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TauCeti.Analysis.Fourier.Hardy.Coburn

Coburn's alternative for Toeplitz operators #

For a nonzero essentially bounded symbol a on the unit circle, either the Toeplitz operator T_a or its adjoint T_a† = T_ā is injective. Consequently a Fredholm Toeplitz operator of index zero is invertible; this is how invertibility of T_a for a continuous symbol a reduces to the computation of its Fredholm index.

The proof is the Hardy product argument. A Hardy function f lies in the kernel of T_a exactly when a f has only negative Fourier modes (TauCeti.toeplitz_apply_eq_zero_iff). Suppose that T_a f = 0 and T_ā g = 0. The conjugate conj g of a Hardy function has only nonpositive modes, so the integrable product a f conj g has only negative modes. Its complex conjugate ā g conj f has only negative modes for the same reason, so a f conj g also has only positive modes. Hence all Fourier coefficients of a f conj g vanish and a f conj g = 0 almost everywhere. If f and g are nonzero, then by boundary uniqueness they are nonzero almost everywhere, which forces a = 0.

Main results #

References #

theorem TauCeti.eq_zero_of_toeplitz_apply_eq_zero_of_toeplitz_star_apply_eq_zero {a : ↥(MeasureTheory.Lp ℂ ⊤ AddCircle.haarAddCircle)} {f g : HardySpace} (hf₀ : f ≠ 0) (hg₀ : g ≠ 0) (hf : (toeplitz a) f = 0) (hg : (toeplitz (star a)) g = 0) :
a = 0

The Hardy product argument. If the Toeplitz operators T_a and T_ā = T_a† have nonzero kernel vectors f and g, then the symbol a is zero.

Coburn's alternative. For a nonzero essentially bounded symbol a, either the Toeplitz operator T_a or its adjoint has trivial kernel.

A Fredholm Toeplitz operator of index zero is invertible.