TheoremBase

The representation clause produces a tangent field pairing with all gradients of test functions as the given field does; the difference pairs to zero with every derivative of a test function, hence with every mean-zero test function, hence with every test function once a scaling argument against test functions of arbitrarily large mass kills the remaining constant, and a field pairing to zero with all test functions vanishes.

Proof

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. By Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii) the triple (R,B(R),μ)(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu) is a probability space and L2(μ;R)L^{2}(\mu;\mathbb{R}) is the space of classes of square-integrable random variables on it, with ⟨ξ,η⟩μ=∫Rξη dμ\langle\xi,\eta\rangle_{\mu}=\int_{\mathbb{R}}\xi\eta\,d\mu and ∥ξ∥μ2=∫Rξ2 dμ\lVert\xi\rVert_{\mu}^{2}=\int_{\mathbb{R}}\xi^{2}\,d\mu, by claim 1 of One-Dimensional Test Functions: Scalars, Derivatives, and the Difference Quotient of the Derivative. For ψ∈Cc∞(R)\psi\in C_{c}^{\infty}(\mathbb{R}) claim 2 of that lemma gives ∇ψ=ψ′\nabla\psi=\psi' and ⟨ξ,∇ψ⟩μ=∫Rξ ψ′ dμ\langle\xi,\nabla\psi\rangle_{\mu}=\int_{\mathbb{R}}\xi\,\psi'\,d\mu. Write λ1\lambda_{1} for the Lebesgue measure on B(R)\mathcal{B}(\mathbb{R}).

Since TμT_{\mu} is a subset of L2(μ;R)L^{2}(\mu;\mathbb{R}) 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 (i), it suffices to prove the reverse inclusion. Let η∈L2(μ;R)\eta\in L^{2}(\mu;\mathbb{R}), and fix a Borel representative of it, again written η\eta.

Step 1 (A tangent field with the same pairings). Define ℓ:Cc∞(R)→R\ell:C_{c}^{\infty}(\mathbb{R})\to\mathbb{R} by ℓ(ψ)=⟨η,∇ψ⟩μ\ell(\psi)=\langle\eta,\nabla\psi\rangle_{\mu}. It is linear: for ψ,ϕ∈Cc∞(R)\psi,\phi\in C_{c}^{\infty}(\mathbb{R}) and a,b∈Ra,b\in\mathbb{R}, The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure (iii) gives ∇(aψ+bϕ)=a∇ψ+b∇ϕ\nabla(a\psi+b\phi)=a\nabla\psi+b\nabla\phi, and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives the linearity of the integral. By claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm, applied on the probability space (R,B(R),μ)(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu) to the square-integrable random variables η\eta and ∇ψ\nabla\psi,

∣ℓ(ψ)∣≤∥η∥μ ∥∇ψ∥μfor every ψ∈Cc∞(R).|\ell(\psi)|\le\lVert\eta\rVert_{\mu}\,\lVert\nabla\psi\rVert_{\mu}\qquad\text{for every }\psi\in C_{c}^{\infty}(\mathbb{R}).

Hence Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients (iv), applied with C=∥η∥μC=\lVert\eta\rVert_{\mu}, provides ξ∈Tμ\xi\in T_{\mu} with ⟨ξ,∇ψ⟩μ=ℓ(ψ)\langle\xi,\nabla\psi\rangle_{\mu}=\ell(\psi) for every ψ∈Cc∞(R)\psi\in C_{c}^{\infty}(\mathbb{R}).

Put w=η−ξ∈L2(μ;R)w=\eta-\xi\in L^{2}(\mu;\mathbb{R}), with a Borel representative again written ww. By linearity of the inner product,

∫Rw ψ′ dμ=0for every ψ∈Cc∞(R).(∗)\int_{\mathbb{R}}w\,\psi'\,d\mu=0\qquad\text{for every }\psi\in C_{c}^{\infty}(\mathbb{R}). \tag{$*$}

If ∥w∥μ=0\lVert w\rVert_{\mu}=0 then ww is the zero class and η=ξ∈Tμ\eta=\xi\in T_{\mu}, as required; so assume 0<∥w∥μ0<\lVert w\rVert_{\mu}.

Note that ww is integrable with respect to μ\mu, with ∫R∣w∣ dμ≤∥w∥μ\int_{\mathbb{R}}|w|\,d\mu\le\lVert w\rVert_{\mu}: this is claim 1 of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm applied to ∣w∣|w| and the constant random variable 11, whose mean-square norm is 11 since μ\mu is a probability measure.

Step 2 (Pairing with mean-zero test functions). Let φ∈Cc∞(R)\varphi\in C_{c}^{\infty}(\mathbb{R}) with ∫Rφ dλ1=0\int_{\mathbb{R}}\varphi\,d\lambda_{1}=0. By Test Functions on the Real Line: Unit Mass, Large Mass, Mean-Zero Correction, and the Primitive of a Mean-Zero Test Function (v) there is ψ∈Cc∞(R)\psi\in C_{c}^{\infty}(\mathbb{R}) with ψ′=φ\psi'=\varphi, so (∗)(*) gives

∫Rw φ dμ=0.\int_{\mathbb{R}}w\,\varphi\,d\mu=0 .

