陶哲轩博客写数学问题时，通常先把对象定义清楚，再给直觉、反例和证明轮廓。把「The Hahn-Banach theorem, Menger’s theorem, and Helly’s theorem」改写成可阅读的中文笔记，重点是：问题在问什么、已知到哪一步、下一步最容易走偏在哪。原站广告、分享条和导航已去掉。

问题在问什么

In the previous post , I discussed how an induction on dimension approach could establish Hilbert’s nullstellensatz , which we interpreted as a result describing all the obstructions to solving a system of polynomial equations and inequations over an algebraically closed field . Today, I want to point out that exactly the same approach also gives the Hahn-Banach theorem (at least in finite dimensions), which we interpret as a result describing all the obstructions to solving

To simplify the exposition we shall only work in finite dimensions and with finite complexity objects, such as finite systems of linear inequalities, or convex polytopes with only finitely many sides. The results can be extended to the infinite complexity setting but this requires a bit of care, and can distract from the main ideas, so I am ignoring all of these extensions here.

已知结果和反例

Let us first phrase a formulation of the Hahn-Banach theorem – namely, Farkas’ lemma – which is deliberately chosen to mimic that of the nullstellensatz in the preceding post. We consider systems of linear inequalities of the form

where lies in a finite-dimensional real vector space, and are affine -linear functionals. We are interested in the classic linear programming problem of whether such a system admits a solution. One obvious obstruction would be if the above system of inequalities are inconsistent in the sense that they imply . More precisely, if we can find non-negative reals such that , then the above system is not solvable. Farkas’ lemma asserts that this is in fact the only obstruction:

证明或构造的主线

Farkas’ lemma . Let be affine-linear functionals. Then exactly one of the following statements holds:

As in the previous post, we prove this by induction on d. The trivial case d=0 could be used as the base case, but again it is instructive to look at the d=1 case first before starting the induction.

阅读时建议盯住的点

If d=1, then each inequality can be rescaled into one of three forms: , , or , where , , or is a real number. The latter inequalities are either trivially true or trivially false, and can be discarded in either case. As for the inequalities of the first and second type, they can be solved so long as all of the which appear here are less than equal to all of the which appear. If this is not the case, then we have for some , which allows us to fashion -1 as a non-negative linea

Now suppose that and the claim has already been proven for . As in the previous post, we now split for and . Each linear inequality can now be rescaled into one of three forms: , , and .

值得单独记下的条目

• The system of inequalities has a solution .
• There exist non-negative reals such that .
• (Alice can expect to win at least ) There exists an optimal strategy for Alice such that for all q;
• (Bob can expect to lose at most ) There exists an optimal strategy for Bob such that for all p.
• the net flow at v (the total inflow minus total outflow) is -1, and the net flow at w is +1;
• for any other vertex u, the net flow at u is zero, and total inflow or outflow at u is at most .

阅读和落地时建议先做的 5 件事

1. 用自己的语言重写定义和结论，不看原文能不能说清对象是什么。
2. 找一个最小反例或边界情形，确认假设少一条会怎样。
3. 把证明拆成可独立检验的引理，每步只保留一个新想法。
4. 若涉及计算或形式化，先写可复现的小例子，再谈一般情形。
5. 记下尚未解决的缺口：缺估计、缺构造，还是缺正确的范畴。

和智能体、形式化工具怎么接

龙虾PRO做 OpenClaw 落地时，数学笔记最有用的部分往往是「可检验的步骤」：定义、反例、引理边界。智能体适合帮忙展开计算和检索，不适合代替你决定哪条假设能扔。

本文侧重全链路风控方法论。落地时请用自身业务单据做回放验证，不要把示例阈值直接当生产策略。 相关：风控体检 · 方案资源