ULA invariant laws have uniform exponential moments on the needed step ranges

Fix $\gamma$. Dropping the indicator in [F2] gives a global drift $R_{\gamma }V\le e^{-\gamma }V+b_{\gamma }$ with finite $b_{\gamma }$. Each sufficiently large V-sublevel is contained in a compact ball by [F3]. On the product of that starting ball and any fixed positive-radius target ball, the continuous positive density [F1] has a positive minimum: explicitly its Gaussian exponent is bounded because both balls and the continuous gradient are bounded. Restriction to the target ball and normalization of its positive finite volume give a common minorization. These are precisely the drift and minorization hypotheses for the weight V.

By [F6], $g(x_{\star })=0$, and therefore Lipschitz continuity gives $\mathrm{\parallel }g(x)\mathrm{\parallel }\le L\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }$. If $r=\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }$ is positive and at least the radial radius, Cauchy–Schwarz and the radial hypothesis give $\mathrm{\parallel }g(x)\mathrm{\parallel }r\ge ⟨g(x),x-x_{\star }⟩\ge \rho r^{\alpha }$, so $\mathrm{\parallel }g(x)\mathrm{\parallel }\ge \rho r^{\alpha -1}$. For $h=y-x$, integrate the derivative along $x+th$ and subtract $⟨g(x),h⟩$ to obtain $U(y)-U(x)-⟨g(x),h⟩={\int }_0^1⟨g(x+th)-g(x),h⟩dt\le \frac{L}{2}\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{.}$

Completing the square, [F7] gives for $c<1\mathrm{/}2$ and scalar v the identity $(2\pi {)}^{-1\mathrm{/}2}{\int }_{\mathbb{R}}{e}^{-(1\mathrm{/}2-c)z^2+vz}dz=(1-2c{)}^{-1\mathrm{/}2}{e}^{v^2\mathrm{/}(2(1-2c))}\mathrm{.}$ Finite multiplication and Tonelli therefore give $\mathbb{E}{e}^{c\mathrm{\parallel }Z{\mathrm{\parallel }}^2+⟨v,Z⟩}=(1-2c{)}^{-d\mathrm{/}2}{e}^{\mathrm{\parallel }v{\mathrm{\parallel }}^2\mathrm{/}(2(1-2c))}$ for $Z\sim N_d(0,I_d)$.

Apply [F4] to the fixed-step kernel and weight V verified in step 1.1. It supplies a unique invariant probability ${\pi }_{\gamma }$ with finite V moment.

Put $g=\mathrm{\nabla }U(x)$ and $H_s=e^{sU}$ for $0<s<1$ with $2sL\gamma <1$. Apply step 1.2 to $Y=x-\gamma g+\sqrt{2\gamma }Z$, multiply by s, and exponentiate. Step 1.3, with quadratic coefficient $sL\gamma$ and linear coefficient $s\sqrt{2\gamma }(1-L\gamma )g$, gives $\frac{R_{\gamma }{H}_s(x)}{H_s(x)}\le (1-2sL\gamma {)}^{-d\mathrm{/}2}\exp \text{\hspace{0.17em}⁣}(-\frac{s\gamma (1-s-L\gamma \mathrm{/}2)}{1-2sL\gamma }\mathrm{\parallel }g{\mathrm{\parallel }}^2)\mathrm{.}$ The coefficient follows by combining $s\gamma (-1+L\gamma \mathrm{/}2)$ with $s^2\gamma (1-L\gamma {)}^2\mathrm{/}(1-2sL\gamma )$ over the common denominator. All Gaussian integrals are finite under $2sL\gamma <1$.

