[proofplan] We first prove the multiplication-operator statement directly. The operator $M_\varphi-\lambda I$ is multiplication by $\varphi-\lambda$, so it is invertible precisely when $\varphi-\lambda$ is bounded away from $0$ outside a null set. Applying this to the [Borel functional calculus](/theorems/2696) for a normal operator, the resolvent of $T-\lambda I$ is multiplication by the bounded function $z\mapsto (z-\lambda)^{-1}$ exactly when $\lambda$ lies outside the support of the spectral measure. Conversely, if $\lambda$ is in the support, spectral projections of arbitrarily small neighbourhoods of $\lambda$ give vectors on which $T-\lambda I$ is arbitrarily small, so no bounded inverse exists. [/proofplan]

[step:Identify the spectrum of a multiplication operator with the closed essential range]Let $(X,\mathcal{A},\mu)$ be a semi-finite [measure space](/page/Measure%20Space) and let $\varphi:X\to\mathbb{C}$ be bounded and $\mathcal{A}$-measurable. Define the bounded multiplication operator $M_\varphi:L^2(X,\mu)\to L^2(X,\mu)$ by $M_\varphi f=\varphi f$. For $\lambda\in\mathbb C$, the operator $M_\varphi-\lambda I$ is the multiplication operator $M_{\varphi-\lambda}$. Assume first that $\lambda\notin\operatorname{ess\,ran}(\varphi)$. Then there exists $\varepsilon>0$ such that \begin{align*} \mu(\{x\in X:|\varphi(x)-\lambda|<\varepsilon\})=0. \end{align*} Define a bounded measurable function $\psi:X\to\mathbb C$ as follows: for $x\in X$ with $|\varphi(x)-\lambda|\ge \varepsilon$, set $\psi(x)=(\varphi(x)-\lambda)^{-1}$, and for $x\in X$ with $|\varphi(x)-\lambda|<\varepsilon$, set $\psi(x)=0$. Then $\|\psi\|_\infty\le \varepsilon^{-1}$ and \begin{align*} \psi(\varphi-\lambda)=1 \end{align*} $\mu$-a.e. Hence $M_\psi M_{\varphi-\lambda}=I$ and $M_{\varphi-\lambda}M_\psi=I$ on $L^2(X,\mu)$. Therefore $\lambda\in\rho(M_\varphi)$. Conversely, assume that $\lambda\in\operatorname{ess\,ran}(\varphi)$. Suppose, for contradiction, that $M_{\varphi-\lambda}$ is invertible. Let $S\in\mathcal{L}(L^2(X,\mu))$ denote its inverse, and define $C:=\|S\|_{\mathcal{L}(L^2(X,\mu))}$. For every $f\in L^2(X,\mu)$, \begin{align*} \|f\|_{L^2(X,\mu)}\le C\|(\varphi-\lambda)f\|_{L^2(X,\mu)}. \end{align*} For each $n\in\mathbb N$, define \begin{align*} A_n:=\{x\in X:|\varphi(x)-\lambda|<1/n\}. \end{align*} By the definition of essential range, $\mu(A_n)>0$. Since $(X,\mathcal{A},\mu)$ is semi-finite, there exists a measurable set $B_n\subset A_n$ with $0<\mu(B_n)<\infty$. Define $f_n:=\mathbb{1}_{B_n}$. Then $f_n\in L^2(X,\mu)$ and $f_n\ne 0$. Since $|\varphi-\lambda|<1/n$ on $B_n$, \begin{align*} \|(\varphi-\lambda)f_n\|_{L^2(X,\mu)}^2 = \int_{B_n}|\varphi(x)-\lambda|^2\,d\mu(x) \le \frac{1}{n^2}\mu(B_n) = \frac{1}{n^2}\|f_n\|_{L^2(X,\mu)}^2. \end{align*} Thus \begin{align*} \|f_n\|_{L^2(X,\mu)} \le \frac{C}{n}\|f_n\|_{L^2(X,\mu)}. \end{align*} Choosing $n>C$ gives a contradiction because $\|f_n\|_{L^2(X,\mu)}>0$. Hence $M_{\varphi-\lambda}$ is not invertible, so $\lambda\in\sigma(M_\varphi)$. Therefore \begin{align*} \sigma(M_\varphi)=\operatorname{ess\,ran}(\varphi). \end{align*}[/step]

