Stability and Approximation of Statistical Limit Laws for Multidimensional Piecewise Expanding Maps

Abstract

The unpredictability of chaotic nonlinear dynamics leads naturally to statistical descriptions, including probabilistic limit laws such as the central limit theorem and large deviation principle. A key tool in the Nagaev–Guivarc’h spectral method for establishing statistical limit theorems is a “twisted” transfer operator. In the abstract setting of Keller and Liverani (Annali della Scuola Normale Superiore di Pisa-Classe di Scienze, 28:141–152, 1999), we prove that derivatives of all orders of the leading eigenvalues and eigenprojections of the twisted transfer operators with respect to the twist parameter are stable when subjected to a broad class of perturbations. As a result, we demonstrate stability of the variance in the central limit theorem and the rate function from a large deviation principle with respect to deterministic and stochastic perturbations of the dynamics and perturbations induced by numerical schemes. We apply these results to piecewise expanding maps in one and multiple dimensions, including new convergence results for Ulam projections on quasi-Hölder spaces.

Appendices

The Proofs of Lemmas 5.12 and 5.13

Before discussing our strategy for proving Lemmas 5.12 and 5.13, we must discuss the relationship between the space $\mathbb {V}_{\beta }(\Omega )$ and the seminorm $\left|\cdot \right|_\beta$. It is noted in [28] that while $\mathbb {V}_{\beta }(\Omega )$ is independent of $\eta _0$, the seminorm $\left|\cdot \right|_\beta$ is not. However, changing $\eta _0$ preserves the topology induced by the relevant seminorm, which will be critical to proofs in this appendix. The following lemma gives the relevant bounds.

Lemma A.1

For $\zeta > 0$ and $f \in L^1(\mathbb {R}^d)$ let

$$\begin{aligned} \left|f\right|_{\beta ,\zeta } = \sup _{0 < \eta \le \zeta } \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \;\mathrm{d}x. \end{aligned}$$

If $0 < t \le s$, then

$$\begin{aligned} \left|\cdot \right|_{\beta ,t} \le \left|\cdot \right|_{\beta ,s} \le S(t,s) \left|\cdot \right|_{\beta ,t}, \end{aligned}$$

where S(t, s) denotes the minimal number of balls of radius t required to cover (up to a set of measure 0) a ball of radius s.

Proof

The inequality $\left|\cdot \right|_{\beta ,t} \le \left|\cdot \right|_{\beta ,s}$ is trivial. Let $f \in L^1(\mathbb {R}^d)$. If

$$\begin{aligned} \sup _{0< \eta \le t} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \,\mathrm{d}x = \sup _{0 < \eta \le s} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \;\mathrm{d}x, \end{aligned}$$

(41)

then, as $S(t,s) \ge 1$, we clearly have $\left|f\right|_{\beta ,s} \le S(t,s) \left|f\right|_{\beta ,t}$. Alternatively, if (41) does not hold, then

$$\begin{aligned} \sup _{0< \eta \le t} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \,\mathrm{d}x< \sup _{t < \eta \le s} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \;\mathrm{d}x. \end{aligned}$$

By the definition of S(t, s), there exists $\{c_i\}_{i=1}^{S(t,s)} \subseteq \mathbb {R}^d$ and a set N of measure 0 such that

$$\begin{aligned} B_s(x) {\setminus } (N + x) \subseteq \bigcup _{i=1}^{S(t,s)} B_t(x+c_i) \end{aligned}$$

for every $x \in \mathbb {R}^d$. Hence, for any $\eta \in (t, s]$ and $x \in \mathbb {R}^d$ we have

$$\begin{aligned} {\text {osc}} \left( f, B_\eta (x)\right) \le {\text {osc}} \left( f, B_s(x)\right) \le \sum _{i=1}^{S(t,s)} {\text {osc}} \left( f,B_t(x+c_i)\right) . \end{aligned}$$

After integrating, taking the supremum and applying the definition of $\left|\cdot \right|^t_\beta$, we obtain

$$\begin{aligned} \begin{aligned}&\sup _{t < \eta \le s} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_\eta (x)\right) \,\mathrm{d}x \le S(t,s) t^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_t(x)\right) \;\mathrm{d}x \\&\quad \le S(t,s) \left|f\right|_{\beta ,t}, \end{aligned} \end{aligned}$$

completing the proof. $\square$

We obtain Lemmas 5.12 and 5.13 as corollaries to the following result.

