Limiting Absorption Principle for Discrete Schrödinger Operators with a Wigner–von Neumann Potential and a Slowly Decaying Potential

Abstract

We consider discrete Schrödinger operators on $\mathbb {Z}^d$ for which the perturbation consists of the sum of a long-range-type potential and a Wigner–von Neumann-type potential. Still working in a framework of weighted Mourre theory, we improve the limiting absorption principle (LAP) that was obtained in Mandich (J Funct Anal 272(6):2235–2272, 2017). To our knowledge, this is a new result even in the one-dimensional case. The improvement is twofold. It weakens the assumptions on the long-range potential and provides better LAP weights. Both upgrades include logarithmic terms. The proof relies on the fact that some particular functions that contain logarithmic terms, are operator monotone. This fact is proved using Loewner’s theorem and Nevanlinna functions.

Appendices

Appendix A. Nevanlinna Functions, Operator Monotone Functions and Loewner’s Theorem

We revisit Loewner’s theorem on matrix operator monotone functions, see, e.g., [42,43,44,45]. This wonderful theorem makes a striking connection between the operator monotone functions and the Nevanlinna functions. Let $\mathbb {C}_+$ (resp. $\mathbb {C}_-$) denote the complex numbers with strictly positive (resp. negative) imaginary part. A Nevanlinna function (also known as Herglotz, Pick or R function) is an analytic function on $\mathbb {C}_+$ that maps $\mathbb {C}_+$ to $\overline{\mathbb {C}_+}$. A function $f : \mathbb {C}_+ \mapsto \mathbb {C}$ is Nevanlinna if and only if it admits a representation

$$\begin{aligned} f(z) = \alpha + \beta z + \int _{\mathbb {R}} \left( \frac{1}{\lambda - z} - \frac{\lambda }{\lambda ^2+1} \right) d \mu (\lambda ), \quad z \in \mathbb {C}^+ \end{aligned}$$

(7.1)

where $\alpha \in \mathbb {R}$, $\beta \geqslant 0$, and $\mu$ is a positive Borel measure on $\mathbb {R}$ satisfying $\int _{\mathbb {R}} (\lambda ^2+1) ^{-1} d\mu (\lambda ) < \infty$. We refer to [42, Theorem 1 of Chapter II] for a proof of this wonderful result. The integral representation is unique. The measure $\mu$ is recovered from f by the Stieltjes inversion formula

$$\begin{aligned} \mu \left( (\lambda _1, \lambda _2] \right) = \lim \limits _{\delta \downarrow 0} \lim \limits _{\varepsilon \downarrow 0} \frac{1}{\pi } \int _{\lambda _1 + \delta } ^{\lambda _2 + \delta } \mathrm {Im} \left( f(\lambda + \mathrm {i}\varepsilon ) \right) d\lambda .\end{aligned}$$

Standard examples of Nevanlinna functions given in the literature include $z^{p}$ for $0 \leqslant p \leqslant 1$, $-z^{p}$ for $-1 \leqslant p \leqslant 0$, and the logarithm $\ln (z)$ with the branch cut $(-\infty ,0]$.

Notation. Let (a, b) be an open interval (finite or infinite). Denote P(a, b) the set of Nevanlinna functions that continue analytically across (a, b) into $\mathbb {C}_-$ and where the continuation is by reflection.

Functions in P(a, b) are real-valued on (a, b) and their measure satisfies $\mu \left( (a,b) \right) = 0$; see [42, Lemma 2, Chapter II]. Functions in P(a, b) are strictly increasing on (a, b), unless they are constant. Indeed, if f is not a constant function $\beta$ and $\mu$ cannot be simulaneously zero and so

$$\begin{aligned} f'(x) = \beta + \int _{\mathbb {R}} \frac{d\mu (\lambda )}{(\lambda -x)^2} > 0, \quad x \in (a,b).\end{aligned}$$

Let $\mathcal {M}_n(\mathbb {C})$ be the set of $n \times n$ matrices with entries in $\mathbb {C}$, and consider a function $f : (a,b) \mapsto \mathbb {R}$.

Definition. f is matrix monotone of order n in (a, b) if $f(T) \leqslant f(S)$ holds whenever T, S in $\mathcal {M}_n(\mathbb {C})$ are hermitian matrices with spectrum in (a, b) and $T \leqslant S$.

