${{{r^{\ast}_{\text{opt}}}}}\neq 0$ in the wideband regime. Because, otherwise, if ${r_{\text{opt}}}\to 0$ and consequently ${\alpha_{\text{opt}}}\rightarrow0$, the left-hand-side of (\[eq:conditionb\]) becomes zero, while the right-hand-side is different from zero. So, employing (\[eq:stab\]) and taking the limit of both sides of (\[eq:conditionb\]) as $\zeta \to 0$, we can derive that $$\label{eq:conditionbrev}
\frac{p_z({{\alpha^{\ast}_{\text{opt}}}})N_0\log_e2}{{\bar{P}}}\left(1-e^{-\frac{\theta
T{\bar{P}}}{N_0\log_e2}{{\alpha^{\ast}_{\text{opt}}}}}\right)=\theta Te^{-\frac{\theta
T{\bar{P}}}{N_0\log_e2}{{\alpha^{\ast}_{\text{opt}}}}}P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}$$ which, after rearranging, yields $$\label{eq:subsalphab}
\frac{\theta T{\bar{P}}}{N_0\log_e2}{{\alpha^{\ast}_{\text{opt}}}}=\log_e\left(1+\frac{\theta
T{\bar{P}}}{N_0\log_e2}\frac{P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}}{p_z({{\alpha^{\ast}_{\text{opt}}}})}\right).$$ Denoting $\delta=\frac{\theta T{\bar{P}}}{N_0\log_e2}$, we obtain the condition (\[eq:stacondwideband\]) stated in the theorem.

Combining (\[eq:conditionbrev\]) and (\[eq:dotalpha\]) with (\[eq:ddotproof2\]) gives us $$\begin{aligned}
\ddot{{{\sf{R}}}}_E(0)&=-\frac{N_0\log_e^22}{\theta T{\bar{P}}}\frac{{{{r^{\ast}_{\text{opt}}}}}^2
p_z({{\alpha^{\ast}_{\text{opt}}}})(1-e^{-\theta T{{{r^{\ast}_{\text{opt}}}}}})}{1-P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}\big(1-e^{-\theta
T{{{r^{\ast}_{\text{opt}}}}}}\big)}=-\frac{{{{r^{\ast}_{\text{opt}}}}}^2P\{z>{{\alpha^{\ast}_{\text{opt}}}}\} e^{-\theta
T{{{r^{\ast}_{\text{opt}}}}}}\log_e2}{1-P\{z>{{\alpha^{\ast}_{\text{opt}}}}\}\big(1-e^{-\theta T{{{r^{\ast}_{\text{opt}}}}}}\big)}\label{ddotec}\end{aligned}$$ Substituting (\[ddotec\]) and the expression for $\dot{{{\sf{R}}}}_E(0)$ in (\[eq:ebminb\]) into (\[eq:ebresult\]), we obtain (\[eq:s0wb\]). $\square$

The following result shows that in the wideband regime, $\frac{E_b}{N_0}\Big|_{{{\sf{R}}}= 0}$ is the indeed the minimum bit energy.

\[theo:minenergywideband\] In the wideband regime, the bit energy required at zero spectral efficiency (i.e., bit energy required as $B \to \infty$ or equivalently as $\zeta \to 0$) is the minimum bit energy, i.e., $$\begin{gathered}
\frac{E_b}{N_0}\bigg|_{{{\sf{R}}}= 0}=\frac{-\delta\log_e2}{\log_e\xi} =
\frac{E_b}{N_0}_{\min}.\end{gathered}$$

*Proof:* Since $\frac{E_b}{N_0} =
\frac{\frac{{\bar{P}}}{N_0}}{\frac{{{\sf{R}}}_E(\zeta)}{\zeta}}$, we can show the result by proving that ${{\sf{R}}}_E(\zeta)/\zeta$ monotonically decreases with increasing $\zeta$, and hence achieves its maximum as $\zeta \to 0$. We first evaluate the first derivative of ${{\sf{R}}}_E(\zeta)/\zeta$ with respect to $\zeta$: 