Improving simulated annealing through derandomization

Abstract

We propose and study a version of simulated annealing (SA) on continuous state spaces based on $(t,s)_R$-sequences. The parameter $R\in \bar{\mathbb {N}}$ regulates the degree of randomness of the input sequence, with the case $R=0$ corresponding to IID uniform random numbers and the limiting case $R=\infty$ to (t, s)-sequences. Our main result, obtained for rectangular domains, shows that the resulting optimization method, which we refer to as QMC-SA, converges almost surely to the global optimum of the objective function $\varphi$ for any $R\in \mathbb {N}$. When $\varphi$ is univariate, we are in addition able to show that the completely deterministic version of QMC-SA is convergent. A key property of these results is that they do not require objective-dependent conditions on the cooling schedule. As a corollary of our theoretical analysis, we provide a new almost sure convergence result for SA which shares this property under minimal assumptions on $\varphi$. We further explain how our results in fact apply to a broader class of optimization methods including for example threshold accepting, for which to our knowledge no convergence results currently exist. We finally illustrate the superiority of QMC-SA over SA algorithms in a numerical study.

Appendices

Appendix 1: Proof of Lemma 1

Let $n\in \mathbb {N}$, $(\tilde{\mathbf {x}},\mathbf {x}^{\prime })\in \mathcal {X}^2$, $\delta _{\mathcal {X}}=0.5$ and $\delta \in (0,\delta _{\mathcal {X}}]$. Then, by Assumption (A1), $F_K^{-1}(\tilde{\mathbf {x}},\mathbf {u}_1^n)\in B_{\delta }(\mathbf {x}^{\prime })$ if and only if $\mathbf {u}_1^n\in F_K(\tilde{\mathbf {x}},B_{\delta }(\mathbf {x}^{\prime }))$. We now show that, for $\delta$ small enough, there exists a closed hypercube $W(\tilde{\mathbf {x}},\mathbf {x}^{\prime }, \delta )\subset [0,1)^d$ such that $W(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )\subseteq F_K(\mathbf {x},B_{\delta }(\mathbf {x}^{\prime }))$ for all $\mathbf {x}\in B_{v_K(\delta )}(\tilde{\mathbf {x}})$, with $v_K(\cdot )$ as in the statement of the lemma.

To see this note that, because $K(\mathbf {x},\mathrm {d}\mathbf {y})$ admits a density $K(\mathbf {y}|\mathbf {x})$ which is continuous on the compact set $\mathcal {X}^2$, and using Assumption (A2), it is easy to see that, for $i\in 1:d$, $K_i(y_i|y_{1:i-1},\mathbf {x})\ge \tilde{K}$ for all $(\mathbf {x},\mathbf {y})\in \mathcal {X}^2$ and for a constant $\tilde{K}>0$. Consequently, for any $\delta \in [0,0.5]$ and $(\mathbf {x},\mathbf {y})\in \mathcal {X}^2$,

$$\begin{aligned} F_{K_i}\left( x^{\prime }_i+\delta |\mathbf {x}, y_{1:i-1}\right) -F_{K_i}\left( x^{\prime }_i-\delta |\mathbf {x}, y_{1:i-1}\right) \ge \tilde{K}\delta ,\quad \forall i\in 1:d \end{aligned}$$

(6)

where $F_{K_i}(\cdot {}|\mathbf {x}, y_{1:i-1})$ denotes the CDF of the probability measure $K_i(\mathbf {x},y_{1:i-1},\mathrm {d}y_i)$, with the convention that $F_{K_i}(\cdot {}|\mathbf {x}, y_{1:i-1})=F_{K_1}(\cdot {}|\mathbf {x})$ when $i=1$. Note that the right-hand side of (6) is $\tilde{K}\delta$ and not $2\tilde{K}\delta$ to encompass the case where either $x_i^{\prime }-\delta \not \in [0,1]$ or $x_i^{\prime }+\delta \not \in [0,1]$. (Note also that because $\delta \le 0.5$ we cannot have both $x_i^{\prime }-\delta \not \in [0,1]$ and $x_i^{\prime }+\delta \not \in [0,1]$.)

For $i\in 1:d$ and $\delta ^{\prime }>0$, let

$$\begin{aligned} \omega _i(\delta ^{\prime })=\sup _{ \begin{array}{c} (\mathbf {x},\mathbf {y})\in \mathcal {X}^2,\,(\mathbf {x}^{\prime },\mathbf {y}^{\prime })\in \mathcal {X}^2\\ \Vert \mathbf {x}-\mathbf {x}^{\prime }\Vert _{\infty }\vee \Vert \mathbf {y}-\mathbf {y}^{\prime }\Vert _{\infty }\le \delta ^{\prime } \end{array}}|F_{K_i}\left( y_i|\mathbf {x}, y_{1:i-1}\right) -F_{K_i}\left( y_i^{\prime }|\mathbf {x}^{\prime }, y^{\prime }_{1:i-1}\right) | \end{aligned}$$

be the (optimal) modulus of continuity of $F_{K_i}(\cdot |\cdot )$. Since $F_{K_i}$ is uniformly continuous on the compact set $[0,1]^{d+i}$, the mapping $\omega _i(\cdot )$ is continuous and $\omega _i(\delta ^{\prime })\rightarrow 0$ as $\delta ^{\prime }\rightarrow 0$. In addition, because $F_{K_i}\left( \cdot |\mathbf {x}, y_{1:i-1}\right)$ is strictly increasing on [0, 1] for all $(\mathbf {x},\mathbf {y})\in \mathcal {X}^2$, $\omega _i(\cdot )$ is strictly increasing on (0, 1]. Let $\tilde{K}$ be small enough so that, for $i\in 1:d$, $0.25\tilde{K}\delta _{\mathcal {X}}\le w_i(1)$ and let $\tilde{\delta }_i(\cdot )$ be the mapping $z\in (0,\delta _{\mathcal {X}}]\longmapsto \tilde{\delta }_i(z)=\omega _i^{-1} (0.25\tilde{K}z)$. Remark that the function $\tilde{\delta }_i(\cdot )$ is independent of $(\tilde{\mathbf {x}},\mathbf {x}^{\prime })\in \mathcal {X}^2$, continuous and strictly increasing on $(0,\delta _{\mathcal {X}}]$ and such that $\tilde{\delta }_i(\delta ^{\prime })\rightarrow 0$ as $\delta ^{\prime }\rightarrow 0$.