In 1934, Karl Loewner proved the following remarkable theorem that characterizes the matrix monotone functions:

Theorem 7.1

[43] Let $f : (a,b) \mapsto \mathbb {R}$, where (a, b) is a finite or infinite open interval. Then f is matrix operator monotone of order n in (a, b) for all $n \in \mathbb {N}$ if and only if f admits an analytic continuation that belongs to P(a, b).

Loewner’s theorem is a truly wonderful result and has been reproved in several different ways, see, e.g., [42, 44, Theorem I of Chapter VII and Chapter IX for the converse] or [45] and references therein for a concise historical exposition. For the purpose of this article, we need a version of Loewner’s theorem that applies to unbounded self-adjoint operators. In Simon’s book [44, Chapter 2], it is explicitly discussed how Loewner’s theorem extends to unbounded operators. We propose below yet another proof of the extension to the semi-bounded operators (which is the case for $\langle A \rangle$ and $\langle N \rangle$). For self-adjoint operators T, S which are bounded from below, the inequality $T \leqslant S$ means $\mathrm {Dom}[|S|^{1/2}]\subset \mathrm {Dom}[|T|^{1/2}]$ and $\langle \psi , T \psi \rangle \leqslant \langle \psi , S \psi \rangle$ for all $\psi \in \mathrm {Dom}[|S|^{1/2}]$.

Definition. f is operator monotone in $(a,b) \subset \mathbb {R}$ if $f(T) \leqslant f(S)$ holds whenever T, S (possibly unbounded) are self-adjoint operators in $\mathcal {H}$ with spectrum contained in (a, b) and $T \leqslant S$.

Assuming $f \in P(a,b)$, let $\mu$ be the measure associated with f, see (7.1), and denote the supremum of the support of $\mu$ by $\varSigma _{\mu }$. In what follows, the discussion is for general self-adjoint T, S, but one should keep in mind that the results are applied to $T =\langle A \rangle$ and $S = \sqrt{c_d} \langle N \rangle$, cf. Lemma 4.13. We start with a lemma:

Lemma 7.2

Let f belong to $P(a,+\infty )$. Then

• (1) $\varSigma _{\mu } \leqslant a$,

• (2) given T a self-adjoint operator with $\inf (\sigma (T))> a >0$, we have f(T) bounded from below and $\mathrm {Dom}[T] \subseteq \mathrm {Dom}[|f(T)|]$ with equality if and only if $\beta >0$.

Proof

For (1), we start by noting that f belongs to $P(a,+\infty )$ so f admits an integral representation as in (7.1). Thus, $\varSigma _{\mu } \leqslant a$ holds thanks to the Stieltjes inversion formula and the fact that f is real-valued on $(a,+\infty )$. For (2), adding a constant to f does not alter the assumptions of the lemma and leads to the same conclusion, so we may assume $f(\inf (\sigma (T))) >0$, where $\inf (\sigma (T)) > a$ (recall f is non-decreasing on $(a,+\infty )$). We start by showing that $\lim f(x)/x = \beta$, as $x\rightarrow +\infty$, or equivalently

$$\begin{aligned} \lim \limits _{x \rightarrow +\infty } \int _{-\infty } ^{\varSigma _{\mu }} \frac{1+\lambda x}{x(\lambda -x)(\lambda ^2+1)} d\mu (\lambda ) = 0.\end{aligned}$$

We wish to exchange the order of the limit and integration. We have

$$\begin{aligned} \big | (1+\lambda x)x^{-1}(\lambda -x)^{-1} \big | \leqslant 1, \quad \forall (\lambda ,x) \in (-\infty ,0] \times [1,+\infty ) \cup [0,a] \times [a+\sqrt{a^2+1}, + \infty ). \end{aligned}$$

(7.2)

We may apply the dominated convergence theorem, and the above limit follows. This limit implies that f(x)/x is a bounded function on $(a,+\infty )$ and hence $\mathrm {Dom}[T] \subset \mathrm {Dom}[|f(T)|]$. For the reverse inclusion, x/f(x) is a well-defined bounded function on $(\inf (\sigma (T)),+\infty )$ iff $\beta >0$. $\square$

We are now ready to prove the extension of Theorem 7.1 to semi-bounded operators:

Theorem 7.3

Let $f : (a,b) \mapsto \mathbb {R}$, where $0< a < b \leqslant +\infty$. Then f is operator monotone in (a, b) if and only if f admits an analytic continuation that belongs to P(a, b).

