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If a function $g:\mathbb{R}\rightarrow \mathbb{R}$ which is continuous, can be uniformly approximated by polynomials on the real numbers $\mathbb{R}$, then it is required to prove that this function can be nothing but a polynomial.

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If $\{P_n\}$ is a sequence of polynomials which converges uniformly to $g$ on the real line, then $\{P_{n+1}-P_n\}$ is a sequence of polynomials which converges uniformly to $0$ on the real line. So for $n$ large enough, says larger than $n_0$, $P_{n+1}-P_n$ is bounded on the real line, therefore equal to a constant says, $c_n$. Hence $P_n=\sum_{k=n_0}^{n-1}c_j+P_{n_0}$. Since $\{P_n \}$ is Cauchy, the series $\sum_{k\geq n_0}c_k$ is convergent. Denote $c$ the limit; we get that $g=c+P_{n_0}$, a polynomial.

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If a sequence of polynomials $\{Q_n\}$ converges to $0$, then it's supremum norm is finite for $n$ large enough. If a polynomial is not constant it cannot be bounded on the real line. – Davide Giraudo Apr 25 '12 at 20:15
What is the supremum norm by the way? – M.Krov Apr 25 '12 at 20:20
$\lVert f\rVert_{\infty}=\sup_{x\in\mathbb R}|f(x)|$; it's a norm for the uniform convergence. – Davide Giraudo Apr 25 '12 at 20:22
If a polynomial is of the form $Q(x)=x^d+\sum_{k=0}^{d-1}a_jx^j$, with $d\geq 1$ what can you say about $\lim_{x\to +\infty}Q(x)$? – Davide Giraudo Apr 25 '12 at 20:31
$lim_{x \to \infty }Q(x)=lim_{x \to \infty }x^{d}=\infty$ – M.Krov Apr 25 '12 at 20:34

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