x > B$_{1}$ + B$_{2}$ , we have For (72) to hold for large x , λ must satisfy λ ≥ 1 $_{2}$. To finish the proof, it suffices to show that λ < 1 2 for any γ > 0 , so that contradiction arises, which implies that S$_{µ}$$_{0}$ must be countably infinite. For fixed q ∈ (0 , 1) , denote the Lagrange multiplier in (58) as λ ( γ ) . Denote C$_{G}$ ( γ ) = 1 2 log(1 + γ ) , which is the channel capacity of a Gaussian channel with the average power constraint only. By the envelope theorem [19], λ ( γ ) is the derivative of C ( γ, q ) w.r.t. γ . Since C (0 , q ) = C$_{G}$ (0) = 0 and the derivative of C$_{G}$ ( γ ) at γ = 0 is 1 $_{2}$, we have λ (0) ≤ 1 $_{2}$, otherwise we could find a small enough γ such that C ( γ, q ) would exceed C$_{G}$ ( γ ) which is obviously impossible. Next we show that C ( γ, q ) is strictly concave for γ ≥ 0 . Suppose µ$_{1}$ and µ$_{2}$ are the capacity-achieving input distributions of (5) for different power constraints γ$_{1}$ and γ$_{2}$ , respectively. Due to Property (b) in Theorem 1, µ$_{1}$ and µ$_{2}$ must be different. Define µ$_{θ}$ = θµ$_{1}$ + (1 − θ ) µ$_{2}$ for θ ∈ (0 , 1) . It is easy to see that µ$_{θ}$ satisfies that the duty cycle is no greater than 1 − q and the average input power is no greater than θγ$_{1}$ + (1 − θ ) γ$_{2}$ . Now we have where (77) is due to the strict concavity of I ( µ ) . Therefore, the strict concavity of C ( γ, q ) for γ ≥ 0 follows, which implies that λ ( γ ) < λ (0) = 1 2 for all γ > 0 . C ( θγ$_{1}$ + (1 − θ ) γ$_{2}$, q ) ≥ I ( µ$_{θ}$ ) (76) > θI ( µ$_{1}$ ) + (1 − θ ) I ( µ$_{2}$ ) (77) = θC ( γ$_{1}$, q ) + (1 − θ ) C ( γ$_{2}$, q ) , (78) 0 ≥ − ∫ ∞ −∞ φ ( y − x ) log p$_{Y}$ ( y ; µ$_{0}$ )d y − λx 2 − ( c + λ ) (72) ≥ ∫ x + B$_{2}$ x − B$_{2}$ φ ( y − x ) 1 2 ( y − B$_{1}$ ) $^{2}$d y − λx 2 − ( c + λ ) (73) = ∫ B$_{2}$ B$_{2}$ φ ( t ) 1 2 ( x − B$_{1}$ + t ) $^{2}$d t − λx 2 − ( c + λ ) (74) ≥ 1 2 ( x − B$_{1}$ ) $^{2}$(1 − ϵ ) − λx 2 − ( c + λ ) . (75) 18