# L1 analog of Bernstein's inequality

Let $p(x)$ be a degree $n$ polynomial over $[-1, 1]$, and let $q(x) = p'(x) \sqrt{1-x^2}$. Is it true that $$\|q\|_1 \leq O(n) \|p\|_1$$ where we define $\|f\|_p := \left(\int_{-1}^1 |f(x)|^pdx\right)^{1/p}$?

For reference, Bernstein's inequality shows that $$\|q\|_\infty \leq n\|p\|_\infty$$ with equality at the $n$th Chebyshev polynomial, and looking at Legendre polynomials shows that $$\|q\|_2 \leq \sqrt{n(n+1)}\|p\|_2$$ with equality at the $n$th Legendre polynomial.

should be a good source for your question. In particular, (A.4.22) gives $$\|P'\|_p\leq cn^2\|P\|_p,$$ for every polynomial $P$ of degree $n$ and $0<p<\infty$. Apparently, finding the best possible constant $c$ is still an open problem.
• Interesting, thanks. Unfortunately, I don't see techniques there that would even let me prove the $L_2$ version, which I know is true---I really want that $P'$ is $O(n)$, not $O(n^2)$, for most of the interval. – Eric Price Jul 1 '16 at 18:41
• The inequality $$\|P'_n(t)(1-t^2)^{1/2}\|_p\leq Cn\|P_n\|_p$$, $0<p<\infty$, indeed holds true. This is a particular case of Theorem 5 in P. Nevai, Bernsteĭn's inequality in $L^p$ for $0<p<1$,. J. Approx. Theory 27 (1979), no. 3, 239–243. – user111 Jul 1 '16 at 20:45