Proof of Approximability by Test Data Under Negation

Throughout, $U$, $u$, $x_{0}$, $p$ and $X$ are as in the statement, and for $f:U\to\mathbb{R}$ we write $-f$ for the function whose value at $z\in U$ is $-f(z)$.

Step 1 (Sign facts). We record the elementary facts used below. First, $|-1|=1$: by claim 2 of Properties of the Absolute Value in an Ordered Field we have $|-1|=|1|$, and by claim 1 of that lemma $|1|$ equals $1$ or $-1$ and satisfies $0\le|1|$, while $0<1$ by claim 6 of Elementary Order Arithmetic in an Ordered Field and hence $-1<0$ by claim 4 of that lemma; so $|1|=1$.

For a real number $t$, $-(-t)=t$, additive inverses being unique in a field; consequently $-(-f)=f$ for every $f:U\to\mathbb{R}$, the two functions having the same value at every point of $U$.

For $z,z'\in\mathbb{R}^{n}$ write $-z=(-1)z$. Since $\mathbb{R}^{n}$ is a real vector space by Euclidean Space $\mathbb{R}^n$ is a Real Vector Space, we have $(-1)\bigl((-1)z\bigr)=\bigl((-1)(-1)\bigr)z=1z=z$, so $-(-z)=z$; and $(-z)-(-z')=(-1)z-(-1)z'=(-1)(z-z')=-(z-z')$, the scalar multiple distributing over the difference. Hence, by 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,

$$\lVert(-z)-(-z')\rVert=\lVert-(z-z')\rVert=|-1|\,\lVert z-z'\rVert=\lVert z-z'\rVert .$$

For real matrices $A,B$ of the same size write $-A=(-1)A$. The scalar multiple and the difference of matrices are formed entrywise, so $\bigl(-(-A)\bigr)_{ij}=-(-A_{ij})=A_{ij}$ and $\bigl((-A)-(-B)\bigr)_{ij}=-A_{ij}-(-B_{ij})=-(A_{ij}-B_{ij})=\bigl(-(A-B)\bigr)_{ij}$ for all indices; since two real matrices of the same size are equal exactly when all their entries agree, as recorded in Euclidean Space and Real Matrices: Standing Notation (ii), this gives $-(-A)=A$ and $(-A)-(-B)=-(A-B)$. If $A,B\in\mathcal{S}(n)$ then $-A,-B\in\mathcal{S}(n)$ by Second-Order Equations on Euclidean Open Sets (iii), and claim 5 of Properties of the Norm of a Symmetric Real Matrix gives

$$d_{\mathcal{S}(n)}(-A,-B)=\lVert(-A)-(-B)\rVert=\lVert-(A-B)\rVert=|-1|\,\lVert A-B\rVert=d_{\mathcal{S}(n)}(A,B).$$

Finally, for $h,z\in\mathbb{R}^{n}$ and $A\in\mathcal{S}(n)$ we have $(-z)\cdot h=-(z\cdot h)$ by claim 4 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite, and $h\cdot\bigl((-A)h\bigr)=h\cdot\bigl(-(Ah)\bigr)=-\bigl(h\cdot(Ah)\bigr)$ by claim 1 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum followed by claim 5 of The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite.

Step 2 (From above for $u$ to below for $-u$). We show: if $\bigl(x_{0},u(x_{0}),p,X\bigr)$ is approximable by test data from above for $u$, then $\bigl(x_{0},(-u)(x_{0}),-p,-X\bigr)$ is approximable by test data from below for $-u$.

Let $\varepsilon\in\mathbb{R}$ be positive. By Quadratic Test Functions, Limits, Translation and Locality for Approximability by Test Data (i) there are $y\in U$, $r\in\mathbb{R}^{n}$, $Y\in\mathcal{S}(n)$ and a positive $\delta\in\mathbb{R}$ such that

$$d_{E}(y,x_{0})<\varepsilon,\qquad|u(y)-u(x_{0})|<\varepsilon,\qquad\lVert r-p\rVert<\varepsilon,\qquad d_{\mathcal{S}(n)}(Y,X)<\varepsilon$$

and

$$u(z)\le u(y)+r\cdot(z-y)+\tfrac{1}{2}(z-y)\cdot\bigl(Y(z-y)\bigr)\qquad\text{for every }z\in U\text{ with }d_{E}(z,y)<\delta .$$

We check that $y$, the vector $-r$, the matrix $-Y$ and the same $\delta$ witness the criterion of Quadratic Test Functions, Limits, Translation and Locality for Approximability by Test Data (i) for approximability from below of $\bigl(x_{0},(-u)(x_{0}),-p,-X\bigr)$ for $-u$. The first inequality is unchanged. For the second, claim 2 of Properties of the Absolute Value in an Ordered Field gives

$$\bigl|(-u)(y)-(-u)(x_{0})\bigr|=\bigl|-\bigl(u(y)-u(x_{0})\bigr)\bigr|=\bigl|u(y)-u(x_{0})\bigr|<\varepsilon .$$

The third and fourth read $\lVert(-r)-(-p)\rVert=\lVert r-p\rVert<\varepsilon$ and $d_{\mathcal{S}(n)}(-Y,-X)=d_{\mathcal{S}(n)}(Y,X)<\varepsilon$, by Step 1. Finally let $z\in U$ satisfy $d_{E}(z,y)<\delta$. Applying claim 4 of Elementary Order Arithmetic in an Ordered Field to the displayed inequality above reverses it after multiplication by $-1$, and by Step 1 together with the field arithmetic of $\mathbb{R}$,

$$-\Bigl(u(y)+r\cdot(z-y)+\tfrac{1}{2}(z-y)\cdot\bigl(Y(z-y)\bigr)\Bigr)=(-u)(y)+(-r)\cdot(z-y)+\tfrac{1}{2}(z-y)\cdot\bigl((-Y)(z-y)\bigr),$$

so that

$$(-u)(z)\ \ge\ (-u)(y)+(-r)\cdot(z-y)+\tfrac{1}{2}(z-y)\cdot\bigl((-Y)(z-y)\bigr),$$

which is the required inequality.

Step 3 (From below for $u$ to above for $-u$). Symmetrically, if $\bigl(x_{0},u(x_{0}),p,X\bigr)$ is approximable by test data from below for $u$, then $\bigl(x_{0},(-u)(x_{0}),-p,-X\bigr)$ is approximable by test data from above for $-u$. The argument is that of Step 2 with the criterion of Quadratic Test Functions, Limits, Translation and Locality for Approximability by Test Data (i) for approximability from below used in place of the one from above, and the two quadratic inequalities interchanged.

Step 4 (Conclusion). In claim 1, the implication from left to right is Step 2. Conversely, assume $\bigl(x_{0},(-u)(x_{0}),-p,-X\bigr)$ is approximable by test data from below for $-u$. Applying Step 3 with $-u$, $-p$ and $-X$ in place of $u$, $p$ and $X$ shows that $\Bigl(x_{0},\bigl(-(-u)\bigr)(x_{0}),-(-p),-(-X)\Bigr)$ is approximable by test data from above for $-(-u)$; by Step 1 this is exactly the assertion that $\bigl(x_{0},u(x_{0}),p,X\bigr)$ is approximable by test data from above for $u$.

Claim 2 follows in the same way, with Step 3 giving the implication from left to right and Step 2, applied to $-u$, $-p$ and $-X$, giving the converse.