Proposition 6

Let $u$ be a normal partial operator on a Hilbert space E. Let $(f_n)_{n\in\mathbf{N}}$ be a sequence in $\mathscr{L}_u$(Sp($u$)) which converges simply to $f\in \mathscr{L}_u$(Sp($u$)) and such that there exists $g\in \mathscr{L}_u$(Sp($u$)) satisfying $|f_n|\leqslant g$ for all $n\in \mathbf{N}$. Then dom($g(u)$)$\subset$ dom($f_n(u)$) for all $n\in \mathbf{N}$ and dom($g(u)$)$\subset$ dom($f(u)$). Moreover, for every element $y$ of the domain of $g(u)$, one has

$f(u)(y) =$ lim$_{n\rightarrow+\infty}f_n(u)(y)$.

In particular, if $f_n\in \mathscr{L}_u^{\infty}$(Sp($u$)) for all $n\in \mathbf{N}$ and if the functions $f_n$ are uniformly bounded, then $f_n(u)$ converges to $f(u)$ in the space $\mathscr{L}(E)$ endowed with the topology of simple convergence.

Let us denote by $\mu_{x,y}$ (resp. $\mu_x$) the spectral measure of $(x, y)$ (resp. of $x$) relative to $u$.

Let $y\in$ dom($g(u)$), so that $g\in \mathscr{L}^2$(Sp($u$)$, \mu_y$). The condition $|f_n|\leqslant g$ implies that $f_n\in \mathscr{L}^2$(Sp($u$)$, \mu_y$), hence $y\in$ dom($f_n(u)$).

Since $(f_n)$ converges simply to $f$ and $|f_n|\leqslant g$, one has $f\in \mathscr{L}^2$(Sp($u$)$, \mu_y$) by Lebesgue's theorem (INT, IV, p. 137, § 3, n$^o7$, th. 6), hence $y\in$ dom($f(u)$). Moreover, the sequence $(f_n)$ converges to $f$ in $\mathscr{L}^2$(Sp($u$)$, \mu_y$), hence the norm of $f_n(u)y$ converges to the norm of $f(u)y$.

Let $x\in E$. The functions $f$ and $g$, as well as the functions $f_n$ for every $n\in \mathbf{N}$, belong to $\mathscr{L}^1$(Sp($u$)$, \mu_{x,y}$) (prop. 3). By Lebesgue's theorem (INT, IV, loc. cit.), the sequence $(f_n)$ converges to $f$ in $\mathscr{L}^1$(Sp($u$)$, \mu_{x,y}$), whence

Sp($u$) Sp($u$)

It follows that $f_n(u)(y)$ converges to $f(u)(y)$ (EVT, V, p. 17, prop. 10).