Proof of The Almost-Everywhere Gradient of a Lipschitz Semiconvex or Semiconcave Function Is Tangent at Every Absolutely Continuous Measure

Each result cited below is universally quantified over the data in its own statement. Elementary order and arithmetic in $\mathbb{R}$, including rules for finite sums and for squares of nonnegative numbers, are provided by The Real Numbers: Standing Notation and Background §background and are not cited individually. Steps 1 to 4 use only that $w$ is bounded and Lipschitz with constant $L$.

Step 0 (Conventions). By Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §distance the Euclidean distance is $d_{E}(x,y)=\lVert x-y\rVert$, and the metric of The Absolute Value Metric on the Real Line is $|s-t|$; so by Lipschitz Map Between Metric Spaces

$$|w(x)-w(y)|\le L\,\lVert x-y\rVert\qquad\text{for all }x,y\in\mathbb{R}^{d}.\tag{0.1}$$

The set $\mathbb{R}^{d}$ is open (every open ball about one of its points lies in it) and convex (it contains every point $tx+(1-t)y$). For $x\in E_{w}$ let $J$ be a derivative matrix of $w$ at $x$; by claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique its entries are $J_{1i}=\partial_{i}w(x)$, so by the definitions of the matrix-vector product and of the dot product $Jh$ has the single coordinate $Dw(x)\cdot h$, and the Euclidean norm of a point of $\mathbb{R}^{1}$ is the absolute value of its coordinate by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §square. Thus Differentiability at a Point for Maps Between Euclidean Spaces (with $U=\mathbb{R}^{d}$) gives: for $x\in E_{w}$ and every positive $\varepsilon$ there is a positive $\delta$ with

$$\bigl|w(x+h)-w(x)-Dw(x)\cdot h\bigr|\le\varepsilon\lVert h\rVert\qquad\text{whenever }0<\lVert h\rVert<\delta .\tag{0.2}$$

Step 1 (Continuity). Let $x\in\mathbb{R}^{d}$, let $\varepsilon$ be positive and put $\delta=\tfrac{\varepsilon}{L+1}$. By (0.1), every $y\in\mathbb{R}^{d}$ with $\lVert x-y\rVert<\delta$ satisfies $|w(y)-w(x)|\le L\delta=\tfrac{L\varepsilon}{L+1}<\varepsilon$. So $w$ is continuous on $\mathbb{R}^{d}$, in the sense fixed in Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §extrema.

Step 2 (The gradient bound). Let $x\in E_{w}$ and $v=Dw(x)$. If $v=0_{\mathbb{R}^{d}}$, then $\lVert v\rVert=0\le L$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §vanishing. Otherwise $0<\lVert v\rVert$. Let $\varepsilon$ be positive, let $\delta$ be as in (0.2), and put $t=\tfrac{\delta}{2\lVert v\rVert}$ and $h=tv$. Then $\lVert h\rVert=t\lVert v\rVert=\tfrac{\delta}{2}$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §homogeneity, and $v\cdot h=t\lVert v\rVert^{2}$ by Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ and Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §square. By (0.2) and (0.1),

$$t\lVert v\rVert^{2}=v\cdot h\le w(x+h)-w(x)+\varepsilon\lVert h\rVert\le(L+\varepsilon)\,t\lVert v\rVert ,$$

and dividing by the positive number $t\lVert v\rVert$ gives $\lVert v\rVert\le L+\varepsilon$. As $\varepsilon$ was arbitrary, $\lVert Dw(x)\rVert\le L$. Since $\nabla w$ equals $Dw$ on $E_{w}$ and the origin, of norm $0$, off $E_{w}$,

$$\lVert\nabla w(x)\rVert\le L\qquad\text{for every }x\in\mathbb{R}^{d}.\tag{2.1}$$

