[proofplan]
We verify that the map $\Psi: M^\perp \to (X/M)^*$, defined by $\Psi(f)(x + M) = f(x)$, is (1) well-defined, (2) linear and bounded, (3) isometric, and (4) surjective. Well-definedness reduces to showing that $f \in M^\perp$ forces $f$ to be constant on cosets. The isometry follows from the definition of the quotient norm: $\|x + M\|_{X/M} = \inf_{m \in M} \|x + m\|_X$, which connects the operator norms on both sides. Surjectivity uses the composition of an arbitrary functional on $X/M$ with the quotient map.
[/proofplan]
[step:Verify $\Psi$ is well-defined: $f \in M^\perp$ implies $\tilde{f}$ is constant on cosets]
Let $f \in M^\perp$, so $f \in X^*$ and $f(m) = 0$ for every $m \in M$. Define
\begin{align*}
\tilde{f}: X/M &\to \mathbb{R} \\
x + M &\mapsto f(x).
\end{align*}
We must verify this is independent of the coset representative: if $x + M = y + M$, then $x - y \in M$, so $f(x) - f(y) = f(x - y) = 0$ (since $f \in M^\perp$). Hence $\tilde{f}(x + M) = \tilde{f}(y + M)$, and $\tilde{f}$ is well-defined.
[/step]
[step:Show $\tilde{f}$ is a bounded linear functional on $X/M$ with $\|\tilde{f}\|_{(X/M)^*} \le \|f\|_{X^*}$]
**Linearity.** For $x + M, y + M \in X/M$ and $\lambda \in \mathbb{R}$:
\begin{align*}
\tilde{f}((x + M) + \lambda(y + M)) &= \tilde{f}((x + \lambda y) + M) = f(x + \lambda y) = f(x) + \lambda f(y) \\
&= \tilde{f}(x + M) + \lambda \tilde{f}(y + M).
\end{align*}
**Boundedness.** For any $x + M \in X/M$:
\begin{align*}
|\tilde{f}(x + M)| = |f(x)| = |f(x + m)| \quad \text{for every } m \in M,
\end{align*}
since $f(m) = 0$. Therefore $|f(x + m)| \le \|f\|_{X^*} \|x + m\|_X$, and taking the infimum over $m \in M$:
\begin{align*}
|\tilde{f}(x + M)| \le \|f\|_{X^*} \inf_{m \in M} \|x + m\|_X = \|f\|_{X^*} \|x + M\|_{X/M}.
\end{align*}
Hence $\tilde{f} \in (X/M)^*$ with $\|\tilde{f}\|_{(X/M)^*} \le \|f\|_{X^*}$.
[guided]
**Linearity** is inherited directly from the linearity of $f$ and the vector space operations on $X/M$: addition of cosets is $(x + M) + (y + M) = (x + y) + M$, and scalar multiplication is $\lambda(x + M) = \lambda x + M$.
**Boundedness.** The quotient norm is $\|x + M\|_{X/M} = \inf_{m \in M} \|x + m\|_X$. Since $f$ vanishes on $M$, the value $f(x)$ equals $f(x + m)$ for every $m \in M$. So
\begin{align*}
|\tilde{f}(x + M)| = |f(x)| = |f(x + m)| \le \|f\|_{X^*} \|x + m\|_X.
\end{align*}
This holds for every $m \in M$. Taking the infimum over $m$:
\begin{align*}
|\tilde{f}(x + M)| \le \|f\|_{X^*} \inf_{m \in M} \|x + m\|_X = \|f\|_{X^*} \|x + M\|_{X/M}.
\end{align*}
Taking the supremum over all $x + M$ with $\|x + M\|_{X/M} \le 1$: $\|\tilde{f}\|_{(X/M)^*} \le \|f\|_{X^*}$.
[/guided]
[/step]
[step:Prove the reverse inequality $\|\tilde{f}\|_{(X/M)^*} \ge \|f\|_{X^*}$ to establish the isometry]
We must show $\|f\|_{X^*} \le \|\tilde{f}\|_{(X/M)^*}$. Let $\pi: X \to X/M$ denote the quotient map $\pi(x) = x + M$. Observe that $\tilde{f} \circ \pi = f$: for every $x \in X$,
\begin{align*}
(\tilde{f} \circ \pi)(x) = \tilde{f}(x + M) = f(x).
\end{align*}
The quotient map $\pi$ has operator norm $\|\pi\| \le 1$: for any $x \in X$,
\begin{align*}
\|\pi(x)\|_{X/M} = \|x + M\|_{X/M} = \inf_{m \in M} \|x + m\|_X \le \|x + 0\|_X = \|x\|_X.
\end{align*}
For the reverse inequality, we use the identity $f = \tilde{f} \circ \pi$. For any $x \in X$ with $\|x\|_X \le 1$:
\begin{align*}
|f(x)| = |\tilde{f}(\pi(x))| \le \|\tilde{f}\|_{(X/M)^*} \|\pi(x)\|_{X/M} \le \|\tilde{f}\|_{(X/M)^*} \|x\|_X \le \|\tilde{f}\|_{(X/M)^*}.
