一道凸函数积分估值题
2018.06.14
设\(f(x)\)是\([0,1]\)上的实值连续凸函数，\(f(0)=0,f(1)=1\),且\(g(f(x))=x\),证明\(\int_0^1{f\left( x \right) g\left( x \right) dx}\le \frac{1}{3}.\)
\(solution:\)
易见\(g\left( x \right) =f^{-1}\left( x \right)\),只要证\(f^{-1}\left( x \right) f\left( x \right) \le x^2\),\(\Longleftrightarrow \frac{f\left( x \right)}{x}\le \frac{x}{f^{-1}\left( x \right)}\)
根据凸函数的性质\(\frac{f\left( x \right) -f\left( x \right)}{x-0}\le \frac{f\left( t \right) -f\left( 0 \right)}{t-0}=\frac{x}{f^{-1}\left( x \right)},x=f\left( t \right) ,t\in \left[ x,1 \right] .\)
凸函数的几点性质：
- \(f_{-}^{'}\left( x \right) \le f_{+}^{'}\left( x \right)\),且\(f_{-}^{'}\left( x \right) ,f_{+}^{'}\left( x \right)\)单调递增
- \(f\left( x \right) \ge f_{+}^{'}\left( x_0 \right) \left( x-x_0 \right) +f\left( x_0 \right) ,x>x_0\)
- \(f\left( x \right) \ge f_{-}^{'}\left( x_0 \right) \left( x-x_0 \right) +f\left( x_0 \right) ,x<x_0\)
- (Hadamad)\(\left( b-a \right) f\left( \frac{a+b}{2} \right) \le \int_a^b{f\left( x \right) dx}\le \left( b-a \right) \frac{f\left( a \right) +f\left( b \right)}{2}\)