bounded as $\epsilon \to 0$ by some Schwartz norm of $\varphi$. By (b) the term $(\mathcal{G}_\epsilon, \epsilon\varphi)$ in (C.8) drops out when $\epsilon \to 0$, and by (a) the rest tends to $(\mathcal{G}^M, \{it\partial_{y^1} + iy^1\partial_t\}\varphi)$. So we conclude that $(\mathcal{G}^M, \{it\partial_{y^1} + iy^1\partial_t\}\varphi) = 0$ which expresses invariance of $\mathcal{G}^M$ under the 01 Lorentz transformation.

Let us proceed to show the rest of Parts 2,3. It will be crucial that $\mathcal{G}$ can be written as a “Fourier-Laplace transform”:

$$ \mathcal{G}(\epsilon + is, \mathbf{y}) = \int \frac{dE \, d\mathbf{p}}{(2\pi)^{d(n-1)}} g(E, \mathbf{p})e^{-(\epsilon+is)E-i\mathbf{p}\cdot\mathbf{y}}, \quad (C.9) $$

where $g(E, \mathbf{p})$, $E \in \mathbb{R}^{n-1}$, $\mathbf{p} \in (\mathbb{R}^{d-1})^{n-1}$ is a tempered distribution, called “spectral function”, supported at $E \ge 0$ (by which we mean all $E_k \ge 0$) [later this will be improved to $E \ge |\mathbf{p}|$]. We are omitting the indices, thus $\epsilon E = \sum_k \epsilon_k E_k$, etc. The equality in (C.9) is understood in the sense of distributions, with the r.h.s. being the inverse Fourier transform of the tempered distribution $g(E, \mathbf{p})e^{-\epsilon E}$. In other words, what this means is that

$$ \int ds \, dy \, \mathcal{G}(\epsilon + is, \mathbf{y}) f(-s, -\mathbf{y}) = \int \frac{dE \, d\mathbf{p}}{(2\pi)^{d(n-1)}} g(E, \mathbf{p})e^{-\epsilon E} \hat{f}(E, \mathbf{p}), \quad (C.10) $$

for any Schwartz test function $f$, and $\hat{f}$ its Fourier transform.

Let us show (C.9). Notice first that for every $\epsilon > 0$ we can write

$$ \mathcal{G}(\epsilon + is, \mathbf{y}) = \int \frac{dE \, d\mathbf{p}}{(2\pi)^{d(n-1)}} g_{\epsilon}(E, \mathbf{p})e^{-isE - i\mathbf{p}\cdot\mathbf{y}}, \quad (C.11) $$

where $g_\epsilon$ is the Fourier transform of $\mathcal{G}(\epsilon + is, \mathbf{y})$ with respect to $s, \mathbf{y}$. This Fourier transform exists as a tempered distribution, since $\mathcal{G}(\epsilon + is, \mathbf{y})$ is itself a tempered distribution in $s, \mathbf{y}$ (being a real-analytic function, bounded by a power at infinity). In addition, $\mathcal{G}(\epsilon + is, \mathbf{y})$ is differentiable in $\epsilon$ and $s$ and satisfies the Cauchy-Riemann equations. From here it's easy to show that $g_\epsilon$ as a distribution is differentiable in $\epsilon$ and satisfies the differential equations:

$$ \frac{\partial g_{\epsilon}}{\partial \epsilon_k} + E_k g_{\epsilon} = 0 \quad (k = 1, \dots, d-1). \quad (C.12) $$

From here we conclude that

$$ g(E, \mathbf{p}) := g_{\epsilon}(E, \mathbf{p})e^{\epsilon E} \quad (C.13) $$

is an $\epsilon$-independent distribution. Substituting $g_\epsilon(E, \mathbf{p}) = g(E, \mathbf{p})e^{-\epsilon E}$ into (C.11), we obtain (C.9). Note that since $g$ and $g_\epsilon$ are related by an exponential factor, we can so far only claim that $g$ is defined as a distribution on test functions of compact support. Let us show next that it is in fact tempered (i.e. extends to Schwartz test functions).

