[proofplan] For a given $f \in H^*$, we identify a candidate representative $u_f$ using the orthogonal decomposition $H = N(f) \oplus N(f)^{\perp}$. We then verify the representation identity $f(v) = (u_f, v)_H$ for all $v \in H$ by decomposing $v$ into components in $N(f)$ and $N(f)^{\perp}$. Uniqueness follows from the non-degeneracy of the inner product, and the isometry $\|f\|_{H^*} = \|u_f\|_H$ follows from Cauchy-Schwarz and a specific test vector. [/proofplan] [step:Identify the candidate representative $u_f$ via the orthogonal complement of $\ker(f)$] Let $f \in H^*$ be given. If $f = 0$, set $u_f = 0$; the conclusion holds since $(0, v)_H = 0 = f(v)$ for all $v$. Assume $f \neq 0$, so the kernel \begin{align*} N(f) := \{v \in H : f(v) = 0\} \end{align*} is a proper closed subspace of $H$ (closed because $f$ is continuous, proper because $f \neq 0$). Since $N(f) \subsetneq H$, the orthogonal complement $N(f)^{\perp}$ is non-trivial: there exists $w \in N(f)^{\perp}$ with $w \neq 0$. In particular, $f(w) \neq 0$ (since $w \notin N(f)$). Define \begin{align*} u_f := \frac{f(w)}{\|w\|_H^2}\, w. \end{align*} (Since $H$ is a real Hilbert space, $\overline{f(w)} = f(w)$.) [guided] Where does the formula $u_f = f(w) w / \|w\|_H^2$ come from? The orthogonal complement $N(f)^{\perp}$ is one-dimensional: since $f$ is a single nonzero linear functional, its kernel $N(f)$ has codimension $1$, so $\dim N(f)^{\perp} = 1$. Any element of $N(f)^{\perp}$ is a scalar multiple of $w$. The representative $u_f$ must satisfy $(u_f, w)_H = f(w)$ (since for any $v \in H$, the $N(f)$-component is killed by $(u_f, \cdot)_H$ and only the $N(f)^{\perp}$-component contributes). Writing $u_f = \alpha w$: \begin{align*} (u_f, w)_H = \alpha(w, w)_H = \alpha\|w\|_H^2 = f(w), \end{align*} which gives $\alpha = f(w)/\|w\|_H^2$. Hence \begin{align*} u_f = \frac{f(w)}{\|w\|_H^2}\, w. \end{align*} The choice of $w \in N(f)^{\perp} \setminus \{0\}$ is arbitrary -- any nonzero element gives the same $u_f$, since $N(f)^{\perp}$ is one-dimensional. Replacing $w$ by $\lambda w$ ($\lambda \neq 0$): $f(\lambda w)\lambda w / \|\lambda w\|_H^2 = \lambda f(w) \cdot \lambda w / (\lambda^2 \|w\|_H^2) = f(w)w/\|w\|_H^2$. [/guided] [/step] [step:Verify the representation $f(v) = (u_f, v)_H$ for all $v \in H$] For arbitrary $v \in H$, decompose $v$ as \begin{align*} v = \underbrace{\left(v - \frac{f(v)}{f(w)}\, w\right)}_{\in\, N(f)} + \frac{f(v)}{f(w)}\, w. \end{align*} The first term lies in $N(f)$: applying $f$ gives $f(v) - \frac{f(v)}{f(w)} f(w) = f(v) - f(v) = 0$. Since $u_f$ is a scalar multiple of $w \in N(f)^{\perp}$, we have $u_f \in N(f)^{\perp}$, so $(u_f, v - \frac{f(v)}{f(w)} w)_H = 0$. Therefore: \begin{align*} (u_f, v)_H &= \left(u_f,\; v - \frac{f(v)}{f(w)}\, w\right)_H + \frac{f(v)}{f(w)}\, (u_f, w)_H \\ &= 0 + \frac{f(v)}{f(w)} \cdot \frac{f(w)}{\|w\|_H^2}\, (w, w)_H \\ &= \frac{f(v)}{f(w)} \cdot f(w) = f(v). \end{align*} [/step] [step:Prove uniqueness of the representative] Suppose $u_f$ and $u_f'$ both satisfy $f(v) = (u_f, v)_H = (u_f', v)_H$ for every $v \in H$. Then $(u_f - u_f', v)_H = 0$ for every $v \in H$. Taking $v = u_f - u_f'$: \begin{align*} \|u_f - u_f'\|_H^2 = (u_f - u_f', u_f - u_f')_H = 0, \end{align*} so $u_f = u_f'$. [/step] [step:Verify the isometry $\|f\|_{H^*} = \|u_f\|_H$] **Upper bound:** By the Cauchy-Schwarz inequality, for every $v \in H$: \begin{align*} |f(v)| = |(u_f, v)_H| \leq \|u_f\|_H \cdot \|v\|_H, \end{align*} so $\|f\|_{H^*} = \sup_{\|v\|_H \leq 1} |f(v)| \leq \|u_f\|_H$. **Lower bound:** Taking $v = u_f$ (assuming $u_f \neq 0$; if $u_f = 0$ then $f = 0$ and the identity holds): \begin{align*} \|f\|_{H^*} \geq \frac{|f(u_f)|}{\|u_f\|_H} = \frac{(u_f, u_f)_H}{\|u_f\|_H} = \frac{\|u_f\|_H^2}{\|u_f\|_H} = \|u_f\|_H. \end{align*} Combining: $\|f\|_{H^*} = \|u_f\|_H$. The map $f \mapsto u_f$ is therefore an isometric bijection $H^* \to H$, establishing $H^* \cong H$. [/step]