Convergence of Continuous Stochastic Processes on Compact Metric Spaces Converging in the Lipschitz Distance

Abstract

We introduce a new distance, a Lipschitz–Prokhorov distanced_LP, on the set $\mathcal {PM}$ of isomorphism classes of pairs (X, P) where X is a compact metric space and P is the law of a continuous stochastic process on X. We show that $(\mathcal {PM}, d_{LP})$ is a complete metric space. For Markov processes on Riemannian manifolds, we study relative compactness and convergence.

Appendix A

Recall that $\mathcal M$ is the set of isometry classes of compact metric spaces and d_L is the Lipschitz distance on $\mathcal M$ (see Section 2). We show the completeness of the metric space $(\mathcal M, d_{L})$.

Proposition A.1

$(\mathcal M, d_{L})$ is a complete metric space.

Proof

Let $\{X_{i}: i \in \mathbb {N}\}$be a d_L-Cauchysequence in $\mathcal M$.It suffices to show that there are a compact metric space$X \in \mathcal M$and ε_i-isometries f_i : X_i → X with ε_i → 0 as i →∞.

The construction of X: Let f_ij : X_i → X_j be an ε_ij-isometryfor i < j where ε_ij → 0 as i,j →∞. Take a subsequence such that ε_i,i+ 1 < 1/2ⁱ. Let ${\tilde f_{ij}}: X_{i} \to X_{j}$ be defined by

$$ {\tilde f_{ij}}=f_{j-1, j} \circ f_{j-2, j-1} \circ \cdot\cdot\cdot \circ f_{i,i + 1} \quad (i<j), $$

(1)

and ${\tilde \varepsilon _{ij}}={\sum }_{l=i}^{j-1}\varepsilon _{l,l + 1}$. Then${\tilde f_{ij}}$is an${\tilde \varepsilon _{ij}}$-isometryand ${\tilde \varepsilon _{ij}} \to 0$asi,j →∞.Since every compact metric space is separable, there is a countable dense subset$\{x^{1}_{\alpha }: \alpha \in \mathbb {N}\} \subset X_{1}$. We define, for any i > 1,

$$x_{\alpha}^{i}={\tilde f_{1i}}(x_{\alpha}^{1}).$$

Since ${\tilde f_{1i}}$is a homeomorphism, the subset $\{x_{\alpha }^{i}: i \in \mathbb {N}\}$is dense in X_i for each i. Fix $\alpha , \beta \in \mathbb {N}$, and consider the sequence of the real numbers

$$ \{d(x^{i}_{\alpha},x_{\beta}^{i}) : i\in \mathbb{N}\}. $$

(2)

Since $\{{\tilde f_{1i}}:i \in \mathbb {N}\}$has a bounded Lipschitz constant and the compact metric space X₁ is bounded, we have that Eq. 2 is a boundedsequence:

$$d(x^{i}_{\alpha},x_{\beta}^{i} )\le \sup_{i}\text{dil}(\tilde{f}_{1i})d(x^{1}_{\alpha},x^{1}_{\beta}) < \infty. $$

Thus we can take a subsequence of Eq. 2 converging to some real number, write r(α,β). We can check that r becomes a metric on $\{\alpha : \alpha \in \mathbb {N}\}$.In fact, if r(α,β) = 0,we have α = β because

$$ 0<\frac{1}{\sup_{i}\text{dil}({\tilde f_{1i}}^{-1})}d(x_{\alpha}^{1}, x_{\beta}^{1}) \le d(x_{\alpha}^{i}, x_{\beta}^{i}). $$

(3)

By definition, r(α,α) = 0and r is symmetric and non-negative. It is easy to see the triangle inequality. Let(X,d) be the completionof the metric space $(\{\alpha : \alpha \in \mathbb {N}\}, r)$. The compactness of (X,d) will be shown later in this proof.

