# Four tests for an action frequency A point of the action spectrum can be detected without constructing a spectral measure: narrow-band vectors become approximate character eigenvectors uniformly on compact group sets. The same points are characterized by lower bounds for every finite-measure integrated operator, or only for Fourier filters. The implications require a compact-tail estimate for measures and the exact Fourier sign in the measure definition. This lesson proves Takesaki II, Lemma XI.1.12 for the arbitrary LCA dual-Banach action of AF0 setting (D). *Programme proof written in Codex (OpenAI), September 2026; restoration and proof expansion, 5 October 2026. New expression is dedicated under CC0 to the extent of rights held. Human review is not asserted.* ## Spectral points yield approximate character vectors Let $G$ be an arbitrary locally compact Hausdorff abelian group, $H=\widehat G$, and $\alpha$ a uniformly bounded normal action on $X=X_*^*$ with norm-continuous predual orbits, as in [AF0](OA-FLOW-AF.md#af-0). Write $(t,p)=p(t)$. The first two tests are equivalent: **(i)** $p\in\operatorname{Sp}(\alpha)$. **(ii)** There is a net of unit vectors $x_i\in X$ such that, for every compact $K\subset G$, $$\sup_{t\in K}\|\alpha_t x_i-(t,p)x_i\|\longrightarrow0.\tag{F1}$$ Assume (i). Given compact $K$ and $\varepsilon>0$, [the earlier narrow-band proof](OA-FLOW-L87.md#oa-flow.band.orbits) supplies an open neighborhood $U\ni p$ such that every $x\in X_\alpha(U)$ has orbit error at most $\varepsilon\|x\|$ on $K$. This space contains a nonzero vector by a direct current-provider argument. LF1 gives a compactly supported filter $c$ in $U$ with $c(p)=1$. Since $p\in h(I(\alpha))$, the operator $\alpha_c$ cannot vanish. Choose $y$ with $\alpha_cy\ne0$. The filter-support theorem BS3 puts its spectrum inside $\operatorname{supp}c\subset U$; normalize that vector. Thus $X_0^\alpha(U)\ne0$, where this notation denotes the closed filtered span of BS4. In particular a nonzero generator $\alpha_fy$, with $\operatorname{supp}f\subset U$, has spectrum inside that compact support. This is the actual nonzero-band proof; the closed filtered span itself can acquire boundary frequencies, as BS4 explains. Direct the pairs $(K,\varepsilon)$ by inclusion of compact sets and decreasing positive error. A finite union of compact sets is compact, so this is a directed set even when $G$ is not sigma compact. Selecting one normalized narrow-band vector for each pair gives the net in (ii). In particular, no sequence is claimed in the absence of a countable compact exhaustion. ## Finite-measure integrated operators For a finite complex regular Borel measure $\mu$ on $G$, use the negative action parameter and the matching negative character sign: $$\widehat\mu(p)=\int_G\overline{(t,p)}\,d\mu(t),\qquad \alpha_\mu x=\int_G\alpha_{-t}x\,d\mu(t). \tag{F2}$$ The finite-measure convention and its complete vector integral are proved in [AF1](OA-FLOW-AF.md#af-1): $\|\mu\|=|\mu|(G)<\infty$, with Radon total variation and its completion. The second integral is a weak-star integral, defined through the specified predual. For $\phi\in X_*$, the orbit $t\mapsto\phi\circ\alpha_{-t}$ is norm continuous and bounded by $C_\alpha\|\phi\|$. AF1 proves strong measurability on a full-measure countable union of compact sets, with separable essential range even when the predual and group are nonseparable. It therefore gives the full norm integral. Equivalently, integrate it in $X_*$ first over compact sets. Regularity of $|\mu|$ provides compact sets whose complements have arbitrarily small mass; the integrals converge in predual norm, independent of the exhaustion. Their bounded operator $B_\mu:X_*\to X_*$ has $\|B_\mu\|\le C_\alpha\|\mu\|$, and $\alpha_\mu=B_\mu^*$. Therefore $$\alpha_\mu\text{ is normal},\qquad \|\alpha_\mu\|\le C_\alpha\|\mu\|. \tag{F3}$$ There is no global sigma-finiteness