Appendix A: Uniformity of Defect Measures

Proposition 2.6 motivates the definition of defect measures, but this definition allows for some undesirable consequences, which show up in the following pathological example. Let \(f\in C^_0(}^d)\) with \(\Vert f\Vert _}^d)}=1\), and define \(\psi _h(x)=f\left( x+ h^\right) .\) It is easy to show that \(\psi _h\) has defect measure equal to the zero measure, corresponding intuitively to the mass having “escaped to infinity.” Indeed, one has for any \(a\in C^_0(T^*}^d)\) supported in a ball of radius R in \(}^\cong T^*}^d\), \(h<(2R)^\), and \(\chi \in C^(}^d)\) such that \(\chi (x)=1\) on \(|x|\geqslant 2R\), \(\chi =0\) on \(|x|\leqslant R\) that

$$\begin \,}}_h(a)\psi _h=\,}}_h(a)\,}}_h(\chi )\psi _h=O(h^)\end$$

(A.1)

which gives

$$\begin \lim _\langle \,}}_h(a)\psi _h,\psi _h\rangle =0. \end$$

(A.2)

Unfortunately, there is no way to quantify the rate of decay of (A.2), as it depends on not just the symbol norms of a but also the location of its support. In addition, the resulting measure is not a probability measure, which defies our intuition of the defect measure representing the classical location of the particle in phase space. We are therefore motivated to make a more restrictive definition of defect measures to prohibit examples such as (A.1). Call \(\mu \) a uniform semiclassical defect measure of \(\rho _h\) if for any \(a\in S\),

$$\begin \lim _}\left( \rho _h\,}}_h(a)\right) =\int _}^d}a\mathop \!\textrm\mu . \end$$

(A.3)

The reason for the name is given in the following proposition. We use the standard multi-index notation. For \(a\in C^\infty (T^* }^d)\) and \(\alpha \in }^\), we let

$$\begin a^ ( x) = \left( \prod _^ \partial _j^ \right) a(x). \end$$

Proposition A.1

Let \(\rho _h\) be a set of density operators with defect measure \(\mu \). The following are equivalent.

(1)

\(\mu \) is a uniform defect measure.

(2)

For all \(\epsilon >0\), there is an \(h_0(\epsilon ) > 0 \) such that if \(0< h<h_0\) and \(a\in C^_0(T^*}^d)\) with \(\Vert a^\Vert _}\leqslant 1\) for all \(|\alpha |<K(d)\) then

$$\begin \left| }\left( \rho _h\,}}_h(a)\right) -\int _}^d}a\mathop \!\textrm\mu \right| \leqslant \epsilon . \end$$

(3)

For all \(\epsilon >0\), there is an \(h_0(\epsilon )>0 \) such that if \(0< h<h_0\) and \(a\in S\) with \(\Vert a^\Vert _}\leqslant 1\) for all \(|\alpha |<K(d)\) then

$$\begin \left| }\left( \rho _h\,}}_h(a)\right) -\int _}^d}a\mathop \!\textrm\mu \right| \leqslant \epsilon . \end$$

(4)

\(\mu \) is a probability measure.

Proof

We trivially see that \((3)\implies (2)\) and \((3)\implies (1)\).

To show \((2)\implies (3)\), let \(a\in S\), and let \(\chi _k\in C^_0(T^*}^d)\) be smooth partition of unity of \(T^*}^d\), chosen as in [19, Theorem 4.23], with the first K(d) derivatives chosen small enough. By the Cotlar-Stein-Knapp lemma, we have the limit in the strong operator topology:

$$\begin \lim _\left( \sum _^\,}}_h(\chi _ka)\right) \psi =\,}}_h(a)\psi \end$$

for every \(\psi \in L^2(}^d)\). We need the following brief lemma.

Lemma A.2

Let \(A_K\) be a sequence of uniformly bounded operators converging to A in the strong topology, and let \(\rho \) be self-adjoint and trace class. Then

$$\begin \lim _}(\rho A_K)=}(\rho A). \end$$

Proof of Lemma A.2

By subtracting we may assume \(A=0\), and without loss of generality let \(\sup _K\Vert A_K\Vert =1\). Let \(\psi _j\) be an orthonormal basis of eigenvectors of the compact operator \(\rho \), so

$$\begin \rho =\sum _j\lambda _j\left| \psi _j\right\rangle \left\langle \psi _j\right| \end$$

with \(\lambda _j\rightarrow 0\). as \(j\rightarrow \infty \). Let \(\epsilon >0\), and choose J such that \(\lambda _j<\varepsilon \) for \(j>J\). Then

$$\begin |}(\rho A_K)|= & \left| \sum _\lambda _j\langle A_K\psi _j,\psi _j\rangle +\sum _\lambda _j\langle A_K\psi _j,\psi _j\rangle \right| \\\leqslant & \sum _\left| \lambda _j\langle A_K\psi _j,\psi _j\rangle \right| +\varepsilon \xrightarrow \varepsilon \end$$

which shows \(}(\rho A_K)\xrightarrow 0\) as desired. \(\square \)

By Lemma A.2, we see that for fixed h,

$$\begin \lim _}\left( \rho _h\sum _^\,}}_h(\chi _ka)\right) =}(\rho _h\,}}_h(a)) \end$$

and similarly as \(\mu \) is a finite measure

$$\begin \lim _\int _}^d}\sum _^\chi _ka\mathop \!\textrm\mu =\int _}^d}a\mathop \!\textrm\mu . \end$$