Proof

If f is operator monotone in (a, b), then in particular it is matrix operator monotone of order n in (a, b) for all $n \in \mathbb {N}$, and so f admits an analytic continuation that belongs to P(a, b) by Loewner’s theorem. This direction is in fact the hard direction in Loewner’s theroem, but the extension is trivial!

For the converse, we suppose that f admits an analytic continuation that belongs to P(a, b). Consider the two separate cases $b < +\infty$ and $b = +\infty$. If $b < +\infty$, then f is matrix operator monotone of order n in (a, b) for all $n \in \mathbb {N}$ by Loewner’s theorem. But, since the interval (a, b) is finite, this is equivalent to being operator monotone in (a, b); see, e.g., [46, Lemma 2.2] or [44, Chapter 2]. Now the case $b=+\infty$. We have $\varSigma _{\mu } \leqslant a$ by Lemma 7.2, and

$$\begin{aligned} f(x) = \alpha + \beta x + \int _{-\infty } ^{\varSigma _{\mu }} \left( \frac{1}{\lambda - x} - \frac{\lambda }{\lambda ^2+1} \right) d \mu (\lambda ), \quad x \in (a,+\infty ) .\end{aligned}$$

Let

$$\begin{aligned} g_r (x) := \alpha + \int _{-r} ^{\varSigma _{\mu }} \left( \frac{1}{\lambda - x} - \frac{\lambda }{\lambda ^2+1} \right) d \mu (\lambda ), \quad g(x) := f(x) - \beta x, \quad x \in (a,+\infty ). \end{aligned}$$

(7.3)

Note that $\{g_r \}_{r \in \mathbb {R}^+}$ is of a sequence of functions of the real variable x that converges pointwise to g(x) as $r \rightarrow + \infty$ for all $x \in (a,+\infty )$. Moreover, since for every fixed r, $g'_r(x) \geqslant 0$, all functions in the sequence are increasing in the variable x. We need a lemma.

Lemma 7.4

For every fixed $\varepsilon > 0$, the subsequence $\{ g_r(x) \}_{r > 1/ \varepsilon }$ has the property that $g_r(x) \nearrow g(x)$ for every $x >\max (\varSigma _{\mu } + \varepsilon , a)$.

Proof

Clearly $g_r(x) \rightarrow g(x)$ as $r\rightarrow +\infty$ for all $x \in (a,+\infty )$. What needs to be shown is that the sequence $g_r(x)$ is increasing pointwise. Let $\varepsilon > 0$ be given and fix $x > \max (\varSigma _{\mu } + \varepsilon , a)$. Recall a is assumed to be strictly positive. The integrand in (7.3) is equal to

$$\begin{aligned} \frac{1+\lambda x}{(\lambda -x) (\lambda ^2+1)}.\end{aligned}$$

On the one hand, $x > \varSigma _{\mu }+ \varepsilon$ implies that $(\lambda -x)$ is negative for all $\lambda \in \mathrm {supp} \ \mu$. On the other hand, the assumption $x> a > 0$ implies that $(1+\lambda x)$ is negative for all $\lambda < -1/x$. Thus, for all $x >\max (\varSigma _{\mu } + \varepsilon , a)$ and for all $\lambda < -1/a$, the integrand in (7.3) is strictly positive. Thus, as r increases above 1/a, the value of the integral in (7.3) strictly increases. This completes the proof of the lemma. $\square$

Continuation of the Proof of Theorem7.3. Now fix T, S self-adjoint operators with spectrum contained in $(a,+\infty )$ and $T \leqslant S$. Let $\psi \in \mathcal {H}$. $T \leqslant S$ implies $\langle \psi , (\lambda - T)^{-1} \psi \rangle \leqslant \langle \psi , (\lambda - S)^{-1} \psi \rangle$ for all $\lambda \leqslant \varSigma _{\mu }$, e.g., [47, Theorem VI.2.21]. We integrate over the compact $[-r,\varSigma _{\mu }]$ :

$$\begin{aligned} \int _{-r}^{\varSigma _{\mu }} \langle \psi , (\lambda - T)^{-1} \psi \rangle \ d\mu (\lambda ) \leqslant \int _{-r}^{\varSigma _{\mu }} \langle \psi , (\lambda - S)^{-1} \psi \rangle \ d\mu (\lambda ), \quad \text {for every finite} \ r > |\varSigma _{\mu }|.\end{aligned}$$

