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Let $f:\Bbb R\to \Bbb R$ be a function which is the uniform limit of polynomials. I want to show that $f$ is a polynomial.

I mean this seems a bit trivial... If it's the uniform limit of the set of polynomials doesn't that guarantee it's a polynomial?

If I define $f$ to be the set of all polynomial do I define a uniform limit $L$ s.t $d(f,L)<\epsilon$? or approach via contradiction?

any help would be great

leo
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user146296
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    It seems trivial if you think about it in the right way. However, $\sin x$ is a limit of polynomials and that is not a finite polynomial. – user164587 Oct 07 '14 at 03:52
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    Any continuous function on a closed bounded (thus, compact) interval is the uniform limit of a sequence of polynomials. Thus, the claim you try to prove is false, @user164587 – Timbuc Oct 07 '14 at 03:56
  • How do you know how to define a limit of polynomials? Is there a different way to phrase this that would make it easier to understand? – user146296 Oct 07 '14 at 03:57
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    @Timbuc The point is that this is on all of $\mathbb R$, not on a compact interval. – Robert Israel Oct 07 '14 at 03:59
  • I see , @Robert. – Timbuc Oct 07 '14 at 04:38
  • @user146296 I would like to know where to find this exercise since I am studying this subject. – Diego Fonseca May 27 '17 at 01:18
  • @DiegoFonseca This is late, but for future readers this is from Marsden Elementary Classical Analysis 2'nd ed., pg 322, exercise 50 – Peter_Pan Feb 07 '20 at 03:19

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Hint: if $f$ is a non-constant polynomial, then $f$ is unbounded. What does this say about $f_n(z) - f_m(z)$, if $f_n$ is a sequence of polynomials that converges uniformly on $\mathbb R$?

Robert Israel
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    If fn and fm are both sequences of polynomials then the difference will be a polynomial right. I guess I'm struggling with writing up a proof for this. – user146296 Oct 07 '14 at 04:19
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    @leo We don't know yet that $f(z)$ is a polynomial. That's what we're trying to prove. – Robert Israel Oct 07 '14 at 06:53
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    @user146296 It's just one sequence of polynomials, not two. $f_n$ and $f_m$ are two members of the sequence. If it converges uniformly, then for every $\epsilon > 0$ there is $N$ such that for all $m$ and $n$ both $> N$, ... – Robert Israel Oct 07 '14 at 06:55
  • You are right. ${}$ – leo Oct 07 '14 at 15:43
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    So if I have |fn-fm| and fix n and let m got to infinity the fm=Limit. And if I supposed the Limit to be sinx then I get fn-sinx. Am I on the right track..? any more help and reasoning would really help me – user146296 Oct 07 '14 at 22:20
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    The point is that if the sequence converges uniformly, and $m$ is large enough, then $|f_n - f_m|$ is uniformly bounded for $n > m$, and since $f_n - f_m$ is a polynomial this implies that $f_n - f_m$ is a constant function. Now you take the limit as $n \to \infty$, and find that $f = f_m + \text{constant}$, which is a polynomial. – Robert Israel Oct 07 '14 at 22:27