For $\mathbf {x}\in \mathcal {X}$, $\delta ^{\prime }>0$ and $\delta ^{\prime }_i>0$, $i\in 1:d$, let

$$\begin{aligned} B^i_{\delta ^{\prime }}(\tilde{\mathbf {x}})=\{\mathbf {x}\in [0,1]^i:\, \Vert \mathbf {x}-\tilde{x}_{1:i}\Vert _{\infty }\le \delta ^{\prime }\}\cap [0,1]^i \end{aligned}$$

and

$$\begin{aligned} B_{\delta ^{\prime }_{1:i}}(\tilde{\mathbf {x}})=\{\mathbf {x}\in [0,1]^i:\, |x_j-\tilde{x}_{j}|\le \delta ^{\prime }_j,\,j\in 1:i\}\cap [0,1]^i. \end{aligned}$$

Then, for any $\delta ^{\prime }>0$ and for all $(\mathbf {x},y_{1:i-1})\in B_{\tilde{\delta }_i(\delta )}(\tilde{\mathbf {x}})\times B^{i-1}_{\tilde{\delta }_i(\delta )}(\mathbf {x}^{\prime })$, we have

$$\begin{aligned}&|F_{K_i}\left( x^{\prime }_i+\delta ^{\prime }|\mathbf {x}, y_{1:i-1}\right) -F_{K_i}\left( x^{\prime }_i+\delta ^{\prime }|\tilde{\mathbf {x}}, x^{\prime }_{1:i-1}\right) |\le 0.25\tilde{K}\delta \end{aligned}$$

(7)

$$\begin{aligned}&|F_{K_i}\left( x^{\prime }_i-\delta ^{\prime }|\mathbf {x}, y_{1:i-1}\right) -F_{K_i}\left( x^{\prime }_i-\delta ^{\prime }|\tilde{\mathbf {x}}, x^{\prime }_{1:i-1}\right) |\le 0.25\tilde{K}\delta . \end{aligned}$$

(8)

For $i\in 1:d$ and $\delta ^{\prime }\in (0,\delta _{\mathcal {X}}]$, let $\delta _i(\delta ^{\prime })=\tilde{\delta }_i(\delta ^{\prime })\wedge \delta ^{\prime }$ and note that the function $\delta _i(\cdot )$ is continuous and strictly increasing on $(0,\delta _{\mathcal {X}}]$. Let $\delta _d=\delta _d(\delta )$ and define recursively $\delta _{i}=\delta _{i}(\delta _{i+1})$, $i\in 1:(d-1)$, so that $\delta \ge \delta _d\ge \dots \ge \delta _1>0$. For $i\in 1:d$, let

$$\begin{aligned} \underline{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})= \sup _{(\mathbf {x},y_{1:i-1})\in B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })} F_{K_i}\left( x^{\prime }_i-\delta _i|\mathbf {x}, y_{1:i-1}\right) \end{aligned}$$

and

$$\begin{aligned} \bar{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})= \inf _{(\mathbf {x},y_{1:i-1})\in B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })} F_{K_i}\left( x^{\prime }_i+\delta _i|\mathbf {x}, y_{1:i-1}\right) . \end{aligned}$$

Then, since $F_{K_i}(\cdot |\cdot )$ is continuous and the set $B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })$ is compact, there exists points $(\underline{\mathbf {x}}^i,\underline{y}^{i}_{1:i-1})$ and $(\bar{\mathbf {x}}^i,\bar{y}^{i}_{1:i-1})$ in $B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })$ such that

$$\begin{aligned} \underline{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})=F_{K_i}(x^{\prime }_i-\delta _{i}|\underline{\mathbf {x}}^i, \underline{y}^{i}_{1:i-1}),\quad \bar{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime }, \delta _{1:i})=F_{K_i}\left( x^{\prime }_i+\delta _i|\bar{\mathbf {x}}^i, \bar{y}^{i}_{1:i-1}\right) . \end{aligned}$$

In addition, by the construction of the $\delta _i$’s, $B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })\subseteq B_{\tilde{\delta }_i(\delta _{i})}(\tilde{\mathbf {x}})\times B^i_{\tilde{\delta }_i(\delta _{i})}(\mathbf {x}^{\prime })$ for all $i\in 1:d$. Therefore, using (6)–(8), we have, for all $i\in 1:d$,

$$\begin{aligned} \bar{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})-\underline{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})&=F_{K_i}(x^{\prime }_i+\delta _i|\bar{\mathbf {x}}^i, \bar{y}^i_{1:i-1})-F_{K_i}(x^{\prime }_i-\delta _i|\underline{\mathbf {x}}^i, \underline{y}^i_{1:i-1})\\&\ge F_{K_i}\left( x^{\prime }_i+\delta _i|\tilde{\mathbf {x}}, x^{\prime }_{1:i-1}\right) -F_{K_i}\left( x^{\prime }_i-\delta _i|\tilde{\mathbf {x}}, x^{\prime }_{1:i-1}\right) -0.5\tilde{K}\delta _i\\&\ge 0.5\tilde{K}\delta _i\\&>0. \end{aligned}$$

Consequently, for all $i\in 1:d$ and for all $(\mathbf {x},y_{1:i-1})\in B_{\delta _{1}}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })$,

$$\begin{aligned} \Big [\underline{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i}), \bar{v}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})\Big ]\subseteq \Big [F_{K_i}(x^{\prime }_i-\delta _i|\mathbf {x}, y_{1:i-1}),F_{K_i}(x^{\prime }_i+\delta _i|\mathbf {x}, y_{1:i-1})\Big ]. \end{aligned}$$

Let $\underline{S}_{\delta }=0.5\tilde{K}\delta _1$. Then, this shows that there exists a closed hypercube $\underline{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )$ of side $\underline{S}_{\delta }$ such that

$$\begin{aligned} \underline{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )\subseteq F_K\big (\mathbf {x},B_{\delta _{1:d}}(\mathbf {x}^{\prime })\big )\subseteq F_K\big (\mathbf {x},B_{\delta }(\mathbf {x}^{\prime })\big ),\quad \forall \mathbf {x}\in B_{v_K(\delta )}(\tilde{\mathbf {x}}) \end{aligned}$$

where we set $v_K(\delta )=\delta _1$. Note that $v_K(\delta )\in (0,\delta ]$ and thus $v_K(\delta )\rightarrow 0$ as $\delta \rightarrow 0$, as required. In addition, $v_K(\cdot )=\delta _1\circ \dots \delta _d(\cdot )$ is continuous and strictly increasing on $(0,\delta _{\mathcal {X}}]$ because the functions $\delta _i(\cdot )$, $i\in 1:d$, are continuous and strictly increasing on this set. Note also that $v_K(\cdot )$ does not depend on $(\tilde{\mathbf {x}},\mathbf {x}^{\prime })\in \mathcal {X}^2$.

To conclude the proof, let

$$\begin{aligned} k_{\delta }=\big \lceil t+d-d\log (\underline{S}_{\delta }/3)/\log b\big \rceil \end{aligned}$$

(9)

and note that, if $\delta$ is small enough, $k_{\delta }\ge t+d$ because $\underline{S}_{\delta }\rightarrow 0$ as $\delta \rightarrow 0$. Let $\bar{\delta }_K$ be the largest value of $\delta ^{\prime }\le \delta _{\mathcal {X}}$ such that $k_{\delta ^{\prime }}\ge t+d$. Let $\delta \in (0,\bar{\delta }_K]$ and $t_{\delta ,d}\in t:(t+d)$ be such that $(k_{\delta }-t_{\delta ,d})/d$ is an integer. Let $\{E(j,\delta )\}_{j=1}^{b^{k_{\delta }-t_{\delta ,d}}}$ be the partition of $[0,1)^d$ into elementary intervals of volume $b^{t_{\delta ,d}-k_{\delta }}$ so that any closed hypercube of side $\underline{S}_{\delta }$ contains at least one elementary interval $E(j,\delta )$ for a $j\in 1:b^{k_{\delta }-t_{\delta ,d}}$. Hence, there exists a $j_{\tilde{\mathbf {x}},\mathbf {x}^{\prime }}\in 1:b^{k_{\delta }-t_{\delta ,d}}$ such that

$$\begin{aligned} E(j_{\tilde{\mathbf {x}},\mathbf {x}^{\prime },},\delta )\subseteq \underline{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )\subseteq F_K(\mathbf {x},B_{\delta }(\mathbf {x}^{\prime })),\quad \forall \mathbf {x}\in B_{v(\delta )}(\tilde{\mathbf {x}}). \end{aligned}$$