Moreover, since the integrand is norm continuous, we infer

$$\begin{aligned} \langle \psi , g_r(T) \psi \rangle \leqslant \langle \psi , g_r(S) \psi \rangle , \quad \text {for every finite} \ r > |\varSigma _{\mu }|. \end{aligned}$$

(7.4)

Next we choose in Lemma 7.4$\varepsilon$ small enough so that $\max (\varSigma _\mu +\varepsilon , a)<\inf (\sigma (T))\leqslant \inf (\sigma (S))$. We obtain $g_r(x) \nearrow g(x)$ (for $r>1/\varepsilon$) as r goes to infinity. In turn, the monotone convergence theorem for forms, e.g., [47, Theorem VIII.3.11], ensures that $\langle \psi , g_r(T) \psi \rangle$ converges to $\langle \psi , g(T)\psi \rangle$ as $r \rightarrow +\infty$ for all $\psi \in \mathrm {Dom}[|g(T)|^{1/2}]$, and similarly for S. Taking limits in (7.4) gives $\mathrm {Dom}[|g(S)|^{1/2}]\subset \mathrm {Dom}[|g(T)|^{1/2}]$ and

$$\begin{aligned} \langle \psi , g(T) \psi \rangle \leqslant \langle \psi , g(S) \psi \rangle , \quad \psi \in \mathrm {Dom}[|g(S)|^{1/2}].\end{aligned}$$

Thus, if $\beta = 0$ we have

$$\begin{aligned} \langle \psi , f(T) \psi \rangle \leqslant \langle \psi , f(S) \psi \rangle , \quad \psi \in \mathrm {Dom}[|g(S)|^{1/2}]= \mathrm {Dom}[|f(S)|^{1/2}],\end{aligned}$$

whereas if $\beta > 0$, we use the fact that $0 < T\leqslant S$, and $\mathrm {Dom}[S^{1/2}] = \mathrm {Dom}[|f(S)|^{1/2}]$, $\mathrm {Dom}[T^{1/2}]= \mathrm {Dom}[|f(T)|^{1/2}]$, by Lemma 7.2, which yields

$$\begin{aligned} \langle \psi , f(T) \psi \rangle \leqslant \langle \psi , f(S) \psi \rangle , \quad \psi \in \mathrm {Dom}[S^{1/2}] = \mathrm {Dom}[|f(S)|^{1/2}].\end{aligned}$$

This gives the result. $\square$

To close this section, we have a remark about results in the literature on order relations for general self-adjoint operators. In [48], Olson introduced the spectral order for self-adjoint operators: $T \preceq S$ if and only if $E_{(-\infty ,t]}(T) \geqslant E_{(-\infty ,t]}(S)$ for all $t \in \mathbb {R}$. Here, $E_{(-\infty ,t]}(T)$ and $E_{(-\infty ,t]}(S)$ are the spectral resolutions of the identity for T and S. He showed that $T^n \leqslant S^n$ for every $n \in \mathbb {N}$ is equivalent to $T \preceq S$, see also [49, Proposition 5]. Furthermore, it is shown in [50] that this order relation is equivalent to $f(T) \leqslant f(S)$ for any continuous monotone non-decreasing function f defined on an interval which contains $\sigma (T) \cup \sigma (S)$. For the purpose of this article, we do not know if $\langle A \rangle ^n \leqslant (\sqrt{c_d} \langle N \rangle )^n$ holds $\forall n \in \mathbb {N}$ but if it does it would considerably simplify this article. To check this inequality by brute force seems to be unbearable.

