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Proof of The Structure Condition Implies Degenerate Ellipticity

propositionprop:structure-condition-implies-elliptic-2026a
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· 12,533 chars · 19 deps · depth 23 Reason: First publication. Proof that the structure condition implies degenerate ellipticity: a quadratic estimate places the perturbed pair X, Y+sigma I inside the matrix hypothesis of the structure condition at a suitable alpha, and applying the condition at y = x - p/alpha with an explicit choice of alpha and sigma contradicts a strict failure of ellipticity.

Given XYX\preceq Y, the perturbed pair XX, Y+σIY+\sigma I satisfies the two-sided quadratic bound of the structure condition once 3α3\alpha dominates three explicit constants; applying the structure condition at y=xα1py=x-\alpha^{-1}p, so that α(xy)=p\alpha(x-y)=p, and letting α\alpha be large and σ\sigma small contradicts a strict inequality F(x,r,p,X)<F(x,r,p,Y)F(x,r,p,X)<F(x,r,p,Y).

Proof

Conventions. The order \le and the arithmetic of R\mathbb{R} are those of the ordered field of real numbers. Three elementary consequences of the ordered field axioms are used freely. First, multiplication by a nonnegative real number preserves \le: if aba\le b and 0λ0\le\lambda, then either a=ba=b, and the products are equal, or a<ba<b, and then λaλb\lambda a\le\lambda b by claim 10 of Elementary Order Arithmetic in an Ordered Field when 0<λ0<\lambda, while λ=0\lambda=0 makes both products 00. Second, two inequalities may be added: if aba\le b and cdc\le d then a+cb+cb+da+c\le b+c\le b+d by the compatibility of \le with addition and transitivity. Third, if a<ba<b and 0<λ0<\lambda, then λa<λb\lambda a<\lambda b by claim 10 of Elementary Order Arithmetic in an Ordered Field. We write 3=1+1+13=1+1+1, so that 131\le3 and hence α3α\alpha\le3\alpha for every nonnegative α\alpha. Of the properties of Ω\Omega supplied by Bounded Open Domain in Euclidean Space, only its openness is used below; that setting is adopted because the structure condition is stated in it.

Fix xΩx\in\Omega, rRr\in\mathbb{R}, pRnp\in\mathbb{R}^{n} and X,YS(n)X,Y\in\mathcal{S}(n) with XYX\preceq Y. We must show that F(x,r,p,Y)F(x,r,p,X)F(x,r,p,Y)\le F(x,r,p,X).

Throughout, σR\sigma\in\mathbb{R} denotes a positive number, and we put Yσ=Y+σInY_{\sigma}=Y+\sigma I_{n}, which lies in S(n)\mathcal{S}(n) by Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §symmetric. By claim 7 of Elementary Order Arithmetic in an Ordered Field the inverse σ1\sigma^{-1} exists and is positive.

Step 1: an upper quadratic bound for the perturbed pair. We claim that for all ξ,ηRn\xi,\eta\in\mathbb{R}^{n},

ξ(Xξ)η(Yση)  (1+σ1Y)Yξη2.\xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta)\ \le\ \bigl(1+\sigma^{-1}\lVert Y\rVert\bigr)\lVert Y\rVert\,\lVert\xi-\eta\rVert^{2}.

Put ζ=ξη\zeta=\xi-\eta, so that ξ=η+ζ\xi=\eta+\zeta. By claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claims 2 and 5 of Bilinearity and Symmetry of the Dot Product on Rn\mathbb{R}^n,

ξ(Yξ)=η(Yη)+η(Yζ)+ζ(Yη)+ζ(Yζ),\xi\cdot(Y\xi)=\eta\cdot(Y\eta)+\eta\cdot(Y\zeta)+\zeta\cdot(Y\eta)+\zeta\cdot(Y\zeta),

and ζ(Yη)=η(Yζ)\zeta\cdot(Y\eta)=\eta\cdot(Y\zeta): indeed claim 5 of Elementary Properties of the Transpose of a Real Matrix gives ζ(Yη)=(Yζ)η\zeta\cdot(Y\eta)=(Y^{\top}\zeta)\cdot\eta, while Y=YY^{\top}=Y by Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §symmetric and the dot product is symmetric by claim 1 of Bilinearity and Symmetry of the Dot Product on Rn\mathbb{R}^n. Since XYX\preceq Y gives ξ(Xξ)ξ(Yξ)\xi\cdot(X\xi)\le\xi\cdot(Y\xi) by Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §ordering, we obtain

ξ(Xξ)η(Yη)  2(η(Yζ))+ζ(Yζ),\xi\cdot(X\xi)-\eta\cdot(Y\eta)\ \le\ 2\bigl(\eta\cdot(Y\zeta)\bigr)+\zeta\cdot(Y\zeta),

where 2u2u abbreviates u+uu+u.

