Then with this notation, | T$_{2}$ | = ∣ ( ̂ ∂ ) ′ ̂ Ω ̂ ∂ − ∂ ′ ̂ Ω ∂ ∣ = | ( ̂ ∂ + ∂ ) ′ ̂ Ω( ̂ ∂ − ∂ ) | ≤ ‖ ( ̂ ∂ + ∂ ) $^{′}$‖‖ ̂ Ω ‖‖ ( ̂ ∂ − ∂ ) ‖ . (110) ‖ ̂ ∂ − ̂ ∂ ‖ ≤ ‖ ( ˜ B ′ w ˜ B$_{w}$/n ) − − I$_{K}$ ‖ ( ‖ ̂ ∂ − ∂ ‖ + ‖ ∂ ‖ ) . (111) | T$_{3}$ | ≤ ‖ ̂ h − h$_{0}$ ‖ 2 ∞ ,w × 1 n n ∑ i =1 v$_{K}$ ( X$_{i}$ ) 2 ‖ v ∗ K ‖ 2 sd = o$_{p}$ (1) × O$_{p}$ (1) = o$_{p}$ (1) (112) : by the triangle inequality definition of ̂ v K and v $_{K}$: | T$_{4}$ | ≤ ‖ ̂ h − h$_{0}$ ‖ 2 ∞ ,w × ( 1 n n ∑ i =1 ̂ v ∗ $_{K}$( X$_{i}$ ) 2 ‖ v ∗ K ‖ 2 sd + 1 n n ∑ i =1 v ∗ $_{K}$( X$_{i}$ ) 2 ‖ v ∗ K ‖ 2 sd ) (113) = o$_{p}$ (1) × ( ̂ ∂ $^{′}$̂ Ω ̂ ∂ + ∂ ′ ̂ Ω ∂ ) ≤ o$_{p}$ (1) × ‖ ̂ Ω ‖ × ( ‖ ̂ ∂ ‖ 2 + ‖ ∂ ‖ 2 ) . (114) where ‖ ̂ h − h$_{0}$ ‖ ∞ ,w = o$_{p}$ (1) by hypothesis and n − 1 ∑ n i =1 v$_{K}$ ( X$_{i}$ ) $^{2}$/ ‖ v ∗ K ‖ 2 sd by Markov’s inequality and the fact that ‖ v ∗ K ‖ L $^{2}$( X ) ≍ ‖ v ∗ K ‖ sd under Assumption 2(ii)(iv). Control of T$_{4}$ : by the triangle inequality definition of v ∗ and v ∗ $_{K}$: Moreover, ‖ ̂ Ω ‖ = O$_{p}$ (1) by Assumption 8, ‖ ∂ ‖ ≍ 1 by Assumption 2(ii)(iv), and ‖ ̂ ∂ ‖ ≤ ‖ ̂ ∂ − ̂ ∂ ‖ + ‖ ̂ ∂ − ∂ ‖ + ‖ ∂ ‖ = O$_{p}$ (1) by Assumption 5 and 9(iv). It follows that | T$_{4}$ | = o$_{p}$ (1). Control of T$_{5}$ : By the inequality 2 | a | ≤ 1 + a $^{2}$, we have where ‖ ̂ h − h$_{0}$ ‖ ∞ ,w = o$_{p}$ (1) by hypothesis, n − 1 ∑ n i =1 ϵ 2 $_{i}$v ∗ $_{K}$( X$_{i}$ ) $^{2}$/ ‖ v ∗ K ‖ 2 sd → p 1 by Assumption 8, and the remaining term is O$_{p}$ (1) by the arguments for T$_{3}$ . Control of T$_{6}$ : The proof is essentially the same as that for T$_{2}$ , except we replace ̂ Ω by the matrix 44 | T$_{5}$ | ≤ ‖ ̂ h − h$_{0}$ ‖ ∞ ,w 1 n ∑ i =1 (1 + ϵ 2 $_{i}$) v ∗ $_{K}$( X$_{i}$ ) 2 ‖ v ∗ K ‖ 2 sd = o$_{p}$ (1) × O$_{p}$ (1) = o$_{p}$ (1) (115) Note that ‖ ̂ Ω ‖ = O$_{p}$ (1) by Assumption 8(ii). By the triangle inequality and definition of ∂ , ̂ ∂ , and ̂ ∂ , Assumption 5 implies ‖ ( ˜ B ′ w ˜ B$_{w}$/n ) − − I$_{K}$ ‖ = o$_{p}$ (1); ‖ ̂ ∂ − ∂ ‖ = o$_{p}$ (1) by Assumption 9(iv), because ̂ h ∈ N$_{K,n}$ wpa1; and ‖ ∂ ‖ ≍ 1 because ‖ v ∗ K ‖ L $^{2}$( X ) ≍ ‖ v ∗ K ‖ sd under Assumption 2(ii)(iv). Therefore, ∂ ∂ = o$_{p}$ (1), ∂ + ∂ = O$_{p}$ (1), and so T$_{2}$ = o$_{p}$ (1). ‖ ̂ ∂ − ̂ ∂ ‖ = o$_{p}$ (1), ‖ ̂ ∂ + ̂ ∂ ‖ = O$_{p}$ (1), and so | T$_{2}$ | Control of T$_{3}$ : First note that