这个性质早就知道了，但并不太清楚严谨的证明是什么。这也是《Introduction to linear optimization》书中第三章课后题的第一题。这篇博客给出严谨的证明。
Exercise 3.1 (Local minimum of convex functions)
Let f : R n → R f: \mathcal{R}^n \rightarrow \mathcal{R} f:Rn→R be a convex function and let S ∈ R n S \in \mathcal{R}^n S∈Rn be a convex set. Let x ∗ x^\ast x∗ be an element of S S S. Suppose that x ∗ x^\ast x∗ is a local optimum for the problem of minimizing f ( x ) f(x) f(x) over S S S; that is, there exists some ϵ > 0 \epsilon>0 ϵ>0 such that f ( x ∗ ) ≤ f ( x ) f(x^\ast)\leq f(x) f(x∗)≤f(x) for all x ∈ S x\in S x∈S for which ∣ x − x ∗ ∣ ≤ ϵ |x - x^\ast|\leq \epsilon ∣x−x∗∣≤ϵ. Prove that x ∗ x^\ast x∗ is globally optimal; that is, f ( x ∗ ) ≤ f ( x ) f(x^\ast)\leq f(x) f(x∗)≤f(x) for all x ∈ S x\in S x∈S.
Solution:
We prove this problem by contradiction (反证法).
Proof.
Suppose that there exists a point x 0 ∈ S x_0\in S x0∈S in which f ( x 0 ) < f ( x ∗ ) f(x_0)<f(x^\ast) f(x0)<f(x∗). Since f ( x ) f(x) f(x) is a convex function, for any λ ∈ [ 0 , 1 ] \lambda\in [0,1] λ∈[0,1],
f ( λ x ∗ + ( 1 − λ ) x 0 ) ≤ λ f ( x ∗ ) + ( 1 − λ ) f ( x 0 ) . f(\lambda x^\ast + (1-\lambda)x_0)\leq \lambda f(x^\ast) + (1-\lambda )f(x_0). f(λx∗+(1−λ)x0)≤λf(x∗)+(1−λ)f(x0).
Since f ( x 0 ) < f ( x ∗ ) f(x_0)<f(x^\ast) f(x0)<f(x∗),
f ( λ x ∗ + ( 1 − λ ) x 0 ) < λ f ( x ∗ ) + ( 1 − λ ) f ( x ∗ ) = f ( x ∗ ) . f(\lambda x^\ast + (1-\lambda)x_0)< \lambda f(x^\ast) + (1-\lambda )f(x^\ast)=f(x^\ast). f(λx∗+(1−λ)x0)<λf(x∗)+(1−λ)f(x∗)=f(x∗).
We can find a λ \lambda λ that makes λ x ∗ + ( 1 − λ ) x 0 \lambda x^\ast + (1-\lambda)x_0 λx∗+(1−λ)x0 locates at the neighbourhood of x ∗ x^\ast x∗, i.e.,
x ∗ − ϵ ≤ λ x ∗ + ( 1 − λ ) x 0 ≤ x ∗ + ϵ ; x^\ast-\epsilon\leq\lambda x^\ast + (1-\lambda)x_0\leq x^\ast+\epsilon; x∗−ϵ≤λx∗+(1−λ)x0≤x∗+ϵ;
i.e., we can make λ \lambda λ satisfies:
− ϵ ≤ ( 1 − λ ) ( x ∗ − x 0 ) ≤ ϵ ; -\epsilon\leq(1-\lambda)(x^\ast-x^0)\leq \epsilon; −ϵ≤(1−λ)(x∗−x0)≤ϵ;
i.e., we can make λ \lambda λ satisfies:
1 − λ ≤ ϵ ∣ x 0 − x ∗ ∣ . 1-\lambda\leq \frac{\epsilon}{|x_0-x^\ast|}. 1−λ≤∣x0−x∗∣ϵ.
This λ \lambda λ does exist, so there will exists a point in the neigourhood of x ∗ x^\ast x∗ but its function value is smaller than f ( x ∗ ) f(x^\ast) f(x∗), which contradicts that x ∗ x^\ast x∗ is a local minimum.
□ \Box □
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