6.4. Contradiction from an eventual sign🔗

If there are no zeros above some T>0, choose σ in {−1,1} such that σg(t)>0 for t>T. The kernel in J_h is nonnegative. The signed tail integral therefore bounds the signed integral over [2T,2T+1] from below. Splitting at T gives

∫[2T,2T+1] σg(t) (P_h(t)/h²)^k cosh(b_h t) dt
  ≤ σJ_h + ∫[0,T] |g(t)| (P_h(t)/h²)^k cosh(b_h t) dt.

Take h→0+. Both displayed bounded-interval integrals converge and J_h→0. Put w(t)=σg(t)cosh(at), R=∫[0,T]|w(t)|dt, and m=∫[2T,2T+1]w(t)dt. Continuity and strict positivity give m>0. For every k≥1,

m (4T²+1/4)^k ≤ R (T²+1/4)^k.

Since 0 < T²+1/4 < 4T²+1/4, this contradicts geometric growth. Shift k by one to use the checked geometric growth lemma. Thus zeros occur above every height, which implies the frozen Hardy statement.