Proof of One-Dimensional Test Functions: Scalars, Derivatives, and the Difference Quotient of the Derivative

Each result cited is universally quantified over the data in its own statement. Throughout, the identification of $\mathbb{R}$ with $\mathbb{R}^{1}$ is that of the preamble; points of $\mathbb{R}^{1}$ are read as $1$-tuples by Euclidean Points as Tuples of Real Numbers (as in Probability Measures on Euclidean Space and Random Vectors: Standing Notation (i)), so that a point $x\in\mathbb{R}^{1}$ has the single coordinate $x_{1}=x$; and a one-term finite sum satisfies $\sum_{k=1}^{1}a_{k}=a_{1}$ by claim 1 of Properties of Finite Sums. Sums, scalar multiples and the dot product on $\mathbb{R}^{1}$ are those fixed in Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation (i), namely Sum of Points of $\mathbb{R}^n$, Scalar Multiple of a Point of $\mathbb{R}^n$ and clause 2 of Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$, all formed coordinatewise. Field arithmetic in $\mathbb{R}$ is that of Field, a quotient $s/t$ for $t\ne0$ meaning $s\,t^{-1}$.

Step 1: claim 1. Let $x,y\in\mathbb{R}^{1}$. Then $x\cdot y=\sum_{i=1}^{1}x_{i}y_{i}=x_{1}y_{1}=xy$. By The Euclidean Norm on $\mathbb{R}^n$, $\lVert x\rVert=\sqrt{\sum_{i=1}^{1}x_{i}^{2}}=\sqrt{x^{2}}$, the nonnegative square root of Existence and Uniqueness of the Nonnegative Square Root. Since $x^{2}=|x|^{2}$ by claim 1 of Nonnegativity of Squares in an Ordered Field and $0\le|x|$ by claim 1 of Properties of the Absolute Value in an Ordered Field, the number $|x|$ is a nonnegative square root of $x^{2}$, so by the uniqueness in that theorem $\lVert x\rVert=|x|$; hence $\lVert x\rVert^{2}=|x|^{2}=x^{2}$. Now let $\mu\in\mathcal{P}(\mathbb{R})$. By Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii), $L^{2}(\mu;\mathbb{R})$ consists of the classes of the Borel maps $\xi:\mathbb{R}^{1}\to\mathbb{R}^{1}$ with $\int_{\mathbb{R}}\lVert\xi\rVert^{2}\,d\mu<\infty$, with $\langle\xi,\eta\rangle_{\mu}=\int_{\mathbb{R}}\xi\cdot\eta\,d\mu$ and $\lVert\xi\rVert_{\mu}=\sqrt{\int_{\mathbb{R}}\lVert\xi\rVert^{2}\,d\mu}$, so that $\lVert\xi\rVert_{\mu}^{2}=\int_{\mathbb{R}}\lVert\xi\rVert^{2}\,d\mu$ (the square of the nonnegative square root of a number being that number, by Existence and Uniqueness of the Nonnegative Square Root). A map into $\mathbb{R}^{1}$ is Borel exactly when it is Borel as a map into $\mathbb{R}$, since $\mathcal{B}(\mathbb{R}^{1})=\mathcal{B}(\mathbb{R})$ (preamble). Substituting $\lVert\xi(x)\rVert^{2}=\xi(x)^{2}$ and $\xi(x)\cdot\eta(x)=\xi(x)\eta(x)$ pointwise gives the description of $L^{2}(\mu;\mathbb{R})$ and the two displayed formulas.