Then letting \(\varepsilon >0\) and picking the \(h_0(\varepsilon )\) from part (2) we get

$$\begin & \left| }\left( \rho _h\,}}_h(a)\right) -\int _}^d}a\mathop \!\textrm\mu \right| \\ & \quad =\lim _\left| }\left( \rho _h\sum _^\,}}_h(\chi _ka)\right) -\int _}^d}\sum _^\chi _ka\mathop \!\textrm\mu \right| \leqslant \epsilon \end$$

To show \((1)\implies (3)\), let \(a\in S\), and let \(\chi \in C^_0(T^*}^d)\) be supported in a ball of radius R, such that \(\mu (T^*}^d\setminus B_R(0))\leqslant \epsilon / 8\). We also note that \(\chi \) may be chosen so its first K(d) derivatives are small (possibly making R larger). Then we have that

$$\begin }(\,}}_h(a)\,}}_h(1-\chi )\rho _h)\leqslant \Vert \,}}_h (a) \Vert }(\,}}_h(1-\chi )\rho _h)\leqslant C }(\,}}_h(1-\chi )\rho _h) \end$$

for C depending on the first K(d) derivatives of a. Next we apply (A.3) with the symbol \(1-\chi \) (which is independent of a) to get that

$$\begin C }(\,}}_h(1-\chi )\rho _h) = C \varepsilon / 8 + o (1) \end$$

as \(h \rightarrow 0\). So that there exists \(h_0 >0\) such that for \(0< h < h_0\) we have \(C }(\,}}_h(1-\chi )\rho _h) < \varepsilon / 4\). With this, the triangle inequality, and symbol calculus, we have for \(h<h_0(\varepsilon )\)

$$\begin \begin \left| }\left( \rho _h\,}}_h(a)\right) -\int _}^d}a\mathop \!\textrm\mu \right|&\leqslant \left| }\left( \rho _h\,}}_h(a)\,}}_h(\chi )\right) -\int _}^d}a\chi \mathop \!\textrm\mu \right| +\frac\\&\leqslant \left| }\left( \rho _h\,}}_h(a\chi )\right) -\int _}^d}a\chi \mathop \!\textrm\mu \right| +\frac \end \end$$

(A.4)

where for the last inequality we have chosen possibly smaller \(h_0(\varepsilon )\) and used the fact that a has bounded derivatives.

We now prove (3) using a compactness argument. Observe that by Calderón-Vaillancourt, if (A.3) holds for \(a\in S\), it must also hold for any \(a\in C^(T^*}^d)\) for some sufficiently large \(K_0\). Given \(h_j\), let \(X_j\) be the set of \(b\in C^(B_R(0))\) such that

$$\begin \left| }(\rho _h\,}}_h(b\chi ))-\int _b\chi \mathop \!\textrm\mu \right| <\frac \end$$

(A.5)

for all \(h<h_j\). Then for \(K_0(d)\) large enough, condition (A.5) makes \(X_j\) to be an open set in \(C^(B_R(0))\), so letting \(h_j\rightarrow 0\), gives \((X_j)_}}\) to be an open cover of \(C^(B_R(0))\). Let

$$\begin Y=\(B_R(0)):\Vert b^\Vert _(B_R(0))}\leqslant 1\quad \forall \,\,|\alpha |\leqslant K_0(d)+1\}, \end$$

and let \(\overline\) denote the closure of Y in \(C^(B_R(0))\). By the Arzela-Ascoli theorem, \(\overline\) is compact in the topology of \(C^(B_R(0))\), so there is a finite subset of the \(X_j's\) that covers \(\overline\), and hence an \(h_0\) such that for \(h<h_0\)

$$\begin \left| }(\rho _h\,}}_h(b))-\int _b\mathop \!\textrm\mu \right| <\frac \end$$

holds for all b with \(\Vert b^\Vert _}\leqslant 1\) for \(|\alpha |\leqslant K_0(d) + 1\). Letting \(K(d)=K_0(d)+1\), and \(b=a|_\) in (A.4) completes the proof of (3).

To show \((3)\implies (4)\), we simply see that

$$\begin \int _}^d}\mathop \!\textrm\mu =\lim _}(\rho _h\,}}_h(1))=\lim _}\rho _h=1. \end$$

Here we use that the function \(f(x) = 1\) is in the symbol class S.

Finally, we show \((4)\implies (3)\). Let \(\chi \in C^_0(T^*}^d)\), so

$$\begin }\left( \rho _h\,}}_h(1-\chi )\right) -\int _}^d}1-\chi \mathop \!\textrm\mu =-\left( }\left( \rho _h\,}}_h(\chi )\right) -\int _}^d}\chi \mathop \!\textrm\mu \right) \rightarrow 0 \end$$

as \(h\rightarrow 0\), so in particular (A.3) holds in the special case of the noncompact symbol \(1-\chi \). But that was the only symbol used in the proof of \((1)\implies (3)\), so in particular shows \((4)\implies (3)\) as well. This shows all equivalences and hence completes the proof. \(\square \)

Appendix B: Coherent States

In this appendix, we prove Propositions 2.9, 2.11, and 2.12, which give examples of \((\delta ,\theta )\)-defect measures.