By A Weighted Young Inequality and the Splitting of a Quadratic Form §young, applied with a=ηa=\eta, b=Yζb=Y\zeta and t=σt=\sigma,

2(η(Yζ))  ση2+σ1Yζ2.2\bigl(\eta\cdot(Y\zeta)\bigr)\ \le\ \sigma\lVert\eta\rVert^{2}+\sigma^{-1}\lVert Y\zeta\rVert^{2}.

By Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §vector-bound we have YζYζ\lVert Y\zeta\rVert\le\lVert Y\rVert\lVert\zeta\rVert; both sides are nonnegative, by claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and claim 1 of Properties of the Norm of a Symmetric Real Matrix, so claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives Yζ2Y2ζ2\lVert Y\zeta\rVert^{2}\le\lVert Y\rVert^{2}\lVert\zeta\rVert^{2}, where Y2=YY\lVert Y\rVert^{2}=\lVert Y\rVert\lVert Y\rVert. Multiplying by the nonnegative σ1\sigma^{-1},

2(η(Yζ))  ση2+σ1Y2ζ2.2\bigl(\eta\cdot(Y\zeta)\bigr)\ \le\ \sigma\lVert\eta\rVert^{2}+\sigma^{-1}\lVert Y\rVert^{2}\lVert\zeta\rVert^{2}.

Also ζ(Yζ)ζ(Yζ)Yζ2\zeta\cdot(Y\zeta)\le\bigl|\zeta\cdot(Y\zeta)\bigr|\le\lVert Y\rVert\lVert\zeta\rVert^{2}, by claim 3 of Properties of the Absolute Value in an Ordered Field and claim 2 of Properties of the Norm of a Symmetric Real Matrix. Adding the last two inequalities,

ξ(Xξ)η(Yη)  ση2+(σ1Y2+Y)ζ2,\xi\cdot(X\xi)-\eta\cdot(Y\eta)\ \le\ \sigma\lVert\eta\rVert^{2}+\bigl(\sigma^{-1}\lVert Y\rVert^{2}+\lVert Y\rVert\bigr)\lVert\zeta\rVert^{2},

using distributivity to collect the two multiples of ζ2\lVert\zeta\rVert^{2}.

Finally, claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of Bilinearity and Symmetry of the Dot Product on Rn\mathbb{R}^n and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §identity give η(Yση)=η(Yη)+ση2\eta\cdot(Y_{\sigma}\eta)=\eta\cdot(Y\eta)+\sigma\lVert\eta\rVert^{2}. Subtracting ση2\sigma\lVert\eta\rVert^{2} from both sides of the previous display and using σ1Y2+Y=(1+σ1Y)Y\sigma^{-1}\lVert Y\rVert^{2}+\lVert Y\rVert=(1+\sigma^{-1}\lVert Y\rVert)\lVert Y\rVert, again by distributivity, proves the claim of Step 1.

Step 2: a lower quadratic bound for the perturbed pair. For all ξ,ηRn\xi,\eta\in\mathbb{R}^{n},

Xξ2Yση2  ξ(Xξ)η(Yση).-\lVert X\rVert\lVert\xi\rVert^{2}-\lVert Y_{\sigma}\rVert\lVert\eta\rVert^{2}\ \le\ \xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta).

Indeed claim 2 of Properties of the Norm of a Symmetric Real Matrix gives ξ(Xξ)Xξ2|\xi\cdot(X\xi)|\le\lVert X\rVert\lVert\xi\rVert^{2}, so Xξ2ξ(Xξ)-\lVert X\rVert\lVert\xi\rVert^{2}\le\xi\cdot(X\xi) by claim 6 of Properties of the Absolute Value in an Ordered Field; the same claims give η(Yση)Yση2\eta\cdot(Y_{\sigma}\eta)\le\lVert Y_{\sigma}\rVert\lVert\eta\rVert^{2}, hence Yση2η(Yση)-\lVert Y_{\sigma}\rVert\lVert\eta\rVert^{2}\le-\eta\cdot(Y_{\sigma}\eta) by claim 4 of Elementary Order Arithmetic in an Ordered Field. Adding the two inequalities gives the display. Moreover

YσY+σ\lVert Y_{\sigma}\rVert\le\lVert Y\rVert+\sigma

by claim 5 of Properties of the Norm of a Symmetric Real Matrix together with σIn=σ=σ\lVert\sigma I_{n}\rVert=|\sigma|=\sigma, which is Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §identity and the definition of the absolute value for the positive σ\sigma.

Step 3: when the structure condition applies to the perturbed pair. Let αR\alpha\in\mathbb{R} be positive and suppose

Xα,Y+σα,(1+σ1Y)Yα.\lVert X\rVert\le\alpha,\qquad\lVert Y\rVert+\sigma\le\alpha,\qquad\bigl(1+\sigma^{-1}\lVert Y\rVert\bigr)\lVert Y\rVert\le\alpha .

