设$f(x)$是$[0,1]$上的非负凹函数，则有

$$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$

传说中的Favard不等式(1933年的)，找了好久没找到证明，只谷歌到一篇介绍推广的文章《Some New Results Related to Favard’s Inequality》

好像在Mathematic in Science and Engineering, vol 187,1992年那本上有，可惜下载不到。后来发现《常用不等式》第4版452页的Berwarld不等式就是这个的推广.（首先没看出来，那个范数不太懂)

解答来着风碎便士同学。参看博士家园

http://www.math.org.cn/forum.php?mod=viewthread&tid=26438&extra=page%3D1

With out loss of generally,we consider the case $f(0)=f(1)=0$,and $f(x)$ has order continuous derivative,Therefore
$$f''(x)\leq 0 $$
Thus
$$ f(x)=-\int_{0}^{1}{K(x,t)f''(t)dt} $$
Where $K(x,t)$ is Green function.
$$ K(x,t)=\left\{
  \begin{array}{ll}
   t(1-x)  & \hbox{$0\leq t\le x\le 1$} \\
    x(1-t)  & \hbox{$0\leq x\le t\le 1$}
  \end{array}
\right.
$$
Then by Minkowski inequality,we have
\begin{align}
\left(\int_{0}^{1}{f^{p}(x)dx}\right)^{\frac{1}{p}}&=\left(\int_{0}^{1}{\left(\int_{0}^{1}{K(x,t)(-f''(t))dt}\right)^{p}dx     }\right)^{\frac{1}{p}}\\
&\leq \int_{0}^{1}{\left(\int_{0}^{1}{K^{p}(x,t)(-f''(t))^{p}dx}\right)dt}\\
&=\frac{1}{(p+1)^{\frac{1}{p}}}\int_{0}^{1}{t(1-t)|f''(t)|dt}
\end{align}
On the other hand
\begin{align}
\int_{0}^{1}{f(x)dx}&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dt}  dx}\\
&=-\int_{0}^{1}{\int_{0}^{1}{K(x,t)f''(t)dx}  dt}\\
&=-\frac{1}{2}\int_{0}^{1}{t(1-t)f''(t)dt} 
\end{align}
Therefore
$$ \int_{0}^{1}{f^{p}(x)dx}\leq \frac{2^p}{p+1}\left(\int_{0}^{1}{f(x)dx}\right)^p $$