Gelfand transform of a Banach algebra need not be isometric

The multiplication $(a+b\epsilon )(c+d\epsilon )=ac+(ad+bc)\epsilon$ is associative, commutative and complex-bilinear, and the norm is submultiplicative: $\mathrm{\parallel }xy\mathrm{\parallel }=\mathrm{\mid }ac\mathrm{\mid }+\mathrm{\mid }ad+bc\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }d\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }\le (\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid })(\mathrm{\mid }c\mathrm{\mid }+\mathrm{\mid }d\mathrm{\mid })=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$; the unit is $1=1+0\epsilon$ with $\mathrm{\parallel }1\mathrm{\parallel }=1$, and the algebra is complete because its two coordinates are controlled by the norm ($\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$, $\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$) and conversely $\mathrm{\parallel }x\mathrm{\parallel }\le 2\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$, so the norm is equivalent to the Euclidean norm on ${\mathbb{C}}^2$; hence $A$ is a nonzero commutative unital complex Banach algebra.

Every character $\chi$ of $A$ satisfies $\chi (\epsilon {)}^2=\chi ({\epsilon }^2)=\chi (0)=0$ by multiplicativity [L1], so $\chi (\epsilon )=0$ since $\mathbb{C}$ is a field; hence $\chi (a+b\epsilon )=a\chi (1)+b\chi (\epsilon )=a$ by linearity and unitality [L1]; in particular $\mathrm{\Delta }(A)$ consists of the single character ${\chi }_0(a+b\epsilon )=a$.

The element $\epsilon =0+1\epsilon$ has norm $\mathrm{\parallel }\epsilon \mathrm{\parallel }=1\mathrm{\ne }0$, while its Gelfand transform vanishes identically: $\widehat{\epsilon }({\chi }_0)={\chi }_0(\epsilon )=0$ by [step 1.2]; hence $\mathrm{\parallel }\widehat{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}=0\mathrm{\ne }1=\mathrm{\parallel }\epsilon \mathrm{\parallel }$, so the Gelfand transform of $A$ is not isometric (and not injective, since $\epsilon \mathrm{\ne }0$ has zero transform).