Gaussian-mixture score identities and a global Lipschitz bound

For fixed $t>0$, positivity in [F1] makes the quotient defining ${\nu }_{t,x}$ meaningful. Its density relative to $\mu$ is measurable and nonnegative, and its integral equals $p_t(x)\mathrm{/}{p}_t(x)=1$, so it is a probability measure. By [F4], it is supported in $\bar{B}(0,R)$ and has finite first and second moments. The regularity conclusion [F2] applies on $[ϵ,T]$. On each compact set of $x$ values, every spatial derivative of ${\varphi }_{v(t)}(x-z)$ is bounded uniformly for $z\in \bar{B}(0,R)$ because $v(t)>0$ is fixed and the Gaussian derivatives are polynomials times a Gaussian. The finite measure $\mu$ supplies an integrable constant majorant. Difference quotients have the same kind of bound on a slightly larger compact set by the fundamental theorem of calculus, so [F3] justifies differentiating the convolution and its bounded-moment numerators under the integral.

For each coordinate $i$, differentiation of $p_t$ gives ${\mathrm{\partial }}_i{p}_t(x)=\int -\frac{x_i-z_i}{v(t)}{\varphi }_{v(t)}(x-z)\mu (dz)=-\frac{x_i-m_{t,x,i}}{v(t)}{p}_t(x)\mathrm{.}$ Dividing by the positive $p_t(x)$ proves $\mathrm{\nabla }\log p_t(x)=-(x-m_{t,x})\mathrm{/}v(t)$.

Differentiate $m_{t,x,j}=p_t(x{)}^{-1}\int z_j{\varphi }_{v(t)}(x-z)\mu (dz)$ in coordinate $i$. The quotient rule and ${\mathrm{\partial }}_i{\varphi }_{v(t)}(x-z)=-(x_i-z_i){\varphi }_{v(t)}(x-z)\mathrm{/}v(t)$ yield ${\mathrm{\partial }}_i{m}_{t,x,j}=\frac{1}{v(t)}(\int z_j{z}_i{\nu }_{t,x}(dz)-m_{t,x,j}{m}_{t,x,i})=(C_{t,x}{)}_{ji}\mathrm{/}v(t)\mathrm{.}$ All terms are finite by [F4], and the derivative is justified by step 1.1. Differentiating the score identity from step 2.1 therefore gives $D_x{s}_t=-I_d\mathrm{/}v(t)+C_{t,x}\mathrm{/}v(t{)}^2$.

The matrix $C_{t,x}$ is positive semidefinite because for every $u\in {\mathbb{R}}^d$, $u^{\mathsf{T}}{C}_{t,x}u=\int (u\cdot (z-m_{t,x}){)}^2{\nu }_{t,x}(dz)\ge 0$. Also $u^{\mathsf{T}}{C}_{t,x}u=\int (u\cdot z{)}^{2}d{\nu }_{t,x}-{(\int u\cdot zd{\nu }_{t,x})}^2\le \int \mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{\mid }z{\mathrm{\mid }}^{2}d{\nu }_{t,x}\le R^2\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{.}$ Thus $\mathrm{\parallel }{C}_{t,x}{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le R^2$. Since $v$ is increasing and $v(t)\ge v(ϵ)>0$, the Jacobian formula in step 3.1 and the triangle inequality for operator norms give the stated uniform derivative bound. The mean satisfies $\mathrm{\mid }{m}_{t,x}\mathrm{\mid }\le \int \mathrm{\mid }z\mathrm{\mid }d{\nu }_{t,x}\le R$, so the score formula in step 2.1 gives the displayed growth bound.

For any $x,y$, the fundamental theorem of calculus along the segment $y+r(x-y)$ gives $s_t(x)-s_t(y)={\int }_0^1{D}_x{s}_t(y+r(x-y))(x-y)dr$. Taking norms and using step 4.1 proves the global Lipschitz bound uniformly for $t\in [ϵ,T]$. Since $g\ge g_{\min }>0$, $v(ϵ)\downarrow 0$ as $ϵ\downarrow 0$; the bound therefore supplies no uniform-in-$ϵ$ constant.