📂Hilbert Space

Proof of the Riesz Representation Theorem

Theorem¹

Let $\left( H, \left\langle \cdot,\cdot \right\rangle \right)$ be a Hilbert space. For a linear functional $f \in H^{ \ast }$ of $H$ and $\mathbf{x} \in H$, there exists a unique $\mathbf{w} \in H$ satisfying $f ( \mathbf{x} ) = \left\langle \mathbf{x} , \mathbf{w} \right\rangle$ and $\| f \|_{H^{\ast}} = \| \mathbf{w} \|_{H}$.

Explanation

Simply put, this amounts to characterizing what every element of the dual space $H^{ \ast }$ of a Hilbert space looks like. Although this may sound simple, considering that a function can be defined however one pleases, the very idea of setting out to prove such a theorem is nothing short of remarkable.

Proof

If $f = 0$, then $\mathbf{y} = \mathbf{0}$ exists satisfying $f ( \cdot ) = \left\langle \cdot , \mathbf{0} \right\rangle$ and $\| f \| = 0$, so there is nothing to prove; hence assume $f \ne 0$.

• Part 1. Existence

  Orthogonal decomposition theorem

  For a closed subspace $W$ of a Hilbert space $H$,

  $$ H = W \oplus W^{\perp} $$

  Let $W= \ker (f) = \left\{ \mathbf{x} \in H : f(\mathbf{x}) = 0 \right\}$. Since $\ker (f)$ is a closed subspace, by the orthogonal decomposition theorem

  $$ H = W \oplus W^{\perp} $$

  and $W^{\perp} \ne \left\{ \mathbf{0} \right\}$. Therefore, choose one $\mathbf{y} \in W^{\perp}$ with $\left\| \mathbf{y} \right\|=1$. Since $W^{\perp}$ is also a vector space, the existence of such a $\mathbf{y}$ is necessarily guaranteed as long as there is at least one element other than $\mathbf{0}$. Now, for an arbitrary $\mathbf{x} \in H$, consider the following vector $\mathbf{z} \in H$.

  $$ \mathbf{z} := f(\mathbf{x})\mathbf{y} - f(\mathbf{y})\mathbf{x} $$

  Note that here $f(\mathbf{x}), f(\mathbf{y})$ are constants. Applying the linear functional $f$ to this $\mathbf{z}$, by linearity we obtain the following.

  $$ f(\mathbf{z}) = f\left( f(\mathbf{x})\mathbf{y} - f(\mathbf{y})\mathbf{x} \right) = f(\mathbf{x})f(\mathbf{y}) - f(\mathbf{y})f(\mathbf{x}) = 0 $$

  Therefore $\mathbf{z} \in W$. Then, since we said $\mathbf{y} \in W^{\perp}$, the inner product of the two is $0$.

  $$ \left\langle \mathbf{z}, \mathbf{y} \right\rangle = \left\langle f(\mathbf{x})\mathbf{y} - f(\mathbf{y})\mathbf{x}, \mathbf{y} \right\rangle = 0 $$

  Expanding the inner product yields the following.

  $$ f(\mathbf{x}) \left\langle \mathbf{y}, \mathbf{y} \right\rangle - f(\mathbf{y})\left\langle \mathbf{x}, \mathbf{y} \right\rangle = f(\mathbf{x}) \left\| \mathbf{y} \right\|^{2} - f(\mathbf{y})\left\langle \mathbf{x}, \mathbf{y} \right\rangle = f(\mathbf{x}) - f(\mathbf{y})\left\langle \mathbf{x}, \mathbf{y} \right\rangle=0 $$

  $$ \implies f(\mathbf{x}) = f(\mathbf{y})\left\langle \mathbf{x}, \mathbf{y} \right\rangle = \left\langle \mathbf{x}, \overline{f(\mathbf{y})}\mathbf{y} \right\rangle $$

  Here, letting $\mathbf{w} = \overline{f(\mathbf{y})}\mathbf{y}$, we have $f(\mathbf{x}) = \left\langle \mathbf{x} , \mathbf{w} \right\rangle$. Moreover, by the Cauchy–Schwarz inequality, $$ \| f \| = \sup_{ \| \mathbf{x} \| =1} | f(\mathbf{x}) | = \sup_{ \| \mathbf{x} \| =1} \left| \left\langle \mathbf{x} , \mathbf{w} \right\rangle \right| \le \sup_{ \| \mathbf{x} \| =1} \| \mathbf{x} \| \cdot \| \mathbf{w} \| = \| \mathbf{w} \| $$

  and, since $\left\| \dfrac{\mathbf{w}}{\left\| \mathbf{w}\right\|} \right\|=1$,

  $$ \| f \| = \sup_{ \| \mathbf{x} \| =1} | f(\mathbf{x}) | = \sup_{ \| \mathbf{x} \| =1} \left| \left\langle \mathbf{x} , \mathbf{w} \right\rangle \right| \ge \left\langle {{ \mathbf{w} } \over { \| \mathbf{w} \| }} , \mathbf{w} \right\rangle = {{ \| \mathbf{w} \|^2 } \over { \| \mathbf{w} \| }} = \| \mathbf{w} \| $$

holds, so $\left\| f \right\| = \left\| \mathbf{w} \right\|$. Therefore, we can see that $\mathbf{w}$ satisfies the conditions of the theorem.

• Part 2. Uniqueness

  Now suppose that $\mathbf{w}^{\prime}$ also satisfies

  $$ f(\mathbf{x}) = \left\langle \mathbf{x}, \mathbf{w}^{\prime} \right\rangle $$

  Lemma

  Let $\left( H, \left\langle \cdot,\cdot \right\rangle \right)$ be a Hilbert space. Suppose that the following holds for $x, y \in H$.

  $$ \left\langle x, y \right\rangle = \left\langle x, z \right\rangle, \quad \forall x \in H $$

  Then $y = z$.

  Then, by the lemma above, we can see that such $\mathbf{w}=\mathbf{w}^{\prime}$.

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