Separable Hilbert Spaces are Isometrically Isomorphic to ℓ2

# Separable Hilbert Spaces are Isometrically Isomorphic to ℓ2

Theorem 1: Every separable Hilbert space $H$ is isometrically isomorphic to $\ell^2$. |

**Proof:**Since $H$ is a separable Hilbert space, $H$ has a countable orthonormal basis by the theorem on the Existence of Orthonormal Bases for Infinite-Dimensional Separable Hilbert Spaces page. Let $(e_n)_{n=1}^{\infty}$ be such a basis.

- Let $T : H \to \ell^2$ be defined for each $h \in H$ by:

\begin{align} \quad T(h) = (\langle e_n, h \rangle)_{n=1}^{\infty} \end{align}

- Note that for every $h \in H$, $T(h) \in \ell^2$ since by Bessel's Inequality for Inner Product Spaces:

\begin{align} \quad \| T(h) \|_2^2 = \| (\langle e_n, h \rangle) \|_2^2 = \sum_{n=1}^{\infty} \langle e_n, h \rangle^2 \leq \| h \|^2 < \infty \end{align}

- And so $\| T(h) \|_2 < \infty$.

- Also observe that $T$ is linear since for all $h, h' \in H$ we have that:

\begin{align} \quad T(h + h') = (\langle e_n, h + h' \rangle) = (\langle e_n, h \rangle + \langle e_n, h' \rangle) = (\langle e_n, h \rangle) + (\langle e_n, h' \rangle) = T(h) + T(h') \end{align}

- And for all $h \in H$ and $\alpha \in \mathbb{R}$:

\begin{align} \quad T(\alpha h) = (\langle e_n, \alpha h \rangle) = (\alpha \langle e_n, h \rangle ) = \alpha (\langle e_n, h \rangle) = \alpha T(h) \end{align}

- We now show that $T$ is isometry. Now since $(e_n)$ is an orthonormal basis for $H$ we have that every $h \in H$ can be uniquely written as $\displaystyle{h = \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n}$. Therefore we have that:

\begin{align} \quad \| h \|^2 = \left \langle \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n, \sum_{n=1}^{\infty} \langle e_n, h \rangle e_n \right \rangle = \sum_{n=1}^{\infty} \langle e_n, h \rangle^2 \langle e_n, e_n \rangle = \sum_{n=1}^{\infty} \langle e_n, h \rangle^2 = \| T(h) \|_2^2 \end{align}

- Squaring both sides sows that $\| T(h) \|_2 = \| h \|$ for every $h \in H$. So $T$ is an isometry and is thus bounded and injective.

- Lastly we show that $T$ is surjective. Let $(x_n)$ be a sequence in $\ell^2$. Let:

\begin{align} \quad h_x = \sum_{n=1}^{\infty} x_ne_n \end{align}

- Note that $h_x$ converges since by The Pythagorean Identity for Inner Product Spaces and since $(x_n) \in \ell^2$:

\begin{align} \quad \| h_x \|^2 = \sum_{n=1}^{\infty} |x_n|^2 < \infty \end{align}

- And we see that:

\begin{align} \quad T(h_x) = (\langle e_n, h_x\rangle ) = \left ( \left \langle e_n, \sum_{n=1}^{\infty} x_ne_n \right \rangle \right ) = \left ( x_n \| e_n \|^2 \right ) = (x_n) \end{align}

- So $T$ is surjective.

- Lastly, since $H$ is a Banach space and $\ell^2$ is a Banach space and from above, $T$ is bounded and bijective, we have that $T^{-1}$ is bounded by one of the Corollaries to the Open Mapping Theorem. Therefore $T$ is an isometric isomorphism. So every separable Hilbert space is isometrically isomorphic to $\ell^2$. $\blacksquare$