Reflection Positive One-Parameter Groups and Dilations

Abstract

The concept of reflection positivity has its origins in the work of Osterwalder–Schrader on constructive quantum field theory. It is a fundamental tool to construct a relativistic quantum field theory as a unitary representation of the Poincaré group from a non-relativistic field theory as a representation of the euclidean motion group. This is the second article in a series on the mathematical foundations of reflection positivity. We develop the theory of reflection positive one-parameter groups and the dual theory of dilations of contractive hermitian semigroups. In particular, we connect reflection positivity with the outgoing realization of unitary one-parameter groups by Lax and Phillips. We further show that our results provide effective tools to construct reflection positive representations of general symmetric Lie groups, including the $ax+b$-group, the Heisenberg group, the euclidean motion group and the euclidean conformal group.

Appendices

Appendix A: Rotation Invariant Reflection Positive Measures on ${\mathbb R}^d$

In this section we take a closer look at rotation invariant tempered measures $\nu$ on ${\mathbb R}^d$ which are reflection positive in the sense that their Fourier transform $\widehat{\nu }$ is a reflection positive distribution w.r.t. (${\mathbb R}^d, {\mathbb R}^d_+,\theta )$, where $\theta (x_0,\mathbf {x}) = (-x_0,\mathbf {x})$. In Euclidean Quantum Field Theories, the Fourier transforms of these measures are the Schwinger distributions $S_2(x,y) = \widehat{\nu }(x-y)$ describing the euclidean $2$-point “functions” (cf. [6, p. 91]).

The following lemma will be useful in our discussion of examples.

Lemma 7.1

For a non-zero positive Borel measure $\rho$ on $[0,\infty [$, we consider the rotation invariant measure $d\nu (x) := \Theta (x)dx$ on ${\mathbb R}^d$ with the density $\Theta (x) := \int _0^\infty \frac{1}{y^2 + x^2}\, d\rho (y)$. Then the following are equivalent:

• (i) There exists a non-zero $x \in {\mathbb R}^d$ with $\Theta (x) < \infty$.

• (ii) $\Theta (x) < \infty$ for every non-zero $x \in {\mathbb R}^d$.

• (iii) $L^2({\mathbb R}^d,\nu ) \not =\{0\}$.

• (iv) $\int _0^\infty \frac{d\rho (y)}{1+y^2} < \infty$. This implies that $\rho$ is tempered.

If these conditions are satisfied, then the following assertions hold:

• (a) $\lim _{\Vert x\Vert \rightarrow \infty } \Theta (x)x^2 = \rho ([0,\infty [)> 0.$

• (b) $L^2({\mathbb R}^d,\nu )$ contains a non-zero polynomial if and only if $d = 1$ and $\nu$ is finite, and then every polynomial in $L^2({\mathbb R}^d,\nu )$ is constant. We have $\nu ({\mathbb R}) = \pi \int _0^\infty \frac{d\rho (y)}{y}$.

Proof

The equivalence of (i)–(iv) is easy to see. Suppose that these conditions are satisfied.

(a) follows from

$$\begin{aligned} \lim _{x \rightarrow \infty } \Theta (x)x^2 = \lim _{x \rightarrow \infty } \int _0^\infty \frac{x^2}{y^2 + x^2}\, d\rho (y) = \int _0^\infty \, d\rho (y) > 0. \end{aligned}$$

(b) Let $f = \sum _{j = 0}^N f_j : {\mathbb R}^d \rightarrow {\mathbb C}$ be a polynomial of degree $N$, where the $f_j$ are homogeneous of degree $j$. For a suitably normalized measure $\sigma$ on the sphere ${\mathbb S}^{d-1}$, we have the integral formula

$$\begin{aligned} \int _{{\mathbb R}^d} F(x)\, dx = \int _0^\infty \int _{{\mathbb S}^{d-1}} F(r x)\, d\sigma (x)\, r^{d-1}\, dr, \end{aligned}$$

so that

$$\begin{aligned} \int _{{\mathbb R}^d} F(x)\, d\nu (x) = \int _0^\infty \int _{{\mathbb S}^{d-1}} F(r x) \, d\sigma (x)\, \Theta (r)r^{d-1}\, dr. \end{aligned}$$

The function $|f|^2$ is a polynomial of degree $2N$, and this implies that $h(r) := \int _{{\mathbb S}^{d-1}} |f(r x)|^2\, d\sigma (x)$ also is a polynomial of degree $2N$. If

$$\begin{aligned} \int _{{\mathbb R}^d} |f(x)|^2\, d\nu (x) = \int _0^\infty h(r)\Theta (r)r^{d-1}\, dr \end{aligned}$$

is finite, (a) implies that $\int _1^\infty h(r)r^{d-3}\, dr < \infty$, so that we obtain $2N + d- 3 < -1$, i.e., $2N + d < 2$. This leaves only the possibilities $d = 1$ and $N= 0$, so that $f$ must be constant. But $1 \in L^2({\mathbb R}^d,\nu )$ is equivalent to the finiteness of the measure $\nu$. In view of

$$\begin{aligned} \nu ({\mathbb R}) = \int _0^\infty \int _{{\mathbb R}} \frac{dx}{y^2 + x^2}\, d\rho (y) = \int _0^\infty \frac{\pi }{y}\, d\rho (y), \end{aligned}$$

$\nu$ is finite if and only if $\int _0^\infty \frac{1}{y}\, d\rho (y) < \infty$.$\square$

Definition 7.2

We call a positive Borel measure $\rho$ on $[0,\infty [$ tame if the conditions (i)–(iv) from the preceding lemma are satisfied.

Proposition 7.3

If $\rho$ is tame, then $\nu = \Theta \cdot dx$ with $\Theta (x) := \int _0^\infty \frac{1}{y^2 + x^2}\, d\rho (y)$ is a tempered measure on ${\mathbb R}^d$ if and only if $d > 2$ or

• (a) $d = 1$ and $\int _0^1 \frac{1}{y}\, d\rho (y) < \infty$.

• (b) $d= 2$ and $\int _0^1 \ln (y^{-1})\, d\rho (y) < \infty$.

In particular, $\rho (\{0\}) = 0$ for $d = 1,2$.

Proof

That $\nu$ is a tempered measure means that, for some $N \in {\mathbb N}$, the integral

$$\begin{aligned} \int _{{\mathbb R}^d} (1 + p^2)^{-N}\, d\nu (p) = \int _0^\infty \int _{{\mathbb R}^d} (1 + p^2)^{-N}\frac{1}{y^2 + p^2}\, dp\, d\rho (y) \end{aligned}$$

is finite. This is equivalent to the finiteness of the integral

$$\begin{aligned} \int _0^\infty \int _0^\infty (1 + r^2)^{-N}\frac{r^{d-1}}{y^2 + r^2}\, dr\, d\rho (y). \end{aligned}$$

(27)

If (27) is finite, we obtain in particular that

$$\begin{aligned} \int _0^\infty \int _1^2(1 + r^2)^{-N}\frac{r^{d-1}}{y^2 + r^2}\, dr\, d\rho (y) < \infty , \end{aligned}$$

and this is equivalent to the tameness of $\rho$.

Now we split the double integral (27) according to $r < 1$ and $r > 1$ to find

$$\begin{aligned}&\int _0^\infty \int _1^\infty (1 + r^2)^{-N}\frac{r^{d-1}}{y^2 + r^2}\, dr\, d\rho (y) = \int _0^\infty I(y)\, d\rho (y) \end{aligned}$$

with

$$\begin{aligned} I(y) = \int _1^\infty (1 + r^2)^{-N}\frac{r^{d-1}}{y^2 + r^2}\, dr \le I(0). \end{aligned}$$

This integral is finite if and only if $2N + 2 > d$. To see if $I$ is $\rho$-integrable, we have to estimate the asymptotics of $I(y)$ for $y \rightarrow \infty$. The Monotone Convergence Theorem implies that $\lim _{y \rightarrow \infty } I(y)= 0$. The asymptotics of $I$ is the same as the asymptotics of the function

$$\begin{aligned} J(y) := \int _1^\infty \frac{r^{d-1-2N}}{y^2 + r^2}\, dr = \frac{1}{y^2} \int _1^\infty \frac{r^{d-1-2N}}{1+ \frac{r^2}{y^2}}\, dr \le \frac{1}{y^2} \int _1^\infty r^{d-1-2N}\, dr, \end{aligned}$$

provided $2N > d$. Now the tameness of $\rho$ implies that $I$ is $\rho$-integrable for $2N > d$.

