Positive polynomials in $\mathbb{R}^+$

Question

Let $P\in \mathbb{R}[X]$ be a polynomial such that $P (x)\geq 0$ for all $x\geq 0$. Show that there exists two polynomials $A, B\in \mathbb{R}[X]$ so that $$P=A^{2}+X\cdot B^{2}$$

Answer 1

Idea: Write $$P(x) = Q(x)+xR(x)$$

where $Q(x) ={1\over 2}(P(x)+P(-x))$ and $R(x) = {P(x)-Q(x)\over x}$ and try to prove that $Q$ and $R$ are perfect square of a real polynomials.

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This is better:

1. step: Suppose all the zeroes of $P$ are real and simple (i.e. first degree). Then $$P(x) = a(x-x_1)(x-x_2)...(x-x_n)$$ Because $P(x)\geq 0$ for $x\geq 0$ they all have to be nonpositive. So $y_i = -x_i$ is positive for each $i$. So we have $$P(x) = a(x+y_1)(x+y_2)...(x+y_n)$$

Now we can prove that such polynomial is of a form we want:

Say $a>0$ and if $M = A^2+xB^2$ then $$(x+a)M = x(A+aB)^2+(aA-xB)^2$$ and we are done.

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2. step Now if $P$ has also nosimple real zeroes, then we can write it as $$P= Q\cdot R^2$$ where $Q$ again has only simple real zeros, so we can use 1. step and we are done.

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3. step Say $P$ has also nonreal zeros. So $P=Q\cdot R$ where $Q$ has only real zeros and $R$ only nonreal (so they are in conjugate pairs). Now we can use step 2. for $Q$ and we can prove that we can write $R$ as $C^2+D^2$. I leave that to you. Only thing is to prove now that product of polynomial $A^2 +xB^2$ and $C^2+D^2$ is again of form $A_1^2+xB_1^2$ and that I also leave to you.