A unit-supported local integral and a quantifier counterexample

Worked example, authored by the operator. Fix a prime p, the field Q_p and its ring Z_p. Normalize multiplicative Haar measure so Z_p^* has measure 1, and set f(x)=1 for x in Z_p^* and 0 otherwise. For an unramified continuous multiplicative character chi (meaning chi is 1 on Z_p^*) consider I(s)=integral over Q_p^* of f(x) chi(x) |x|_p^s d^*x. On the support, both chi(x) and |x|_p^s are 1 for every complex s. Consequently I(s)=measure(Z_p^*)=1: convergence and nonvanishing follow without analytic continuation.

The unramified assumption matters. If chi is nontrivial on the units, choose a unit u with chi(u)!=1. Haar invariance gives J=integral over Z_p^* of chi(x)d^*x=chi(u)J, so J=0. Unit support alone therefore cannot ensure nonvanishing for every character.

This elementary local integral illustrates the support/character issue; it is not the GL(n+1) Whittaker construction. Its quantifier order must also be preserved: let R(w,t) mean w=t on {0,1}. For every t there exists w with R(w,t), but no single w works for both t=0 and t=1. Likewise choosing W after each pi does not establish a single W chosen before every pi. The target still requires the specified Whittaker transformation law, representations, shift matrix and integral conventions.

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