none of the subsets $\mathcal{U}_l^m$ ($l=1,\cdots,L$) is empty, we can expand the above recursive term using (\[eq:FmDef\]). It follows that $$\label{eq:FoldTerm}
Q_1^m = N_0 \Gamma_m \sum_{l=1}^L (1 + \Gamma_m)^{c_l^m} \max_{k \in \mathcal{U}_l^m} \left\{1 / H_m^k \right\},$$ where the exponent $c_l^m$ is defined as $c_1^m=0$ and $c_{l+1}^m=c_l^m+1$. Otherwise, if a subset $\mathcal{U}_l^m = \emptyset$ for some $m$, we have that $Q_l^m=Q_{l+1}^m$, $\max_{k\in\mathcal{U}_l^m} \left\{1/H_m^k\right\}=\max_{k\in \emptyset} \left\{1/H_m^k\right\}=0$, and $c_l^m=c_{l-1}^m$. Eq. (\[eq:FoldTerm\]) still holds true.

Finally, the objective function (\[eq:ObjFun2\]) can be rewritten as $$% \mbox{$\sum_{m=0}^M$} Q_1^m = \mbox{$\sum_{m=0}^M$} N_0 \Gamma_m \mbox{$\sum_{l=1}^L$} (1 + \Gamma_m)^{c_l^m} \max_{k \in \mathcal{U}_l^m} \left\{1 / H_m^k \right\}.
\sum_{m=0}^M N_0 \Gamma_m \sum_{l=1}^L (1 + \Gamma_m)^{c_l^m} \max_{k \in \mathcal{U}_l^m} \left\{1 / H_m^k \right\}.$$ Since $(1+\Gamma_m)>0$, it can be seen that to minimize the total BS power, we need to keep the $c_l^m$’s as low as possible.

### Performance Bounds \[subsec:bounds\]

The reformulation and simplification allow us to derive performance bounds for the total BS power consumption. First, we derive the upper bound for the objective function (\[eq:ObjFun2\]). Define a variable $$%  \overline{G}_m = \max_{l\in\{1,\cdots,L\}} \left\{ \max_{k\in\mathcal{U}_l^m} \left\{\Gamma_m/H_m^k\right\} \right\}.
  \overline{G}_m = \max_{l\in\{1,\cdots,L\}} \max_{k\in\mathcal{U}_l^m} \left\{\Gamma_m/H_m^k\right\},$$ which corresponds to the user with the worst channel condition among all users that connect to BS $m$. 