> [!definition] > > Let $\ch$ be a [[Hilbert Space|Hilbert space]], $\alg$ be a [[Commutative Banach Algebra|commutative]] [[C-Star Algebra|C*-subalgebra]] of $L(\ch, \ch)$ containing $I$, $\Sigma = \sigma(\alg)$ be its [[Multiplicative Functional|spectrum]], and $\Gamma^{-1}: C(\Sigma) \to \alg$ defined by $f \mapsto T_f$ be the [[Gelfand-Naimark Theorem|inverse Gelfand transform]]. If $f \in B(\Sigma)$ be a [[Space of Bounded Borel Functions|bounded Borel function]], then there exists a unique $T_f \in L(\ch, \ch)$ such that > $ > \angles{T_fx, y} = \angles{f, \mu_{x, y}} \quad \forall x, y \in \ch > $ > where $\mu_{x, y}$ is the [[Spectral Measure|spectral measure]]. Moreover, the map $\Gamma^{-1}: B(\Sigma) \to L(\ch, \ch)$ is a [[Bounded Linear Map|bounded linear map]] with $\norm{T_f} \le \norm{f}_u$, known as the **extended inverse Gelfand transform**. > > *Proof*. Fix $x \in \ch$, define > $ > \phi_x: \ch \to \complex \quad y \mapsto {\angles{f, \mu_{x, y}}} > $ > then $\norm{\phi(y)} \le \norm{f}_u \norm{x} \cdot \norm{y}$ and $\phi$ is a bounded *conjugate* linear functional. By the [[Riesz Representation Theorem]], there exists a unique $T_fx \in \ch$ such that > $ > \angles{f, \mu_{x, y}} = \phi(y) = \angles{T_fx, y} > $ > for all $y \in \ch$. Since the map $x \mapsto \phi_x$ is bounded and linear, and the representation map is an isomorphism, $x \mapsto T_fx$ is also bounded and linear with $\norm{T_f} \le \norm{f}_u$. > [!theorem] > > $\Gamma^{-1}: B(\Sigma) \to L(\ch, \ch)$ is a $*$-homomorphism. > > Let $f \in B(\Sigma)$, then > $ > \begin{align*} > \angles{T_{\ol f}x, y} &= \angles{\ol f, \mu_{x, y}} = \ol{\angles{f, \mu_{x, y}}} = \ol{\angles{T_fx, y}} \\ > &= \angles{x, T_fy} = \angles{T_f^*x, y} > \end{align*} > $ > so $T_{\ol f} = T_f^*$ by the uniqueness of $\Gamma^{-1}$. > > Let $f, g \in C(\Sigma)$, then by the [[Gelfand-Naimark Theorem]], > $ > \angles{T_fT_gx, y} = \angles{fg, \mu_{x, y}} = \angles{f, \mu_{T_gx, y}} > $ > As this holds for all $f \in C(\Sigma)$, $d\mu_{T_gx,y} = gd\mu_{x, y}$ by uniqueness of the [[Riesz Representation Theorem (Radon Measure)|Riesz representation]]. Now suppose that $f \in B(\Sigma)$ and $g \in C(\Sigma)$, then $\angles{fg, \mu_{x, y}} = \angles{f, \mu_{T_gx, y}}$, so > $ > \angles{fg, \mu_{x,y}} = \angles{f, \mu_{T_gx, y}} = \angles{T_fT_gx, y} = \angles{T_gx, T_f^*y} = \angles{g, \mu_{x, T^*_fy}} > $ > and $d\mu_{x, T_f^*y} = fd\mu_{x, y}$. Lastly, suppose that $f, g \in B(\Sigma)$, then > $ > \angles{T_fT_gx, y} = \angles{T_gx, T_f^*y} = \angles{g, \mu_{x, T_f^*y}} = \angles{fg, \mu_{x, y}} = \angles{T_{fg}x, y} > $ > so $T_fT_g = T_{fg}$. > [!theorem] > > If $S \in L(\ch, \ch)$ commutes with every $T \in \Gamma^{-1}(C(\Sigma))$, then it commutes with every $T \in \Gamma^{-1}(B(\Sigma))$. > > *Proof*. If $S$ commutes with $T_f$ for all $f \in C(\Sigma)$, then > $ > \angles{f, \mu_{x, S^*y}} = \angles{T_fx, S^*y} = \angles{ST_fx, y} = \angles{T_fSx, y} = \angles{f, \mu_{Sx, y}} > $ > so $\mu_{x, S^*y} = \mu_{Sx, y}$ Now suppose that $f \in B(\Sigma)$, then > $ > \angles{T_fSx, y} = \angles{f, \mu_{Sx, y}} = \angles{f, \mu_{x, S^*y}} = \angles{T_fx, S^*y} = \angles{ST_fx, y} > $ > [!theorem] > > $\Gamma^{-1}$ is [[Continuity|continuous]] with respect to the $M(\Sigma)$-weak topology on $B(\Sigma)$ and the [[Operator Topologies|weak operator topology]] on $L(\ch, \ch)$. > > *Proof*. Let $\seq{f_n} \subset B(\Sigma)$ such that $f_n \to f \in B(\Sigma)$, then by weak convergence, > $ > \angles{T_{f_n}x, y} = \angles{f_n, \mu_{x, y}} \to \angles{f, \mu_{x, y}} = \angles{T_fx, y} > $ > so $T_{f_n}x \to T_fx$ weakly. > [!theorem] > > $\Gamma^{-1}$ is continuous with respect to the $M(\Sigma)$-weak topology on $B(\Sigma)$ and the [[Operator Topologies|strong operator topology]] on $L(\ch, \ch)$. > > *Proof*. Let $U, L^2$ be in the setting of [[Spectral Theorem|Spectral Theorem (II)]]. Let $f \in B(\Sigma)$, then $T_f$ has representation $UT_fU^{-1} = \phi_f$, where $\phi_f = f$ on each $\Sigma_i$. If $\seq{f_n}$ converges under the conditions of the [[Dominated Convergence Theorem]], then $\phi_{f_n} \to \phi_f$ in the condition of the dominated convergence theorem as well. Therefore for any $v \in \ch$, > $ > \norm{T_{f_n}v - T_{f}v} = \norm{\phi_{f_n}Uv - \phi_fUv}_2 \to 0 > $ > as $n \to \infty$.