# Optimal Energy Efficiency Fairness of Nodes in Wireless Powered Communication Networks
## Appendix B  
**Proof of Theorem 2.** In order to evaluate the convexity of \(U_{EE}\) with respect to \(A_{k}\),we need to take derivative of \(A_{k}\) when \(\theta_{k}\) is fixed. To do so, we first write the energy efficiency of UE in (7) function with respect to \(A_{k}\) given below  
\[U_{EE} \tag{15}\] \[=\sum\limits_{k=1}^{K}A_{k}\log\frac{W}{A_{k}N}+\sum\limits_{k=1}^ {K}\{A_{k}\log(\log(1+\theta_{k}))+A_{k}\mathbb{E}[\log(C_{k})]-A_{k}\log( \bar{P}+\beta\mathbb{E}[C_{k}])\},\]  
where \(\bar{P}\) and \(\beta\) are the same as proof of lemma 1. By taking a second order derivative of \(A_{k}\log\frac{W}{A_{k}N}\) with respect to \(A_{k}\), we get  
\[\frac{\partial^{2}\left(A_{k}\log\frac{W}{A_{k}N}\right)}{\partial A_{k}^{2}}= -\frac{1}{A_{k}}<0. \tag{16}\]Notice that the first part of the formulas is concave, we only need to prove that the latter part is the concave function. Hence, we define the remaining part as the function given below  
\[U_{EE,k}=A_{k}\log(\log(1+\theta_{k}))+A_{k}\mathbb{E}[\log(C_{k})]-A_{k}\log( \bar{P}+\beta\mathbb{E}[C_{k}]). \tag{107}\]  
We can obtain from (10) that \(C_{k}\) is a function for \(A_{k}\). Similarly, by taking a second derivative of \(U_{EE,k}\) with respect to \(A_{k}\), we have  
\[\begin{array}{l}\frac{\partial^{2}U_{EE,k}}{\partial A_{k}^{2}}\\ =-2 mN^{{}^{\prime}}(A_{k})-A_{K}mN^{{}^{\prime\prime}}\left(A_{k}\right)+\frac {\beta m\exp(-mN(A_{k}))N^{{}^{\prime}}(A_{k})}{P+\beta\exp(-mN(A_{k}))}-\left\{ \frac{A_{k}\beta\mathbb{E}^{{}^{\prime}}[C_{k}]}{P+\beta\mathbb{E}[C_{k}]} \right\}^{{}^{\prime}},\end{array} \tag{108}\]  
where \(m=\frac{2\pi\mu^{2}P_{k}\lambda_{k}^{\frac{\theta}{2}+1}}{\alpha-2}\theta_{k}\), \(N\left(A_{k}\right)=\frac{\sum_{j=1}^{K}\frac{A_{j}^{\theta+1}}{P_{j}\lambda_ {j}^{\frac{\theta}{2}}}}{A_{k}^{\frac{\theta}{2}+1}}\) which is from the expression of \(\mathbb{E}\left[\log\left(C_{k}\right)\right]\) as (10).  
Therefore, we have \(\mathbb{E}\left[\log\left(C_{k}\right)\right]=-mN\left(A_{k}\right)\), \(\mathbb{E}^{{}^{\prime}}\left[\log\left(C_{k}\right)\right]=-mN^{{}^{\prime}} \left(A_{k}\right)\) and \(\mathbb{E}^{{}^{\prime}}\left[C_{k}\right]=-m\exp\left(-mN\left(A_{k}\right) \right)\)\(N^{{}^{\prime}}\left(A_{k}\right)\). By taking the derivative of \(A_{k}\),we have  
\[\frac{dN(A_{k})}{dA_{k}} = \frac{\alpha+1}{P_{k}\lambda_{k}^{\frac{\theta}{2}}}A_{k}^{\frac{ \theta}{2}-1}-\frac{(\frac{\theta}{2}+1)}{A_{k}^{\frac{\theta}{2}+2}}\sum_{j =1}^{K}\frac{A_{j}^{\alpha+1}}{P_{j}\lambda_{j}^{\frac{\theta}{2}}}, \tag{109}\] \[\frac{d^{2}N(A_{k})}{{d{A_{k}}^{2}}} = -\frac{2(\alpha+1)}{P_{k}\lambda_{k}^{\frac{\theta}{2}}}A_{k}^{ \frac{\theta}{2}-2}+\frac{(\alpha+2)(\alpha+4)}{4A_{k}^{\frac{\theta}{2}+3}} \sum_{j=1}^{K}\frac{A_{j}^{\alpha+1}}{P_{j}\lambda_{j}^{\frac{\theta}{2}}}. \tag{110}\]  
Hence, we can get the further simplified expression of \(U_{EE,k}\) as follows  
\[\frac{\partial^{2}U_{EE,k}}{\partial A_{k}^{2}} = \frac{(\alpha+2)m}{4A_{k}^{\frac{\theta}{2}+2}}\sum_{j=1}^{K} \frac{A_{j}^{\alpha+1}}{P_{j}\lambda_{j}^{\frac{\theta}{2}}}(-\frac{\alpha \bar{P}\exp(mN(A_{k}))}{\bar{P}\exp(mN(A_{k}))+\beta})<0. \tag{111}\]  
Therefore, it is obvious that \(U_{EE,k}\) is concave function, so \(U_{EE}\) is also concave function.