Is a contractive automorphism of a Banach algebra an isometry?

Question

If $A$ and $B$ are C*-algebras and $T:A\to B$ is an injective $^*$-homomorphism, then $T$ is necessarily isometric, that is $\|T(a)\|=\|a\|$, for every $a$ in $A$. In particular, every automorphism of $A$ is automatically isometric.

However this is not necessarily true for Banach algebras: for a counter-example one could take $A$ to be any Banach space with multiplication defined to be identically zero, that is $ab=0$, for every $a$ and $b$ in $A$, and then it is easy to produce a contractive automorphism of $A$ which is not isometric, e.g. $T(a) =a/2$.

In order to avoid such pathologies one could require $A$ to be unital and semi-simple. The question would then be:

Let $A$ be a commutative semisimple Banach algebra with unit and $T$ be an automorphism of $A$ such that $\|Ta\| \leq \|a\|, a \in A$. Is it true that $T$ is an isometry?

Answer 1

Ok, so I believe I've come up with a counter-example. Consider $\ell ^\infty ({\mathbb Z})$ as a Banach algebra with coordinatewise multiplication and norm $$ \|a\| = \sup_ic_i|a_i|, $$ where the $c_i$ are positive real numbers, with $1<c_i<c_{i+1}$, for every $i\in {\mathbb Z}$, and $$ \lim_{i\to -\infty }c_i = 1, \quad\text{and}\quad \lim_{i\to \infty }c_i = 2. $$ It is not hard to see that this is indeed a sub-multiplicative norm due to the fact that the $c_i$ are bigger than one. Moreover, since the $c_i$ are bounded above and below, this norm is equivalent to the usual sup-norm on $\ell ^\infty ({\mathbb Z})$, hence a complete norm. That $\ell ^\infty ({\mathbb Z})$ is semisimple is also clear, since it is isomorphic to a C$^*$-algebra.

The automorphism given by $T(a)_i = a_{i+1}$ is then contractive but its inverse, $ T^{-1}(a)_i = a_{i-1}, $ isn't.