Let $a\in \mathbb {N}$ and note that, by the properties of (t, s)-sequences in base b, the point set $\{\mathbf {u}^n\}_{n=ab^{k_{\delta }}}^{(a+1)b^{k_{\delta }}-1}$ is a $(t,k_{\delta },d)$-net in base b because $k_{\delta }> t$. In addition, since $k_{\delta }\ge t_{\delta ,d}\ge t$, the point set $\{\mathbf {u}^n\}_{n=ab^{k_{\delta }}}^{(a+1)b^{k_{\delta }}-1}$ is also a $(t_{\delta ,d},k_{\delta },d)$-net in base b ([34] Remark 4.3, p. 48). Thus, since for $j\in 1:b^{k_{\delta }-t_{\delta ,d}}$ the elementary interval $E(j,\delta )$ has volume $b^{t_{\delta ,d}-k_{\delta }}$, the point set $\{\mathbf {u}^n\}_{n=ab^{k_{\delta }}}^{(a+1)b^{k_{\delta }}-1}$ therefore contains exactly $b^{t_{\delta _d}}\ge b^t$ points in $E(j_{\tilde{\mathbf {x}},\mathbf {x}^{\prime },},\delta )$ and the proof is complete.

Appendix 2: Proof of Lemma 2

Using the Lipschitz property of $F_{K_i}(\cdot |\cdot )$ for all $i\in 1:d$, conditions (7) and (8) in the proof of Lemma 1 hold with $\tilde{\delta }_i(\delta )=\delta (0.25\tilde{K}/C_K)$, $i\in 1:d$. Hence, we can take $v_K(\delta )=\delta (0.25\tilde{K}/C_K)^{d}\wedge \delta$ and thus $\underline{S}_{\delta }= \delta 0.5\tilde{K}\big (1 \wedge (0.25\tilde{K}/C_K)^{d}\big )$. Then, the expression for $k_{\delta }$ follows using (9) while the expression for $\bar{\delta }_{K}\le 0.5$ results from the condition $k_{\delta }\ge t+d$ for all $\delta \in (0,\bar{\delta }_{K}]$.

Appendix 3: Proof of Lemma 3

We first state and prove three technical Lemmas:

Lemma 6

Let $\mathcal {X}=[0,1]^d$ and $K:\mathcal {X}\rightarrow \mathcal {P}(\mathcal {X})$ be a Markov kernel which verifies Assumptions (A1)-(A2). Then, for any $\delta \in (0,\bar{\delta }_K]$, with $\bar{\delta }_K$ as in Lemma 1, and any $(\tilde{\mathbf {x}},\mathbf {x}^{\prime })\in \mathcal {X}^2$, there exists a closed hypercube $\bar{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )\subset [0,1)^d$ of side $\bar{S}_{\delta }=2.5\bar{K}\delta$, with $\bar{K}=\max _{i\in 1:d}\{\sup _{\mathbf {x},\mathbf {y}\in \mathcal {X}}K_i(y_i|y_{1:i-1},\mathbf {x})\}$, such that

$$\begin{aligned} F_K(\mathbf {x},B_{v_K(\delta )}(\mathbf {x}^{\prime }))\subseteq \bar{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta ),\quad \forall \mathbf {x}\in B_{v_K(\delta )}(\tilde{\mathbf {x}}) \end{aligned}$$

(10)

where $v_K(\cdot )$ is as in Lemma 1.

Proof

The proof of Lemma 6 is similar to the proof of Lemma 1. Below, we use the same notation as in this latter.

Let $\delta \in (0,\bar{\delta }_K]$, $(\tilde{\mathbf {x}},\,\mathbf {x}^{\prime })\in \mathcal {X}^2$ and note that, for any $(\mathbf {x},\mathbf {y})\in \mathcal {X}^2$,

$$\begin{aligned} F_{K_i}(x^{\prime }_i+\delta |\mathbf {x},y_{1:i-1})-F_{K_i}(x^{\prime }_i-\delta |\mathbf {x},y_{1:i-1})\le 2\bar{K}\delta ,\quad i\in 1:d. \end{aligned}$$

(11)

Let $0<\delta _1\le \dots \le \delta _d\le \delta$ be as in the proof of Lemma 1 and, for $i\in 1:d$, define

$$\begin{aligned} \underline{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})=\inf _{(\mathbf {x},\,\mathbf {y})\in B_{v_K(\delta )}(\tilde{\mathbf {x}}),\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })}F_{K_i}(x^{\prime }_i-\delta _i|\mathbf {x},y_{1:i-1}) \end{aligned}$$

and

$$\begin{aligned} \bar{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime }, \delta _{1:i})=\sup _{(\mathbf {x},\,\mathbf {y})\in B_{v_K(\delta )}(\tilde{\mathbf {x}}),\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })}F_{K_i}(x^{\prime }_i+\delta _i|\mathbf {x},y_{1:i-1}). \end{aligned}$$

Let $i\in 1:d$ and $(\underline{\mathbf {x}}^i,\underline{\mathbf {y}}^i),(\bar{\mathbf {x}}^i,\bar{\mathbf {y}}^i)\in B_{v_K(\delta )}(\tilde{\mathbf {x}})\times B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })$ be such that

$$\begin{aligned} \underline{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})=F_{K_i}(x^{\prime }_i-\delta _i|\underline{\mathbf {x}}^i,\underline{y}^i_{1:i-1}),\quad \bar{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )=F_{K_i}(x^{\prime }_i+\delta _i|\bar{\mathbf {x}}^i,\bar{y}^i_{1:i-1}). \end{aligned}$$

Therefore, using (7), (8) and (10), we have, $\forall i\in 1:d$,

$$\begin{aligned} 0<\bar{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})-\underline{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})&=F_{K_i}(x^{\prime }_i+\delta _i|\bar{\mathbf {x}}^i,\bar{y}^i_{1:i-1})-F_{K_i}(x^{\prime }_i-\delta _i|\underline{\mathbf {x}}^i,\underline{y}^i_{1:i-1})\\&\le F_{K_i}(x^{\prime }_i\!+\!\delta _i|\tilde{\mathbf {x}},x^{\prime }_{1:i-1})\!-\!F_{K_i}(x^{\prime }_i-\delta _i|\tilde{\mathbf {x}},x^{\prime }_{1:i-1})+0.5\tilde{K}\delta _i\\&\le \delta _i(2\bar{K}+0.5\tilde{K})\\&\le 2.5\delta _i\bar{K} \end{aligned}$$

where $\tilde{K}\le \bar{K}$ is as in the proof of Lemma 1. (Note that $\bar{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})-\underline{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta _{1:i})$ is indeed strictly positive because $F_{K_i}(\cdot |\mathbf {x},y_{1:i-1},)$ is strictly increasing on [0, 1] for any $(\mathbf {x},\mathbf {y})\in \mathcal {X}^2$ and because $\delta _i>0$.)

