# Normed algebra with multiplicative norm, non-isomorphic to $R$, $C$, $H$.

Is there an infinite dimensional real normed algebra $A$ such that $\|xy\|=\|x\|\cdot \|y\|$ for all $x,y \in A$?

Thanks.

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First of all, any normed algebra with a submultiplicative norm can be completed to a Banach algebra, and if the equality you require holds for all elements in the original algebra, it will hold for all elements in the completion. – user16299 Feb 17 '12 at 22:53
Secondly, by stipulating that $A$ is infinite-dimensional, you rule out the three example you mention. (Your question also seems to tacitly assume you are working with real normed algebras, otherwise the examples you give don't fit.) Could you please edit to clarify these points? – user16299 Feb 17 '12 at 22:55
Related: math.stackexchange.com/q/529/152 – Grigory M Nov 3 '12 at 12:21

It depends on whether you assume $A$ to have a unit element or not and whether $A$ is associative or not.

If $A$ has a unit element then the answer is no, independent of associativity of $A$. This is the main result in the article

Kazimierz Urbanik and Fred B. Wright, Absolute-valued algebras, Proc. Amer. Math. Soc. 11 (1960), 861-866.

Urbanik and Wright give the following example of an infinite-dimensional non-associative and non-unital Banach algebra with the desired property: Fix a bijection $\phi\colon \mathbb{N} \times \mathbb{N} \to \mathbb{N}$. For $x, y \in \ell^2(\mathbb{N})$ define $z = xy$ for $n = \phi(k,l)$ by $z_n = x_k \cdot y_l$. Then $$\lVert z \rVert^2 = \sum_{n=1}^\infty z_n^2 = \sum_{k,l = 1}^\infty x_k^2 y_l^2 = \left(\sum_{k=1}^\infty x_k^2\right)\left(\sum_{l=1}^\infty y_k^2\right) = \lVert x\rVert^2 \lVert y\rVert^2$$ shows that with this multiplication $\ell^2(\mathbb{N})$ has the desired property.

See also this thread on MathOverflow where the above paper is in the answer by Faisal and the answer by Andreas Thom points out that Mazur proved that the only real, associative and unital normed division algebras are the familiar $\mathbb{R},\mathbb{C},\mathbb{H}$.

I don't know whether there is an associative non-unital example or not since the property is destroyed upon passing to the unitization.

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The problem of adjoining a unit (and thus reducing to the unital case) is the possibility of the existence of a non-unit idempotent $e$: if $e^2 = e$ then $(-e,1)(e,0) = (-e^2 + 1e - e0, 0) = (0,0)$ shows that the unitization has zero-divisors. I don't know how to rule the existence of idempotents out or how to prove the (non-)existence of associative non-unital algebras satisfying $\lvert ab \rvert = \lvert a \rvert \lvert b \rvert$. – commenter Nov 3 '12 at 13:04
So, Urbanik-Wright's article gives the answer to this question math.stackexchange.com/questions/222716/…. – Davide Giraudo Nov 3 '12 at 13:51
Yes, but upon completion the other question is answered by the much easier Mazur theorem: the polynomial ring is associative and unital, so the completion of $\mathbb{R}[x]$ with respect to $N$ is a unital Banach division algebra (but from the question it seems that there is a much more elementary argument, provided the fellow student's argument was correct). The present question is a bit unclear to me: is associativity and/or a unit assumed? Maybe I should have posted a comment instead of an answer. – commenter Nov 3 '12 at 14:04