Coercive self-adjoint operators are bijective
ProvedVectorSpaceOpt.coercive_selfadjoint_bijectiveLet H be a real Hilbert space and Q : H →L[ℝ] H a bounded self-adjoint operator. Suppose a constant m > 0 and an upper constant M satisfy the two-sided quadratic estimate
for every x ∈ H. Then Q is bijective. In particular, every right-hand side b has one and only one solution of Q x = b. This result is recorded by Luenberger as Chapter 10, Problem 10 and is used implicitly when the conjugate-gradient theorem names Q⁻¹ b. The milestone makes solvability explicit before any algorithmic convergence claim and is reusable for other coercive quadratic methods.
import Definitions.Def_VectorSpaceOpt_conjugate_gradient
namespace VectorSpaceOpt
/-- The operator-invertibility fact recorded as Chapter 10, Problem 10. -/
theorem coercive_selfadjoint_bijective
{H : Type*} [NormedAddCommGroup H] [InnerProductSpace ℝ H] [CompleteSpace H]
(Q : H →L[ℝ] H) (m M : ℝ) (hm : 0 < m)
(hself : IsRealSelfAdjoint Q) (hbounds : IsCoerciveBetween Q m M) :
Function.Bijective Q := by
sorry
end VectorSpaceOptRead-back
What the Lean code literally says, in plain math · gpt-5
Let be a complete real inner-product space and a continuous real-linear map. For arbitrary , assume , assume is self-adjoint, and assume every satisfies . Then the underlying function of is bijective: it is both injective and surjective. There is no separately stated hypothesis or ; in a nonzero space these are constrained by the quadratic bounds, while the zero-dimensional space is also included.
Confirmed by the mission captain (proposal self-audit).