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Is the norm on a unital semi-simple commutative Banach algebra with $\|I\|=1$, unique? ($I$ denotes the identity element)

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  • $\begingroup$ Obligatory comment for other MO readers, who are algebraists: here semi-simple means "trivial Jacobson radical" $\endgroup$
    – Yemon Choi
    Jan 3, 2017 at 17:23

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Yes, provided that you mean "unique up to equivalence of norms".

In fact this is true even for semisimple noncommutative Banach algebras. This more general result is due to B. E. Johnson in 1967. A second proof, which is very different, was given by B. Aupetit in 1982 with subsequent simplifications by T. J. Ransford. All of this can be found in Section 5.1 of Dales's book Banach Algebras and Automatic Continuity.


EDIT. The OP has clarified that he really wants uniqueness of norms in the "literal" sense (my looser interpretation was based on the original convention of Rickart when he first stated these questions back in the 1950s).

I happen to know that $C^1[0,1]$ has two different but equivalent unital Banach algebra norms (making it a commutative semisimple Banach algebra). Thinking about the proof, and the similar case of Lipschitz algebras, leads to a $2$-dimensional counterexample to your question. However, the example below feels overly complicated and there should be something simpler!

Consider ${\mathbb C}^2$ with pointwise product. The usual max norm is unital and submultiplicative. On the other hand, consider the map $\theta: {\mathbb C}^2 \to M_2({\mathbb C})$ given by $$ \theta(a,b) = \left( \matrix{ a & a-b \\ 0 & b } \right) $$ A little calculation shows that $\theta$ is a unital algebra homomorphism. If we equip $M_2({\mathbb C})$ with the usual operator norm (viewing matrices as maps between Hilbert spaces) then we may define $$ \Vert (a,b)\Vert_\theta = \Vert \theta(a,b)\Vert_{\rm op} $$ Then $\Vert (1,0) \Vert_\theta \geq \sqrt{2} > \Vert(1,0)\Vert_{\rm max}$, so the norms are different.

(I suspect that there is a stronger statement, namely that the Banach spaces $({\mathbb C}^2, \max)$ and $({\mathbb C}^2, \Vert\cdot\Vert_\theta)$ are not isometrically isomorphic; but I haven't had time to think of a proper proof. Probably one can show this by looking at the shapes of the unit spheres for these norms.)

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  • $\begingroup$ It is clear that any two norms on a semi-simple commutative Banach algebra are equivalent, but I want to know, if two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on such a Banach algebra are equal, if they are equal at $I$? $\endgroup$ Jan 3, 2017 at 17:37
  • $\begingroup$ @T.Amdeberhan Normalization may destroy submultiplicativity $\endgroup$
    – Yemon Choi
    Jan 3, 2017 at 17:59
  • $\begingroup$ @T.Amdeberhan The renorming trick that you mention, which all of us Banach algebraists did get taught at one point, says nothing IIRC about whether starting with two different complete Banach space norms you end up with two different renormings. $\endgroup$
    – Yemon Choi
    Jan 3, 2017 at 18:19
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    $\begingroup$ Sorry, I never meant to offend anyone or you. I was trying to be helpful, I did not know your background. I'm not commenting about this problem any more. $\endgroup$ Jan 4, 2017 at 18:18

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