Proof of The Hamilton-Jacobi Operator with Common Noise and Penalty Drift Has Momentum-Continuous Shifts

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

We work in the setting of The Intrinsic Calculus on the Wasserstein Space: Standing Notation; elementary order and arithmetic of real numbers is carried by The Real Numbers: Standing Notation and Background §background, which we adopt for that purpose. By The Discounted Hamilton-Jacobi Equation with Common Noise and a Penalty Drift on the Wasserstein Space §operator, $F$ is a second-order equation operator over $\mathcal{D}_{\Sigma}$ with

$$F(\nu,r,q,Y)=\lambda_{0}\,r-\frac{1}{2}\,\mathrm{tr}\bigl(\Gamma^{\top}\Gamma Y\bigr)+\frac{\theta}{2}\,\lVert q\rVert_{\nu}^{2}+\langle\Sigma(\nu),q\rangle_{\nu}-g(\nu),$$

and its $\delta$-shifts are given by The Bundle of Vector Fields over a Set of Measures, Second-Order Equation Operators on the Wasserstein Space, and Their Delta-Shifts §shifted.

Claim 1. Let $\delta,R,\nu,r,Y,q,q'$ be as in the claim, and write $\sigma=\Sigma(\nu)$, which lies in $T_{\nu}\subseteq L^{2}(\nu;\mathbb{R}^{d})$ by Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair. By Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu, $L^{2}(\nu;\mathbb{R}^{d})$ is a real Hilbert space, in particular a real inner product space with inner product $\langle\cdot,\cdot\rangle_{\nu}$ and norm $\lVert\cdot\rVert_{\nu}$, the norm satisfying $\lVert x\rVert_{\nu}^{2}=\langle x,x\rangle_{\nu}$ by Real Inner Product Space §norm.

Step 1 (an identity for squared norms). For $w,w'\in L^{2}(\nu;\mathbb{R}^{d})$, Elementary Identities in a Real Inner Product Space §bilinear and symmetry of the inner product (condition (a) of Real Inner Product Space §inner-product) give

$$\langle w-w',w+w'\rangle_{\nu}=\langle w,w\rangle_{\nu}+\langle w,w'\rangle_{\nu}-\langle w',w\rangle_{\nu}-\langle w',w'\rangle_{\nu}=\lVert w\rVert_{\nu}^{2}-\lVert w'\rVert_{\nu}^{2}.$$

Step 2 (the difference for $F^{-}_{\delta}$). Put $w=q+\delta\sigma$ and $w'=q'+\delta\sigma$, so that $w-w'=q-q'$ and $w+w'=q+q'+2\delta\sigma$ in the vector space $L^{2}(\nu;\mathbb{R}^{d})$. By The Bundle of Vector Fields over a Set of Measures, Second-Order Equation Operators on the Wasserstein Space, and Their Delta-Shifts §shifted, $F^{-}_{\delta}(\nu,r,q,Y)$ and $F^{-}_{\delta}(\nu,r,q',Y)$ are the values of $F$ at $(\nu,r+\delta\mathcal{E}(\nu),w,Y+\delta H_{\mathcal{E}}(\nu))$ and at $(\nu,r+\delta\mathcal{E}(\nu),w',Y+\delta H_{\mathcal{E}}(\nu))$. The terms $\lambda_{0}(r+\delta\mathcal{E}(\nu))$, $-\tfrac12\mathrm{tr}(\Gamma^{\top}\Gamma(Y+\delta H_{\mathcal{E}}(\nu)))$ and $-g(\nu)$ are the same in both values and cancel in the difference, so by Step 1 and Elementary Identities in a Real Inner Product Space §bilinear,

$$F^{-}_{\delta}(\nu,r,q,Y)-F^{-}_{\delta}(\nu,r,q',Y)=\frac{\theta}{2}\bigl(\lVert w\rVert_{\nu}^{2}-\lVert w'\rVert_{\nu}^{2}\bigr)+\langle\sigma,w-w'\rangle_{\nu}=\frac{\theta}{2}\,\langle q-q',\,q+q'+2\delta\sigma\rangle_{\nu}+\langle\sigma,q-q'\rangle_{\nu}.$$

