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Then with this notation, 

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). 

̂ ̂ ̂ ̂ Control of T$_{3}$ : First note that 

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 ∗ K 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$_{3}$ . 

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Control of T$_{6}$ : The proof is essentially the same as that for T$_{2}$ , except we replace ̂ Ω by the matrix 