This shows that for all $\mathbf {x}\in B_{v_K(\delta )}(\tilde{\mathbf {x}})$ and for all $\mathbf {y}\in B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })$, we have

$$\begin{aligned} \big [F_{K_i}(x^{\prime }_i-\delta _i|\mathbf {x},\mathbf {y}),F_{K_i}(x^{\prime }_i+\delta _i|\mathbf {x},\mathbf {y})\big ]\subseteq \big [\underline{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta ),\bar{u}_i(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )\big ],\quad \forall i\in 1:d \end{aligned}$$

and thus there exists a closed hypercube $\bar{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta )$ of side $\bar{S}_{\delta }=2.5\delta \bar{K}$ such that

$$\begin{aligned} F_K(\mathbf {x},B_{\delta _{1:i-1}}(\mathbf {x}^{\prime }))\subseteq \bar{W}(\tilde{\mathbf {x}},\mathbf {x}^{\prime },\delta ),\quad \forall \mathbf {x}\in B_{v_K(\delta )}(\tilde{\mathbf {x}}). \end{aligned}$$

To conclude the proof of Lemma 6, note that, because $v_K(\delta )\le \delta _i$ for all $i\in 1:d$,

$$\begin{aligned} F_K(\mathbf {x},B_{v_K(\delta )}(\mathbf {x}^{\prime }))\subseteq F_K(\mathbf {x},B_{\delta _{1:i-1}}(\mathbf {x}^{\prime })). \end{aligned}$$

Lemma 7

Consider the set-up of Lemma 3 and, for $(p,a,k)\in \mathbb {N}_+^3$, let

$$\begin{aligned} E^{p}_{a,k}&=\Big \{\exists n\in \{ab^{k},\dots ,(a+1)b^{k}-1\}:\ \mathbf {x}^{n}\ne \mathbf {x}^{ab^{k}-1},\,\,\varphi (\mathbf {x}^{ab^k-1})<\varphi ^*\Big \}\\&\cap \Big \{\forall n\in \{ab^{k},\dots ,(a+1)b^{k}-1\}:\, \mathbf {x}^{n} \in (\mathcal {X}_{\varphi (\mathbf {x}^{ab^k-1})})_{2^{-p}}\Big \}. \end{aligned}$$

Then, for all $k\in \mathbb {N}$, there exists a $p^*_k\in \mathbb {N}$ such that $\mathrm {Pr}\big (\bigcap _{a\in \mathbb {N}}E^{p}_{a,k}\big )=0$ for all $p\ge p^*_k$.

Proof

Let $\epsilon >0$, $a\in \mathbb {N}$ and $l\in \mathbb {R}$ be such that $l<\varphi ^*$, and for $k\in \mathbb {N}$, let $E(k)=\{E(j,k)\}_{j=1}^{k^d}$ be the splitting of $\mathcal {X}$ into closed hypercubes of volume $k^{-d}$.

Let $p^{\prime }\in \mathbb {N}_+$, $\delta =2^{-p^{\prime }}$ and $P^l_{\epsilon ,\delta }\subseteq E(\delta )$ be the smallest coverage of $(\mathcal {X}_{l})_{\epsilon }$ by hypercubes in $E(\delta )$; that is, $|P^l_{\epsilon ,\delta }|$ is the smallest value in $1:\delta ^{-d}$ such that $(\mathcal {X}_{l})_{\epsilon }\subseteq \cup _{W\in P^l_{\epsilon ,\delta }}$. Let $J^l_{\epsilon ,\delta }\subseteq 1:\delta ^{-d}$ be such that $j\in J^l_{\epsilon ,\delta }$ if and only if $E(j,\delta )\in P^l_{\epsilon ,\delta }$. We now bound $|J^l_{\epsilon ,\delta }|$ following the same idea as in [20].

By assumption, there exists a constant $\bar{M}<\infty$ independent of l such that $M(\mathcal {X}_{l})\le \bar{M}$. Hence, for any fixed $w>1$ there exists a $\epsilon ^*\in (0,1)$ (independent of l) such that $\lambda _d\big ((\mathcal {X}_{l})_{\epsilon }\big )\le w M(\mathcal {X}_{l})\epsilon \le w \bar{M}\epsilon$ for all $\epsilon \in (0,\epsilon ^*]$. Let $\epsilon =2^{-p}$, with $p\in \mathbb {N}$ such that $2^{-p}\le 0.5\epsilon ^*$, and take $\delta _{\epsilon }=2^{-p-1}$. Then, we have the inclusions $(\mathcal {X}_{l})_{\epsilon }\subseteq \cup _{W\in P^l_{\epsilon ,\delta _{\epsilon }}} \subseteq (\mathcal {X}_{l})_{2\epsilon }$ and therefore, since $2\epsilon \le \epsilon ^*$,

$$\begin{aligned} |J^l_{\epsilon ,\delta _{\epsilon }}|\le \frac{\lambda _d\big ((\mathcal {X}_{l})_{2\epsilon }\big )}{\lambda _d(E(j,\delta _{\epsilon }))}\le \frac{w \bar{M} (2\epsilon )^{d}}{\delta _{\epsilon }^{d}}\le \bar{C}\delta _{\epsilon }^{-(d-1)},\quad \bar{C}:=w \bar{M} 2^d \end{aligned}$$

(12)

where the right-hand side is independent of l.

Next, for $j\in J^l_{\epsilon ,\delta _{\epsilon }}$, let $\bar{\mathbf {x}}^j$ be the center of $E(j,\delta _{\epsilon })$ and $W^l(j,\delta _{\epsilon })=\cup _{j^{\prime }\in J^l_{\epsilon ,\delta _{\epsilon }}} \bar{W}(\bar{\mathbf {x}}^j,\bar{\mathbf {x}}^{j^{\prime }},\delta _{\epsilon })$, with $\bar{W}(\cdot ,\cdot ,\cdot )$ as in Lemma 6. Then, using this latter, a necessary condition to move at iteration $n+1$ of Algorithm 1 from a point $\mathbf {x}^{n}\in E(j_{n},\delta _{\epsilon })$, with $j_{n}\in J^{l}_{\epsilon ,\delta _{\epsilon }}$, to a point $\mathbf {x}^{n+1}\ne \mathbf {x}^{n}$ such that $\mathbf {x}^{n+1}\in E(j_{n+1},\delta _{\epsilon })$ for a $j_{n+1}\in J^{l}_{\epsilon ,\delta _{\epsilon }}$ is that $\mathbf {u}_R^{n+1}\in W^{l}(j_{n},\delta _{\epsilon })$.