Proposition A.2

If $Q \in \mathcal {P}(\kappa )$ satisfies ${{\,\mathrm{diam}\,}}(Q) < \eta _0$, then

$$\begin{aligned} \left|\mathbb {E}_Q\right|_\beta \le S(\eta _0 - {{\,\mathrm{diam}\,}}(Q), \eta _0)\left( 1 + \frac{\kappa }{\root d \of {\frac{3}{2}} - 1} \right) ^\beta . \end{aligned}$$

(42)

We prove Proposition A.2 by using Lemma A.1 to extend a bound for $\sup _{\left|f\right|_\beta = 1} \left|\mathbb {E}_Q f\right|_{\beta ,\eta _0 - {{\,\mathrm{diam}\,}}(Q)}$ to a bound for $\left|\mathbb {E}_Q\right|_\beta$. We do this by combining two bounds for

$$\begin{aligned} \eta ^{-\beta } \int {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}m \end{aligned}$$

for $\eta \in (0, \eta _0 - {{\,\mathrm{diam}\,}}(Q)]$:

• 1. We obtain the “big” $\eta$ bound by scaling the $\eta$ balls in ${\text {osc}}$ up to $\eta +{{\,\mathrm{diam}\,}}(Q)$ balls. This bound is useful for large $\eta$, but grows unboundedly as $\eta$ vanishes. We obtain this bound in Lemma A.4.

• 2. We obtain the “small” $\eta$ bound by using the geometry of the elements of Q to quantify the decay of the measure of the support of ${\text {osc}} \left( \mathbb {E}_Q f, B_\eta (\cdot )\right)$ as $\eta$ vanishes. This bound is used for $\eta$ arbitrarily close to 0. Obtaining this bound is more complicated and is developed in Lemmas A.5–A.7.

Before proving either of these bounds, we derive an explicit expression for $\int {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}m$.

Lemma A.3

Let $Q \in \mathcal {P}(\kappa )$ define $Q' = Q \cup \{ \Omega ^c\}$, and for each $\eta > 0$ and $x \in \mathbb {R}^d$ let

$$\begin{aligned} N(x,\eta ) = \{ J \in Q' : B_\eta (x) \cap J \ne \emptyset \}. \end{aligned}$$

For each $\eta > 0$, $f \in \mathbb {V}_{\beta }(\Omega )$ and $S \subseteq Q'$ let

$$\begin{aligned} M_S(f) = \max _{J,K \in S} \left|\hat{f}_{J} - \hat{f}_{K}\right| \quad \text {and} \quad A_{S,\eta } = \{ x \in \mathbb {R}^d : N(x, \eta ) = S \}. \end{aligned}$$

Then each $A_{S,\eta }$ is measurable, and for every $x \in \mathbb {R}^d$

$$\begin{aligned} {\text {osc}} \left( \mathbb {E}_Q f,B_\eta (x)\right) = M_{N(x,\eta )}(f). \end{aligned}$$

(43)

Hence,

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x = \sum _{S \subseteq Q' } m(A_{S,\eta }) M_{S}(f). \end{aligned}$$

(44)

Proof

For every $J \in Q$, the equality

$$\begin{aligned} \{x \in \mathbb {R}^d : B_\eta (x) \cap J \ne \emptyset \} = \bigcup _{y \in J} B_\eta (y), \end{aligned}$$

implies that both sets are open and therefore measurable. Recalling from Definition 5.7 that Q is finite and noting the equality

$$\begin{aligned}\begin{aligned} A_{S,\eta } =&\left( \bigcap _{J \in S} \{x \in \mathbb {R}^d : B_\eta (x) \cap J \ne \emptyset \} \right) \\&\bigcap \left( \bigcap _{K \in Q' {\setminus } S} \{x \in \mathbb {R}^d : B_\eta (x) \cap K = \emptyset \} \right) , \end{aligned} \end{aligned}$$

we conclude that each $A_{S,\eta }$ is measurable. Considering the definition of $N(x,\eta )$, we note that the family of sets $\{ A_{S, \eta } : S \subseteq Q'\}$ partitions $\mathbb {R}^d$. Hence,

