Let \(\mathcal H\) be a Hilbert space, let \(U = (u_n)_{n\geq1}\) be an orthonormal sequence in \(\mathcal H\), and let \(\mathcal H_U\subseteq\mathcal H\) be the closure of the linear span of \(\theset{u_n\suchthat n\geq1}\). Then the coefficient map \[ \begin{aligned} \mathcal H_U &\longrightarrow \ell^2(\NN) \\ x &\longmapsto (\inner{x}{u_n})_{n\geq1} \end{aligned} \] is an isometric isomorphism. Equivalently, for every \((a_n)_{n\geq1}\in\ell^2(\NN)\) there is a unique \(x\in\mathcal H_U\) such that \[ a_n=\inner{x}{u_n}\quad\text{for all }n, \qquad\text{and then}\qquad \norm{x}^2=\sum_{n=1}^\infty\abs{a_n}^2. \]
In particular, if \(U\) is complete, so that \(\mathcal H_U=\mathcal H\), then the coefficient map \(\mathcal H\to\ell^2(\NN)\) is an isometric isomorphism.
For \(a=(a_n)\in\ell^2(\NN)\), set \[ x_N\coloneqq\sum_{n=1}^N a_nu_n. \] Orthonormality gives, for \(M<N\), \[ \norm{x_N-x_M}^2=\sum_{n=M+1}^N\abs{a_n}^2, \] so \((x_N)\) is Cauchy and converges, by completeness of \(\mathcal H\), to some \(x\), which lies in \(\mathcal H_U\) because each \(x_N\) lies in the span of \(U\). Continuity of the inner product yields \(\inner{x}{u_n}=a_n\) for every \(n\), and taking limits in \(\norm{x_N}^2=\sum_{n=1}^N\abs{a_n}^2\) gives \[ \norm{x}^2=\sum_{n=1}^\infty\abs{a_n}^2. \]
Conversely, let \(x\in\mathcal H_U\) and put \(a_n\coloneqq\inner{x}{u_n}\). Bessel’s inequality T-4BDE3 gives \(a\in\ell^2(\NN)\). Applying the construction above produces \(y\in\mathcal H_U\) with the same coefficients. Then \(x-y\in\mathcal H_U\) is orthogonal to every \(u_n\), hence to the span of \(U\) and therefore to its closure \(\mathcal H_U\); in particular \(x-y\) is orthogonal to itself, so \(x=y\). This proves bijectivity and the norm identity.