Let $k^{\delta _{\epsilon }}$ be the largest integer such that (i) $b^{k}\le \bar{S}_{\delta _{\epsilon }}^{-d}b^t$, with $\bar{S}_{\delta _{\epsilon }}=2.5\bar{K}\delta _{\epsilon }$, $\bar{K}<\infty$, as in Lemma 6, and (ii) $(k-t)/d$ is a positive integer (if necessary reduce $\epsilon$ to fulfil this last condition). Let $E^{\prime }(\delta _{\epsilon })=\{E^{\prime }(k,\delta _{\epsilon })\}_{k=1}^{b^{k^{\delta _{\epsilon }}-t}}$ be the partition of $[0,1)^d$ into hypercubes of volume $b^{t-k^{\delta _{\epsilon }}}$. Then, for all $j\in J^l_{\epsilon ,\delta _{\epsilon }}$, $W^l(j,\delta _{\epsilon })$ is covered by at most $2^d|J^l_{\epsilon ,\delta _{\epsilon }}|$ hypercubes of $E^{\prime }(\delta _{\epsilon })$.

Let $\epsilon$ be small enough so that $k^{\delta _{\epsilon }}>t+dR$. Then, using the properties of $(t,s)_R$-sequences (see Section 3.1), it is easily checked that, for all $n\ge 0$,

$$\begin{aligned} \mathrm {Pr}\big (\mathbf {u}_R^{n}\in E^{\prime }(k,\delta _{\epsilon })\big )\le b^{t-k^{\delta ^{\epsilon }}+dR},\quad \forall k\in 1:b^{k^{\delta _{\epsilon }}-t}. \end{aligned}$$

(13)

Thus, using (12)–(13) and the definition of $k^{\delta _{\epsilon }}$, we obtain, for all $j\in J^l_{\epsilon ,\delta _{\epsilon }}$ and $n\ge 0$,

$$\begin{aligned} \mathrm {Pr}\big (\mathbf {u}_R^{n}\in W^l(j,\delta _{\epsilon })\big )\le 2^d|J^l_{\epsilon ,\delta _{\epsilon }}|b^tb^{t-k^{\delta ^{\epsilon }}+dR}\le C^*\delta _\epsilon ,\quad C^*=2^d\bar{C}b^{t+1}(2.5\bar{K})^{d}b^{dR}. \end{aligned}$$

Consequently, using the definition of $\epsilon$ and $\delta _{\epsilon }$, and the fact that there exist at most $2^d$ values of $j\in J^l_{\epsilon ,\delta _{\epsilon }}$ such that, for $n\in \mathbb {N}$, we have $\mathbf {x}^{n}\in E(j,\delta _{\epsilon })$, we deduce that, for a $p^*\in \mathbb {N}$ large enough (i.e. for $\epsilon =2^{-p^*}$ small enough)

$$\begin{aligned} \mathrm {Pr}\big (E^{p}_{a,k}|\, \varphi (\mathbf {x}^{ab^k-1})=l\big )\le b^{k} 2^dC^* 2^{-p-1},\quad \forall (a,k)\in \mathbb {N}^2,\quad \forall l<\varphi ^*,\quad \forall p\ge p^* \end{aligned}$$

implying that, for $p\ge p^*$,

$$\begin{aligned} \mathrm {Pr}\big (E^{p}_{a,k}\big )\le b^{k} 2^dC^* 2^{-p-1},\quad \forall (a,k)\in \mathbb {N}^2. \end{aligned}$$

Finally, because the uniform random numbers $\mathbf {z}^n$’s in $[0,1)^s$ that enter into the definition of $(t,s)_R$-sequences are IID, this shows that

$$\begin{aligned} \mathrm {Pr}\big (\cap _{j=a}^{a+m}E_{j,k}^p\big )\le (b^{k}2^dC^*2^{-p-1})^m,\quad \forall (a,m,k)\in \mathbb {N}^3,\quad \forall p\ge p^*. \end{aligned}$$

To conclude the proof, for $k\in \mathbb {N}$ let $\rho _k\in (0,1)$ and $p^*_k\ge p^*$ be such that

$$\begin{aligned} b^{k}2^dC^*2^{-p-1}\le \rho _k,\quad \forall p>p^*_k. \end{aligned}$$

Then, $\mathrm {Pr}\big (\cap _{a\in \mathbb {N}} E^p_{a,k})=0$ for all $p\ge p^*_k$, as required.

Lemma 8

Consider the set-up of Lemma 3. For $k\in \mathbb {N}$, let $\tilde{E}(dk)=\{\tilde{E}(j,k)\}_{j=1}^{b^{dk}}$ be the partition of $[0,1)^d$ into hypercubes of volume $b^{-dk}$. Let $k^{R} \in (dR+t):(dR+t+d)$ be the smallest integer k such $(k-t)/d$ is an integer and such that $(k-t)/d\ge R$ and, for $m\in \mathbb {N}$, let $I_m=\{mb^{k^R},\dots ,(m+1)b^{k^R}-1\}$. Then, for any $\delta \in (0,\bar{\delta }_K]$ verifying $k_\delta > t+d+dR$ (with $\bar{\delta }_K$ and $k_\delta$ as in Lemma 1), there exists a $p(\delta )>0$ such that

$$\begin{aligned} \mathrm {Pr}\big (\exists n\in I_m:\,\, \mathbf {u}_R^{n}\in \tilde{E}(j, k_{\delta }-t_{\delta ,d})\big )\ge p(\delta ),\quad \forall j\in 1: b^{ k_{\delta }-t_{\delta ,d}},\quad \forall m\in \mathbb {N} \end{aligned}$$

where $t_{\delta ,d}\in t:(t+d)$ is such that $(k_{\delta }-t_{\delta ,d})/d\in \mathbb {N}$.

Proof

Let $m\in \mathbb {N}$ and note that, by the properties of $(t,s)_R$-sequence, the point set $\{\mathbf {u}_{\infty }^{n}\}_{n\in I_m}$ is a $(t,{k^R},d)$-net in base b. Thus, for all $j\in 1:b^{k^R-t}$, this point set contains $b^t$ points in $\tilde{E}(j, k^{R}-t)$ and, consequently, for all $j\in 1:b^{dR}$, it contains $b^tb^{k^R-t-dR}=b^{k^R-dR}\ge 1$ points in $\tilde{E}(j,dR)$. This implies that, for all $j\in 1:b^{dR}$, the point set $\{\mathbf {u}_R^{n}\}_{n\in I_m}$ contains $b^{k^R-dR}\ge 1$ points in $\tilde{E}(j,dR)$ where, for all $n\in I_{m_i}$, $\mathbf {u}_R^{n}$ is uniformly distributed in $\tilde{E}(j_{n},dR)$ for a $j_n\in 1:b^{dR}$.

In addition, it is easily checked that each hypercube of the set $\tilde{E}(dR)$ contains

$$\begin{aligned} b^{k_{\delta }-t_{\delta ,d}-dR}\ge b^{k_{\delta }-t-d-dR}> 1 \end{aligned}$$

hypercubes of the set $\tilde{E}(k_{\delta }-t_{\delta ,d})$, where $k_{\delta }$ and $t_{\delta ,d}$ are as in the statement of the lemma. Note that the last inequality holds because $\delta$ is chosen so that $k_{\delta }> t+d+dR$. Consequently,

$$\begin{aligned} \mathrm {Pr}\big (\exists n\in I_m:\,\, \mathbf {u}_R^n\in \tilde{E}(j, k_{\delta }-t_{\delta ,d})\big )\ge p(\delta ):=b^{dR+t_{\delta ,d}-k_{\delta }}>0,\quad \forall j\in 1: b^{ k_{\delta }-t_{\delta ,d}} \end{aligned}$$

and the proof is complete.