1.1 B.1. Coherent States on \(}^d\)

We prove Proposition 2.9. The proof is essentially the same as [19, §5.1 Examples 1 and 2], but is quantitative and keeps track of errors.

Proof of Proposition 2.9

We first consider the case when \(\beta =0\). In this case, we can without loss of generality assume \(x_0=0\), by absorbing it into the definition of g. Then

$$\begin \psi _h(x)=g(x)e^\langle x,\xi _0\rangle }, \end$$

so for \(a\in S_ \cap C_0^\infty (T^* }^d)\):

$$\begin }(\rho _h\,}}_h(a))&=\langle \,}}_h(a)\psi _h,\psi _h\rangle \\&=\int \frac\left( \iint e^\langle x-y,\xi -\xi _0\rangle } g(y)\,a\left( \frac,\xi \right) \mathop \!\textrmy\mathop \!\textrm\xi \right) \overline\mathop \!\textrmx\\&=\int \frac\left( \iint e^\langle y,\xi \rangle } g(x-y)\,a\left( x-\frac,\xi +\xi _0\right) \mathop \!\textrmy\mathop \!\textrm\xi \right) \overline\mathop \!\textrmx. \end$$

We evaluate the inner double integral using the explicit form of stationary phase given in [19, Theorem 3.17].

$$\begin \begin&\iint e^\langle y,\xi \rangle } g(x-y)\,a\left( x-\frac,\xi +\xi _0\right) \mathop \!\textrmy\mathop \!\textrm\xi \\&\quad \sim (2\pi h)^d\left( \sum _^\frac\langle D_y,D_\rangle ^k\left. g(x-\cdot _y)a\left( x-\frac,\cdot _+\xi _0\right) \right| _ \right) \\&\quad =(2\pi h)^d\left( g(x)a(x,\xi _0)+O(h^)\right) \end \end$$

with \(\sim \) referring to asymptotic summation and the last equality coming from a being an element of \(S_\). Then

$$\begin \left| }(\rho _h\,}}_h(a))-\int |g(x)|^2a(x,\xi _0)\mathop \!\textrmx\right| \leqslant O(h^)\int |g(x)|\mathop \!\textrmx=O(h^) \end$$

which gives the result for \(\beta =0\). The case of \(\beta =1\) follows from taking the semiclassical Fourier transform.

We now treat the case when \(0<\beta <1\). Here we have

$$\begin \psi _h(x)=h^}g\left( \frac}\right) e^\langle x,\xi _0\rangle }. \end$$

We again have that \(}(\rho _h \,}}_h (a) ) = \langle \,}}_h(a) \psi _h, \psi _h \rangle \). We now approximate \(\,}}_h(a)\) by its left quantization defined as:

$$\begin \,}}_h^L(a)u(x):=\frac\iint e^\langle x-y,\xi \rangle }a(x,\xi )u(y)\mathop \!\textrmy\mathop \!\textrm\xi . \end$$

By [19, Theorem 4.13], \(\,}}_h(a)=\,}}_h^L(a)+O(h^)_}^d)\rightarrow L^2(}^d)}\). Therefore,

$$\begin&}(\rho _h\,}}_h(a))=\langle \,}}_h^L(a)\psi _h,\psi _h\rangle +O(h^)\\&\quad =\frac\int \overline}\right) }\left( \iint h^g\left( \frac}\right) e^\langle x-y,\xi -\xi _0\rangle }a(x,\xi )\mathop \!\textrmy\mathop \!\textrm\xi \right) \mathop \!\textrmx\\&\qquad +O(h^)\\&\quad =\frac\int \overline}\right) }\\&\qquad \int e^\langle x-x_0,\xi \rangle }a(x,\xi +\xi _0)\left( \int h^g\left( \frac}\right) e^\langle y,\xi \rangle }\mathop \!\textrmy\right) \mathop \!\textrm\xi \mathop \!\textrmx\\&\qquad +O(h^)\\&\quad =\frac\iint \overline}\right) }e^\langle x,\xi \rangle }a(x+x_0,\xi +\xi _0)\hat\left( \frac}\right) \mathop \!\textrm\xi \mathop \!\textrmx+O(h^)\\&\quad = \frac \iint \overline\hat(\xi )e^a\left( h^x+x_0,h^\xi +\xi _0\right) \mathop \!\textrm\xi \mathop \!\textrmx+O(h^). \end$$

Because \(\Vert g \Vert _}^d )} = 1\), we have by the Fourier inversion formula that:

$$\begin \frac \iint \overline\hat(\xi ) e^ \mathop \!\textrm\xi \mathop \!\textrmx = 1 \end$$

so that

$$\begin&|}(\rho _h\,}}_h(a))-a(x_0,\xi _0)| \\&\quad \leqslant \left| \frac \iint \overline\hat(\xi )e^(a\left( h^x+x_0,h^\xi +\xi _0\right) - a(x_0,\xi _0) )\mathop \!\textrm\xi \mathop \!\textrmx \right| \\ &\qquad + O(h^). \end$$

By the mean value theorem and the fact that \(a\in S_\delta \cap C_0^\infty (T^* }^d)\), we have for \(x,\xi \in } ^d\),

$$\begin |a\left( h^x+x_0,h^\xi +\xi _0\right) - a(x_0,\xi _0) |&\leqslant (h^ | x | + h^ | \xi |) \Vert \nabla a \Vert _ \\&\leqslant C (h^ | x | + h^ | \xi | ) h^. \end$$

So, using that g is Schwartz,

$$\begin |}(\rho _h\,}}_h(a))-a(x_0,\xi _0)|&\leqslant C \iint |g(x)||}(\xi ) |x||\xi | ((h^ | x | + h^ | \xi | ) h^)\mathop \!\textrm\xi \mathop \!\textrmx\\&= O(h^)+O(h^)+O(h^). \end$$

Noting that \(1-2\delta =\beta -\delta +1-\beta -\delta \) shows the \(O(h^)\) term to be superfluous and gives the desired result. \(\square \)

1.2 B.2. Coherent States on the Quantized Torus

We provide full proofs of Propositions 2.11 and 2.12, though we do not claim our parameters to be optimal.

