# An Integral Inequality #1

• May 4th 2008, 10:03 PM
Xingyuan
An Integral Inequality #1
Assume $\displaystyle f \in C^{(1)} ([0,1])$,and $\displaystyle f(0)=0,f(1)=1,$ then

$\displaystyle \int_0^1 |f(x)-f'(x)| dx \geq \frac{1}{e}$

thanks very much (Headbang)
• May 5th 2008, 12:12 AM
Isomorphism
Quote:

Originally Posted by Xingyuan
Assume $\displaystyle f \in C^{(1)} ([0,1])$,and $\displaystyle f(0)=0,f(1)=1,$ then

$\displaystyle \int_0^1 |f(x)-f'(x)| dx \geq \frac{1}{e}$

thanks very much (Headbang)

Use the fact that $\displaystyle \forall x \in [0,1], |e^{-x}(f(x) - f'(x))| \leq |f(x) - f'(x)|$

$\displaystyle \Rightarrow \int^1_0 |e^{-x}(f(x) - f'(x))|\, dx \leq \int^1_0|f(x) - f'(x)|\, dx$

$\displaystyle \Rightarrow \int^1_0 d(|e^{-x}f(x)|) \leq \int^1_0|f(x) - f'(x)|\, dx$

$\displaystyle \Rightarrow |e^{-x}f(x)|\bigg{|}_{0}^{1} \leq \int^1_0|f(x) - f'(x)|\, dx$

$\displaystyle \Rightarrow |e^{-1}f(1)| - |e^{-0}f(0)| \leq \int^1_0|f(x) - f'(x)|\, dx$

$\displaystyle \Rightarrow \int^1_0|f(x) - f'(x)|\, dx \geq \frac1{e}$