Proof of Lemma 3:

To prove the lemma we need to introduce some additional notation. Let $\varOmega =[0,1)^{\mathbb {N}}$, $\mathcal {B}([0,1))$ be the Borel $\sigma$-algebra on $[0,1)$, $\mathcal {F}= \mathcal {B}([0,1))^{\otimes \mathbb {N}}$ and $\mathbb {P}$ be the probability measure on $(\varOmega ,\mathcal {F})$ defined by

$$\begin{aligned} \mathbb {P}(A)=\prod _{i\in \mathbb {N}}\lambda _1(A_i),\quad (A_1,\dots ,A_i\dots )\in \mathcal {B}([0,1))^{\otimes \mathbb {N}}. \end{aligned}$$

Next, for $\omega \in \varOmega$, we denote by $\big (\mathbf {U}_R^n(\omega )\big )_{n\ge 0}$ the sequence of points in $[0,1)^{d}$ defined, for all $n\ge 0$, by (using the convention that empty sums are null),

$$\begin{aligned} \mathbf {U}_R^n(\omega )=\big (U_{R,1}^n(\omega ),\dots ,U_{R,d}^n(\omega )),\quad U_{R,i}^n(\omega )=\sum _{k=1}^{R}a_{ki}^nb^{-k}+b^{-R}\omega _{n d+i},\quad i\in 1:s. \end{aligned}$$

Note that, under $\mathbb {P}$, $\big (\mathbf {U}_R^n\big )_{n\ge 0}$ is a $(t,d)_R$-sequence in base b. Finally, for $\omega \in \varOmega$, we denote by $\big (\mathbf {x}^n_\omega \big )_{n\ge 0}$ the sequence of points in $\mathcal {X}$ generated by Algorithm 1 when the sequence $\big (\mathbf {U}_R^n(\omega )\big )_{n\ge 0}$ is used as input.

Under the assumptions of the lemma there exists a set $\varOmega _1\in \mathcal {F}$ such that $\mathbb {P}(\varOmega _1)=1$ and

$$\begin{aligned} \exists \bar{\varphi }_\omega \in \mathbb {R}\text { such that }\lim _{n\rightarrow \infty }\varphi \big (\mathbf {x}^n_\omega \big )=\bar{\varphi }_\omega ,\quad \forall \omega \in \varOmega _1. \end{aligned}$$

Let $\omega \in \varOmega _1$. Since $\varphi$ is continuous, for any $\epsilon >0$ there exists a $N_{\omega , \epsilon }\in \mathbb {N}$ such that $\mathbf {x}^n_\omega \in (\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }$ for all $n\ge N_{\omega ,\epsilon }$, where we recall that $(\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }=\{\mathbf {x}\in \mathcal {X}:\exists \mathbf {x}^{\prime }\in \mathcal {X}_{\bar{\varphi }_\omega }\text { such that } \Vert \mathbf {x}-\mathbf {x}^{\prime }\Vert _{\infty }\le \epsilon \}$. In addition, because $\varphi$ is continuous and $\mathcal {X}$ is compact, there exists an integer $p_{\omega , \epsilon }\in \mathbb {N}$ such that we have both $\lim _{\epsilon \rightarrow 0}p_{\omega ,\epsilon }=\infty$ and

$$\begin{aligned} (\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }\subseteq (\mathcal {X}_{\varphi (x^{\prime })})_{2^{-p_{\omega ,\epsilon }}},\quad \forall x^{\prime }\in (\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }. \end{aligned}$$

(14)

Next, let $\mathbf {x}^*\in \mathcal {X}$ be such that $\varphi (\mathbf {x}^*)=\varphi ^*$, $k^R \in (dR+t):(dR+t+d)$ be as in Lemma 8 and, for $(p,a,k)\in \mathbb {N}_+^3$, let

$$\begin{aligned} \tilde{E}^{p}_{a,k}&=\Big \{\omega \in \varOmega :\,\exists n\in \{ab^{k},\dots ,(a+1)b^{k}-1\}:\ \mathbf {x}_{\omega }^{n}\ne \mathbf {x}^{ab^{k}-1}_{\omega },\,\varphi (\mathbf {x}_{\omega }^{ab^k-1})<\varphi ^*\Big \}\\&\cap \Big \{\omega \in \varOmega :\,\forall n\in \{ab^{k},\dots ,(a+1)b^{k}-1\}:\mathbf {x}^{n}_{\omega }\in \big (\mathcal {X}_{\varphi (\mathbf {x}^{ab^k-1}_{\omega })}\big )_{2^{-p}}\Big \}. \end{aligned}$$

Then, by Lemma 7, there exists a $p^*\in \mathbb {N}$ such that $\mathbb {P}\big (\cap _{a\in \mathbb {N}} \tilde{E}^p_{a,k^R}\big )=0$ for all $p\ge p^*$, and thus the set $\tilde{\varOmega }_2=\cap _{p\ge p^*}\big (\mathcal {X}\setminus \cap _{a\in \mathbb {N}} \tilde{E}^p_{a,k^R}\big )$ verifies $\mathbb {P}(\tilde{\varOmega }_2)=1$. Let $\varOmega _2=\varOmega _1\cap \tilde{\varOmega }_2$ so that $\mathbb {P}(\varOmega _2)=1$.

For $\omega \in \varOmega _2$ let $\epsilon _\omega >0$ be small enough so that, for any $\epsilon \in (0,\epsilon _\omega ]$, we can take $p_{\omega ,\epsilon }\ge p^*$ in (14). Then, for any $\omega \in \varOmega _2$ such that $\bar{\varphi }_\omega <\varphi ^*$, there exists a subsequence $(m_i)_{i\ge 1}$ of $(m)_{m\ge 1}$ such that, for all $i\ge 1$, either

$$\begin{aligned} \mathbf {x}^{n}_\omega =\mathbf {x}^{m_ib^{k^R}-1}_\omega ,\quad \forall n\in I_{m_i}:=\big \{m_i b^{k^R},\dots , (m_i+1)b^{k^R}-1\big \} \end{aligned}$$

or

$$\begin{aligned} \exists n\in I_{m_i}\text { such that }\mathbf {x}^{n}_\omega \not \in \Big (\mathcal {X}_{\varphi (\mathbf {x}^{m_ib^{k^R}-1}_\omega )}\Big )_{2^{-p_{\omega ,\epsilon }}}. \end{aligned}$$

(15)

Assume first that there exist infinitely many $i\in \mathbb {N}$ such that (15) holds. Then, by (14), this leads to a contradiction with the fact that $\omega \in \varOmega _2\subseteq \varOmega _1$. Therefore, for any $\omega \in \varOmega _2$ such that $\bar{\varphi }_{\omega }<\varphi ^*$ there exists a subsequence $(m_i)_{i\ge 1}$ of $(m)_{m\ge 1}$ such that, for a $i^*$ large enough,

$$\begin{aligned} \mathbf {x}_\omega ^{n}=\mathbf {x}_\omega ^{m_ib^{k^R}-1},\quad \forall n\in I_{m_i},\quad \forall i\ge i^*. \end{aligned}$$

(16)

Let $\tilde{\varOmega }_2=\{\omega \in \varOmega _2:\,\bar{\varphi }_{\omega }<\varphi ^*\}\subseteq \varOmega _2$ . Then, to conclude the proof, it remains to show that $\mathbb {P}(\tilde{\varOmega }_2)=0$. We prove this result by contradiction and thus, from henceforth, we assume $\mathbb {P}(\tilde{\varOmega }_2)>0$.

