If $ax^2+bx+c \leq p(x) \leq lx^2+mx+n$ , show that the degree of $p(x)$ is $2$.

Question

If $a,l\neq0$ , $ax^2+bx+c \leq p(x) \leq lx^2+mx+n$ , show that the degree of $p(x)$ is $2$.

How can we exactly say (how to prove) that $p(x)$ is a quadratic ?

What methods can be used to determine this ?

Answer 1

If $a=-1, l = 1 $ $b=c=m=n = 0$ then $x \mapsto 0$ satisfies the equation.

In fact, the function $x \mapsto x^2 \sin x$ satisfies the equation.

If $p$ is a polynomial, then we can establish that the maximum degree of $p$ is $2$. If we divide across by $x^2$, then we get $a \le \lim_{|x| \to \infty} {p(x) \over x^2} \le l$.

Suppose the degree of $p$ is $d>2$, then $\lim_{|x| \to \infty} {p(x) \over x^d} = \hat{p}_d \neq 0$, where $p(x) = \sum_{k=0}^d \hat{p}_k x^k$, then $\lim_{x \to \infty} {p(x) \over x^2} = \lim_{x \to \infty} {p(x) \over x^d} x^{d-2} = \hat{p}_d \lim_{x \to \infty} x^{d-2} = \infty$, which is a contradiction.

Answer 2

The reason which I would like to use in this case is as follows. From the inequalities it is conclusive that $p(x)$ grows no faster than a quadratic polynomial $lx^2+mx+c,\; l\neq 0$. Also it grows no slower than a quadratic polynomial $ax^2+bx+c,\: a\neq0$. Hence it must grow only as a quadratic polynomial.

Answer 3

I suppose $p$ is an polynomial. So there are $a_0,\ldots, a_k$ such that $p(x) = a_0 + \cdots + a_kx^k$ and $a_k \neq 0$. Since
$$\frac{p(x)}{lx^2+mx+n}\leq 1$$ for every $x\in \mathbb R$ such that $lx^2+mx+n\neq 0$ we have $$\lim_{x\to \infty}\frac{p(x)}{lx^2+mx+n}$$ exist. So $k\leq 2$. On the other hand by the same argument $$\lim_{x \to \infty} \frac{ax^2+bx+c}{p(x)}$$ exist. So $k\geq 2$. Therefore the degree of $p$ is 2.