📂Hilbert Space

Proving that Every Separable Hilbert Space is Isometric to the l^2 space

Theorem¹

Every infinite-dimensional separable Hilbert space $H$ is isometrically isomorphic to $\ell^{2}$.

Description

The fact that a separable Hilbert space is isometrically isomorphic to $\ell^{2}$ essentially means that in the study of Hilbert spaces, one can focus solely on $\ell^{2}$.

Proof

Gram-Schmidt Orthogonalization of Separable Hilbert Spaces

Every separable Hilbert space has an orthonormal basis.

Corollary of Bessel’s Inequality

If $\left\{ \mathbf{v}_{k} \right\}_{k \in \mathbb{N}}$ is an orthonormal set in the Hilbert space $H$, the following holds:

For every $\left\{ c_{k} \right\}_{k \in \mathbb{N}} \in \ell^{2}$, the infinite series $\sum_{k \in \mathbb{N}} c_{k} \mathbf{v}_{k}$ converges.

Following the Gram-Schmidt orthogonalization, the Hilbert space $H$ obtains an orthonormal basis $\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}}$, and according to the corollary, the convergence of $\sum_{k \in \mathbb{N}} c_{k} \mathbf{e}_{k}$ is guaranteed for every $\left\{ c_{k} \right\}_{k \in \mathbb{N}} \in \ell^{2}$. Now, consider $\left\{ \delta_{k} \right\}_{k \in \mathbb{N}}$ as the orthonormal basis of $\ell^{2}$ and define the operator $U : H \to \ell^{2}$ as follows:

$$ U \left( \sum_{k \in \mathbb{N}} c_{k} \mathbf{e}_{k} \right) := \sum_{k \in \mathbb{N}} c_{k} \delta_{k} $$

Then, $U$ is a bijective mapping between $H$ and $\ell^{2}$.

Equivalence Conditions of Orthonormal Bases: Assume $H$ is a Hilbert space. For an orthonormal system $\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}} \subset H$ of $H$, the following are equivalent:

• (i): $\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}} \subset H$ is an orthonormal basis of $H$.
• (ii): For every $\mathbf{x}\in H$ $$ \mathbf{x}= \sum_{k \in \mathbb{N}} \langle \mathbf{x}, \mathbf{e}_{k} \rangle \mathbf{e}_{k} $$
• (iv): For every $\mathbf{x}\in H$ $$ \sum_{k \in \mathbb{N}} \left| \langle \mathbf{x}, \mathbf{e}_{k} \rangle \right|^{2} = \left\| \mathbf{x}\right\|^{2} $$

Every $\mathbf{v} \in H$ uniquely decomposes with respect to the orthonormal basis $\left\{ \mathbf{e}_{k} \right\}_{k \in \mathbb{N}}$ as follows:

$$ \mathbf{v} = \sum_{k \in \mathbb{N}} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \mathbf{e}_{k} $$

Therefore,

$$ \begin{align*} \left\| U \mathbf{v} \right\|^{2} =& \left\| \sum_{k \in \mathbb{N}} \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \delta_{k} \right\|^{2} \\ =& \sum_{k \in \mathbb{N}} \left| \left\langle \mathbf{v} , \mathbf{e}_{k} \right\rangle \right|^{2} \\ =& \left\| \mathbf{v} \right\|^{2} \end{align*} $$

and, $U : H \to \ell^{2}$ is an isometric isomorphism.

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