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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,

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