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I'm pretty sure the answer is in the negative.

Can someone show me the proof?

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$-x^2$ is never positive for all $x\in\Bbb R$. – Andrea Mori Aug 11 '12 at 12:17
Did you mean $|P(x)|<a$? – ronno Aug 11 '12 at 12:18
I think he is talking about positive definite P(x) – Seyhmus Güngören Aug 11 '12 at 12:18
yes that's what I meant, sorry for the earlier error. – hollow7 Aug 11 '12 at 12:18
up vote 3 down vote accepted

The answer is no. Write $$ P(x)=ax^n+\text{terms of lower degree} $$ with $a\neq0$ and $n\geq1$. Then $$ \lim_{x\to\infty}P(x)=\pm\infty $$ according to the sign of $a$. This shows that $|P(x)|$ is unbounded.

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Except constant polynomials. – Michael Hardy Aug 11 '12 at 19:49
@MichaelHardy : of course. This is somewhat implicit in my answer since I specified $n\geq1$. Obviously constant polymonials are bounded. – Andrea Mori Aug 11 '12 at 20:20

Only constant polynomial will do the job. If $P$ has degree $d\geq 1$, assuming it WLOG monic, we have $|P(x)|\geq \frac{|x|^d}2$ for $|x|$ large enough, as $\lim_{x\to +\infty}\frac{P(x)}{x^d}=1$.

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(The question has been changed since the answer below was posted, so it no longer applies)

$P(x)=-x^2$ is less than 17 for all real $x$.

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I think there exist such if $n\rightarrow\infty$ where $n$ is the order of the polinomial. For example when you have a cosine and use the taylor series expansion then you will get a polinomial and this polinomial will be bounded by the absolute value. Since the polinomials are of not infinite degree then $|P(x)|$ can not be bounded.

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Taylor polynomials only give an approximation in some interval. – Andrea Mori Aug 11 '12 at 12:25
There are exact polinomial approximations for some bounded functions. And I think if we let the polinomial to be of infinite degree it is possible to claim that there exist a bound. – Seyhmus Güngören Aug 11 '12 at 12:28
A polynomial has finite degree. If it has infinite degree is not a polynomial. Your link reads "Taylor series". – Andrea Mori Aug 11 '12 at 12:31
Ok you are right so I must correct my post as when the degree goes to $\infty$ and else not. – Seyhmus Güngören Aug 11 '12 at 12:35
How does it look like now? I hope I wont get more downvotes since I am trying to be constructive. – Seyhmus Güngören Aug 11 '12 at 12:38

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