1999 Putnam A-5 as a direct result of Archimedean Property I was working on question A-5 from the 1999 Putnam Exam:
Prove that there is a constant C such that, if $p(x)$ is a polynomial of degree 1999, then
$|p(0)|≤C\int_{-1}^1|p(x)|dx$
This seemed like a direct consequence of the Archimedean Property of the reals to me, since the value of the integral must be a finite, positive real number, and the lefthand side is a finite, nonnegative real number. However, all of the solutions (they gave 3: see this page http://kskedlaya.org/putnam-archive/) were much more complicated; is there a case I am missing by merely viewing this as a result of the Archimedean property? Or is there some missing condition that makes it not applicable for this problem?

One way to do this is to let $\mathcal P$ denote the set of polynomials of degree $\le 1999.$ Then $\mathcal P$ is a vector space (say over $\mathbb R$) of dimension $2000.$ On this vector space we can introduce two norms:
$$\|p\|_1 = \sum_{k=0}^{1999} |p(k)|,\,\, \|p\|_2 = \int_{-1}^1 |p(x)|\, dx.$$
It's easy to check that these are indeed norms on $\mathcal P.$ So we have two norms on a finite dimensional vector space. A standard result says any two norms on a finite-dimensional vector space are equivalent. It follows that there is a constant $C$ such that
$$\|p\|_1 \le C\|p\|_2\,\, \text { for all } p\in \mathcal P.$$
Since $|p(0)| \le \|p\|_1,$ we've proved the result.
I wonder if this answer would satisfy the Putnam overlords.