To this end, consider the inverse of (C.11):

$$ g_{\epsilon}(E, \mathbf{p}) = g(E, \mathbf{p})e^{-\epsilon E} = \int ds \, dy \, \mathcal{G}(\epsilon + is, \mathbf{y})e^{isE + i\mathbf{p}\cdot\mathbf{y}}, \quad (C.14) $$

and integrate it against a compactly supported test function $\varphi(E, \mathbf{p})$. We get (compare (C.10)):

$$ \int dE \, d\mathbf{p} \, g(E, \mathbf{p})e^{-\epsilon E} \varphi(E, \mathbf{p}) = \int ds \, dy \, \mathcal{G}(\epsilon + is, \mathbf{y}) \hat{\varphi}(-s, -\mathbf{y}). \quad (C.15) $$
As $\epsilon \to 0$, the l.h.s. tends to the pairing $(g, \varphi)$. Using Part 1 of the theorem, the r.h.s. tends in the same limit to $\int ds dy \mathcal{G}^M(s, y) \hat{\varphi}(-s, -y)$ which exists in the sense of tempered distributions and so is bounded by some Schwartz-space norm $|\hat{\varphi}|_N$. We get

$$|(g, \varphi)| \le \text{const.} |\hat{\varphi}|_N \le \text{const.} |\varphi|_{N'}, \qquad (C.16)$$

where in the second inequality we used that the Fourier transform is continuous in the Schwartz space. This inequality, valid for any compactly supported $\varphi$, means that $g$ extends to a tempered distribution on the whole Schwartz space. The representation (C.9) is thus established.

Next let us show that $g$ is supported at $E \ge 0$. For this we will pass to the large $\epsilon$ limit in (C.13). Supposing that $E_k < 0$ for some $k$, the factor $e^{E\epsilon}$ in (C.13) decreases exponentially as the corresponding $\epsilon_k \to +\infty$. On the other hand $g_\epsilon(E, \mathbf{p})$ is bounded in the same limit by a power of $\epsilon$, because it's the Fourier transform of $\mathcal{G}(\epsilon + is, y)$ which satisfies a powerlaw bound.<fnref n="100" /> This implies that $g(E, \mathbf{p}) = 0$ unless $E \ge 0$.<fnref n="101" />

Consider then the following lemma, proven analogously to, and easier than, lemma C.2 below.

**Lemma C.1.** Let $g(E, \mathbf{p})$ be a tempered distribution supported at $E \ge 0$, and consider the distribution $g(E, \mathbf{p})e^{-\epsilon E}$ ($\epsilon > 0$). This distribution, being initially defined by this formula on compactly supported test functions, extends to a tempered distribution, and moreover $g(E, \mathbf{p})e^{-\epsilon E} \to g(E, \mathbf{p})$ as $\epsilon \to 0$, in the sense of tempered distributions.

Let us now take the $\epsilon \to 0$ limit on both sides of (C.9) (or, which is the same, (C.10)). The l.h.s. has a limit by Part 1, while the r.h.s. has a limit by lemma C.1. We obtain that $\mathcal{G}^M(s, y)$ and $g(E, \mathbf{p})$ are related by the Fourier transform:

$$\mathcal{G}^M(s, y) = \int \frac{dE \, d\mathbf{p}}{(2\pi)^{d(n-1)}} g(E, \mathbf{p}) e^{-isE - i\mathbf{p}y}. \qquad (C.17)$$

We can now complete the proof of Part 2, namely to show the spectral condition. Above we proved that $\mathcal{G}^M(s, y)$ is Lorentz invariant. Since $g(E, \mathbf{p})$ is its Fourier transform, it is also Lorentz invariant, and in particular its support must be a Lorentz-invariant set. We also know that $\text{supp} g \subset \{E \ge 0\}$. These two facts together imply that $\text{supp} g$ must be contained in the product of the forward null cones, i.e. $g(E, \mathbf{p}) = 0$ unless each $E_k \ge |\mathbf{p}_k|$, which is the spectral condition.

Part 3 follows by standard Wightman theory arguments. Namely, let us write $(iy_k^0, y_k) = \xi_k + i\eta_k$ where $\xi_k, \eta_k \in \mathbb{R}^{1,d-1}$ and $\eta_k = (\text{Re } y_k^0, \text{Im } y_k) > 0$. The extension to the forward tube is given by the equation (with $p = (E, \mathbf{p})$)

$$\int dp g(p) e^{i(p, \xi)} e^{(p, \eta)}, \qquad (C.18)$$

which reduces to (C.9) for real $y_k$. It is holomorphic by Part (c) of the following lemma, while Parts (a,b) imply that this extension has the same limit as (C.9).

