content_image/317231.jpg content_image/317230.jpg content_image/317229.jpg
hence so is g$_{λ}$ ( z ; µ ) . According to Lemma 6, each element in the set S$_{µ}$$_{0}$ \{ 0 } is a zero of the function g$_{λ}$ ( z ; µ$_{0}$ ) . 

Next we show that for any bounded interval L of$_{R}$, S$_{µ}$$_{0}$ $^{⋂}$L is a finite set. Suppose, to the contrary, S$_{µ}$$_{0}$ $^{⋂}$L is infinite, then it has a limit point in R by the Bolzano-Weierstrass Theorem [18] and hence, g$_{λ}$ ( z ; µ$_{0}$ ) = 0 on the whole complex plane$_{C}$ by the Identity Theorem [20]. Then, by (16), (54) and (57), for every x ∈$_{R}$ , 

where 

and c = 1 2 log(2 πe ) + I ( µ$_{0}$ ) − q 1 − $_{q}$d (0) − λ ( γ + 1) is a constant. 

<img src='content_image/317231.jpg'>

<img src='content_image/317230.jpg'>

As in the proof of Lemma 2, there exist a, b ∈ R such that | log p$_{Y}$ ( y ; µ$_{0}$ ) | ≤ 1 $_{2}$y 2 + ay + b . As a result, there exist some α, β > 0 such that | r ( y ) | ≤ αy 2 + β . Since the convolution of r ( y ) and the Gaussian density is equal to the zero function by (69), r ( y ) must be the zero function according to [10, Corollary 9 ]. This requires the capacity-achieving output distribution p$_{Y}$ ( y ; µ$_{0}$ ) be Gaussian, which cannot be true unless X is Gaussian, which contradicts the assumption that X has a probability mass at 0. Therefore, S$_{µ}$$_{0}$ $^{⋂}$L must be a finite set for any bounded interval L , which further implies that S$_{µ}$$_{0}$ is at most countable. 

Finally, we show that S$_{µ}$$_{0}$ is countably infinite. Suppose, to the contrary, S$_{µ}$$_{0}$ = { x$_{i}$ } N i =1 is a finite set with µ$_{0}$ ( { x$_{i}$ } ) = p$_{i}$ and | x$_{i}$ | ≤ B$_{1}$ for all i = 1 , 2 , . . . , N . For any y > B$_{1}$ , 

<img src='content_image/317229.jpg'>

For any ϵ > 0 , choose B$_{2}$ > 0 such that ∫ B$_{2}$ − B$_{2}$ φ ( x )d x > 1 − ϵ . By (16), (54), (57) and (58), for any 