The construction of ε_i-isometries f_i: We define a map $f_{i}: \{x_{\alpha }^{i}: \alpha \in \mathbb {N}\} \to X$by

$$f_{i}(x_{\alpha}^{i})=\alpha.$$

Now we extend the map f_i to the whole space X_i.Since $\text {dil}(\tilde {f}_{ij})$ is bounded, we have

$$\begin{array}{@{}rcl@{}} d(f_{i}(x_{\alpha}^{i}), f_{i}(x^{i}_{\beta}))&=& d(\alpha, \beta)=\lim_{j \to \infty}d(x_{\alpha}^{j},x_{\beta}^{j})=\lim_{j \to \infty}d({\tilde{f}_{ij}}(x_{\alpha}^{i}),{\tilde{f}_{ij}} (x_{\beta}^{i})) \\ &\le& \sup_{j}\text{dil}({\tilde f_{ij}})d(x_{\alpha}^{i}, x_{\beta}^{i}) < \infty. \end{array} $$

(4)

Let $x_{\alpha (n)}^{i} \to x^{i} \in X_{i}$asn →∞. By the inequality (4), we have that $\lim _{n \to \infty }f_{i}(x_{\alpha (n)}^{i})$ exists. This limit does not depend on the way of taking sequences converging to xⁱ(use the triangle inequality to check it). Thus we define

$$f_{i}(x^{i})=\lim_{n \to \infty}f_{i}(x_{\alpha(n)}^{i}),$$

and this is well-defined. Thus we have extended the map f_i to the whole space X_i.

Now we check that f_i is bi-Lipschitz. We have

$$\begin{array}{@{}rcl@{}} d(f_{i}(x_{\alpha}^{i}), f_{i}(x^{i}_{\beta}))&=& d(\alpha, \beta)=\lim_{j \to \infty}d(x_{\alpha}^{j},x_{\beta}^{j})=\lim_{j \to \infty}d({\tilde{f}_{ij}}(x_{\alpha}^{i}),{\tilde{f}_{ij}} (x_{\beta}^{i})) \\ &\ge& \frac{1}{\sup_{j}\text{dil}({\tilde f^{-1}_{ij}})}d(x_{\alpha}^{i}, x_{\beta}^{i}). \end{array} $$

(5)

Note that $0<\sup _{j}\text {dil}({\tilde f^{-1}_{ij}})<\infty$. By the inequality (5), we see that f_i is bijective. By the inequality (4) and (5), we have f_i is bi-Lipschitz. Since f_iisa homeomorphism and X_i is compact, we see that X = f_i(X_i)is compact. Thus $X \in \mathcal M$.

Finally we check that f_i is an ε_i-isometryfor some ε_i → 0 as i →∞. We set

$$\varepsilon_{i}=\max\{|\log(\sup_{j}\text{dil}({\tilde f_{ij}}^{-1}))|, |\log(\sup_{j}\text{dil}({\tilde f_{ij}}))|\}.$$

Then, by the inequality (4) and (5), we can see that f_i : X_i → X is an ε_i-isometry with ε_i → 0 as i →∞. Thus we have shown that X is the d_L-limitof X_i. We have completed the proof. □

Note that, in the above proof, we have that

$$\begin{array}{@{}rcl@{}} f_{j}\circ {\tilde f_{ij}}={f_{i}}. \end{array} $$

(6)

We use Eq. 6 in the proof of Theorem 2.9.

Remark A.2

Note that $(\mathcal M, d_{L})$ is not separable. This is because of the following two facts:

(a):

if d_L(X,Y ) < ∞,the Hausdorff dimensions of X and Y must coincide;

(b):

for any non-negative real number d, there is a compact metric space X whose Hausdorff dimension is equal to d.

See, e.g., [7, Proposition 1.7.19] for (a) and [17] for (b). Let$X \in \mathcal M$and$\mathcal M_{X}=\{Y \in \mathcal M: d_{L}(X,Y)<\infty \}$. We also note that there is a $X \in \mathcal M$ such that even when we restrict d_Lto $\mathcal M_{X}$, the metric space $(\mathcal M_{X}, d_{L})$ is not separable. See [22].