Next we turn to the other part of the integral:

$$\begin{aligned} \int _0^\infty \int _0^1 (1 + r^2)^{-N}\frac{r^{d-1}}{y^2 + r^2}\, dr \, d\rho (y). \end{aligned}$$

Since $1 \le 1 + r^2 \le 2$ for $0 \le r \le 1$, it is finite if and only if the integral

$$\begin{aligned} \int _0^\infty \int _0^1 \frac{r^{d-1}}{y^2 + r^2}\, dr\, d\rho (y) = \int _0^\infty K(y)\, d\rho (y) \quad \text{ for } \quad K(y) := \int _0^1 \frac{r^{d-1}}{y^2 + r^2}\, dr, \end{aligned}$$

is finite. The integral $K(y)$ is finite for $y > 0$, and $K(0) = \int _0^1 r^{d-3} \, dr$ is finite if and only if $d > 2$. We thus have to estimate the asymptotics of $K$ for $y \rightarrow \infty$, and, for $d \le 2$, also for $y \rightarrow 0$. In view of

$$\begin{aligned} K(y) \le \int _0^1 \frac{dr}{y^2 + r^2} \le \frac{1}{y^2}, \end{aligned}$$

the finiteness of the integral $\int _1^\infty K(y)\, d\rho (y)$ follows for every $d$ from the tameness of $\rho$. For $d > 2$, the function $K$ is continuous on $[0,\infty [$, hence $\rho$-integrable. This completes the proof for $d > 2$.

For $d = 1$ we have

$$\begin{aligned} K(y) = \int _0^1 \frac{1}{y^2 + r^2}\, dr = \frac{1}{y}\left( \arctan \frac{1}{y}- \arctan 0\right) = \frac{1}{y} \arctan \frac{1}{y} \sim \frac{\pi }{2y} \end{aligned}$$

for $y \rightarrow 0$. We thus find the necessary and sufficient condition (a) for the finiteness of the $\rho$-integral of $I$.

For $d = 2$ we have

$$\begin{aligned} K(y) = \int _0^1 \frac{r}{y^2 + r^2}\, dr = \frac{1}{2} \big (\ln (y^2 + 1) - \ln (y^2)\big ) \sim \ln (y^{-1}) \end{aligned}$$

for $y \rightarrow 0$. Therefore $I$ is $\rho$-integrable over $[0,1]$ if and only if (b) is satisfied.$\square$

The following proposition can also be derived from [6, Prop. 6.2.5]. We include it for the sake of easier reference.

Proposition 7.4

If $\rho$ is a tame measure on $[0,\infty [$ for which the measure $\nu = \Theta \cdot dx$ is tempered, then the distribution $\widehat{\nu } \in C^{-\infty }({\mathbb R}^d)$ is reflection positive for $({\mathbb R}^d, {\mathbb R}^d_+, \theta )$ and $\theta (x_0, \mathbf {x}) = (-x_0, \mathbf {x})$, i.e.,

Proof

Writing

$$\begin{aligned} \nu = \int _0^\infty \nu _y\, d\rho (y) \quad \text{ with } \quad d\nu _y(p)= \frac{dp}{y^2 + p^2}, \end{aligned}$$

we see that it suffices to show that the distributions $\widehat{\nu _y}$ are reflection positive. First we observe that

$$\begin{aligned} \int _{{\mathbb R}^d} \theta \widehat{\psi } \cdot \overline{\widehat{\psi }}\, d\nu _y = \int _{{\mathbb R}^d} \overline{\widehat{\psi }(-p_0, \mathbf {p})} \frac{\widehat{\psi }(p_0, \mathbf {p})}{y^2 + p^2}\, dp \quad \text{ for } \quad p = (p_0, \mathbf {p})\in {\mathbb R}\times {\mathbb R}^{d-1}. \end{aligned}$$

For each $\mathbf {p}\in {\mathbb R}^{d-1}$, the function $h_\mathbf {p}(p_0) := \widehat{\psi } (-p_0, \mathbf {p})$ is a Schwartz function with $\mathop {\mathrm{supp}}\nolimits (\widehat{h}_{\mathbf {p}}) \subseteq ]0, \infty [$, and

$$\begin{aligned} \int _{{\mathbb R}^d} \theta \widehat{\psi } \cdot \overline{\widehat{\psi }}\, d\nu _y = \int _{{\mathbb R}^{d-1}} \int _{\mathbb R}\frac{\overline{h_{\mathbf {p}}(p_0)} h_{\mathbf {p}}(-p_0)}{p_0^2 + y^2 + \mathbf {p}^2}\, dp_0 \, d\mathbf {p}. \end{aligned}$$

If $y = 0$, then $d > 2$, so that $\{0\}$ is a Lebesgue zero set in ${\mathbb R}^{d-1}$. Therefore it suffices to show that, for $f \in \mathcal {S}({\mathbb R})$ with $\mathop {\mathrm{supp}}\nolimits (\widehat{f}) \subseteq {\mathbb R}_+$ and $y > 0$, the measure $d\nu _y^1(x) := \frac{dx}{x^2 + y^2}$ satisfies

$$\begin{aligned} \int _{\mathbb R}f(x) \overline{f(-x)}\, \frac{dx}{x^2 + y^2} = \int _{\mathbb R}f(x) \overline{f(-x)}\, \nu _y^1(x) \, dx \ge 0. \end{aligned}$$

In view of $\widehat{\nu _y^1}(p)= \frac{\pi }{y}\ e^{-y|p|}$, this follows from

$$\begin{aligned}&\int _{\mathbb R}(\widehat{f} * \overline{\widehat{f}})(p) e^{-y |p|}\, dp \!=\! \int _{{\mathbb R}_+} (\widehat{f} * \overline{\widehat{f}})(p) e^{-y p} \, dp \!=\! \mathcal {L}(\widehat{f}* \overline{\widehat{f}})(y) = |\mathcal {L}(\widehat{f})(y)|^2 \!\ge \! 0. \end{aligned}$$

More conceptually, the preceding calculation means that $\widehat{\nu _y^1}$ is a reflection positive function on ${\mathbb R}$, resp., its restriction to the semigroup ${\mathbb R}_+$ is also positive definite with respect to the trivial involution (cf. [28]).$\square$

Note that Proposition 7.8 below implies in particular that, for $d = 1$, the preceding proposition does not cover all reflection positive measures. Example that are not covered arise for $\widehat{\nu }(x) = |x|^{-s}$ (on ${\mathbb R}^\times$) for $s > 1$.

Lemma 7.5

Suppose that $\rho$ is a positive Borel measure on $[0,\infty [$ whose Laplace transform $\mathcal {L}(\rho )$ exists on ${\mathbb R}_+$. Then the function $\mathcal {L}(\rho )$ on ${\mathbb R}_+$ extends to a symmetric distribution on ${\mathbb R}$ if and only if $\rho$ is tempered, i.e., $\int _1^\infty \frac{1}{y^k}\, d\rho (y) < \infty$ for some $k \in {\mathbb N}$. More precisely, for $\ell \in {\mathbb N}_0$,

$$\begin{aligned} \int _0^1 x^\ell \mathcal {L}(\rho )(x)\, dx < \infty \quad \Leftrightarrow \quad \int _1^\infty \frac{1}{y^{\ell + 1}}\, d\rho (y) < \infty . \end{aligned}$$

In particular, $\mathcal {L}(\rho )$ is locally integrable on $[0, \infty [$ if and only if $\int _1^\infty \frac{1}{y}\, d\rho (y) < \infty$.

Proof

Since the Laplace transform of the finite measure $\rho \vert _{[0,1]}$ extends to a continuous function on $[0,\infty [$, we may w.l.o.g. assume that $\rho ([0,1]) = 0$. We put $D^\times (x) = \mathcal {L}(\rho )(|x|)$ for $x \in {\mathbb R}^\times$. In view of [31, Thm. VIII, §VII.4], $D^\times$ extends to a distribution on ${\mathbb R}$ if and only if there exists an $\ell \in {\mathbb N}_0$ with

$$\begin{aligned} \int _0^1 x^\ell D^\times (x)\, dx < \infty . \end{aligned}$$

(28)

We rewrite this integral as follows

$$\begin{aligned} \int _0^1 x^\ell D^\times (x)\, dx&= \int _0^1 x^\ell \int _1^\infty e^{-y x}\, d\rho (y) dx = \int _1^\infty \int _0^1 x^\ell e^{-y x}\, dx\, d\rho (y) \\&= \int _1^\infty \int _0^y \frac{u^\ell }{y^\ell } e^{-u}\, \frac{du}{y}\, d\rho (y) = \int _1^\infty \int _0^y u^\ell e^{-u}\, du\, \frac{1}{y^{\ell +1}}\, d\rho (y). \end{aligned}$$

In view of $0 < \int _0^\infty u^\ell e^{-u}\,du < \infty$, it follows that $\int _0^1 x^\ell D^\times (x)\, dx$ is finite if and only if $\int _1^\infty \frac{1}{y^{\ell +1}}\, d\rho (y)$ is finite.$\square$

Theorem 7.6

The reflection positive distributions on ${\mathbb R}$ which are represented by a locally integrable function are the Fourier transforms $D = \widehat{\nu }$ of measures of the form

$$\begin{aligned} \nu = c \delta _0 + \Theta \cdot dx, \qquad \Theta (x) = \frac{1}{\pi } \int _0^\infty \frac{y}{y^2 + y^2}\, d\rho (y), \end{aligned}$$

(29)

where $c \ge 0$ and $\rho$ is a positive Radon measure on ${\mathbb R}_+$ satisfying

(30)

Proof

Let $D \in C^{-\infty }({\mathbb R})$ be reflection positive. Then its restriction to ${\mathbb R}_+$ is positive definite w.r.t. the involution $x^* = x$, so that [28, Thm. 4.13] implies $D\vert _{{\mathbb R}_+}$ is represented by an analytic function which is the Laplace transform $\mathcal {L}(\rho )$ of a positive Radon measure $\rho$ on $[0,\infty [$. If $D$ is represented by a locally integrable function, $\mathcal {L}(\rho )$ is locally integrable, so that Lemma 7.5 leads to

$$\begin{aligned} \int _1^\infty \frac{1}{y}\, d\rho (y) < \infty . \end{aligned}$$

We now have almost everywhere on ${\mathbb R}$

$$\begin{aligned} D(x) = \mathcal {L}(\rho )(|x|) = \int _0^\infty e^{-y|x|}\, d\rho (y) = \widehat{\nu }(x) \quad \text{ for } \quad \nu = \rho (\{0\})\delta _0 + \Theta \cdot dx, \end{aligned}$$

provided $\nu$ is a tempered measure. Here we used that $e^{-y|x|}$ is the Fourier transform of the measure $\frac{1}{\pi }\frac{y \cdot dx}{y^2 + x^2}$. In view of Proposition 7.3, the temperedness of $\nu$ is equivalent to (30).