Since α3α\alpha\le3\alpha, each of the three constants is at most 3α3\alpha. Multiplying the first by the nonnegative ξ2\lVert\xi\rVert^{2}, the second (together with YσY+σ\lVert Y_{\sigma}\rVert\le\lVert Y\rVert+\sigma and transitivity) by the nonnegative η2\lVert\eta\rVert^{2}, reversing signs by claim 4 of Elementary Order Arithmetic in an Ordered Field and adding, Step 2 gives

3α(ξ2+η2)  ξ(Xξ)η(Yση);-3\alpha\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)\ \le\ \xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta);

and multiplying the third by the nonnegative ξη2\lVert\xi-\eta\rVert^{2}, Step 1 gives

ξ(Xξ)η(Yση)  3αξη2,\xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta)\ \le\ 3\alpha\lVert\xi-\eta\rVert^{2},

both for all ξ,ηRn\xi,\eta\in\mathbb{R}^{n}. So the pair XX, YσY_{\sigma} satisfies, at this α\alpha, the hypothesis imposed on a pair of symmetric matrices by the structure condition.

Step 4: the contradiction. Suppose, contrary to what is to be proved, that F(x,r,p,Y)F(x,r,p,X)F(x,r,p,Y)\le F(x,r,p,X) fails. By the totality of \le, recorded in the definition of a total order, we then have F(x,r,p,X)F(x,r,p,Y)F(x,r,p,X)\le F(x,r,p,Y) and the two values differ, that is, F(x,r,p,X)<F(x,r,p,Y)F(x,r,p,X)<F(x,r,p,Y). Put

θ=F(x,r,p,Y)F(x,r,p,X),θ=θ21,\theta=F(x,r,p,Y)-F(x,r,p,X),\qquad\theta'=\theta\cdot2^{-1},

so that θ\theta is positive by claim 1 of Elementary Order Arithmetic in an Ordered Field, and θ\theta' is positive with θ+θ=θ\theta'+\theta'=\theta by claim 8 there.

Since FF is continuous at (x,r,p,Y)(x,r,p,Y) in the sense of Continuity of a Second-Order Equation Operator §at-point, there is a positive δR\delta\in\mathbb{R} such that every yΩy\in\Omega and ZS(n)Z\in\mathcal{S}(n) with dE(y,x)<δd_{E}(y,x)<\delta and dS(n)(Z,Y)<δd_{\mathcal{S}(n)}(Z,Y)<\delta satisfy

F(y,r,p,Z)F(x,r,p,Y)<θ\bigl|F(y,r,p,Z)-F(x,r,p,Y)\bigr|<\theta'

(the remaining two conditions of that definition hold trivially, since rr=0<δ|r-r|=0<\delta and pp=0<δ\lVert p-p\rVert=0<\delta). Since ω\omega is a modulus of continuity, condition 2 of Modulus of Continuity, applied with θ\theta', provides a positive δωR\delta_{\omega}\in\mathbb{R} such that every tTt\in T with tδωt\le\delta_{\omega} satisfies ω(t)θ\omega(t)\le\theta'. Since Ω\Omega is open and xΩx\in\Omega, there is a positive ρR\rho\in\mathbb{R} such that every zRnz\in\mathbb{R}^{n} with dE(z,x)<ρd_{E}(z,x)<\rho lies in Ω\Omega. Let μ\mu be the least of δ\delta and ρ\rho, positive by claim 9 of Elementary Order Arithmetic in an Ordered Field.

Now fix σ=δ21\sigma=\delta\cdot2^{-1}, which is positive and satisfies σ<δ\sigma<\delta by claim 8 of Elementary Order Arithmetic in an Ordered Field. Since YσY=σInY_{\sigma}-Y=\sigma I_{n}, Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §norm and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix §identity give dS(n)(Yσ,Y)=σIn=σ<δd_{\mathcal{S}(n)}(Y_{\sigma},Y)=\lVert\sigma I_{n}\rVert=\sigma<\delta.

Put S=p2+pS=\lVert p\rVert^{2}+\lVert p\rVert, a nonnegative number by claim 1 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. By claim 7 of Elementary Order Arithmetic in an Ordered Field the inverses μ1\mu^{-1} and δω1\delta_{\omega}^{-1} exist and are positive, so the five numbers

X,Y+σ,(1+σ1Y)Y,pμ1,Sδω1\lVert X\rVert,\quad\lVert Y\rVert+\sigma,\quad\bigl(1+\sigma^{-1}\lVert Y\rVert\bigr)\lVert Y\rVert,\quad\lVert p\rVert\mu^{-1},\quad S\delta_{\omega}^{-1}

are all nonnegative. By the totality of \le just cited, of any two real numbers one is at least the other; applying this four times along the list yields one of the five numbers, say MM, that is at least each of them. In particular 0M0\le M, since MM is itself one of the five. Put α=M+1\alpha=M+1. By claim 6 of Elementary Order Arithmetic in an Ordered Field and claim 1 there, M<αM<\alpha; combining this with 0M0\le M by claim 2 there gives 0<α0<\alpha, so α1\alpha^{-1} exists and is positive by claim 7 there.

