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Consider the operator $M$ acting on the space $\mathbb{R}[X]$ of real polynomials by $Mp(x)=xp(x)$. We equip $\mathbb R[X]$ with the $L^2$ norm $$ \|p\|^2=\int p(x)^2d\mu(x), $$ where $\mu$ is a Radon measure having all its moments, namely $\mathbb R[X]\subset L^2(\mu)$. The question is : Can we find a nice formula for the operator norm $\|M\|_{op}$ ?

EDIT : The answer may be $+\infty$ for some $\mu$, see the comment below from Davide Giraudo.

In order to have a chance to obtain a formula in this general setting, I change the question by asking if one can find a ($n,\mu$-dependent) expression for $$ \sup \int x^2p(x)^2d\mu(x) $$ where the supremun is taken over all polynomials of degree less that $n$ satisfying $$ \int p(x)^2d\mu(x)\leq 1. $$

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Are you sure that $M$ is continuous? If we take $\mu =e^{-|x|}d\lambda$, where $\lambda$ is Lebesgue measure, then $||p(x^k)||^2=||x^{k+1}||^2=\int x^{2k+2}e^{-|x|}dx=2\int_0^{+\infty}x^{2k+2}e^{-x}dx=2(2k+2)!$ but $||x^k||^2=2(2k)!$. – Davide Giraudo Dec 21 '11 at 12:27
Good point, let me change my question then. – Student Dec 21 '11 at 12:31
You can construct an orthonormal basis $\{p_k\}$ for $L^(\mu)$, with $deg p_k=k$. If $deg p\leq n$, $p=\sum_{j=0}^na_jp_j$, and $||p||\leq 1$ is equivalent to $\sum_{j=0}^na_j^2\leq 1$. $\int x^2p(x)^2d\mu=\sum_{1\leq i,j\leq n}\int xa_ip_ixa_jp_j\leq (\sum_{i=1}^n|a_i|\sqrt{\int x^2p_i^2d\mu})^2\leq \sum_{i=0}^n\int p_i(x)^2x^2d\mu.$ – Davide Giraudo Dec 21 '11 at 12:59
Thx Davide, I think I was looking for something like that. If you post it as an answer, I'll take it. – Student Dec 21 '11 at 13:08
Ok, I will do that. – Davide Giraudo Dec 21 '11 at 13:09
up vote 3 down vote accepted

We consider $M_n\colon\mathbb R_n[X]\to\mathbb R[X]$ defined by $M_n(p)(x)=xp(x)$. We want to get a bound for $\lVert M_n\rVert=\sup_{\lVert p\rVert=1,p\in\mathbb R_n[X]}\lVert M_n\rVert$. We can construct $\{p_k\}$, an orthonormal sequence of polynomials such that $\operatorname{deg}p_k=k$. Let $p\in\mathbb R_n[X]$, $\lVert p\rVert=1$. We can write $p=\sum_{j=0}^na_jp_j$ where $a_j$ are real numbers, and $||p||=1$ means $\sum_{j=0}^na_j^2=1$. We have \begin{align*} ||M_n(p)||^2&=\int_{\mathbb R}x^2p(x)^2d\mu(x)\\ &=\sum_{i=0}^n\sum_{j=0}^na_ia_j\int_{\mathbb R}xp_i(x)xp_j(x)d\mu(x)\\ &\leq \sum_{i=0}^n\sum_{j=0}^n|a_i||a_j|\left(\int_{\mathbb R}x^2p_i(x)^2d\mu\right)^{1/2}\left(\int_{\mathbb R}x^2p_j(x)^2d\mu(x)\right)^{1/2}\\ &=\left(\sum_{j=0}^n|a_j|\left(\int_{\mathbb R}x^2p_j(x)^2d\mu\right)^{1/2}\right)^2\\ &\leq \sum_{j=0}^n|a_j|^2\sum_{j=1}^n\int_{\mathbb R}x^2p_j(x)^2d\mu\\ &=\sum_{j=0}^n\int_{\mathbb R}x^2p_j(x)^2d\mu, \end{align*} which proves that $||M_n||\leq \sum_{j=0}^n\int_{\mathbb R}x^2p_j(x)^2d\mu$.

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