Proving $\int_{a}^{b}\vert P^{\prime}_{n}(x)\vert\text{d}x\leq 2n \max_{a\leqslant x\leqslant b}\vert P_{n}(x)\vert $ So Assume that $P_{n}(x)$ is an algebraic polynomial of degree n. Prove that
$$\int_{a}^{b}\vert P^{\prime}_{n}(x)\vert\;\text{d}x\leq 2n\cdot\max_{a\leqslant x\leqslant b}\vert P_{n}(x)\vert$$
Please use the method of variable limit integral
I know A.A.Markoff$\;\;$Theorem——The above conditions are the same
$$\vert P^{\prime}_{n}(x)\vert\leq\dfrac{\displaystyle 2\max_{a\leqslant x\leqslant b}\vert P_{n}(x)\vert\cdot n^2}{b-a}
$$
So when you integrate the inequality on both sides from a to b, you can get
$$\int_{a}^{b}\vert P^{\prime}_{n}(x)\vert\;\text{d}x\leq 2n^2\cdot\max_{a\leqslant x\leqslant b}\vert P_{n}(x)\vert$$
That is so close to the final integral, but it's not
So please help me,by the method of variable limit integral(That's the hint of the question)
Thank you very much
 A: I'm not sure exactly what you mean by "the method of variable limit integral", but you may be able to modify this proof to do it that way:
The degree of $P_n'(t)$ is $n-1$; let $\{r_1,r_2,\dots,r_{n-1}\}$ be the complete set of roots (some may be repeated). Now let $t_0=a$, let $t_1<t_2<\dots<t_{m-1}$ be the ordered list of the distinct elements in $\{r_1,r_2,\dots,r_{n-1}\}\cap(a,b)$, and let $t_m=b$. Since $P_n$ is continuous, $t_1,\dots,t_{m-1}$ are exactly the critical values of $P_n$ in $(a,b)$ and therefore $P_n$ is monotone on each interval $(t_i,t_{i+1})$. That is, there is a constant $C\in\{-1,1\}$ such that both $|P_n(t_{i+1})-P_n(t_i)|=C\cdot[P_n(t_{i+1})-P_n(t_i)]$ and $|P_n'(t)|=C\cdot P_n'(t)$ hold for all $t\in(t_i,t_{i+1})$, and hence 
\begin{align}
\int_{t_i}^{t_{i+1}}|P_n'(t)|\,dt&=C\cdot\int_{t_i}^{t_{i+1}}P_n'(t)\,dt\\
&=C\cdot[P_n(t_{i+1})-P_n(t_i)]\\
&=|P_n(t_{i+1})-P_n(t_i)|\\
&\leq|P_n(t_{i+1})|+|P_n(t_i)|\leq 2\cdot\max_{a\leq x\leq b}|P_n(x)|.
\end{align}
This common bound holds for every $i\in\{0,1,\dots,m-1\}$, so we finally have $$\int_a^b|P_n'(t)|\,dt=\sum_{i=0}^{m-1}\int_{t_i}^{t_{i+1}}|P_n'(t)|\,dt\leq m\cdot2\cdot\max_{a\leq x\leq b}|P_n(x)|\leq 2n\cdot\max_{a\leq x\leq b}|P_n(x)|.$$
