$M$ is a closed subspace of a Banach space $X$
part (a) of theorem 4.9 proves $M^*=X^*/M^{\perp}$ ("=" means isometric isomorphism)
He constructed a map $\sigma:M^*\to X^*/M^{\perp}$
$$\sigma (m^*)=x^*+M^{\perp}$$
and has already proved $\sigma$ is a isomorphism, we need to show $\sigma$ is isometric, the proof is
Fix $m^* \in M^*$. If $x^* \in X^*$ extends $m^*$, it is obvious that $||m^*|| \le ||x^*||$
The greatest lower bound of the numbers $||x^*||$ so obtained is $||x^* + M^{\perp}||$, by the definition of the quotient norm. Hence $||m^*|| \le || \sigma (m^*) || \le ||x^*||$ But Theorem 3.3 furnishes an extension $x^*$ of $m^*$ with $||x^*|| = ||m^*||$. It follows that $|| \sigma( m^*)|| = ||m^*||$.
I don't understand "The greatest lower bound of the numbers $||x^*||$ so obtained is $||x^* + M^{\perp}||$", what does "so obtained" mean? and how to get $||m^*|| \le || \sigma (m^*) || \le ||x^*||$ from the definition of quotient norm?
(quotient norm : $\|\sigma(m^*)\|=\|x^*+M^{\perp}\|=\inf\{\|x^*-y\|,y\in M^{\perp}\}$)
(if my description of the theorem is unclear, your can refer to this post Question about a proof in Rudin's book - annihilators)
