Show that if $\forall x \in [0,1], P(x) \ge 0$ therefore $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

Question

Let $P \in \Bbb R[X]$. Show that if $\forall x \in [0,1], P(x) \ge 0$ therefore $\exists A,B,C,D \in \Bbb R[X], P=A^2+XB^2+(1-X)C^2+X(1-X)D^2$.

I showed that this is true for $\deg P=2$ but I don't see how to generalize my proof. I tried to introduce coefficients for the polynomials, also tried Lagrange interpolation without success. Does someone have any hint?

If we consider $P \ge 0$ on $\Bbb R$ a classical problem shows that $P=A^2+XB^2$. Adapting the proof of this result seems difficult.

Edit: I found this reference (a very similar problem with solution): George Pólya and Gábor Szegő, Problems and Theorems in Analysis II (problem $47$, p.$78$).

Answer 1

HINT:

The interval $[0,1]$ is defined by the inequality $x(1-x)\ge 0$.

It is better ( and easier in fact) to show a stronger statement: every polynomial that is $\ge 0$ on $[0,1]$ is of the form

$$P(x)= A^2(x) + x(1-x)D^2(x)$$

( the terms $B$, $C$ may be made $0$).

It's easier to show by induction since the multiplicative property is immediate. Let's see how we can prove it.

A real polynomial that is positive on $[0,1]$ is a product of polynomial of this form

1. $x^2 + b x + c$, with non-real roots $b^2 - 4 c< 0$

2. $(x-\alpha)$ where $\alpha \le 0$

3. $-(x-\beta)$, where $\beta \ge 1$.

4. $Q^2\ $ a square

Now you need to show that every polynomials of each of the four types has a writing as above.

Obs: Any polynomial of form $x B^2(x) + (1-x) C^2(x)$ is clearly positive on $[0,1]$. So it will have a representation with $A$, $D$. And conversely! So we have two possible representations for polynomials positive on $[0,1]$.