Show that the set \(\ts{ u_k(j) \da \delta_{kj} } \subseteq \ell^2(\ZZ)\) forms an orthonormal system.
\(\theset{u_k}\) is an orthonormal basis of \(\ell^2(\ZZ)\): its span is the space of finitely supported sequences, which is dense, since the truncations of \(a \in \ell^2(\ZZ)\) to \(|j| \le N\) converge to \(a\).
Solution
1. \(u_k = (\delta_{kj})_{j \in \ZZ}\) lies in \(\ell^2(\ZZ)\) and \(\|u_k\|_2 = 1\).
Exactly one entry of \(u_k\), at \(j = k\), is nonzero, and it equals \(1\), so \(\|u_k\|_2^2 = \sum_j |\delta_{kj}|^2 = 1\).
2. \(\inner{u_k}{u_m} = 0\) for \(k \neq m\).
\(\inner{u_k}{u_m} = \sum_j \delta_{kj}\delta_{mj}\), and each term vanishes because \(j\) cannot equal both \(k\) and \(m\).
3. Q.E.D.
Steps 1 and 2 are the two conditions for \(\theset{u_k}_{k \in \ZZ}\) to be orthonormal.