Property of a polynomial with no positive real roots

Question

The following is an exercise (Exercise #3 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.

Let $P(x)$ be a nonzero polynomial with real coefficients. Show that the following two conditions are equivalent:

• There exists a nonzero polynomial $Q(x)$ with real coefficients such that all coefficients of $P(x) Q(x)$ are nonnegative.

• There does not exist a real number $a > 0$ such that $P(a) = 0$.

That the first item implies the second is straightforward (since if $a$ is a positive real root of $P(x)$ then it is a positive real root of $P(x)Q(x)$, but since $P(x)Q(x)$ is nonzero with all coefficients nonnegative, this is impossible). However, I can't seem to find a way to prove that the second item implies the first. I would very much like to see a proof if someone can provide it.

Answer 1

The equivalent statement of the second statement implies the first statement is there exists a nonzero polynomial $Q(x)$ with real coefficients such that at least one coefficient of $P(x)Q(x)$ is negative implies there exists a real number $a>0$ such that $P(a)=0$. But this statement is false since if we take $P\left( x \right) = {x^2} + 1$ and $Q\left( x \right) = - x$ then the coefficients of $P(x)Q(x)$ are all negative. But $P(x)$ has no positive real root.

Answer 2

Call $S$ the set of polynomial satisfying the first property.

It is straightforward to show that if $P,Q$ are two nonzero polynomials, $PQ \in S \iff P \in S $ and $Q \in S$.

Since $\Bbb R[x]$ has the unique factorisation property, it is enough to understand who is in $S$ by checking what irreducible polynomial is in $S$.

Obviously, any polynomial with nonnegative coefficient is in $S$ and any one with a positive real root isn't, so we are left with checking polynomials of the form $X^2-aX+b$ for $a,b>0$ and $\delta = a^2/b < 4$.

Let $p(x) = x^2+ax+b$.

We try to show it is in $S$ by multiplying it with $p(-x)$ :
$p(x)p(-x) = x^4 + (2b-a^2)x^2 + b^2$.

It turns out that this polynomial is "less bad" than $p$ : the new polynomial has $\delta' = a'^2/b' = (2-\delta)^2$

If $\delta < 2$ then $p(x)p(-x)$ has nonnegative coefficients, so it is in $S$, which proves that $p$ is in $S$. So values of $\delta$ close to $0$ are good, while values close to $4$ are bad.

The map $f : \delta \mapsto \delta' = (2- \delta)^2$ has two fixed points, $1$ and $4$. Checking the derivatives show that the map is repulsive at $4$. More precisely, $f'(\delta) \in [2 \sqrt 2 ; 4]$ for $\delta \in [2+\sqrt 2 ; 4]$, which shows that $f^n(\delta)$ eventually gets lower than $2$ after some $n = \Theta(-\log(4 - \delta))$ steps.

This shows that every degree $2$ irreducible polynomial $x^2+ax+b$ is in $S$, and now we know exactly which irreducible polynomials are not in $S$ (the $x-a$ for $a>0$)

So $p$ is in $S$ if and only if it has no positive real root.

Answer 3

Another proof, due to Polya (1928):

Suppose that $f(x) = \sum_{j=0}^d f_j x^j$ is positive on $[0, \infty)$. We claim that, for $N$ sufficiently large, $(1+x)^N f(x)$ has positive coefficients.

Proof: Suppose to the contrary that there is an infinite sequence of pairs $(k_i, N_i)$ such that the coefficient of $x^{k_i}$ in $(1+x)^{N_i} f(x)$ is nonpositive, with $N_i \to \infty$. Passing to a subsequence, we may assume that $\lim_{i \to \infty} k_i/N_i$ exists; call this limit $\alpha$. It will be convenient to assume $\alpha \neq 1$; if $\alpha=1$ then replace $f$ by its reversal $\sum_{j=0}^d f_{d-j} x^j$, which will replace $k_i$ by $N_i-k_i$ and hence $\alpha$ by $1-\alpha$.

Now, the coefficient of $x^{k_i}$ in $(1+x)^{N_i} f(x)$ is $$\sum_{j=0}^d f_j \binom{N_i}{k_i-j} = \binom{N_i}{k_i} \sum_{j=0}^d f_j \frac{k_i}{(N_i-k_i+1)} \frac{(k_i-1)}{(N_i-k_i+2)} \cdots \frac{(k_i-j)}{(N_i-k_i+j+1)} .$$ So the sum on the right hand side is $\leq 0$ for all $i$.

Sending $i \to \infty$, the right hand sum approaches $\sum_{j=0}^d f_j \left( \frac{\alpha}{1-\alpha} \right)^j = f\left( \frac{\alpha}{1-\alpha} \right) >0$. (We wanted $\alpha \neq 1$ so $\tfrac{\alpha}{1-\alpha}$ is well defined.) So we have a sequence of nonpositive numbers aproaching a positive limit, a contradiction.