Proof That a Hilbert Space Is Reflexive
Theorem
A Hilbert space $H$ is reflexive: $$ H^{\ast \ast} \approx H $$
- $X^{\ast}$ is the dual space of $X$, and $X^{\ast \ast}$ denotes the double dual.
- $X \approx Y$ means that $X$ and $Y$ are isometric.
Explanation
Though short and simple, the fact that there is no need to consider anything larger than the dual space when studying Hilbert spaces is a very good thing.
Proof
Part 1. $(H^{ \ast } , \| \cdot \| )$ is a Hilbert space
Riesz representation theorem: Let $H$ be a Hilbert space. For a linear functional $f \in H^{ \ast }$ of $H$ and $\mathbf{x} \in H$, there uniquely exists $\mathbf{y} \in H$ satisfying $f ( \mathbf{x} ) = \left\langle \mathbf{x} , \mathbf{y} \right\rangle$ and $\| f \| = \| \mathbf{y} \|$.
Define the function $\left\langle \cdot , \cdot \right\rangle^{ \ast } : H^{ \ast } \times H^{ \ast } \to \mathbb{C}$ as $\displaystyle \left\langle f, g \right\rangle^{ \ast } : = \left\langle \mathbf{y}_{g}, \mathbf{y}_{f} \right\rangle = f ( \mathbf{y}_{g} )$. Here, $\mathbf{y}_{f}, \mathbf{y}_{g} \in H$ are the elements satisfying $\| f \| = \| \mathbf{y}_{f} \|$ and $\| g \| = \| \mathbf{y}_{g} \|$ from the Riesz representation theorem.
Then $\left\langle \cdot , \cdot \right\rangle^{ \ast }$ satisfies the following three conditions and thus becomes an inner product on $H^{ \ast }$.
(i): $\left\langle \lambda f_{1} + f_{2} , g \right\rangle^{ \ast } = ( \lambda f_{1} + f_{2} ) ( \mathbf{y}_{g} )= \lambda f_{1}( \mathbf{y}_{g} ) + f_{2} ( \mathbf{y}_{g} ) = \lambda \left\langle f_{1}, g \right\rangle^{ \ast } + \left\langle f_{2}, g \right\rangle^{ \ast }$
(ii): $\left\langle f , g \right\rangle^{ \ast } = \left\langle \mathbf{y}_{g} , \mathbf{y}_{f} \right\rangle = \overline{ \left\langle \mathbf{y}_{f} , \mathbf{y}_{g} \right\rangle } = \overline{ \left\langle g , f \right\rangle^{ \ast } }$
(iii): $\left\langle f , f \right\rangle^{ \ast } = \left\langle \mathbf{y}_{f}, \mathbf{y}_{f} \right\rangle = \| \mathbf{y}_{f} \|^{2} \ge 0$ and $\left\langle f , f \right\rangle^{ \ast } = \| \mathbf{y}_{f} \|^{2} = 0 \iff \mathbf{y}_{f} = 0 \iff f = 0$
Properties of linear operators: If $Y$ is a Banach space, then $(B(X,Y), \left\| \cdot \right\| )$ is a Banach space.
Here, since $Y=\mathbb{C}$ is a Banach space, $(B(X,Y), \left\| \cdot \right\|) = (H^{\ast}, \left\| \cdot \right\|)$ is also a Banach space. Since $H^{\ast}$ is a complete space equipped with an inner product, it is a Hilbert space.
Part 2.
We define a function $\Phi$. Using the function $\phi_{\mathbf{x}} \in H^{\ast \ast}$ that substitutes $\mathbf{x} \in H$ into a function $f \in H^{ \ast }$, define $\Phi : H \to H^{\ast \ast}$ so that $\Phi (\mathbf{x}) := \phi_{\mathbf{x}} (f) = f(\mathbf{x})$.
