Polynomial inequalities of the form $\int_0^1 |f(x)|^2 x \,dx \geq C \int_0^1 |f(x)|^2 \,dx$

Question

Let $P_n$ denote all (real or complex) polynomials $f(x)=\sum_{k=0}^n a_k x^k$. I'm interested in inequalities of the form $$ \int_0^1 |f(x)|^2 x \,dx \geq C \int_0^1 |f(x)|^2 \, dx, \quad \text{for all } f \in P_n. $$ Are there results about the behavior of $C$, say, for increasing $n$? Or are there explicit expressions for such a $C$ (assuming we work with the best possible $C$)?

Answer 1

It is about roots of Legendre polynomials. The constants in real and complex cases are equal. I redenote the space of polynomials of degree at most $n$ by $\mathcal{P}_n$, since $P_n$ are standard notations for Legendre polynomials.

Let $L_{n+1}$ be a monic polynomial of degree $n+1$, which is orthogonal to $\mathcal{P}_n$ with weight 1 (i.e., $\langle L_{n+1},g\rangle:=\int_0^1 L_{n+1}(x)\overline{g(x)}dx=0$ for all $g\in \mathcal{P}_n$). Then your $C$ is the minimal root of $L_{n+1}$. In more standard notations, we have $L_{n+1}(x)=c_{n+1}P_{n+1}(2x-1)$, where $P_n$ is Legendre polynomial of degree $n+1$ and $c_{n+1}$ is a positive constant which we do not care on. Then, if $\xi_1<\xi_2<\ldots<\xi_{n+1}\in [-1,1]$ are the roots of $P_{n+1}$, your $C$ equals $(\xi_1+1)/2$.

Fix $\int_0^1 |f|^2(x)dx=1$ and look at $f$ for which $\int_0^1 x|f|^2(x)dx$ is minimal, it exists by compactness. By homogeneity, this $f$ minimizes the ratio $\Phi(f):=\frac{\int_0^1 x|f|^2(x)dx}{\int_0^1 |f|^2(x)dx}$ over the set $\mathcal{P}_n\setminus\{0\}$. Thus, for every $h\in \mathcal{P}_n$, we have $(\frac{d}{dt}\Phi(f+th))|_{t=0}=0$.

If $h\in \mathcal{P}_n$ is orthogonal to $f$, i.e. $\int_0^1f(x)\overline{h(x)}dx=0$, then the derivative (with respect to $t$) of $\int_0^1 |f+th|^2(x)dx$ at $t=0$ equals 0. Therefore, the derivative of $\int_0^1 x|f+th|^2(x)dx$ also must be equal to 0, and we have $\Re \langle xf(x),h\rangle=0$. Replacing here $h$ to $ih$ we get $\Im \langle xf(x),h\rangle=0$. So, $h\perp xf(x)$.

That is, orthogonal hyperplanes to elements $f$ and $xf$ in $\mathcal{P}_n$ coincide. But $xf$ possibly does not belong to $\mathcal{P}_n$, and we can not conclude that $xf$ and $f$ are proportional. To resolve this, let $A$ be a coefficient of $x^n$ in $f$, then the polynomial $xf-AL_{n+1}\in \mathcal{P}_n$ has the same orthogonal complement in $\mathcal{P}_n$ as $f$. Thus, we have $xf-AL_{n+1}=\lambda f$ for certain complex $\lambda$, $AL_{n+1}=f(x)(x-\lambda)$, and thus $\lambda$ is a root of $L_{n+1}$, in particular $\lambda\in (0,1)$. Finally, we have $x|f|^2=A\bar{f}L_{n+1}+\lambda |f|^2$, thus $\int_0^1 x|f|^2dx=\lambda$. Therefore, the answer is the minimal root of $L_{n+1}$.

Answer 2

I think it is all about convex hulls of space curves :), and for the special case of the moment curve $\gamma(t) = (t,t^{2},\ldots, t^{2n+1})$ you get the smallest positive root of Legendre polynomial of degree $n+1$.

Every interior point of the convex hull of the set $\gamma([0,1])$ can be written as a convex combination of $n+1$ points of $\gamma([0,1])$, say $\gamma(t_{1}), \ldots, \gamma(t_{n+1})$. Moreover, $n+1$ is the smallest such number, also known as Carathéodory number of the convex set $\mathrm{conv(}\gamma([0,1])$). Notice that Carathéodory's theorem would give at most $2n+2$ points, but for this special moment curve one gets twice less points because it has totally positive torsion, see this paper for details. Additionally, such a representation is aways unique and also $0<t_{1}<t_{2}<\ldots<t_{n+1}<1$. In particular, we can write \begin{align*} \int_{0}^{1}\gamma(t)dt = \sum_{k=1}^{n+1} \alpha_{k} \gamma(t_{k}) \quad(*) \end{align*} for some $\alpha_{k} \in (0,1)$ with $\sum \alpha_{k}=1$ and $0<t_{1}<\ldots<t_{n+1}<1$. The identity (*) implies that $\int_{0}^{1}Q(t)dt = \sum_{k=1}^{n+1}\alpha_{k} Q(t_{k})$ for any polynomial $Q(t)$ of degree at most $2n+1$. In particular, applying it to $n+1$ degree orthogonal polynomial $L_{n+1}$ we get $$ 0=\int_{0}^{1}t^{\ell}L_{n+1}(t)dt=\sum_{k=1}^{n+1}\alpha_{k}t_{k}^{\ell}L_{n+1}(t_{k})=0, \quad \ell=0, 1, \ldots, n $$ Since $\det(\{t_{k}^{\ell}\}_{k, \ell=1}^{n+1})\neq 0$ we get $L_{n+1}(t_{k})=0$, i.e., $t_{k}$'s are the roots of $n+1$ degree Legendre polynomials.

This is exactly, what is called, Gauss–Legendre quadrature, where $\alpha_{k}$ are called weights, and $t_{k}$ - the nodes. In particular, if $P(t)$ has degree at most $n$ we have \begin{align*} \frac{\int_{0}^{1}x|P(x)|^{2}dx}{\int_{0}^{1}|P(x)|^{2}dx} = \sum_{k=1}^{n+1}t_{k}\frac{\alpha_{k}|P(t_{k})|^{2}}{\sum_{j=1}^{n+1}\alpha_{j}|P(t_{j})|^{2}}\geq t_{1}. \end{align*}

Finally, choosing any not identically zero degree $n$ polynomial $P$ with $P(t_{2})=\ldots=P(t_{n+1})=0$ gives the sharpness of the result.

The above argument works verbatim if you replace $dt$ by any probability measure $d\mu(t)$ with bounded support in $[0, \infty)$ and $\int_{0}^{\infty} \gamma(t)d\mu(t) \in \mathrm{int}( \mathrm{conv(}\gamma(\mathrm{conv}(\mathrm{supp}(\mu))))$. In that case $t_{1}$ will be smallest positive zero of $n+1$ degree orthogonal polynomial $L_{n+1}$, i.e., orthogonal polynomials with respect to the measure $d\mu(t)$.