symbol X , which follows distribution µ$_{X}$$_{1}$ , and the corresponding output Y . The following properties hold: The proof of Theorem 2 is relegated to Section V. Evidently, increasing the input power by scaling the input linearly not only maintains its duty cycle, but also increases the mutual information. Therefore, the optimal input distribution must exhaust the power budget γ . This section is devoted to a proof of Theorem 1 for the case of the idealized duty cycle constraint ( q, 0) . The conditional probability density function (pdf) of the output given the input of the AWGN channel (1) is where is the standard Gaussian pdf. With the idealized constraint, the capacity of the AWGN channel is achieved by an i.i.d. process and the duty cycle constraint reduces to a per symbol cost constraint. For given input distribution µ , the pdf of the output exists and is expressed as 8 p$_{Y}$ ( y ; µ ) = ∫ p$_{Y}$$_{|}$$_{X}$ ( y | x ) µ (d x ) = E$_{µ}$ { φ ( y − X ) } . (15) φ ( t ) = 1 √ 2 π e − t 2 2 (14) p$_{Y}$$_{|}$$_{X}$ ( y | x ) = φ ( y − x ) (13) L ( µ $^{∗}$) = I ( X ; Y ) − I ( B$_{1}$ ; B$_{2}$ ) . (12) IV. PROOF OF THEOREM 1 (THE CASE OF IDEALIZED DUTY CYCLE CONSTRAINT) a) L ( µ ) is a lower bound of the channel capacity; b) The maximum of L ( · ) is achieved by a discrete first-order Markov process, denoted by µ $^{∗}$; c) µ ∗ satisfies the following property: Define B$_{i}$ = 1$_{{}$$_{X}$$_{i}$$_{̸}$$_{=0}$$_{}}$ , i = 1 , 2 , . . . . Then for every i , conditioned on B$_{i}$ and B$_{i}$$_{+1}$ , the variables X$_{i}$ and X$_{i}$$_{+1}$ are independent, and