# Convergence rate of Newton method with Armijo's Rule

Actually, I prove the convergence rate but I did not use the condition "compare $$f(x^k + \beta^l sd^k)$$ with $$f(x^k + d^k)$$ and let $$\alpha_k$$ be either $$s\beta^l$$ or 1, whichever is lesser in terms of their value." Since I think if I can prove the convergence rate of $$\alpha_k$$, then for the choice of 1, it can only be better. So I just show the convergence rate for $$\alpha_k$$ and did not consider the case that $$\alpha_k = 1$$

So I want to ask if there is something wrong for my proof.

Suppose $$0 < mI \leq \nabla^2 f(x) \leq MI$$ for all $$x$$, and consider Newton's method with Armijo's line search rule as followings,

$$f(x^k) - f(x^k + \beta^l sd^k) \geq -\sigma \beta^l s \nabla f(x^k)^T d^k$$, where $$d^k = -(\nabla^2 f(x^k))^{-1}\nabla f(x^k)$$.

compare $$f(x^k + \beta^l sd^k)$$ with $$f(x^k + d^k)$$ and let $$\alpha_k$$ be either $$s\beta^l$$ or 1, whichever is lesser in terms of their value.

I prove that the convergence rate is linear in this way,