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Is there any sequence of polynomials which converge to $|x|$ uniformly on $\mathbb{R}$? I'm trying to prove that the space of all polynomial functions equipped with sup-norm is not complete. And I know that by Weierstrass-approximation, every continuous function is a uniform limit of a sequence of polynomials. I think my problem now is to find an Cauchy sequence of polynomial functions which converges to $|x|$.

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  • $\begingroup$ The domain cannot be $\mathbb R$: The only bounded polynomials are constant. $\endgroup$ – user99914 May 6 '15 at 1:06
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    $\begingroup$ @avid19 That question seems to be about pointwise, not uniform, convergence (actually, the question does not specify, but the answers all treat that convergence.) $\endgroup$ – Travis Willse May 6 '15 at 1:10
  • $\begingroup$ math.stackexchange.com/a/418660 is an answer to a related question, which is just a more verbose version of orangeskid's answer. $\endgroup$ – Jonas Meyer May 6 '15 at 2:35
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Hint Any polynomial of degree $\geq 2$ asymptotically grows faster than $|x|$.

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Hint: For $m$,$n\ge n_1$ we have $|P_n- P_m| \le 1$ on $\mathbb{R}$ so a constant polynomial. So from index $n_1$ on, only the free term may change, and so the limit must be a polynomial.

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For $p(x)=\sum_{j=0}^{n}A_jx^j$ with $n>0$ and $A_n\ne 0$ we have,for $x\ne 0$,$$p(x)=A_nx^n (1+\sum_{j=0}^{n-1}(A_j/A_n)/x^{n-j}).$$ For all sufficiently large values of $|x|$ we have $$|\sum_{j=0}^{n-1}(A_j/A_n)/x^{n-j})|<1/2 $$ $$\text {and }p(x)=A_nx^n (1+q(x)) \text { where } 1+q(x)>1/2.$$ From this it should be clear that if $p$ is ANY real polynomial then $\{|(p(x)-|x|)| : x\in R\}$ has no upper bound. So we cannot uniformly approximate $|x|$ on all of $R$ by a polynomial.

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