Introduction to Banach Spaces and Algebras, by Allan, Exercise 4.6

Question

I am reading Introduction to Banach Spaces and Algebras, by Allan,
Unfortunately I do not succeed in completing, Exercise 4.6

Let $(A, \|\cdot\|)$ be a normed algebra. A Banach algebra $(B, \||\cdot|\|)$ is a Banach extension of $(A, \|\cdot\|)$, if there is an isometric embedding of $(A, \|\cdot\|)$ to $(B, \||\cdot|\|)$

Let $A$ and $B$ be Banachalgebra , let $I$ be a closed ideal in $B$ and $\phi: A \to B$ be a continuous homomorphism. set $U=A\oplus I$, whit the norm $$\|(a, x)\|=\|a\|+\|x\|$$ $a\in A , x\in I$

and define a product on $U$ by setting

$$(a_1, a_2)(b_1, b_2)=(a_1 a_2,b_1 b_2+\phi(a_1)b_2+b_1\phi(a_2) )$$ $a_1, a_2\in A , b_1, b_2\in I$

prove that $(U, \|\cdot\|)$ is a Banach extension of $A$

Any comment or response is appreciated.

Answer 1

In the comments, you indicate that your difficulty is finding an isometric embedding $A \to U$.

To do this, note that since $I$ is an ideal of $B$, it is at least an additive subgroup and hence contains $0$. We can then define $i: A \to U$ by setting $i(a) = (a,0)$. It is clear that this map is an isometric embedding by definition of $(U, \|\cdot\|_U)$.