Lemma 4: Let θ, a, b ∈ (0 , 1] . Define then Proof: It is easy to check that f (1) = b log b $_{a}$, f (0 $^{+}$) = b − a and Define g ( θ ) = θ ( b − a ) − a log $^{(}$1 − θ + b $_{a}$θ ) for θ ∈ (0 , 1] , then we have Since g (0 $^{+}$) = 0 , g ( θ ) ≥ 0 for all θ ∈ (0 , 1] . According to (42), we have f $^{′}$( θ ) = g ( θ ) θ 2 ≥ 0 . It follows that for all θ ∈ (0 , 1] , and hence Lemma 4 is thus established. Applying Lemma 4 with a = p$_{Y}$ ( y ; µ$_{0}$ ) and b = p$_{Y}$ ( y ; µ ) , we have where the right hand side is an integrable function of y by the result that − ∫ ∞ −∞ p$_{Y}$ ( y ; µ$_{2}$ ) log p$_{Y}$ ( y ; µ$_{1}$ )d y < ∞ for any µ$_{1}$, µ$_{2}$ ∈ Λ( q ) . In fact, as in the proof of Lemma 2 (see (28)), there exist a, b ∈$_{R}$ such that 13 ∣ ∣ 1 θ p$_{Y}$ ( y ; µ$_{θ}$ ) log p$_{Y}$ ( y ; µ$_{θ}$ ) p$_{Y}$ ( y ; µ$_{0}$ )$^{∣}$ ∣ ∣ ≤ p$_{Y}$ ( y ; µ ) + p$_{Y}$ ( y ; µ$_{0}$ ) − p$_{Y}$ ( y ; µ ) log p$_{Y}$ ( y ; µ ) − p$_{Y}$ ( y ; µ ) log p$_{Y}$ ( y ; µ$_{0}$ ) (47) | f ( θ ) | ≤ max$^{{}$ | b − a | ,$^{∣}$ ∣ ∣ b log b a$^{∣}$ ∣ ∣ } (45) ≤ b + a − b log b − b log a. (46) b − a = f (0 $^{+}$) ≤ f ( θ ) ≤ f (1) = b log b a , (44) g $^{′}$( θ ) = θ ( b − a ) 2 (1 − θ ) a + θb ≥ 0 . (43) f $^{′}$( θ ) = b − a θ − a θ 2 log$^{(}$1 − θ + b a θ$^{)}$ . (42) | f ( θ ) | ≤ b + a − b log b − b log a . (41) f ( θ ) = (1 − θ ) a + θb θ log (1 − θ ) a + θb a , (40)