Appendix B. Polylogarithms of Positive Order are Nevanlinna Functions

That logarithm with the standard branch cut is a Nevanlinna function follows from the identity $\ln (z) = \ln (r) + \mathrm {i}\theta$, where $z = r e^{\mathrm {i}\theta }$, $r>0$, and $\theta \in (-\pi ,\pi )$. The integral representation of the logarithm is

$$\begin{aligned} \ln (z) = \int _{-\infty } ^0 \left( \frac{1}{\lambda -z} - \frac{\lambda }{\lambda ^2+1} \right) d\lambda . \end{aligned}$$

The composition of Nevanlinna functions produces another Nevanlinna function. So, for example, $\ln ^p(z)$ is Nevanlinna for $0 \leqslant p \leqslant 1$. What about higher powers of the logarithm? Certainly the square and cube of the logarithm are not Nevannlina functions. Indeed writing

$$\begin{aligned} \ln ^2(z) = \ln ^2(r) - \theta ^2 + \mathrm {i}2 \theta \ln (r), \quad \text {and} \quad \ln ^3(z) = \ln ^3(r) - 3\theta ^2 \ln (r) + \mathrm {i}\theta ( 3\ln ^2(r) -\theta ^2)\end{aligned}$$

reveals that these functions do not map $\mathbb {C}_+$ to $\overline{\mathbb {C}_+}$. In spite of this, there are functions that are Nevanlinna and are “almost” equal to the logarithms. To motivate the idea, we note that the Stieltjes inversion formula gives

$$\begin{aligned} \lim \limits _{\delta \downarrow 0} \lim \limits _{\varepsilon \downarrow 0} \frac{1}{\pi } \int _{\lambda _1 + \delta } ^{\lambda _2 + \delta } \mathrm {Im} \left( \ln ^2(\lambda + \mathrm {i}\varepsilon ) \right) d\lambda = {\left\{ \begin{array}{ll} 0 &{} \lambda _1 \geqslant 0, \\ \int _{\lambda _1} ^{\lambda _2} 2\ln (|\lambda |) d\lambda &{} \lambda _2 \leqslant 0 \\ \end{array}\right. } \end{aligned}$$

when applied to $\ln ^2(z)$ and

$$\begin{aligned} \lim \limits _{\delta \downarrow 0} \lim \limits _{\varepsilon \downarrow 0} \frac{1}{\pi } \int _{\lambda _1 + \delta } ^{\lambda _2 + \delta } \mathrm {Im} \left( \ln ^3(\lambda + \mathrm {i}\varepsilon ) \right) d\lambda = {\left\{ \begin{array}{ll} 0 &{} \lambda _1 \geqslant 0, \\ \int _{\lambda _1} ^{\lambda _2} \left( 3\ln ^2(| \lambda |) - \pi ^2 \right) d\lambda &{} \lambda _2 \leqslant 0 \\ \end{array}\right. } \end{aligned}$$

when applied to $\ln ^3(z)$. This suggests to calculate the Nevanlinna functions corresponding to the measures $\mathrm {d}\mu (\lambda ) = \varvec{1}_{\{\lambda < -1\}} \ln (-\lambda ) d\lambda$ and $\mathrm {d}\mu (\lambda ) = \varvec{1}_{\{\lambda < -1\}} \ln ^2(-\lambda ) d\lambda$.

We introduce polylogarithms. We refer to [51, 52] for formulas and a detailed exposition. The polylogarithm of order $\sigma \in \mathbb {C}$ is defined by the power series

$$\begin{aligned} \mathrm {Li}_{\sigma }(z) := \sum _{k=1} ^{\infty } \frac{z^k}{k^{\sigma }}.\end{aligned}$$

The definition is valid for complex $|z| <1$ and is extended to the complex plane by analytic continuation. For the purpose of this article, we are interested in the polylogarithms with $\sigma > 2$, or $\sigma = 3$ if we want to simplify by taking the smallest integer above 2. The standard branch cut is $[1,+\infty )$ for $\mathrm {Li}_1(z)$ and $(1,+\infty )$ for $\mathrm {Li}_2(z)$ and $\mathrm {Li}_3(z)$. The polylogarithm of order 1 can be written in terms of a logarithm as $\mathrm {Li}_1(z) = - \ln (1-z)$. The polylogarithm of order 2 is called the dilogarithm or Spence function, while the polylogarithm of order 3 is called the trilogarithm. On p. 494 of [52], the following integral representation is given without proof:

$$\begin{aligned} \mathrm {Li}_{\sigma +1} (z) = \frac{z}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \frac{\ln ^{\sigma } (\lambda ) }{\lambda (\lambda -z)} d\lambda , \end{aligned}$$

