5
$\begingroup$

I'm having a bit of trouble showing that the Gelfand transform $A \rightarrow C(\operatorname{sp}(A))$ is isometric iff $\|x^2\| = \|x\|^2$ for a general unital commutative Banach algebra. For a $C^*$ algebra, where we know $xx^*$ is closed in $A$ then this task is simple however I can't see a way of showing this for a general Banach algebra.

Could someone please give me a push in the right direction?

For reference, the $C^*$ algebra proof I know relies on using $y = xx^*$ to show something along the lines of $\|\hat{x}\|^2 = \|\hat{x}\hat{x}^*\| = \|x\hat{x}^*\| =\|\hat{y}\| = \|y\| = \|xx^*\| = \|x^2\|$.

$\endgroup$
3
$\begingroup$

Ah, it's a wonder what stepping away from the problem for a minute can do!

For all those interested, a solution is as follows: $$\|\hat{x}^2\| = \|\hat{x}\|^2$$ is always true by the properties of the Gelfand transform, and so if the Gelfand transform is an isometry it must be the case that $\|x\|^2 = \|x^2\|$

If $\|x^2\| = \|x\|^2$, then we must have $\|x^{2^n}\| = \|x\|^{2^n}$, by repeated application of this. Since we also have $\|x\| = r(x)$, where $r(x)$ is is the spectral radius of $x$. However, because the spectrum $\sigma(x)$ must be the same as the range of the Gelfand transform of $x$, we also have $r(x) = \|\hat{x}\|$, so $\|x\| = \|\hat{x}\|$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.