Proof of Proposition 2.11

For this proof, we work in the orthonormal basis \((Q_n)\). Then for \(a\in S_(}^)\), \(\,}}_N(a)\) has a representation as a matrix A with

$$\begin A_=\sum _}^d}\hat(l,j-m-rN)(-1)^e^}. \end$$

(B.1)

Let \(M=\lfloor N^\rfloor \) for \(\gamma \in (\delta ,1)\). We see that if \(|j-m|>M\) (in distance \(}N\)), then for any \(k\in }_\)

$$\begin |A_|&\leqslant \sum _}^d}|\hat(l,j-m-rN)| \nonumber \\&\leqslant C\sum _}^d}\langle l \rangle ^\Vert (\partial _x^+1)\partial _^ka\Vert _}| j-m-rN|^ \nonumber \\&\leqslant C N^ \sum _}^d} |j - m - rN|^. \end$$

(B.2)

Note that there exists a large constant \(M_0 > 0 \) (independent of N) such that

$$\begin \sum _ r \in }^d\\ |r| > M_0 \end} |j - m - rN|^ \leqslant 1. \end$$

Therefore:

$$\begin \sum _}^d} |j - m - rN|^ \leqslant M_0 M^ + 1 \leqslant C N^ \end$$

for sufficiently large N. Therefore,

$$\begin \begin |A_|&\leqslant C N^N^N^=CN^ \end \end$$

where \(C=C(a,\delta ,k)\) is independent of N. Taking k arbitrarily large gives that \(|A_|=O(N^)\), so (as with pseudodifferential operators on \(}^d\)), the mass of the kernel is concentrated on the diagonal and falls off super-polynomially away from it. We also note that even when j and m are close, we may remove one of the sums in (B.1) up to an \(O(N^)\) error. Specifically we see that when \(|j-m|\leqslant M\), \(k\in }\)

$$\begin \left| A_-\sum _}^d}\hat(l,j-m)e^}\right| =O(N^). \end$$

Indeed

$$\begin \left| A_-\sum _}^d}\hat(l,j-m)e^}\right|&=\left| \sum _ l,r\in }^d \\ r \ne 0 \end}\hat(l,j-m-rN)(-1)^e^} \right| \\&\leqslant \sum _ l,r\in }^d \\ r \ne 0 \end} |\hat(l,j-m-rN)|. \end$$

Note that there exists \(c_d > 0 \) such that for all \(r \ne 0\),

$$\begin |j - m -rN| \geqslant c_d N^\gamma |r| \end$$

therefore for each \(k\in }_\), using that \(a \in S_\delta (}^)\),

$$\begin \sum _ l,r\in }^d \\ r \ne 0 \end} |\hat(l,j-m-rN)|&\leqslant C \sum _ l,r\in }^d \\ r \ne 0 \end} \langle l \rangle ^|j - m -rN|^ \Vert (\partial _x^+1)\partial _^ka\Vert _} \\&\leqslant C N^ \sum _}^d_}c_d N^ \langle r \rangle ^ \end$$

by the same reasoning as in (B.2). By Fourier inversion in the first variable, we have in that case

$$\begin A_=}_2a\left( \frac,j-m\right) +O(N^) \end$$

(B.3)

where \(}_2\) is the Fourier coefficient in the second component (so \(}_2a\) acts on \(}^d\times }^d\)).

Then we compute

$$\begin&}_(\rho _N\,}}_N(a))=C_N\left\langle \psi (N)\right| \,}}_N(a)\left| \psi (N)\right\rangle \nonumber \\&\quad =C_N\sum _}/N})^d}\frac\overlineA_g(j/N)e^\nonumber \\&\quad =C_N\sum _\frac\overlineg(j/N)\nonumber \\&\quad \quad \int _}^d}a\left( \frac,\xi \right) e^\mathop \!\textrm\xi +O(N^)\nonumber \\&\quad =C_N\sum _\frac\overlineg(j/N)\nonumber \\&\quad \quad \int _}^d}a\left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi +O(N^) \end$$

(B.4)

But when \(|j-m|\leqslant M\), we have

$$\begin \left| a\left( \frac,\xi +\xi _0\right) -a\left( \frac,\xi +\xi _0\right) \right| \leqslant MN^\Vert \partial a\Vert _}\leqslant CMN^ \end$$

and

$$\begin \left| g\left( \frac\right) -g\left( \frac\right) \right| \leqslant CMN^, \end$$

so

$$\begin & \overlineg(j/N)\int _}^d}a\left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi \nonumber \\ & \quad =|g(m/N)|^2\int _}^d}a\left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi +O(MN^)\qquad \end$$

(B.5)