(8.1)

for $| \mathrm {arg} (1-z) | < \pi , \mathrm {Re}(\sigma ) > -1$, or for $z=1, \mathrm {Re}(\sigma ) > 0$. Here, $\varGamma$ is the gamma function. Obviously (8.1) is equivalent to

$$\begin{aligned} \mathrm {Li}_{\sigma +1} (z) = - \frac{1}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \frac{\ln ^{\sigma } (\lambda ) }{\lambda (\lambda ^2+1)} d\lambda + \frac{1}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \left( \frac{1}{\lambda -z} - \frac{\lambda }{ \lambda ^2+1} \right) \ln ^{\sigma } (\lambda ) d\lambda . \end{aligned}$$

(8.2)

This means that $\mathrm {Li}_{\sigma +1}(z)$ is a Nevanlinna function for $\mathrm {Re}(\sigma ) > -1$. Although not difficult to prove, it is not clear where a Proof of (8.1) can be found in the literature. Thus, we prove it:

Proposition 8.1

(8.1) is true.

Proof

Let $\lambda \geqslant 1$. Writing $1/(\lambda (\lambda -z))$ as a power series in z we have $1/(\lambda (\lambda -z)) = \sum _{k=0} ^{\infty } \lambda ^{-k-2} z^k$ for $|z| < 1$. Then the rhs of (8.1) is equal to

$$\begin{aligned} \frac{z}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \sum _{k=0} ^{\infty } \frac{\ln ^{\sigma } (\lambda ) }{\lambda ^{k+2}} z^k d\lambda = \sum _{k=1} ^{\infty } \frac{z^k}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \frac{\ln ^{\sigma } (\lambda ) }{\lambda ^{k+1}} d\lambda = \sum _{k=1} ^{\infty } \frac{z^k}{k^{\sigma +1}}, \quad \text {Re}(\sigma ) >-1.\end{aligned}$$

To evaluate the last integral the change of variable $k \ln (\lambda ) = t$ was performed, followed by the definition of the gamma function. Thus, the rhs of (8.1) is equal to $\mathrm {Li}_{\sigma +1}(z)$ for $|z|<1$ and $\text {Re}(\sigma ) >-1$. The result follows by the uniqueness of the analytic continuation. $\square$

While we are at it, we note that $\mathrm {Li}_{0} (z) = z/(1-z)$ is also a Nevanlinna function.

Definition

For $\sigma \in \mathbb {C}$,

$$\begin{aligned} \varPhi _{\sigma } (z) := - \mathrm {Li}_{\sigma }(-z), \quad z \in \mathbb {C}{\setminus } (-\infty ,-1]. \end{aligned}$$

(8.3)

Clearly (8.2) implies that the $\varPhi _{\sigma }$ are Nevanlinna for $\text {Re}(\sigma ) >-1$ with integral representations given by:

$$\begin{aligned} \varPhi _{\sigma +1} (z) = \frac{1}{\varGamma (\sigma +1)} \int _{1} ^{\infty } \frac{\ln ^{\sigma } (\lambda ) }{\lambda (\lambda ^2+1)} d\lambda + \frac{1}{\varGamma (\sigma +1)} \int _{-\infty } ^{-1} \left( \frac{1}{\lambda -z} - \frac{\lambda }{ \lambda ^2+1} \right) \ln ^{\sigma } (-\lambda ) \mathrm{d}\lambda , \end{aligned}$$

for $|\text {arg}(1+z) | < \pi$. Finally, the other reason we resort to polylogarithms is because they decay at the same rate as the logarithms, at least for positive integer order (this follows directly from the inversion/reflection formula [51, (6) of Appendix A.2.7] together with $\mathrm {Li}_n(0)=0$), namely

$$\begin{aligned} \lim \limits _{x \rightarrow +\infty } \frac{\varPhi _n(x)}{\ln ^ n (x)} = \frac{1}{n!}, \quad n \in \mathbb {N}. \end{aligned}$$

(8.4)

Appendix C. Almost Analytic Extensions and Helffer–Sjöstrand Calculus

We refer to [9, 28, 38, 53, 54] for more details. Let $\rho \in \mathbb {R}$ and denote by $\mathcal {S}^{\rho }(\mathbb {R})$ the class of functions $\varphi$ in $C^{\infty }(\mathbb {R})$ such that

$$\begin{aligned} |\varphi ^{(k)} (x)| \leqslant C_k \langle x \rangle ^{\rho -k}, \quad \text {for all} \ k \in \mathbb {N}. \end{aligned}$$