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x = \sum _{S \subseteq Q' } \int _{A_{S,\eta }} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x, \end{aligned}$$

where we note that the sum on the right-hand side is well defined as only finitely many terms are ever nonzero. Thus, in order to prove (44) it suffices to prove (43). If $N(x,\eta ) = S$, then

$$\begin{aligned} (\mathbb {E}_Q f)(B_\eta (x)) = \left\{ \hat{f}_{J} : J \cap B_\eta (x) \ne \emptyset \right\} = \left\{ \hat{f}_{J} : J \in S \right\} , \end{aligned}$$

which is finite as $Q'$ is finite. By applying the definition of ${\text {osc}}$, we find that

$$\begin{aligned} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) = \max _{J,K \in S} \left|\hat{f}_{J} - \hat{f}_{K}\right| = M_S(f), \end{aligned}$$

which is exactly (44). $\square$

We may now obtain the “big” $\eta$ bound.

Lemma A.4

If $Q \in \mathcal {P}(\kappa )$ then for each $f \in \mathbb {V}_{\beta }(\Omega )$ and $\eta > 0$, we have

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x \le \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_{\eta +{{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x. \end{aligned}$$

(45)

Furthermore, if ${{\,\mathrm{diam}\,}}(Q) < \eta _0$ and $\eta \in (0, \eta _0 - {{\,\mathrm{diam}\,}}(Q)]$ then

$$\begin{aligned} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x \le \left( 1 +\frac{{{\,\mathrm{diam}\,}}(Q)}{\eta }\right) ^\beta \left|f\right|_\beta . \end{aligned}$$

(46)

Proof

Fix $x \in \mathbb {R}^d$. By Lemma A.3, we have

$$\begin{aligned} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) = \max _{J,K \in N(x, \eta )} \left|\hat{f}_{J} -\hat{f}_{K}\right|. \end{aligned}$$

Suppose that $y \in I$ for some $I \in N(x, \eta ) {\setminus } \{\Omega ^c\}$. By the definition of $N(x, \eta )$ there exists $z \in I$ such that $\left|z - x\right| < \eta$. Since $\left|z-y\right| \le {{\,\mathrm{diam}\,}}(Q)$ we have $\left|y-x\right| < \eta + {{\,\mathrm{diam}\,}}(Q)$. Hence,

$$\begin{aligned} \bigcup _{I \in N(x,\eta ) {\setminus } \{\Omega ^c\}} I \subseteq B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x). \end{aligned}$$

(47)

Now suppose that $J,K \in N(x,\eta )$. In the case where $J,K \in N(x,\eta ){\setminus } \{\Omega ^c\}$ the inclusion (47) implies that for almost every $(y_1, y_2) \in J \times K$ we have

$$\begin{aligned} \left|f(y_1) - f(y_2)\right| \le {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) . \end{aligned}$$

By taking expectations with respect to $y_1$ over J and $y_2$ over K, we obtain

$$\begin{aligned} \left|\hat{f}_{J} - \hat{f}_{K}\right| \le {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) . \end{aligned}$$

(48)

Alternatively, if either J or K is equal to $\Omega ^c$ then

$$\begin{aligned} \begin{aligned} \left|\hat{f}_{J} - \hat{f}_{K}\right| = \max \left\{ \left|\hat{f}_{J}\right|, \left|\hat{f}_{K}\right|\right\}&\le \max _{I \in N(x, \eta ) {\setminus } \Omega ^c} \left|\hat{f}_{I}\right| \\&\le \mathop {{{\,\mathrm{ess\,sup}\,}}}\limits _{y \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}} \left|f(y)\right|, \end{aligned} \end{aligned}$$

(49)

where we obtain the last inequality by using (47). Noting that the set $B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x) \cap \Omega ^c$ has nonzero measure, we have

$$\begin{aligned} \mathop {{{\,\mathrm{ess\,sup}\,}}}\limits _{y \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)} \left|f(y)\right| \le {\text {osc}} \left( f,B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) . \end{aligned}$$

(50)

By combining (49) and (50), we obtain (48) for the case where either J or K is equal to $\Omega ^c$. As J and K were arbitrary elements of $N(x,\eta )$, this implies that

$$\begin{aligned} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) = \max _{I,J \in N(x,\eta )} \left|\hat{f}_{I} -\hat{f}_{J}\right| \le {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) . \end{aligned}$$

