Algebra norm

An algebra norm is a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra in which the original algebra embeds densely.

Let $S$ be a bounded multiplicatively closed subset of a unital Banach algebra. The displayed auxiliary norm $p$ is equivalent to the original norm, and each left multiplication $L_s$ is a contraction for $s\in S$. Taking the operator norm of the faithful left regular representation produces an equivalent algebra norm with $\Vert 1{\Vert }_S=1$ and $\Vert s{\Vert }_S\le 1$. If $M=\max (\Vert 1\Vert ,{\sup }_S\Vert s\Vert )$, then $\Vert a\Vert /M\le \Vert a{\Vert }_S\le M\Vert a\Vert$. The auxiliary norm is itself submultiplicative, since $p(ab)\le p(a)\Vert b\Vert \le p(a)p(b)$, but $p(1)=M$ can exceed one. The operator norm enforces the required unital normalization.

For a finite commuting family in a unital Banach algebra, divide each $x_j$ by $r(x_j)+\epsilon /2$. The spectral radius formula makes each normalized element's powers bounded. Commutativity lets every product be reordered into one power of each generator, so the entire generated semigroup is bounded. Renorming a bounded multiplicative semigroup gives the displayed bounds simultaneously. Applying this to two commuting elements proves submultiplicativity and subadditivity of their spectral radius.

An algebra norm is submultiplicative when it satisfies the displayed product inequality. Induction gives $\Vert a_1\cdots a_n\Vert \le {\prod }_j\Vert a_j\Vert$, and in particular $\Vert a^n\Vert \le \Vert a{\Vert }^n$. The operator norm has this property because $\Vert STv\Vert \le \Vert S\Vert \Vert T\Vert \Vert v\Vert$. The supremum norm for bounded functions with pointwise multiplication is another example. This estimate supports the Neumann series and the spectral radius formula.

Take disjoint disks escaping to infinity and continuous cutoff functions $h_n$ supported in them and equal to one on smaller disks. The locally finite function $f={\sum }_{n}n{h}_n$ is continuous. Choose nonzero $g_n$ supported in the smaller disks; then $f{g}_n=n{g}_n$. Any algebra norm would give $n\Vert g_n\Vert =\Vert f{g}_n\Vert \le \Vert f\Vert \Vert g_n\Vert$, forcing $n\le \Vert f\Vert$ for every positive integer. This is impossible even without a completeness assumption.

The displayed formula is a submultiplicative norm on the algebra of entire functions: definiteness follows from the identity theorem. No submultiplicative norm on the whole algebra can be complete. The coordinate function $Z(z)=z$ has every complex number in its algebraic spectrum of an element, because $Z-\lambda$ vanishes somewhere and has no entire inverse. This contradicts compactness of spectra in a Banach algebra.

The displayed supremum norm is finite on every entire function, positive on every nonzero one by the identity theorem, and submultiplicative under pointwise multiplication. It is not complete: partial geometric sums ${\sum }_{j=0}^n(z/2{)}^j$ are Cauchy on the disk but their limit there extends meromorphically with a pole at two. More generally, no complete algebra norm exists on all entire functions because the coordinate function has algebraic spectrum of an element equal to the whole complex plane.

Any submultiplicative algebra norm on all of $C(K)$ dominates its supremum norm, even when the new norm is incomplete. Every evaluation character extends continuously to the completion: the restrictions of the completion's character space form a closed subset of $K$, and an omitted point would give a nonzero function annihilated by an invertible function in the completion.