Putnam 1999, Problem A-5

Prove that there is a constant \(C\) such that, if \(p(x)\) is a polynomial of degree 1999, then
\[
|p(0)| \leq C \int_{-1}^1 |p(x)| \, dx.
\]

Solution:
First solution: (by Reid Barton) Let $r_1, \ldots, r_{1999}$ be the roots of $P$. Draw a disc of radius $\varepsilon$ around each $r_i$, where $\varepsilon < 1/3998$; this disc covers a subinterval of $[-1/2,1/2]$ of length at most $2\varepsilon$, and so of the 2000 (or fewer) uncovered intervals in $[-1/2,1/2]$, one, which we call $I$, has length at least $\delta = (1-3998\varepsilon)/2000 > 0$. We will exhibit an explicit lower bound for the integral of $|P(x)|/|P(0)|$ over this interval, which will yield such a bound for the entire integral.

Note that
$$\frac{|P(x)|}{|P(0)|} = \prod_{i=1}^{1999} \frac{|x-r_i|}{|r_i|}.$$

Also note that by construction, $|x-r_i| \geq \varepsilon$ for each $x \in I$. If $|r_i| \leq 1$, then we have $\frac{|x-r_i|}{|r_i|} \geq \varepsilon$. If $|r_i| > 1$, then
$$\frac{|x-r_i|}{|r_i|} = |1-x/r_i| \geq 1-|x/r_i| \geq 1-1/2 = 1/2 > \varepsilon.$$

We conclude that $\int_I |P(x)/P(0)| dx \geq \delta\varepsilon$, independent of $P$.

Second solution: It will be a bit more convenient to assume $P(0) = 1$ (which we may achieve by rescaling unless $P(0) = 0$, in which case there is nothing to prove) and to prove that there exists $D > 0$ such that $\int_{-1}^1 |P(x)| dx \geq D$, or even such that $\int_0^1 |P(x)| dx \geq D$.

We first reduce to the case where $P$ has all of its roots in $[0,1]$. If this is not the case, we can factor $P(x)$ as $Q(x)R(x)$, where $Q$ has all roots in the interval and $R$ has none. Then $R$ is either always positive or always negative on $[0,1]$; assume the former. Let $k$ be the largest positive real number such that $R(x) - kx \geq 0$ on $[0,1]$; then
$$\int_{-1}^1 |P(x)| dx = \int_{-1}^1 |Q(x)R(x)| dx$$
$$> \int_{-1}^1 |Q(x)(R(x)-kx)| dx,$$
and $Q(x)(R(x)-kx)$ has more roots in $[0,1]$ than does $P$ (and has the same value at 0). Repeating this argument shows that $\int_0^1 |P(x)| dx$ is greater than the corresponding integral for some polynomial with all of its roots in $[0,1]$.

Under this assumption, we have
$$P(x) = c \prod_{i=1}^{1999} (x-r_i)$$
for some $r_i \in (0,1]$. Since
$$P(0) = -c \prod r_i = 1,$$
we have
$$|c| \geq \prod |r_i^{-1}| \geq 1.$$

Thus it suffices to prove that if $Q(x)$ is a monic polynomial of degree 1999 with all of its roots in $[0,1]$, then $\int_0^1 |Q(x)| dx \geq D$ for some constant $D > 0$. But the integral of $\int_0^1 \prod_{i=1}^{1999} |x-r_i| dx$ is a continuous function for $r_i \in [0,1]$. The product of all of these intervals is compact, so the integral achieves a minimum value for some $r_i$. This minimum is the desired $D$.

Third solution (by Abe Kunin): It suffices to prove the stronger inequality
$$\sup_{x\in[-1,1]} |P(x)| \leq C \int_{-1}^1 |P(x)| dx$$
holds for some $C$. But this follows immediately from the following standard fact: any two norms on a finite-dimensional vector space (here the polynomials of degree at most 1999) are equivalent. (The proof of this statement is also a compactness argument: $C$ can be taken to be the maximum of the L1-norm divided by the sup norm over the set of polynomials with L1-norm 1.)

Note: combining the first two approaches gives a constructive solution with a constant that is better than that given by the first solution, but is still far from optimal. I don't know offhand whether it is even known what the optimal constant and/or the polynomials achieving that constant are.