To this end, let $\mathbf {x}^*\in \mathcal {X}$ be such that $\varphi (\mathbf {x}^*)=\varphi ^*$, $\mathbf {x}\in \mathcal {X}$ and $\delta \in (0,\bar{\delta }_K]$, with $\bar{\delta }_K$ as in Lemma 1. Then, using this latter, a sufficient condition to have $F_K^{-1}(\mathbf {x},\mathbf {U}_R^{n}(\omega ))\in B_{\delta }(\mathbf {x}^*)$, $n\ge 1$, is that $\mathbf {U}_R^{n}(\omega )\in \underline{W}(\mathbf {x},\mathbf {x}^*,\delta )$, with $\underline{W}(\cdot ,\cdot ,\cdot )$ as in Lemma 1. From the proof of this latter we know that the hypercube $\underline{W}(\mathbf {x},\mathbf {x}^*,\delta )$ contains at least one hypercube of the set $\tilde{E}(k_{\delta }-t_{\delta ,d})$, where $t_{\delta ,d}\in t:(t+d)$ is such that $(k_{\delta }-t_{\delta ,d})/d\in \mathbb {N}$ and, for $k\in \mathbb {N}$, $\tilde{E}(dk)$ is as in Lemma 8. Hence, by this latter, for any $\delta \in (0,\delta ^*]$, with $\delta ^*$ such that $k_{\delta ^*}>t+d+dR$ (where, for $\delta >0$, $k_\delta$ is defined in Lemma 1), there exists a $p(\delta )>0$ such that

$$\begin{aligned} \mathbb {P}\Big (\omega \in \varOmega :\, \exists n\in I_m,\, F_K^{-1}\big (\mathbf {x}, \mathbf {U}_R^n(\omega )\big )\in B_{\delta }(\mathbf {x}^*)\Big )\ge p(\delta ),\quad \forall (\mathbf {x},m)\in \mathcal {X}\times \mathbb {N} \end{aligned}$$

and thus, using (16) and under Assumption (A2), it is easily checked that, for any $\delta \in (0,\delta ^*]$,

$$\begin{aligned} \mathbb {P}_2\Big (\omega \in \tilde{\varOmega }_2: F_K^{-1}\big (\mathbf {x}_{\omega }^{n-1}, \mathbf {U}_R^n(\omega )\big )\in B_{\delta }(\mathbf {x}^*)\text { for infinitly many }n\in \mathbb {N}\Big )=1 \end{aligned}$$

where $\mathbb {P}_2$ denotes the restriction of $\mathbb {P}$ on $\tilde{\varOmega }_2$ (recall that we assume $\mathbb {P}(\tilde{\varOmega }_2)>0$).

For $\delta >0$, let

$$\begin{aligned} \varOmega ^{\prime }_{\delta }=\Big \{\omega \in \tilde{\varOmega }_2:\,\, F_K^{-1}(\mathbf {x}_{\omega }^{n-1}, \mathbf {U}_R^n(\omega ))\in B_{\delta }(\mathbf {x}^*)\text { for infinitly many }n\in \mathbb {N}\Big \} \end{aligned}$$

and let $\tilde{p}^*\in \mathbb {N}$ be such that $2^{-\tilde{p}^*}\le \delta ^*$. Then, the set $\varOmega ^{\prime }=\cap _{\tilde{p}\ge \tilde{p}^*}\varOmega ^{\prime }_{2^{-\tilde{p}}}$ verifies $\mathbb {P}_2(\varOmega ^{\prime })=1$.

To conclude the proof let $\omega \in \varOmega ^{\prime }$. Then, because $\varphi$ is continuous and $\bar{\varphi }_\omega <\varphi ^*$, there exists a $\tilde{\delta }_{\bar{\varphi }_\omega }>0$ so that $\varphi (\mathbf {x})>\bar{\varphi }$ for all $\mathbf {x}\in B_{\tilde{\delta }_{\bar{\varphi }_\omega }}(\mathbf {x}^*)$. Let $\delta _{\bar{\varphi }_\omega }:=2^{-\tilde{p}_{\omega ,\epsilon }}\ge \tilde{\delta }_{\bar{\varphi }_\omega }\wedge \bar{\delta }_K$ for an integer $\tilde{p}_{\omega ,\epsilon }\ge \tilde{p}^*$. Next, take $\epsilon$ small enough so that we have both $B_{\delta _{\bar{\varphi }_\omega }}(\mathbf {x}^*)\cap (\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }=\varnothing$ and $\varphi (\mathbf {x})\ge \varphi (\mathbf {x}^{\prime })$ for all $(\mathbf {x},\mathbf {x}^{\prime })\in B_{\delta _{\bar{\varphi }_\omega }}(\mathbf {x}^*)\times (\mathcal {X}_{\bar{\varphi }_\omega })_{\epsilon }$.

Using above computations, the set $B_{\tilde{\delta }_{\bar{\varphi }_\omega }}(\mathbf {x}^*)$ is visited infinitely many time and thus $\varphi (\mathbf {x}_\omega ^n)>\bar{\varphi }_\omega$ for infinitely many $n\in \mathbb {N}$, contradicting the fact that $\varphi (\mathbf {x}_\omega ^n)\rightarrow \bar{\varphi }_\omega$ as $n\rightarrow \infty$. Hence, the set $\varOmega ^{\prime }$ is empty. On the other hand, as shown above, under the assumption $\mathbb {P}(\tilde{\varOmega }_2)>0$ we have $\mathbb {P}_2(\varOmega ^{\prime })=1$ and, consequently, $\varOmega ^{\prime }\ne \varnothing$. Therefore, we must have $\mathbb {P}(\tilde{\varOmega }_2)=0$ and the proof is complete.

Appendix 4: Proof of Theorem 2

Using Lemmas 4 and 5, we know that $\varphi (x^n)\rightarrow \bar{\varphi }\in \mathbb {R}$ and thus it remains to show that $\bar{\varphi }=\varphi ^*$.

Assume that $\bar{\varphi }\ne \varphi ^*$ and, for $\epsilon =2^{-p}$, $p\in \mathbb {N}_+$, let $N_\epsilon \in \mathbb {N}$, $p_\epsilon$ and $\delta _{\bar{\varphi }}>0$ be as in the proof of Lemma 3 (with the dependence of $N_\epsilon$, $p_\epsilon$ and of $\delta _{\bar{\varphi }}$ on $\omega \in \varOmega$ suppressed in the notation for obvious reasons).