[guided]The point of the multiplication-operator case is that invertibility can be read pointwise, up to null sets. Let $M_\varphi:L^2(X,\mu)\to L^2(X,\mu)$ denote multiplication by the bounded measurable function $\varphi:X\to\mathbb C$ on the semi-finite measure space $(X,\mathcal{A},\mu)$. For a scalar $\lambda\in\mathbb C$, subtracting $\lambda I$ gives \begin{align*} (M_\varphi-\lambda I)f=(\varphi-\lambda)f, \end{align*} so $M_\varphi-\lambda I=M_{\varphi-\lambda}$. First suppose $\lambda$ is not in the closed essential range of $\varphi$. By definition, there is an $\varepsilon>0$ such that the set on which $\varphi$ gets within distance $\varepsilon$ of $\lambda$ is null: \begin{align*} \mu(\{x\in X:|\varphi(x)-\lambda|<\varepsilon\})=0. \end{align*} This means that outside a null set, the function $\varphi-\lambda$ never approaches $0$. We may therefore define its reciprocal a.e. as follows: for $x\in X$ with $|\varphi(x)-\lambda|\ge \varepsilon$, set $\psi(x)=(\varphi(x)-\lambda)^{-1}$, and for $x\in X$ with $|\varphi(x)-\lambda|<\varepsilon$, set $\psi(x)=0$. The function $\psi:X\to\mathbb C$ is measurable because it is obtained from the measurable function $\varphi-\lambda$ by a Borel operation, and it is essentially bounded because $|\psi(x)|\le\varepsilon^{-1}$ outside the null set. Hence $M_\psi$ is a bounded operator on $L^2(X,\mu)$. Moreover, \begin{align*} \psi(\varphi-\lambda)=1 \end{align*} $\mu$-a.e., so multiplication gives \begin{align*} M_\psi M_{\varphi-\lambda}=I \end{align*} and \begin{align*} M_{\varphi-\lambda}M_\psi=I. \end{align*} Thus $M_\varphi-\lambda I$ is invertible, and $\lambda\in\rho(M_\varphi)$. Now suppose $\lambda\in\operatorname{ess\,ran}(\varphi)$. We prove that $M_\varphi-\lambda I$ cannot have a bounded inverse. Assume toward a contradiction that $M_{\varphi-\lambda}$ is invertible, and let $S\in\mathcal{L}(L^2(X,\mu))$ be its inverse. Define \begin{align*} C:=\|S\|_{\mathcal{L}(L^2(X,\mu))}. \end{align*} Since $S M_{\varphi-\lambda}=I$, every $f\in L^2(X,\mu)$ satisfies \begin{align*} \|f\|_{L^2(X,\mu)} = \|S M_{\varphi-\lambda}f\|_{L^2(X,\mu)} \le C\|(\varphi-\lambda)f\|_{L^2(X,\mu)}. \end{align*} This is a bounded-below estimate for multiplication by $\varphi-\lambda$. The definition of essential range says that for every $n\in\mathbb N$, the set \begin{align*} A_n:=\{x\in X:|\varphi(x)-\lambda|<1/n\} \end{align*} has positive measure. The semi-finiteness of $(X,\mathcal{A},\mu)$ says that every measurable set of positive measure contains a measurable subset of finite positive measure. Applying this to $A_n$, choose a measurable subset $B_n\subset A_n$ with $0<\mu(B_n)<\infty$, and define $f_n:=\mathbb{1}_{B_n}$. Then $f_n\in L^2(X,\mu)$ and \begin{align*} \|f_n\|_{L^2(X,\mu)}^2=\mu(B_n)>0. \end{align*} Because $B_n\subset A_n$, we have $|\varphi(x)-\lambda|<1/n$ for every $x\in B_n$. Therefore \begin{align*} \|(\varphi-\lambda)f_n\|_{L^2(X,\mu)}^2 = \int_{B_n}|\varphi(x)-\lambda|^2\,d\mu(x) \le \frac{1}{n^2}\mu(B_n) = \frac{1}{n^2}\|f_n\|_{L^2(X,\mu)}^2. \end{align*} Substituting this into the bounded-below estimate gives \begin{align*} \|f_n\|_{L^2(X,\mu)} \le \frac{C}{n}\|f_n\|_{L^2(X,\mu)}. \end{align*} For $n>C$, this is impossible because $\|f_n\|_{L^2(X,\mu)}>0$. Hence $M_{\varphi-\lambda}$ is not invertible. Therefore every $\lambda$ in the closed essential range belongs to $\sigma(M_\varphi)$, and every $\lambda$ outside the closed essential range belongs to $\rho(M_\varphi)$. This proves \begin{align*} \sigma(M_\varphi)=\operatorname{ess\,ran}(\varphi). \end{align*}[/guided]