Step 3 (Claim 1). We apply Integrating a Semiconvex Viscosity Subsolution of the Penalty-Drift Equation against a Measure of Finite Relative Fisher Information §borel with the following data from its Data paragraph: the open convex set $\mathbb{R}^{d}$ (Step 0) in place of its $D$; $U$ the zero function on $\mathbb{R}^{d}$, which is smooth by claim 2 of Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set and hence of class $C^{2}$ by Smooth Map on a Euclidean Open Set; $\lambda=\kappa=1$ and $\theta'=0$; $\tilde{g}$ the zero function on $\mathbb{R}^{d}$, which is bounded, its extension $\bar{g}$ being the zero function on $\mathbb{R}^{d}$, which is continuous (being constant) and hence Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps; the bounded function $w$; and $A=L^{2}$, $B=0$, $p_{0}=0$. With these data the set $E_{w}$ and the map $\nabla w$ of that lemma are those of the present statement. Its gradient hypothesis holds: for $x\in E_{w}$, $0\le\lVert Dw(x)\rVert\le L$ by (2.1), so $\lVert Dw(x)\rVert^{2}\le L^{2}=A+B\bigl(U(x)-p_{0}\bigr)$. As $w$ is continuous on $\mathbb{R}^{d}$ (Step 1), the clause gives $E_{w}\in\mathcal{B}(\mathbb{R}^{d})$ and that $\nabla w$ is Borel. Together with (2.1) this gives the measurability statements and the bound of claim 1. The function $x\mapsto\lVert\nabla w(x)\rVert^{2}$ is Borel, as the composition of $\nabla w$ with the Borel map $z\mapsto\lVert z\rVert^{2}$ (Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pairs and Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps), and it is bounded by $L^{2}$ by (2.1). Let $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$. The function $x\mapsto\lVert\nabla w(x)\rVert^{2}$ is integrable with respect to $\mu$ by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §measures, that is $\int_{\mathbb{R}^{d}}\lVert\nabla w\rVert^{2}\,d\mu<\infty$, and the class of $\nabla w$ belongs to $L^{2}(\mu;\mathbb{R}^{d})$ by Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu, for every $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$.

Step 4 (Claim 1: the non-differentiability set is null for absolutely continuous measures). Let $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ be absolutely continuous. By Rademacher's Theorem in $\mathbb{R}^n$ §ae (with $n=d$ and $f=w$, Lipschitz with constant $L$ by (0.1)) there is $N\in\mathcal{B}(\mathbb{R}^{d})$ with $\lambda_{d}(N)=0$ such that every $x\in\mathbb{R}^{d}\setminus N$ is a point as in that clause, at which $w$ is differentiable by Rademacher's Theorem in $\mathbb{R}^n$ §derivative. Hence $\mathbb{R}^{d}\setminus E_{w}\subseteq N$. The set $\mathbb{R}^{d}\setminus E_{w}$ belongs to $\mathcal{B}(\mathbb{R}^{d})$, a $\sigma$-algebra, by Step 3. Since $\mu$ is absolutely continuous and $N$ is a Borel set with $\lambda_{d}(N)=0$, $\mu(N)=0$; by claim 2 of Basic Properties of a Measure, $\mu(\mathbb{R}^{d}\setminus E_{w})\le\mu(N)=0$, and by claim 3 there, as $\mu(\mathbb{R}^{d})=1$,

$$\mu(\mathbb{R}^{d}\setminus E_{w})=0,\qquad\mu(E_{w})=1 .\tag{4.1}$$

As $\mu$ was an arbitrary absolutely continuous member of $\mathcal{P}_{2}(\mathbb{R}^{d})$, this completes the proof of claim 1.

Step 5 (Claim 2). Suppose that $w$ is semiconvex with constant $K$, fix an absolutely continuous $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$, so that (4.1) holds for $\mu$ by claim 1 (Step 4), and let $\phi:\mathbb{R}^{d}\to\mathbb{R}$, $\phi(x)=w(x)+\tfrac{K}{2}\lVert x\rVert^{2}$, which is convex on $\mathbb{R}^{d}$ by Semiconvex Function on a Convex Subset of $\mathbb{R}^n$. Let $x\in E_{w}$. For $h\in\mathbb{R}^{d}$, Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ §square and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ give $\lVert x+h\rVert^{2}=\lVert x\rVert^{2}+2\,x\cdot h+\lVert h\rVert^{2}$, hence

$$\phi(x+h)-\phi(x)-\bigl(Dw(x)+Kx\bigr)\cdot h=\bigl(w(x+h)-w(x)-Dw(x)\cdot h\bigr)+\tfrac{K}{2}\lVert h\rVert^{2}.$$

Given a positive $\varepsilon$, let $\delta_{0}$ be given by (0.2) for $\tfrac{\varepsilon}{2}$ and let $\delta$ be the lesser of $\delta_{0}$ and $\tfrac{\varepsilon}{K+1}$. For $0<\lVert h\rVert<\delta$ the absolute value of the left-hand side is at most $\tfrac{\varepsilon}{2}\lVert h\rVert+\tfrac{K}{2}\cdot\tfrac{\varepsilon}{K+1}\lVert h\rVert\le\varepsilon\lVert h\rVert$. So $\phi$ is differentiable at $x$ with derivative matrix the row whose $i$th entry is the $i$th coordinate of $Dw(x)+Kx$, and the subdifferential at a point of differentiability (with $U=\mathbb{R}^{d}$ and $f=\phi$) gives

$$\partial_{\mathbb{R}^{d}}\phi(x)=\{Dw(x)+Kx\}\qquad(x\in E_{w}),\tag{5.1}$$

$\partial_{\mathbb{R}^{d}}$ being the subdifferential relative to $\mathbb{R}^{d}$.

