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
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 $$\int_a^b|P_n'(t)|\,dt=\sum_{i=0}^{m-1}\int_{t_i}^{t_{i+1}}|P_n'(t)|\,dt\leq m\cdot2\cdot\max_{a\leq x\leq b}|P_n(x)|\leq 2n\cdot\max_{a\leq x\leq b}|P_n(x)|.$$