Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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 $m \ge n$.

share|improve this answer
    
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
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.