$f(x)$ is a bounded polynomial. Prove that $f$ must be constant.

Question

Question: Suppose $f(x)$ is a bounded polynomial, in other words, there is an $M$ such that $|f(x)|\le M$ for all $x\in R$. Prove that $f$ must be a constant.

I think the question assumes $M\in R$. I can't think of any polynomial that is bounded by some real number unless the polynomial degree is zero, so $f$ should be constant. But, I am not sure about how to prove this.

My attempt: Prove the statement by contrapositive.

If $f$ is not constant, then $f$ should be unbounded. Let $f(x)=b_dx^d+b_{d-1}x^{d-1}+...+b_1x+b_0$ for $d>0$. Since polynomial is infinitely differentiable (explanation), we can differentiate $f$ for $d$ times. Then, we get $b_d\in R$. That is, $f$ is increasing or decreasing continuously. Therefore, if $f$ is bounded, $f$ should be constant.

Is it okay? If not, could you give some hint?

Thank you in advance!

Answer 1

Assume that $f(x)=a_{n}x^{n}+\cdots+a_{1}x+a_{0}$ for $a_{n}\ne 0$ and $n\geq 1$, then \begin{align*} \lim_{x\rightarrow\infty}\dfrac{f(x)}{a_{n}x^{n}}&=\lim_{x\rightarrow\infty}\left(1+\cdots+\dfrac{a_{1}}{a_{n}x^{n-1}}+\dfrac{a_{0}}{a_{n}x^{n}}\right)\\ &=1. \end{align*} On the other hand, we have \begin{align*} \left|\dfrac{f(x)}{a_{n}x^{n}}\right|\leq\dfrac{M}{|a_{n}||x|^{n}}\rightarrow 0, \end{align*} so \begin{align*} \lim_{x\rightarrow\infty}\dfrac{f(x)}{a_{n}x^{n}}=0, \end{align*} a contradiction, so we must have $a_{n}=0$. Proceed in the similar fashion we get all $a_{n-1}=\cdots=a_{1}=0$, and hence $f$ is constant.

Answer 2

If you differentiate $d$ times, you get $d! \cdot b_d$. This is a real number, but remember, it's the only the first derivative that determines increasing or decreasing, not the $d$th derivative! So no, unfortunately, your proof doesn't work.

Here's what I would try: consider the polynomial divided by $x^d$. If $d > 0$ and the polynomial were bounded, this function should go to $0$ as $x \to \infty$ (you need to prove this). However...