(9.1)

Lemma 9.1

[38, 53] Let $\varphi \in \mathcal {S}^{\rho }(\mathbb {R})$, $\rho \in \mathbb {R}$. Then for every $N \in \mathbb {Z}^+$ and every $c>0$, there exists a smooth function $\tilde{\varphi }_N : \mathbb {C}\rightarrow \mathbb {C}$, called an almost analytic extension of $\varphi$, satisfying:

$$\begin{aligned} \tilde{\varphi }_N(x+\mathrm {i}0)= & {} \varphi (x), \quad \forall x \in \mathbb {R};\nonumber \\ \mathrm {supp} \ (\tilde{\varphi }_N) \subset \varOmega:= & {} \{x+\mathrm {i}y : |y| \leqslant c \langle x \rangle \};\nonumber \\ \tilde{\varphi }_N(x+\mathrm {i}y)= & {} 0, \quad \forall y \in \mathbb {R}\ \mathrm {whenever} \ \varphi (x) = 0;\nonumber \\ \forall \ell \in \mathbb {N}\cap [0,N], \Bigg | \frac{\partial \tilde{\varphi }_N}{\partial \overline{z}}(x+\mathrm {i}y) \Bigg |\leqslant & {} c_{\ell } \langle x \rangle ^{\rho -1-\ell } |y|^{\ell } \ \mathrm {for \ some \ constants} \ c_{\ell } >0.\nonumber \\ \end{aligned}$$

(9.2)

Now let Q be a self-adjoint operator.

Lemma 9.2

Let $\rho < 0$ and $\varphi \in \mathcal {S}^{\rho }(\mathbb {R})$. Then for all $k \in \mathbb {N}$ and $N \in \mathbb {N}$:

$$\begin{aligned} \varphi ^{(k)}(Q) = \frac{\mathrm {i}(k!)}{2\pi } \int _{\mathbb {C}} \frac{\partial \tilde{\varphi }_N}{\partial \overline{z}} (z) (z-Q)^{-1-k} \mathrm{d}z \wedge \mathrm{d}\overline{z} \end{aligned}$$

(9.3)

where the integral exists in the norm topology. For $\rho \geqslant 0$, the following limit exists:

$$\begin{aligned} \varphi ^{(k)}(Q)f = \lim \limits _{R \rightarrow \infty } \frac{\mathrm {i}(k!)}{2\pi } \int _{\mathbb {C}} \frac{\partial (\tilde{\varphi \theta _R})_N}{\partial \overline{z}} (z) (z-Q)^{-1-k} f \mathrm{d}z \wedge \mathrm{d}\overline{z}, \quad \text {for all} \ f \in \mathrm {Dom}[\langle Q \rangle ^{\rho }]. \end{aligned}$$

(9.4)

In particular, if $\varphi \in \mathcal {S}^{\rho }(\mathbb {R})$ with $0 \leqslant \rho < k$ and $\varphi ^{(k)}$ is a bounded function, then $\varphi ^{(k)}(Q)$ is a bounded operator and (9.3) holds (with the integral converging in norm).

Proposition 9.3

[28]. Let T be a bounded self-adjoint operator satisfying $T \in \mathcal {C}^1(Q)$. Then:

$$\begin{aligned}{}[T,(z-Q)^{-1}]_{\circ } = (z-Q)^{-1}[T,Q]_{\circ }(z-Q)^{-1}, \end{aligned}$$

(9.5)

and for any $\varphi \in \mathcal {S}^{\rho }(\mathbb {R})$ with $\rho < 1$, $T \in \mathcal {C}^1(\varphi (Q))$ and

$$\begin{aligned}{}[T,\varphi (Q)]_{\circ } = \frac{\mathrm {i}}{2\pi }\int _{\mathbb {C}} \frac{\partial \tilde{\varphi }_N}{\partial \overline{z}} (z-Q)^{-1}[T,Q]_{\circ }(z-Q)^{-1} \mathrm{d}z\wedge \mathrm{d}\overline{z}. \end{aligned}$$

(9.6)