Proof of A Strictly Proper Second-Order Equation Operator is Proper

Let $\gamma\in\mathbb{R}$ with $0<\gamma$ be such that $F$ is strictly proper with constant $\gamma$, and let $\mathcal{S}(n)$ be the set of symmetric real $n\times n$ matrices. We verify the two conditions of Proper Second-Order Equation Operator.

Condition 1 of that definition, degenerate ellipticity of $F$, is condition 1 of Strictly Proper Second-Order Equation Operator and therefore holds by hypothesis.

For condition 2, let $x\in U$, $p\in\mathbb{R}^n$, $X\in\mathcal{S}(n)$, and let $r,s\in\mathbb{R}$ satisfy $r\le s$. Applying condition 2 of Strictly Proper Second-Order Equation Operator with $s$ in the role of its first real argument and $r$ in the role of its second (which is legitimate because $r\le s$) gives

$$\gamma\,(s-r)\le F(x,s,p,X)-F(x,r,p,X).$$

Since $r\le s$, claim 3 (translation) of Elementary Arithmetic in an Ordered Field gives $0\le s-r$, and then claim 5 (multiplication by a nonnegative element) of that lemma, applied with the nonnegative element $\gamma$, gives $\gamma\cdot 0\le\gamma\,(s-r)$, that is, $0\le\gamma\,(s-r)$. By transitivity of $\le$ in the ordered field,

$$0\le F(x,s,p,X)-F(x,r,p,X),$$

and claim 3 (translation) of Elementary Arithmetic in an Ordered Field, read in the other direction, yields

$$F(x,r,p,X)\le F(x,s,p,X).$$

As $x$, $p$, $X$ and the pair $r\le s$ were arbitrary, condition 2 of Proper Second-Order Equation Operator holds, and $F$ is proper.