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I am stuck with the following problem:

With $\omega: [-1,1]\rightarrow \mathbb{R}$, $\omega\in C^n(-1,1)$. Suppose that $\omega$ has a finite number of zeroes $t_1<t_2<\cdots <t_n$ (i.e. $\omega(t_i)=0,\forall i$) on $[-1,1]$. Prove that $$\left\vert\int_{-1}^1 \omega(t) dt \right\vert \leq 2^n \int_{-1}^1\vert \omega^{(n)}(t)\vert dt$$

I think I should show it inductively, but I can figure out how to do it. If someone could give me some hints that would be greatly appreciated.

Luc M
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    If $\omega(t) = 1$, then $\omega^{(n)}(t) = 0$ for $n \ge 1$. Thus, $\displaystyle\left|\int_{-1}^{1}\omega(t),dt\right| = 2$, but $2^n\displaystyle\int_{-1}^{1}\left|\omega^{(n)}(t)\right|,dt = 0$. Are you sure the problem statement is correct, or am I missing something? – JimmyK4542 Jun 09 '14 at 20:14
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    Perhaps $\omega$ is supposed to have at least one zero? (But even that does not help against polynomials of degree less than $n$.) – Harald Hanche-Olsen Jun 09 '14 at 20:18
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    Say $\omega(t) = t^2$. Then $\omega^{(n)}(t) = 0$ for $n \ge 3$, and we get the same problem. Also, if $\omega(t)$ is a polynomial of degree $d$, then $\omega^{(n)}(t) = 0$ for $n \ge d+1$, so having more zeros won't help. – JimmyK4542 Jun 09 '14 at 20:24
  • I have an other statement of the problem which adds $t_i\in[-1,1]$ such that $t_1<t_2<\cdots <t_n$ and $\omega(t_i)=0,\forall i$. – Luc M Jun 09 '14 at 20:37
  • I had an idea to start with the case $n=1$: if $\omega$ has a finite number of zero, there is at least one $x\in[-1,1]$ such that $\omega(x)\ne 0$. Therefore if the closest zero to $x$ is $t_i>x$ then there exist $c\in (x,t_i)$ such that $\omega'(c)\ne 0$. – Luc M Jun 09 '14 at 20:49
  • @JimmyK4542 you are right, there is a problem with the statement I gave. I think there must be exactly $n$ zeroes. Does that make more sense ? – Luc M Jun 09 '14 at 20:52

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Applying Rolle's theorem an awful lot of times, you will find that all derivatives of $\omega$ of orders $\le n-1$ have at least one zero in $[-1,1]$. This allows you to run the following, outrageously wasteful, chain of inequalities for $k=0,1,\dots, n-1 $:

$$\int_{-1}^1 |\omega^{(k)}(x)|\,dx \le 2 \sup_{[-1,1]} |\omega^{(k)}(x)| \le 2\int_{-1}^1 |\omega^{(k+1)}(x)|\,dx$$ where the last step uses the fundamental theorem of calculus: $\omega^{(k)}(x)$ is given by the integral of $\omega^{(k+1)} $ from some zero of $\omega^{(k)}$ to $x$.