Suppose, conversely, that (30) is satisfied. Then Proposition 7.3 implies that $\nu$ is tempered, so that $D := \widehat{\nu }$ is a positive definite distribution. That it is reflection positive follows from the reflection positivity of the functions $\widehat{\nu }_y(x) = e^{-y|x|}$.$\square$

Remark 7.7

The integral representation from Theorem 7.6 does not extend to all reflection positive distributions on ${\mathbb R}$. Any distribution of the form $E := P\left( {1\over i}{d\over dx}\right) \delta _0$, where $P$ is a non-negative even polynomial, is reflection positive. In this case $E \vert _{{\mathbb R}_+} = 0$.

We also conclude that, for any reflection positive distribution $D$ on ${\mathbb R}$, the distribution $D + E$ is another extension of $D\vert _{{\mathbb R}_+}$, so that reflection positive extensions are not unique.

Proposition 7.8

For every tempered measure $\rho$ on $[0,\infty [$, there exists a reflection positive distribution $D$ on ${\mathbb R}$ with $D\vert _{{\mathbb R}_+} = \mathcal {L}(\rho )$.

Proof

In view of Lemma 7.5, we may assume that $\int _1^\infty \frac{1}{\lambda ^{4N}}\, d\rho (\lambda ) < \infty .$ Then the measure

$$\begin{aligned} d\widetilde{\rho }(\lambda ) := \frac{1}{1 + \lambda ^{4N}}\, d\rho (\lambda ) \end{aligned}$$

on $[0,\infty [$ is finite. Its Laplace transform therefore defines a continuous reflection positive function $\widetilde{D}(x) := \mathcal {L}(\widetilde{\rho })(|x|)$ ([28, Cor. 3.3]). For the polynomial $P(x) := 1 + x^{4N}$, we now obtain the positive definite distribution $D := P({1\over i}{d\over dx}) \widetilde{D},$ and its restriction to ${\mathbb R}_+$ is given by

$$\begin{aligned} D(x) \!=\! P\Big ({1\over i}{d\over dx}\Big ) \mathcal {L}(\widetilde{\rho })(x) \!=\! \int _0^\infty (1 + \lambda ^4)e^{-\lambda x}\, d\widetilde{\rho }(\lambda ) = \int _0^\infty e^{-\lambda x}\, d\rho (\lambda ) = \mathcal {L}(\rho )x). \end{aligned}$$

Therefore $D$ is a reflection positive extension of $\mathcal {L}(\rho )$.$\square$

Appendix B: Lorentzian Invariant Tempered Measures on the Light Cone

For the following theorem, we recall the $L^{\mathop {\uparrow }}$-invariant measures $\mu _m$ from Definition 6.10. It is our version of the Lehmann Spectral Formula from Quantum Field Theory ([6, Thm. 6.2.4]) describing the $2$-point functions of Poincaré invariant field theories. The proof given below provides many details skipped in [6].

Theorem 8.1

For every $L^{\mathop {\uparrow }}$-invariant Borel measure $\mu$ on the closed forward light cone $\overline{V_+}$, there exists a $\sigma$-finite Borel measure $\rho$ on $[0,\infty [$ and a constant $c \ge 0$ so that

$$\begin{aligned} \mu = c \delta _0 + \int _0^\infty \mu _m\, d\rho (m), \end{aligned}$$

(31)

where $\rho (\{0\}) = 0$ for $d = 1$.

• (a) The measure $\mu$ is tempered if and only if the following conditions are satisfied:

  • (C1) $\rho$ is a tempered measure on $[0,\infty [$.

  • (C2) $\int _0^1 \frac{1}{m}\, d\rho (m) < \infty$ for $d = 1$.

  • (C3) $\int _0^1 \ln (m^{-1})\, d\rho (m) < \infty$ for $d = 2$. In particular, $\rho (\{0\}) = 0$.

• (b) Let $\mathop {\mathrm{pr}}\nolimits : {\mathbb R}^d \rightarrow {\mathbb R}^{d-1}, (p_0, \mathbf {p}) \mapsto \mathbf {p}$ be the projection map. Then the measure $\widetilde{\mu }= \mathop {\mathrm{pr}}\nolimits _*\mu$ on ${\mathbb R}^{d-1}$ is tempered if and only if $\mu$ is tempered and, in addition,

  $$\begin{aligned} \int _1^\infty \frac{1}{m}\, d\rho (m) < \infty . \end{aligned}$$

  If this condition is not satisfied, then every Borel subset $E \subseteq {\mathbb R}^{d-1}$ satisfies $\widetilde{\mu }(E) \in \{ 0, \infty \}$, so that $L^2({\mathbb R}^{d-1}, \widetilde{\mu }) = \{0\}$.

Proof

In [30, Thm. IX.33], the decomposition (31) is stated only in the case $d = 4$, but the proof works in the general case, where it leads to an integral representation of $\mu$ in terms of a measure $\rho$ on $[0,\infty [$, whose restriction to $]0,\infty [$ is a Radon measure with possibly infinite mass for the interval $[0,1]$ (see also [2, Lemma 9.1.2/3] for a description of the ergodic $\mathop {\mathrm{O}}\nolimits _{1,d-1}({\mathbb R})$-invariant measures on ${\mathbb R}^d$).

(a) Now the problem consists in characterizing the temperedness of $\mu$ in terms of properties of the measure $\rho$. To this end, we may w.l.o.g. assume that $c = 0$. Temperedness of $\mu$ is equivalent to the existence of an $N \in {\mathbb N}$ such that the following integral is finite

$$\begin{aligned} \int _{{\mathbb R}^d} \frac{1}{(1 + p^2)^N}\, d\mu (p)&= \int _0^\infty \int _{{\mathbb R}^d} \frac{1}{(1 + p^2)^N}\, d\mu _m(p)\, d\rho (m) \\&= \int _0^\infty \int _{{\mathbb R}^{d-1}} \frac{1}{(1 + (m^2 + \mathbf {p}^2) + \mathbf {p}^2)^N} \frac{1}{\sqrt{m^2 + \mathbf {p}^2}}\, d\mathbf {p}\, d\rho (m). \end{aligned}$$

For $d = 1$, the preceding formula simplifies to

$$\begin{aligned} \int _{{\mathbb R}} \frac{1}{(1 + p^2)^N}\, d\mu (p)&= \int _0^\infty \frac{1}{(1 + m^2)^N}\frac{1}{m}\, d\rho (m). \end{aligned}$$

There exists an $N \in {\mathbb N}_0$ for which such an integral is finite if and only if (a) is satisfied and $\rho$ is a tempered.

From now on we assume that $d > 1$. Evaluating the integral over ${\mathbb R}^{d-1}$ in polar coordinates, we see that this is equivalent to the finiteness of the double integral

$$\begin{aligned} \int _0^\infty \int _0^\infty \frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m). \end{aligned}$$

(32)

The finiteness of (32) implies in particular the finiteness of the integral

$$\begin{aligned} \int _0^\infty \int _1^2\frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m), \end{aligned}$$

which is equivalent to the finiteness of the integral

$$\begin{aligned} \int _0^\infty \frac{1}{(1 + m^2)^N}\frac{1}{\sqrt{m^2 + 1}}\, d\rho (m) = \int _0^\infty \frac{1}{(1 + m^2)^{N+ \frac{1}{2}}}\, d\rho (m). \end{aligned}$$

This implies that $\rho ([0,M]) < \infty$ for every $M > 0$ and that $\rho$ is tempered.

Suppose, conversely, that $\rho$ is tempered. We discuss the integral (32) by splitting it into the two pieces corresponding to $r < 1$ and $r > 1$. For the $r >1$-part we obtain for $N \ge N_0$ and $N_0 > \frac{d-1}{2}$:

$$\begin{aligned}&\int _1^\infty \frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr \\&\le \frac{1}{\sqrt{1 + m^2}} \int _1^\infty \frac{1}{(1 + m^2 + 2r^2)^{N_0}(1 + m^2 + 2r^2)^{N-N_0}} r^{d-2}\, dr \\&\le \frac{1}{\sqrt{1 + m^2}} \frac{1}{ (3 + m^2)^{N-N_0}} \int _1^\infty \frac{1}{(1 + 2r^2)^{N_0}} r^{d-2}\, dr. \end{aligned}$$

Our condition on $N_0$ ensures that this integral is finite, and now the temperedness of $\rho$ implies that

$$\begin{aligned} \int _0^\infty \int _1^\infty \frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m) < \infty \end{aligned}$$

if $N$ is sufficiently large.

