Under what conditions is a norm on an algebra equivalent to a submultiplicative norm? I've heard that the reason normed algebras are defined as having a submultiplicative norm is because there is no loss of generality, since every norm on an algebra is equivalent to a submultiplicative norm.
I can't quite figure out why this would be the case, or whether it only applies to specific types of algebras.
Not having much luck searching on the net. Anyone know anything about this?
 A: Lemma.  Let $X$ be a Banach space equipped with a separately continuous  bilinear map
$$
  \mu :X\times X\to X.
  $$
Then there exists a constant $C$ such that
$$
  \|\mu (x,y)\|\leq C\|x\|\|y\|, \quad \forall x,y\in X.
  $$
Proof. For each $a$ in $X$,  let
$$
  L_a : x\in  X\mapsto  \mu (a,x)\in  X,
$$ and $$
  R_a : x\in  X\mapsto  \mu (x, a)\in  X.
  $$
Each $L_a$ and each $R_a$ is bounded by assumption and we claim that
$$
  \sup_{\|a\|\leq 1}\|L_a\|<\infty .
  $$
By the uniform boundedness principle it
is enough to show that, for every fixed $x$ in $X$, one has that
$$
  \sup_{\|a\|\leq 1}\|L_a(x)\|<\infty ,
  \tag{1}
  $$
but this follows easily from
$$
  \|L_a(x)\| =   \|ax\| =   \|R_x(a)\| \leq  $$$$ \leq\|R_x\|\|a\| \leq  \|R_x\|.
  $$
Letting $C$ be the supremum in (1), we get
$$
  \|\mu(x,y)\| =
  \|x\|\left\|\mu\left({x\over \|x\|},y\right)\right\| = $$$$ =
 \|x\|\|L_{{x\over \|x\|}}(y)\|   \leq C\|x\|\|y\|.
  $$
QED
Corollary. If $A$ is a Banach space as well as an algebra, such that the multiplication operation  is separately continuous, then
there exists a constant $C$, such that the norm $|||\cdot|||$ defined  on $A$ by
$$
  |||a||| := C\|a\|,
  $$
is submultiplicative.
Proof.   By the Lemma there  is a constant $C$ such that
$$
  \|xy\|\leq C\|x\|\|y\|, \quad \forall x,y\in X,
  $$
so
$$
  C\|xy\|\leq (C\|x\|)(C\|y\|).
  $$
QED