As there are \(O(N^dM^d)\) pairs (j, m) with \(|m-j|\leqslant M\), the errors (B.5) sum in (B.4) to give

$$\begin & }_(\rho _N\,}}_N(a))\nonumber \\ & \quad =\sum _}/N})^d}\frac|g(m/N)|^2\sum _\int _}^d}a\nonumber \\ & \qquad \left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi +O(M^N^) \end$$

(B.6)

where we have used Remark 1 to remove the \(C_N\). The inner sum in (B.6) is just a partial Fourier series, and as \(\gamma >\delta \), it differs from the full Fourier series by order \(O(N^)\). This gives

$$\begin }_(\rho _N\,}}_N(a))=\sum _}/N})^d}\frac|g(m/N)|^2a\left( \frac, \xi _0\right) +O(M^N^).\nonumber \\ \end$$

(B.7)

But (B.7) is just a Riemann sum, and we have the estimate

$$\begin & \left| \sum _}/N})^d}\frac|g(m/N)|^2a\left( \frac,\xi _0\right) -\int _}^d}a(x,\xi _0)|g(x)|^2\mathop \!\textrmx\right| =O(N^). \end$$

Therefore,

$$\begin & \left| }_(\rho _N\,}}_N(a))-\int _}^d}a(x,\xi _0)|g(x)|^2\mathop \!\textrmx\right| \leqslant O(M^N^)=(N^) \end$$

for all \(\gamma >\delta \). \(\square \)

Proof of Proposition 2.12

Let \(a\in S_(}^)\) and let \(A=\,}}_N(a)\) given in the basis \((Q_n)\). Once again choose \(M=\lfloor N^\rfloor \) for \(\gamma \in (\delta ,1)\). We have by (B.2) and (B.3) that (writing \(\psi :=\psi _N\) for convenience and letting subscripts refer to the indices)

$$\begin \begin }(\rho _N\,}}_N(a))&=C_N'\sum _}/N})^d}\overlineA_\psi _j\\&=C_N'\sum _\overline\int _}^d}a\left( \frac,\xi \right) e^\mathop \!\textrm\xi \psi _j+O(N^). \end \end$$

(B.8)

But as \(g\in }(}^d)\), it has super-polynomial decay, and we get

$$\begin \psi _m&=N^}\sum _}^d}g_N\left( \frac+k\right) \nonumber \\&=N^}\sum _}^d}g\left( N^\left( \frac+k-x_0\right) \right) e^\nonumber \\&=N^}g\left( N^\left( \frac-x_0\right) \right) e^+O(N^) \end$$

(B.9)

by throwing away all but the first term in the sum. In fact, for many m we may even throw out the first term. Given any \(\epsilon >0\), let \(R_=\lfloor N^\rfloor \). Then if \(|m-Nx_0|\leqslant R_\), we have by (B.9) that

$$\begin |\psi _m|=O(N^). \end$$

(B.10)

We then approximate terms in the summand as in Proposition 2.11, and get when \(|j-m|\leqslant M\) that

$$\begin \left| a\left( \frac,\xi +\xi _0\right) -a\left( \frac,\xi +\xi _0\right) \right| \leqslant MN^\Vert \partial a\Vert _}\leqslant CMN^ \end$$

and

$$\begin \left| g\left( N^\left( \frac-x_0\right) \right) -g\left( N^\left( \frac-x_0\right) \right) \right| \leqslant CMN^ \end$$

so in that case (using \(\delta \leqslant \beta \))

$$\begin \begin \overline\left( \frac-x_0\right) \right) }g\left( N^ \left( \frac-x_0\right) \right) \int _}^d}a\left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi \\=\left| g\left( N^\left( \frac-x_0\right) \right) \right| ^2\int _}^d}a \left( \frac,\xi -\xi _0\right) e^\mathop \!\textrm\xi +O(MN^) \end\nonumber \\ \end$$

(B.11)

But by (B.8), (B.9), and (B.10) we have \(}(\rho _N\,}}_N(a))\) to be equal to

$$\begin \begin&C_N'N^\sum _}\sum _\overline\left( \frac-x_0\right) \right) }g\left( N^ \left( \frac-x_0\right) \right) \cdots \\&\quad \cdots \int _}^d}a\left( \frac,\xi \right) e^\mathop \!\textrm\xi +O(N^) \end\nonumber \\ \end$$

(B.12)

The sum in (B.12) has \(O(M^dR_^d)\) terms, so adding up all the errors given in (B.11) gives

$$\begin&}(\rho _N\,}}_N(a))\\&\quad =N^\sum _}\left| g\left( N^\left( \frac-x_0\right) \right) \right| ^2\\ &\quad \sum _\int _}^d}a\left( \frac,\xi +\xi _0\right) e^\mathop \!\textrm\xi \\ &\qquad +O(M^N^R_^d)\\&\quad =N^\sum _}\left| g\left( N^\left( \frac-x_0\right) \right) \right| ^2a \left( \frac,\xi _0\right) \\ &\qquad +O(M^N^N^) \end$$

where in the last line we approximated the inner sum with a Fourier series as in (B.7), obtaining an \(O(N^)\) error, and applied Remark 1 to ignore the \(C_N'\). At this point, we (up to \(O(N^)\) error) add back in the remaining m terms and compute the Riemann sum to get