By integrating with respect to x over $\mathbb {R}^d$, we obtain (45). To prove (46), suppose that ${{\,\mathrm{diam}\,}}(Q) < \eta _0$. If $\eta \in (0, \eta _0 -{{\,\mathrm{diam}\,}}(Q)]$ then $\eta + {{\,\mathrm{diam}\,}}(Q) \in (0, \eta _0]$ and so the definition of $\left|\cdot \right|_\beta$ implies that

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_{\eta +{{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x \le (\eta + {{\,\mathrm{diam}\,}}(Q))^{\beta } \left|f\right|_\beta . \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_{\eta }(x)\right) \;\mathrm{d}x&\le \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_{\eta +{{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x \\&\le \left( 1 +\frac{{{\,\mathrm{diam}\,}}(Q)}{\eta }\right) ^\beta \left|f\right|_\beta . \end{aligned} \end{aligned}$$

$\square$

We will now pursue the “small” $\eta$ bound.

Lemma A.5

Let $Q \in \mathcal {P}(\kappa )$ and let $S \subseteq Q'$ satisfy $\left|S\right| > 1$. If $I \in S {\setminus } \{ \Omega ^c\}$ and $\eta > 0$, then $A_{S, \eta } \subseteq B_\eta (\partial I)$.

Proof

The claim is trivially true if $A_{S,\eta }$ is empty; henceforth, we assume that it is not. Let $x \in A_{S, \eta }$. We distinguish between two cases: either $x \in I$ or $x \notin I$. Suppose that $x \in I$. As $\left|S\right| > 1$ and $N(x,\eta ) = S$ there exists some $J \in Q' {\setminus } \{I\}$ such that $B_\eta (x) \cap J \ne \emptyset$. Actually, as the closure of the interior of J is J, we have $B_\eta (x) \cap {{\,\mathrm{int}\,}}(J) \ne \emptyset$. In this case, let $y \in B_\eta (x) \cap {{\,\mathrm{int}\,}}(J)$; as J and I are convex elements of a measurable partition we have $J \cap I \subseteq \partial J \cap \partial I$ and so $y \notin I$. Alternatively, if $x \notin I$, then let $y \in B_\eta (x) \cap I$, which is non-empty by a similar argument. In both cases, we have a pair of points in $A_{S, \eta }$: one in I and the other not. Recalling that elements of $Q'$ have non-empty interior and then considering the line segment that joins x and y, it is straightforward to verify that there exists some $z \in \partial I$ on this line segment. Clearly, $\left|x-z\right| < \eta$ and so $x \in B_\eta (\partial I)$, which completes the proof. $\square$

Lemma A.6

Let $Q \in \mathcal {P}(\kappa )$. If $\eta > 0$ and $S \subseteq Q'$ is such that $\left|S\right| > 1$ and $m(A_{S,\eta }) > 0$, then for each $f \in \mathbb {V}_{\beta }(\Omega )$ we have

$$\begin{aligned} \begin{aligned} M_S(f)&\le \max _{\begin{array}{c} J,K \in S \\ J\ne K \end{array}}\Bigg (\int _J \frac{{\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(J)}\;\mathrm{d}x \\&\quad +\, \int _K \frac{ {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(K)}\;\mathrm{d}x \Bigg ). \end{aligned} \end{aligned}$$

Proof

Let $J,K \in S$ be partition elements satisfying

$$\begin{aligned} M_S(f) = \left|\hat{f}_{J} - \hat{f}_{K}\right|. \end{aligned}$$

We may assume that $J \ne K$, as this case does not contribute to the maximum. Let us first consider the case where $\Omega ^c \in \{J,K\}$; without loss of generality, let $K = \Omega ^c$. For every $j \in J$, we have $J \subseteq B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)$. Hence, as $B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j) \cap \Omega ^c$ has non-empty interior, and therefore nonzero measure, for almost every $j, j' \in J$ and $k' \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j) \cap \Omega ^c$ we have

$$\begin{aligned} \left|f(j')\right| = \left|f(j') - f(k')\right| \le {\text {osc}} \left( f,B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)\right) . \end{aligned}$$

