Example 2

Let E be a Hilbert space over K and $B = (e_i)_{i\in I}$ an orthonormal basis of E. Let $(\lambda_i)_{i\in I}$ be a family of elements of K. Let D be the vector subspace of E consisting of the elements $x\in E$ such that the family $(\lambda_i\langle e_i|x\rangle )_{i\in I}$ is square-summable in K. The space D is dense in E since it contains the vector $e_i$ for every $i\in I$. The partial operator $u$ with domain D given by

is called a diagonal partial operator in the basis B, and $(\lambda_i)_{i\in I}$ is called the family of eigenvalues of $u$.

The operator $u$ is closed. In fact, let $(x_n, u(x_n))_{n\in\mathbf{N}}$ be a sequence of elements of the graph of $u$ which converges in $E\times E$, and let $(x, y)$ be its limit. Then $\langle e_i|x_n\rangle \rightarrow \langle e_i|x\rangle$ for every $i\in I$ and

for every $i\in I$. Consequently, $\lambda_i\langle e_i|x\rangle =\langle e_i|y\rangle$ for every $i\in I$, which proves that $x\in D$ and $u(x) =y$, that is to say that $u$ is closed.

This example is in fact a particular case of the preceding one, applied to the topological space X = I endowed with the discrete topology and the counting measure $\mu$, since E identifies with the space $\ell^2(I) = L^2(I, \mu)$ by the mapping $x\mapsto (\langle e_i|x\rangle )_{i\in I}$ (EVT, V, p. 23, cor. 2) and $u$ then identifies with the multiplication operator $m_{\lambda}$, where $\lambda$ is the function $i\mapsto \lambda_i$.