### 60th Putnam 1999

Problem A5

Find a constant k such that for any polynomial f(x) of degree 1999, we have |f(0)| ≤ k ∫-11 |f(x)| dx.

Solution

The basic idea is simple. Divide the interval [-1, 1] into 2000 subintervals of length 1/1000, then there must be at least one subinterval K such that there is no root of f(x) whose real part falls in K. Take K' to be the central 1/3 of K. Then every point of K' is a distance at least 1/3000 from any root of f(x). Now |f(x)| = ∏|x - α| ≥ (1/3000)1999. Hence the integral is at least (1/3000)2000.

That would work if we were looking for a lower bound for the ratio of the integral to the leading coefficient. But we were asked for the ratio of the integral to the modulus of constant term (which is ∏ |α| ), so we need to estimate ∏|x - α|/|α|. If f(0) = 0, then any constant k works, so we can assume that all the roots α ≠ 0.

If |α| ≥ 1, then |x| < 1 - 1/3000 (K' is the central third of a subinterval length 1/1000 of [-1, 1], so its endpoint is at least 1/3000 from ±1). Hence |x - α| ≥ |α| - |x| ≥ |α| - 1 + 1/3000, and |x - α|/|α| > 1/3000. If |α| < 1, then |x - α|/|α| > |x - α| > 1/3000.

So we have established that |f(x)|/|f(0)| > (1/3000)1999 on the interval K'. Hence the integral is at least (1/3000)2000 |f(0)| and we have established the result with k = (1/3000)2000.