Suppose that $X$ be a compact space and $\left|\cdot\right|$ be an algebra norm on $C(X)$

Is every algebra norm $\left|\cdot\right|$ on $C(X)$ equivalent to uniform norm $\left|\cdot\right|_X$?

I don't know where to start. Any clues?

  • $\begingroup$ By "algebra norm" do you mean "Banach algebra norm"? $\endgroup$ Jan 25 '19 at 15:05
  • $\begingroup$ @Robert Israel, not necessary Banach $\endgroup$
    – user62498
    Jan 25 '19 at 15:06
  • $\begingroup$ So just submultiplicative? $\endgroup$ Jan 25 '19 at 15:10
  • $\begingroup$ @ Robert Israel , yes $\endgroup$
    – user62498
    Jan 25 '19 at 15:15

I am going to use $\|\cdot\|$ and $\|\cdot\|_X$ for the two norms. The identity map from $(C(X), \|\cdot\|)$ to $(C(X), \|\cdot\|_X)$ is nonexpansive. That is, $|f(x)| \leq \|f\|$ for all $f \in C(X)$ and $x \in X$; this is just because evaluating at $x$ is a complex homomorphism and hence must take $f$ to some point in its spectrum, which is contained in the disk of radius $\|f\|$ about the origin by the spectral radius formula.

Thus, the identity map fails to be an equivalence if and only if the reverse map from $(C(X), \|\cdot\|_X)$ to $(C(X), \|\cdot\|)$ is unbounded. By the Banach isomorphism theorem, this is the same as asking whether there exists an incomplete algebra norm on $C(X)$. If $X$ is any infinite compact Hausdorff space, then Dales and Esterle showed that the continuum hypothesis implies the existence of an incomplete algebra norm on $C(X)$, and Woodin showed that $\neg$CH plus Martin's axiom implies that there is no incomplete algebra norm on $C(X)$. I believe both results are covered in the book An Introduction to Independence for Analysts by Dales and Woodin.


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