[proofplan]
We prove existence, uniqueness, and isometry of the Riesz representer for a bounded linear functional $f$ on a real [Hilbert space](/page/Hilbert%20Space). Uniqueness follows from positive definiteness of the inner product. For existence, we decompose $H = \ker(f) \oplus \ker(f)^\perp$ using the Orthogonal Projection Theorem, pick a unit vector $e$ in $\ker(f)^\perp$, and construct $u = f(e)\,e$. The verification reduces to showing that $v - \frac{f(v)}{f(e)}e \in \ker(f)$ and using orthogonality. The isometry $\|f\|_{H^*} = \|u\|_H$ follows from Cauchy--Schwarz and a direct evaluation.
[/proofplan]
[step:Prove uniqueness of the representer via positive definiteness]
Suppose $u_1, u_2 \in H$ both satisfy $f(v) = (u_1, v) = (u_2, v)$ for all $v \in H$. By linearity of the inner product:
\begin{align*}
(u_1 - u_2, v) = 0 \quad \text{for all } v \in H.
\end{align*}
Choosing the test vector $v = u_1 - u_2$:
\begin{align*}
\|u_1 - u_2\|_H^2 = (u_1 - u_2, u_1 - u_2) = 0.
\end{align*}
By the positive definiteness of the norm, $u_1 = u_2$.
[/step]
[step:Handle the trivial case $f = 0$]
If $f$ is the zero functional, set $u = 0$. Then $(0, v) = 0 = f(v)$ for all $v \in H$, and $\|f\|_{H^*} = 0 = \|u\|_H$.
[/step]
[step:Decompose $H$ along the kernel of $f$ using the Orthogonal Projection Theorem]
Assume $f \ne 0$. Define $M := \ker(f) = \{v \in H : f(v) = 0\}$. Since $f: H \to \mathbb{R}$ is a [bounded](/page/Bounded%20Linear%20Operator) (hence [continuous](/page/Continuity)) linear functional, $M = f^{-1}(\{0\})$ is a closed subspace of $H$. Since $f \ne 0$, the subspace $M$ is proper: $M \ne H$.
By the [Orthogonal Projection Theorem](/theorems/???), $H = M \oplus M^\perp$. Since $M \ne H$, the orthogonal complement $M^\perp$ is non-trivial: there exists $z \in M^\perp$ with $z \ne 0$. Define the unit vector:
\begin{align*}
e := \frac{z}{\|z\|_H} \in M^\perp, \quad \|e\|_H = 1.
\end{align*}
Since $e \in M^\perp$ and $e \ne 0$, we have $e \notin M$, hence $f(e) \ne 0$.
[guided]
Why does the proof hinge on the kernel? Every bounded linear functional $f: H \to \mathbb{R}$ either vanishes identically (handled in the previous step) or has a kernel of codimension one. The Orthogonal Projection Theorem gives us a clean decomposition $H = \ker(f) \oplus \ker(f)^\perp$, where $\ker(f)^\perp$ is one-dimensional. Any vector in $\ker(f)^\perp$ is "detected" by $f$, and by normalising such a vector we can construct the representer explicitly.
We define $M := \ker(f)$. Since $f$ is a bounded linear functional, $M = f^{-1}(\{0\})$ is the preimage of a closed set under a continuous map, hence $M$ is closed in $H$. Since $f \ne 0$, there exists $v_0 \in H$ with $f(v_0) \ne 0$, so $M \ne H$.
The [Orthogonal Projection Theorem](/theorems/???) applies to any closed subspace of a Hilbert space: it guarantees $H = M \oplus M^\perp$ with $M^\perp \ne \{0\}$ (since $M \ne H$). Pick any non-zero $z \in M^\perp$ and normalise:
\begin{align*}
e := \frac{z}{\|z\|_H} \in M^\perp, \quad \|e\|_H = 1.
\end{align*}
Since $e \in M^\perp \setminus \{0\}$ and $M^\perp \cap M = \{0\}$, we have $e \notin M$, so $f(e) \ne 0$. This non-vanishing is essential: we will divide by $f(e)$ in the construction.
