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Testing a coercive weak solution with itself gives the energy bound

Statement

Let H be a real or complex Hilbert space, a a bounded coercive sesquilinear form with constant α>0, F a bounded conjugate-linear functional on H, and u∈H a solution of a(u,v)=F(v) for all v∈H. Then α∥u∥2≤Re⁡a(u,u)=Re⁡F(u)≤∥F∥ ∥u∥,hence α∥u∥≤∥F∥. The bound is a priori in the sense that it uses only the equation, coercivity and the norm of the datum, not the construction of u; it applies directly to homogeneous Dirichlet solutions u∈H01(Ω) of Weak Dirichlet solutions for a divergence-form operator after substituting their coercivity constants. For an inhomogeneous Dirichlet solution, first subtract a lifting to obtain a solution in H01(Ω) and use its residual datum; the original solution need not itself be an admissible test.

Facts & Assumptions

Given: A real or complex Hilbert space H; a bounded coercive sesquilinear form a with coercivity constant α>0; a bounded conjugate-linear functional F with ∥F∥=sup⁡∥v∥≤1∣F(v)∣; and a vector u∈H with a(u,v)=F(v) for every v∈H.

[F1]

Coercivity: Re⁡a(u,u)≥α∥u∥2; and Re⁡F(u)≤∣F(u)∣≤∥F∥ ∥u∥, since Re⁡z≤∣z∣ (Bounded, coercive and symmetric sesquilinear forms, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, Real and imaginary parts, complex conjugation, and modulus, A bounded linear operator between normed spaces, Hilbert space).

[F2]

The equation with the test v=u reads a(u,u)=F(u) (The Lax--Milgram theorem gives existence and uniqueness if Countable Choice is additionally assumed; here the identity uses only the assumed equation).

Proof

1.1F2given

Testing with the solution: substitute v=u in the assumed equation, obtaining a(u,u)=F(u) and hence, taking real parts, Re⁡a(u,u)=Re⁡F(u).

2.1F1step 1.1algebra∎

Two-sided bound: by coercivity, α∥u∥2≤Re⁡a(u,u)=Re⁡F(u)≤∣F(u)∣≤∥F∥ ∥u∥. If u≠0, divide by ∥u∥ to obtain α∥u∥≤∥F∥; if u=0, the same inequality holds trivially. The estimate uses only the equation, coercivity and the datum norm, not any construction of u.

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