[proofplan]
We prove that the map $\Theta: X^*/M^\perp \to M^*$, defined by $\Theta(f + M^\perp) = f|_M$, is an isometric isomorphism. Well-definedness follows from the definition of $M^\perp$: two functionals in the same coset agree on $M$. The isometry requires showing $\|f + M^\perp\|_{X^*/M^\perp} = \|f|_M\|_{M^*}$, where the key tool is the [Norm-Preserving Extension](/theorems/880) (Hahn-Banach), which guarantees that every functional on $M$ extends to $X$ without increasing the norm. Surjectivity also follows from Hahn-Banach extension.
[/proofplan]
[step:Verify $\Theta$ is well-defined: functionals in the same coset restrict identically to $M$]
Let $f, g \in X^*$ with $f + M^\perp = g + M^\perp$, i.e., $f - g \in M^\perp$. Then for every $m \in M$:
\begin{align*}
f(m) - g(m) = (f - g)(m) = 0,
\end{align*}
since $f - g \in M^\perp$ means $(f - g)(m) = 0$ for all $m \in M$. Hence $f|_M = g|_M$, so $\Theta(f + M^\perp) = f|_M$ does not depend on the choice of representative.
[guided]
The annihilator $M^\perp = \{h \in X^* : h(m) = 0 \text{ for all } m \in M\}$ is a closed subspace of $X^*$ (closed because it is the intersection $\bigcap_{m \in M} \ker(\hat{m})$ of [closed sets](/page/Closed%20Set), where $\hat{m}: X^* \to \mathbb{R}$ is the evaluation functional $\hat{m}(f) = f(m)$, which is [continuous](/page/Continuity) in the norm topology of $X^*$).
The quotient $X^*/M^\perp$ is therefore a [Banach space](/page/Banach%20Space) with the quotient norm $\|f + M^\perp\|_{X^*/M^\perp} = \inf_{h \in M^\perp} \|f + h\|_{X^*}$.
Two elements $f, g$ lie in the same coset of $M^\perp$ precisely when $f - g$ vanishes on $M$, which means $f$ and $g$ have the same restriction to $M$. This is exactly what makes $\Theta$ well-defined: the restriction $f|_M$ depends only on the coset $f + M^\perp$.
[/guided]
[/step]
[step:Show $\Theta$ is linear and bounded with $\|\Theta(f + M^\perp)\|_{M^*} \le \|f + M^\perp\|_{X^*/M^\perp}$]
**Linearity.** For $f, g \in X^*$ and $\lambda \in \mathbb{R}$:
\begin{align*}
\Theta((f + M^\perp) + \lambda(g + M^\perp)) &= \Theta((f + \lambda g) + M^\perp) = (f + \lambda g)|_M = f|_M + \lambda g|_M \\
&= \Theta(f + M^\perp) + \lambda \Theta(g + M^\perp).
\end{align*}
**Boundedness.** For any $h \in M^\perp$ and any $m \in M$ with $\|m\|_X \le 1$:
\begin{align*}
|f(m)| = |(f + h)(m)| \le \|f + h\|_{X^*} \|m\|_X \le \|f + h\|_{X^*},
\end{align*}
using $h(m) = 0$ in the first equality. Taking the supremum over $\|m\| \le 1$:
\begin{align*}
\|f|_M\|_{M^*} \le \|f + h\|_{X^*}.
\end{align*}
Since this holds for every $h \in M^\perp$, taking the infimum over $h$:
\begin{align*}
\|f|_M\|_{M^*} \le \inf_{h \in M^\perp} \|f + h\|_{X^*} = \|f + M^\perp\|_{X^*/M^\perp}.
\end{align*}
[guided]
**Linearity** follows from the linearity of restriction and the vector space operations on the quotient: $(f + M^\perp) + \lambda(g + M^\perp) = (f + \lambda g) + M^\perp$.
**Boundedness.** The key identity is $f(m) = (f + h)(m)$ for any $h \in M^\perp$ and $m \in M$, since $h(m) = 0$. This means we can replace $f$ by any element of its coset when evaluating on $M$. The norm bound $|(f + h)(m)| \le \|f + h\|_{X^*}$ (for $\|m\| \le 1$) then gives
\begin{align*}
\|f|_M\|_{M^*} = \sup_{\substack{m \in M \\ \|m\| \le 1}} |f(m)| = \sup_{\substack{m \in M \\ \|m\| \le 1}} |(f+h)(m)| \le \|f + h\|_{X^*}.
\end{align*}
Taking the infimum over all $h \in M^\perp$ gives $\|f|_M\|_{M^*} \le \|f + M^\perp\|_{X^*/M^\perp}$.
[/guided]
[/step]
[step:Prove the reverse inequality $\|f + M^\perp\|_{X^*/M^\perp} \le \|f|_M\|_{M^*}$ using Hahn-Banach extension]
We show that for any $f \in X^*$, the infimum $\inf_{h \in M^\perp} \|f + h\|_{X^*}$ is at most $\|f|_M\|_{M^*}$.