Next we note that,

$$\begin{aligned}&\int _1^\infty \int _0^1 \frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m) \\&\le \int _1^\infty \int _0^1 \frac{1}{(1 + m^2)^N} \frac{1}{m} \, dr\, d\rho (m) = \int _1^\infty \frac{1}{(1 + m^2)^N}\frac{1}{m}\, d\rho (m) < \infty . \end{aligned}$$

This already shows that $\int _1^\infty \mu _m\, d\rho (m)$ is tempered if $\rho$ is tempered.

It remains to consider the integral

$$\begin{aligned} \int _0^1 \int _0^1 \frac{1}{(1 + m^2 + 2r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m), \end{aligned}$$

which is finite if and only if

$$\begin{aligned} \int _0^1 \int _0^1 \frac{r^{d-2}}{\sqrt{m^2 + r^2}}\, dr\, d\rho (m) < \infty . \end{aligned}$$

Let

$$\begin{aligned} I(m) := \int _0^1 \frac{r^{d-2}}{\sqrt{m^2 + r^2}} \, dr, \end{aligned}$$

and observe that this defines a continuous function for $m > 0$ with $I(0)= \int _0^1 r^{d-3}\, dr < \infty$ if and only if $d > 2$. Since $\rho ([0,1]) < \infty$, we conclude that, for $d\ge 3$, $\mu$ is a tempered measure if and only if $\rho$ has this property.

This leaves the case $d = 2$. Then

$$\begin{aligned} I(m) = \int _0^1 \frac{1}{\sqrt{m^2 + r^2}}\, dr = \ln (1 + \sqrt{1 + m^2}) - \ln (m). \end{aligned}$$

We thus find condition (b), which is needed to ensure that $\mu$ is a tempered measure.

(b) The measure $\widetilde{\mu }$ is of the form $\widetilde{\Theta }(\mathbf {p})\, d\mathbf {p}$ for

$$\begin{aligned} \widetilde{\Theta }(\mathbf {p}) = \int _0^\infty \frac{1}{\sqrt{m^2 + \mathbf {p}^2}}\, d\rho (m). \end{aligned}$$

(33)

For $d = 1$, the measure $\widetilde{\mu }$ is a point measure which is tempered if and only if it is finite. This is equivalent to

$$\begin{aligned} \widetilde{\Theta }(0) = \int _0^\infty \frac{1}{m}\, d\rho (m) < \infty . \end{aligned}$$

Suppose that $d > 1$. If $\widetilde{\mu }$ is finite on a set of positive Lebesgue measure, then there exists a non-zero $\mathbf {p}\in {\mathbb R}^{d-1}$ with $\widetilde{\Theta }(\mathbf {p}) < \infty$. This implies that compact subsets of $[0,\infty [$ have finite $\rho$-measure ($\rho$ is a Radon measure), and that

$$\begin{aligned} \int _1^\infty \frac{1}{m}\, d\rho (m) < \infty . \end{aligned}$$

(34)

In particular, $\rho$ is tempered. If (34) is not satisfied, then $\widetilde{\Theta }(\mathbf {p}) = \infty$ implies that $\widetilde{\mu }(E) = \infty$ for every Borel subset $E \subseteq {\mathbb R}^{d-1}$ of positive Lebesgue measure and $\widetilde{\mu }(E) = 0$ if $E$ is a Lebesgue zero set.

Let us assume that $\rho$ is a Radon measure on $[0,\infty [$ satisfying (34). Temperedness of $\widetilde{\mu }$ is equivalent to the existence of an $N \in {\mathbb N}$ such that

$$\begin{aligned} \int _{{\mathbb R}^{d-1}} \frac{1}{(1 + \mathbf {p}^2)^N}\, d\widetilde{\mu }(p)&= \int _0^\infty \int _{{\mathbb R}^{d-1}} \frac{1}{(1 + \mathbf {p}^2)^N} \frac{1}{\sqrt{m^2 + \mathbf {p}^2}}\, d\mathbf {p}\, d\rho (m) < \infty . \end{aligned}$$

Evaluating the integral over ${\mathbb R}^{d-1}$ in polar coordinates, we see that this is equivalent to the finiteness of the double integral

$$\begin{aligned} \int _0^\infty \int _0^\infty \frac{1}{(1 + r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m). \end{aligned}$$

(35)

We discuss the integral (35) by splitting it into the two pieces corresponding to $r < 1$ and $r > 1$. For the $r >1$-part we obtain for $N > \frac{d-1}{2}$:

$$\begin{aligned} \int _1^\infty \frac{1}{(1 + r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr \le \frac{1}{\sqrt{1 + m^2}} \int _1^\infty \frac{r^{d-2}}{(1 +r^2)^{N}}\, dr. \end{aligned}$$

Our condition on $N$ ensures that this integral is finite, and now the finiteness of $\int _0^\infty \, \frac{1}{\sqrt{1+m^2}}\, d\rho (m)$ implies the finiteness of the double integral

$$\begin{aligned} \int _0^\infty \int _1^\infty \frac{1}{(1 + r^2)^N} \frac{1}{\sqrt{m^2 + r^2}} r^{d-2}\, dr\, d\rho (m). \end{aligned}$$

Next we note that,

$$\begin{aligned} \int _1^\infty \int _0^1 \frac{1}{(1 + r^2)^N} \frac{ r^{d-2}}{\sqrt{m^2 + r^2}}\, dr\, d\rho (m) \le \int _1^\infty \frac{1}{m} \, d\rho (m) \cdot \int _0^1 \frac{r^{d-2}}{(1 + r^2)^N}\, dr < \infty . \end{aligned}$$

It remains to consider the integral

$$\begin{aligned} \int _0^1 \int _0^1 \frac{1}{(1 + r^2)^N} \frac{r^{d-2}}{\sqrt{m^2 + r^2}} \, dr\, d\rho (m), \end{aligned}$$

which is finite if and only if

$$\begin{aligned} \int _0^1 \int _0^1 \frac{r^{d-2}}{\sqrt{m^2 + r^2}}\, dr\, d\rho (m) < \infty . \end{aligned}$$

In (a) we have seen that this integral is finite if $d > 2$, and that, for $d = 2$, its finiteness is equivalent to (C3). This proves (b).$\square$

Remark 8.2

For the measure $\nu = \Theta \cdot dx$ on ${\mathbb R}^d$, where $\Theta (x) = \frac{1}{\pi } \int _0^\infty \frac{1}{m^2 + x^2}\, d\rho (m)$, the projection $\widetilde{\nu }$ to ${\mathbb R}^{d-1}$ under $\mathop {\mathrm{pr}}\nolimits (x_0,\mathbf {x}) := \mathbf {x}$, is of the form $\widetilde{\Theta }(\mathbf {x})\, d\mathbf {x}$, where

$$\begin{aligned} \widetilde{\Theta }(\mathbf {x})&= \frac{1}{\pi } \int _0^\infty \int _{\mathbb R}\frac{1}{m^2 + \mathbf {x}^2 + x_0^2}\, dx_0\, d\rho (m)\\&= \int _0^\infty \frac{1}{\sqrt{m^2 + \mathbf {x}^2}} \frac{1}{\pi } \int _{\mathbb R}\frac{\sqrt{m^2 + \mathbf {x}^2}}{m^2 + \mathbf {x}^2 + x_0^2}\, dx_0\, d\rho (m) \\&= \int _0^\infty \frac{1}{\sqrt{m^2 + \mathbf {x}^2}}\, d\rho (m). \end{aligned}$$

This implies that $\widetilde{\nu }= \widetilde{\mu }$ [cf. (33)].

Appendix C: Positive Definite Functions

In this appendix we collect some definitions and results concerning positive definite functions and kernels.

Definition 9.1

Let $X$ be a set and $\mathcal {F}$ be a complex Hilbert space.

• (a) A function $K : X \times X \rightarrow B(\mathcal {F})$ is called a $B(\mathcal {F})$ -valued kernel. A $B(\mathcal {F})$-valued kernel $K$ on $X$ is said to be positive definite if, for every finite sequence $(x_1, v_1), \ldots , (x_n,v_n)$ in $X \times \mathcal {F}$,

  $$\begin{aligned} \sum _{j,k = 1}^n \langle K(x_j, x_k)v_k, v_j \rangle \ge 0. \end{aligned}$$

• (b) If $(S,*)$ is an involutive semigroup, then a function $\varphi : S \rightarrow B(\mathcal {F})$ is called positive definite if the kernel $K_\varphi (s,t) := \varphi (st^*)$ is positive definite.

