Proof of The Structure Condition Implies Degenerate Ellipticity

Conventions. The order $\le$ and the arithmetic of $\mathbb{R}$ 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 $a\le b$ and $0\le\lambda$, then either $a=b$, and the products are equal, or $a<b$, and then $\lambda a\le\lambda b$ by claim 10 of Elementary Order Arithmetic in an Ordered Field when $0<\lambda$, while $\lambda=0$ makes both products $0$. Second, two inequalities may be added: if $a\le b$ and $c\le d$ then $a+c\le b+c\le b+d$ by the compatibility of $\le$ with addition and transitivity. Third, if $a<b$ and $0<\lambda$, then $\lambda a<\lambda b$ by claim 10 of Elementary Order Arithmetic in an Ordered Field. We write $3=1+1+1$, so that $1\le3$ and hence $\alpha\le3\alpha$ for every nonnegative $\alpha$. Of the properties of $\Omega$ 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 $x\in\Omega$, $r\in\mathbb{R}$, $p\in\mathbb{R}^{n}$ and $X,Y\in\mathcal{S}(n)$ with $X\preceq Y$. We must show that $F(x,r,p,Y)\le F(x,r,p,X)$.

Throughout, $\sigma\in\mathbb{R}$ denotes a positive number, and we put $Y_{\sigma}=Y+\sigma I_{n}$, which lies in $\mathcal{S}(n)$ by Euclidean Space and Real Matrices: Standing Notation (iii). By claim 7 of Elementary Order Arithmetic in an Ordered Field the inverse $\sigma^{-1}$ exists and is positive.

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

$$\xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta)\ \le\ \bigl(1+\sigma^{-1}\lVert Y\rVert\bigr)\lVert Y\rVert\,\lVert\xi-\eta\rVert^{2}.$$

Put $\zeta=\xi-\eta$, so that $\xi=\eta+\zeta$. By claim 3 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum and claims 2 and 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite,

$$\xi\cdot(Y\xi)=\eta\cdot(Y\eta)+\eta\cdot(Y\zeta)+\zeta\cdot(Y\eta)+\zeta\cdot(Y\zeta),$$

and $\zeta\cdot(Y\eta)=\eta\cdot(Y\zeta)$: indeed claim 5 of Elementary Properties of the Transpose of a Real Matrix gives $\zeta\cdot(Y\eta)=(Y^{\top}\zeta)\cdot\eta$, while $Y^{\top}=Y$ by Euclidean Space and Real Matrices: Standing Notation (iii) and the dot product is symmetric by claim 1 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite. Since $X\preceq Y$ gives $\xi\cdot(X\xi)\le\xi\cdot(Y\xi)$ by Euclidean Space and Real Matrices: Standing Notation (iv), we obtain

$$\xi\cdot(X\xi)-\eta\cdot(Y\eta)\ \le\ 2\bigl(\eta\cdot(Y\zeta)\bigr)+\zeta\cdot(Y\zeta),$$

where $2u$ abbreviates $u+u$.

By A Weighted Young Inequality and the Splitting of a Quadratic Form (i), applied with $a=\eta$, $b=Y\zeta$ and $t=\sigma$,

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

By Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (iii) we have $\lVert Y\zeta\rVert\le\lVert Y\rVert\lVert\zeta\rVert$; both sides are nonnegative, by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square 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 $\lVert Y\zeta\rVert^{2}\le\lVert Y\rVert^{2}\lVert\zeta\rVert^{2}$, where $\lVert Y\rVert^{2}=\lVert Y\rVert\lVert Y\rVert$. Multiplying by the nonnegative $\sigma^{-1}$,

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

Also $\zeta\cdot(Y\zeta)\le\bigl|\zeta\cdot(Y\zeta)\bigr|\le\lVert Y\rVert\lVert\zeta\rVert^{2}$, 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,

$$\xi\cdot(X\xi)-\eta\cdot(Y\eta)\ \le\ \sigma\lVert\eta\rVert^{2}+\bigl(\sigma^{-1}\lVert Y\rVert^{2}+\lVert Y\rVert\bigr)\lVert\zeta\rVert^{2},$$

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

Finally, claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum, claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi) give $\eta\cdot(Y_{\sigma}\eta)=\eta\cdot(Y\eta)+\sigma\lVert\eta\rVert^{2}$. Subtracting $\sigma\lVert\eta\rVert^{2}$ from both sides of the previous display and using $\sigma^{-1}\lVert Y\rVert^{2}+\lVert Y\rVert=(1+\sigma^{-1}\lVert Y\rVert)\lVert Y\rVert$, again by distributivity, proves the claim of Step 1.