[step:Show that points outside the spectral support lie in the resolvent set]Let $S:=\operatorname{supp}E$. Assume $\lambda\in\mathbb C\setminus S$. If $\lambda\notin\sigma(T)$, then the closedness of $\sigma(T)$ gives $\varepsilon>0$ such that $\sigma(T)\cap B(\lambda,\varepsilon)=\varnothing$, and hence $E(\sigma(T)\cap B(\lambda,\varepsilon))=0$. If $\lambda\in\sigma(T)\setminus S$, then the definition of $S=\operatorname{supp}E$ gives $\varepsilon>0$ such that \begin{align*} E(\sigma(T)\cap B(\lambda,\varepsilon))=0. \end{align*} Thus in either case there exists $\varepsilon>0$ with $E(\sigma(T)\cap B(\lambda,\varepsilon))=0$. Define the bounded Borel function $g_\lambda:\sigma(T)\to\mathbb C$ by setting $g_\lambda(z)=(z-\lambda)^{-1}$ for $z\notin B(\lambda,\varepsilon)$ and $g_\lambda(z)=0$ for $z\in B(\lambda,\varepsilon)$. Then $\|g_\lambda\|_\infty\le\varepsilon^{-1}$. By the Borel functional calculus rules from [Borel Functional Calculus Rules]([citetheorem:8413]), the operator \begin{align*} G_\lambda:=g_\lambda(T)=\int_{\sigma(T)}g_\lambda\,dE \end{align*} is bounded, and \begin{align*} (T-\lambda I)G_\lambda = ((z-\lambda)g_\lambda)(T). \end{align*} The bounded Borel function $(z-\lambda)g_\lambda$ equals $1$ outside $\sigma(T)\cap B(\lambda,\varepsilon)$ and differs from $1$ only on an $E$-null Borel set. By the null-set invariance of spectral integrals, which follows from the matrix-coefficient formula for the Borel functional calculus, changing a bounded Borel function on a Borel set $\Delta\subset\sigma(T)$ with $E(\Delta)=0$ does not change its spectral integral. Therefore its spectral integral is $I$, so \begin{align*} (T-\lambda I)G_\lambda=I. \end{align*} The same multiplication rule gives \begin{align*} G_\lambda(T-\lambda I)=I. \end{align*} Thus $T-\lambda I$ is invertible, and $\lambda\in\rho(T)$.[/step]

[guided]Let $S:=\operatorname{supp}E$. We prove that every $\lambda\in\mathbb C\setminus S$ belongs to the resolvent set $\rho(T)$. There are two cases. If $\lambda\notin\sigma(T)$, then $\lambda\in\rho(T)$ by the definition of spectrum. If $\lambda\in\sigma(T)\setminus S$, then the definition of $S$ gives an $\varepsilon>0$ such that \begin{align*} E(\sigma(T)\cap B(\lambda,\varepsilon))=0. \end{align*} Define $g_\lambda:\sigma(T)\to\mathbb C$ by setting $g_\lambda(z)=(z-\lambda)^{-1}$ for $z\notin B(\lambda,\varepsilon)$ and $g_\lambda(z)=0$ for $z\in B(\lambda,\varepsilon)$. This is a bounded Borel function and $\|g_\lambda\|_\infty\le\varepsilon^{-1}$. By [citetheorem:8413], the operator $G_\lambda:=g_\lambda(T)=\int_{\sigma(T)}g_\lambda\,dE$ is bounded, and multiplication in the functional calculus gives \begin{align*} (T-\lambda I)G_\lambda=((z-\lambda)g_\lambda)(T). \end{align*} The function $(z-\lambda)g_\lambda$ equals $1$ except on the Borel set $\sigma(T)\cap B(\lambda,\varepsilon)$, whose spectral projection is zero. By the null-set invariance of spectral integrals, changing a bounded Borel function on a Borel set $\Delta\subset\sigma(T)$ with $E(\Delta)=0$ does not change its spectral integral; this follows by testing the spectral integral against matrix coefficients. Therefore its spectral integral is $I$. The same multiplication rule gives $G_\lambda(T-\lambda I)=I$. Hence $T-\lambda I$ has the bounded inverse $G_\lambda$, so $\lambda\in\rho(T)$.[/guided]