\end{align*}
Taking the supremum over $\|x\| \le 1$: $\|f\|_{X^*} \le \|\tilde{f}\|_{(X/M)^*}$.
Combined with the previous step: $\|\tilde{f}\|_{(X/M)^*} = \|f\|_{X^*}$. Hence $\Psi$ is an isometry.
[guided]
The key observation is that $f$ factors through the quotient: $f = \tilde{f} \circ \pi$. This factorisation is the defining property of $\tilde{f}$.
**Why does this give the reverse bound?** We have two bounded [linear maps](/page/Linear%20Map): $\tilde{f}: X/M \to \mathbb{R}$ and $\pi: X \to X/M$. Their composition is $f: X \to \mathbb{R}$. The submultiplicativity of operator norms gives $\|f\| = \|\tilde{f} \circ \pi\| \le \|\tilde{f}\| \cdot \|\pi\|$. But we can say more: $\|\pi\| \le 1$ (from $\|\pi(x)\| = \inf_{m} \|x + m\| \le \|x\|$), so
\begin{align*}
\|f\|_{X^*} \le \|\tilde{f}\|_{(X/M)^*} \cdot 1 = \|\tilde{f}\|_{(X/M)^*}.
\end{align*}
Combined with the bound $\|\tilde{f}\| \le \|f\|$ from the previous step, we get $\|\tilde{f}\|_{(X/M)^*} = \|f\|_{X^*}$.
This argument reveals why the isometry works: $\tilde{f}$ is really "$f$ modulo $M$", and the quotient norm is designed precisely so that the factorisation $f = \tilde{f} \circ \pi$ preserves norms.
[/guided]
[/step]
[step:Show $\Psi$ is linear and injective]
**Linearity.** For $f, g \in M^\perp$ and $\lambda \in \mathbb{R}$, and any $x + M \in X/M$:
\begin{align*}
\Psi(f + \lambda g)(x + M) = (f + \lambda g)(x) = f(x) + \lambda g(x) = \Psi(f)(x + M) + \lambda \Psi(g)(x + M).
\end{align*}
Hence $\Psi(f + \lambda g) = \Psi(f) + \lambda \Psi(g)$.
**Injectivity.** Since $\Psi$ is an isometry ($\|\Psi(f)\|_{(X/M)^*} = \|f\|_{X^*}$), it preserves norms. If $\Psi(f) = 0$, then $\|f\|_{X^*} = \|\Psi(f)\|_{(X/M)^*} = 0$, so $f = 0$. Hence $\Psi$ is injective.
[/step]
[step:Prove surjectivity: every $\varphi \in (X/M)^*$ arises as $\Psi(f)$ for some $f \in M^\perp$]
Let $\varphi \in (X/M)^*$. Define
\begin{align*}
f: X &\to \mathbb{R} \\
x &\mapsto \varphi(x + M).
\end{align*}
That is, $f = \varphi \circ \pi$, where $\pi: X \to X/M$ is the quotient map. Then $f$ is linear (composition of linear maps) and bounded: $|f(x)| = |\varphi(\pi(x))| \le \|\varphi\|_{(X/M)^*} \|\pi(x)\|_{X/M} \le \|\varphi\|_{(X/M)^*} \|x\|_X$, so $f \in X^*$.
For any $m \in M$, $f(m) = \varphi(m + M) = \varphi(0 + M) = 0$ (since $m + M = M = 0 + M$ is the zero element of $X/M$). Hence $f \in M^\perp$.
Finally, $\Psi(f) = \varphi$: for any $x + M \in X/M$, $\Psi(f)(x + M) = f(x) = \varphi(x + M)$.
This completes the proof that $\Psi: M^\perp \to (X/M)^*$ is an isometric isomorphism.
[guided]
**Construction.** Given $\varphi \in (X/M)^*$, the natural candidate for $f$ is the composition $f = \varphi \circ \pi$: apply $\varphi$ to the coset of $x$. This is the reverse of the construction $\tilde{f} = \Psi(f)$: there we went from $f$ on $X$ to $\tilde{f}$ on $X/M$; here we go from $\varphi$ on $X/M$ to $f$ on $X$.
**$f$ is bounded:** $|f(x)| = |\varphi(\pi(x))| \le \|\varphi\| \cdot \|\pi(x)\| \le \|\varphi\| \cdot \|x\|$, using $\|\pi\| \le 1$.
**$f \in M^\perp$:** For $m \in M$, $\pi(m) = m + M = 0_{X/M}$ (the zero coset), so $f(m) = \varphi(0_{X/M}) = 0$. This is exactly the annihilator condition.
**$\Psi(f) = \varphi$:** By definition, $\Psi(f)(x + M) = f(x) = \varphi(\pi(x)) = \varphi(x + M)$ for all $x + M \in X/M$.
The proof is now complete. The map $\Psi: M^\perp \to (X/M)^*$ is bijective (injective by isometry, surjective by the construction above), linear, and isometric, hence an isometric isomorphism.
[/guided]
[/step]