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

$$-\lVert X\rVert\lVert\xi\rVert^{2}-\lVert Y_{\sigma}\rVert\lVert\eta\rVert^{2}\ \le\ \xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta).$$

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

$$\lVert Y_{\sigma}\rVert\le\lVert Y\rVert+\sigma$$

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

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

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

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

$$-3\alpha\bigl(\lVert\xi\rVert^{2}+\lVert\eta\rVert^{2}\bigr)\ \le\ \xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta);$$

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

$$\xi\cdot(X\xi)-\eta\cdot(Y_{\sigma}\eta)\ \le\ 3\alpha\lVert\xi-\eta\rVert^{2},$$

both for all $\xi,\eta\in\mathbb{R}^{n}$. So the pair $X$, $Y_{\sigma}$ satisfies, at this $\alpha$, 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 $F(x,r,p,Y)\le F(x,r,p,X)$ fails. By the totality of $\le$, recorded in the definition of a total order, we then have $F(x,r,p,X)\le F(x,r,p,Y)$ and the two values differ, that is, $F(x,r,p,X)<F(x,r,p,Y)$. Put

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

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

Since $F$ is continuous at $(x,r,p,Y)$ in the sense of Continuity of a Second-Order Equation Operator (i), there is a positive $\delta\in\mathbb{R}$ such that every $y\in\Omega$ and $Z\in\mathcal{S}(n)$ with $d_{E}(y,x)<\delta$ and $d_{\mathcal{S}(n)}(Z,Y)<\delta$ satisfy

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

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

Now fix $\sigma=\delta\cdot2^{-1}$, which is positive and satisfies $\sigma<\delta$ by claim 8 of Elementary Order Arithmetic in an Ordered Field. Since $Y_{\sigma}-Y=\sigma I_{n}$, Euclidean Space and Real Matrices: Standing Notation (v) and Vector, Entry and Comparison Bounds for the Norm of a Symmetric Real Matrix (vi) give $d_{\mathcal{S}(n)}(Y_{\sigma},Y)=\lVert\sigma I_{n}\rVert=\sigma<\delta$.

Put $S=\lVert p\rVert^{2}+\lVert p\rVert$, a nonnegative number by claim 1 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square 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 $\mu^{-1}$ and $\delta_{\omega}^{-1}$ exist and are positive, so the five numbers

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

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 $M$, that is at least each of them. In particular $0\le M$, since $M$ is itself one of the five. Put $\alpha=M+1$. By claim 6 of Elementary Order Arithmetic in an Ordered Field and claim 1 there, $M<\alpha$; combining this with $0\le M$ by claim 2 there gives $0<\alpha$, so $\alpha^{-1}$ exists and is positive by claim 7 there.

Each of the five numbers is at most $M$ and $M<\alpha$, so each is less than $\alpha$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, and in particular at most $\alpha$. For the first three this is exactly the hypothesis of Step 3, so the pair $X$, $Y_{\sigma}$ satisfies the matrix hypothesis of the structure condition at this $\alpha$. From $\lVert p\rVert\mu^{-1}<\alpha$, multiplying by the positive $\alpha^{-1}\mu$ and simplifying gives $\alpha^{-1}\lVert p\rVert<\mu$; likewise from $S\delta_{\omega}^{-1}<\alpha$, multiplying by the positive $\alpha^{-1}\delta_{\omega}$ gives $\alpha^{-1}S<\delta_{\omega}$.

Put $y=x-\alpha^{-1}p$, where $\alpha^{-1}p$ is the scalar multiple of $p$ by $\alpha^{-1}$. Then $y-x$ is the scalar multiple of $p$ by $-\alpha^{-1}$, so claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square gives

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

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

$$\alpha\lVert x-y\rVert^{2}+\lVert x-y\rVert=\alpha\bigl(\alpha^{-1}\lVert p\rVert\bigr)^{2}+\alpha^{-1}\lVert p\rVert=\alpha^{-1}\lVert p\rVert^{2}+\alpha^{-1}\lVert p\rVert=\alpha^{-1}S,$$

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

The structure condition, applied with these $x$, $y$, $r$, $\alpha$ and the pair $X$, $Y_{\sigma}$, therefore gives

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

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

On the other hand $d_{E}(y,x)<\delta$ and $d_{\mathcal{S}(n)}(Y_{\sigma},Y)<\delta$, so the continuity estimate applies with $Z=Y_{\sigma}$ and gives $|F(y,r,p,Y_{\sigma})-F(x,r,p,Y)|<\theta'$, whence $F(x,r,p,Y)-\theta'<F(y,r,p,Y_{\sigma})$ 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 $F(x,r,p,X)$ from both sides and using $\theta'+\theta'=\theta$,

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

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

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