$$\begin&}(\rho _N\,}}_N(a))\\&\quad =N^\sum _}/N})^d}\left| g\left( N^\left( \frac-x_0\right) \right) \right| ^2a\left( \frac,\xi _0\right) \\ &\qquad +O(M^N^N^)\\&\quad =\int N^|g(N^(x-x_0))|^2a(x,\xi _0)\mathop \!\textrmx+O(N^)+O(N^)\\ &\qquad +O(M^N^N^)\\&\quad =\int N^|g(N^(x-x_0))|^2a(x,\xi _0)\mathop \!\textrmx+O(M^N^N^) \end$$

with the last line using that \(\delta \leqslant \beta \). We substitute and use \(\int |g|^2=1\) to obtain

$$\begin \begin&\left| \int _}^d} N^|g(N^(x-x_0))|^2a(x,\xi _0)\mathop \!\textrmx-a(x_0,\xi _0)\right| \\&\quad \leqslant \left| \int _}^d}|g(y)|^2\left( a(N^y+x_0,\xi _0) -a(x_0,\xi _0)\right) dy\right| \\&\qquad +\left| \int _}^d\setminus N^}^d}|g(y)|^2\mathop \!\textrmy\right| \\&\quad \leqslant C\left| \int _}^d}|g(y)|^2N^N^|y|\mathop \!\textrmy\right| +O(N^)\leqslant O(N^). \end \end$$

This gives

$$\begin |}(\rho _N\,}}_N(a))-a(x_0,\xi _0)|\leqslant O(N^)+O(M^N^N^) \end$$

which completes the proof by taking \(\varepsilon \) and \(\gamma -\delta \) small enough. \(\square \)

Appendix C: Estimating Derivatives of a Flow

Here we present a proof of Proposition 2.15 which we restate here for convenience.

Proposition C.1

Suppose \(\phi ^t: }^ \rightarrow }^\) is a smooth dynamical system on \(}^\) with Lyapunov exponent \(\Lambda \) where t either takes integer values, or values in \(}_\). Let \(a\in C^\infty (}^)\). Then for all \(\varepsilon >0\) and \(\alpha \in }^\), there exists \(C = C(\varepsilon ,\alpha )\) such that:

$$\begin |\partial ^\alpha (a\circ \phi ^t ) (x) | \leqslant C e^ \Vert a \Vert _} . \end$$

Here \(C^k\) is the norm defined as:

$$\begin \Vert a \Vert _ = \sum _ \Vert \partial ^\alpha a\Vert _}^)}. \end$$

This proof of Proposition C.1 essentially follows [25, §5.2] but in the simpler setting of the torus. To our knowledge, such a proof in this setting is not present in the current literature. In an effort to improve readability, we include examples throughout the proof, at the cost of being rather voluminous.

Proof

We prove Proposition C.1 by strong induction on the order of \(\alpha \). We trivially have the claim when \(|\alpha | = 0\). If \(|\alpha | = 1\), let i be such that \(\partial ^\alpha = \partial _i\). Then we have:

$$\begin \partial ^\alpha _x ( a \circ \phi ^t)(x) = \sum _^ (\partial _j a)\circ (\phi ^t (x))\partial _i (\phi _t(x))_j \end$$

where \((\phi ^t(x))_j\) is the jth component of \(\phi ^t(x)\). By the definition of the Lyapunov exponent, for any \(\epsilon > 0\), there exists \(C = C(\epsilon )> 0\) such that:

$$\begin |\partial _i \phi ^t (x) | \leqslant C e^, \end$$

so that

$$\begin |\partial ^\alpha _x ( a \circ \phi ^t)(x)| \leqslant C e^ \Vert a \Vert _. \end$$

To ease notation in the remained of the proof, define \(\lambda := \Lambda + \varepsilon \).

Before proving the inductive step, we show how to go from \(\alpha \) of order 1 to order 2. We will use the following notation for Jacobians and tensors. If \(f: }\rightarrow }^\), then Df will be the vector with ith component \(\partial _i f\). Note that Df will be a (0, 1) tensor. We define \(D^2f\) as the (0, 2) tensor with i, j entry \((D^2 f )_ = \partial _i \partial _j f\). If \(\phi : }^ \rightarrow }^\), then we let \(D\phi \) be the (1, 1) tensor with i, j entry \((D\phi )_^i = \partial _j \phi _i \). We let \(D^2 \phi \) denote the (1, 2) tensor with i, j, k entry \((D^2\phi )_^i = (\partial _j \partial _k \phi _i)\).

Observe that:

$$\begin (D(a \circ \phi ^n) (x))_i= \partial _i (a\circ \phi ^n)(x) = \sum _j \partial _j ((a\circ \phi ^) \circ \phi ^1(x) ) (\partial _i \phi _j^1)(x) \end$$

so that

$$\begin D(a \circ \phi ^n)(x) =( D(a\circ \phi ^) \circ \phi ^1(x) ) D\phi ^1(x) \end$$

where the product of two tensors is the contraction of the upper and lower components. That is, \(( D(a\circ \phi ^) \circ \phi ^1 )\) is a (0, 1) tensor which can be written \(A_\), and \(D\phi ^1 \) is a (1, 1) tensor which can be written \(B_^\nu \). In this case AB is the (0, 1) tensor contracting the lower component of A with the upper component of B: \((AB)_\gamma = \sum _\nu A_\nu B^\nu _\gamma \).