Step 2: claim 2. Let $\psi\in C_{c}^{\infty}(\mathbb{R})$. By Test Functions on Euclidean Space, Their Gradient Maps and Laplacians (ii), $\psi$ is of class $C^{1}$ and of class $C^{2}$ on $\mathbb{R}^{1}$ in the sense of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation (ii), so by clause 1 of Functions of Class Cᵏ on an Open Subset of ℝⁿ, read through its clause 3, the partial derivative $\partial_{1}\psi(x)$ exists at every $x\in\mathbb{R}^{1}$, and $\nabla\psi(x)=D\psi(x)$ is the point of $\mathbb{R}^{1}$ with single coordinate $\partial_{1}\psi(x)$ by Gradient of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$, that is, $\nabla\psi(x)=\partial_{1}\psi(x)$. By claim 2 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line, $\psi$ is differentiable at every $x\in\mathbb{R}$ with $\psi'(x)=\partial_{1}\psi(x)$; thus $\psi'=\partial_{1}\psi=\nabla\psi$. Since $\psi$ is of class $C^{2}$, clause 2 of Functions of Class Cᵏ on an Open Subset of ℝⁿ makes $\partial_{1}\psi=\psi'$ of class $C^{1}$ on $\mathbb{R}^{1}$, so $\partial_{1}\partial_{1}\psi(x)$ (clause 4 there) exists at every $x$, and claim 2 of One-Dimensional Derivatives, Partial Derivatives, and Smoothness on the Real Line, applied to $f=\psi'$, shows that $\psi'$ is differentiable at every $x$ with $(\psi')'(x)=\partial_{1}\partial_{1}\psi(x)$. By The Laplacian of a Twice Continuously Differentiable Function (i), $\Delta\psi(x)=\sum_{i=1}^{1}\partial_{i}\partial_{i}\psi(x)=\partial_{1}\partial_{1}\psi(x)$. Hence $(\psi')'=\partial_{1}\partial_{1}\psi=\Delta\psi$. By The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (i), applied with $q=1$ and $i=1$, the function $\partial_{1}\psi=\psi'$ is continuous and bounded, and by The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (ii) so is $\Delta\psi$; continuity there is that of Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation (iv), namely continuity for the Euclidean distance of $\mathbb{R}^{1}$, which is $d_{\mathbb{R}}$, into $(\mathbb{R},d_{\mathbb{R}})$, exactly the reading fixed in the preamble. Finally, let $\mu\in\mathcal{P}(\mathbb{R})$ and $\xi\in L^{2}(\mu;\mathbb{R})$; the class of $\nabla\psi=\psi'$ belongs to $L^{2}(\mu;\mathbb{R})$ by Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (iii), and the two displayed formulas are those of claim 1 with $\eta=\nabla\psi=\psi'$, respectively with $\xi=\nabla\psi=\psi'$.

Step 3: claim 3. Let $\psi\in C_{c}^{\infty}(\mathbb{R})$. Since $\Delta\psi$ is bounded (Step 2), by Bounded Real-Valued Function on a Set there is a real $L\ge0$ with $|\Delta\psi(t)|\le L$ for every $t\in\mathbb{R}$; now let $L$ be any real number with this property; then $0\le|\Delta\psi(0)|\le L$ by claim 1 of Properties of the Absolute Value in an Ordered Field and claim 2 of Elementary Order Arithmetic in an Ordered Field, so $L$ is a bound for $\Delta\psi$ in the sense of Bounded Real-Valued Function on a Set. Apply The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane to the function $\psi'$, in the role of the function $\phi$ of that lemma: by Step 2, $\psi'$ is differentiable at every point of $\mathbb{R}$, its derivative $(\psi')'=\Delta\psi$ is continuous as a map from $(\mathbb{R},d_{\mathbb{R}})$ to $(\mathbb{R},d_{\mathbb{R}})$, and $|\Delta\psi(t)|\le L$ for every $t$. The function $F$ of that lemma is given by $F(x,y)=\bigl(\psi'(x)-\psi'(y)\bigr)/(x-y)$ for $x\ne y$ and $F(x,x)=(\psi')'(x)=\Delta\psi(x)$, where a point of $\mathbb{R}^{2}$ is read there as the pair $(x,y)$ of its two coordinates, which is the point $\iota(x,y)$ by the definition of the concatenation map in Concatenation Identifies a Product of Euclidean Spaces with a Euclidean Space; so $F=F_{\psi}$. By The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane (i), $|F_{\psi}(x,y)|\le L$ and $F_{\psi}(x,y)=F_{\psi}(y,x)$ for all $x,y\in\mathbb{R}$; by The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane (ii), $F_{\psi}$ is continuous on $\mathbb{R}^{2}$; and by The Difference Quotient of a Function with Bounded Continuous Derivative is Bounded, Symmetric and Continuous on the Plane (iii), $F_{\psi}$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{2})$ and $\mathcal{B}(\mathbb{R})$, that is, Borel in the sense of Probability Measures on Euclidean Space and Random Vectors: Standing Notation (ii). Now let $\rho\in\mathcal{P}(\mathbb{R}^{2})$. By Probability Measures on Euclidean Space and Random Vectors: Standing Notation (iii), $\rho$ is a Borel measure on the metric space $(\mathbb{R}^{2},d_{E})$ with $\rho(\mathbb{R}^{2})=1$, and $\mathcal{B}(\mathbb{R}^{2})$ is the Borel $\sigma$-algebra of that metric space (Probability Measures on Euclidean Space and Random Vectors: Standing Notation (i)); so claim 6(b) of Borel Measurability and Bounded Integration on a Metric Space, applied to the Borel function $F_{\psi}$ with the bound $L\ge0$, shows that $F_{\psi}$ is integrable with respect to $\rho$ and $\bigl|\int_{\mathbb{R}^{2}}F_{\psi}\,d\rho\bigr|\le L\,\rho(\mathbb{R}^{2})=L$.

