Why is $\Vert xy\Vert>\Vert x\Vert \Vert y\Vert$ not allowed?

Question

If we look at the usual norm on $\mathbb R$, i.e. $\vert\cdot\vert$, then we see that $\vert xy\vert=\vert x\vert\vert y\vert$. Untill now I've just assumed that this propperty also holds for norms in general, but after giving it a closer look I don't see why. So the question is:

If we have defined a norm on an algebra $X$, what can we say about the relation between $\Vert xy\Vert$ and $\Vert x\Vert\Vert y\Vert$ and why?

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Edit: I've read in the comments that we should have something like $\Vert xy\Vert\leq\Vert x\Vert\Vert y\Vert$, however I'm not yet fully convinced that this is a necessity. For example: what would go wrong if we allow for some $x,y$ that: $\Vert xy\Vert>\Vert x\Vert\Vert y\Vert$

Answer 1

You can think of $\| xy \| \le \| x \| \| y \|$ as a multiplicative version of the triangle inequality $\| x + y \| \le \| x \| + \| y \|$; in that sense, it's as natural to use as the triangle inequality. It's one of the axioms of a normed algebra for several reasons, including but not limited to:

• It guarantees that multiplication is continuous in the norm topology (and might, if my memory serves, in fact be equivalent to multiplication being continuous in the norm topology and the norm of the multiplicative identity being $1$).
• It lets you prove convenient bounds such as $\| \sum a_n x^n \| \le \sum |a_n| \| x \|^n$, which are important in the setting of Banach algebras for understanding how power series in Banach algebras behave.
• It's satisfied by the operator norm on bounded endomorphisms of a normed vector space, and so is a necessary condition for a normed algebra to embed isometrically into an algebra of bounded endomorphisms.

Like any axiom, you can drop it if you want - you can do anything if you want - but there's no guarantee that the result will be interesting. Axioms aren't laws. They're rules of games that people have found interesting to play.

Answer 2

To add to Qiaochu's answer, if $(A,\|\cdot \|)$ is a normed algebra such that multiplication is continuous, then there is an equivalent norm $\|\cdot\|'$ on $A$ such that $$ \|xy\|' \leq \|x\|'\|y\|' $$ In fact, by the uniform boundedness principle, one has that multiplication is jointly continuous. ie. $\exists M \geq 1$ such that $$ \|xy\| \leq M \|x\|\|y\| $$ Define $$ \|x\|' := \sup\{\|\lambda x + xy\| : |\lambda| + \|y\| \leq 1, \lambda\in \mathbb{C}, y\in A\} $$ Then one can see that $$ \|x\| \leq \|x\|' \leq M \|x\| $$ so $\|\cdot \|'$ is equivalent to $\|\cdot \|$. Also, $$ \|xz\|' = \sup\{\|\lambda xz + xzy\| : |\lambda| + \|y\| \leq 1 \} $$ But if $|\lambda| + \|y\| \leq 1$ then $\|\lambda z + yz\| \leq \|z\|'$. So $$ \|xz\|' \leq \sup\{\|\lambda xz + xzy\| : \|\lambda z + yz\| \leq \|z\|'\} $$ $$ \leq \|z\|'\sup\{\|xw\| : \|w\| \leq 1 \} $$ $$ \leq \|x\|'\|z\|' $$ So assuming that the norm is submultiplicative is, for all intents and purposes, the same as assuming that multiplication is continuous, which is obviously desirable when dealing with topological algebras.