I'm pretty sure the answer is in the negative.
Can someone show me the proof?
I'm pretty sure the answer is in the negative.
Can someone show me the proof?
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.
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 $\lim_{x\to +\infty}\frac{P(x)}{x^d}=1$.
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.
(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 $x$.