For $s=1\mathrm{/}2$, step 2.2 reduces to $R_{\gamma }V\le V\exp \text{\hspace{0.17em}⁣}(\frac{d}{2}[-\log (1-L\gamma )]-\frac{\gamma }{4}\mathrm{\parallel }g{\mathrm{\parallel }}^2)\mathrm{.}$ If $\gamma \le {\gamma }_0$, then $-\log (1-L\gamma )\le 2L\gamma$, since $-\log (1-z)={\int }_0^z(1-u{)}^{-1}du\le 2z$ for $z\le 1\mathrm{/}2$. Choose a radius R at least the radial radius and large enough that ${\rho }^2{R}^{2(\alpha -1)}\mathrm{/}4\ge dL+1$. Outside this ball step 1.2 makes the exponent at most $-\gamma$. Inside it let $M={\sup }_{\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }\le R}V(x)<\mathrm{\infty }$. Since $e^{a\gamma }-e^{-\gamma }={\int }_{-\gamma }^{a\gamma }{e}^{u}du\le (a+1)\gamma e^{a{\gamma }_0}$ for $a=dL$, the preceding bound gives $R_{\gamma }V\le e^{-\gamma }V+b\gamma ,b=M(dL+1)e^{dL{\gamma }_0},$ uniformly for small steps. Integrate against the finite-V invariant law from step 2.1 to obtain ${\pi }_{\gamma }V\le b\gamma \mathrm{/}(1-e^{-\gamma })\le b{e}^{{\gamma }_0}$, using $1-e^{-\gamma }={\int }_0^{\gamma }{e}^{-u}du\ge \gamma e^{-{\gamma }_0}$. This proves the first uniform bound.

For $s=1\mathrm{/}4$ and $0<\gamma <L^{-1}$, the logarithm of the prefactor in step 2.2 is at most $dL\gamma \mathrm{/}2$, by the same integral estimate with $z=L\gamma \mathrm{/}2<1\mathrm{/}2$. Moreover $s(1-s-L\gamma \mathrm{/}2)\mathrm{/}(1-2sL\gamma )\ge 1\mathrm{/}16$, because the numerator is at least $1\mathrm{/}16$ and the denominator is at most one. Choose R' at least the radial radius such that ${\rho }^2(R^{\mathrm{\prime }}{)}^{2(\alpha -1)}\mathrm{/}16\ge dL\mathrm{/}2+1$. Outside this ball $R_{\gamma }W\le e^{-\gamma }W$; inside use $M^{\mathrm{\prime }}={\sup }_{\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }\le R^{\mathrm{\prime }}}W(x)$ and the exponential-difference estimate from step 3.1 with $a=dL\mathrm{/}2$ and upper horizon $L^{-1}$. Thus $R_{\gamma }W\le e^{-\gamma }W+b^{\mathrm{\prime }}\gamma ,b^{\mathrm{\prime }}=M^{\mathrm{\prime }}(dL\mathrm{/}2+1)e^{d\mathrm{/}2},$ for the whole step range. Since $W\le V$, its invariant moment is finite by step 2.1. Integrating and using $1-e^{-\gamma }\ge \gamma e^{-1\mathrm{/}L}$ gives ${\pi }_{\gamma }W\le b^{\mathrm{\prime }}{e}^{1\mathrm{/}L}$, uniformly for every allowed step.

Put $r=\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }$. For $r\ge 1$, [F3] gives $V(x)\ge e^{-C\mathrm{/}2}{e}^{c{r}^{\alpha }\mathrm{/}2}\ge e^{-C\mathrm{/}2}{e}^{cr\mathrm{/}2}$ because $\alpha >1$. By [F5], $r^2\mathrm{/}{e}^{cr\mathrm{/}2}$ is bounded on $[1,\mathrm{\infty })$; on $[0,1]$ use $r^2\le 1\le V$. Thus $r^2\le AV$ for some finite A. The Lipschitz bound in step 1.2 gives $\mathrm{\parallel }\mathrm{\nabla }U(x){\mathrm{\parallel }}^2\le L^2{r}^2$. Taking $M_g=(1+L^2)A$, $M_V=b{e}^{{\gamma }_0}$ and $M_W=b^{\mathrm{\prime }}{e}^{1\mathrm{/}L}$ proves all asserted estimates. All constants depend only on the fixed potential and its displayed parameters.