Taking expectations with respect to $j'$ and j over J yields

$$\begin{aligned} M_S(f) = \left|\hat{f}_{J}\right| \le \int _J \frac{{\text {osc}} \left( f,B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(J)}\;\mathrm{d}x, \end{aligned}$$

which implies the required conclusion. Alternatively suppose that neither J nor K is equal to $\Omega ^c$. Fix $j \in J$ and $k \in K$. For any $j' \in J$, we have $\left|j - j'\right| \le {{\,\mathrm{diam}\,}}(Q)$ and so $j' \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)$. Similarly, for every $k' \in K$ we have $k' \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(k)$. As $m(A_{S,\eta }) > 0$, we know that $A_{S,\eta } \ne \emptyset$. For $z \in A_{S,\eta }$, the intersection $B_\eta (z) \cap J$ is non-empty and so $z \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)$. Similarly, $z \in B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(k)$. Hence, for almost every $j' \in J$ and $k' \in K$,

$$\begin{aligned} \begin{aligned} \left|f(j') - f(k')\right|&\le \left|f(j') - f(z)\right| + \left|f(k') - f(z)\right|\\&\le {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)\right) + {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(k)\right) . \end{aligned} \end{aligned}$$

By taking the expectation with respect to $j'$ over J and $k'$ over K, we find

$$\begin{aligned} \left|\hat{f}_{J} - \hat{f}_{K}\right| \le {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(j)\right) + {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(k)\right) . \end{aligned}$$

(51)

Since (51) holds for every $j \in J$ and $k \in K$, we may take expectations again to obtain

$$\begin{aligned} \left|\hat{f}_{J} - \hat{f}_{K}\right| \le \int _J \frac{{\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(J)} \;\mathrm{d}x + \int _K \frac{{\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(K)} \;\mathrm{d}x. \end{aligned}$$

We obtain the required inequality by taking the maximum over all distinct pairs of $J,K \in S$. $\square$

Combining the previous two results yields the “small” $\eta$ bound of Lemma A.7.

Lemma A.7

Let $Q \in \mathcal {P}(\kappa )$. If ${{\,\mathrm{diam}\,}}(Q) < \eta _0$, $\eta \in (0, \eta _0 - {{\,\mathrm{diam}\,}}(Q)]$ and $f \in \mathbb {V}_{\beta }(\Omega )$, then

$$\begin{aligned} \eta ^{-\beta } \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \,\mathrm{d}x \le \left( \max _{I \in Q} \frac{m(B_\eta (\partial I))}{m(I)} \right) \left( 1 + \frac{{{\,\mathrm{diam}\,}}(Q)}{\eta }\right) ^\beta \left|f\right|_\beta . \end{aligned}$$

Proof

By Lemma A.3, we have

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x = \sum _{S \subseteq Q' } m(A_{S,\eta }) M_{S}(f). \end{aligned}$$

(52)

Let $G = \{ S \subseteq Q' : \left|S\right|> 1, m(A_{S,\eta }) > 0 \}$. Since $m(A_{S,\eta }) M_{S}(f) = 0$ if $S \notin G$, we may restrict the sum in (52):

$$\begin{aligned} \int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x = \sum _{S \in G} m(A_{S,\eta }) M_{S}(f). \end{aligned}$$

(53)

Applying Lemma A.6 to each of the terms in (53) yields

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {R}^d} {\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \;\mathrm{d}x \\&\begin{aligned}&\quad \le \sum _{S \in G} m(A_{S,\eta }) \max _{J,K \in S, J \ne K} \biggl ( \int _J \frac{{\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(J)} \;\mathrm{d}x \\&\qquad +\, \int _K \frac{{\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) }{m(K)} \;\mathrm{d}x \biggr ). \end{aligned} \end{aligned} \end{aligned}$$

(54)

By rearranging the terms in (54) to sum over elements of Q, we obtain

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {R}^d}{\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \mathrm {d}x \\&\quad \le \sum _{I \in Q} \frac{\sum _{S \in G, I \in S} m(A_{S,\eta })}{m(I)} \int _I {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x \\&\quad \le \left( \max _{I \in Q} \frac{\sum _{ S\in G, I \in S} m(A_{S,\eta })}{m(I)} \right) \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x, \end{aligned} \end{aligned}$$

