Manuscript XVII-A@1 · XVII-A.tex · section · 711d7ee4

The boundary smearing

Exact reconstruction of the named observable › The boundary smearing · lines 511–538

XVII-A@1#eq:boundary-representative

After fixing the objects above, define

Q_{B} = \frac{1}{2} \int_{- π /2}^{π /2} β (t) d t, O_{B} = \frac{1}{2} (1 + sin (λ_{*} Q_{B})) .

The integral is an operator-norm integral of a continuous finite-matrix function.

It belongs to $A (B)$ even though its displayed scalar weight has endpoints at $\partial I_{B}$ . To see this, choose smooth $χ_{j}$ compactly supported in $I_{B}^{\circ}$ with $χ_{j} \to 1$ in $L^{1} (I_{B})$ . Then

\frac{1}{2} \int_{I_{B}} (1 - χ_{j} (t)) β (t) d t_{\infty} \leq \frac{1}{2} t \in I_{B} sup ∥ β (t) ∥_{\infty} ∥ 1 - χ_{j} ∥_{L^{1} (I_{B})} ⟶ 0.

Norm closure and functional calculus give $Q_{B}, O_{B} \in A (B)$ .

Displayed formulas in this passage

XVII-A@1#eq:boundary-representative · line 514
$Q_{B} = \frac{1}{2} \int_{- π /2}^{π /2} β (t) d t, O_{B} = \frac{1}{2} (1 + sin (λ_{*} Q_{B})) .$
unlabeled · line 528
$\frac{1}{2} \int_{I_{B}} (1 - χ_{j} (t)) β (t) d t_{\infty} \leq \frac{1}{2} t \in I_{B} sup ∥ β (t) ∥_{\infty} ∥ 1 - χ_{j} ∥_{L^{1} (I_{B})} ⟶ 0.$

The boundary smearing

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