Let $X$ be a Banach space with dual space $X^*$, and let $M$ be a closed subspace of $X$. Then $M^\perp=\{x^*\in X^*: x*(m)=0 \text{ for all } m\in M\}$ is a closed subspace in $X^*$. I read the following statement in Garnett's Bounded Analytic Functions:
The Hahn-Banach theorem gives us the isometric isomorphism $(X/M)^*\simeq M^\perp$.
My question is whether the Hahn-Banach theorem is required to show the $\it{isomorphism}$ part of the claim. Consider the following argument.
Let $\pi:X\rightarrow X/M$ be the natural projection, and define $T:(X/Y)^*\rightarrow X^*$ by $Tf=f\circ \pi$. $T$ is readily seen to be an injective homomorphism. What remains to show is that the image of $T$ is $M^\perp$. If $f$ is in the image of $T$, then there is some $g\in (X/M)^*$ such that $Tg=f$. Then for $m\in M$, $f(m)=Tg(m)=g\circ\pi (m)=g(m+M)=0$, and $f\in M^\perp$.
Conversely, suppose $f\in M^\perp$. Define $\bar f:X/M\rightarrow \mathbb{C}$ by $\bar f(x+M)=f(x)$. Then $\bar f$ is linear. To show $\bar f$ is continuous, let $\epsilon>0$ be given. We will show $\bar f$ is continuous at $0+M$. Since $f$ is linear and continuous, it is uniformly continuous. So choose $\delta >0$ such that $$||x-y||<\delta \implies |f(x)-f(y)|<\epsilon.$$ Then if $||x+M||<\delta$, so that there is some $m\in M$ for which $||x-m||<\delta$, then $|\bar f(x+M)|=|f(x)|\leq |f(x)-f(m)|+|f(m)|<\epsilon+0$. Thus $\bar f\in (X/M)^*$ and $T\bar f=\bar f\circ \pi=f$, which shows that the image of $T$ is $M^\perp$.
Is this reasoning correct? Is the Hahn-Banach theorem only invoked for the isometric part? Thank you.
Your reasoning is correct.
In fact, Hahn-Banach is not needed for the isometric part either.
On the other hand, it is needed to show that the dual of a subspace $M\subseteq X$ is isometric to the quotient space $X^*/M^\perp$.