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I am trying this problem:

If $p(x)$ is a bounded polynomial for all $x\in \mathbb R$, then $p(x)$ must be a constant.

I am trying to prove it by contradition. So I assume that $p(x)$ is bounded for all $x\in \mathbb R$ and $p(x)$ is not constant. So I have that $$p(x)= a_nx^n+ a_{n-1}x^{n-1}+\cdots + a_1x+ a_0$$ with $a_n\neq 0$ for $n\geq 1$. I have figured out that $$|p(x)|\leq |x|^n\cdot C$$ with $C=|a_n|+ |a_{n-1}|+\cdots + |a_1|+ |a_0|$.

But now I am not sure how to proceed and use the fact that $p(x)$ is bounded to find a contradiction.

Any ideas or hints would be appreciated. :)

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  • $\begingroup$ Hint: $|p(x)| \geq |a_n| |x|^n - \sum_{i=0}^{n-1} |a_i| |x|^i$; from this bound you can conclude that $|p(x)|$ is unbounded by considering large enough $|x|$. ("Large enough" in this context means it needs to exceed two different thresholds...can you figure out what they are?) $\endgroup$
    – Ian
    Commented Apr 2, 2022 at 3:28
  • $\begingroup$ Do you get that inequality from the fact that $|a_n||x|^n\leq \sum_{i=0}^{n-1}|a_i| |x|^i $? My first guess for a threshold is to consider an expression like $(\text{something here})/|a_n|$, so when I plug it in $|a_n||x|^n$ I get an expression like $(\text{something here})/|a_n|^{n-1}$. $\endgroup$
    – learningmaths
    Commented Apr 2, 2022 at 4:21
  • $\begingroup$ This is the "reverse triangle inequality", which comes from writing $a_n x^n$ as $p(x) - \sum_{i=0}^{n-1} a_i x^i$ and then applying the normal triangle inequality. Anyway, the first threshold is $1$; if you exceed that threshold then you can conclude $|p(x)| \geq |a_n||x|^n - |x|^{n-1} \sum_{i=0}^{n-1} |a_i|=|x|^{n-1}(|a_n||x| - \sum_{i=0}^{n-1} |a_i|)$. $\endgroup$
    – Ian
    Commented Apr 2, 2022 at 15:41
  • $\begingroup$ Ok, now I understand that first inequality. So, first assuming that $|x|>1$, we obtain $|p(x)|\geq |x|^{n-1}(|a_n||x|-\sum_{i=0}^{n-1}|a_i|)$. This means that the second threshold shold be $|x|>\left(\sum_{i=0}^{n-1}|a_i|\right)/|a_n|$. From here we can conclude that $|p(x)|$ is unbounded. Correct me if I am wrong :) Thanks for your help! $\endgroup$
    – learningmaths
    Commented Apr 3, 2022 at 3:49
  • $\begingroup$ That's right. Once $|x|$ exceeds 1 and exceeds that ratio by at least $c$ you get $|p| \geq c|x|^{n-1}$. (Note for $n=1$ you'll need to make this $c$ big too). $\endgroup$
    – Ian
    Commented Apr 3, 2022 at 11:26

3 Answers 3

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Yeah, let $n \ge 1$ be the degree of $p$, and WLOG suppose the leading coefficient is positive. Then $\lim_{x \rightarrow \infty} p(x) = \lim_{x \rightarrow \infty} x^n = \infty$, contradicting the fact that $p$ is bounded. So that $n=0$.

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You could also try the following: for $x \neq 0$ and $a_n\neq0$, we have

$p(x) = a_nx^n+...+a_0 = x^n(a_n+a_{n-1}x^{-1}+...+a_0x^{-n})$

Now just consider the limit as $x\to+\infty$, all the terms in parentheses but the first one tend to zero, so the part in parentheses tends to $a_n$, as $x^n$ tends to $+\infty$, we have that the polynomial tends to $\pm\infty$ (depending on the sign of $a_n$), and the polynomial cannot be bounded.

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This has been already asked and answered in this forum.

A direct proof is given to show that every coefficient $a_1, a_2, \ldots, a_n$ must be zero in the expansion $$ f(x) = a_n x^n + a_{n - 1} x^{n - 1} + \cdots + a_1 x + a_0 $$ if $f(x$ is bounded. This ultimately results in $f(x) \equiv a_0$, a constant.

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

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    $\begingroup$ If you find that a question has been asked and asked before then you can flag the question as a duplicate. There is no reason to repeat an existing answer. $\endgroup$
    – Martin R
    Commented Apr 2, 2022 at 4:41

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