Positive definite kernels can be characterized as those for which there exists a Hilbert space $\mathcal {H}$ and a function $\gamma : X \rightarrow B(\mathcal {H},\mathcal {F})$ such that

$$\begin{aligned} K(x,y) = \gamma (x)\gamma (y)^* \quad \text{ for } \quad x,y \in X \end{aligned}$$

(36)

(cf. [26, Thm. I.1.4]). Here one may assume that the vectors $\gamma (x)^*v$, $x \in X, v \in \mathcal {F}$, span a dense subspace of $\mathcal {H}$. If this is the case, then the pair $(\gamma ,\mathcal {H})$ is called a realization of $K$. The map $\Phi : \mathcal {H}\rightarrow \mathcal {F}^X, \Phi (v)(x) := \gamma (x)v$, then realizes $\mathcal {H}$ as a Hilbert subspace of $\mathcal {F}^X$ with continuous point evaluations $\mathop {\mathrm{ev}}\nolimits _x : \mathcal {H}\rightarrow \mathcal {F}, f \mapsto f(x)$. Then $\Phi (\mathcal {H})$ is the unique Hilbert space in $\mathcal {F}^X$ with continuous point evaluations $\mathop {\mathrm{ev}}\nolimits _x$, for which $K(x,y) = \mathop {\mathrm{ev}}\nolimits _x \mathop {\mathrm{ev}}\nolimits _y^*$ for $x,y \in X$. We write $\mathcal {H}_K \subseteq \mathcal {F}^X$ for this subspace and call it the reproducing kernel Hilbert space with kernel $K$.

Example 9.2

(Vector-valued GNS construction) (cf. [26, Sect. 3.1]) Let $(\pi , \mathcal {H})$ be a representation of the unital involutive semigroup $(S,*)$, $\mathcal {F}\subseteq \mathcal {H}$ be a closed subspace for which $\pi (S)\mathcal {F}$ is total in $\mathcal {H}$ and $P : \mathcal {H}\rightarrow \mathcal {F}$ denote the orthogonal projection. Then $\varphi (s) := P\pi (s)P^*$ is a $B(\mathcal {F})$-valued positive definite function on $S$ with $\varphi (\mathbf {1}) = \mathbf {1}_\mathcal {F}$ because $\gamma (s) := P\pi (s) \in B(\mathcal {H},\mathcal {F})$ satisfies

$$\begin{aligned} \gamma (s)\gamma (t)^* = P\pi (st^*)P^* = \varphi (st^*). \end{aligned}$$

The map

$$\begin{aligned} \Phi : \mathcal {H}\rightarrow \mathcal {F}^S, \quad \Phi (v)(s) = \gamma (s)v = P \pi (s)v \end{aligned}$$

is an $S$-equivariant realization of $\mathcal {H}$ as the reproducing kernel space $\mathcal {H}_\varphi \subseteq \mathcal {F}^S$, on which $S$ acts by right translation, i.e., $(\pi _\varphi (s)f)(t) = f(ts)$.

Conversely, let $S$ be a unital involutive semigroup and $\varphi : S \rightarrow B(\mathcal {F})$ be a positive definite function with $\varphi (\mathbf {1}) = \mathbf {1}_\mathcal {F}$. Write $\mathcal {H}_\varphi \subseteq \mathcal {F}^S$ for the corresponding reproducing kernel space and $\mathcal {H}_\varphi ^0$ for the dense subspace spanned by $\mathop {\mathrm{ev}}\nolimits _s^*v, s \in S, v \in \mathcal {F}$. Then $(\pi _\varphi (s)f)(t) := f(ts)$ defines a $*$-representation of $S$ on $\mathcal {H}_\varphi ^0$. We say that $\varphi$ is exponentially bounded if all operators $\pi _\varphi (s)$ are bounded, so that we actually obtain a representation of $S$ by bounded operators on $\mathcal {H}_\varphi$. As $\mathbf {1}_\mathcal {F}= \varphi (\mathbf {1}) = \mathop {\mathrm{ev}}\nolimits _\mathbf {1} \mathop {\mathrm{ev}}\nolimits _\mathbf {1}^*$, the map $\mathop {\mathrm{ev}}\nolimits _\mathbf {1}^* : \mathcal {F}\rightarrow \mathcal {H}$ is an isometric inclusion, so that we may identify $\mathcal {F}$ with a subspace of $\mathcal {H}$. Then $\mathop {\mathrm{ev}}\nolimits _\mathbf {1} : \mathcal {H}\rightarrow \mathcal {F}$ corresponds to the orthogonal projection onto $\mathcal {F}$ and $\mathop {\mathrm{ev}}\nolimits _\mathbf {1} \circ \pi _\varphi (s) = \mathop {\mathrm{ev}}\nolimits _s$ leads to

$$\begin{aligned} \varphi (s) = \mathop {\mathrm{ev}}\nolimits _s \mathop {\mathrm{ev}}\nolimits _\mathbf {1}^* =\mathop {\mathrm{ev}}\nolimits _\mathbf {1} \pi _\varphi (s) \mathop {\mathrm{ev}}\nolimits _\mathbf {1}^*. \end{aligned}$$

(37)

If $S = G$ is a group with $s^* = s^{-1}$, then $\varphi$ is always exponentially bounded and the representation $(\pi _\varphi , \mathcal {H}_\varphi )$ is unitary.

Lemma 9.3

Let $(S,*)$ be a unital involutive semigroup and $\varphi : S \rightarrow B(\mathcal {F})$ be a positive definite function with $\varphi (\mathbf {1}) = \mathbf {1}$. We write $(\pi _\varphi , \mathcal {H}_\varphi )$ for the representation on the corresponding reproducing kernel Hilbert space $\mathcal {H}_\varphi \subseteq \mathcal {F}^S$ by $(\pi _\varphi (s)f)(t) := f(ts)$. Then the inclusion

$$\begin{aligned} \iota : \mathcal {F}\rightarrow \mathcal {H}_\varphi , \quad \iota (v)(s) := \varphi (s)v \end{aligned}$$

is surjective if and only if $\varphi$ is multiplicative, i.e., a representation.

Proof

If $\varphi$ is multiplicative, then $(\pi (s) \iota (v))(t) = \varphi (ts)v = \varphi (t)\varphi (s)v \in \iota (\mathcal {F})$. Therefore the $S$-cyclic subspace $\iota (\mathcal {F})$ is invariant, which implies that $\iota$ is surjective.

Suppose, conversely, that $\iota$ is surjective. This means that each $f \in \mathcal {H}_\varphi$ satisfies $f(s) = \varphi (s)f(\mathbf {1})$. For $v \in \mathcal {F}$ and $t,s \in S$, this leads to

$$\begin{aligned} \varphi (st)v = \pi (t)(\iota (v))(s) = \varphi (s) (\pi (t)\iota (v))(\mathbf {1}) = \varphi (s) \iota (v)(t) = \varphi (s) \varphi (t)v. \end{aligned}$$

Therefore $\varphi$ is multiplicative.$\square$

Remark 9.4

The preceding lemma can also be expressed without referring to positive definite functions and the corresponding reproducing kernel space. In this context it asserts the following. Let $\pi : S \rightarrow B(\mathcal {H})$ be a $*$-representation of a unital involutive semigroup $(S,*)$, $\mathcal {F}\subseteq \mathcal {H}$ be a closed cyclic subspace and $P : \mathcal {H}\rightarrow \mathcal {F}$ the orthogonal projection. Then the function

$$\begin{aligned} \varphi : S \rightarrow B(\mathcal {F}), \quad \varphi (s) := P \pi (s)P^* \end{aligned}$$

is multiplicative if and only if $\mathcal {F}= \mathcal {H}$.

The following lemma is an abstraction of [6, Thm. 6.2.2].

Lemma 9.5

Let $V$ be a vector spaces over ${\mathbb K}\in \{{\mathbb R},{\mathbb C}\}$ and $\beta : V \times V\rightarrow {\mathbb K}$ be a hermitian form on $V$ (for ${\mathbb K}= {\mathbb R}$ this means that it is symmetric and bilinear). Then the kernel $e^\beta$ is positive definite if and only if $\beta$ is positive semidefinite.

Proof

If $\beta$ is positive semidefinite, then $\beta$ defines a positive definite kernel on $V \times V$. Hence the kernels $\beta ^n(x,y) := \beta (x,y)^n$ are also positive definite, see [26, Rem. I.17(b)], and therefore $e^\beta = \sum _{n = 0}^\infty \frac{\beta ^n}{n!}$ is positive definite.

If, conversely, $e^\beta$ is a positive definite kernel, let $x_1,\ldots ,x_N\in V$ and $c_1,\ldots ,c_N\in {\mathbb K}$. For $t\in {\mathbb R}$ define

$$\begin{aligned} y_1=tx_1,\ldots , y_N=t x_N,\quad y_{N+1}=0,\ldots ,y_{2N}=0 \end{aligned}$$

and

$$\begin{aligned} d_1=c_1,\ldots , d_N=c_N,\quad d_{N+1}=-c_1,\ldots ,d_{2N}=-c_N. \end{aligned}$$

Then

$$\begin{aligned} 0 \le \sum _{i,j=1}^{2N}e^{\beta (y_i,y_j)}d_i\overline{d_j}= \sum _{i,j=1}^{N}(e^{t^2\beta (x_i,x_j)}-1)c_i\overline{c_j} \end{aligned}$$

and letting $t$ tend to zero after dividing by $t^2$ yields the claim.$\square$

Appendix D: Distribution Vectors and Tempered Measures

In this appendix we take a closer look at distribution vectors for representations of vector spaces by multiplication operators on $L^2$-spaces.

