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How to prove that : there is no function $N\colon \mathbb{R}[X] \rightarrow \mathbb{R}$, such that : $N$ is a norm of $\mathbb{R}$-vector space and $N(PQ)=N(P)N(Q)$ for all $P,Q \in \mathbb{R}[X]$.

Once, my teacher asked if there is a multipicative norm on $\mathbb{R}[X]$, and one of my classmate proved that there was none. But I can't remember the proof (all I remember is that he was using integration somewhere...).

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    Remark: Such $N$ is completely determined by the values $N(X+a)$ with $a\in\mathbb R$ and $N(X^2+bX+c)$ with $b^2<4c$.2012-10-28
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    Here is a (perhaps stupid) idea: Assume by contradiction that such a norm exists. Show that the completion of $\mathbb R[X]$ is a Banach algebra isomorphic to the Banach algebra of continuous functions on a compact Hausdorff space, and derive a contradiction from that.2012-11-03
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    @Pierre-YvesGaillard: I think the idea with completion is a good one. Instead of trying to prove that $\mathbb{R}[X]$ is an algebra of continuous functions (which is non-trivial, see e.g. [this American Math Monthly paper](http://kaltonmemorial.missouri.edu/docs/amm2007.pdf)) it would probably be easier to appeal to Mazur's theorem stating that the only (associative, unital) real Banach division algebras are $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$. Davide pointed out that [my answer](http://math.stackexchange.com/q/110443/) also answers the present question. It seems far too complicated...2012-11-03
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    @Pierre-YvesGaillard: I'm not sure if your argument uses the fact that the norm is multiplicative. It seems that it only requires the norm to be sub-multiplicative.2012-11-04
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    @commenter I agree the approach to the problem you linked works here and seems too complicated. At least, it shows that such a norm doesn't exist, so we are looking to a quite simple argument showing that.2012-11-04
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    Dear @Haskell: Thanks. My "idea" was just to complete $\mathbb R[X]$ to get a Banach algebra. I know that there are many theorems about Banach algebras (that's almost the only thing I know about Banach algebras). I thought that there might be a way to derive a contradiction from these theorems. One might indeed expect the multiplicativity of the norm to be one the main ingredients used to derive such a contradiction... The main point of my comment was to make the key words "Banach algebra" appear in this thread...2012-11-04

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