Similarly

$$\begin (D^2 (a\circ \phi ^n) (x))_&= \partial _i \partial _j (a\circ \phi ^n) (x)\\&= \sum _ ((\partial _k \partial _\ell (a\circ \phi ^))\circ \phi ^1(x) ) \partial _j\phi ^1_\ell (x) \partial _i \phi ^1_k(x) \\&\quad + \sum _k (\partial _k (a \circ \phi ^) \circ \phi ^1(x) )\partial _j \partial _i \phi ^1_k(x) \end$$

so that

$$\begin D^2 (a\circ \phi ^n )(x)&=( D^2 (a\circ \phi ^) \circ \phi ^1(x) ) (D\phi ^1(x) \otimes D\phi ^1(x)) \\&\quad + (D(a \circ \phi ^) \circ \phi ^1(x)) D^2 \phi ^1(x) \end$$

Therefore

$$\begin \underbrace D( a \circ \phi ^)(x) \\ D^2 ( a \circ \phi ^)(x) \end^t}_(x)^t} = \begin D(a \circ \phi ^n)\circ \phi ^1 (x) \\ D^2 (a \circ \phi ^n)\circ \phi ^1 (x) \end ^t\underbrace D \phi ^1 (x) &\quad D^2 \phi ^1 (x) \\ 0 &\quad D\phi ^1 (x) \otimes D\phi ^1 (x) \end}_ , \end$$

so that

$$\begin \begin A_n(x)&= A_(\phi ^1 (x)) M(x)= A_ (\phi ^2(x)) M(\phi ^1 (x)) M(x))\\&= A_0 (\phi ^n(x)) M(\phi ^(x)) M(\phi ^(x)) \cdots M(x). \end \end$$

(C.1)

By the upper triangular structure of M, it is easy to check that

$$\begin M(\phi ^(x)) M(\phi ^(x)) \cdots M(x)= \begin M_1(x) &\quad M_2(x) \\ 0 &\quad M_3(x) \end \end$$

(C.2)

where:

$$\begin M_1(x)&:=\prod _^( (D\phi ^1 )\circ ( \phi ^( x)) ), M_2(x) := \sum _^n (M_2)_j(x), \\ &M_3(x) = \prod _^( ((D\phi ^1 )\circ \phi ^(x) )\otimes ((D\phi ^1 )\circ \phi ^(x) ) ) \end$$

with \((M_2)_j (x)\) defined as

$$\begin \left( \prod _^( (D\phi ^1 )\circ ( \phi ^( x)) )\right) ((D^2 \phi ^1 )\circ \phi ^(x))\left( \prod _^( (D\phi ^1 )\circ ( \phi ^( x)) )^\right) \end$$

(C.3)

where \(a^: = a \otimes a\).

We then have, by (C.1) and (C.2),

$$\begin \begin |D^2 (a \circ \phi ^n)(x) |&= |(Da\circ \phi ^n (x))M_2(x) + (D^2 a \circ \phi ^n(x) )M_3(x)|\\&\leqslant C \Vert a \Vert _ |M_2(x)| + C\Vert a \Vert _ | M_3(x) | \end \end$$

(C.4)

where the absolute value of a tensor denotes the maximum of the absolute value of all entries.

But now observe that

$$\begin \prod _^( (D\phi ^1 )\circ ( \phi ^( x))&= D(\phi ^) \circ \phi ^(x),\\ \prod _^( (D\phi ^1 )\circ ( \phi ^( x)) )^&= (D(\phi ^)(x))^ \end$$

so that

$$\begin | M_3(x) |&= | D(\phi ^n )(x) |^2 \leqslant C e^ \end$$

(C.5)

by the inductive hypothesis and setting a to be coordinate functions. And we also have, by (C.3), that

$$\begin | (M_2 )_j (x)|&\leqslant C e^ e^\Vert D^2 \phi ^1 \Vert _ =C' e^ \end$$

where we use that \(\phi ^1\) is a smooth function on a compact set so that \(\Vert D^2 \phi ^1 \Vert _<\infty \). But we stress that this norm only comes up once in each \((M_2)_j\) term so that the constant at the end is independent of n. We therefore have that:

$$\begin |M_2(x) | \leqslant \sum _^n |(M_2)_j(x)| \leqslant C e ^ \sum _^n e^\leqslant C e^(1- e^)^ \end$$

(C.6)

We can absorb the \((1- e^)^\) term into the constant C to get, using (C.4), (C.5), and (C.6), that

$$\begin |D^2 (a \circ \phi ^n)(x) | \leqslant C \Vert a \Vert _ e^. \end$$

Now if the flow is continuous in time and \(t\in }_\), we let \(n\in }\) be such that \(n \leqslant t \leqslant n+1 \) and observe that:

$$\begin |D^2 (a\circ \phi ^t) (x)| = |D^2 (a\circ \phi ^ \circ \phi ^n )(x) |\leqslant C \Vert a \circ \phi ^ \Vert _ e^. \end$$

Note that \(s:= t-n \in [0,1]\) which is a compact set, so that:

$$\begin \Vert a \circ \phi ^ \Vert _ \leqslant \Vert a \Vert _ \max _\Vert \phi ^s \Vert _. \end$$

(C.7)

We can then absorb \(\max _\Vert \phi ^s \Vert _\) into the constant C to get the result for continuous time.

