Proposition 22

Let $g$ be a $\mu$-measurable function on X.

a) The spectrum of $m_g$ is the essential $\mu$-image S of $g$;

b) Let $\lambda \in \mathbf{C}-$ Sp($m_g$). The resolvent $R(m_g, \lambda )$ is the multiplication operator $m_h$, where $h$ is the function on X defined by $h(x) = 0$ if $g(x) =\lambda$ and $h(x) = (\lambda -g(x))^{-1}$ otherwise.

Let us prove that $\mathbf{C}-$ S is contained in the resolvent set of $m_g$. Let $\lambda \in \mathbf{C}-$ S. There exists an open neighbourhood U of $\lambda$ such that the subset $Y =\overset{-1}{g}(U)$ of X is locally $\mu$-negligible. The function $k$ defined on X by $k(x) = (\lambda -g(x))^{-1}$ if $x \notin Y$ and $k(x) = 0$ if $x\in Y$ then belong to $\mathscr{L}^{\infty}(X, \mu)$ (Lemma 5 of IV, p. 184); the multiplication operator by $k$ is therefore an endomorphism of $L^2(X, \mu)$.

Since $|gk|\leqslant 1 +|\lambda k|$, we have

for $f\in \mathscr{L}^2(X, \mu)$, which implies that the image of $m_k$ is contained in the domain of $m_g$. Conversely, let $f\in \mathscr{L}^2(X, \mu)$ whose class $\widetilde{f}$ belongs to the domain of $m_g$. Outside the locally $\mu$-negligible set Y, we have $f(x) =k(x)(\lambda -g(x))f(x)$, hence $\widetilde{f}$ is in the image of $m_k$. The same formula proves that $\lambda$ belongs to the resolvent set of $m_g$ and that $R(m_g, \lambda ) =m_k$. Since the set Y is locally $\mu$-negligible, the multiplication operator $m_k$ coincides with the operator $m_h$ described in assertion b).

Let us prove conversely that $\mathbf{C}-$ Sp($m_g$) is contained in $\mathbf{C}-$ S. Let $\lambda \in \mathbf{C}-$ Sp($m_g$). Let $M>\|R(m_g, \lambda )\|$ be a real number. Let us denote by Y the set of $x\in X$ such that $|\lambda -g(x)|<M^{-1}$. Let us prove that Y is locally $\mu$-negligible, which will imply that $\lambda$ does not belong to S, and will complete the proof.

Let K be a compact subset of X. Let $\varphi$ be the characteristic function of $Y\cap K$; it is an element of $\mathscr{L}^2(X, \mu)$, and we denote by $\widetilde{\varphi}$ its class in $L^2(X, \mu)$. Let $\psi$ be a function in $\mathscr{L}^2(X, \mu)$ whose class in $L^2(X, \mu)$ is $R(m_g, \lambda )(\widetilde{\varphi})$. We have $R(m_g, \lambda )(\widetilde{\varphi})\in$ dom($m_g$) and $(\lambda -m_g)(R(m_g, \lambda )(\widetilde{\varphi})) =\widetilde{\varphi}$, hence $(\lambda -g(x))\psi (x) = 1$ for $\mu$-almost all $x\in Y\cap K$. This implies

In view of the choice of M, this means that $\varphi$ is zero $\mu$-almost everywhere. Thus, the set $Y\cap K$ is $\mu$-negligible. Therefore the set Y is locally $\mu$-negligible (INT, IV, p. 172, § 5, n$^o2$, Prop. 5).