<footnote n="100">¹⁰⁰This is the only place where we use the powerlaw bound on $\mathcal{G}(\epsilon + is, y)$ for large rather than small $\epsilon$.</footnote>

<footnote n="101">¹⁰¹If unhappy with this intuitive reasoning, the argument may be made more rigorous in its integrated version: show that $g$ vanishes on test functions supported in the complement of $E \ge 0$.</footnote>
**Lemma C.2.** Let $g(p)$ be a tempered supported at $p \succeq 0$ (closed forward light cone). Consider the distribution $g_\eta(p) = g(p)e^{-(p,\eta)}$, initially defined by this formula on compactly supported test functions. Then

(a) $g_\eta$ for $\eta \succ 0$ extends to a tempered distribution;

(b) $g_\eta \to g$ as $\eta \to 0$ from inside the forward light cone $\eta \succ 0$, in the sense of tempered distributions;

(c) The Fourier transform $\hat{g}_\eta(\xi)$ of the distribution $g_\eta(p)$ is a holomorphic function of $\xi + i\eta$ for $\eta \succ 0$.

*Proof.* Let $\omega(p)$ be a $C^\infty$ function which is identically 1 on the forward light cone $\overline{V_+}$, and zero as soon as $\text{dist}(p, \overline{V_+}) \ge 1$ where $\text{dist}$ is the Euclidean distance. We can choose this function so that all its derivatives are uniformly bounded by a constant depending only on the derivative order: $|\omega^{(\alpha)}(p)| \le C_\alpha$ for any $p$.

Consider the family of $C^\infty$ functions parametrized by $\xi, \eta \in \mathbb{R}^{1,d-1}$:

$$ \Omega_{\xi, \eta}(p) = e^{i(p, \xi)} e^{(p, \eta)} \omega(p). \qquad (C.19) $$

It is not hard to check that $\Omega_{\xi,\eta}$ is a Schwartz function for $\eta \succ 0$ and any $\xi$.

Let us define $g_\eta$ paired with a Schwartz function $\varphi(p)$ via

$$ (g_\eta, \varphi) = (g, \Omega_{0,\eta}\varphi). \qquad (C.20) $$

We know that $\Omega_{0,\eta}\varphi$ is a Schwartz function for $\eta \succ 0$, so this definition makes sense. Furthermore it is not hard to check that $\Omega_{0,\eta}\varphi \to \omega\varphi$ in the Schwartz space topology as $\eta \to 0, \eta \succ 0$. This proves Parts (a),(b).

Next, let us define

$$ F(\xi, \eta) = (g, \Omega_{\xi, \eta}), \quad \xi, \eta \in \mathbb{R}^{1,d-1}. \qquad (C.21) $$

We know that $\Omega_{\xi,\eta}$ is a Schwartz function for $\eta \succ 0$, so $F(\xi, \eta)$ is a function. Moreover it is not hard to show that the family $\Omega_{\xi,\eta}$ is continuous and continuously differentiable in the Schwartz space topology. It also obviously satisfies the Cauchy-Riemann equations: $(\partial_\xi + i\partial_\eta)\Omega_{\xi,\eta} = 0$. This implies that $F(\xi, \eta)$ is a holomorphic function in $\xi + i\eta$. It remains to show that $F(\xi, \eta) = \hat{g}_\eta(\xi)$. It's enough to check this integrated against a compactly supported test function $\chi(\xi)$:

$$ \begin{aligned} \int F(\xi, \eta)\chi(\xi) d\xi &= \int (g, \Omega_{\xi, \eta})\chi(\xi) d\xi \\ &= \left( g, \int d\xi \chi(\xi)\Omega_{\xi, \eta} \right) = (g, \Omega_{0, \eta}\hat{\chi}) = (g_\eta, \hat{\chi}) = (\hat{g}_\eta, \chi). \end{aligned} \qquad (C.22) $$

The proof is complete.