Now we present the general inductive step, but the reader should keep in mind that morally the idea is the same as above, but with more notation. Suppose we have for all \(k'\leqslant k-1\) and \(|\alpha | = k'\):

$$\begin | \partial ^\alpha (a \circ \phi ^t ) (x)| \leqslant C \Vert a \Vert _} e^. \end$$

We then fix \(n \in }\), and aim to bound \(D^ (a\circ \phi ^n)(x)\). Define \(A_n(x)\) as the \(1\times k\) vector of tensors whose ith entry is \(D^i (a \circ \phi ^n)(x)\). By writing

$$\begin D^i(a\circ \phi ^n) = D^( (D (a \circ \phi ^ ) \circ \phi ^1) D\phi ^1 )(x), \end$$

(C.8)

and repeatedly applying the product rule, we can find a \(k \times k\) matrix of tensors, M(x) such that \(A_n(x) = A_(\phi ^1(x)) M(x)\). Note that M(x) will be upper triangular. Indeed, for any \(i\leqslant k\), \(D^i(a \circ \phi ^n)\) is a linear combination of \(D^j (a\circ \phi ^) \circ \phi ^1\), for \(j \leqslant i\). Also, for each \(i,j = 1, \dots , k\), the i, j entry of M(x) must be a (i, j) tensor because it will be paired with a (0, i) tensor to get a (0, j) tensor. Each component of \((M(x))_\) will contain j derivatives of the components of \(\phi ^1(x)\) with a coefficient depending on k. Lastly note that for each \(i = 1,\dots , k\), \((M(x))_ = (D\phi ^1 (x))^\) as this is the term in (C.8) when all derivatives fall on \(D(a\circ \phi ^\circ \phi ^1 )\).

We then recursively use the relation of \(A_n\) to \(A_\) to get that:

$$\begin A_n(x) = A_0 (\phi ^n(x))\left( \prod _^n M(\phi ^(x))\right) . \end$$

For \(i = 1,\dots ,k\), let \(M_i\) be the (i, k) entry of \(\prod _^n M(\phi ^)\), so that:

$$\begin D^k (a\circ \phi ^n)(x) = \sum _^k (D^j(a)\circ \phi ^n(x) ) M_j(x). \end$$

(C.9)

It now suffices to provide appropriate estimates of \(M_j(x)\). For \(i = 1,\dots ,n\) let \(B^i: = M(\phi ^)(x)\). Then for each fixed \(j = 1, \dots , k\):

$$\begin M_j = \sum _} B^1_ B^2_ \cdots B^n_,k}. \end$$

(C.10)

Because \(B^i\) are upper triangular, this sum is nonzero only if \(\ell _m \leqslant \ell _\) for each \(m = 1,\dots , n-1\), and \(j \leqslant \ell _1\). Each term in the sum can be expressed as a path on a grid of \(k\times (n+1)\) nodes. Labeling the bottom left node (1, 1), every path must begin at (j, 1) and end at \((k,n+1)\). The path can only go to the right or up. That is (m, n) can only go to \((m',n+1)\) for \(m' \geqslant m\) (provided that the path terminates at \((k,n+1)\)). The edges of the path give the index of \(B^i\) in the sum. That is, if the nodes \((m,i),(n,i+1)\) are on the path, then we include the term \(B^i_\) in the sum.

For example, if \(k = 4\) and \(n= 5\), then the following path represents the term \(B^1_B^2_B_^3B^4_B^5_\).

Each path can be represented as a vector whose ith component is m where (m, i) is on the path, e.g., the example path is written as (1, 2, 2, 2, 4, 4).

Each path will consist of flat sections (where the i and \(i-1\) components of the vector representation are the same) and jumps (where the component of the vector representation is not flat). In the above example, the jumps are at indices 2 and 5. The corresponding entry of \(B^i\) for the jumps will be individually norm-bounded. Recall that \(B^i_\) will be a \((j,\ell )\) tensor where all terms are derivatives of \(\phi ^1\). As long as \((j,\ell ) \ne (1,k)\), at most \(k-1\) derivatives are taken of \(\phi ^1\). Note that \(B_\) can only appear once in any term in the sum (C.10), and this is bounded by the \(C^k\) norm of \(\phi ^1\). We therefore can apply the inductive hypothesis to get that \(|B^i_(x)| \leqslant C e^\). Next, we bound the flat sequences. Suppose such a sequence is \(\prod _^B_^i\). Recall that \(B_^i= (D\phi ^1)^\circ (\phi ^(x))\), so that

$$\begin \left| \prod _^B_^i (x)\right| = | (D(\phi ^)\circ \phi ^ (x))^|\leqslant C e^. \end$$

(C.11)

This follows by the inductive hypothesis by setting a to be coordinate functions. Now suppose we have a term of the sum (C.10) whose path has a vector representation \((a_1,a_2,\dots ,a_) \) with jumps at indices \(\ell _1,\ell _2,\dots , \ell _s\). We assume that there is a jump at index 2 and the path ends in a flat section (as in the example), but all cases can be treated similarly. We can then bound this term of the sum (using (C.11)) by:

$$\begin&| B_,a_}^| \cdots | B_,a_}^| \left| \prod _^B^i_,a_} \right| \cdots \left| \prod _+1}^B^i_},a_}} \right| \\&\quad \leqslant Ce^ + \cdots + a_)} e^ (\ell _2 - \ell _1 -1)} \cdots e^}(n+1 - \ell _-1)}\\&\quad = C e^ (\ell _2 - \ell _1