$$\begin{aligned}
\Gamma_{a1}&=\overline{C}_\Sigma-\underline{C}^{I}\\
&\leq C(h_2^2P)+C(h_3^2P)\\
&\leq 2C(h_2^2P)\\
&\leq \log(3/2)\end{aligned}$$ where we used $h_3^2P\leq h_2^2P\leq1/2$. Therefore, if $h_2^2P\leq1/2$ we can write (by combining $\Gamma_m$ and $\Gamma_{a1}$) $$\begin{aligned}
\max\left\{\overline{C}_\Sigma-\log\left(\frac{3}{2}\right),\frac{\overline{C}_\Sigma}{3}\right\}&\leq C_g\leq\overline{C}_\Sigma.\end{aligned}$$

Case $h_2^2P>1/2$
-----------------

We call the gap for this case $\Gamma_{a2}$. Notice that using $h_2^2P>1/2$ in $\underline{C}^{III}$ leads to $$\begin{aligned}
C_g&\geq\underline{C}^{III}=2C\left(h_2^2P-\frac{1}{2}\right).\end{aligned}$$

Now we bound $\Gamma_{a2}$ by bounding the difference between the upper bound $\overline{C}_{\Sigma g}$ and the lower bound $\underline{C}^{III}$. We obtain $$\begin{aligned}
\Gamma_{a2}&= \overline{C}_{\Sigma g}-\underline{C}^{III}\\
&\leq C(h_2^2P+h_3^2P)+C((|h_2|+|h_3|)^2P)\nonumber\\
&\quad-2C\left(h_2^2P-1/2\right)\\
&\leq C(2h_2^2P)+C(4h_2^2P)-2C\left(h_2^2P-1\right)\\
&\leq 2C(2h_2^2P)+1/2-2C\left(h_2^2P-1\right)\\
&= \log\left(2+\frac{1}{h_2^2P}\right)+\frac{1}{2}\triangleq\overline{\Gamma}_{a2}\end{aligned}$$ where we used $h_3^2\leq h_2^2$. Thus the gap is upper bounded by $\overline{\Gamma}_{a2}$ which approaches 3/2 as $P\to\infty$. Moreover, using $h_2^2P>1/2$ we have $$\begin{aligned}
\Gamma_{a2}&\leq \frac{5}{2}.\end{aligned}$$ As a result, for $h_2^2P>1/2$ we have $$\begin{aligned}
\max\left\{\overline{C}_{\Sigma g}-\frac{5}{2},\frac{\overline{C}_\Sigma}{3}\right\}\leq C_g\leq \min\{\overline{C}_{\Sigma g},\overline{C}_\Sigma\}.\end{aligned}$$

Thus, we have bound the gap between our sum-capacity upper and lower bounds by a constant independent of the channel coefficients. Notice that the multiplicative gap is important for the case of low power, especially when the additive gap becomes larger than the upper bound. Let us now consider the symmetric Y-channel, where $h_1=h_2=h_3=1$. In this case, given $P=P_r$, we can show that the gap between the upper and lower bounds is always less than 1 bit.

Gap Calculation 