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Let $f(x)$ be a polynomial function.

It is known that for every $x$:

$$ f'(x) \leq f(x) $$


For every $x$:

$$ f(x) \geq 0 $$


Suppose by contradiction that $f(z)<0$ for some $z$. Then $f'(z)<0$ too, so $f$ must go down and down to $-\infty$.

Moreover, the rate of decrease must by at least exponential because:

$$ |f'(x) \geq f(x)| $$

and we know that equality holds for the exponential function.

This contradict the fact that $f(x)$ is polynomial.

MY QUESTIONS: Is my intuition true? If so, how to formalize it? If not, then what is the correct answer?

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Yes, your intuition is correct. However, I don't see a really easy way to formalize it and prove the assertion from there. I find it easier to argue with the degree of the polynomials, which disallows $\lvert f'(x)\rvert \geqslant \lvert f(x)\rvert$ for $\lvert x\rvert$ large. – Daniel Fischer Dec 2 '13 at 12:06
up vote 1 down vote accepted

Your intuition seems quite good. Here are a couple notes on some ways to make it rigorous:

1) Since $f$ is a polynomial, $\displaystyle \lim_{x \to \pm \infty} |f'(x)|/|f(x)| = 0$ (unless $f \equiv 0$, but in that case you're already done.)

2) If for some $x_0$, $f(x_0)<0$, then since $f$ is continuous, there is a certain interval $(a,b)$ containing $x_0$ such that $f(x) < 0$ for all $x \in (a,b)$. This means that $f'(x)<0$ for every $x \in (a,b)$ implying that $f$ is strictly decreasing on $(a,b)$. In particular, you can deduce that $f(b) < f(x_0)<0$. Now, let $$A = \sup\{ y \geq b \mid f(x)<0 \text{ for every } x \in (a,y) \}$$ If $A \neq \infty$ then by the same work as before you get that $f(A) < f(x_0) < 0$, implying there is some $\delta > 0$ such that for all $x \in (a, A+\delta)$, $f(x)<0$ contradicting the choice of $A$, hence $A = \infty$.

3) From the previous mark, if $f(x_0)<0$, then $\forall x \geq x_0$, $f(x) < 0$ and hence $f'(x) \leq f(x) < 0$. Now, as you pointed out, this means $\forall x \geq x_0$, $|f'(x)|/|f(x)| \geq 1$. But this contradicts the first point.

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Just to make sure I understand: point 2 is the formalization of "If $f(x_0)<0$ for some $x_0$ then $f'(x_0)<0$ too so $f$ must decrease to $-\infty$"? – Erel Segal-Halevi Dec 3 '13 at 15:53
In point 3, shouldn't it be "$\forall x \geq x_0$"? – Erel Segal-Halevi Dec 3 '13 at 15:55
@ErelSegalHalevi Sorry, I was working from right to left, not left to right.. a strange bout of dyslexia I suppose.. I've edited for the fix. – Tom Dec 3 '13 at 16:19
In point 2, shouldn't it be "$A=sup\{y≥a...\}$"? – Erel Segal-Halevi Dec 4 '13 at 7:42
@ErelSegalHalevi $(a,b)$ was chosen so that $f(x) < 0$ for every $x \in (a,b)$, so $y \geq b$ is "$A$"-ok. – Tom Dec 4 '13 at 13:20

Assume that $f<0$ somewhere. There are only two cases: (i) $f$ attains a critical point with a strictly negative value. But then, if $x^*$ is this point, $f(x^*) \geq f'(x^*) =0$, a contradiction. The other case is that $f<0$ definitely. As you noticed, considering either $f(x)e^{-x}$ or $f(x)e^x$, you can prove that $f$ goes to $-\infty$ exponentially fast, and this is impossible since $f$ is a polynomial. I think this approach can be made completely rigorous.

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I don't understand why there are only these two cases? – Erel Segal-Halevi Dec 3 '13 at 15:42

$f(x)$ from your assumptions is a polynomial, not just a function with polynomial growth.

From your conditions, on all critical points, the function is non-negative. after its last critical point, it must be decreasing, but $\deg f'=\deg(f) - 1$ so $\lim_{x\to\infty}\frac{f}{f'} > 0$.

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Yes, I mean that. I will correct the answer, thank you very much for pointing it out. – rewritten Dec 3 '13 at 15:46
"after its last critical point, it must be decreasing" - I didn't understand, why is this? Also, why must the limit be positive? (I didn't understand which fact is generally true and which fact is true only because of the question conditions). – Erel Segal-Halevi Dec 3 '13 at 15:50

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