Theorem: Bound on Total Variation of Scalar-Valued Measure associated with Resolution of the Identity

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B {\HH}$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
Let $\norm {\EE_{x, y} }$ denote the total variation of $\EE_{x, y}$. 
Then:
:$\norm {\EE_{x, y} } \le \norm x \norm y$

Proof:
Let $\map P X$ be the set of finite partitions of $X$ into $\map \BB X$-measurable sets.
We have:
{{begin-eqn}}
{{eqn	| l = \norm {\EE_{x, y} }
		| r = \map {\cmod {\EE_{x, y} } } X
		| c = {{Defof|Total Variation/Measure Theory/Complex Measure|Total Variation: Measure Theory: Complex Measure}}
}}
{{eqn	| r = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map {\EE_{x, y} } {S_j} } : \set {S_1, \ldots, S_n} \in \map P X}
}}
{{end-eqn}}
Let $\set {S_1, S_2, \ldots, S_n} \in \map P X$.
We have:
:$\ds \sum_{j \mathop = 1}^n \cmod {\map {\EE_{x, y} } {S_j} } = \sum_{j \mathop = 1}^n \cmod {\innerprod {\map \EE {S_j} x} y}$
Let $\alpha_j \in \C$ be such that $\cmod {\alpha_j} = 1$ and:
:$\cmod {\innerprod {\map \EE {S_j} x} y} = \alpha_j \innerprod {\map \EE {S_j} x} y$
We then have:
{{begin-eqn}}
{{eqn	| l = \sum_{j \mathop = 1}^n \cmod {\innerprod {\map \EE {S_j} x} y}
		| r = \sum_{j \mathop = 1}^n \alpha_j \innerprod {\map \EE {S_j} x} y
}}
{{eqn	| r = \innerprod {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} y
}}
{{eqn	| o = \le
		| r = \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} \norm y
		| c = Cauchy-Bunyakovsky-Schwarz Inequality: Inner Product Spaces
}}
{{end-eqn}}
Since $\set {S_1, S_2, \ldots, S_n}$ is a partition, we have $S_i \cap S_j = \O$ for $i \ne j$. 
Hence from Orthogonality of Resolution of the Identity evaluated at Disjoint Sets, we have:
:$\innerprod {\map \EE {S_i} x} {\map \EE {S_j} x} = 0$ for $i \ne j$.
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x}^2
		| r = \sum_{j \mathop = 1}^n \norm {\alpha_j \map \EE {S_j} x}^2
		| c = Pythagoras's Theorem (Inner Product Space)
}}
{{eqn	| r = \sum_{j \mathop = 1}^n \norm {\map \EE {S_j} x}^2
		| c = $\cmod {\alpha_j} = 1$
}}
{{eqn	| r = \sum_{j \mathop = 1}^n \map {\EE_{x, x} } {S_j}
}}
{{eqn	| r = \map {\EE_{x, x} } {\bigcup_{j \mathop = 1}^n S_j} 
		| c = $(5)$ in {{Defof|Resolution of the Identity}}
}}
{{eqn	| r = \map {\EE_{x, x} } X
		| c = {{Defof|Set Partition}}
}}
{{eqn	| r = \innerprod {\map \EE X x} x
}}
{{eqn	| r = \innerprod x x
		| c = $(1)$ in {{Defof|Resolution of the Identity}}
}}
{{eqn	| r = \norm x^2
		| c = {{Defof|Inner Product Norm}}
}}
{{end-eqn}}
and hence:
:$\ds \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} = \norm x$ 
Hence we have:
:$\ds \norm {\sum_{j \mathop = 1}^n \alpha_j \map \EE {S_j} x} \norm y = \norm x \norm y$
Hence we have:
:$\ds \sum_{j \mathop = 1}^n \cmod {\map {\EE_{x, y} } {S_j} } \le \norm x \norm y$
Taking the supremum over $\set {S_1, \ldots, S_n} \in \map P X$ we obtain:
:$\norm {\EE_{x, y} } \le \norm x \norm y$
{{qed}}
Category:Resolutions of the Identity
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