· 9,744 chars · 15 deps · depth 24 Reason: First publication. Proof for the viscous Hamilton-Jacobi operator: strict properness from the monotonicity of the trace, the structure condition by reduction to the example of an elliptic operator with a continuous inhomogeneity, and continuity by an explicit estimate of the four terms.
Strict properness reduces to the monotonicity of the trace for the semidefinite ordering, and the structure condition follows by applying the example of an elliptic operator with a continuous inhomogeneity. Continuity is checked directly, by bounding each of the four terms of F separately: a trace norm bound, the Cauchy-Schwarz inequality for the quadratic term, and the continuity of f.
Proof
Conventions. The order ≤ and the arithmetic of R are those of the ordered field of real numbers. Multiplication by a nonnegative real number preserves ≤: if a≤b and 0≤λ, then either a=b, and the two products are equal, or a<b, and then λa≤λb by claim 10 of Elementary Order Arithmetic in an Ordered Field when 0<λ, while λ=0 makes both products 0. Adding a fixed real number to both sides of an inequality also preserves it, by the compatibility of ≤ with addition. Finally, 0<2−1 by claims 8 and 7 of Elementary Order Arithmetic in an Ordered Field, so 2κ=κ⋅2−1 is nonnegative and 21=1⋅2−1 is positive.
For condition 2, let x∈Ω, p∈Rn, X∈S(n) and r,s∈R with s≤r. The three terms −2κtr(X), 21∥p∥2 and −f(x) are the same in F(x,r,p,X) and in F(x,s,p,X), so
F(x,r,p,X)−F(x,s,p,X)=γr−γs=γ(r−s)
by distributivity. In particular γ(r−s)≤F(x,r,p,X)−F(x,s,p,X), and claim 2 is proved.
Proof of claim 1. Fix x0∈Ω, r0∈R, p0∈Rn and X0∈S(n), and let ε∈R be positive. Put 4=2+2; then 0<2 by claim 8 of Elementary Order Arithmetic in an Ordered Field and hence 0<4 by claim 3 there, so 4−1 exists and is positive by claim 7 there, and ε′=ε⋅4−1 is positive by claim 5 there. Moreover ε′+ε′+ε′+ε′=(1+1+1+1)ε′=4(ε⋅4−1)=ε by distributivity, commutativity and the defining property of the multiplicative inverse.
If moreover ∥h∥≤1, then multiplying by the nonnegative ∥h∥ gives ∥h∥2≤∥h∥, so that by distributivity
∥q∥2−∥p0∥2≤(2∥p0∥+1)∥h∥.
(iv) Since Ω⊆Ω and f is continuous at x0 relative to Ω, there is a positive δf∈R such that every y∈Ω with dE(x0,y)<δf satisfies ∣f(y)−f(x0)∣<ε′; in particular this holds for every y∈Ω.
Choice of δ. Put
c1=γ,c2=2κβn+1,c3=21(2∥p0∥+1)+1.
Each ci is positive: c1=γ is positive by hypothesis, while 2κβn and 21(2∥p0∥+1) are nonnegative, being products of nonnegative numbers — for βn this is recorded in The Trace as a Sum of Quadratic Forms, its Monotonicity and a Norm Bound, where 1≤βn — so c2 and c3 are at least 1 and 0<1 by claim 6 of Elementary Order Arithmetic in an Ordered Field. By claim 7 there each ci−1 exists and is positive, and by claim 5 there each ε′ci−1 is positive. Applying claim 9 of Elementary Order Arithmetic in an Ordered Field repeatedly to the five positive numbers 1, δf, ε′c1−1, ε′c2−1 and ε′c3−1 produces a real number δ that is at most each of them and equal to one of them, hence positive.
Conclusion. Let y∈Ω, s∈R, q∈Rn and Y∈S(n) satisfy
dE(y,x0)<δ,∣s−r0∣<δ,∥q−p0∥<δ,dS(n)(Y,X0)<δ,
and put h=q−p0, so that ∥h∥<δ≤1. Grouping the four terms of F,
We bound each summand by ε′. By (i) and claim 10 of Elementary Order Arithmetic in an Ordered Field, γ∣s−r0∣<γδ, and multiplying δ≤ε′c1−1 by the nonnegative γ=c1 gives γδ≤ε′; hence the first summand is less than ε′. By (ii), 2κβn≤c2, so multiplying by the nonnegative dS(n)(Y,X0) and using (ii) again gives 2κ∣tr(Y)−tr(X0)∣≤c2dS(n)(Y,X0)<c2δ≤ε′, the last two steps by claim 10 and by multiplying δ≤ε′c2−1 by c2. By (iii), and since ∥h∥≤1, the same reasoning with c3 gives 21∣∥q∥2−∥p0∥2∣≤c3∥h∥<c3δ≤ε′. By (iv) and dE(y,x0)<δ≤δf, the fourth summand is less than ε′.