=x^2 is lipschitz continuous on x=[0,1].