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

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?

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