Let $\mathrm{id}$ be the identity map of $\mathbb{R}^{d}$; by Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §identity it is Borel, its class belongs to $L^{2}(\mu;\mathbb{R}^{d})$, and $\mathrm{id}\in T_{\mu}$. Let $T=\nabla w+K\,\mathrm{id}$, the pointwise sum. As $\nabla w$ and $\mathrm{id}$ are Borel and square-integrable against $\mu$ (Step 3), The Space of Square-Integrable Random Vectors §space, applied on the probability space $(\mathbb{R}^{d},\mathcal{B}(\mathbb{R}^{d}),\mu)$ as in Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu, shows that $T$ is Borel with $\int_{\mathbb{R}^{d}}\lVert T\rVert^{2}\,d\mu<\infty$, and by The Space of Square-Integrable Random Vectors §classes its class is $\nabla w+K\,\mathrm{id}$. By (5.1), $\partial_{\mathbb{R}^{d}}\phi(x)=\{T(x)\}$ for every $x\in E_{w}$. We apply A Square-Integrable Selection of the Subdifferential of a Convex Potential Belongs to the Tangent Space §tangent with $G=\mathbb{R}^{d}$ (open and convex by Step 0), the convex function $\phi$, its Borel set $D$ taken to be $E_{w}$ (Step 3), which satisfies $E_{w}\subseteq\mathbb{R}^{d}$ and $\mu(E_{w})=1$ by (4.1), and this $T$: it gives $T\in T_{\mu}$. By Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §closed, $T_{\mu}$ is a linear subspace of $L^{2}(\mu;\mathbb{R}^{d})$, so it contains $T+(-K)\,\mathrm{id}$, which is the class of $\nabla w$ by The Space of Square-Integrable Random Vectors §classes, since $T(x)-Kx=\nabla w(x)$ for every $x$. Hence $\nabla w\in T_{\mu}$; as $\mu$ was an arbitrary absolutely continuous member of $\mathcal{P}_{2}(\mathbb{R}^{d})$, claim 2 is proved.

Step 6 (Claim 3). Suppose that $-w$ is semiconvex with constant $K$. The function $-w$ is bounded, since $|-w(x)|=|w(x)|$, and Lipschitz with constant $L$, since $|(-w)(x)-(-w)(y)|=|w(x)-w(y)|$; so Steps 1 to 5 apply to $-w$ in place of $w$. Let $x\in\mathbb{R}^{d}$. If $w$ is differentiable at $x$ with derivative matrix $J$, let $-J$ be the row with entries $-J_{1i}$; then $(-J)h=-(Jh)$ for $h\in\mathbb{R}^{d}$ by Matrix-Vector Product and Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ (the single coordinate of $Jh$ being the dot product of the vector of entries of $J$ with $h$, as in Step 0), so $(-w)(x+h)-(-w)(x)-(-J)h=-\bigl(w(x+h)-w(x)-Jh\bigr)$ has the same absolute value, and $-w$ is differentiable at $x$ with derivative matrix $-J$. Applying this to $-w$, whose negative is $w$, gives the converse; so $E_{-w}=E_{w}$. By claim 1 of A Derivative Matrix is the Jacobian Matrix, and is Unique, $\partial_{i}(-w)(x)=-J_{1i}=-\partial_{i}w(x)$ for $x\in E_{w}$, so $D(-w)(x)=-Dw(x)$ there, and hence $\nabla(-w)(x)=-\nabla w(x)$ for every $x\in\mathbb{R}^{d}$ (both sides being the origin off $E_{w}$). Let $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ be absolutely continuous. Claim 2 for $-w$ gives $\nabla(-w)\in T_{\mu}$. By The Space of Square-Integrable Random Vectors §classes the class of $\nabla w$ is $(-1)$ times that of $\nabla(-w)$, and $T_{\mu}$ is a linear subspace by Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §closed; hence $\nabla w\in T_{\mu}$, which proves claim 3.