Putnam 2016, Problem A-6

Find the smallest constant \( C \) such that for every real polynomial \( P(x) \) of degree 3 that has a root in the interval \( [0, 1] \),
\[
\int_0^1 |P(x)| \, dx \leq C \max_{x \in [0,1]} |P(x)|.
\]

Solution:
We prove that the smallest such value of $C$ is $5/6$.

**First solution:** (based on a suggestion of Daniel Kane)

We first reduce to the case where $P$ is nonnegative in $[0,1]$ and $P(0) = 0$. To achieve this reduction, suppose that a given value $C$ obeys the inequality for such $P$. For $P$ general, divide the interval $[0,1]$ into subintervals $I_1,\ldots,I_k$ at the roots of $P$. Write $\ell(I_i)$ for the length of the interval $I_i$; since each interval is bounded by a root of $P$, we may make a linear change of variable to see that

$$\int_{I_i} |P(x)|dx \leq C\ell(I_i) \max_{x \in I_i} |P(x)| \quad (i = 1,\ldots,k).$$

Summing over $i$ yields the desired inequality.

Suppose now that $P$ takes nonnegative values on $[0,1]$, $P(0) = 0$, and $\max_{x \in [0,1]} P(x) = 1$. Write $P(x) = ax^3 + bx^2 + cx$ for some $a,b,c \in \mathbb{R}$; then

$$\int_0^1 P(x)dx = \frac{1}{4}a + \frac{1}{3}b + \frac{1}{2}c = \frac{2}{3}\left(\frac{1}{8}a + \frac{1}{4}b + \frac{1}{2}c\right) + \frac{1}{6}(a+b+c) = \frac{2}{3}P\left(\frac{1}{2}\right) + \frac{1}{6}P(1) \leq \frac{2}{3} + \frac{1}{6} = \frac{5}{6}.$$

Consequently, the originally claimed inequality holds with $C = 5/6$. To prove that this value is best possible, it suffices to exhibit a polynomial $P$ as above with $\int_0^1 P(x)dx = 5/6$; we will verify that

$$P(x) = 4x^3 - 8x^2 + 5x$$

has this property, by interpolating between the constraints $P(1/2) \leq 1$ and $P(1) \leq 1$.

This still leaves the matter of correctly guessing the optimal polynomial. If one supposes that it should be extremized both at $x = 1$ and at an interval value of the disc, it is forced to have the form $P(x) = 1 + (x-1)(cx-1)^2$ for some $c > 0$; the interpolation property then pins down $c$ uniquely.

**Second solution:** (by James Merryfield, via AoPS) As in the first solution, we may assume that $P$ is nonnegative on $[0,1]$ and $P(0) = 0$. Since $P$ has degree at most 3, Simpson's rule for approximating $\int_0^1 P(x)dx$ is an exact formula:

$$\int_0^1 P(x)dx = \frac{1}{6}(P(0) + 4P\left(\frac{1}{2}\right) + P(1)).$$

This immediately yields the claimed inequality for $C = 5/6$. Again as in the first solution, we obtain an example showing that this value is best possible.