I'm reading the proof of the following theorem in Rudin's "Functional Analysis":
Let $M$ be a closed subspace of a Banach space $X$. The Hahn-Banach theorem extends each $m^* \in M^*$ to a functional $x^* \in X^*$. Define $\sigma (m^*) = x^* + M^{\perp}$. Then $\sigma$ is an isometric isomorphism of $M^*$ onto $X^*/M^{\perp}$.
My problem is this part of the proof:
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 $|| \sigma( m^*)|| = ||m^*||$.
3.3 Theorem
Suppose $M$ is a subspace of a vector space $X$, $p$ is a seminorm on $X$, and $f$ is a linear functional on $M$ such that $|f(x)| \le p(x) \ \ (x \in M)$. Then $f$ extends to a linear functional $\Lambda$ on $X$ that satisfies $| \Lambda x | \le p(x)$ $ \ \ (x \in X)$.
What I don't see is how theorem 3.3 implies the equality of norms: $||x^*|| = ||m^*||$ and why is $||m^*|| \le ||x^*||$ (in the first line). It would be clear to me if $||x^*||$ was a regular norm of a linear mapping (we take the supremum over a bigger set) but I don't see it if it is a quotient norm.
Could you explain that to me?
Thank you