(55)

where we omit the case of $I = \Omega ^c$, as it does not contribute to the sum. Since the sets $\{A_{S,\eta }\}_{S \subseteq Q'}$ are disjoint, Lemma A.5 implies that

$$\begin{aligned} \sum _{ S \in G, I \in S } m(A_{S,\eta }) \le m(B_\eta (\partial I)). \end{aligned}$$

Thus,

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^d}&{\text {osc}} \left( \mathbb {E}_Q f, B_\eta (x)\right) \mathrm {d}x \\&\le \left( \max _{I \in Q} \frac{m(B_\eta (\partial I))}{m(I)} \right) \int _{\mathbb {R}^d} {\text {osc}} \left( f, B_{\eta + {{\,\mathrm{diam}\,}}(Q)}(x)\right) \;\mathrm{d}x. \end{aligned} \end{aligned}$$

The required inequality follows by applying the definition of $\left|\cdot \right|_\beta$. $\square$

Before proving Proposition A.2, we require a technical lemma for an inequality from convex geometry. For $U,V \subseteq \mathbb {R}^d$ the Minkowski sum of U and V is denoted by $U + V$ and equal to $\{u + v : u \in U, v \in V\}$; for basic properties, we refer to [19, Section 6.1].

Lemma A.8

If I is a compact convex polytope, then for every $\eta > 0$ we have

$$\begin{aligned} m(B_\eta (\partial I) \cap I) \le m(B_\eta (\partial I) \cap I^c). \end{aligned}$$

Proof

Let $m_{d-1}$ denote $d-1$-dimensional Lebesgue measure. By Steiner’s formula [19, Theorem 6.6], there exists a polynomial $p_I$ with positive coefficients and of degree d such that $m(B_\eta (I)) = p_I(\eta )$. The constant coefficient of $p_I$ is clearly m(I), while the coefficient of the linear term is $m_{d-1}(\partial I)$, i.e. the surface area of I. Note that $m(B_\eta (\partial I) \cap I^c) = p_I(\eta ) - m(I)$. We will prove that $m(B_\eta (\partial I) \cap I) \le \eta m_{d-1}(\partial I)$. Since $p_I$ has degree greater than or equal to 2 and positive coefficients, it follows that

$$\begin{aligned} m(B_\eta (\partial I) \cap I) \le \eta m_{d-1}(\partial I) \le p_I(\eta ) - m(I) \le m(B_\eta (\partial I) \cap I^c), \end{aligned}$$

and would therefore complete the proof.

Let $\mathcal {F}(I)$ denote the set of set of facets of I. Clearly $m_{d-1}(\partial I) = \sum _{F \in \mathcal {F}(I)} m_{d-1}(F)$. Let $y \in B_\eta (\partial I) \cap I$ and denote by F the (possibly not unique) facet in $\mathcal {F}(I)$ that minimises the distance from y to $\partial I$. Let x be the point on F attaining said minimum. If $x-y$ is not normal to F, then the ball $B_{\left|x-y\right|}(y)$ is not tangent to F and so there exists $z \in B_{\left|x-y\right|}(y) \cap I^c$. The line segment from y to z must intersect $\partial I$ at some point that is strictly closer to y than x, which contradicts x minimising the distance from y to $\partial I$. Hence, $x-y$ must be normal to F and so $y \in F + [0,\eta ] \times n_F$, where $n_F$ is the inward-facing unit normal vector to F. This implies that

$$\begin{aligned} B_\eta (\partial I) \cap I \subseteq \bigcap _{F \in \mathcal {F}(I)} F + [0,\eta ] \times n_F \end{aligned}$$

and so $m(B_\eta (\partial I) \cap I) \le \eta \sum _{F \in \mathcal {F}(I)} m_{d-1}(F) = \eta m_{d-1}(\partial I)$ as required. $\square$

The proof of Proposition A.2

We begin by bounding

$$\begin{aligned} \sup _{\left|f\right|_\beta = 1} \left|\mathbb {E}_Q f\right|_{\beta ,\eta _0 - {{\,\mathrm{diam}\,}}(Q)}. \end{aligned}$$

Let $b: \mathbb {R}\rightarrow \mathbb {R}$ be defined by

$$\begin{aligned} b(\eta ) = \left( 1 + \frac{{{\,\mathrm{diam}\,}}(Q)}{\eta }\right) ^\beta . \end{aligned}$$