Step 4: claim 4. Let $\psi,\phi\in C_{c}^{\infty}(\mathbb{R})$ and $a,b\in\mathbb{R}$, and put $\theta=a\psi+b\phi$. By The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (iii), $\theta\in C_{c}^{\infty}(\mathbb{R})$, and $\nabla\theta(x)=a\,\nabla\psi(x)+b\,\nabla\phi(x)$ and $\Delta\theta(x)=a\,\Delta\psi(x)+b\,\Delta\phi(x)$ for every $x\in\mathbb{R}$. The sum and scalar multiples on the right of the first identity are formed on the single coordinate, so by Step 2, $\theta'(x)=\nabla\theta(x)=a\,\psi'(x)+b\,\phi'(x)$ for every $x$. Let $x,y\in\mathbb{R}$. If $x\ne y$, then $x-y\ne0$ has the inverse $(x-y)^{-1}$, and by distributivity, associativity and commutativity of multiplication in the field $\mathbb{R}$,

$$F_{\theta}(x,y)=\bigl(a\,\psi'(x)+b\,\phi'(x)-a\,\psi'(y)-b\,\phi'(y)\bigr)(x-y)^{-1}=a\,\bigl(\psi'(x)-\psi'(y)\bigr)(x-y)^{-1}+b\,\bigl(\phi'(x)-\phi'(y)\bigr)(x-y)^{-1}=a\,F_{\psi}(x,y)+b\,F_{\phi}(x,y).$$

If $x=y$, then $F_{\theta}(x,x)=\Delta\theta(x)=a\,\Delta\psi(x)+b\,\Delta\phi(x)=a\,F_{\psi}(x,x)+b\,F_{\phi}(x,x)$. Hence $F_{\theta}=a\,F_{\psi}+b\,F_{\phi}$ pointwise on $\mathbb{R}^{2}$. Let $\rho\in\mathcal{P}(\mathbb{R}^{2})$. By Step 3, $F_{\psi}$ and $F_{\phi}$ are integrable with respect to $\rho$, so by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, applied on the measure space $(\mathbb{R}^{2},\mathcal{B}(\mathbb{R}^{2}),\rho)$, $a\,F_{\psi}+b\,F_{\phi}=F_{\theta}$ is integrable and $\int_{\mathbb{R}^{2}}F_{\theta}\,d\rho=a\int_{\mathbb{R}^{2}}F_{\psi}\,d\rho+b\int_{\mathbb{R}^{2}}F_{\phi}\,d\rho$.