By Lemma 3, the linearity of E$_{µ}$ $^{{}$X 2 − γ } with respect to (w.r.t.) µ and Condition (i), J ′ $_{µ$_{0}$}$( µ ) can be easily calculated as Therefore, Condition (ii) is equivalent to E$_{µ}$ { f$_{λ}$ ( X ; µ$_{0}$ ) } ≤ 0 for all µ ∈ Λ( q ) . Thus Lemma 5 follows. We call x ∈$_{R}$ a point of increase of a measure µ if µ ( O ) > 0 for every open subset O of$_{R}$ containing x . Let S$_{µ}$ be the set of points of increase of µ . Based on Lemma 5, we derive another sufficient and necessary condition for the optimal input distribution, which will be used to prove Property (c) of Theorem 1 in Section IV-C. Lemma 6: Let Then µ$_{0}$ ∈ Λ( γ, q ) achieves the capacity if and only if there exists λ ≥ 0 such that for every x ∈$_{R}$ , Furthermore, g$_{λ}$ ( x ; µ$_{0}$ ) = 0 for every x ∈ S$_{µ}$$_{0}$ \{ 0 } . Proof: The necessity part is shown as follows. Suppose µ$_{0}$ achieves the capacity, then by Lemma 5, there exists λ ≥ 0 such that λ E$_{µ}$$_{0}$ $^{{}$X 2 − γ } = 0 and E$_{µ}$ { f$_{λ}$ ( X ; µ$_{0}$ ) } ≤ 0 for all µ ∈ Λ( q ) . For any x ∈$_{R}$\{ 0 } , choose µ such that µ ( { 0 } ) = q and µ ( { x } ) = 1 − q , so by the fact that µ ∈ Λ( q ) , we have Due to the continuity of d ( x ; µ$_{0}$ ) by Lemma 2, f$_{λ}$ ( x ; µ$_{0}$ ) is also continuous so that (59) holds for all x ∈$_{R}$ , i.e., g$_{λ}$ ( x ; µ$_{0}$ ) ≤ 0 for every x ∈$_{R}$ . To finish proving the necessity, it suffices to show that g$_{λ}$ ( x ; µ$_{0}$ ) = 0 for all x ∈ S$_{µ}$$_{0}$ \{ 0 } . Evidently, g$_{λ}$ (0; µ$_{0}$ ) = f$_{λ}$ (0; µ$_{0}$ ) and by (17) and λ E$_{µ}$$_{0}$ $^{{}$X 2 − γ } = 0 , 15 0 ≥ E$_{µ}$ { f$_{λ}$ ( X ; µ$_{0}$ ) } = qf$_{λ}$ (0; µ$_{0}$ ) + (1 − q ) f$_{λ}$ ( X ; µ$_{0}$ ) . (59) ∫ f$_{λ}$ ( x ; µ$_{0}$ ) µ$_{0}$ (d x ) = 0 . (60) J ′ $_{µ$_{0}$}$( µ ) = E$_{µ}$ { f$_{λ}$ ( X ; µ$_{0}$ ) } . (56) g$_{λ}$ ( x ; µ ) = qf$_{λ}$ (0; µ ) + (1 − q ) f$_{λ}$ ( x ; µ ) . (57) g$_{λ}$ ( x ; µ$_{0}$ ) ≤ 0 . (58)