[/guided]
[/step]
[step:Construct the representer $u = f(e)\,e$ and verify $f(v) = (u, v)$]
Define:
\begin{align*}
u := f(e)\,e \in M^\perp.
\end{align*}
Let $v \in H$ be arbitrary. Decompose $v$ as $v = w + \frac{f(v)}{f(e)}\,e$, where:
\begin{align*}
w := v - \frac{f(v)}{f(e)}\,e.
\end{align*}
We verify $w \in M$:
\begin{align*}
f(w) = f(v) - \frac{f(v)}{f(e)}\,f(e) = f(v) - f(v) = 0.
\end{align*}
Since $w \in M$ and $e \in M^\perp$, the inner product $(e, w) = 0$. Now compute:
\begin{align*}
(u, v) &= \left(f(e)\,e,\; w + \frac{f(v)}{f(e)}\,e\right) \\
&= f(e)\,(e, w) + f(e) \cdot \frac{f(v)}{f(e)}\,(e, e) \\
&= 0 + f(v) \cdot \|e\|_H^2 \\
&= f(v).
\end{align*}
[guided]
The construction $u = f(e)\,e$ is natural once we recognise that $M^\perp$ is one-dimensional. To see this: $f: H \to \mathbb{R}$ is surjective (since $f \ne 0$), so the first isomorphism theorem gives $H / \ker(f) \cong \mathbb{R}$, hence $\dim(H / M) = 1$. Since $H = M \oplus M^\perp$, we have $\dim(M^\perp) = \dim(H/M) = 1$, so $M^\perp$ is spanned by $e$. Therefore the representer must be a scalar multiple of $e$. The scalar is determined by the requirement $(u, e) = f(e)$: substituting $u = \lambda e$ gives $\lambda \|e\|_H^2 = f(e)$, so $\lambda = f(e)$ since $\|e\|_H = 1$.
To verify the representation for a general $v \in H$, we write $v$ as the sum of its $M$-component and its $M^\perp$-component. Define:
\begin{align*}
w := v - \frac{f(v)}{f(e)}\,e.
\end{align*}
We check $w \in M = \ker(f)$:
\begin{align*}
f(w) = f(v) - \frac{f(v)}{f(e)}\,f(e) = f(v) - f(v) = 0.
\end{align*}
So $v = w + \frac{f(v)}{f(e)}\,e$ with $w \in M$ and $\frac{f(v)}{f(e)}\,e \in M^\perp$. This is the orthogonal decomposition of $v$. Now:
\begin{align*}
(u, v) &= (f(e)\,e,\; w) + \left(f(e)\,e,\; \frac{f(v)}{f(e)}\,e\right) \\
&= f(e)\,(e, w) + f(v)\,(e, e).
\end{align*}
The first term vanishes because $e \in M^\perp$ and $w \in M$. The second term equals $f(v) \cdot 1 = f(v)$ because $\|e\|_H = 1$. Hence $(u, v) = f(v)$ for all $v \in H$.
[/guided]
[/step]
[step:Establish the isometry $\|f\|_{H^*} = \|u\|_H$]
For the upper bound, the [Cauchy–Schwarz inequality](/page/Hilbert%20Space) applied to the inner product $(\cdot, \cdot)$ gives for all $v \in H$:
\begin{align*}
|f(v)| = |(u, v)| \le \|u\|_H \,\|v\|_H.
\end{align*}
Taking the supremum over $\|v\|_H \le 1$ yields $\|f\|_{H^*} \le \|u\|_H$.
For the lower bound, evaluate $f$ at $v = u$ (assuming $u \ne 0$):
\begin{align*}
f(u) = (u, u) = \|u\|_H^2.
\end{align*}
Therefore:
\begin{align*}
\|f\|_{H^*} \ge \frac{|f(u)|}{\|u\|_H} = \frac{\|u\|_H^2}{\|u\|_H} = \|u\|_H.
\end{align*}
Combining gives $\|f\|_{H^*} = \|u\|_H$, establishing that $f \mapsto u$ is an isometric isomorphism between $H^*$ and $H$.
[/step]
