Uncategorized
Uncategorized
Show, that these conditions are equivalent:
\|\nabla f(x) - \nabla f(z) \| \le L \|x-z\|
and
f(z) \le f(x) + \nabla f(x)^T(z-x) + \frac L 2 \|z-x\|^2
We say that the function belongs to the class f \in C^{k,p}_L (Q) if it is k times continuously differentiable on Q, and the p derivative has a Lipschitz constant L.
\|\nabla^p f(x) - \nabla^p f(y)\| \leq L \|x-y\|, \qquad \forall x,y \in Q
The most commonly used C_L^{1,1}, C_L^{2,2} for \mathbb{R}^n. Notice that:
- p \leq k
- If q \geq k, then C_L^{q,p} \subseteq C_L^{k,p}. The higher is the order of the derivative, the stronger is the limitation (fewer functions belong to the class).
Prove that the function belongs to the class C_L^{2,1}. \subseteq C_L^{1,1} if and only if \forall x \in \mathbb{R}^n:
\|\nabla^2 f(x)\| \leq L
Prove that the last condition can be rewritten in the form without loss of generality:
-L I_n \preceq \nabla^2 f(x) \preceq L I_n
Show that for gradient descent with the following stepsize selection strategies:
- constant step h_k = \dfrac{1}{L}
- Dropping sequence h_k = \dfrac{\alpha_k}{L}, \quad \alpha_k \to 0.
you can get the estimation of the function decrease at the iteration of the view:
f(x_k) - f(x_{k+1}) \geq \dfrac{\omega}{L}\|\nabla f(x_k)\|^2
\omega > 0 - some constant, L - Lipschitz constant of the function gradient.