# Gelfand transform is an isometry

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 $\|\hat{x}\|^2 = \|\hat{x}\hat{x}^*\| = \|x\hat{x}^*\| =\|\hat{y}\| = \|y\| = \|xx^*\| = \|x^2\|$.

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 $\|x\| = \|\hat{x}\|$