Lemma \[lem.strictFischer\] shows, that the gap in  and  is non zero in general, i.e. $\Delta>0$, thus it is not possible to reach the upper capacity bounds.

With $$\begin{aligned}
\nonumber
\mathbf{W}_{\mathrm{Off}}=
\left[
  \begin{array}{cc}
     \mathbf{\tilde{H}}^H\mathbf{\tilde{H}} & \mathbf{\tilde{H}}^H\mathbf{\tilde{H}}_e \\
     \mathbf{\tilde{H}}_e^H\mathbf{\tilde{H}} & \mathbf{\tilde{H}}_e^H\mathbf{\tilde{H}}_e \\
  \end{array}
\right] - \mathbf{W}_D,\end{aligned}$$ we can rewrite $$\begin{aligned}
& \det\left(\mathbf{I}_{n_T} + \frac{\rho}{n_T} \left[
  \begin{array}{cc}
     \mathbf{\tilde{H}}^H\mathbf{\tilde{H}} & \mathbf{\tilde{H}}^H\mathbf{\tilde{H}}_e \\
     \mathbf{\tilde{H}}_e^H\mathbf{\tilde{H}} & \mathbf{\tilde{H}}_e^H\mathbf{\tilde{H}}_e \\
  \end{array}
\right]\right) \nonumber \\
& = \det\left(\mathbf{I}_{n_T}
+\frac{\rho}{n_T}(\mathbf{W}_{\mathrm{Off}}+\mathbf{W}_{D})\right) \nonumber \\
& =\det\left(\mathbf{I}_{n_T} +\frac{\rho}{n_T}\mathbf{W}_{D}\right)\times \nonumber\\
& \det\left(\mathbf{I}_{n_T} +\left[\mathbf{I}_{n_T}
+\frac{\rho}{n_T}\mathbf{W}_{D}\right]^{-1}\frac{\rho}{n_T}\mathbf{W}_{\mathrm{Off}}\right)\nonumber\end{aligned}$$ to arrive at $$\begin{aligned}
\nonumber
\Delta=-\frac{1}{2} \mathbb{E}\Bigg[\log_2\det\Bigg(\mathbf{I}_{n_T} +\underbrace{\left[\mathbf{I}_{n_T}
+\frac{\rho}{n_T}\mathbf{W}_{D}\right]^{-1}\frac{\rho}{n_T}\mathbf{W}_{\mathrm{Off}}}_{\mathbf{A}}\Bigg)\Bigg].\end{aligned}$$ Using $$\begin{aligned}
\nonumber
\det(\mathbf{I}_{n_T}+\mathbf{A})=\exp\left(\sum_{k=1}^{L_1}\ln(1+\mu_{k})\right)\end{aligned}$$ yields $$\begin{aligned}
\nonumber
\Delta \leq \frac{1}{2\ln(2)}\mathbb{E}\left[\sum\limits_{k=1}^{L_1}\mu_{k}^2\right]\end{aligned}$$ where the inequality follows from Taylor series expansion $x-\frac{1}{2}x^2\leq \ln(1+x)$ around $x=0$ and the fact that $\mathrm{tr}(\mathbf{A})=0$, since $\mathbf{A}$ has zero block matrices on its diagonal. Its off-diagonal blocks have the form $\mathbf{B}=\left[\mathbf{I}_{n_T}
  +\frac{\rho}{n_T}\mathbf{\tilde{H}}^H\mathbf{\tilde{H}}\right]^{-1}
  \frac{\rho}{n_T}\mathbf{\tilde{H}}^H\mathbf{\tilde{H}_e}$ and $\mathbf{B}_e=\left[\mathbf{I}_{n_T}
  +\frac{\rho}{n_T}\mathbf{\tilde{H}_e}^H\mathbf{\tilde{H}_e}\right]^{-1}
  \frac{\rho}{n_T}\mathbf{\tilde{H_e}}^H\mathbf{\tilde{H}}$, respectively. Note that the matrices in brackets have the same eigenvalues. This implies that each eigenvalue of $\mathbf{A}$ appears twice, i.e. $\mu_{k}=\mu_{k+\nicefrac{L_1}{2}}$, $1 \leq k \leq \nicefrac{L_1}{2}$. Additionally applying the inequality [@HornJohnson (5.6.Ex.26)] $\mathrm{tr}(\mathbf{A}^2)\le ||\mathbf{A}||^2$ we obtain $\sum\limits_{k=1}^{L_1}\mu_{k}^2
 =2\sum\limits_{k=1}^{L_1/2}\mu_{k}^2
 \le 2 ||\mathbf{B}||_F^2$. Further we have $$||\mathbf{B}||_F^2=\mathrm{tr}\left\{
  \left(\frac{\rho}{n_T}\right)^2 \mathbf{\tilde{H_e}}\mathbf{\tilde{H_e}}^H
  \mathbf{\tilde{H}}
    \left[\mathbf{I}_{n_T}
      +\frac{\rho}{n_T}\mathbf{\tilde{H}}^H\mathbf{\tilde{H}}\right]^{-2}
  \mathbf{\tilde{H}}^H
\right\}$$ which can be interpreted as the trace of a 