we have $$\begin{aligned}
\frac{\c^{\Tr} \x^{k+t} -z^*} {\c^{\Tr} \x^k -z^*}
\le  \left( 1 - \frac{\delta}{m\gamma} \right)^t.
\label{myIneq2}\end{aligned}$$ Since $| B^k |=m$ and $$\c^{\Tr} \x^k -z^* = \x^{k^{\Tr}} \s^* 
= \sum_{j \in B^k} x_j^k s_j^*,$$ there must have a $\bar{j} \in B^k$ such that $$x_{\bar{j}}^k s_{\bar{j}}^* \ge \frac{1}{m} (\c^{\Tr} \x^k -z^* ).$$ Using Assumption 4, $\gamma \ge x_{\bar{j}}^k > 0$, we have $$\begin{aligned}
s_{\bar{j}}^* \ge \frac{1}{mx_{\bar{j}}^k } (\c^{\Tr} \x^k -z^* )
\ge \frac{1}{m \gamma } (\c^{\Tr} \x^k -z^* ).
\label{myIneq3}\end{aligned}$$ Moreover, for any integer $t >0$, we have $$\c^{\Tr} \x^{k+t} -z^* = \s^{*^{\Tr}}\x^{k+t} 
\ge x_{\bar{j}}^{k+t} s_{\bar{j}}^*.$$ this gives $$\begin{aligned}
x_{\bar{j}}^{k+t} \le  \frac{\c^{\Tr} \x^{k+t} -z^* }{s_{\bar{j}}^*}.
\label{myIneq4}\end{aligned}$$ Substituting (\[myIneq3\]) and (\[myIneq2\]) into (\[myIneq4\]) gives $$\begin{aligned}
x_{\bar{j}}^{k+t} 
\le m \gamma \frac{\c^{\Tr} \x^{k+t} -z^* }
{\c^{\Tr} \x^k -z^* }
\le m \gamma \left( 1 - \frac{\delta}{m\gamma} \right)^t.
\label{myIneq5}\end{aligned}$$ Substituting (\[tIter\]) into (\[myIneq5\]) and using the identity $x^{\log_b y}= y^{\log_b x}$ and the inequality $\log(1-x) \le -x$ for all $x \le  1$, we have $$\begin{aligned}
x_{\bar{j}}^{k+t} & \le &  m \gamma \left( 1 - \frac{\delta}{m\gamma} \right)^{m \frac{\gamma}{\delta}
\log\left( m \frac{\gamma}{\delta} \right)}
\nonumber \\
& = & m \gamma \left[ \left( 1 - \frac{\delta}{m\gamma} 
\right)^{\log\left( m \frac{\gamma}{\delta} \right)}
\right]^{m \frac{\gamma}{\delta}}
\nonumber \\
& = & m \gamma 