Consider the set of polynomials with real coefficients as a vector space with the following inner-product: $\langle f, g \rangle = \int_{a}^{b} f(x)g(x) dx$.

Hilbert showed, in a paper from 1894, that the norm (with respect to this inner-product) of a non-zero polynomial in $\mathbb{Z}[X]$ can get arbitrarily small when $b-a < 4$. In other words, $\min_{0 \neq p \in \mathbb{Z}[X]} \int_{a}^{b} p^2(x) dx = 0$.

My questions are:

  1. What happens when $b-a \ge 4$? Can the norm get arbitrarily small in this case? Or is there some (positive) lower bound for $\min_{0 \neq p \in \mathbb{Z}[X]} \int_{a}^{b} p^2(x) dx$?

  2. When $b-a < 4$, is there an explicit construction of a sequence $p_n \in \mathbb{Z}[X],n\ge 1$, with norm tending to 0?

I posted similar questions in MSE and got no responses.

(Hilbert's proof was as follows: Minimizing the norm for (non-zero) polynomials of degree less than $n$ is equivalent to minimizing a certain positive-definite quadratic form. The corresponding matrix $A_n$ has entries $a_{i,j} = \langle x^i, x^j \rangle = \frac{b^{i+j+1} - a^{i+j+1}}{i+j+1}, 0 \le i,j \le n-1$. A calculation using an orthonormal basis for our inner-product space shows that $\det A_n = (\frac{b-a}{4})^{n^2}n^{-1/4} (2 \pi)^n c_n$ where $c_n$ converges to a positive constant. A result by Minkowski shows that in general, the minimal value of a positive quadratic form $\langle v, Av \rangle$ in $n$ variables is at most $n (\det A)^{1/n}$. Since $\lim_{n} n (\det A_n)^{1/n} = 0$ for $b-a<4$, the result follows.)

  • $\begingroup$ This problem might be answered with potential theory methods .note that logarithmic capacity of a segment is a quarter of its length. $\endgroup$
    – BigM
    Mar 20 '14 at 18:24
  • $\begingroup$ Dear @user48438, can you please add some details? Thank you very much. $\endgroup$ Mar 23 '14 at 20:02
  • $\begingroup$ Honestly I am not aware how to relate the posted problem to potential theory machinery. however I have the feeling that it can be done via potential theory machinery. you can take a look at books.google.com/books/about/… $\endgroup$
    – BigM
    Mar 31 '14 at 20:42

For Question 1: Once $b-a \geq 4$, the norm cannot get arbitrarily small.

Suppose $p(x)$ has leading term $c_n x^n$. The minimum of $\int_a^b p^2(x) \, dx$ over all polynomials $p(x) = c_n x^n + O(x^{n-1})$ with real coefficients is a multiple of $P_n(l(x))$, where $P_n$ is the $n$-th Legendre polynomial and $l$ is the affine-linear transformation $l(x) = (2x-(a+b))/(b-a)$ that takes $(a,b)$ to $(-1,1)$. Now $P(x/2)$ has leading coefficient $$ \lambda_n := \frac12 \frac34 \frac56 \cdots \frac{2n-1}{2n} \sim \frac1{\sqrt{\pi n}} $$ and norm $\int_{-2}^2 P(x/2)^2 \, dx = 4/(2n+1)$. Thus in general $$ \int_a^b p^2(x) \, dx \geq \frac4{(2n+1)\lambda_n^2} \left(\frac{b-a}{4}\right)^{\!2n+1} c_n^2, $$ with the factor $r_n := 4/(2n+1)\lambda_n^2$ approaching $2\pi$ as $n \rightarrow \infty$. Since $|c_n| \geq 1$, we deduce a lower bound on $\int_a^b p^2(x) \, dx$ that approaches $2\pi$ when $b-a=4$, and increases exponentially with $n$ when $b-a > 4$. [Added later, in connection with the recent MO Question 188807: since $\{ r_n \}_{n \geq 0}$ is an increasing sequence and the $n=0$ bound $b-a$ is attained by the integer polynomial $p(x) = 1$ (or $p(x) = -1$), it also follows that $b-a$ is the minimum over all nonzero $p \in {\bf Z}[X]$ once $b-a \geq 4$.]

It follows also that if $b-a>4$ then for any $M$ there are only finitely many $p \in {\bf Z}[X]$ for which $\int_a^b p^2(x) \, dx < M$. When $b-a = 4$ this is not true when $a,b$ are integers; I do not know what happens otherwise. [For the integer case: it is enough to prove it for $(a,b) = (-2,+2)$. let $T_n$ be the $n$-th Chebyshev polynomial; then $p(x) = 2T_n(x/2)$ has integer coefficients and $$ \int_{-2}^2 p(x)^2 \, dx \leq \int_{-2}^2 2^2 \, dx = 16. $$ (In fact for large $n$ the integral approaches $8$.)]


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