1.1 D.1 Representations of Cones

Let $V$ be a finite-dimensional real vector space, $\tau : V \rightarrow V$ be a linear involution and $\Omega \subseteq V$ be an open convex cone invariant under the involution $v \mapsto v^\sharp = -\tau (v)$. The cone

$$\begin{aligned} \widehat{\Omega } = \{ \alpha \in V_{\mathbb C}^* : (\forall x = x^\sharp \in \Omega )\ \alpha (x) \ge 0\} \end{aligned}$$

parametrizes the bounded characters of $\Omega$ by assigning to $\alpha \in \widehat{\Omega }$ the character $e_\alpha (v) := e^{-\alpha (v)}$.

Any $\sigma$-finite measure $\mu$ on $\widehat{\Omega }$ defines a contraction representation of $\Omega$ on $\mathcal {H}:= L^2(\widehat{\Omega }, \mu )$ by

$$\begin{aligned} (\pi _\mu (v)f)(\alpha ) := e^{-\alpha (v)} f(\alpha ) \quad \text{ for } \quad v \in \Omega , \alpha \in \widehat{\Omega }. \end{aligned}$$

(38)

On the subspace $V^c := i V^{-\tau } \oplus V^{\tau } \subseteq V_{\mathbb C}$ formula (38) defines a unitary representation $\pi _\mu ^c$ because every $\alpha \in \widehat{\Omega }$ is purely imaginary on $V^c$.

The following lemma is a supplement to the generalized Bochner–Schwartz Theorem from [28] because it tells us which measures $\mu$ on $\widehat{\Omega }$ actually have Fourier–Laplace transforms defining distributions on $\Omega$.

Lemma 10.1

Let $V$ be a finite-dimensional vector space, $\tau \in \mathop {\mathrm{GL}}\nolimits (V)$ an involution and $\Omega \subseteq V$ be an open convex cone invariant under the involution $v \mapsto v^\sharp = - \tau (v)$. Further, let $\mu$ be a positive Borel measure on $\widehat{\Omega }$ such that, for every $\varphi \in C^\infty _c(\Omega )$ the Fourier–Laplace transform $\widehat{\varphi }(\alpha ) := \int _{\Omega } \varphi (x) e^{-\alpha (x)}\, dx$ is $\mu$-integrable on $\widehat{\Omega }$. Then $\widehat{\mu }(\varphi ) := \mu (\overline{\widehat{\varphi }})$ defines a distribution on $\Omega$. In the special case $V = \Omega$ and $\tau = \mathop {\mathrm{id}}\nolimits _V$, the measure $\mu$ is tempered.

Proof

We consider the linear functional

$$\begin{aligned} E : C^\infty _c(\Omega ) \rightarrow {\mathbb C}, \quad E(\varphi ) := \mu (\widehat{\varphi }). \end{aligned}$$

Then, for every test function $\psi$ and $\psi ^\sharp := \psi ^* \circ \tau$, the regularized functional $E_\psi (\varphi ) := E(\psi * \varphi * \psi ^\sharp ) = \mu (|\widehat{\psi }|^2 \widehat{\varphi })$ is continuous because it is the Fourier transform of the bounded measure $|\widehat{\psi }|^2 \mu$.

Now we consider a sequence $0 \le \psi _n \in C^\infty _c(\Omega )$ with $\widehat{\psi }_n(0) = \int _{\Omega } \psi _n(x)\, d\mu _V(x) = 1$ and $\mathop {\mathrm{supp}}\nolimits (\psi _n) \rightarrow \{0\}$ (a $\delta$-sequence). Then

$$\begin{aligned} \widehat{\psi }_n(\alpha ) = \int _{V} \psi _n(x) e^{-\alpha (x)}\, d\mu _V(x) \rightarrow e^{-\alpha (0)} = 1 \end{aligned}$$

holds pointwise. We also have $\Vert \widehat{\psi }_n\Vert _\infty \le \Vert \psi _n\Vert _1 = 1$ for the sup-norm of $\widehat{\Omega }$, so that Dominated Convergence implies that $E_{\psi _n} \rightarrow E$ pointwise on $C^\infty _c(V)$.

The Uniform Boundedness Theorem, applied to the restriction of $E_{\psi _n}$ to the Fréchet spaces $C^\infty _K(\Omega )$, $K \subseteq \Omega$ compact, now implies that $E$ is continuous on every subspace $C^\infty _K(\Omega )$, hence continuous on $C^\infty _c(\Omega )$. This means that $E$ is a positive definite distribution.

For $V = \Omega$ and $\tau = \mathop {\mathrm{id}}\nolimits _V$, the Bochner–Schwartz Theorem ([31, Thm. XVIII, §VII.9]) further implies that $\mu$ is tempered.$\square$

Remark 10.2

Suppose that $D \in C^{-\infty }(\Omega )$ is a positive definite distribution, so that we can use the generalized Bochner–Schwartz Theorem ([28, Thm. 4.11]) to write it as the Fourier–Laplace transform of a measure $\mu$ in the sense of

$$\begin{aligned} D(\varphi ) = \int _{\widehat{\Omega }} \overline{\widehat{\varphi }}\, d\mu \quad \text{ for } \quad \varphi \in C^\infty _c(\Omega ), \quad \text{ where } \quad \widehat{\varphi }(\alpha ) := \int _{\Omega } \varphi (x) e^{-\alpha (x)}\, dx \end{aligned}$$

is the Fourier–Laplace transform of $\varphi$. Then we obtain an isomorphism

$$\begin{aligned} \Gamma : L^2(\widehat{\Omega }, \mu ) \rightarrow \mathcal {H}_D \subseteq C^{-\infty }(\Omega ), \quad \Gamma (f)(\varphi ) = \langle f, \widehat{\varphi } \rangle , \end{aligned}$$

under which $D$ corresponds to the constant function $1$. In particular, $\widehat{\varphi } \in L^2(\widehat{\Omega }, \mu )$ for $\varphi \in C^\infty _c(\Omega )$.

Remark 10.3

In the special case $\tau = -\mathop {\mathrm{id}}\nolimits _V$ the integrability of the functions $\widehat{\varphi }$ is equivalent to the existence of the Laplace transform

$$\begin{aligned} \widehat{\mathcal {L}}(\mu )(x) = \int _{\widehat{\Omega }} e^{-\alpha (x)}\, d\mu (\alpha ) \quad \text{ for } \quad x \in \Omega . \end{aligned}$$

This follows from

$$\begin{aligned} \int _{\widehat{\Omega }} \widehat{\varphi }\, d\mu = \int _\Omega \mathcal {L}(\mu )(x) \varphi (x)\, dx \end{aligned}$$

because $\mathcal {L}(\mu )$ is continuous whenever it exists (by the Dominated Convergence Theorem).

Example 10.4

The constant function $1$ on $\widehat{\Omega }$ need not be a distribution vector for the representation $\pi _\mu ^c$ of $V^c$, even if $\widehat{\mu }$ defines a distribution on $\Omega$. A simple example is provided for $V = {\mathbb R}$, $\tau = -\mathop {\mathrm{id}}\nolimits _V$ and $\Omega = {\mathbb R}_+$ by the measure $d\mu (t) := e^{\sqrt{t}} \, dt$ on $\widehat{\Omega } = [0,\infty [$. Then the Laplace transform $\mathcal {L}(\mu )$ is defined on $\Omega$, so that each function $\widehat{\varphi }$, $\varphi \in C^\infty _c(\Omega )$, is integrable, but $\mathcal {L}(\mu )$ does not extend to a distribution on ${\mathbb R}$ because $\int _1^\infty x^{-\ell }\, d\mu (x) = \infty$ for every $\ell \in {\mathbb N}$ (cf. Lemma 7.5).

Example 10.5

We now turn to the special case $V = {\mathbb R}^d$, $\Omega = {\mathbb R}^d_+$ and $\tau (x) = (-x_0, \mathbf {x})$ for $x = (x_0, \mathbf {x})$.

(a) Identifying $\widehat{\Omega }$ with $[0,\infty [ \times {\mathbb R}^{d-1}$, the contraction representation of $\Omega$ on $L^2(\widehat{\Omega },\mu )$ is given by

$$\begin{aligned} (\pi _\mu (x)f)(p) = e^{-x_0 p_0} e^{-i \mathbf {x}\mathbf {p}} f(p) \quad \text{ for } \quad p = (p_0, \mathbf {p}). \end{aligned}$$

Here we use the embedding

$$\begin{aligned} \iota : [0,\infty [ \times {\mathbb R}^{d-1} \rightarrow \widehat{\Omega } \subseteq V_{\mathbb C}^* \cong {\mathbb C}^d, \quad \iota (p_0, \mathbf {p})(x) := p_0 x_0 + i \mathbf {x}\mathbf {p}. \end{aligned}$$

The corresponding unitary representation of $V^c = i{\mathbb R}\oplus {\mathbb R}^{d-1} \cong {\mathbb R}^d$ is given by

$$\begin{aligned} (\pi _\mu ^c(x)f)(p) = e^{-i xp} f(p) = e^{-ix_0 p_0} e^{-i \mathbf {x}\mathbf {p}} f(p) \quad \text{ for } \quad x = (x_0, \mathbf {x}), p = (p_0,\mathbf {p}). \end{aligned}$$

In view of Corollary 10.8, the function $1$ is a distribution vector for $\pi ^c_\mu$ if and only if $\mu$ is a tempered measure on ${\mathbb R}^d$.