By taking the minimum of the bounds in Lemmas A.4 and A.7, we have

$$\begin{aligned} \left|\mathbb {E}_Q f\right|_{\beta ,\eta _0 - {{\,\mathrm{diam}\,}}(Q)} \le \sup _{0 < \eta \le \eta _0} \min \left\{ \max _{I \in Q} \frac{m(B_\eta (\partial I))}{m(I)}, 1\right\} b(\eta ) \left|f\right|_\beta . \end{aligned}$$

(56)

We will now bound $\max _{I \in Q} \frac{m(B_\eta (\partial I))}{m(I)}$. Lemma A.8 implies that for any $I \in Q$ we have

$$\begin{aligned} \frac{m(B_{\eta }(\partial I))}{m(I)}\le \frac{2m(B_{\eta }(\partial I) \cap I^c)}{m(I)}. \end{aligned}$$

Noting that $B_{\eta }(\partial I) \cap I^c = B_{\eta }(I) {\setminus } I$ and $B_\eta (I) = I + (\eta /2)B_{1}(0)$, we obtain

$$\begin{aligned} \frac{m(B_{\eta }(\partial I))}{m(I)} \le 2 \frac{m(I + (\eta /2)B_{1}(0)) -m(I)}{m(I)}. \end{aligned}$$

(57)

Let $B_I$ be a ball inscribed in I of maximal volume. Then, by scaling and possibly translating by some vector $v_I \in \mathbb {R}^d$, we find that $B_{1}(0) \subseteq \frac{2}{{{\,\mathrm{diam}\,}}(B_I)} I + v_I$. Consequently

$$\begin{aligned} m\left( I + \frac{\eta }{2}B_{1}(0)\right) \le m\left( I + \frac{\eta }{{{\,\mathrm{diam}\,}}(B_I)}I\right) = \left( 1 + \frac{\eta }{{{\,\mathrm{diam}\,}}(B_I)} \right) ^d m(I). \end{aligned}$$

(58)

Applying (58) to (57), and recalling that $1/ {{\,\mathrm{diam}\,}}(B_I) \le \kappa /{{\,\mathrm{diam}\,}}(Q)$ as $Q \in \mathcal {P}(\kappa )$, we find that

$$\begin{aligned} \frac{m(B_{\eta }(\partial I))}{m(I)} \le 2 \left( 1 + \frac{\eta }{{{\,\mathrm{diam}\,}}(B_I)} \right) ^d - 2 \le 2 \left( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 2. \end{aligned}$$

(59)

By applying (59) to (56), we obtain

$$\begin{aligned} \begin{aligned}&|\mathbb {E}_Q f |_{\beta ,\eta _0 - {{\,\mathrm{diam}\,}}(Q)} \\&\quad \le \sup _{0 < \eta \le \eta _0} \min \left\{ \left( 2 \left( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 2\right) b(\eta ), b(\eta )\right\} \left|f\right|_\beta . \end{aligned} \end{aligned}$$

(60)

It is clear that b is monotonically decreasing. Note that

$$\begin{aligned} \begin{aligned}&2 \Bigg (\Bigg ( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \Bigg )^d - 1\Bigg )b(\eta ) \\&\quad = 2 \eta ^{-\beta }\left( \left( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 1\right) (\eta + {{\,\mathrm{diam}\,}}(Q))^\beta . \end{aligned} \end{aligned}$$

(61)

The map $\eta \mapsto (\eta + {{\,\mathrm{diam}\,}}(Q))^\beta$ is clearly monotonically increasing on $(0, \eta _0]$. As $d \ge 2$ and $\beta \in (0, 1]$, the map

$$\begin{aligned} \eta \mapsto \eta ^{-\beta } \left( \left( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 1\right) \end{aligned}$$

is monotonically increasing on $(0, \eta _0]$ too. Thus, the left-hand side of (61) is monotonically increasing. Since both b and the left-hand side of (61) are continuous on $(0, \eta _0]$, b is monotonically decreasing and the left-hand side of (61) is monotonically increasing, it follows that if $\eta ' \in (0, \infty )$ solves

$$\begin{aligned} 2 \left( 1 + \frac{\kappa \eta '}{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 2 = 1, \end{aligned}$$