Each of the five numbers is at most MM and M<αM<\alpha, so each is less than α\alpha by claim 2 of Elementary Order Arithmetic in an Ordered Field, and in particular at most α\alpha. For the first three this is exactly the hypothesis of Step 3, so the pair XX, YσY_{\sigma} satisfies the matrix hypothesis of the structure condition at this α\alpha. From pμ1<α\lVert p\rVert\mu^{-1}<\alpha, multiplying by the positive α1μ\alpha^{-1}\mu and simplifying gives α1p<μ\alpha^{-1}\lVert p\rVert<\mu; likewise from Sδω1<αS\delta_{\omega}^{-1}<\alpha, multiplying by the positive α1δω\alpha^{-1}\delta_{\omega} gives α1S<δω\alpha^{-1}S<\delta_{\omega}.

Put y=xα1py=x-\alpha^{-1}p, where α1p\alpha^{-1}p is the scalar multiple of pp by α1\alpha^{-1}. Then yxy-x is the scalar multiple of pp by α1-\alpha^{-1}, so claim 5 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n gives

dE(y,x)=yx=α1p=α1p<μ,d_{E}(y,x)=\lVert y-x\rVert=|-\alpha^{-1}|\,\lVert p\rVert=\alpha^{-1}\lVert p\rVert<\mu ,

using claim 2 of Properties of the Absolute Value in an Ordered Field and the definition of the absolute value. Since μρ\mu\le\rho this gives yΩy\in\Omega, and since μδ\mu\le\delta it gives dE(y,x)<δd_{E}(y,x)<\delta. Moreover xyx-y is the scalar multiple of pp by α1\alpha^{-1}, so α(xy)=p\alpha(x-y)=p, and

αxy2+xy=α(α1p)2+α1p=α1p2+α1p=α1S,\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert=\alpha\bigl(\alpha^{-1}\lVert p\rVert\bigr)^{2}+\alpha^{-1}\lVert p\rVert=\alpha^{-1}\lVert p\rVert^{2}+\alpha^{-1}\lVert p\rVert=\alpha^{-1}S,

by the commutativity and associativity of multiplication, the defining property of the multiplicative inverse and distributivity. This number lies in TT, being the product of two nonnegative numbers.

The structure condition, applied with these xx, yy, rr, α\alpha and the pair XX, YσY_{\sigma}, therefore gives

F(y,r,p,Yσ)F(x,r,p,X)  ω(α1S)  θ,F(y,r,p,Y_{\sigma})-F(x,r,p,X)\ \le\ \omega\bigl(\alpha^{-1}S\bigr)\ \le\ \theta',

the second inequality because α1S<δω\alpha^{-1}S<\delta_{\omega}, so that α1Sδω\alpha^{-1}S\le\delta_{\omega}.

On the other hand dE(y,x)<δd_{E}(y,x)<\delta and dS(n)(Yσ,Y)<δd_{\mathcal{S}(n)}(Y_{\sigma},Y)<\delta, so the continuity estimate applies with Z=YσZ=Y_{\sigma} and gives F(y,r,p,Yσ)F(x,r,p,Y)<θ|F(y,r,p,Y_{\sigma})-F(x,r,p,Y)|<\theta', whence F(x,r,p,Y)θ<F(y,r,p,Yσ)F(x,r,p,Y)-\theta'<F(y,r,p,Y_{\sigma}) by claim 9 of Properties of the Absolute Value in an Ordered Field and claim 1 of Elementary Order Arithmetic in an Ordered Field. Subtracting F(x,r,p,X)F(x,r,p,X) from both sides and using θ+θ=θ\theta'+\theta'=\theta,

θ=θθ<F(y,r,p,Yσ)F(x,r,p,X)  θ,\theta'=\theta-\theta'<F(y,r,p,Y_{\sigma})-F(x,r,p,X)\ \le\ \theta',

so θ<θ\theta'<\theta' by claim 2 of Elementary Order Arithmetic in an Ordered Field, which is impossible.

Therefore F(x,r,p,Y)F(x,r,p,X)F(x,r,p,Y)\le F(x,r,p,X). Since xΩx\in\Omega, rRr\in\mathbb{R}, pRnp\in\mathbb{R}^{n} and the pair XYX\preceq Y in S(n)\mathcal{S}(n) were arbitrary, FF is degenerate elliptic.

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