(b) Note that $\pi _\mu (x)1 \in \mathcal {H}$ for every $x \in \Omega$ is equivalent to the existence of the Fourier–Laplace transform of $\mu$ as a function on the open half space $\Omega$. In view of $|e^{-\alpha (x)}| = e^{-p_0 x_0}$ for $\alpha = (p_0, i \mathbf {p})$, this in turn is equivalent to the square-integrability of all functions $e_x(\alpha ) := e^{-\alpha (x)}$, $x \in \Omega$, on $\widehat{\Omega }$. This in turn is equivalent to the integrability of all functions $e^{-t p_0}$, $t > 0$. If $\mu$ is supported by the forward light cone $\overline{V_+} \subseteq {\mathbb R}^d$, then the temperedness of $\mu$ implies the existence of its Laplace transform on $\Omega$.

The following lemma makes the condition from Lemma 10.1 more explicit in terms of the product decomposition $\Omega = {\mathbb R}_+ \times {\mathbb R}^{d-1}$.

Lemma 10.6

For $V = {\mathbb R}^d$, $\Omega = {\mathbb R}^d_+$ and $\tau (x) = (-x_0, \mathbf {x})$, a positive Borel measure $\mu$ on $\widehat{\Omega }$ defines a distribution $\widehat{\mu }$ on the open half space $\Omega = {\mathbb R}^d_+$ if and only if all functions $e_t(p_0,\mathbf {p}) := e^{-tp_0}$, $t > 0$, on $\widehat{\Omega }$ are distribution vectors for the representation $\pi _\mu ^c\vert _{{\mathbb R}^{d-1}}$ of the subgroup $\{0\} \times {\mathbb R}^{d-1}$ on $L^2(\widehat{\Omega }, \mu )$. This in turn is equivalent to the temperedness of the measures $\nu _t := \mathop {\mathrm{pr}}\nolimits _*(e_t \mu )$ on ${\mathbb R}^{d-1}$, where $\mathop {\mathrm{pr}}\nolimits (x_0,\mathbf {x}) = \mathbf {x}$.

Proof

In view of Lemma 10.1, $\widehat{\mu } \in C^{-\infty }(\Omega )$ is equivalent to the integrability of all function $\widehat{\psi }$, $\psi \in C^\infty _c(\Omega )$. Consider such a function of the form $\psi (x_0,\mathbf {x}) = \psi _0(x) \psi _1(\mathbf {x})$ with $\psi _0 \in C^\infty _c({\mathbb R})$ and $\psi _1 \in C^\infty _c({\mathbb R}^{d-1})$. Then $\widehat{\psi }(p_0,\mathbf {p}) = \widehat{\psi }_0(p_0) \widehat{\psi }_1(\mathbf {p})$ is $\mu$-integrable on $\widehat{\Omega }$. If $\psi _0 \ge 0$ is supported in the interval $[a,b] \subseteq {\mathbb R}_+$, then

$$\begin{aligned} \widehat{\psi }_0(p_0) = \int _0^\infty e^{-p_0 x_0}\psi _0(x_0)\, dx_0 \ge e^{-p_0 b} \int _0^\infty \psi _0(x_0)\, dx_0 \end{aligned}$$

implies that $e^{-p_0 b} \widehat{\psi }_1(\mathbf {p}) \in L^2(\widehat{\Omega }, \mu )$ for every $b > 0$ and $\psi _1 \in C^\infty _c({\mathbb R}^{d-1})$. According to Corollary 10.8, this means that, for every $b > 0$, the function $e^{-b p_0}$ is a distribution vector for the representation of ${\mathbb R}^{d-1}$ on $L^2(\widehat{\Omega }, \mu )$. In view of Lemma 10.7, this is equivalent to the temperedness of the measure $\mathop {\mathrm{pr}}\nolimits _*(e^{-b p_0} \mu )$ on ${\mathbb R}^{d-1}$.

Suppose, conversely, that this condition is satisfied and let $\varphi \in C^\infty _c(\Omega )$ and $0 < a < b$ with

$$\begin{aligned} \mathop {\mathrm{supp}}\nolimits (\varphi ) \subseteq [a,b] \times {\mathbb R}^{d-1}. \end{aligned}$$

For $\varphi _{x_0}(\mathbf {x}) := \varphi (x_0,\mathbf {x})$ and

$$\begin{aligned} \widehat{\varphi }(p_0,\mathbf {p}) = \int _0^\infty \int _{{\mathbb R}^{d-1}} \varphi (x_0, \mathbf {x}) e^{-p_0 x_0} e^{-i \mathbf {p}\mathbf {x}}\, d\mathbf {x}dx_0 = \int _0^\infty \widehat{\varphi }_{x_0}(\mathbf {p}) e^{-p_0 x_0} dx_0, \end{aligned}$$

we obtain the estimate $|\widehat{\varphi }(p_0,\mathbf {p})| \le e^{-p_0 a} \int _a^b |\widehat{\varphi }_{x_0}(\mathbf {p})|\, dx_0.$ Therefore

$$\begin{aligned} \int _{\widehat{\Omega }} |\widehat{\varphi }(p_0,\mathbf {p})| \, d\mu (p)&\le \int _{\widehat{\Omega }} e^{-p_0 a} \int _a^b |\widehat{\varphi }_{x_0}(\mathbf {p})|\, dx_0 \, d\mu (p)\\&= \int _{{\mathbb R}^{d-1}} \int _a^b |\widehat{\varphi }_{x_0}(\mathbf {p})|\, dx_0 \, d p_*(e^{-p_0 a}\mu )(\mathbf {p}) \\&= \int _a^b \int _{{\mathbb R}^{d-1}} |\widehat{\varphi }_{x_0}(\mathbf {p})| \, d p_*(e^{-p_0 a}\mu )(\mathbf {p})\, dx_0. \end{aligned}$$

To see that this integral is finite, we first observe that the function ${\mathbb R}_+ \rightarrow C^\infty _c({\mathbb R}^{d-1}), x_0 \mapsto \varphi _{x_0}$ is continuous. Since the measure $\nu := p_*(e^{-p_0 a}\mu )$ on ${\mathbb R}^{d-1}$ is tempered, the map $\gamma : C^\infty _c({\mathbb R}^{d-1}) \rightarrow L^1({\mathbb R}^{d-1},\nu ), \psi \mapsto \widehat{\psi }$ is continuous, and hence

$$\begin{aligned} \int _{\widehat{\Omega }} |\widehat{\varphi }(p_0,\mathbf {p})| \, d\mu (p) \le \int _a^b \Vert \gamma (\varphi _{x_0})\Vert _1\, dx_0 < \infty . \end{aligned}$$

$\square$

1.2 D.2 Representations of Vector Groups

We have seen above that the temperedness of a measure on ${\mathbb R}^d$ is closely connected with distribution vectors. The following lemma provides a useful criterion to check this condition. Its main ingredient is the automatic continuity result from Lemma 10.1.

Lemma 10.7

Let $(X,{\mathfrak {S}},\mu )$ be a measure space. We write $M(X,{\mathbb C})$ for the vector space of measurable functions $X \rightarrow {\mathbb C}$. For $H_j \in M(X,{\mathbb R})$, $j = 1,\ldots , d$, we consider the continuous unitary representation of ${\mathbb R}^d$ on $L^2(X,\mu )$, given by

Put $R := \sqrt{\sum _{j = 1}^d H_j^2}.$ Then

$$\begin{aligned} \mathcal {H}^{-\infty } \cong \Big \{h \in M(X,{\mathbb C}) : (\exists n \in {\mathbb N}) \ \Vert (1+ R^{2})^{-n} f \Vert _2 < \infty \Big \}, \end{aligned}$$

where the pairing $\mathcal {H}^{-\infty } \times \mathcal {H}^\infty \rightarrow {\mathbb C}$ is given by $(h,f) \mapsto \int _X h \overline{f} \, d\mu$.

The following assertions are equivalent:

• (i) The constant function $1$ is a distribution vector.

• (ii) For the measurable map $\eta := (H_1, \ldots , H_d) : X \rightarrow {\mathbb R}^d$, the measure $\eta _*\mu$ on ${\mathbb R}^d$ is tempered.

• (iii) $\widehat{\varphi } \circ \eta \in L^2(X,\mu )$ for every $\varphi \in C^\infty _c({\mathbb R}^d)$.

If these conditions are satisfied, then the corresponding distribution on ${\mathbb R}^d$ is given by the Fourier transform of $\eta _*\mu$.

Proof

Put $\mathcal {H}:= L^2(X,\mu )$.

