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The sectorial form angle controls the numerical range of its operator

Statement

Let a be a closed sectorial form on V⊆H with constants M,θ and associated operator A (Closed sectorial form and its associated operator), and write Sθ‾:={ζ∈C:∣arg⁡ζ∣≤θ}∪{0} for the closed sector of half-angle θ around the positive real axis. Then for every u∈D(A):

  1. ⟨Au,u⟩=−a(u,u);
  2. ⟨(A−M)u,u⟩∈−Sθ‾.

In particular the normalized quadratic form values ⟨(A−M)u,u⟩/∥u∥H2 for u∈D(A)∖{0} lie in −Sθ‾ and those of A lie in its translate by M; when A is bounded this is the containment of the numerical range (Numerical range and numerical radius). No choice principle is used.

Facts & Assumptions

Given: A closed sectorial form a on the dense subspace V⊆H with constants M≥0 and θ∈[0,π/2) and associated operator A (Closed sectorial form and its associated operator); the closed sector Sθ‾={ζ:Re⁡ζ≥0, ∣Im⁡ζ∣≤tan⁡θ Re⁡ζ}; and a vector u∈D(A) with f:=Au.

[L1]

D(A)={u∈V:∃f∈H, a(u,v)=−(f,v) ∀v∈V} and Au:=f, and the form satisfies Re⁡a(u,u)≥−M∥u∥H2 and ∣Im⁡a(u,u)∣≤tan⁡θ (Re⁡a(u,u)+M∥u∥H2) for all u∈V; the pairing is linear in the first argument (Closed sectorial form and its associated operator).

[L2]

The inner product of a complex Hilbert space satisfies ⟨u,v⟩=⟨v,u⟩‾, ⟨v,v⟩≥0 with equality only for v=0, and is linear in the first argument; in particular ⟨v,v⟩=∥v∥H2 is real and nonnegative (Real and complex inner-product spaces and their induced length, Hilbert space).

Proof

technique · direct
1.1L1givenalgebra

Claim 1. For u∈D(A) the defining relation with f=Au gives a(u,v)=−(Au,v) for every v∈V; testing with v=u gives a(u,u)=−(Au,u), that is ⟨Au,u⟩=−a(u,u).

1.2L1L2givenalgebra

The shifted form and its sector. Define b(u,v):=a(u,v)+M⟨u,v⟩ on V×V. Then b is sesquilinear and, for u∈V, Re⁡b(u,u)=Re⁡a(u,u)+M∥u∥H2≥0 by [L1] and [L2], while Im⁡b(u,u)=Im⁡a(u,u) and ∣Im⁡a(u,u)∣≤tan⁡θ (Re⁡a(u,u)+M∥u∥H2), so ∣Im⁡b(u,u)∣≤tan⁡θRe⁡b(u,u); by the description of Sθ‾ in the givens this says b(u,u)∈Sθ‾.

2.1step 1.1step 1.2L2algebra

Claim 2. For u∈D(A), ⟨(A−M)u,u⟩=⟨Au,u⟩−M⟨u,u⟩=−a(u,u)−M∥u∥H2=−(a(u,u)+M⟨u,u⟩)=−b(u,u) by [step 1.1], [step 1.2] and [L2], and −b(u,u)∈−Sθ‾ by [step 1.2].

3.1step 1.1step 2.1L2given∎

Normalized consequences. If u∈D(A)∖{0} then ∥u∥H2>0 by [L2], and multiplying by the positive real scalar ∥u∥H−2 preserves the closed sector Sθ‾, so ⟨(A−M)u,u⟩/∥u∥H2∈−Sθ‾ and ⟨Au,u⟩/∥u∥H2∈M−Sθ‾; for bounded A on a nonzero H, these normalized values are exactly the numerical ranges of A−M and A, respectively. On H={0} both operators are zero and their numerical ranges are {0} by the convention in Numerical range and numerical radius; the containments still hold since 0∈−Sθ‾ and 0=M−M∈M−Sθ‾. Claims 1 and 2 are [step 1.1] and [step 2.1], and the argument fixed the arbitrary vector u and used no selection, so no choice principle was used.

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