Norms on unitization of nonunital Banach algebra Let $A$ be a nonunital Banach algebra and denote by $A^+$ the unitization of $A$. One commonly used Banach algebra norm on $A^+$ is given by $||(a,\lambda)||=||a||+|\lambda|$ (where $a\in A,\lambda\in\mathbb{C}$). Another possibility is $||(a,\lambda)||=\sup_{b\in A,||b||\leq 1}||ab+\lambda b||$, which is a $C^*$ norm when $A$ is a $C^*$-algebra. 
Are the two norms equivalent? And if $\pi:A^+\rightarrow\mathbb{C}$ is the natural homomorphism, and $A^+$ is given the second norm, is it still true that $||\pi||\leq 1$ when $A$ is a Banach (not necessarily $C^*$) algebra?
 A: Let $\| \cdot \|$ be the norm on $A$, but the other two norms let be
$$ \| (a,\lambda)\|_1=\| a\|+|\lambda|\quad \text{and}\quad \| (a,\lambda)\|_2=\sup\{ \| ab+\lambda b\|; b\in A, \| b\| \leq 1\}. $$
It is obvious that $\| ab+\lambda b\| \leq (\| a\|+|\lambda|)\| b\|$ which gives $\|(a,\lambda)\|_2\leq \|(a,\lambda)\|_1$. Hence, the identity operator $I:(a,\lambda) \mapsto (a,\lambda)$ is a bounded bijection from $(A^+, \| \cdot \|_1)$ to $(A^+, \| \cdot \|_2)$.
By the Open Mapping Theorem, the inverse is bounded as well, i.e., there exists a constant $c>0$ such that $\|(a,\lambda)\|_1\leq c \|(a,\lambda)\|_2$. This proves the equivalence of the norms.
Let $\pi:(A^+,\| \cdot\|) \to {\mathbb C}$ be given by $\pi(a,\lambda)=\lambda$, where $\| \cdot \|$ is an arbitrary norm which makes $A^+$ a unital Banach algebra. It is
easy to see that $\pi$ is a nonzero homomorphism. Since $\pi$ is multiplicative it maps invertible elements into invertible elements (=non-zero numbers). Denote by $\sigma(\cdot)$ the spectrum and by $r(\cdot)$ the spectral radius.
It is obvious that $\sigma(\pi(a,\lambda))=\{\lambda\}$ and $\lambda\in \sigma((a,\lambda))$.
Hence $|\lambda|=r(\pi(a,\lambda))\leq r((a,\lambda))$. It is a standard fact that in any  unital Banach algebra the spectral radius is smaller than the norm. Thus we conclude that
$$ |\lambda|=r(\pi((a,\lambda))\leq r((a,\lambda))\leq\|(a,\lambda)\|. $$
Hence $\| \pi\| \leq 1$. Actually, $\| \pi\|=1$ as $\pi((0,1))=1$.