(a) For $d = 1$ and the measurable function $H : X \rightarrow {\mathbb R}$, the subspace of smooth vectors is

$$\begin{aligned} \mathcal {H}^\infty = \{ f \in \mathcal {H}: (\forall n \in {\mathbb N}) \ \Vert H^n f \Vert _2 < \infty \} \end{aligned}$$

and the natural Fréchet topology on this space is defined by the seminorms $p_n(f) := \Vert (1 + H^2)^n f\Vert _2$. From

$$\begin{aligned} \Vert fh\Vert _1 = \Vert f(1+ H^{2})^n (1 + H^{2})^{-n}h\Vert _1 \le p_n(f) \Vert (1 + H^{2})^{-n}h\Vert _2 \end{aligned}$$

it follows that the space of distribution vectors can be identified with

$$\begin{aligned} \mathcal {H}^{-\infty } = \Big \{ h \in M(X,{\mathbb C}) : (\exists n \in {\mathbb N}) \ \Vert (1+ H^{2})^{-n} f \Vert _2 < \infty \Big \}. \end{aligned}$$

(b) In the general case it follows from (a) that the subspace of smooth vectors is

$$\begin{aligned} \mathcal {H}^\infty = \{ f \in \mathcal {H}: (\forall \mathbf {n}\in {\mathbb N}_0^d) \ \Vert H_1^{n_1} \cdots H_d^{n_d}f \Vert _2 < \infty \} \end{aligned}$$

and the natural Fréchet topology on this space is defined by the seminorms $p_n(f) := \Vert (1 + R^{2})^n f\Vert _2.$ Therefore

$$\begin{aligned} \mathcal {H}^{-\infty } \cong \Big \{ h \in M(X,{\mathbb C}) : (\exists n \in {\mathbb N}) \ \Vert (1+ R^{2})^{-n} f \Vert _2 < \infty \Big \}. \end{aligned}$$

In particular, $1$ is a distribution vector if and only if some function $(1 + R^{2})^{-n}$ is integrable, which is equivalent to the integrability of the function $(1 + \Vert x\Vert ^2)^{-n}$ with respect to the measure $\eta _*\mu$ on ${\mathbb R}^d$, and this means that $\eta _*\mu$ is tempered. This proves the equivalence of (i) and (ii).

(ii) $\Rightarrow$ (iii) The integrated representation of $L^1({\mathbb R}^d)$ on $\mathcal {H}$ is given by $U(\varphi ) f= (\widehat{\varphi } \circ (-\eta )) \cdot f$ and the integrated representation $U^{-\infty }$ on $\mathcal {H}^{-\infty }$ is given by the same formula. Hence (iii) follows from $\widehat{\varphi } \circ (-\eta ) = U^{-\infty }(\varphi ) 1 \in L^2(X,\mu )$ for $\varphi \in C^\infty _c({\mathbb R}^d)$.

(iii) First we use the Dixmier–Malliavin Theorem ([3, Thm. 3.1]) to see that every test function on ${\mathbb R}^d$ is a finite sum of convolution products $\varphi * \psi$. Therefore (iii) implies that, for every $\varphi \in C^\infty _c({\mathbb R}^d)$, the function $\widehat{\varphi } \circ \eta$ is $\mu$-integrable, resp., $\widehat{\varphi }$ is integrable w.r.t. $\eta _*\mu$. Now Lemma 10.1, applied to $\Omega = {\mathbb R}^d$ and $\tau = \mathop {\mathrm{id}}\nolimits _{{\mathbb R}^d}$, implies that $\eta _*\mu$ is tempered.

For the last assertion, we simply calculate:

$$\begin{aligned} D(\varphi ) := \langle 1, U^{-\infty }(\varphi ) 1 \rangle = \int _X \overline{\widetilde{\varphi }(\eta (x))}\, d\mu (x) = \int _{{\mathbb R}^d} \overline{\widetilde{\varphi }(x)}\, d(\eta _*\mu )(x) = (\eta _*\mu )^{\wedge }{}(\varphi ). \end{aligned}$$

$\square$

Applying Lemma 10.7 to $X = V^* \cong {\mathbb R}^d$ and $\eta (x) = -x$, we obtain:

Corollary 10.8

Let $V$ be a finite-dimensional real vector space. For a positive Borel measure on $V^*$, the following are equivalent:

• (i) $\mu$ is tempered.

• (ii) $1 \in L^2(V^*,\mu )^{-\infty }$ for the representation $(U_v f)(\alpha ) = e^{-i\alpha (v)} f(\alpha )$.

• (iii) $\widehat{\varphi } \in L^2(V^*,\mu )$ for every $\varphi \in C^\infty _c(V)$.

Remark 10.9

The first part of Lemma 10.7 remains true for vector-valued $L^2$-spaces $L^2(X,Q,\mu ;\mathcal {H})$, where $Q: X \rightarrow \mathop {\mathrm{Herm}}\nolimits _+(\mathcal {H})$ is an operator-valued density,

$$\begin{aligned} \langle f,g \rangle \!=\! \int _X \langle Q(x) f(x), g(x)\rangle \, d\mu (x) \quad \text{ and } \quad U_{\mathbf {t}}(f) \!:=\! e^{i \sum _{j = 1}^d t_j H_j} f, \, \mathbf {t}\!=\! (t_1,\ldots , t_d). \end{aligned}$$

Remark 10.10

(Projectable measures and “time-zero” subspaces)

(a) Let $\mu$ be a measure on ${\mathbb R}^d$ and $\mathop {\mathrm{pr}}\nolimits : {\mathbb R}^d \rightarrow {\mathbb R}^{d-1}, p = (p_0, \mathbf {p}) \mapsto \mathbf {p}$ be the projection. We call $\mu$ projectable if $\mathop {\mathrm{pr}}\nolimits _*\mu$ is tempered. In view of Corollary 10.8, this is equivalent to the integrability of all Fourier transforms $\widehat{\varphi } \circ \mathop {\mathrm{pr}}\nolimits$, $\varphi \in C^\infty _c({\mathbb R}^{d-1})$, which in turn means that $1$ is a distribution vector for the canonical action of ${\mathbb R}^{d-1}$ on $L^2({\mathbb R}^d,\mu )$. From Lemma 10.7 we now see that $1$ is also a distribution vector for ${\mathbb R}^d$, which means that $\mu$ is tempered.

In particular, we obtain an isometric embedding

$$\begin{aligned} \mathop {\mathrm{pr}}\nolimits ^* : L^2({\mathbb R}^{d-1}, \mathop {\mathrm{pr}}\nolimits _*\mu ) \hookrightarrow L^2({\mathbb R}^d,\mu ), \quad f \mapsto f \circ \mathop {\mathrm{pr}}\nolimits \end{aligned}$$

which maps onto the subspace of those functions not depending on the first argument $p_0$. Accordingly their Fourier transform is supported in the $x_0 = 0$ hyperplane. Therefore $L^2({\mathbb R}^{d-1}, \mathop {\mathrm{pr}}\nolimits _*\mu )$ is also called the time-zero subspace of $L^2({\mathbb R}^d,\mu )$.

(b) For the distribution $D := \widehat{\mu }$, the projectability of $\mu$ means that $D$ can be restricted to the subspace ${\mathbb R}^{d-1}$ by

$$\begin{aligned} D_r(\varphi ) := \int _{{\mathbb R}^d} \overline{\widehat{\varphi }(\mathbf {p})}\, d\mu (p) = (\mathop {\mathrm{pr}}\nolimits _*\mu )^{\wedge }{}(\varphi ), \end{aligned}$$

respectively extended to the space $C^\infty _c({\mathbb R}^{d-1})$. The scalar product $\langle \varphi , \psi \rangle := D_r(\varphi ^* * \psi )$ on $C^\infty _c({\mathbb R}^{d-1})$ leads to $\mathcal {H}_{D_r} \cong L^2({\mathbb R}^{d-1}, \mathop {\mathrm{pr}}\nolimits _*\mu )$, which is naturally realized as a subspace of $L^2({\mathbb R}^d,\mu )$. Accordingly, we see that, as a space of distributions on ${\mathbb R}^d$, we may consider $\mathcal {H}_{D_r}$ as a subspace of $\mathcal {H}_D$.

(c) The Lorentz invariant measures $\mu _m$ are interesting examples of measures $\mu$ on ${\mathbb R}^d$ for which the inclusion $L^2({\mathbb R}^{d-1}, \mathop {\mathrm{pr}}\nolimits _*\mu ) \hookrightarrow L^2({\mathbb R}^{d}, \mu )$ is actually surjective.

(d) For $d = 1$, a measure $\mu$ is projectable if and only if it is finite, i.e., if its Fourier transform $\widehat{\mu }$ is a continuous function. Then $L^2({\mathbb R}^0, \mathop {\mathrm{pr}}\nolimits _*\mu )$ is the subspace of constant functions in $L^2({\mathbb R},\mu )$.

(e) A similar picture prevails for operator-valued measures, distributions and $L^2$-spaces such as those discussed in Sect. 5. In this context the subspace $\mathcal {H}\subseteq L^2({\mathbb R},Q;\mathcal {H})$ of constant functions plays the role of the time-zero subspace. This matches well with Proposition 5.2(iii), where ${\mathcal {E}}_0 \cong \mathcal {H}$ is identified as consisting of those functions whose Fourier transform is supported in $\{0\}$.