Một bài khá quen thuộc ,cần dùng thêm định lí Fubini để giải:
Bài 50:Cho hàm $f:[0;1]\to [0,+\infty)$ khả vi liên tục trên miền xác định.Đặt $M=\displaystyle \max_{x\in [0;1]} |f'(x)|$Chứng minh rằng:$$\left |\int_0^1 f^3\left(x\right)dx-f^2\left(0\right)\int_0^1 f\left(x\right)dx\right|\leq M \left(\int_0^1 f\left(x\right)dx\right)^2$$
Với $t\in [0;1]$ ta có:
\[ - M \le f'\left( t \right) \le M\]
\[ \Leftrightarrow - M.f\left( t \right) \le f'\left( t \right).f\left( t \right) \le M.f\left( t \right)\]
\[ \Leftrightarrow - M.\int\limits_0^x {f\left( t \right)dt} \le \int\limits_0^x {f'\left( t \right).f\left( t \right)dt} \le M.\int\limits_0^x {f\left( t \right)dt} \,\,\forall x \in \left( {0;1} \right)\]
\[ \Leftrightarrow - M.\int\limits_0^x {f\left( t \right)dt} \le \frac{1}{2}\left[ {{f^2}\left( x \right) - {f^2}\left( 0 \right)} \right] \le M.\int\limits_0^x {f\left( t \right)dt} \]
\[ \Rightarrow - M.f\left( x \right)\int\limits_0^x {f\left( t \right)dt} \le \frac{1}{2}\left( {{f^3}\left( x \right) - {f^2}\left( 0 \right).f\left( x \right)} \right) \le M.f\left( x \right)\int\limits_0^x {f\left( t \right)dt} \]
\[ \Rightarrow - M.\int\limits_0^1 {\left( {f\left( x \right)\int\limits_0^x {f\left( t \right)dt} } \right)dx} \le \frac{1}{2}\int\limits_0^1 {\left( {{f^3}\left( x \right) - {f^2}\left( 0 \right).f\left( x \right)} \right)dx} \le M.\int\limits_0^1 {\left( {f\left( x \right)\int\limits_0^x {f\left( t \right)dt} } \right)dx} \]
Mặt khác ta có:
\[\int\limits_0^1 {\left( {f\left( x \right)\int\limits_0^x {f\left( t \right)dt} } \right)dx} = \int\limits_0^1 {\left( {\int\limits_0^x {f\left( t \right)dt} } \right)d\left( {\int\limits_0^x {f\left( t \right)dt} } \right) = \frac{1}{2}} {\left( {\int_0^1 f \left( x \right)dx} \right)^2}\]
Nên ta có đpcm.