The two-user Gaussian interference capacity region to within one bit

First suppose both prescribed powers are positive; zero powers are handled separately below. Write $a={\mathrm{S}\mathrm{N}\mathrm{R}}_1$, $b={\mathrm{S}\mathrm{N}\mathrm{R}}_2$, $c={\mathrm{I}\mathrm{N}\mathrm{R}}_1$, $d={\mathrm{I}\mathrm{N}\mathrm{R}}_2$ and $L(x)={\log }_{2}x$. The weak regime is $c<b$, $d<a$, the mixed regimes have exactly one reversed comparison, and the strong regime is $c\ge b$, $d\ge a$. Positive powers make these comparisons equivalent to the gain comparisons in [F1]. In the strong regime [F1] supplies the exact region, so there is no loss. We treat one mixed regime and obtain the other by exchanging all user indices.

In the weak regime set $k_c=\max (1,c)$, $k_d=\max (1,d)$, $q_1=\min (1,d)$, $q_2=\min (1,c)$, $t_1=a\mathrm{/}{k}_d$, $t_2=b\mathrm{/}{k}_c$, $N_1=1+q_2$, and $N_2=1+q_1$. These correspond to the nominal private powers $P_{1p}=P_1\mathrm{/}{k}_d$ and $P_{2p}=P_2\mathrm{/}{k}_c$, including zero cross gains without division by an INR. Let $A_i,D_i,E_i,G_i$ be [F5]'s quantities for this split. Then $1\le N_i\le 2$, $t_1\ge a\mathrm{/}(1+d)$, $t_2\ge b\mathrm{/}(1+c)$, $(N_1+t_1)(1+d)\ge 1+a$, and $(N_2+t_2)(1+c)\ge 1+b$: for example $(1+d)\mathrm{/}{k}_d\ge 1$ and $N_1\ge 1$. These elementary inequalities will compare all facets uniformly, including INR below one. Denote by $O_1,O_2,O_3$ the three weak outer sum right-hand sides in their displayed order and by $W_1,W_2$ the outer $2R_1+R_2$ and $R_1+2R_2$ right-hand sides.

In the mixed regime $c\ge b$, $d<a$, take $q_2=0$, $t_2=0$, $N_1=1$, $q_1=\min (1,d)$, $N_2=1+q_1$, $t_1=a\mathrm{/}\max (1,d)$. Thus user 2 is nominally all common, and the private user 1 signal has unintended received power $\min (1,d)$. The seven-facet quantities become $A_1=L(1+t_1)$, $D_1=L(1+a)$, $E_1=L(1+c+t_1)$, $G_1=L(1+c+a)$, $A_2=0$, $D_2=L(1+b\mathrm{/}{N}_2)$, $E_2=L((1+d)\mathrm{/}{N}_2)$, and $G_2=L((1+d+b)\mathrm{/}{N}_2)$. Let $O=L(1+a)+L(1+b\mathrm{/}(1+d))$ and $V=L(1+b+d)+L(1+c+a\mathrm{/}(1+d))+L(1+b\mathrm{/}(1+c))$, the mixed outer first sum and weighted right sides. The remaining sum bound is $G_1$. Since $(1+d)\mathrm{/}\max (1,d)\ge 1$, we have $(1+t_1)(1+d)\ge 1+a$ and $(1+c+t_1)(1+d)\ge 1+c+a$.

It remains to justify clipping a coordinate at zero; a negative-coordinate polygon argument would not do this. If both capacity coordinates are at most one, use the zero code. Suppose $R_1\le 1<R_2$ and write $B=L(1+b)$. The capacity individual bound gives $R_2\le B$, so it suffices to achieve $(0,\max (0,B-1))$ on the axis. Shut user 1 off. In a weak nominal split, if $c<1$ user 2 is entirely private and achieves $B$. If $c\ge 1$, its desired private power is $b\mathrm{/}c$ and its unintended private power is one. Its common layer can be decoded at the unintended receiver while treating its private layer as noise at any rate up to $C=L((1+c)\mathrm{/}2)$. At its intended receiver the private rate bound is $A=L(1+b\mathrm{/}c)$, the common bound is $H=L(1+b-b\mathrm{/}c)$, and the total bound is $B$. The product identities $(1+b\mathrm{/}c)(1+b-b\mathrm{/}c)\ge 1+b$ and $(1+b\mathrm{/}c)(1+c)\ge 1+b$ give $A+H\ge B$ and $A+C\ge B-1$. For a total rate $T\le \min (B,A+C)$, allocate private rate $\min (A,T)$ and common rate $\max (0,T-A)$; this common rate is at most both $C$ and $H$. Every intended MAC subset inequality and the unintended common decoding inequality therefore holds, with closure for equality, so [F3] achieves at least $\max (0,B-1)$ with this same nominal split. At $c=1$ the common power and $C$ are zero and the calculation still holds.

