Proof of The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane

Each result cited is universally quantified over the data in its own statement. Continuity of a real-valued function on a subset $A\subseteq\mathbb{R}^{2}$ is that of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation (iv): continuity at each point of $A$ relative to $A$ in the sense of Continuous Map Between Metric Spaces, from $(\mathbb{R}^{2},d_{E})$ to $(\mathbb{R},d_{\mathbb{R}})$. For $(x,y),(a,b)\in\mathbb{R}^{2}$, $d_{E}((x,y),(a,b))=\lVert(x-a,y-b)\rVert$ (claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, the difference of points of $\mathbb{R}^{2}$ being coordinatewise by clause 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$), so $|x-a|\le d_{E}((x,y),(a,b))$ and $|y-b|\le d_{E}((x,y),(a,b))$ by claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. By The Increment Form of the Derivative, and Differentiability Implies Continuity, $\phi$ is continuous at every point of $\mathbb{R}$, as a map from $(\mathbb{R},d_{\mathbb{R}})$ to itself. Elementary identities of the absolute value are taken from Properties of the Absolute Value in an Ordered Field (claim 2 for $|s-t|=|t-s|$, claim 5 for the triangle inequality, used twice for three summands) and of the field from Zero Products and Elementary Identities in a Field (claim 2: $(-s)t=-(st)$, in particular $-t=(-1)t$).

Step 1: the mean value representation. Let $x,y\in\mathbb{R}$ with $x\ne y$, and let $u<v$ be the two numbers $x,y$ in increasing order. By Basic Facts about Intervals of the Real Line (v), $[u,v]$ is an interval of which every point $w$ with $u<w<v$ is an interior point. The restriction $\phi|_{[u,v]}$ is continuous on $[u,v]$ by claim 1 of Restriction Stability of Continuity and of the Derivative, and differentiable at every point of $(u,v)$ with derivative $\phi'$ there by claim 2 of the same lemma. By Mean Value Theorem on a Closed Real Interval and on Any Interval there is $c$ with $u<c<v$ and

$$\phi'(c)=\frac{\phi(v)-\phi(u)}{v-u}.$$

Now $\frac{\phi(u)-\phi(v)}{u-v}=\frac{\phi(v)-\phi(u)}{v-u}$: indeed $\phi(u)-\phi(v)=-(\phi(v)-\phi(u))$ and $u-v=-(v-u)$ by claim 6 of Additive Cancellation and Elementary Additive Identities in a Field, and for real numbers $p$ and $w\ne0$ one has $(-p)(-w)^{-1}=p\,w^{-1}$: multiplying $p\,w^{-1}$ by $(-w)(-w)^{-1}=1$ and using claim 2 of Zero Products and Elementary Identities in a Field twice, $p\,w^{-1}=p\,w^{-1}(-w)(-w)^{-1}=-(p\,w^{-1}w)(-w)^{-1}=(-p)(-w)^{-1}$. Therefore $F(x,y)=F(y,x)=\phi'(c)$, with $c$ strictly between $x$ and $y$, that is, $u<c<v$.

Step 2: claim 1. Symmetry for $x\ne y$ is contained in Step 1, and is trivial for $x=y$. For the bound: if $x=y$ then $|F(x,y)|=|\phi'(x)|\le L$; if $x\ne y$ then $|F(x,y)|=|\phi'(c)|\le L$ with $c$ as in Step 1.

Step 3: continuity at a diagonal point $(a,a)$. Let $\varepsilon>0$. Since $\phi'$ is continuous at $a$, there is $\eta>0$ with $|\phi'(c)-\phi'(a)|<\varepsilon$ whenever $|c-a|<\eta$. Let $(x,y)\in\mathbb{R}^{2}$ with $d_{E}((x,y),(a,a))<\eta$; then $|x-a|<\eta$ and $|y-a|<\eta$. If $x=y$, then $F(x,y)-F(a,a)=\phi'(x)-\phi'(a)$, which has absolute value less than $\varepsilon$. If $x\ne y$, then $F(x,y)=\phi'(c)$ with $u<c<v$, where $u<v$ are $x,y$ in increasing order (Step 1); by claim 9 of Properties of the Absolute Value in an Ordered Field and claim 1 of Elementary Order Arithmetic in an Ordered Field, $|x-a|<\eta$ and $|y-a|<\eta$ mean $a-\eta<x<a+\eta$ and $a-\eta<y<a+\eta$, so $a-\eta<u<c<v<a+\eta$ and thus $a-\eta<c<a+\eta$ by transitivity of the strict order, i.e. $|c-a|<\eta$ by the same two claims, and $|F(x,y)-F(a,a)|=|\phi'(c)-\phi'(a)|<\varepsilon$. Hence $F$ is continuous at $(a,a)$ relative to $\mathbb{R}^{2}$.