Let $x^*\in \mathcal {X}$ be a global maximizer of $\varphi$ and $n=a_nb^{k_{\delta _{\bar{\varphi }}}}-1$ with $a_n\in \mathbb {N}$ such that $n>N_{\epsilon }$. For $k\in \mathbb {N}$, let $E(k)=\{E(j,k)\}_{j=1}^{k}$ be the splitting of [0, 1] into closed hypercubes of volume $k^{-1}$. Then, by Lemma 6, a necessary condition to have a move at iteration $n^{\prime }+1\ge 1$ of Algorithm 1 from $x^{n^{\prime }}\in (\mathcal {X}_{\bar{\varphi }})_{\epsilon }$ to $x^{n^{\prime }+1}\ne x^{n^{\prime }}$, $x^{n^{\prime }+1}\in (\mathcal {X}_{\bar{\varphi }})_{\epsilon }$ is that

$$\begin{aligned} u_{\infty }^{n^{\prime }}\in \bar{W}(\epsilon ):=\bigcup _{j,j^{\prime }\in J^{\bar{\varphi }}_{\epsilon , \epsilon /2}} \bar{W}(\bar{x}^j,\bar{x}^{j^{\prime }},\epsilon /2) \end{aligned}$$

where, for $j\in 1:(\epsilon /2)^{-d}$, $\bar{x}^j$ denotes the center of $E(j,\epsilon /2)$, $J^{\bar{\varphi }}_{\epsilon , \epsilon /2}$ is as in the proof of Lemma 7 and $\bar{W}(\cdot ,\cdot ,\cdot )$ is as in Lemma 6. Note that, using (12) with $d=1$, $|J^{\bar{\varphi }}_{\epsilon , \epsilon /2}|\le C^*$ for a constant $C^*<\infty$ (independent of $\epsilon$).

Let $b^{k^{\delta _{\epsilon }}}$ be the largest integer $k\ge t$ such that $b^{t-k}\ge \bar{S}_{\epsilon /2}^{d}$, with $\bar{S}_{\epsilon /2}$ as in Lemma 6, and let $\epsilon$ be small enough so that $b^{k^{\delta _{\epsilon }}}>2^dC^*b^t$. The point set $\{u_{\infty }^{n^{\prime }}\}_{n^{\prime }=a_nb^{k^{\delta _{\epsilon }}}}^{ (a_n+1)b^{k^{\delta _{\epsilon }}}-1}$ is a $(t, k^{\delta _{\epsilon }},d)$-net in base b and thus the set $\bar{W}(\epsilon )$ contains at most $2^dC^* b^t$ points of this points set. Hence, if for $n>N_{\epsilon }$ only moves inside the set $(\mathcal {X}_{\bar{\varphi }})_{\epsilon }$ occur, then, for a $\tilde{n}\in a_nb^{k^{\delta _{\epsilon }}}:\big ((a_n+1)b^{k^{\delta _{\epsilon }}}-\eta _{\epsilon }-1)\big )$, the point set $\{x^{n^{\prime }}\}_{n^{\prime }=\tilde{n}}^{\tilde{n}+\eta _{\epsilon }}$ is such that $x^{n^{\prime }}=x^{\tilde{n}}$ for all $n\in \tilde{n}:(\tilde{n}+\eta _{\epsilon })$, where $\eta _{\epsilon }\ge \lfloor \frac{b^{k^{\delta _{\epsilon }}}}{2^dC^*{}^2b^t}\rfloor$; note that $\eta _{\epsilon }\rightarrow \infty$ as $\epsilon \rightarrow 0$.

Let $k^{\epsilon }_0$ be the largest integer which verifies $\eta _{\epsilon }\ge 2b^{k^{\epsilon }_0}$ so that $\{u_{\infty }^n\}_{n=\tilde{n}}^{\tilde{n}+\eta _{\epsilon }}$ contains at least one $(t,k^{\epsilon }_0,d)$-net in base b. Note that $k^{\epsilon }_0\rightarrow \infty$ as $\epsilon \rightarrow 0$, and let $\epsilon$ be small enough so that $k^{\epsilon }_0\ge k_{\delta _{\bar{\varphi }}}$, with $k_{\delta }$ as in Lemma 1. Then, by Lemma 1, there exists at least one $n^*\in (\tilde{n}+1):(\tilde{n}+\eta _{\epsilon })$ such that $\tilde{y}^{n^*}:=F_K^{-1}(x^{\tilde{n}},u_{\infty }^{n^*})\in B_{\delta _{\bar{\varphi }}}(x^*)$. Since, by the definition of $\delta _{\bar{\varphi }}$, for all $(x,x^{\prime })\in B_{\delta _{\bar{\varphi }}}(x^*)\times (\mathcal {X}_{\bar{\varphi }})_{\epsilon }$, and for $\epsilon$ small enough, we have $\varphi (x)>\varphi (x^{\prime })$, it follows that $\varphi (\tilde{y}^{n^*})>\varphi (x^{\tilde{n}})$. Hence, there exists at least one $n\in \tilde{n}:(\tilde{n}+\eta _{\epsilon })$ such that $x^n\ne x^{\tilde{n}}$, which contradicts the fact that $x^n=\tilde{x}$ for all $n\in \tilde{n}:(\tilde{n}+\eta _{\epsilon })$. This shows that $\bar{\varphi }$ is indeed the global maximum of $\varphi$.

Appendix 5: Additional figures for the example of Sect. 6.1

Fig. 3

Minimization of $\tilde{\varphi }_1$ defined by (2) for 1000 starting values sampled independently and uniformly on $\mathcal {X}_1$. Results are presented for the Cauchy kernel $(K_n^{(2)})_{n\ge 1}$ with $(T_n^{(1)})_{n\ge 1}$ (top plots) and for $(T_n^{(2)})_{n\ge 1}$. For $m=1,2$, simulations are done for the smallest (left plots) and the highest (right) value of $T_n^{(m)}$ given in (4). The plots show the minimum number of iterations needed for SA (white boxes) and QMC-SA to find a $\mathbf {x}\in \mathcal {X}_1$ such that $\tilde{\varphi }_1(\mathbf {x})<10^{-5}$. For each starting value, the Monte Carlo algorithm is run only once and the QMC-SA algorithm is based on the Sobol’ sequence

Fig. 4

Minimization of $\tilde{\varphi }_1$ defined by (2) for 1000 starting values sampled independently and uniformly on $\mathcal {X}_1$. Results are presented for the Gaussian kernel $(K_n^{(3)})_{n\ge 1}$ with $(T_n^{(1)})_{n\ge 1}$ (top plots) and for $(T_n^{(3)})_{n\ge 1}$. For $m=\{1,3\}$, simulations are done for the smallest (left plots) and the highest (right) value of $T_n^{(m)}$ given in (4). The plots show the minimum number of iterations needed for SA (white boxes) and QMC-SA to find a $\mathbf {x}\in \mathcal {X}_1$ such that $\tilde{\varphi }_1(\mathbf {x})<10^{-5}$. For each starting value, the Monte Carlo algorithm is run only once and the QMC-SA algorithm is based on the Sobol’ sequence