一道积分不等式的最优估计探索

此题由网友“时间都知道”向我提问,为忽悠死他某年的考研题,原题如下
设$f(x)$在区间$[0,1]$上二阶连续可微,则$${\int }_0^1|f^{\prime }(x)|dx\le 9{\int }_0^1|f(x)|dx+{\int }_0^1|f^{\prime \prime }(x)|dx$$
\qquad原题倒不算难,考虑在$[0,\frac{1}{3}]$与$[\frac{2}{3},1]$中各取一点$x_1$,$x_2$,利用$Lagrange$放缩,具体不予以赘述.
\qquad关于此题,9显然不是最优秀的,如何寻找最优秀的那个估计系数才是值得思考的.后来我拿这题请教专门研究$pde$方向的白老师,他说这是$pde$中内插不等式一个常用结论,看来$pde$确实是一门有趣的学科,有时间一定要好好学学.
\qquad下面是个人对最优系数探索的思考.
$Solve$:
\qquad首先待定$x_0$:$$|f^{\prime }(x)|=|f^{\prime }(x_0)+{\int }_{x_0}^x{f}^{\prime \prime }(t)dt|\le |f^{\prime }(x_0)|+{\int }_0^1|f^{\prime \prime }(x)|dx$$\qquad寻找$\alpha$:$|f^{\prime }(x_0)|\le \alpha {\int }_0^1|f(x)|dx$达到最优
\qquad为了使$\alpha$尽可能小,自然$|f^{\prime }(x_0)|$越小越好
\qquad不妨设$|f^{\prime }(x_0)|=\underset{[0,1]}{\min }|f^{\prime }(x)|$
\qquad再欲用$Lagrange$引进$f(x)$,待定$x_1$$$|f(x)|=|f(x_1)+f^{\prime }(\xi )(x-x_1)|$$\qquad希望$|f(x_1)+f^{\prime }(\xi )(x-x_1)|=|f(x_1)|+|f^{\prime }(\xi )(x-x_1)|$
\qquad这样才有$|f(x)|\ge |f^{\prime }(\xi )||x-x_1|\ge |f^{\prime }(x_0)||x-x_1|$
\qquad只需取$|f(x_1)|=\underset{[0,1]}{\min }|f(x)|$即可
\qquad此时$${\int }_0^1|f(x)|dx\ge |f^{\prime }(x_0)|{\int }_0^1|x-x_1|dx=|f^{\prime }(x_0)|(\frac{x_1^2}{2}+\frac{(1-x_1{)}^2}{2})$$\qquad而$$\frac{x_1^2}{2}+\frac{(1-x_1{)}^2}{2}\ge \frac{1}{4}$$\qquad所以$\alpha =4$.又$f(x)=x-1/2$时有等号成立,故$\alpha =4$确实最优.