<figure><img src="image_17.png" /></figure>

## D Intuition about lemma 4.2

The proof of lemma 4.2 in section 4.1 was by contradiction. To help intuition, we will give here a constructive argument of a special case of lemma 4.2, namely $d = 1$ and $n = 2$.
I.e. we will show how any Schwartz function $f \in S(\mathbb{R})$ can be approximated by Schwartz functions $g$ which for $E \ge 0$ agree with Laplace transform:

$$ \mathcal{L}(\varphi)(E) = \int_{0}^{\infty} dt \, \varphi(t)e^{-Et}, \qquad (D.1) $$

$\varphi \in C_0^\infty(\mathbb{R}_+)$ (compactly supported with support strictly inside $(0, +\infty)$), while for $E < 0$, $g(E)$ is extended arbitrarily. Recall that the Schwartz space topology is given by the family of norms

$$ |f|_n = \sup_{E \in \mathbb{R}, m \le n} (1+E^2)^{n/2} |f^{(m)}(E)|, \qquad (D.2) $$

and we need to find a sequence $\{g_r\}_{r=1}^\infty$ such that $|f - g_r|_n \to 0$ as $r \to \infty$ for any $n$ (we stress that one sequence $g_r$ should work for any $n$).

We will also consider the Schwartz space $S(\mathbb{R}_+)$, consisting of $C^\infty$ functions on $E \ge 0$ (not necessarily vanishing at $E = 0$) with topology given by the family of norms $|f|_{n,+}$ defined by the same equations as (D.2) but with sup taken over $E \ge 0$. It will be sufficient to arrange that for any $n$

$$ |f - \mathcal{L}(\varphi_r)|_{n,+} \to 0 \quad (r \to \infty). \qquad (D.3) $$

This is because there exists an extension operator which takes a function $h \in S(\mathbb{R}_+)$ and provides a function $\mathcal{E}(h) \in S(\mathbb{R})$ such that $\mathcal{E}(h) = h$ for $E \ge 0$ (which is why it called an extension operator), and in addition

$$ |\mathcal{E}(h)|_n \le C_n |h|_{n,+} \qquad (D.4) $$

for all $n$ with some finite constants $C_n$ independent of $h$. E.g., Seeley's linear extension operator [135, 136] has this property. Then, given (D.3), we put

$$ g_r = f + \mathcal{E}(\mathcal{L}(\varphi_r) - f), \qquad (D.5) $$

which, on the one hand satisfies $g_r(E) = \mathcal{L}(\varphi_r)(E)$ for $E \ge 0$ and on the other hand by (D.3) and (D.4) has $|g_r - f|_n \le C_n |\mathcal{L}(\varphi_r) - f|_{n,+} \to 0$ which is what we need.

So let us focus on satisfying (D.3). By a map $x = \frac{1}{1+E}$ the half-line $[0, +\infty)$ is mapped to the interval $(0, 1]$ and the function $f(E)$ is mapped to a function $F(x) = f(\frac{1}{x} - 1)$ which is a $C^\infty$ function vanishing at $x = 0$ faster than any power of $x$. For any $\varepsilon$ and any $N$ we can find, by the Weierstrass theorem, a polynomial $Q(x)$ such that

$$ |F^{(N)}(x) - Q(x)| \le \varepsilon \quad (0 \le x \le 1). \qquad (D.6) $$

Let $P(x)$ be the polynomial such that $P^{(N)}(x) = Q(x)$ and $P(0) = \dots = P^{(N-1)}(0) = 0$. Then $P(x) = O(x^N)$ and it is not hard to see that

$$ |F^{(n)}(x) - P^{(n)}(x)| \le \varepsilon x^{N-n} \quad (0 \le x \le 1). \qquad (D.7) $$

