under which all messages can be decoded by Rx3.

The achievablity follows from ${\mathcal{C}}_2$ in Theorem \[thm:cap\_set2\] by evaluating *Set2* and ${\mathcal{C}}_2$ with zero mean jointly Gaussian channel inputs $X_1$, $X_2$ and $X_3$. In fact, $X_1\sim{\mathcal{N}}(0,P_1)$, $X_2\sim{\mathcal{N}}(0,P_2)$ and $X_3\sim{\mathcal{N}}(0,P_3)$, where $E(X_1X_2)=0$, $E(X_1X_3)=\rho_1\sqrt{P_1P_3}$, and $E(X_2X_3)=\rho_2\sqrt{P_2P_3}$. The converse proof is based on the similar reasoning as in [@Sato81] and is provided in Appendix \[app:thm:Gaus\_converse\].

Note that, the channel parameters, i.e., $h_{tr}$, $P_t$ for $t,r\in\{1,2,3\}$, must satisfy (\[eqn:set1\_I\_Gaus\])-(\[eqn:set1\_IV\_Gaus\]) for all $-1\leq\rho_1,\rho_2\leq 1:\rho_1^2+\rho_2^2\leq 1$, to numerically evaluate the ${\mathcal{C}}_1^G$ using (\[eqn:Gaus\_Cap1\_I\])-(\[eqn:Gaus\_Cap1\_III\]). Here, we choose $P_1=P_3=3$, $P_2=6$, $h_{11}=h_{22}=h_{33}=1$, $h_{31}=h_{32}=\sqrt{1.5}$, $h_{12}=7$, $h_{13}=3$, $h_{21}=5$, and $h_{23}=15$, which satisfy (\[eqn:set1\_I\_Gaus\])-(\[eqn:set1\_IV\_Gaus\]); hence, the regions are derived under strong interference conditions $Set_G$.

![Capacity region for the Gaussian three-user C-IFC for fixed $\rho_1=\rho_2$.[]{data-label="fig:Gaus1"}](Case1.eps){width="11cm"}

Fig. \[fig:Gaus1\] shows the capacity region for the Gaussian three-user C-IFC of Theorem \[thm:Gaus\_cap\_set1\], for the above parameter selection, where $\rho_1=\rho_2$ is fixed in each surface. $\rho_1=\rho_2=0$ region corresponds to the *no cooperation* case, where the channel inputs are independent. It can be seen that as $\rho_1=\rho_2$ increases, the bound on $R_3$ becomes more restrictive while the sum-rate bounds become looser; because Tx3 dedicates parts of its power for cooperation. The capacity for this channel is the union 