[step:Use spectral projections near support points to rule out invertibility]Assume $\lambda\in S$. For every $n\in\mathbb N$, define the Borel set \begin{align*} \Delta_n:=\sigma(T)\cap B(\lambda,1/n). \end{align*} By the definition of $S$, $E(\Delta_n)\ne 0$. Choose $x_n\in H$ such that $E(\Delta_n)x_n\ne 0$, and define \begin{align*} u_n:=\frac{E(\Delta_n)x_n}{\|E(\Delta_n)x_n\|_H}. \end{align*} Here $\operatorname{Range}(E(\Delta_n)):=\{E(\Delta_n)y:y\in H\}$. Then $\|u_n\|_H=1$, $u_n\in\operatorname{Range}(E(\Delta_n))$, and $E(\Delta_n)u_n=u_n$. Since $T=\int_{\sigma(T)}z\,dE(z)$ by [citetheorem:8412], the Borel functional calculus gives \begin{align*} T-\lambda I=\int_{\sigma(T)}(z-\lambda)\,dE(z). \end{align*} Since $E(\Delta_n)u_n=u_n$, the multiplication rule in the Borel functional calculus gives \begin{align*} (T-\lambda I)u_n=((z-\lambda)\mathbb{1}_{\Delta_n})(T)u_n. \end{align*} The operator norm bound for spectral integrals gives \begin{align*} \|(T-\lambda I)u_n\|_H \le \sup_{z\in\Delta_n}|z-\lambda|\,\|u_n\|_H \le \frac{1}{n}. \end{align*} If $T-\lambda I$ were invertible, with inverse $R\in\mathcal{L}(H)$, then \begin{align*} 1=\|u_n\|_H = \|R(T-\lambda I)u_n\|_H. \end{align*} Since $R$ is bounded, \begin{align*} \|R(T-\lambda I)u_n\|_H \le \|R\|_{\mathcal{L}(H)}\|(T-\lambda I)u_n\|_H \le \frac{\|R\|_{\mathcal{L}(H)}}{n}. \end{align*} Taking $n>\|R\|_{\mathcal{L}(H)}$ gives a contradiction. Hence $T-\lambda I$ is not invertible, and $\lambda\in\sigma(T)$.[/step]

[guided]Assume $\lambda\in S=\operatorname{supp}E$. For each $n\in\mathbb N$, set \begin{align*} \Delta_n:=\sigma(T)\cap B(\lambda,1/n). \end{align*} Because $\lambda$ lies in the support of $E$, the projection $E(\Delta_n)$ is not the zero operator. Choose $x_n\in H$ with $E(\Delta_n)x_n\ne0$, and define \begin{align*} u_n:=\frac{E(\Delta_n)x_n}{\|E(\Delta_n)x_n\|_H}. \end{align*} Then $\|u_n\|_H=1$. Also $u_n\in\operatorname{Range}(E(\Delta_n)):=\{E(\Delta_n)y:y\in H\}$, so $E(\Delta_n)u_n=u_n$ because $E(\Delta_n)$ is a projection. By [citetheorem:8412], \begin{align*} T=\int_{\sigma(T)}z\,dE(z). \end{align*} Hence the Borel functional calculus gives \begin{align*} T-\lambda I=\int_{\sigma(T)}(z-\lambda)\,dE(z). \end{align*} Since $E(\Delta_n)u_n=u_n$, the multiplication rule in [citetheorem:8413] gives \begin{align*} (T-\lambda I)u_n=((z-\lambda)\mathbb{1}_{\Delta_n})(T)u_n. \end{align*} The operator norm of a spectral integral is bounded by the supremum norm of its integrand, so \begin{align*} \|(T-\lambda I)u_n\|_H \le \sup_{z\in\Delta_n}|z-\lambda|\,\|u_n\|_H \le \frac{1}{n}. \end{align*} Thus the unit vectors $u_n$ are approximate eigenvectors for $T$ at $\lambda$. If $T-\lambda I$ were invertible, let $R\in\mathcal{L}(H)$ denote its inverse. Then for every $n\in\mathbb N$, \begin{align*} 1=\|u_n\|_H = \|R(T-\lambda I)u_n\|_H \le \|R\|_{\mathcal{L}(H)}\|(T-\lambda I)u_n\|_H \le \frac{\|R\|_{\mathcal{L}(H)}}{n}. \end{align*} Choosing $n>\|R\|_{\mathcal{L}(H)}$ contradicts this inequality. Therefore $T-\lambda I$ is not invertible, and $\lambda\in\sigma(T)$.[/guided]

[step:Conclude that the spectrum is exactly the support] The previous two steps prove both inclusions. If $\lambda\notin\operatorname{supp}E$, then $\lambda\in\rho(T)$, so $\lambda\notin\sigma(T)$. If $\lambda\in\operatorname{supp}E$, then $T-\lambda I$ is not invertible, so $\lambda\in\sigma(T)$. Therefore \begin{align*} \sigma(T)=\operatorname{supp}E. \end{align*} The multiplication-operator statement was proved in the first step. The spectral-support equality for $T$ follows from the two direct spectral-projection inclusions above. [/step]