Unitization of a nonunital Banach algebra

Associativity: expanding both sides of the associativity identity for $((\lambda ,a)(\mu ,b))(\nu ,c)$ and $(\lambda ,a)((\mu ,b)(\nu ,c))$ by bilinearity gives the common value $(\lambda \mu \nu ,\lambda \mu c+\lambda \nu b+\mu \nu a+\lambda (bc)+\mu (ac)+\nu (ab)+(ab)c):$ the left side produces $\lambda \mu c+\nu (\lambda b+\mu a+ab)+(\lambda b+\mu a+ab)c$ and the right side produces $\lambda (\mu c+\nu b+bc)+\mu \nu a+a(\mu c+\nu b+bc)$, and the two agree because scalars may be moved across the product, the multiplication of $A$ is bilinear, and $a(bc)=(ab)c$ by associativity.

The element $(1,0)$ is a two-sided identity: $(1,0)(\mu ,b)=(\mu ,b+0+0)=(\mu ,b)$ and $(\lambda ,a)(1,0)=(\lambda ,0+a+0)=(\lambda ,a)$. Submultiplicativity holds because $\mathrm{\parallel }(\lambda ,a)(\mu ,b)\mathrm{\parallel }=\mathrm{\mid }\lambda \mu \mathrm{\mid }+\mathrm{\parallel }\lambda b+\mu a+ab\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{\parallel }b\mathrm{\parallel }+\mathrm{\mid }\mu \mathrm{\mid }\mathrm{\parallel }a\mathrm{\parallel }+\mathrm{\parallel }a\mathrm{\parallel }\mathrm{\parallel }b\mathrm{\parallel }=(\mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\parallel }a\mathrm{\parallel })(\mathrm{\mid }\mu \mathrm{\mid }+\mathrm{\parallel }b\mathrm{\parallel })=\mathrm{\parallel }(\lambda ,a)\mathrm{\parallel }\mathrm{\parallel }(\mu ,b)\mathrm{\parallel }$ by [L1], and $\mathrm{\parallel }(1,0)\mathrm{\parallel }=1$.

Completeness: a sequence $({\lambda }_n,a_n)$ is Cauchy in the sum norm exactly when $({\lambda }_n)$ is Cauchy in $\mathbb{C}$ and $(a_n)$ is Cauchy in $A$ (the two inequalities $\mathrm{\mid }\lambda \mathrm{\mid },\mathrm{\parallel }a\mathrm{\parallel }\le \mathrm{\parallel }(\lambda ,a)\mathrm{\parallel }\le \mathrm{\mid }\lambda \mathrm{\mid }+\mathrm{\parallel }a\mathrm{\parallel }$ compare the norm with the maximum of the coordinate norms); since $\mathbb{C}$ and $A$ are complete by [L1], the coordinates converge and their pair is the limit; so $\tilde{A}$ is a complex Banach algebra.

The map $j(a):=(0,a)$ is isometric and multiplicative: $j(ab)=(0,ab)=(0,a)(0,b)=j(a)j(b)$, and $\mathrm{\parallel }j(a)\mathrm{\parallel }=0+\mathrm{\parallel }a\mathrm{\parallel }$; its image is a two-sided ideal because $(\lambda ,b)(0,a)=(0,\lambda a+ba)$ and $(0,a)(\mu ,b)=(0,\mu a+ab)$, and it is closed as the kernel of the continuous scalar projection $(\lambda ,a)\mapsto \lambda$.

By [L2] applied in $\tilde{A}$, the spectrum of $a\in A$ is the set of $\lambda$ with $(\lambda ,0)-(0,a)=(\lambda ,-a)$ not invertible, which is the convention displayed in the statement.