(62)

then

$$\begin{aligned} \sup _{0 < \eta \le \eta _0} \min \left\{ \left( 2 \left( 1 + \frac{\kappa \eta }{{{\,\mathrm{diam}\,}}(Q)} \right) ^d - 2\right) b(\eta ), b(\eta )\right\} \le b(\eta '). \end{aligned}$$

Solving (62) yields

$$\begin{aligned} \frac{{{\,\mathrm{diam}\,}}(Q)}{\eta '} = \frac{\kappa }{\root d \of {\frac{3}{2}} - 1}. \end{aligned}$$

By substituting this into (60), we obtain the bound

$$\begin{aligned} \left|\mathbb {E}_Q f\right|_{\beta ,\eta _0 - {{\,\mathrm{diam}\,}}(Q)} \le \left( 1 + \frac{\kappa }{\root d \of {\frac{3}{2}} - 1} \right) ^\beta \left|f\right|_\beta . \end{aligned}$$

Applying Lemma A.1 yields the required bound. $\square$

With Proposition A.2 in hand, we may now prove Lemmas 5.12 and 5.13.

Proof of Lemma 5.12

As $\lim _{\epsilon \rightarrow 0} {{\,\mathrm{diam}\,}}(Q_\epsilon ) = 0$ there exists $\epsilon _2 >0$ such that for every $\epsilon \in (0, \epsilon _2]$ we have ${{\,\mathrm{diam}\,}}(Q_\epsilon ) < \eta _0$ and

$$\begin{aligned} 1 + {{\,\mathrm{diam}\,}}(Q_\epsilon )/(\eta _0 - {{\,\mathrm{diam}\,}}(Q_\epsilon )) < \sqrt{d/(d-1)}. \end{aligned}$$

By [7, Section 8.5, page 236], this implies

$$\begin{aligned} S(1, 1 + {{\,\mathrm{diam}\,}}(Q_\epsilon )/(\eta _0 - {{\,\mathrm{diam}\,}}(Q_\epsilon ))) = S(\eta _0 - {{\,\mathrm{diam}\,}}(Q_\epsilon ), \eta _0) \le 2d. \end{aligned}$$

The desired conclusion follows by Proposition A.2. $\square$

Proof of Lemma 5.13

If ${{\,\mathrm{diam}\,}}(Q) < \eta _0$, then $\left|\mathbb {E}_Q\right|_\beta < \infty$ by Proposition A.2. Alternatively, if ${{\,\mathrm{diam}\,}}(Q) \ge \eta _0$, then repeatedly applying Lemma A.1 yields

$$\begin{aligned}\begin{aligned} \left|\mathbb {E}_Q\right|_\beta&= \sup _{\left|f\right|_\beta \le 1 } \left|\mathbb {E}_Q f\right|_\beta \\&\le \sup \{ \left|\mathbb {E}_Q f\right|_{\beta ,2 {{\,\mathrm{diam}\,}}(Q)} : \left|f\right|_{\beta ,2{{\,\mathrm{diam}\,}}(Q)} \le S(\eta _0, 2{{\,\mathrm{diam}\,}}(Q)) \}\\&\le S(\eta _0, 2{{\,\mathrm{diam}\,}}(Q)) \left|\mathbb {E}_Q\right|_{\beta ,2 {{\,\mathrm{diam}\,}}(Q)}, \end{aligned} \end{aligned}$$

which is finite by Proposition A.2 applied to the seminorm $\left|\cdot \right|_{\beta ,2 {{\,\mathrm{diam}\,}}(Q)}$ (i.e. when $\eta _0 = 2 {{\,\mathrm{diam}\,}}(Q)$). In either case, we have $\left|\mathbb {E}_Q\right|_\beta < \infty$ and so, as $\left|\mathbb {E}_Q \right|_{L^1} =1$, we have $\left\| \mathbb {E}_Q\right\| _\beta < \infty$ too. As Q partitions $\Omega$, for every $f \in \mathbb {V}_{\beta }(\Omega )$ the support of $\mathbb {E}_Q f$ is a subset of $\Omega$. Hence $\mathbb {E}_Q f \in \mathbb {V}_{\beta }(\Omega )$ for every $f \in \mathbb {V}_{\beta }(\Omega )$ and so $\mathbb {E}_Q \in L(\mathbb {V}_{\beta }(\Omega ))$. $\square$

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1.1 Variance

1.2 Rate Function