The complete weak facet comparison table is $\begin{array}{ccc}\text{inner right side}& \text{capacity upper bound}& \text{upper bound minus inner right side}\\ D_1& L(1+a)& \le L{N}_1\le 1\\ D_2& L(1+b)& \le L{N}_2\le 1\\ A_1+G_2& O_1& \le L(N_1{N}_2)\le 2\\ A_2+G_1& O_2& \le L(N_1{N}_2)\le 2\\ E_1+E_2& O_3& \le L(N_1{N}_2)\le 2\\ A_1+G_1+E_2& W_1& \le L(N_1^2{N}_2)\le 3\\ A_2+G_2+E_1& W_2& \le L(N_2^2{N}_1)\le 3\end{array}$. Here the first two comparisons follow from $D_1=L((N_1+a)\mathrm{/}{N}_1)$ and its reflection. For $A_1+G_2$, cancel the common factor $1+d+b$ in the numerator and use $(N_1+t_1)(1+d)\ge 1+a$; the denominator is $N_1{N}_2$. The reflected inequality proves the next row. For $E_1+E_2$, each of its numerator factors $1+c+t_1$ and $1+d+t_2$ dominates the corresponding outer factor by step 2.1. For the $W_1$ row cancel $1+a+c$ and compare the remaining inner factors $(N_1+t_1)(1+d+t_2)$ with the outer factors $((1+a)\mathrm{/}(1+d))(1+d+b\mathrm{/}(1+c))$, factor by factor using step 2.1; the denominator is $N_1^2{N}_2$. Reflection gives the final row. This verifies every entry of the table by multiplication of nonnegative factors.

The full mixed comparison table is $\begin{array}{ccc}\text{inner right side}& \text{capacity upper bound}& \text{upper bound minus inner right side}\\ D_1& L(1+a)& 0\\ D_2& L(1+b)& \le L{N}_2\le 1\\ A_1+G_2& O& \le L{N}_2\le 1\\ A_2+G_1& G_1& 0\\ E_1+E_2& G_1& \le L{N}_2\le 1\\ A_1+G_1+E_2& L(1+a)+G_1& \le L{N}_2\le 1\\ A_2+G_2+E_1& V& \le L{N}_2+L(1+b\mathrm{/}(1+c))\le 2\end{array}$. The third row follows by canceling $1+d+b$ and using $(1+t_1)(1+d)\ge 1+a$. The fifth uses $(1+c+t_1)(1+d)\ge 1+c+a$. The sixth uses the former inequality after canceling $G_1$; its capacity upper bound is the sum of the individual $R_1$ bound and the MAC sum bound, hence is a valid bound on $2R_1+R_2$. In the last row $G_2$ has the same output numerator as the first term of $V$, and $1+c+t_1\ge 1+c+a\mathrm{/}(1+d)$; the sole additional outer factor is $1+b\mathrm{/}(1+c)\le 2$ because $b\le c$. These calculations prove all rows, and each gap is no larger than its facet's sum of coefficients. They include $d<1$, $d=1$, $d=0$, $b=0$, and $c=b$ directly. Therefore for mixed capacity pairs with both coordinates at least one, $(R_1-1,R_2-1)$ obeys all seven achievable facets. Reflecting indices proves the other mixed regime.

If a weak-regime capacity pair has $R_1,R_2\ge 1$, put $r_i=R_i-1$. Subtracting one in each coordinate reduces the linear functionals with coefficient pairs $(1,0)$, $(0,1)$, $(1,1)$, $(2,1)$, and $(1,2)$ by respectively $1,1,2,3,3$. The table therefore places $(r_1,r_2)$ in all seven facets of [F5], which is achievable with the nominal split and its authorized auxiliary-removal alternatives. There are no missing INR subcases: if $c<1$ then $q_2=c$ and user 2's common power is zero; if $d<1$ the reflected fact holds. The formulas remain exact, and facets made redundant by singleton streams do not need to be imposed anew. At $c=1$ or $d=1$ the formulas from the two sides coincide. If $c,d<1$, both common layers vanish and the seven-facet region reduces to the rectangle $r_1\le L(1+a\mathrm{/}(1+c))$, $r_2\le L(1+b\mathrm{/}(1+d))$; each individual gap is at most one. Thus the same calculation covers both sub-noise INRs, either one, and their zero and unit boundaries.

For a mixed axis, the nominally all-common strongly interfering user has unintended link at least as strong as its intended link; with the other user shut off its entire intended single-user rate is decodable at both receivers. The other user's axis is covered by the calculation in step 2.3, including the entirely-private case when its INR is below one. Reflecting indices handles $R_2\le 1<R_1$. Singleton layers and boundary rates follow from [F3]'s closure; the superposition and fallback regions use the explicit slack-power selection in [F5]. Thus for every weak, mixed, or strong capacity pair with positive prescribed powers there is an achievable pair whose coordinates are at least $\max (0,R_i-1)$. All changes of layer labels and powers in the non-axis seven-facet construction are exactly the permitted auxiliary removal of [F5], so the full coordinatewise claim and nominal power policy in the statement both hold.

Finally let a prescribed power vanish. If $P_1=0$, the average nonnegative energy constraint forces user 1's codeword to vanish almost surely, so $R_1=0$ and $a=d=0$; the scalar individual converse gives $R_2\le L(1+b)$. If also $P_2=0$, both rates are zero. Otherwise, when $c\ge b$ the nominal all-common user 2 code achieves its exact single-user rate at both receivers, since $\mathrm{\mid }{h}_{21}{\mathrm{\mid }}^2{P}_2\ge \mathrm{\mid }{h}_{22}{\mathrm{\mid }}^2{P}_2$; there is no condition on user 1's gains to check because it sends no information. When $c<b$ the nominal private user 2 split achieves at least $\max (0,L(1+b)-1)$ by the direct axis construction of step 2.3. The case $P_2=0<P_1$ is reflected. All deterministic zero-power layers are omitted, and no argument divides by a gain or a zero power. This proves the asserted full-channel claim for nonnegative finite powers as well as the positive-power cases.