最优化理论|为什么凸问题的解集是凸集
💡 问题描述
假设需要求解的凸问题表示如下：
\[ \begin{equation}
\begin{aligned}
&\underset{x}{\min}~f(x) \\
&{\rm s.t.}~~x \in S.
\end{aligned}
\end{equation}
\]
\(f:\mathbb{R}^n \to \mathbb{R}\)为凸函数, \(S\)为凸集.若 \(C\) 是(1)的解集, 证明：\(C\)是凸集？
🚩 证明:
任取\(x_1,x_2 \in C\)均是凸问题(1)的解, 即\(f(x_1) = f(x_2)\), \(\forall \theta \in [0,1]\), 对于
\[ \theta x_1 + (1-\theta)x_2.
\]
由于\(f(x)\)是凸函数, 所以
\[ f\Big(\theta x_1 + (1-\theta)x_2\Big) \le \theta f(x_1) + (1-\theta)f(x_2) = f(x_1).
\]
所以\(\theta x_1 + (1-\theta)x_2 \in C\), 故而\(C\)为凸集.
英文版本：
🚩 Proof:
Let $ x_1, x_2 \in C $ be any solutions to the convex problem (1), i.e., $ f(x_1) = f(x_2) $. For any $ \theta \in [0,1] $, consider the point
\[ \theta x_1 + (1-\theta)x_2.
\]
Since $ f(x) $ is a convex function, it follows that
\[ f\Big(\theta x_1 + (1-\theta)x_2\Big) \le \theta f(x_1) + (1-\theta)f(x_2) = f(x_1).
\]
Thus, $ \theta x_1 + (1-\theta)x_2 \in C $, which implies that $ C $ is a convex set.
