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Complex sesquilinear coercivity differs from bilinear positivity

Example

On H=C define a(u,v):=uv‾ and b(u,v):=uv. Then a is sesquilinear in the convention of Bounded, coercive and symmetric sesquilinear forms (linear in the first argument, conjugate-linear in the second), bounded with M=1, and coercive with constant α=1 because a(u,u)=∣u∣2. The expression b is bilinear, not conjugate-linear in the second argument, and it fails the coercivity condition: b(i,i)=−1, so Re⁡b(u,u)≥α∣u∣2 fails at u=i for every α>0. More generally b(u,u)=u2 is not real for u∉R∪iR, and its real part is negative, for example, at u=1+2i, since (1+2i)2=−3+4i. Hence the conjugation in the second slot is not cosmetic: the complex Lax--Milgram hypotheses cannot be applied to this bilinear pairing b, and the real bilinear convention of Bounded, coercive and symmetric sesquilinear forms is genuinely a different hypothesis. This tests exactly the convention on which The Lax--Milgram theorem and A bounded form is represented by a unique bounded operator depend, and complements the real-form sources [Si] and [H], which state real bilinear versions.

Facts & Assumptions

Given: The Hilbert space H=C with its usual inner product and ∣z∣ the complex modulus; the pairings a(u,v)=uv‾ and b(u,v)=uv.

[F1]

Sesquilinearity, boundedness and coercivity definitions: a is linear in the first slot and conjugate-linear in the second with ∣a(u,v)∣≤M∣u∣∣v∣, and coercive with constant α when Re⁡a(u,u)≥α∣u∣2 (Bounded, coercive and symmetric sesquilinear forms, Real and complex inner-product spaces and their induced length, Hilbert space).

[F2]

Scalar facts: uu‾=∣u∣2≥0, ∣uv‾∣=∣u∣∣v∣, i‾=−i and i2=−1 (Real and imaginary parts, complex conjugation, and modulus).

[F3]

Lax--Milgram and the form-to-operator lemma are stated for sesquilinear forms in the conjugate-linear-second-slot convention (The Lax--Milgram theorem, A bounded form is represented by a unique bounded operator).

Proof

1.1F1F2

The sesquilinear form a: for scalars λ, a(λu,v)=λuv‾=λa(u,v) and a(u,λv)=uλv‾=λ‾ a(u,v), so a is linear in the first argument and conjugate-linear in the second; ∣a(u,v)∣=∣u∣∣v∣≤1⋅∣u∣∣v∣ gives the bound M=1, and a(u,u)=∣u∣2 gives coercivity with α=1.

1.2F2F1

The bilinear pairing b is not of this type: b(λu,v)=λb(u,v)=b(u,λv), so b is bilinear; but with u=v=1 and λ=i, b(1,i)=i≠−i=i‾ b(1,1), so b is not conjugate-linear in the second slot.

2.1F2step 1.2algebra

b fails coercivity: b(i,i)=i2=−1 has real part −1, while α∣i∣2=α>0 for every α>0; hence Re⁡b(u,u)≥α∣u∣2 fails at u=i for every α. More generally b(u,u)=u2 is not real unless u∈R∪iR.

3.1F1F2F3step 1.2step 2.1algebra∎

Consequences: neither Lax--Milgram nor the representation lemma applies to this b, since it is not sesquilinear. More generally, a complex form that is both bilinear and sesquilinear satisfies ic(u,v)=c(u,iv)=−ic(u,v), hence is the zero form. The zero form satisfies the bounded sesquilinear hypotheses of the representation lemma; on a nonzero space it cannot be coercive, but on H={0} it is coercive with every α>0 and satisfies the Lax--Milgram form hypotheses. Thus the conjugation convention matters, with this zero-form exception.

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