Step 3 (the estimate). By the triangle inequality The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle and homogeneity Elementary Identities in a Real Inner Product Space §homogeneity, using $\delta>0$,

$$\lVert q+q'+2\delta\sigma\rVert_{\nu}\le\lVert q\rVert_{\nu}+\lVert q'\rVert_{\nu}+2\delta\lVert\sigma\rVert_{\nu}\le2R+2\delta R=2(1+\delta)R .$$

Taking absolute values in Step 2, with the triangle inequality for the absolute value and the Cauchy-Schwarz inequality in $L^{2}(\nu;\mathbb{R}^{d})$ applied to each inner product, and using $\theta>0$ and $\lVert\sigma\rVert_{\nu}\le R$,

$$\bigl|F^{-}_{\delta}(\nu,r,q,Y)-F^{-}_{\delta}(\nu,r,q',Y)\bigr|\le\frac{\theta}{2}\,\lVert q-q'\rVert_{\nu}\cdot2(1+\delta)R+R\,\lVert q-q'\rVert_{\nu}=\bigl(\theta(1+\delta)+1\bigr)R\,\lVert q-q'\rVert_{\nu}.$$

Since $0<\delta<1$ and $\theta>0$, we have $\theta(1+\delta)+1\le2\theta+1$, and as $R\lVert q-q'\rVert_{\nu}\ge0$ the right-hand side is at most $(2\theta+1)R\lVert q-q'\rVert_{\nu}$. This is the first inequality.

Step 4 (the difference for $F^{+}_{\delta}$). Put $w=q-\delta\sigma$ and $w'=q'-\delta\sigma$, so that $w-w'=q-q'$ and $w+w'=q+q'-2\delta\sigma$. By The Bundle of Vector Fields over a Set of Measures, Second-Order Equation Operators on the Wasserstein Space, and Their Delta-Shifts §shifted the two values are those of $F$ at $(\nu,r-\delta\mathcal{E}(\nu),w,Y-\delta H_{\mathcal{E}}(\nu))$ and $(\nu,r-\delta\mathcal{E}(\nu),w',Y-\delta H_{\mathcal{E}}(\nu))$; exactly as in Step 2,

$$F^{+}_{\delta}(\nu,r,q,Y)-F^{+}_{\delta}(\nu,r,q',Y)=\frac{\theta}{2}\,\langle q-q',\,q+q'-2\delta\sigma\rangle_{\nu}+\langle\sigma,q-q'\rangle_{\nu},$$

and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §triangle with Elementary Identities in a Real Inner Product Space §homogeneity gives $\lVert q+q'-2\delta\sigma\rVert_{\nu}\le2(1+\delta)R$. The estimate of Step 3, with Cauchy-Schwarz, gives the second inequality.

Claim 2. We verify Equation Operators on the Wasserstein Space with Momentum-Continuous Shifts §momentum, whose hypotheses on $F$ hold since $F$ is a second-order equation operator over $\mathcal{D}_{\Sigma}$. Let $\delta,R,\eta\in\mathbb{R}$ with $0<\delta<1$, $0<R$ and $0<\eta$ be given; then $(2\theta+1)R$ is positive, and we choose

$$\rho=\frac{\eta}{(2\theta+1)R},$$

which is positive. Let $\nu\in\mathcal{D}_{\Sigma}$ with $\lVert\Sigma(\nu)\rVert_{\nu}\le R$, $r\in\mathbb{R}$ with $|r|\le R$, $q,q'\in L^{2}(\nu;\mathbb{R}^{d})$ with $\lVert q\rVert_{\nu}\le R$, $\lVert q'\rVert_{\nu}\le R$ and $\lVert q-q'\rVert_{\nu}<\rho$, and $Y\in\mathcal{S}(d)$ with $\lVert Y\rVert\le R$. Claim 1 applies to these data (the bounds on $r$ and $Y$ are not needed) and gives

$$\bigl|F^{\mp}_{\delta}(\nu,r,q,Y)-F^{\mp}_{\delta}(\nu,r,q',Y)\bigr|\le(2\theta+1)R\,\lVert q-q'\rVert_{\nu}<(2\theta+1)R\,\rho=\eta$$

for both shifts, the strict inequality because $(2\theta+1)R$ is positive and $\lVert q-q'\rVert_{\nu}<\rho$. Hence $F$ has momentum-continuous shifts relative to the pair. $\blacksquare$