$p_{\Omega}$ for $p_{\Omega,V}$.

Recall the Feynman-Kac formula (cf. [@BS]) for non-negative, measurable potentials $V:\Omega\rightarrow {\mathbb{R}}^+,$ $$p_{\Omega,V}(x,y;t) = p_{{\mathbb{R}}^m}(x,y;t) {\mathbb{E}}[e^{- \int_0^t V(\beta(s))ds} \big(\Pi_{s\in[0,t]}{\bf1}_{\Omega}(\beta(s))\big) \, : \,  \beta(0)=x, \beta(t) = y ],$$ where $\beta$ is a Brownian bridge. Hence if $V_1: \Omega \to {\mathbb{R}}^+$ and $V_2: \Omega \to {\mathbb{R}}^+$ are measurable functions with $0 \le V_2 \le  V_1$, then

$$\label{e26}
0\le p_{\Omega,V_1}(x,y;t)\le p_{\Omega,V_2}(x,y;t),\qquad \forall x\in \Omega, \forall y\in \Omega, \forall t>0.$$

Since $V\ge 0$ we then conclude that $$0 \le v_{\Omega, V}(x) \le v_{\Omega}(x)\le \mathfrak{c}_m\lambda_1(\Omega)^{-1}, \qquad  \forall x \in \Omega,$$ with $\mathfrak{c}_m$ given by .

\[lem1\] If $0 < |\Omega|<\infty$, and if $V: \Omega \to {\mathbb{R}}^+$ is measurable, then $$\label{e29}
\lambda_{1}(\Omega, V)^{-1} \le
\|v_{\Omega,V}\|_{\infty}
\le (4+3m\log 2)\lambda_{1}(\Omega, V)^{-1}.$$

Lemma \[lem1\] implies that $$\label{e29a}
\mathfrak{d}_m:=\sup\{\lambda_1(\Omega,V)\|v_{\Omega,V}\|_{\infty}:\Omega\,\textup{ open in }\,{\mathbb{R}}^m,\,0<|\Omega|<\infty,\, V\in \mathcal{V}_{\Omega} \}<\infty,$$ where $\mathcal{V}_{\Omega}$ is the set of measurable functions from $\Omega$ into ${\mathbb{R}}^+$. By choosing $V=0$ in the expression under the supremum in the right-hand side of we see that $\mathfrak{c}_m\le \mathfrak{d}_m$.

Throughout this paper we denote by $B(p;R)=\{x\in{\mathbb{R}}^m:|x-p|<R\}$ the open ball centred at $p$ with radius $R$, $B_R=B(0;R)$, and $$\omega_m=|B_1|.$$

[*Proof of Lemma* ]{} To prove the upper bound in we note that Lemma 1 and its proof in [@vdBC] hold with 