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When are homomorphisms between Banach algebras contractions?

I recall from my student days that there are results which show that a positive answer to the above question holds under very general conditions (on the algebras). I have tried a literature search but only turned up the fact that it is true for $\ast$-homomorphisms between $C^\ast$-algebras (Proposition 3.2.2 in Dales et al., “Banach spaces of continuous functions as dual spaces”). Does any forum member have a reference for a definitive account of this theme?

I should add that I know that this not true for ANY pair of algebras, and I know that there are substantial results on classes of algebras for which it IS true. As a non-expert on complete normed algebras, I am looking for some references on the present state of the art.

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    $\begingroup$ I am not sure what results you are thinking of since there are plenty of automorphisms of $M_2({\bf C})$ which send $I_2$ to $I_2$ but have norm strictly greater than 1... $\endgroup$ – Yemon Choi Oct 16 at 7:10
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    $\begingroup$ Characters on commutative Banach algebras are contractive, that much is true... $\endgroup$ – Yemon Choi Oct 16 at 7:11
  • $\begingroup$ Again: can you help us to narrow down what classes of algebras were mentioned in the results that you are thinking of? As I said in my answer, either there are some missing conditions on the homomorphisms, or the class of range algebras must be restricted quite heavily. $\endgroup$ – Yemon Choi Oct 16 at 13:56
  • $\begingroup$ I figured "automatically nonexpansive" would be a good search term, but when I googled it almost all the results were things I have written ... so I guess the term isn't in common use. Googling "automatically contractive" brought up a few other sources but I didn't see anything like a general reference on the subject. $\endgroup$ – Nik Weaver Oct 16 at 16:28
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About the only positive result that comes to mind is the fact that homomorphisms cannot increase the spectral radius, so that if the range space is a uniform algebra then homomorphisms are necessarily contractive.

In my view and in my experience, at the level of generality considered by this question, the assertion/hope that "a positive answer to the above question holds under very general conditions" is not backed up by evidence. The fact that the answer is negative even for bijective homomorphisms ${\bf M}_2 \to {\bf M_2}$, with both sides carrying the natural ${\rm C}^*$-norm, is one reason to doubt that much can be said in the noncommutative setting, without extra restrictions on the nature of the homomorphism.

(To get such homomorphisms, let $$ s_t= \left(\matrix{ 1 & t \\ 0 & 1 } \right)$$ and consider the automorphism of ${\bf M}_2$ given by $x \mapsto s_t^{-1} x s_t$. A simple calculation shows that the norm-1 element $$p=\left(\matrix{1 & 0 \\ 0 & 0 }\right)$$ satisfies $$s_t^{-1}ps_t = \left(\matrix{ 1 & t \\ 0 & 0} \right)$$ and the latter matrix has norm $ > |t|$.)

One can also find commutative unital Banach algebras with trivial Jacobson radical (and hence for which the spectral radius does at least see every element) such that there are continuous unital endomorphisms of the algebra with norm strictly bigger than $1$. For instance, take $A_+=\ell^1({\bf Z}_+)$ with convolution product (a.k.a. the completion of the polynomial ring ${\bf C}[z]$ in the natural $\ell^1$-norm). Continuous unital endomorphisms of $A_+$ are uniquely determined by where they send the generating element $\delta_1$ (thought of as the variable $z$) and conversely every power-bounded element $a\in A_+$ definess a continuous unital endomorphism of $A_+$ which sends $\delta_1\mapsto a$. It now remains to note that there exist power-bounded elements of $A_+$ which have norm $>1$; see

MR0241980 (39 #3315) Reviewed D. J. Newman, Homomorphisms of $l_+$. Amer. J. Math. 91 (1969), 37–46. https://mathscinet.ams.org/mathscinet-getitem?mr=241980

which provides the example $a= (\delta_0+\delta_1-\delta_2)/\sqrt{5}$ among others.

To my mind, if a property of pairs of Banach algebras ("every continuous homomorphism from $A$ to $B$ is contractive") fails for $A=B={\bf M}_2$ or $A=B=\ell^1({\bf Z}_+)$ with convolution, it is hard to justify a hope or claim that it holds for a wide class of algebras. Being by training a Banach algebraist rather than a ${\rm C}^*$-algebraist, I just don't see why one would expect homomorphisms to be automatically contractive.

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