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Assume $f(x)\in\Bbb{R}[x]$ is a polynomial of degree $n$.

Question. If $\int_{-1}^1f^2(x)\,dx=1$, is it true that $$\vert f(x)\vert\leq \frac1{\sqrt2}(n+1), \qquad \text{for $\vert x\vert\leq1$}\,\,\,?$$

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    $\begingroup$ You would get some sort of trivial bound if you knew the sup norm of the Legendre polynomials, which I don't know but I assume must be known. I wonder if it recovers your bound? $\endgroup$ Sep 7, 2017 at 16:33
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    $\begingroup$ @NateEldredge math.stackexchange.com/questions/417999 $\endgroup$ Sep 7, 2017 at 18:21
  • $\begingroup$ Alternative reformulation: $\sum_{k=0}^n P_k^2\leqslant \frac{n+1}2$ on $[-1,1]$, where $P_i$ are orthonormal Legendre polynomials. $\endgroup$ Sep 7, 2017 at 19:49
  • $\begingroup$ @NateEldredge Why are legendre polynomials extremal? $\endgroup$
    – Igor Rivin
    Sep 7, 2017 at 22:05
  • $\begingroup$ @IgorRivin: All I mean is that since they are orthonormal, we can write $f = \sum_{k=0}^n a_k f_k$ where $\sum_{k=0}^n a_k^2 = 1$, so for instance we get trivial bounds like $\|f\|_{\infty} \le \sum_{k=0}^n \|f_k\|_{\infty}$ or $\|f\|_\infty^2 \le \sum_{k=0}^n \|f_k\|_\infty^2$. $\endgroup$ Sep 7, 2017 at 22:07

3 Answers 3

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This is problem VI.103 in volume 2 of Polya and Szego, where they also characterize the extremal polynomials.

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    $\begingroup$ These two incredible volumes by Polya and Szegö deserve to be better known! $\endgroup$ Sep 7, 2017 at 23:09
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Inequalities between different $L^{p}$ norms of the type $\|q_{n}\|_{p}\leq C_{n}\|q_{n}\|_{q}$, $0<q<p\leq\infty$, where $q_{n}$ is a polynomial of degree $n$ are sometimes called Nikolskii's inequalities, see p.102 of

R.A. DeVore; G.G. Lorentz, Constructive approximation, Springer-Verlag, Berlin, 1993.

For the norm $$\|f\|_{p}=\left(\int_{a}^{b}|f|^{p}dx\right)^{1/p},$$ on an interval $[a,b]$, one has $$\|q_{n}\|_{p}\leq(2(q+1)n^{2}/(b-a))^{1/q-1/p}\|q_{n}\|_{q}.$$ The proof essentially uses Markov inequality. For $[a,b]=[-1,1]$, $q=2$ and $p=\infty$, it would give $$\|q_{n}\|_{\infty}\leq\sqrt{3}n\|q_{n}\|_{2}.$$ About the Legendre polynomials discussed in the comments, it is known that the sum $\sum_{k=0}^{n}P_{k}^{2}(x)$, that is the reciprocal of the so-called Christoffel function, behaves, for $n$ large, like $n/(\pi\sqrt{1-x^{2}})$ for $x\in(-1,1)$, and behaves like $Cn^{2}$, $C$ some constant, at the endpoints $\pm1$.

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This may not be of interest to you, but the weaker bound $|f(x)|\le (2+o(1))n$ has an easy proof, based on Bernstein's inequality $$ |f'(x)| \le \frac{n}{\sqrt{1-x^2}} \|f\|_{\infty} . \quad\quad\quad\quad (B) $$ This puts limits on how fast $f$ can decay near its maximum, and the worst case scenario occurs when the maximum is at $\pm 1$ (let's say at $x=1$). If we write $(1-x^2)^{-1/2}=(x+1)^{1/2}(1-x)^{1/2}$ and approximate the first factor by $\sqrt{2}$ near $x=1$, then integrating (B) gives that $|f(x)|\ge n\|f\|_{\infty}\sqrt{2(x-1+1/2n^2)}$ on $1-1/2n^2\le x\le 1$. This makes a contribution $\ge \|f\|_{\infty}^2/(4n^2)$ to the $L^2$ norm, so $\|f\|_{\infty}\le 2n$, as claimed. (Doing it properly produces the $o(1)$ term.)

The constant $2$ could be improved somewhat since (B) also holds with $n^2$ in place of $n/\sqrt{1-x^2}$, but I don't think I could get close to the conjectured optimal value $1/\sqrt{2}$ in this way.

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