已知$f(x)$在$[0,1]$上二阶可导，求证：$\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x$
证明:
对任意$0<\xi<\dfrac{1}{3}$和$\dfrac{2}{3}<\eta<1$,则存在$\lambda\in(\xi,\eta)$,使得
\begin{equation*}
 |f'(\lambda)|=\left|\frac{f(\eta)-f(\xi)}{\eta-\xi}\right|\leq 3|f(\xi)|+3|f(\eta)|
\end{equation*}
因此对任意的$x\in(0,1)$成立
\begin{equation*}
    |f'(x)|=\left|f'(\lambda)+\int_\lambda^x f''(t)dt\right|\leq 3|f(\xi)|+3|f(\eta)|+\int_0^1|f''(t)|dt
\end{equation*}
分别对$\xi$在$(0,\dfrac{1}{3})$上和对$\eta$在$(\dfrac{2}{3},1)$上积分以上不等式，得
\begin{align*}
  \frac{1}{9}|f'(x)|&\leq \int_0^{\frac{1}{3}}|f(\xi)|d\xi+\int_{\frac{2}{3}}^1 |f(\eta)|d\eta+
  \frac{1}{9}\int_0^1|f''(t)|dt\\
  &\leq \int_0^1|f(t)|dt+\frac{1}{9}\int_0^1|f''(t)|dt
\end{align*}
于是
\begin{equation*}
    |f'(x)|\leq 9\int_0^1|f(t)|dt+\int_0^1|f''(t)|dt,\hspace{10mm}x\in[0,1]
\end{equation*}
对上式两边在$[0,1]$积分,得到
\begin{equation*}
    \int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x
\end{equation*}
$\square$
问题似乎可以加强到
\[\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 4\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x\]