Proof of The Gradient of a $C^2$ Function on Euclidean Space is One-Sided Lipschitz on Bounded Sets

Each result cited is universally quantified over the data in its own statement.

Throughout, $\mathbb{R}^{n}$ carries the Euclidean distance $d_{E}$, a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$, with $d_{E}(x,y)=\lVert x-y\rVert$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$; by Second-Order Equations on Euclidean Open Sets §space, which imports it from Real Matrices, Symmetric Matrices and the Semidefinite Ordering: Standing Notation §numbers, boundedness of $B$ means boundedness in $(\mathbb{R}^{n},d_{E})$. By Second-Order Equations on Euclidean Open Sets §test-functions and Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §derivatives, $\partial_{i}V$ is the partial derivative, also written $\frac{\partial V}{\partial x_{i}}$, $\partial_{j}\partial_{i}V$ is the iterated partial derivative of clause 4 of C^k Maps on a Euclidean Open Set, and $DV(z)$ is the gradient, the point of $\mathbb{R}^{n}$ with coordinates $\partial_{1}V(z),\dots,\partial_{n}V(z)$.

Step 1 (a compact convex set containing $B$). By Bounded Subset of a Metric Space there are a point $p\in\mathbb{R}^{n}$ and a real number $R>0$ with $d_{E}(p,z)\le R$ for every $z\in B$. Let $K=\bar{B}_{d_{E}}(p,R)$ be the closed ball $\{z\in\mathbb{R}^{n}:d_{E}(p,z)\le R\}$. Then $B\subseteq K$, and $K$ is nonempty because $d_{E}(p,p)=0\le R$ by the axioms of a metric. By claim 1 of A Closed Euclidean Ball is Convex and Compact the set $K$ is convex, and by claim 2 there it is compact in $\mathbb{R}^{n}$ with the metric topology of $d_{E}$.

Step 2 (a uniform bound on the second partial derivatives over $K$). Fix $i,j\in\{1,\dots,n\}$. Since $V$ is of class $C^{2}$ on the open set $\mathbb{R}^{n}$ (open by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), clause 2 of C^k Maps on a Euclidean Open Set, read through the scalar convention of clause 3 there, shows that $\partial_{i}V:\mathbb{R}^{n}\to\mathbb{R}$ is of class $C^{1}$ on $\mathbb{R}^{n}$; by clause 1 there, applied to $\partial_{i}V$, the partial derivative of $\partial_{i}V$ with respect to the $j$th variable exists at every point, and the resulting function, which is $g_{ij}=\partial_{j}\partial_{i}V$ by clause 4 there, is continuous in the Euclidean sense at every point of $\mathbb{R}^{n}$. By claim 1 of Euclidean Continuity Agrees with Metric Continuity for Real-Valued Functions (with $E=\mathbb{R}^{n}$), $g_{ij}$ is continuous at every point relative to $\mathbb{R}^{n}$ as a map into $\mathbb{R}$ with the metric $d_{\mathbb{R}}(s,t)=|s-t|$ of The Absolute Value Metric on the Real Line. Hence the restriction of $g_{ij}$ to $K$ has the continuity property required in Extreme Value Theorem on a Compact Subset of a Metric Space: given $z\in K$ and $\varepsilon>0$, the $\delta>0$ supplied at $z$ by continuity relative to $\mathbb{R}^{n}$ works for all points of $K\subseteq\mathbb{R}^{n}$ within distance $\delta$ of $z$. As $K$ is nonempty and compact by Step 1, Extreme Value Theorem on a Compact Subset of a Metric Space gives $a_{ij},b_{ij}\in K$ with $g_{ij}(a_{ij})\le g_{ij}(z)\le g_{ij}(b_{ij})$ for all $z\in K$. Put

which is nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field and claim 2 of Elementary Arithmetic in an Ordered Field. For $z\in K$, claim 3 of Properties of the Absolute Value in an Ordered Field and the nonnegativity of the absolute values give $g_{ij}(z)\le g_{ij}(b_{ij})\le|g_{ij}(b_{ij})|\le M_{ij}$ and, using sign reversal (claim 4 of Elementary Order Arithmetic in an Ordered Field), $-M_{ij}\le-|g_{ij}(a_{ij})|\le g_{ij}(a_{ij})\le g_{ij}(z)$; so $|g_{ij}(z)|\le M_{ij}$ by claim 6 of Properties of the Absolute Value in an Ordered Field. Now let

