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通过加和减项(y+1)αLg(hy)Lg(h)并再次使用不等式(u+v)≤ 2u+2v,wegetGh(y)=(y+1)αLg(h(y+1))Lg(h)- (y+1)αLg(hy)Lg(h)+(y+1)αLg(hy)Lg(h)- yαLg(hy)Lg(h)(0,1/h)-1) (y)≤ 2(y+1)2αLg(h(y+1))- Lg(hy)Lg(h)(0,1/h)-1) (y)+ 2.(y+1)α- yαLg(hy)Lg(h)(0,1/h)-1) (y).我们记得Lg:=infx∈(0,1]Lg(x)>0乘以(A1),所以Lg(h(y+1))- Lg(hy)Lg(h)(0,1/h)-1) (y)≤Lg | Lg(h(y+1))- Lg(hy)|1(0,1/h)-1) (y)。利用中值定理和Lgfrom(A1)的导数的界,我们观察到| Lg(h(y+1))- Lg(hy)|=|Lg(ξ)|h(y+1)- hy|≤ hC1 +ξ≤ Ch+y,式中ξ=ξ(y,h)∈ [hy,h(y+1)]。注意到约束y<h- 1相当于h<y+1,我们进一步得到Lg(h(y+1))- Lg(hy)Lg(h)(0,1/h)-1) (y)≤CLgh+y(0,1/h)-1) (y)≤CLgy+1+y≤CLgy+1,作为y≥ 1,然后我们用它来推导2(y+1)2αLg(h(y+1))- Lg(hy)Lg(h)(0,1/h)-1) (y)≤18CLg(y+1)2(α-1).此外,我们观察到,通过(4.1)和(4.2),(y+1)α- yαLg(hy)Lg(h)(0,1/h)-1) (y)≤ 2CΔαy2(α-1+δ),我们选择δ∈ (0,-α) ,确保2(α-1 + δ) < -1.
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