📝 题目
例 2 设 $f\left( x\right) ,g\left( x\right)$ 为(a, b)上的凹函数,求证:
$$ h\left( x\right) \overset{\text{ 定义 }}{ = }\max \left( {f\left( x\right) ,g\left( x\right) }\right) $$
也是(a, b)上的凹函数.
💡 答案与解析
证 设 ${t}_{1},{t}_{2} > 0,{t}_{1} + {t}_{2} = 1$ ,则对 $\forall {x}_{1},{x}_{2} \in \left( {a,b}\right)$ ,有
$$ f\left( {{t}_{1}{x}_{1} + {t}_{2}{x}_{2}}\right) \leq {t}_{1}f\left( {x}_{1}\right) + {t}_{2}f\left( {x}_{2}\right) \leq {t}_{1}h\left( {x}_{1}\right) + {t}_{2}h\left( {x}_{2}\right) , $$
$$ g\left( {{t}_{1}{x}_{1} + {t}_{2}{x}_{2}}\right) \leq {t}_{1}g\left( {x}_{1}\right) + {t}_{2}g\left( {x}_{2}\right) \leq {t}_{1}h\left( {x}_{1}\right) + {t}_{2}h\left( {x}_{2}\right) , $$
由此推出
$$ h\left( {{t}_{1}{x}_{1} + {t}_{2}{x}_{2}}\right) = \max \left( {f\left( {{t}_{1}{x}_{1} + {t}_{2}{x}_{2}}\right) ,g\left( {{t}_{1}{x}_{1} + {t}_{2}{x}_{2}}\right) }\right) $$
$$ \leq {t}_{1}h\left( {x}_{1}\right) + {t}_{2}h\left( {x}_{2}\right) . $$
由凹函数定义,即知 $h\left( x\right)$ 是(a, b)上的凹函数.
\subsubsection{二、函数作图}