Step 4: continuity at a point $(a,b)$ with $a\ne b$. Let $D=\{(x,y)\in\mathbb{R}^{2}:x\ne y\}$.

(a) $F|_{D}$ is continuous on $D$ relative to $D$. The coordinate projections $\mathrm{pr}_{1},\mathrm{pr}_{2}:\mathbb{R}^{2}\to\mathbb{R}$, $\mathrm{pr}_{1}(x,y)=x$ and $\mathrm{pr}_{2}(x,y)=y$, are continuous on $\mathbb{R}^{2}$ (given $\varepsilon$, take $\delta=\varepsilon$, by the coordinate bound recalled above). By claim 3 of Semicontinuity and Continuity Under Composition with a Continuous Map, $\phi\circ\mathrm{pr}_{1}$ and $\phi\circ\mathrm{pr}_{2}$ are continuous on $\mathbb{R}^{2}$, so by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space the functions $N(x,y)=\phi(x)-\phi(y)=\phi(\mathrm{pr}_{1}(x,y))+(-1)\phi(\mathrm{pr}_{2}(x,y))$ and $Q(x,y)=x-y=\mathrm{pr}_{1}(x,y)+(-1)\mathrm{pr}_{2}(x,y)$ are continuous on $\mathbb{R}^{2}$, hence their restrictions to $D$ are continuous on $D$ relative to $D$ by claim 1 of Restriction Stability of Continuity and of the Derivative. The restriction $Q|_{D}$ vanishes nowhere on $D$, so by claim 2 of Continuity of the Reciprocal of a Nonvanishing Real-Valued Function on a Metric Space the function $(x,y)\mapsto(x-y)^{-1}$ is continuous on $D$ relative to $D$, and by claim 5 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space so is the product $(x,y)\mapsto(\phi(x)-\phi(y))(x-y)^{-1}$, which is $F|_{D}$ by the definition of $F$ on $D$.

(b) Every point of $D$ has a ball around it inside $D$, and continuity relative to $D$ upgrades to continuity relative to $\mathbb{R}^{2}$. Let $(a,b)\in D$ and put $\rho=\tfrac{|a-b|}{2}$, positive by claim 1 of Properties of the Absolute Value in an Ordered Field and claim 8 of Elementary Order Arithmetic in an Ordered Field. If $d_{E}((x,y),(a,b))<\rho$, then $|x-a|<\rho$ and $|y-b|<\rho$, so by the triangle inequality applied to $a-b=(a-x)+(x-y)+(y-b)$, $|a-b|\le|x-a|+|x-y|+|y-b|<|x-y|+|a-b|$ (claim 3 of Elementary Order Arithmetic in an Ordered Field and $\rho+\rho=|a-b|$ by claim 8 there), whence $0<|x-y|$ by claim 1 there, so $x\ne y$ by claim 1 of Properties of the Absolute Value in an Ordered Field, and $(x,y)\in D$. Now let $\varepsilon>0$. By (a) there is $\delta_{0}>0$ such that every $(x,y)\in D$ with $d_{E}((x,y),(a,b))<\delta_{0}$ satisfies $|F(x,y)-F(a,b)|<\varepsilon$. Let $\delta$ be the least of $\delta_{0}$ and $\rho$. Every $(x,y)\in\mathbb{R}^{2}$ with $d_{E}((x,y),(a,b))<\delta$ lies in $D$ and satisfies $|F(x,y)-F(a,b)|<\varepsilon$. Hence $F$ is continuous at $(a,b)$ relative to $\mathbb{R}^{2}$.

Step 5: conclusion. Steps 3 and 4 show that $F$ is continuous at every point of $\mathbb{R}^{2}$ relative to $\mathbb{R}^{2}$, that is, continuous on $\mathbb{R}^{2}$; this is claim 2. Since $d_{\mathbb{R}}$ is the Euclidean distance on $\mathbb{R}=\mathbb{R}^{1}$ by $\mathbb{R}^1$ with the Euclidean Distance Is Isometric to the Real Line, $F$ is continuous from $(\mathbb{R}^{2},d_{E})$ to $\mathbb{R}$ with its Euclidean distance, so by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, $F$ is measurable with respect to the $\sigma$-algebra $\mathcal{B}_{2}$ there and $\mathcal{B}(\mathbb{R})$, and $\mathcal{B}_{2}=\mathcal{B}(\mathbb{R}^{2})$ by claim 5 of the same lemma. This proves claim 3.