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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 $b^2<4c$. –  Hagen von Eitzen Oct 28 '12 at 14:43
    
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. –  Pierre-Yves Gaillard Nov 3 '12 at 12:25
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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) 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 also answers the present question. It seems far too complicated... –  commenter Nov 3 '12 at 14:41
    
@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. –  Haskell Curry Nov 4 '12 at 11:33
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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. –  Davide Giraudo Nov 4 '12 at 12:03

1 Answer 1

Using the article Absolute-Valued Algebras mentioned by commenter, a proof would look like this.

Firstly, recall

Lemma If $ (E,\| \cdot \|) $ is a normed $ \mathbb{R} $-vector space such that $ \| x + y \|^{2} + \| x - y \|^{2} \geq 4 \| x \| \| y \| $ holds for all $ x,y \in E $, then the norm $ \| \cdot \| $ is induced by some inner product $ \langle \cdot,\cdot \rangle $ on $ E $.

Note Anyone who has a link to a proof of the lemma above is warmly welcome.

Suppose that $ \| \cdot \| $ is a multiplicative norm on $ \mathbb{R}[X] $. Then for any $ x,y \in \mathbb{R}[X] $, we have \begin{align} \| x + y \|^{2} + \| x - y \|^{2} &= \| (x + y)^{2} \| + \| (x - y)^{2} \| \quad (\text{By the multiplicativity of $ \| \cdot \| $.}) \\ &\geq \| (x + y)^{2} - (x - y)^{2} \| \quad (\text{By the Triangle Inequality.}) \\ &= \| 4xy \| \\ &= 4 \| x \| \| y \|. \quad (\text{By the multiplicativity of $ \| \cdot \| $.}) \end{align} The lemma now says that $ \| \cdot \| $ is induced by some inner product $ \langle \cdot,\cdot \rangle $ on $ \mathbb{R}[X] $.

By applying the Gram-Schmidt orthogonalization procedure to the linearly independent set $ \{ 1,X \} $, we obtain a non-zero $ a \in \mathbb{R}[X] $ that is orthogonal to $ 1 $ (namely, $ a = X - \frac{\langle 1,X \rangle}{\langle 1,1 \rangle} \cdot 1 $). Then \begin{align} \| 1 - a^{2} \| &= \| (1 - a)(1 + a) \| \\ &= \| 1 - a \| \| 1 + a \| \quad (\text{By the multiplicativity of $ \| \cdot \| $.}) \\ &= \| 1 - a \|^{2} \quad (\text{By orthogonality, $ \| 1 + a \| = \| 1 - a \| $.}) \\ &= \langle 1 - a,1 - a \rangle \\ &= \| 1 \|^{2} + \| a \|^{2} - 2 \langle 1,a \rangle \\ &= \| 1 \|^{2} + \| a \|^{2} \quad (\text{By orthogonality once again.}) \\ &= \| 1 \| + \| a^{2} \| \quad (\text{By the multiplicativity of $ \| \cdot \| $.}) \\ &= \| 1 \| + \| - a^{2} \|. \end{align} In summary, $ \| 1 - a^{2} \| = \| 1 \| + \| - a^{2} \| $ holds. Squaring both sides of this equation gives us \begin{align} \| 1 - a^{2} \|^{2} &= (\| 1 \| + \| - a^{2} \|)^{2}, \\ \langle 1 - a^{2},1 - a^{2} \rangle &= \| 1 \|^{2} + \| - a^{2} \|^{2} + 2 \| 1 \| \| - a^{2} \|, \\ \| 1 \|^{2} + \| - a^{2} \|^{2} + 2 \langle 1,- a^{2} \rangle &= \| 1 \|^{2} + \| - a^{2} \|^{2} + 2 \| 1 \| \| - a^{2} \|, \\ \langle 1,- a^{2} \rangle &= \| 1 \| \| - a^{2} \|. \end{align} Hence, we get equality when we apply the Cauchy-Schwarz Inequality to the vectors $ 1 $ and $ - a^{2} $. Equality holds if and only if both $ 1 $ and $ - a^{2} $ lie in the same $ 1 $-dimensional subspace of $ \mathbb{R}[X] $; as $ 1,- a^{2} \neq 0 $, this says that both are non-zero scalar multiples of each other. However, $ \langle 1,- a^{2} \rangle = \| 1 \| \| - a^{2} \| > 0 $, so more precisely, $ 1 $ and $ - a^{2} $ are positive scalar multiples of each other. We therefore obtain $ a^{2} \in \mathbb{R}_{<0} $, which is a contradiction. ////

Conclusion There can be no multiplicative norm on $ \mathbb{R}[X] $.

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A variant of the lemma (assuming $|x| = |y| = 1$) is a theorem due to Schoenberg, A remark on M. M. Day's characterization of inner-product spaces and a conjecture of L. M. Blumenthal, Proc. Amer. Math. Soc. 3 (1952), 961-964, Theorem 2. There are some squares missing in the last displayed equation, I think you want $|1-a^2|^{2} = \cdots = |1|^2 + |\pm a^2|^2$. –  commenter Nov 8 '12 at 17:52
    
@commenter That last equation seems just fine as it is? –  WimC Nov 8 '12 at 21:21
    
@WimC: Oh, yes, you're right. Too much switching back and forth between squares and square roots :-) Sorry for the false alarm. –  commenter Nov 8 '12 at 21:54
    
Hi folks. The recent edit was made in order to address the issue of clarity mentioned in the comments here. –  Haskell Curry Dec 21 '12 at 6:26

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