We also put $p(E) = P(\frac{1}{1+E})$. From $f(E) = F(\frac{1}{1+E})$ we know that

$$ |f^{(n)}(E)| \le B_n \max_{m \le n} \left| F^{(m)} \left( \frac{1}{1+E} \right) \right|. \qquad (D.8) $$
So combining this with eq. (D.7), and going up to $n = N/2$ we may conclude that

$$|f - p|_{N/2,+} \le B'_N \epsilon. \qquad (D.9)$$

Now, by construction $p$ has the form

$$p(E) = \sum_{N \le n \le M} a_n \frac{1}{(1+E)^n}. \qquad (D.10)$$

Since $\frac{1}{(1+E)^n} = \frac{1}{(n-1)!} \int_0^\infty t^{n-1}e^{-(1+E)t} dt$, we see that $p(E)$ is the Laplace transform of a function $\psi(t)$:

$$p = \mathcal{L}(\psi), \quad \psi(t) = \sum_{N \le n \le M} \frac{a_n}{(n-1)!} t^{n-1} e^{-t}. \qquad (D.11)$$

Now we can finish the argument as follows. For $r = 1, 2, 3, \dots$ we apply the above argument with $N = 2r$ and $\epsilon = 1/(B'_N r)$ to find $\psi_r$ such that, by (D.9),

$$|f - \mathcal{L}(\psi_r)|_{r,+} \le 1/r. \qquad (D.12)$$

The function $\psi_r$ is not in $C_0^\infty(0, \infty)$ although $\psi_r^{(k)} = 0$ for $k = 0, \dots, 2r-2$, and it vanishes at $\infty$ exponentially. We can therefore approximate $\psi_r$ by a $C_0^\infty(0, \infty)$ function $\varphi_r$ so that $|\psi_r - \varphi_r|_{2r-2,+}$ is arbitrarily small, where the order $2r-2$ of the norm is related to the order of the vanishing of $\psi_r$ at $t = 0$. Furthermore we have the following lemma:

**Lemma D.1.** Let $\chi$ be a $C^\infty$ function on $[0, +\infty)$ which exponentially vanishes at infinity and

$$\chi^{(k)}(0) = 0, \quad k = 0, \dots, n-1. \qquad (D.13)$$

Then, with some constant $D_n$ independent of $\chi$,

$$|\mathcal{L}(\chi)|_{n,+} \le D_n |\chi|_{n+2,+}. \qquad (D.14)$$

*Proof.* We use the following elementary properties of Laplace transform:

$$\left(\frac{d}{dE}\right)^m \mathcal{L}(\chi)(E) = \mathcal{L}[\chi(t)(-t)^m](E), \\ E^n \mathcal{L}(\chi)(E) = \mathcal{L}[\chi^{(n)}(t)](E), \qquad (D.15)$$

where the second equation is derived by integration by parts and is valid under (D.13) and exponential decay. So we have (where $\lesssim$ denotes $\le$ with some $n$-dependent but function-independent constant)

$$|\mathcal{L}(\chi)|_{n,+} \lesssim \sum_{m=0}^{n} \sup_{E \ge 0} (1+E^n) |\mathcal{L}(\chi)^{(m)}(E)| \\ \le \sum_{m=0}^{n} \sup_{E \ge 0} |\mathcal{L}[\chi(t)t^m](E)| + |\mathcal{L}[(\chi(t)t^m)^{(n)}](E)|, \qquad (D.16)$$

Using further the elementary bound $|\mathcal{L}(f)(E)| \lesssim \sup_{t \ge 0} |(1+t^2)f(t)|$ we deduce (D.14). $\square$
We use this lemma with $n = r$ and $\chi = \psi_r - \varphi_r$, which satisfies $\chi^{(k)} = 0$ up to $k = 2r - 2 \ge r - 1$, so (D.13) is satisfied. By (D.14), we have

$$|\mathcal{L}(\psi_r) - \mathcal{L}(\varphi_r)|_{r,+} \le D_r |\psi_r - \varphi_r|_{r+2,+} \le D_r |\psi_r - \varphi_r|_{2r-2,+} \quad (D.17)$$

as long as $r \ge 4$ so that $2r - 2 \ge r + 2$. As mentioned above $|\psi_r - \varphi_r|_{2r-2,+}$ can be made arbitrarily small. Combining with (D.12), we can arrange so that $|f - \mathcal{L}(\varphi_r)|_{r,+} \le 2/r \to 0$ as $r \to \infty$, which in particular implies (D.3).

**Open Access.** This article is distributed under the terms of the Creative Commons Attribution License (CC-BY 4.0), which permits any use, distribution and reproduction in any medium, provided the original author(s) and source are credited.

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