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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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    $\begingroup$ Remark: Such $N$ is completely determined by the values $N(X+a)$ with $a\in\mathbb R$ and $N(X^2+bX+c)$ with $b^2<4c$. $\endgroup$ – Hagen von Eitzen Oct 28 '12 at 14:43
  • $\begingroup$ 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. $\endgroup$ – Pierre-Yves Gaillard Nov 3 '12 at 12:25
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    $\begingroup$ @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... $\endgroup$ – commenter Nov 3 '12 at 14:41
  • $\begingroup$ @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. $\endgroup$ – Haskell Curry Nov 4 '12 at 11:33
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    $\begingroup$ @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. $\endgroup$ – Davide Giraudo Nov 4 '12 at 12:03
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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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    $\begingroup$ 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 $|1-a^2|^{2} = \cdots = |1|^2 + |\pm a^2|^2$. $\endgroup$ – commenter Nov 8 '12 at 17:52
  • $\begingroup$ @commenter That last equation seems just fine as it is? $\endgroup$ – WimC Nov 8 '12 at 21:21
  • $\begingroup$ @WimC: Oh, yes, you're right. Too much switching back and forth between squares and square roots :-) Sorry for the false alarm. $\endgroup$ – commenter Nov 8 '12 at 21:54
  • $\begingroup$ Hi folks. The recent edit was made in order to address the issue of clarity mentioned in the comments here. $\endgroup$ – Haskell Curry Dec 21 '12 at 6:26
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This question (and much more) is settled by the Gelfand–Mazur theorem for normed algebras over the ground field $\mathbb{F} \in \{\mathbb{R},\mathbb{C}\}$. For our purposes, let us say that an algebra over a field $\mathbb{F}$ is a vector space over $\mathbb{F}$ which is at the same time a ring (with unit).¹ Furthermore, a normed algebra is an algebra over $\mathbb{F} \in \{\mathbb{R},\mathbb{C}\}$ equipped with a vector space norm which is submultiplicative ($\lVert ab\rVert \leq \lVert a\rVert\lVert b\rVert$). Finally, a division algebra is a non-zero algebra with unit in which every non-zero element is invertible.

(¹: Normed algebras are not usually assumed to have a unit in the literature, but in the context of the Gelfand–Mazur theorem we will need a unit anyway.)

We use the following version of the Gelfand–Mazur theorem:

Theorem (Gelfand–Mazur). Let $A$ be a real normed division algebra. Then $A$ is isomorphic with $\mathbb{R}$, $\mathbb{C}$, or the algebra of quaternions $\mathbb{H}$.

For a proof, see theorem 14.7 in [F.F. Bonsall, J. Duncan, Complete Normed Algebras, Springer–Verlag, Berlin Heidelberg New York 1973].

In order to apply the above theorem to the question at hand, note that we can extend any multiplicative norm $N$ on $\mathbb{R}[X]$ to a multiplicative vector space norm $N' : \mathbb{R}(X) \to \mathbb{R}_{\geq 0}$, where $\mathbb{R}(X)$ denotes the field of rational functions over $\mathbb{R}$. This is done by defining $$ N'\left(\frac{f(X)}{g(X)}\right) \: := \: \frac{N(f(X))}{N(g(X))}. $$ It is not very hard to prove that $N'$ is well defined, multiplicative, and a vector space norm (use that $N$ is multiplicative). But now $\mathbb{R}(X)$ would be an infinite-dimensional normed division algebra, which is impossible by the Gelfand–Mazur theorem. We conclude that there is no multiplicative norm on $\mathbb{R}[X]$.


More generally, a slight modification of this argument shows that there does not exist a multiplicative norm on any non-zero $\mathbb{R}$-algebra $A$ satisfying both of the following properties:

  1. $A$ is a commutative ring with unit;
  2. $A$ is not isomorphic with $\mathbb{R}$ or $\mathbb{C}$.

Proof. If $A$ has non-trivial zero divisors ($a,b\neq 0$, $ab = 0$), then there is no hope of defining a multiplicative norm on $A$, for any norm on $A$ satisfies $\lVert ab\rVert = 0$ but $\lVert a\rVert \lVert b\rVert \neq 0$. So assume that $A$ is an integral domain. We distinguish two cases:

  • If $A$ is finite-dimensional, then we know from abstract algebra that $A$ is a field. But then $A$ is a finite field extension of $\mathbb{R}$, so it must be isomorphic with $\mathbb{R}$ or $\mathbb{C}$, contrary to our assumption.
  • If $A$ is infinte-dimensional, then we use the argument from before. Any multiplicative norm on $A$ extends to a multiplicative norm on $\text{Frac}(A)$, the field of fractions of $A$. Furthermore, the natural map $A \to \text{Frac}(A)$, $a \mapsto \frac{a}{1}$ is injective and $\mathbb{R}$-linear, so we find that $\text{Frac}(A)$ is infinite-dimensional as well. Again it follows from the Gelfand–Mazur theorem that there does not exist a multiplicative norm on $\text{Frac}(A)$, so we conclude that does not exist a multiplicative norm on $A$ either. $\quad\Box$

The result might even extend to the non-commutative case (with the added assumption that $A$ is not isomorphic with $\mathbb{H}$), since the Gelfand–Mazur theorem does not assume commutativity. However, I'm not too well acquainted with “division rings of fractions” in the non-commutative setting, and I hear that their theory features more than a few subtleties.


Despite these more general results, I can't help but feel that there must be a more elementary proof for the special case $A = \mathbb{R}[X]$ from the original question...

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