assumption on Haar measure and no assertion that arbitrary $X$-orbits are Bochner integrable in norm. ## Compact-tail control of a measure Assume (ii), and fix $\mu$. For a compact $K\subset G$, a unit vector $x_i$, and $t\in K$, compare $\alpha_{-t}x_i$ with $\overline{(t,p)}x_i=(-t,p)x_i$. The triangle inequality and the uniform action bound give $$\begin{aligned} \|\alpha_\mu x_i-\widehat\mu(p)x_i\| &\le |\mu|(K) \sup_{s\in-K}\|\alpha_sx_i-(s,p)x_i\|\\ &\quad +(C_\alpha+1)|\mu|(G\setminus K). \end{aligned}\tag{F4}$$ For every unit predual functional, integrate its continuous scalar pairing and bound it by the same compact supremum. Taking the norming supremum proves (F4) even when the X-valued orbit has no norm integral. The compact part follows by integration of the norm bound; on the complement, $\|\alpha_{-t}x_i\|+\|x_i\|\le C_\alpha+1$. For fixed $K$, the first term tends to zero by (ii), since $-K$ is compact. Given $\delta>0$, regularity supplies $K$ with $|\mu|(G\setminus K)<\delta/(C_\alpha+1)$; then take the net index large. Consequently $\alpha_\mu x_i-\widehat\mu(p)x_i\to0$ in norm, and $\|x_i\|=1$ implies $$|\widehat\mu(p)|\le\|\alpha_\mu\| \quad\text{for every finite complex regular measure }\mu. \tag{F5}$$ This is test (iii). The two limits are ordered: first choose a compact set controlling the measure tail, then use compact-uniform approximation on that set. No uniform approximation over all of a noncompact $G$ is required. ## Fourier filters and the reverse implication Test (iv) is the corresponding statement only for Fourier functions: $$|f(p)|\le\|\alpha_f\| \quad\text{for every }f\in A(H). \tag{F6}$$ Condition (iii) implies (iv) with an explicit sign check. Write $f=\mathcal Fa$ using the positive Fourier convention, and set $d\mu(s)=a(-s)\,ds$. [L24’s inversion proof](OA-FLOW-L24.md#oa-flow.grp.translations) gives Haar inversion invariance on an LCA group; [HR3](OA-FLOW-HR.md#hr-03) and AF1 give the finite regular density measure at a Borel representative of the kernel. Changing variables $t=-s$ in (F2) gives $$\alpha_\mu=\alpha_f, \qquad\widehat\mu(p) =\int_G\overline{(s,p)}a(-s)\,ds =\int_G(t,p)a(t)\,dt=f(p). \tag{F7}$$ Finally assume (iv). If $p\notin\operatorname{Sp}(\alpha)=h(I(\alpha))$, some $f\in I(\alpha)$ has $f(p)\ne0$. But $\alpha_f=0$, contradicting (F6). Thus (iv) implies (i). Together with the preceding sections, (i)–(iv) are equivalent. $\square$ All four tests hold also in AF0 setting (B). Use AF1’s norm integral on every vector for (F2), giving the same bound (F3) without a normality assertion; (F4) is the norm integral inequality there. The band proof, hull implication and all scalar sign calculations are unchanged. Thus the dual-space theorem is retained and the ordinary Banach generality has its actual complete construction. **Problem.** Let $\alpha_t=\operatorname{diag}(e^{it\lambda_1},e^{it\lambda_2})$ on $\mathbb C^2$. Identify its action spectrum and the norm $\|\alpha_\mu\|$. **Solution.** The two coordinate lines have frequencies $\lambda_1,\lambda_2$, so $\operatorname{Sp}(\alpha)=\{\lambda_1,\lambda_2\}$, with a repeated frequency listed once. Formula (F2) makes $\alpha_\mu$ diagonal with entries $\widehat\mu(\lambda_1)$ and $\widehat\mu(\lambda_2)$, hence $\|\alpha_\mu\|=\max_j|\widehat\mu(\lambda_j)|$. Thus (F5) holds at either spectral frequency. For $p$ outside the finite set, a local Fourier cutoff can be one at $p$ and zero at both $\lambda_j$, violating (F6) and therefore (F5). $\square$ The mathematical source is M. Takesaki, *Theory of Operator Algebras II*, Lemma XI.1.12, printed pages 322–323 ([edition record](https://doi.org/10.1007/978-3-662-10451-4)). Equations (F2)–(F7) use the same negative character sign as the negative action parameter. The narrow-band proof and all finite-measure domains are supplied at exact earlier locators; the compact-tail order of limits retains arbitrary LCA groups.