that the integral $\int_{{{B_d(\vy, \frac 1 2)}}} \exp(-{s}\lambda g(\vx))\;{\mathrm{d}}\vx$ is within $(1+ 2\exp(-d))$ of $\int_{{{B_d(\vy, \frac 1 4)}}} \exp(-{s}\lambda g(\vx))\;{\mathrm{d}}\vx$ for ${s}\lambda > 16d$. This implies that the term $(A)$ above is at most $(1+2\exp(-d))\widehat{Z_{\lambda}^f}$. To bound the second term $(B)$ above, we argue that a ball of radius $\frac 1 8$ around any single vertex $\vy$ of the hypercube contributes significantly more than than the term $(B)$. Indeed $$\begin{aligned}
  \int_{{B_d(\vy, \frac 1 8)}} \exp(-\lambda f(\vx)) \;{\mathrm{d}}\vx &\geq \int_{{B_d(\vy, \frac 1 8)}} \exp(-\lambda (h(\vy) + \frac{{s}}{8})) \;{\mathrm{d}}\vx\\
  &\geq \exp(-\lambda (d+\frac{M}{4})) \cdot (\frac 1 8)^d \int_{{B_d(\mathbf{0}, 1)}} \;{\mathrm{d}}\vx
  &\geq \exp(-\lambda (\frac{3M}{4})) \cdot (\frac 1 8)^d \int_{{B_d(\mathbf{0}, 1)}} \;{\mathrm{d}}\vx\end{aligned}$$ Whereas, $$\begin{aligned}
  \int_{{B_d(0, R)}} \exp(-\lambda M) \;{\mathrm{d}}\vx &= \exp(-\lambda M) \cdot R^d \cdot \int_{{B_d(\mathbf{0}, 1)}} \;{\mathrm{d}}\vx.\end{aligned}$$ Thus $(B)/\widehat{Z_{\lambda}^f}$ is at most $\exp(-\lambda M/4 + d \ln 8R)$. Under the assumptions on $\lambda$, it follows that (B) is at most $\exp(-\Omega(d))$ times $\widehat{Z_\lambda^h}$. In other words, we have shown that $Z_{\lambda}^f$ is at most $(1+\exp(-\Omega(d))) \widehat{Z_{\lambda}^h}$.

\[rem:sampling\_wasserstein\] The equivalence of sampling extends immediately to Wasserstein distance. Indeed given a sampler for ${\mathcal{D}}_h^{\lambda, {\mathbb{H}_{d}}}$, one gets a Wasserstein sampler for $\widehat{{{\mathcal{D}}_f^{\lambda}}}$ by sampling from a simple isotropic log-concave distribution. A Wasserstein sampler for a ball 