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This is a question from Bergman's companion to Rudin.

a) Show that the only polynomials which are bounded as functions $\mathbb{R} \rightarrow \mathbb{R}$ are constant functions.

(I can do this) Also done here

b)Deduce that if a sequence of polynomials $P_n:\mathbb{R} \rightarrow \mathbb{R}$ converges uniformly on $\mathbb{R}$ to $f$ then $f$ is a polynomial.

I figure that the uniform convergence implies at some point (for large n) the polynomials must have the same highest power because otherwise large values of $\mathbb{R}$ would destroy any hope of uniform convergence. Then eventually the second highest power must be equal as well by a similar argument...Then I guess you could make a similar argument for the co-efficients by plugging in large values of x, the difference in each co-efficient must be quite small in order to maintain the uniform convergence.

I would like some help understanding if/why this means that the limit actually is a polynomial.

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up vote 16 down vote accepted

Hint: if $f_n$ converges uniformly, there exists $n$ such that $|f_n - f_m| \le 1$ for all $m \ge n$.

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Ok thanks, so the difference of two polynomials is a polynomial which by a.) is unbounded unless it is constant. By the hint uniform convergence implies that the difference is constant. So, the limit differs from a polynomial only by a constant which can be made arbitrarily small so it really is a polynomial. – user9352 Jul 8 '11 at 13:42
@user9352: you don't need to know that the constant can be made arbitrarily small. Once you know that it differs from a polynomial by a constant, it's already a polynomial. – Qiaochu Yuan Jul 8 '11 at 14:29

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