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]]. > [!theorem] > > Let $x, y \in \ch$, then there exists a [[Bounded Linear Functional|bounded linear functional]] defined on $C(\Sigma)$, characterised by > $ > f \mapsto \angles{T_fx, y} \quad \abs{\angles{T_fx, y}} \le \norm{T_f} \cdot \norm{x} \cdot \norm{y} = \norm{f}_u \cdot \norm{x} \cdot \norm{y} > $ > By the [[Riesz Representation Theorem (Radon Measure)|Riesz Representation Theorem]], there exists a unique [[Space of Radon Measures|complex Radon measure]] $\mu_{x, y}$ on $C(\Sigma)$ such that > $ > \angles{T_fx, y} = \angles{f, \mu_{x, y}}\quad \norm{\mu_{x, y}} \le \norm{x} \cdot \norm{y} > $ > The mapping $\ch^2 \to M(X)$ defined by $(x, y) \mapsto \mu_{x, y}$ is a bounded sesquilinear map, such that $\mu_{x, y} = \ol{\mu}_{y, x}$, and $\mu_{x, x}$ is a positive measure for all $x \in \ch$. > > *Proof*. Sesquilinearity comes from the inner product. Since $\alg$ is a commutative $C^*$-algebra, it is symmetric, and the Gelfand transform maps adjoints into complex conjugates. This allows writing > $ > \begin{align*} > \angles{f, \mu_{x, y}} &= \angles{T_fx, y} = \angles{x, T_f^*y} = \ol{\angles{T^*_fy, x}} \\ > &= \ol{\angles{\ol f, \mu_{y, x}}} = \angles{f, \ol\mu_{y, x}} > \end{align*} > $ > Let $f \in C(\Sigma)$ with $f \ge 0$, and $g = \sqrt{f}$, then $g \in C(\Sigma)$ with $T^*_gT_g = T^2_g = T_{t^2} = T_f$, with > $ > \angles{f, \mu_{x, x}} = \angles{T_fx, x} = \angles{T_g^*T_gx, x} = \norm{T_gx}^2 \ge 0 > $ > so $\mu_{x, x} \ge 0$.