I am reading
L. Grippo and M. Sciandrone. "On the convergence of the block nonlinear Gauss–Seidel method under convex constraints." Operations Research Letters (2000).
And have some trouble proving the Proposition 1. Here I reorganize it in my own words
Suppose $f:R^n\rightarrow R$ is continuously differentiable, and $X\subset R^n$ is a convex closed set, and $\{z_k\}\subset X$ is a convergent sequence such that $lim_{k\rightarrow\infty} z_k=z$.
For each $z_k$, let $d_k$ be a descent direction of $f$ at $z_k$, i.e., $\nabla f(z_k)^T d_k<0 $, and satisfing $\forall k,||d_k||\le M$.
Consider the line search algorithm: let $\delta\in(0,1),r\in(0,1)$, step size $a_k:=max_{n\ge 1}\{\delta^n:f(z_k+\delta^n d_k)\le f(z_k)+r\delta^n\nabla f(z_k)^T d_k\}$
Prove: if $lim_{k\rightarrow \infty}f(z_k)-f(z_k+\delta^n d_k)=0$, then $lim_{k\rightarrow \infty}f(z_k)^T d_k=0$
Note that here $z_{k+1}$ is not obatained via line search along $d_k$, in fact $\{z_k\}$ is a predefined sequence .
I konw it's obvious from the definetion of $a_k$ that $0\le -ra_kf(z_k)^T d_k\le f(z_k)-f(z_k+a_k d_k)$, thus $lim_{k\rightarrow \infty}=a_kf(z_k)^Td_k=0$. But i find it hard to "separate" $a_k$ and $f(z_k)^Td_k$ or to bound $a_k$ in terms of $f(z_k)^Td_k$ , I've tried but faild.
Another way i've tried is to prove $a_k->0$, if so from $f(z_k+a_k d_k)\le f(z_k)+ra_k\nabla f(z_k)^T d_k$ i can get $[f(z_k+a_k d_k)-f(z_k)]/a_k\le r\nabla f(z_k)^T d_k$, using mean theorem i have $\nabla f(z_k+\theta a_kd_k )^T d_k\le r\nabla f(z_k)^T d_k$, then the result follows by taking the limits. But i can't prove $a_k->0$
Can anyone share some ideas?