# 没有不好的老师，只有不好的领导~

This is an in-mail from TYUST.

(接前：21 20 18) 命题 3.1 的证明
---- 在Step5, Bv 是主角, Γv 是主力.
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Step 6. By the previous step, there exists α  (0, 1) depending only on d, eps, n, Q such that Δ:= αBv + (1 - α)Λ  0.
---- 由 Bv 和 Λ 做凸组合, 生成 Δ.
---- α 的存在性及 Λ 的非负性待考.(?)
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Then, by Lemma 2.3, (V, Δ) is eps'-lc where eps' = αeps because (V, Bv) is sub-eps-lc and (V, Λ) is lc.
---- 上一句 “做相”, 这一句配对、定性.
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Moreover, by Step 4, a(T, V, Δ) = α a(T, V, Bv) + (1 - α) a(T, V, Λ) ≤ α + (1 - α) = 1.
---- 这一句是“做盘”.
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On the other hand, replacing H, we can assume H - Λ is ample.
---- 替换 H, 使得 H - Λ 丰.
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Moreover, by construction, Bv~R -Kv ~R Ωv, SuppΩv ⊆ Λ, and the coefficients of Ωv are bounded from below and above, hence we can assume H - Bv is ample as well.
---- 之前认为 Kv = - Ωv, 实际上要弱一些.
---- Bv 和 Ωv 的关系不明显.(?)
---- SuppΩv ⊆ Λ (这是哪来的?).
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---- H 相当于“侯”.(V 空间上).
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This in turn implies H - Δ = α(H - Bv) + (1 - α)(H - Λ) is ample too.
---- 差的凸组合中, 公共的被减量有两项可以抵消.
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In addition, there is a natural number r depending only on d, eps, n, Q such that H ≤ r.
---- 暂时看不出.(?)
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H
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V       Δ

符号大全上下标.|| 常用：↑↓ πΓΔΛΘΩμφΣ∈  ∪ ∩ ⊆ ⊇ ⊂ ⊃ ≤ ≥ ⌊ ⌋ ⌈ ⌉ ≠ ≡ ⁰ ¹ ² ³ ᵈ ₀ ₁ ₂ ₃ ᵢ .

#### Glossary(AG)

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Introduction
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http://blog.sciencenet.cn/blog-315774-1186296.html

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