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Consider the space $P^n[0,1]$ of polynomials of degree $\leq n$ on $[0,1]$, equipped with the sup norm. Now, this is a finite dimensional space, so all linear operators have to be continuous, hence bounded.

My question is: what is the norm of the differential operator $d/dx : P^n[0,1] \rightarrow P^{n-1}[0,1]$? I can't see the relation between the supremum of a polynomial $p(x)=a_0+\dots+a_nx^n$ and the supremum of $p'(x)=a_1 + 2a_2 x +\dots+n a_n x^{n-1}$.

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  • $\begingroup$ Do you want the exact norm or some bound? $\endgroup$ Jan 22 '14 at 18:35
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    $\begingroup$ Such results are called Bernstein inequalities. For trigonometric polyomials the norm is $n$, and the only inequality that I remember is $|P'(x)| \leq ||P|| n /(\sqrt{x(1-x)})$. That does not give any bounds, but, maybe, it would be helpfull. Bernstein inequalities (and thus concrete norms) are known in a large number of cases. and I can not google the one you need :) $\endgroup$
    – user68061
    Jan 22 '14 at 18:58
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The answer is given here for the interval $[-1,1]$ by Markov's inequality for polynomials: The norm is $n^2$ and $||P'||_\infty=n^2||P||_\infty$ for the $n$-th Chebyshev polynomial $T_n$. For the interval $[0,1]$, we have to use scaling: If $p$ is a polynomial considered on $[0,1]$, then $q(x)=p(\frac{x+1}2)$ is a polynomial considered on $[-1,1]$. Hence $||q'||_{\infty,[-1,1]}\leq n^2||q||_{\infty,[-1,1]}$. Since $q'(x)=\frac12 p'(\frac{x+1}2)$, we find $||p'||_{\infty,[0,1]}\leq 2n^2||p||_{\infty,[0,1]}$ and equality is attained for $t_n(x)=T_n(2x-1)$.

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