Consider the restriction $f|_M \in M^*$. By the [Norm-Preserving Extension of Linear Functionals](/theorems/880), applied to the normed space $X$ and its subspace $M$: since $f|_M \in M^*$, there exists an extension $\tilde{g} \in X^*$ with $\tilde{g}|_M = f|_M$ and $\|\tilde{g}\|_{X^*} = \|f|_M\|_{M^*}$.
The hypotheses of the [Norm-Preserving Extension](/theorems/880) are: $X$ is a real [normed vector space](/page/Normed%20Vector%20Space) and $M \subset X$ is a subspace — both hold.
Now define $h := f - \tilde{g} \in X^*$. For any $m \in M$:
\begin{align*}
h(m) = f(m) - \tilde{g}(m) = f|_M(m) - f|_M(m) = 0.
\end{align*}
Hence $h \in M^\perp$. Therefore:
\begin{align*}
\|f + M^\perp\|_{X^*/M^\perp} \le \|f - h\|_{X^*} = \|\tilde{g}\|_{X^*} = \|f|_M\|_{M^*}.
\end{align*}
Combined with the previous step: $\|\Theta(f + M^\perp)\|_{M^*} = \|f|_M\|_{M^*} = \|f + M^\perp\|_{X^*/M^\perp}$. Hence $\Theta$ is an isometry.
[guided]
This is the step where Hahn-Banach is essential. The question is: can we find an element of the coset $f + M^\perp$ whose $X^*$-norm is exactly $\|f|_M\|_{M^*}$? If so, the infimum defining the quotient norm is attained at this value.
**Strategy.** Start with $f|_M \in M^*$ and extend it back to all of $X$ using the [Norm-Preserving Extension of Linear Functionals](/theorems/880). This theorem guarantees the existence of $\tilde{g} \in X^*$ with:
- $\tilde{g}|_M = f|_M$ (agreement on $M$),
- $\|\tilde{g}\|_{X^*} = \|f|_M\|_{M^*}$ (norm preservation).
The hypotheses of the [Norm-Preserving Extension](/theorems/880) require that $X$ is a real normed vector space and $M$ is a subspace. Both conditions are given in the theorem statement ($X$ is a Banach space, hence a normed space, and $M \subset X$ is a closed subspace, hence a subspace).
**Constructing the annihilator element.** Set $h = f - \tilde{g}$. Then $h|_M = f|_M - \tilde{g}|_M = 0$, so $h \in M^\perp$. This means $\tilde{g} = f - h \in f + M^\perp$, i.e., $\tilde{g}$ is a representative of the coset $f + M^\perp$.
**Norm computation.**
\begin{align*}
\|f + M^\perp\|_{X^*/M^\perp} = \inf_{h \in M^\perp} \|f + h\|_{X^*} \le \|f - (f - \tilde{g})\|_{X^*} = \|\tilde{g}\|_{X^*} = \|f|_M\|_{M^*}.
\end{align*}
So the infimum is not just bounded above — it is attained by the Hahn-Banach extension. This is a crucial point: the quotient norm $\|f + M^\perp\|$ equals the norm of the norm-preserving extension of $f|_M$, which equals $\|f|_M\|_{M^*}$.
Without Hahn-Banach, we would only have the inequality $\|f|_M\| \le \|f + M^\perp\|$ (from the previous step), and the reverse inequality would fail.
[/guided]
[/step]
[step:Prove surjectivity: every bounded linear functional on $M$ arises as a restriction from $X^*$]
Let $\varphi \in M^*$. By the [Norm-Preserving Extension of Linear Functionals](/theorems/880), applied to the subspace $M$ of the normed space $X$, there exists $F \in X^*$ with $F|_M = \varphi$ and $\|F\|_{X^*} = \|\varphi\|_{M^*}$.
Then $\Theta(F + M^\perp) = F|_M = \varphi$. Hence $\Theta$ is surjective.
[guided]
Surjectivity also relies on the [Norm-Preserving Extension](/theorems/880). Given any $\varphi \in M^*$, we need to find a coset $f + M^\perp \in X^*/M^\perp$ with $\Theta(f + M^\perp) = \varphi$, i.e., $f|_M = \varphi$.
The Norm-Preserving Extension theorem provides $F \in X^*$ with $F|_M = \varphi$. Setting $f = F$, we have $\Theta(F + M^\perp) = F|_M = \varphi$.
Note that the specific choice of extension $F$ does not matter: any two extensions $F_1, F_2 \in X^*$ of $\varphi$ satisfy $(F_1 - F_2)|_M = 0$, so $F_1 - F_2 \in M^\perp$, and $F_1 + M^\perp = F_2 + M^\perp$. The coset is unique even though the extension is not.
[/guided]
[/step]
[step:Conclude that $\Theta$ is an isometric isomorphism]
Collecting the results: $\Theta: X^*/M^\perp \to M^*$ is linear (Step 2), isometric (Steps 2-3 combined), and surjective (Step 4). An isometry is automatically injective (if $\Theta(f + M^\perp) = 0$, then $\|f + M^\perp\| = \|\Theta(f + M^\perp)\| = 0$, so $f + M^\perp = 0 + M^\perp$, i.e., $f \in M^\perp$). Therefore $\Theta$ is a bijective isometry, hence an isometric isomorphism $X^*/M^\perp \cong M^*$.
[/step]