Since all $M_{ij}$ are nonnegative, $M$ is nonnegative and $M-M_{ij}$, a finite sum of the remaining nonnegative terms, is nonnegative (claim 2 of Elementary Arithmetic in an Ordered Field), so $M_{ij}\le M$ by claim 3 there. Consequently

Step 3 (two first-order Taylor estimates). Let $x,y\in B$; then $x,y\in K$ by Step 1. Apply claim (ii) of Multivariate Taylor Expansion with Uniform Second-Order Remainder with $W=\mathbb{R}^{n}$ (open by claim 1 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), $f=V$ (of class $C^{2}$, hence of class $C^{1}$ by claim 2 of Euclidean Space is Open in Itself, and $C^k$ Maps are Continuous), $M_{2}=M$, and with the lemma's pair of points taken to be $(y,x)$, so that the lemma's increment is $x-y$, whose $i$th coordinate is $x_{i}-y_{i}$ by clause 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$. The segment consists of the points $y+\tau(x-y)$ with $\tau\in[0,1]$; it lies in $W=\mathbb{R}^{n}$, and each such point equals $\tau x+(1-\tau)y$ (coordinatewise real arithmetic, the operations being those of the vector space $\mathbb{R}^{n}$), which lies in $K$ by the convexity of $K$ (Step 1, Convex Subset of $\mathbb{R}^n$ with $t=\tau$). So the bound of Step 2 holds on the segment. The lemma's $|h|$ is $d_{E}(y,x)=d_{E}(x,y)=\lVert x-y\rVert$ by symmetry of the metric (Euclidean Distance is a Metric on $\mathbb{R}^n$) and claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. Writing $\rho=\lVert x-y\rVert\,\lVert x-y\rVert$, we obtain

Applying the same claim with the pair $(x,y)$ instead (increment $y-x$ with coordinates $y_{i}-x_{i}$; segment points $x+\tau(y-x)=\tau y+(1-\tau)x\in K$ by convexity; $|h|=d_{E}(x,y)=\lVert x-y\rVert$), we obtain

Step 4 (adding the estimates). In $P+Q$ the terms $V(x)-V(y)$ and $V(y)-V(x)$ cancel, and since $-(y_{i}-x_{i})=x_{i}-y_{i}$, field arithmetic (distributivity and rearrangement of finite sums) gives

By the description of the gradient recalled at the start and clause 1 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, the $i$th coordinate of $DV(x)-DV(y)$ is $\partial_{i}V(x)-\partial_{i}V(y)$ and that of $x-y$ is $x_{i}-y_{i}$, so clause 2 there identifies the right-hand side with the dot product:

By the triangle inequality (claim 5 of Properties of the Absolute Value in an Ordered Field) and the two estimates of Step 3, added using claims 2 and 3 of Elementary Arithmetic in an Ordered Field (from $0\le b-a$ and $0\le d-c$ follows $0\le(b+d)-(a+c)$, so weak inequalities add), and $\tfrac12 s+\tfrac12 s=s$ for real $s$ (with $\tfrac12=2^{-1}$, which exists by claim 8 of Elementary Order Arithmetic in an Ordered Field),

Set $c=nM$. It is nonnegative: $0\le n$ as $1\le n$ and $0\le1$ (claim 1 of Elementary Arithmetic in an Ordered Field), and $0\le M$, so claim 5 there gives $0=M\cdot0\le M\,n$. Finally, claim 3 of Properties of the Absolute Value in an Ordered Field gives $-|P+Q|\le P+Q$, and sign reversal (claim 4 of Elementary Order Arithmetic in an Ordered Field) applied to $|P+Q|\le c\rho$ gives $-c\rho\le-|P+Q|$. Hence

Since $c$ depends only on $V$ and $B$ (through $K$ and $M$) and not on $x,y$, and $x,y\in B$ were arbitrary, this proves the lemma.