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](http://link.springer.com/article/10.1007%252FBF02418278), 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](https://math.stackexchange.com/questions/299780/understanding-a-proof-about-hilbert-matrix) 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.)



Hilbert's Theorem on $L_2$ norm of polynomials in $\mathbb{Z}[X]$ - Explicit construction and a converse?
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](http://en.wikipedia.org/wiki/Legendre_polynomials) 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](https://mathoverflow.net/questions/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](http://en.wikipedia.org/wiki/Chebyshev_polynomials); 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 $8$.)]


