# Energy Efficient Transceiver Design for NOMA VLC Downlinks with Finite-Alphabet Inputs
## Appendix A Proof of Theorem 1  
We need to establish the following two propositions to prove Theorem 1.  
**Proposition A1**.: _For any Mpositive real-valued numbers \(p_{i}\) and positive integers \(K_{i}\), if \(0\leq p_{1}\leq p_{2}\leq\cdots\leq p_{M}\) and \(D>0\), then, we have \(h_{M}p_{1}\geq D\) and \(h_{M}p_{i}\geq D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\)._  
**Proof of Proposition A1:** Let \(\mathcal{G}_{M}=\{h_{M}\sum_{i=1}^{M}p_{i}s_{i}:s_{i}\in\mathcal{S}_{i}\}\). Since \(D>0\), we can rearrange all the \(2^{\sum_{j=1}^{M}K_{j}}\) elements of \(\mathcal{G}_{M}\) in a strictly increasing order such that \(g_{0}<g_{1}<\cdots<g_{\frac{2^{\sum_{j=1}^{M}K_{j}}}{2^{\sum_{j=1}^{M}K_{j}}}-1}\). Now, it is more clear to see that if we reduce the power of \(\mathcal{S}_{i}\) such that \(g_{k}=kD\), then \(\min_{g\
eq\tilde{g},\tilde{g},\tilde{g}\in\mathcal{G}_{M}}|g-\tilde{g}|=\min _{g\
eq\tilde{g},g\
otin\mathcal{G}_{M}}|g-\tilde{g}|=D\), where \(\mathcal{G}_{M}=\{h_{M}\sum_{i=1}^{M}\tilde{p}_{i}s_{i}:s_{i}\in\mathcal{S}_{ i}\}=\{kD\}_{k=0}^{k=2^{\sum_{j=1}^{i-1}K_{j}}-1}\), and the consumed peak power of each user for \(\mathcal{G}_{M}\) is not larger than that for \(\mathcal{G}_{M}\). In addition, it holds that \(p_{i}\geq\tilde{p}_{i}\) for \(1\leq i\leq M\). Then, according to Proposition A1 in [23], if we let \(\mathcal{X}_{i}=\{h_{M}\tilde{p}_{i}k_{i}:k_{i}\in\{k\}_{k=0}^{k=2^{K_{i}}-1}\},0\leq\tilde{p}_{1}\leq\tilde{p}_{2}\leq\cdots\leq\tilde{p}_{M}\) and \(\{h_{M}\sum_{i=1}^{M}\tilde{p}_{i}k_{i}:k_{i}\in\{k\}_{k=0}^{k=2^{K_{i}}-1}\}= \{D\times k\}_{k=0}^{k=2^{\sum_{i=1}^{M}K_{i}}-1}\), then we have \(\{h_{M}\tilde{p}_{1}k:k\in\{k\}_{k=0}^{k=2^{K_{1}}-1}\}=\{D\times k\}_{k=0}^{k= 2^{K_{1}}-1}\) and \(\{h_{M}\tilde{p}_{i}k:k\in\{k\}_{k=0}^{k=2^{K_{i}}-1}\}=\{D\times k\times 2^{ \sum_{j=1}^{i-1}K_{j}}\}_{k=0}^{k=2^{K_{i}}-1}\) for \(2\leq i\leq M\), i.e., \(h_{M}\tilde{p}_{1}=D\) and \(h_{M}\tilde{p}_{u}=D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq U\). Therefore, it indeed holds that \(h_{M}p_{1}\geq h_{M}\tilde{p}_{1}=D\) and \(h_{M}p_{i}\geq h_{M}\tilde{p}_{i}=D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\), and the proof of Proposition A1 is complete.  
**Proposition A2**.: _For any given minimum Euclidean distance \(D\), if \(h_{M}p_{1}\geq D\), \(h_{M}p_{i}\geq D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\) and \(\sum_{i=1}^{M}p_{i}(2^{K_{i}}-1)=1\), then we have \(D\leq\tilde{D}\)._  
**Proof of Proposition A2:** To prove it by contradiction, we assume that under the condition that \(h_{M}p_{1}\geq D\), \(h_{M}p_{i}\geq D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\) and \(\sum_{i=1}^{M}p_{i}(2^{K_{i}}-1)=1\), there exists at least one Dsuch that \(D>\tilde{D}\). Then, we can get that \(h_{M}p_{1}>\tilde{D}\) and \(h_{M}p_{i}>\tilde{D}\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\), and thus, we have:  
\[h_{M}\sum_{i=1}^{M}p_{i}(2^{K_{i}}-1)> \tilde{D}(2^{K_{1}}-1)+\sum_{i=2}^{M}\tilde{D}\times 2^{\sum_{j=1}^{i-1}K_{j}} (2^{K_{i}}-1)= \tilde{D}(2^{\sum_{j=1}^{M}K_{j}}-1) \tag{1}\]  
Then, combing this result with \(\tilde{D}=\frac{h_{M}}{2^{\sum_{j=1}^{M}K_{j}}-1}\) in Theorem 1, we can obtain that \(\sum_{i=1}^{M}p_{i}(2^{K_{i}}-1)>1\), which contradicts the condition \(\sum_{i=1}^{M}p_{i}(2^{K_{i}}-1)=1\), and thus, our assumption isnot true. In addition, we can find that if and only if \(h_{M}p_{1}=D\) and \(h_{M}p_{i}=D\times 2^{\sum_{j=1}^{i-1}K_{j}}\) for \(2\leq i\leq M\), we have \(D=\tilde{D}\). This completes the proof of Proposition A2.  
Then, according to Proposition A2, we have \(\tilde{D}=\frac{h_{M}}{2^{\sum_{j=1}^{M}K_{j}}-1},\tilde{p}_{1}=\frac{\tilde{D }}{h_{M}}\) and \(\tilde{p}_{i}=\frac{\tilde{D}\times 2^{\sum_{j=1}^{i-1}K_{j}}}{h_{M}}\) for \(i=2,3,\cdots,M\). Therefore, the proof of Theorem 1 is complete.