Part 3. $\Phi$ is linear
Since $f \in H^{ \ast }$, $\Phi ( \lambda \mathbf{x} + y) = f ( \lambda \mathbf{x} + y) = \lambda f ( \mathbf{x} ) + f ( y) = \lambda \Phi ( \mathbf{x} ) + \Phi ( y)$
Part 4. $\Phi$ is injective
Suppose $\mathbf{x} \in \ker \Phi$. Then $\mathbb{0} = \Phi (\mathbf{x}) = f(\mathbf{x})$, so $\mathbf{x} \in \ker f$. Taking the substitution function on the $f$ obtained from the Riesz representation theorem gives $\phi_x (f) = f(\mathbf{x}) = \left\langle \mathbf{x} , \mathbf{y}_{f} \right\rangle = \mathbb{0}$. Since this must hold regardless of $\mathbf{y}_{f}$, we must have $\mathbf{x} = \mathbb{0}$.
Properties of the kernel: $\ker \Phi = \left\{ \mathbb{0} \right\} \iff \Phi$ is injective.
Since $\ker \Phi = \left\{ \mathbb{0} \right\}$, by the properties of the kernel, $\Phi$ is injective.
Part 5. $\Phi$ is surjective.
Since we already showed in Part 1. that $H^{ \ast }$ is a Hilbert space, we can use the Riesz representation theorem.
Part 5-1.
For $F \in H^{\ast \ast} ( F : H^{\ast \ast} \to \mathbb{C})$, there uniquely exists $g_{F} \in H^{ \ast }$ satisfying $F( \cdot ) = \left\langle \cdot , g_{F} \right\rangle^{ \ast }$ and $\| F \| = \| g_{F} \|$.
Part 5-2.
For $g_{F} \in H^{ \ast } ( g_{F} : H^{ \ast } \to \mathbb{C})$, there uniquely exists $\mathbf{x}_{g_{F}} \in H$ satisfying $g_{F} ( \cdot ) = \left\langle \cdot , \mathbf{x}_{g_{F}} \right\rangle$ and $ \| g_{F} \| = \| \mathbf{x}_{g_{F}} \| $.
$$ \begin{align*} F(f) =& \left\langle f , g_{F} \right\rangle^{ \ast } & \text{by Part 5-1.} \\ =& \left\langle f , g_{F} \right\rangle^{ \ast } &\text{by definition of } \left\langle \cdot , \cdot \right\rangle^{ \ast } \\ =& \left\langle \mathbf{x}_{g_{F}} , \mathbf{y}_{f} \right\rangle^{ \ast } &\text{by definition of }\left\langle \cdot , \cdot \right\rangle \\ =& f \left( \mathbf{x}_{g_{F}} \right) & \text{by Riesz Representation Theorem} \\ =& \Phi ( \mathbf{x}_{g_{F}} ) &\text{by definition of } \Phi \end{align*} $$
Therefore, for every $F(f) \in H^{\ast \ast}$, there exists $\mathbf{x}_{g_{F}} \in H$ satisfying $\Phi ( \mathbf{x}_{g_{F}} ) = F(f)$, so $\Phi$ is surjective.
Part 6. $\Phi$ preserves the norm.
From the definition of $\Phi$, since $\Phi ( \mathbf{x}_{g_{F}} ) = \phi_{ \mathbf{x}_{g_{F}} } (f)$,
$$ \Phi ( \mathbf{x}_{g_{F}} ) = \phi_{ \mathbf{x}_{g_{F}} } (f) = F(f) $$
In other words,
$$ \left\| \Phi ( \mathbf{x}_{g_{F}} ) \right\| = \left\| \phi_{ \mathbf{x}_{g_{F}} } \right\| = \left\| F \right\| $$
But since $\| F \| = \| g_{F} \|$ from Part 5-1. and $\| g_{F} \| = \| \mathbf{x}_{g_{F}} \|$ from Part 5-2.,
$$ \left\| \Phi ( \mathbf{x}_{g_{F}} ) \right\| = \| \mathbf{x}_{g_{F}} \| $$
Putting together Part 2. through Part 6., we can see that $\Phi$ is an isometry.
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