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Problem:

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.2012-04-25
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    What is the supremum norm by the way?2012-04-25
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    $\lVert f\rVert_{\infty}=\sup_{x\in\mathbb R}|f(x)|$; it's a norm for the uniform convergence.2012-04-25
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    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 $\lim_{x\to +\infty}Q(x)$?2012-04-25
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    $lim_{x \to \infty }Q(x)=lim_{x \to \infty }x^{d}=\infty$2012-04-25
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    Ok, I think now I can see why $Q_{n}(x)$ has to be constant, but I am still not able to deduce from this that $g$ is polynomial? Can you give me the next hint please?2012-04-25
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    $g$ doesn't need to be constant, but only a polynomian (for example we can have $P_n=P+\frac 1n$, where $P$ is a polynomial). We can write $P_n=\sum_{k=n_0}^{n-1}c_k+P_{n_0}$, where $n_0$ is like in the answer.2012-04-25
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    I realize I didn't defined $n_0$, now it's done.2012-04-25