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Consider $E$ - the vector space of all real polynomials of one variable.

I need to prove that it is not complete under any norm.

I was thinking I could use the fact that certain functions, for example $ \exp x$ can be approximated by a sequence of polynomials $P(x) = (1 + \frac{x}{n})^n, \ \ n \in \mathbb{N}$.

This approach doesn't seem to depend of any particular norm, does it?

Could you tell me if I'm right or if I'm missing something?

Thanks a lot!

Bilbo
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    $E$ has a countable basis, but the dimension of a Banach space is always uncountable: http://math.stackexchange.com/questions/217516/let-x-be-an-infinite-dimensional-banach-space-prove-that-every-basis-of-x-is. – Martín-Blas Pérez Pinilla Oct 14 '14 at 06:15
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    The fact that a Banach space cannot have dimension $\aleph_0$ is well known. One should ponder on the fact that $\mathcal{P}$ with the norms $L^1$ norm $p(\cdot) \mapsto \int_I |p(t)| dt $ or $L^{\infty}$ norm $p(\cdot) \mapsto \sup_{t \in I} |p(t)|$ has certain completions $\ne \mathcal{P}$ – orangeskid Oct 14 '14 at 06:38
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    Probably good to add that the norm is supposed to extend the standard norm on $\mathbb{R}$. Otherwise, there are counterexamples, like the trivial norm. – Andrew Dudzik Oct 14 '14 at 07:00
  • @Slade Could you explain to me what you mean by extending the standard norm on $\mathbb{R}$? – Bilbo Oct 14 '14 at 09:17
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    @Bilbo I mean, presumably you have decided in advance that the constant polynomial $c$ satisfies $| c | = |c|$, i.e. its norm is the usual absolute value in $\mathbb{R}$. Otherwise, answers like David's don't make sense, because they assume that $| j!| = j!$. The trivial norm is $| x | = 1$ for all $x\neq 0$. This is a complete norm that you can put on any ring. It's also quite pathological, for a number of reasons, and probably shouldn't appear near the word "Banach" (unless you know what a graded reduction of affinoid algebras is). – Andrew Dudzik Oct 14 '14 at 09:29
  • @orangeskid No, I don't. So if I assume that $||c|| = |c|$ for any norm $|| \cdot ||$ on the ring of polynomials, then my approach is correct? I mean, I can say that there is a sequence of polynomials which doesn't converge to a polynomial? – Bilbo Oct 14 '14 at 09:32
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    @Bilbo Sure. David's answer is lacking details, but should basically be correct. Use the usual proof that the power series for $e^x$ is absolutely convergent. The point is that, regardless of what $| X|$ is, $n!$ grows much faster than $| X|^n$. – Andrew Dudzik Oct 14 '14 at 11:18
  • Ups, I've just noticed I addressed my answer to your comment incorrectly. And as for $e^x$, I'll try that. Thanks :) – Bilbo Oct 14 '14 at 12:15

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i'm a bit out of my depth here, so apologies if this is nonsense, but i think your idea works. suppose $\|x\|=N$ and define $f_n=\sum_{j=0}^n \frac{x^j}{j!}$. this sequence converges absolutely for all $N$, but is not a polynomial.

David Holden
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  • Well, that's exactly what I thought. But I would be grateful if someone else approved this answer, too :) – Bilbo Oct 14 '14 at 09:15