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← 笔记Notes2026.09.21 · 2 分钟min

割线引理,以及我为什么总在重推它The secant-slope lemma, and why I keep re-deriving it

The secant-slope lemma, and why I keep re-deriving it割线引理,以及我为什么总在重推它

数学Mathematics笔记Note 02

一行凸性,三个有用的不等式。写在这里,好让我不用每次做优化论证时都从头搭一遍。

One line of convexity, three useful inequalities. Written down here so I stop rebuilding it from scratch at the start of every optimisation argument.

  • analysis
  • convexity
  • optimization
目录Contents

优化里的凸性论证,几乎总会归结为同三个关于割线斜率的不等式。我从头推过它们至少四次——比该有的次数多了三次——所以推导现在住在这里。

命题

设 ff 在区间 II 上凸,x<y<zx \lt y \lt z 是 II 中的三点。则

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引理 1 —— 凸函数的割线斜率对两个端点都单调不减。

中间那一项是这个命题有用的原因:它说整段上的平均斜率落在两个部分平均斜率之间。

证明

把 yy 写成两个端点的凸组合。由 x<y<zx \lt y \lt z 可知

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由 ff 的凸性:

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两边减去 f(x)f(x),并用 t f(z)−t f(x)=t (f(z)−f(x))t\,f(z) - t\,f(x) = t\,(f(z) - f(x)):

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除以严格为正的 y−x=t (z−x)y - x = t\,(z - x),就得到引理 1 的左半:

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引理 1 的右半,把目标不等式两边乘以正量 (z−x)(z−y)(z - x)(z - y) 再整理同类项,剩下的是

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而这又是凸性——权重之和为 1。所以两个不等式是同一个命题从相反方向读出来的,不需要另给一个论证。□\square

为什么值得写下来

这条引理是平凡的。它的价值在于把「由凸性可知」——含糊其辞通常正是从这里开始——换成一个你能指着看的具体不等式。我见过的大多数塌掉的论证,都是因为某一步默认了这条引理却没说出口,然后又把它用到了一个不凸的函数上。

归档于 数学 · 2026.09.21

目录Contents

Convexity arguments in optimisation almost always reduce to the same three inequalities about secant slopes. I have derived them from scratch at least four times, which is three times too many, so the derivation lives here now.

Statement

Let ff be convex on an interval II, and let x<y<zx \lt y \lt z be points of II. Then

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Lemma 1 — secant slopes of a convex function are non-decreasing in both endpoints.

The middle term is what makes this useful: it says the average slope over the whole interval lies between the two partial average slopes.

Proof

Write yy as a convex combination of the endpoints. Since x<y<zx \lt y \lt z,

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By convexity of ff:

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Subtract f(x)f(x) from both sides and use t f(z)−t f(x)=t (f(z)−f(x))t\,f(z) - t\,f(x) = t\,(f(z) - f(x)):

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Dividing by y−x=t (z−x)y - x = t\,(z - x), which is strictly positive, gives the left inequality of Lemma 1:

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For the right inequality of Lemma 1, multiply out the target inequality by the positive quantity (z−x)(z−y)(z - x)(z - y) and collect terms; what is left is

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which is convexity again, with the weights summing to one. So the two inequalities are the same statement read in opposite directions, and no separate argument is needed. □\square

Why it is worth writing down

The lemma is trivial. Its value is that it converts "by convexity" — which is where hand-waving usually starts — into a specific inequality you can point at. Most of the arguments I have watched fall apart did so because a step assumed this lemma without stating it, and then applied it to a function that was not convex.

Filed under Mathematics · 2026.09.21