[proofplan]
We decompose the Hilbert space using the kernel of $f$ and its orthogonal complement, then construct the representing element from a unit vector in $\ker(f)^\perp$. The orthogonal decomposition forces every $v \in H$ to split into a kernel component and a multiple of this unit vector, yielding the representation $f(v) = (u_f, v)_H$. Uniqueness follows from non-degeneracy of the inner product, the isometry $\|f\|_{H^*} = \|u_f\|_H$ from Cauchy--Schwarz and a test-function argument, and the full isometric isomorphism property of $\Phi$ from linearity, injectivity, and surjectivity.
[/proofplan]
[step:Dispose of the trivial case $f = 0$]
If $f = 0$, set $u_f = 0 \in H$. Then $f(v) = 0 = (0, v)_H$ for every $v \in H$ and $\|f\|_{H^*} = 0 = \|u_f\|_H$. Uniqueness is immediate: if $(u, v)_H = 0$ for all $v \in H$, taking $v = u$ gives $\|u\|_H^2 = 0$, so $u = 0$. For the remainder of the proof, assume $f \ne 0$.
[/step]
[step:Decompose $H$ via the kernel of $f$ and construct the representing element]
Since $f \in H^*$ is bounded and linear, its kernel
\begin{align*}
N := \ker f = \{v \in H : f(v) = 0\}
\end{align*}
is a closed linear subspace of $H$. The subspace $N$ is proper because $f \ne 0$. By the [Orthogonal Decomposition Theorem](/theorems/241), $H = N \oplus N^\perp$ with $N^\perp \ne \{0\}$. Fix $z \in N^\perp$ with $\|z\|_H = 1$ and $f(z) \ne 0$ (such $z$ exists because $N^\perp \ne \{0\}$ and every nonzero element of $N^\perp$ lies outside $N = \ker f$).
For any $v \in H$, define
\begin{align*}
w := v - \frac{f(v)}{f(z)}\,z \in H.
\end{align*}
Applying $f$ gives $f(w) = f(v) - \frac{f(v)}{f(z)} f(z) = 0$, so $w \in N$. Since $z \in N^\perp$, we have $(w, z)_H = 0$, and therefore
\begin{align*}
(v, z)_H = \left(w + \frac{f(v)}{f(z)}\,z,\, z\right)_H = (w, z)_H + \frac{f(v)}{f(z)}\|z\|_H^2 = \frac{f(v)}{f(z)}.
\end{align*}
Rearranging,
\begin{align*}
f(v) = f(z)\,(v, z)_H = (v,\, \overline{f(z)}\,z)_H.
\end{align*}
Set $u_f := \overline{f(z)}\,z \in H$. Then $f(v) = (v, u_f)_H = (u_f, v)_H$ for every $v \in H$, where the last equality uses symmetry of the real inner product (or conjugate symmetry in the complex case, with the convention that $f(v) = (u_f, v)_H$).
[guided]
The goal is to find $u_f \in H$ such that $f(v) = (u_f, v)_H$ for all $v$. Since $f$ is a nonzero [continuous](/page/Continuity) linear functional, its kernel $N = \ker f$ is a closed hyperplane (codimension one). The [Orthogonal Decomposition Theorem](/theorems/241) splits $H = N \oplus N^\perp$, where $N^\perp$ is one-dimensional. Fix a unit vector $z \in N^\perp$ with $f(z) \neq 0$.
Every $v \in H$ decomposes as $v = w + \lambda z$ with $w \in N$ and $\lambda \in \mathbb{R}$. The element
\begin{align*}
w := v - \frac{f(v)}{f(z)}\,z
\end{align*}
is the $N$-component (we check: $f(w) = f(v) - \frac{f(v)}{f(z)}f(z) = 0$). Since $w \perp z$ and $\|z\|_H = 1$:
\begin{align*}
(v, z)_H = (w, z)_H + \frac{f(v)}{f(z)}\|z\|_H^2 = 0 + \frac{f(v)}{f(z)} \cdot 1 = \frac{f(v)}{f(z)}.
\end{align*}
Therefore $f(v) = f(z)(v, z)_H = (v, \overline{f(z)}\,z)_H$. Setting $u_f := \overline{f(z)}\,z$ gives the representation $f(v) = (u_f, v)_H$ for every $v \in H$.
The construction is independent of the choice of $z \in N^\perp \setminus \{0\}$: replacing $z$ by $\lambda z$ yields $f(\lambda z)(\lambda z)/\|\lambda z\|_H^2 = \lambda f(z) \cdot \lambda z / (\lambda^2) = f(z)z/1$, the same $u_f$.
[/guided]
[/step]
[step:Prove uniqueness of the representing element]
Suppose $u, u' \in H$ both satisfy $f(v) = (u, v)_H = (u', v)_H$ for every $v \in H$. Then $(u - u', v)_H = 0$ for every $v \in H$. Taking $v = u - u'$ gives $\|u - u'\|_H^2 = 0$, so $u = u'$.
[/step]
[step:Verify the isometry $\|f\|_{H^*} = \|u_f\|_H$]
By the Cauchy--Schwarz inequality, for every $v \in H$:
\begin{align*}
|f(v)| = |(u_f, v)_H| \le \|u_f\|_H\,\|v\|_H,
\end{align*}
so $\|f\|_{H^*} = \sup_{\|v\|_H \le 1} |f(v)| \le \|u_f\|_H$. For the reverse inequality, test against $v = u_f$:
\begin{align*}
f(u_f) = (u_f, u_f)_H = \|u_f\|_H^2,
\end{align*}
so $\|f\|_{H^*} \ge \frac{|f(u_f)|}{\|u_f\|_H} = \|u_f\|_H$ (when $u_f \ne 0$; if $u_f = 0$ then $f = 0$ and both sides vanish). Hence $\|f\|_{H^*} = \|u_f\|_H$.
[/step]
[step:Verify that $\Phi$ is an isometric isomorphism]
**Linearity.** For $f, g \in H^*$ and $\lambda \in \mathbb{R}$ (or $\mathbb{C}$), the element $u_{\lambda f + g}$ satisfies $(\lambda f + g)(v) = \lambda(u_f, v)_H + (u_g, v)_H = (\lambda u_f + u_g, v)_H$ for every $v \in H$. By uniqueness, $\Phi(\lambda f + g) = \lambda \Phi(f) + \Phi(g)$.
**Injectivity.** If $\Phi(f) = 0$, then $\|f\|_{H^*} = \|\Phi(f)\|_H = 0$, so $f = 0$.
**Surjectivity.** For any $u \in H$, define the functional $f_u: H \to \mathbb{R}$ (or $\mathbb{C}$) by $f_u(v) := (u, v)_H$. The Cauchy--Schwarz inequality gives $|f_u(v)| \le \|u\|_H \|v\|_H$, so $f_u \in H^*$ with $\|f_u\|_{H^*} \le \|u\|_H$. By uniqueness of the representing element, $\Phi(f_u) = u$.
**Isometry.** $\|\Phi(f)\|_H = \|u_f\|_H = \|f\|_{H^*}$ for every $f \in H^*$.
Since $\Phi$ is a bijective bounded [linear map](/page/Linear%20Map) with $\|\Phi(f)\|_H = \|f\|_{H^*}$, the map $\Phi: H^* \to H$ is an isometric isomorphism.
[/step]