Step 3 (Pairing with an arbitrary test function). Let θ\theta be as in Test Functions on the Real Line: Unit Mass, Large Mass, Mean-Zero Correction, and the Primitive of a Mean-Zero Test Function (ii), so θ∈Cc∞(R)\theta\in C_{c}^{\infty}(\mathbb{R}) is nonnegative with ∫Rθ dλ1=1\int_{\mathbb{R}}\theta\,d\lambda_{1}=1, and put a=∫Rw θ dμa=\int_{\mathbb{R}}w\,\theta\,d\mu, a real number because θ\theta is bounded by A Continuous Compactly Supported Function on Rn\mathbb{R}^n is Bounded and Integrable and ww is integrable.

Let ψ∈Cc∞(R)\psi\in C_{c}^{\infty}(\mathbb{R}) and put c=∫Rψ dλ1c=\int_{\mathbb{R}}\psi\,d\lambda_{1}, a real number by Test Functions on the Real Line: Unit Mass, Large Mass, Mean-Zero Correction, and the Primitive of a Mean-Zero Test Function (i). By Test Functions on the Real Line: Unit Mass, Large Mass, Mean-Zero Correction, and the Primitive of a Mean-Zero Test Function (iv) the function φ=ψ−c θ\varphi=\psi-c\,\theta is a test function with ∫φ dλ1=0\int\varphi\,d\lambda_{1}=0, so Step 2 and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion give

0=∫Rw φ dμ=∫Rw ψ dμ−c a,that is∫Rw ψ dμ=a∫Rψ dλ1.(∗∗)0=\int_{\mathbb{R}}w\,\varphi\,d\mu=\int_{\mathbb{R}}w\,\psi\,d\mu-c\,a,\qquad\text{that is}\qquad \int_{\mathbb{R}}w\,\psi\,d\mu=a\int_{\mathbb{R}}\psi\,d\lambda_{1}. \tag{$**$}

Step 4 (The constant vanishes). Let ε\varepsilon be a positive real number and put C=∥w∥μ ε−1C=\lVert w\rVert_{\mu}\,\varepsilon^{-1}, a positive real number by claim 7 of Elementary Order Arithmetic in an Ordered Field and claim 5 of that lemma. By Test Functions on the Real Line: Unit Mass, Large Mass, Mean-Zero Correction, and the Primitive of a Mean-Zero Test Function (iii) there is χ∈Cc∞(R)\chi\in C_{c}^{\infty}(\mathbb{R}) with 0≤χ≤10\le\chi\le1 and C≤∫Rχ dλ1C\le\int_{\mathbb{R}}\chi\,d\lambda_{1}. Applying (∗∗)(**) to χ\chi, and using ∣wχ∣≤∣w∣|w\chi|\le|w| pointwise together with claims 1 and 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion,

∣a∣ C≤∣a∣∫Rχ dλ1=∣∫Rw χ dμ∣≤∫R∣w∣ dμ≤∥w∥μ,|a|\,C\le|a|\int_{\mathbb{R}}\chi\,d\lambda_{1}=\Bigl|\int_{\mathbb{R}}w\,\chi\,d\mu\Bigr|\le\int_{\mathbb{R}}|w|\,d\mu\le\lVert w\rVert_{\mu},

the first inequality by claim 5 of Elementary Arithmetic in an Ordered Field with the nonnegative factor ∣a∣|a| and claim 4 of Properties of the Absolute Value in an Ordered Field. Multiplying by the positive C−1C^{-1} gives ∣a∣≤∥w∥μC−1=ε|a|\le\lVert w\rVert_{\mu}C^{-1}=\varepsilon. As ε\varepsilon was an arbitrary positive real number, Comparison of Real Numbers with Arbitrary Positive Slack (iii) gives ∣a∣=0|a|=0, hence a=0a=0 by claim 1 of Properties of the Absolute Value in an Ordered Field.

Step 5 (Conclusion). By (∗∗)(**) with a=0a=0,

∫Rw ψ dμ=0for every ψ∈Cc∞(R).\int_{\mathbb{R}}w\,\psi\,d\mu=0\qquad\text{for every }\psi\in C_{c}^{\infty}(\mathbb{R}).

Since ww is Borel and integrable with respect to μ∈P(R)\mu\in\mathcal{P}(\mathbb{R}), Test Functions Approximate Bounded Continuous Functions Pointwise, Determine a Finite Borel Measure, and Detect a Vanishing Density (iii), applied in dimension d=1d=1 with g=wg=w, gives μ({x∈R:w(x)≠0})=0\mu(\{x\in\mathbb{R}:w(x)\ne0\})=0. Hence ww is the zero class of L2(μ;R)L^{2}(\mu;\mathbb{R}) by Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation (ii), contradicting 0<∥w∥μ0<\lVert w\rVert_{\mu}; so that case does not occur, and η=ξ∈Tμ\eta=\xi\in T_{\mu}.

Therefore L2(μ;R)⊆TμL^{2}(\mu;\mathbb{R})\subseteq T_{\mu}, and with the reverse inclusion noted at the outset, Tμ=L2(μ;R)T_{\mu}=L^{2}(\mu;\mathbb{R}).

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