[proofplan] We first represent the frequency-localized propagator as convolution against an oscillatory kernel, which will be formally defined in the first step as a function of the spatial displacement. The main point is a uniform $L^\infty$ estimate for that kernel with decay rate $|t|^{-(n-1)/2}$. Away from the characteristic cone, repeated [integration by parts](/theorems/210) in frequency gives rapid decay; near the cone, polar coordinates reduce the problem to stationary phase on the sphere, producing the factor $|t|^{-(n-1)/2}$. The convolution estimate then gives the $L^1 \to L^\infty$ bound, and density of $\mathcal{S}(\mathbb{R}^n)$ in $L^1(\mathbb{R}^n)$ gives the unique extension. [/proofplan] [step:Represent $P_t$ as convolution with an oscillatory kernel] Define, for each $t \in \mathbb{R}$, the kernel $K_t: \mathbb{R}^n \to \mathbb{C}$ by \begin{align*} K_t(z) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} e^{i(z \cdot \xi + t|\xi|)} a(\xi)\, d\mathcal{L}^n(\xi). \end{align*} Since $a \in C_c^\infty(\mathbb{R}^n \setminus \{0\})$, this integral is absolutely convergent and defines a smooth bounded function of $z$. Let $\phi \in \mathcal{S}(\mathbb{R}^n)$. Substituting the [Fourier transform](/page/Fourier%20Transform) into the definition of $P_t$ gives \begin{align*} P_t\phi(x) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i((x-y)\cdot \xi + t|\xi|)}a(\xi)\phi(y)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(\xi). \end{align*} The function $(y,\xi) \mapsto a(\xi)\phi(y)$ is integrable with respect to $\mathcal{L}^n \otimes \mathcal{L}^n$, because $a$ is compactly supported and $\phi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$. Hence the integrand $(y,\xi) \mapsto e^{i((x-y)\cdot \xi + t|\xi|)}a(\xi)\phi(y)$ is absolutely integrable on the product [measure space](/page/Measure%20Space). The [Fubini theorem](/theorems/513) for absolutely integrable product Lebesgue integrals therefore permits the exchange of the two integrations and yields \begin{align*} P_t\phi(x) = \int_{\mathbb{R}^n} K_t(x-y)\phi(y)\, d\mathcal{L}^n(y). \end{align*} Thus $P_t\phi = K_t * \phi$. [guided] The normalization matters here. With the symmetric Fourier transform convention \begin{align*} \widehat{\phi}(\xi) = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} \phi(y)e^{-iy \cdot \xi}\, d\mathcal{L}^n(y), \end{align*} the two factors of $(2\pi)^{-n/2}$ combine to give the kernel normalization $(2\pi)^{-n}$. Define $K_t: \mathbb{R}^n \to \mathbb{C}$ by \begin{align*} K_t(z) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} e^{i(z \cdot \xi + t|\xi|)}a(\xi)\, d\mathcal{L}^n(\xi). \end{align*} This is a legitimate function because $a$ is smooth and compactly supported, so the integrand is bounded by $|a(\xi)|$, which is integrable with respect to $\mathcal{L}^n$. For $\phi \in \mathcal{S}(\mathbb{R}^n)$, substitute the definition of $\widehat{\phi}$: \begin{align*} P_t\phi(x) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i(x \cdot \xi + t|\xi|)}a(\xi)\phi(y)e^{-iy \cdot \xi}\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(\xi). \end{align*} Combining the exponentials gives \begin{align*} P_t\phi(x) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n}\int_{\mathbb{R}^n} e^{i((x-y)\cdot \xi + t|\xi|)}a(\xi)\phi(y)\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(\xi). \end{align*} We may interchange the integrals by the Fubini theorem for absolutely integrable product Lebesgue integrals. Its hypothesis is that the absolute value of the integrand be integrable on the product measure space. That hypothesis holds because $a$ is compactly supported and $\phi \in \mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$, so \begin{align*} \int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|a(\xi)\phi(y)|\, d\mathcal{L}^n(y)\, d\mathcal{L}^n(\xi) < \infty. \end{align*} Therefore the theorem applies to the product measure space $(\mathbb{R}^n \times \mathbb{R}^n, \mathcal{B}(\mathbb{R}^n) \otimes \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n \otimes \mathcal{L}^n)$, the iterated integrals may be exchanged, and \begin{align*} P_t\phi(x) = \int_{\mathbb{R}^n} K_t(x-y)\phi(y)\, d\mathcal{L}^n(y). \end{align*} This is exactly the convolution identity $P_t\phi = K_t * \phi$. [/guided] [/step] [step:Reduce the theorem to a uniform kernel estimate] It suffices to prove that there exists $C_a > 0$ such that, for every $t \in \mathbb{R}$ with $|t| \geq 1$, \begin{align*} \|K_t\|_{L^\infty(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}. \end{align*} Indeed, if $\phi \in \mathcal{S}(\mathbb{R}^n)$, then for every $x \in \mathbb{R}^n$, \begin{align*} |P_t\phi(x)| \leq \int_{\mathbb{R}^n} |K_t(x-y)||\phi(y)|\, d\mathcal{L}^n(y). \end{align*} Using the kernel bound inside the integral gives \begin{align*} |P_t\phi(x)| \leq \|K_t\|_{L^\infty(\mathbb{R}^n)}\|\phi\|_{L^1(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}\|\phi\|_{L^1(\mathbb{R}^n)}. \end{align*} Taking the supremum over $x \in \mathbb{R}^n$ gives the desired estimate on $\mathcal{S}(\mathbb{R}^n)$. [/step] [step:Obtain rapid decay away from the characteristic cone] Choose numbers $0 < r_0 < r_1 < \infty$ such that \begin{align*} \operatorname{supp} a \subset \{\xi \in \mathbb{R}^n : r_0 \leq |\xi| \leq r_1\}. \end{align*} For $z \in \mathbb{R}^n$ and $t \neq 0$, define the phase $\Phi_{z,t}: \mathbb{R}^n \setminus \{0\} \to \mathbb{R}$ by \begin{align*} \Phi_{z,t}(\xi) = z \cdot \xi + t|\xi|. \end{align*} Its frequency gradient is \begin{align*} \nabla_\xi \Phi_{z,t}(\xi) = z + t\frac{\xi}{|\xi|}. \end{align*} Assume first that either $|z| \leq |t|/2$ or $|z| \geq 2|t|$. Then for every $\xi \in \operatorname{supp} a$, \begin{align*} |\nabla_\xi \Phi_{z,t}(\xi)| \geq \frac{1}{2}\max\{|z|, |t|\}. \end{align*} Define the integration-by-parts operator $L_{z,t}: C^\infty(\mathbb{R}^n \setminus \{0\}) \to C^\infty(\mathbb{R}^n \setminus \{0\})$ by \begin{align*} L_{z,t} b(\xi) = \frac{1}{i|\nabla_\xi \Phi_{z,t}(\xi)|^2}\nabla_\xi \Phi_{z,t}(\xi) \cdot \nabla_\xi b(\xi). \end{align*} Then $L_{z,t}(e^{i\Phi_{z,t}}) = e^{i\Phi_{z,t}}$. Let $L_{z,t}^*$ denote the formal adjoint of $L_{z,t}$ with respect to [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal{L}^n$ on the fixed annulus $\{\xi \in \mathbb{R}^n : r_0 \leq |\xi| \leq r_1\}$. Since $a$ vanishes outside a compact subset of this annulus, [integration by parts](/theorems/2098) produces no boundary term and gives \begin{align*} K_t(z) = \frac{1}{(2\pi)^n}\int_{\mathbb{R}^n} e^{i\Phi_{z,t}(\xi)}(L_{z,t}^*)^N a(\xi)\, d\mathcal{L}^n(\xi). \end{align*} Write \begin{align*} A_{z,t}(\xi) = \frac{\nabla_\xi\Phi_{z,t}(\xi)}{|\nabla_\xi\Phi_{z,t}(\xi)|^2}. \end{align*} Then $L_{z,t}b = i^{-1}A_{z,t}\cdot \nabla_\xi b$, and $L_{z,t}^*$ is a finite sum of terms obtained by differentiating products of components of $A_{z,t}$ and derivatives of the amplitude. For every multi-index $\alpha$, the identities \begin{align*} \nabla_\xi\Phi_{z,t}(\xi)=z+t\frac{\xi}{|\xi|} \end{align*} and $r_0 \leq |\xi| \leq r_1$ imply \begin{align*} |\partial_\xi^\alpha \nabla_\xi\Phi_{z,t}(\xi)| \leq C_{\alpha,r_0,r_1}(|z|+|t|) \end{align*} for $|\alpha|=0$, and \begin{align*} |\partial_\xi^\alpha \nabla_\xi\Phi_{z,t}(\xi)| \leq C_{\alpha,r_0,r_1}|t| \end{align*} for $|\alpha|\geq 1$. Together with $|\nabla_\xi\Phi_{z,t}|\geq \frac12\max\{|z|,|t|\}$ on $\operatorname{supp}a$, the quotient rule gives, for each multi-index $\alpha$, \begin{align*} |\partial_\xi^\alpha A_{z,t}(\xi)| \leq C_{\alpha,r_0,r_1}(1+|z|+|t|)^{-1}. \end{align*} Thus each application of $L_{z,t}^*$ introduces one differentiated coefficient of this form and hence one factor bounded by $C_{r_0,r_1}(1+|z|+|t|)^{-1}$, while the remaining derivatives fall on $a$. Induction on $N$ therefore gives \begin{align*} \|(L_{z,t}^*)^N a\|_{L^\infty(\mathbb{R}^n)} \leq C_{a,N}(1+|z|+|t|)^{-N}. \end{align*} Integrating over the fixed compact support of $a$ gives \begin{align*} |K_t(z)| \leq C_{a,N}(1+|z|+|t|)^{-N} \end{align*} for every $N \in \mathbb{N}$, where $C_{a,N}$ depends on $r_0$, $r_1$, $N$, and finitely many derivatives of $a$. Taking $N \geq (n-1)/2$ and using $|t| \geq 1$ gives \begin{align*} |K_t(z)| \leq C_{a,N}|t|^{-(n-1)/2}. \end{align*} [/step] [step:Use stationary phase on the sphere near the characteristic cone] It remains to estimate $K_t(z)$ when \begin{align*} \frac{|t|}{2} < |z| < 2|t|. \end{align*} If $n = 1$, this region is handled in the next step, so assume $n \geq 2$. Use polar coordinates $\xi = \rho\omega$, where $\rho \in (0,\infty)$ and $\omega \in S^{n-1}$. Let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) restricted to the unit sphere $S^{n-1}$. The Lebesgue measure decomposes as \begin{align*} d\mathcal{L}^n(\xi) = \rho^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(\rho). \end{align*} Define $b_\rho: S^{n-1} \to \mathbb{C}$ by \begin{align*} b_\rho(\omega) = \rho^{n-1}a(\rho\omega). \end{align*} Then \begin{align*} K_t(z) = \frac{1}{(2\pi)^n}\int_{0}^{\infty} e^{it\rho}\int_{S^{n-1}} e^{i\rho z \cdot \omega}b_\rho(\omega)\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(\rho). \end{align*} Because $\operatorname{supp} a \subset \{r_0 \leq |\xi| \leq r_1\}$, the $\rho$-integral is restricted to $[r_0,r_1]$. For each fixed $\rho \in [r_0,r_1]$, the phase on $S^{n-1}$ is $\omega \mapsto \rho z \cdot \omega$. Its stationary points are $\omega = z/|z|$ and $\omega = -z/|z|$, and its Hessian restricted to the tangent space of $S^{n-1}$ at either stationary point is nondegenerate with size comparable to $\rho |z|$. [claim:Uniform stationary phase on the sphere] Let $M_n = 2n+4$. There exists a constant $C_n > 0$ with the following property. If $\lambda \geq 1$, $\nu \in S^{n-1}$, and $b \in C^{M_n}(S^{n-1})$, then \begin{align*} \left|\int_{S^{n-1}} e^{i\lambda \nu \cdot \omega} b(\omega)\, d\mathcal{H}^{n-1}(\omega)\right| \leq C_n \lambda^{-(n-1)/2}\max_{0 \leq m \leq M_n}\|\nabla_{S^{n-1}}^m b\|_{L^\infty(S^{n-1})}. \end{align*} Here $\nabla_{S^{n-1}}^m b$ denotes the $m$th covariant derivative on the unit sphere. [/claim] [proof] Rotate the sphere so that $\nu$ becomes the first coordinate vector; the measure $\mathcal{H}^{n-1}$ and the covariant derivative bounds are invariant under this rotation. Cover $S^{n-1}$ by finitely many coordinate charts: two small charts around the critical points $\pm \nu$ and finitely many charts where the tangential gradient of $\omega \mapsto \nu \cdot \omega$ is bounded below. On the nonstationary charts, repeated integration by parts using the tangential gradient gives a bound $C_{n,M_n}\lambda^{-M_n}$ times the displayed derivative norm. On each critical chart, the finite-dimensional normal form for a nondegenerate critical point gives local coordinates in which the phase is a constant plus a nondegenerate quadratic form, with coordinate changes whose derivatives are uniformly bounded because the critical points and Hessians vary over a compact family. In those coordinates, Taylor expansion of the amplitude to order $2n+4$ and integration by parts in the quadratic remainder give the Euclidean oscillatory integral bound $C_n\lambda^{-(n-1)/2}$ times the same derivative norm. Summing over the finite chart cover proves the claim. [/proof] Apply the claim with $\lambda = \rho |z|$, $\nu = z/|z|$, and $b=b_\rho$. The derivative norms of $b_\rho$ up to order $M_n$ are bounded uniformly for $\rho \in [r_0,r_1]$ by finitely many derivatives of $a$ on the fixed annulus. If $\rho |z| < 1$, the same bound follows from the absolute-value estimate \begin{align*} \left|\int_{S^{n-1}} e^{i\rho z \cdot \omega}b_\rho(\omega)\, d\mathcal{H}^{n-1}(\omega)\right| \leq \mathcal{H}^{n-1}(S^{n-1})\|b_\rho\|_{L^\infty(S^{n-1})} \end{align*} after increasing the constant, since $(1+\rho |z|)^{-(n-1)/2}$ is then bounded below by $2^{-(n-1)/2}$. Therefore \begin{align*} \left|\int_{S^{n-1}} e^{i\rho z \cdot \omega}b_\rho(\omega)\, d\mathcal{H}^{n-1}(\omega)\right| \leq C_a(1+\rho |z|)^{-(n-1)/2}. \end{align*} Here $C_a$ depends only on finitely many derivatives of $a$ and on the fixed annulus $r_0 \leq \rho \leq r_1$. Since $\rho \in [r_0,r_1]$ and $|z| \simeq |t|$ in the present region, there exists $C_{r_0,r_1} > 0$ such that \begin{align*} (1+\rho |z|)^{-(n-1)/2} \leq C_{r_0,r_1}|t|^{-(n-1)/2}. \end{align*} Integrating over the compact interval $[r_0,r_1]$ gives \begin{align*} |K_t(z)| \leq C_a |t|^{-(n-1)/2}. \end{align*} [guided] Near the cone, the full frequency gradient may vanish, so integration by parts in all frequency variables is no longer the right tool. The geometric observation is that the only degeneracy is radial: after writing $\xi = \rho\omega$, oscillation in the angular variable still has nondegenerate critical points. Use the polar coordinate substitution \begin{align*} \xi = \rho\omega, \end{align*} where $\rho \in (0,\infty)$ and $\omega \in S^{n-1}$. Let $\mathcal{H}^{n-1}$ denote $(n-1)$-dimensional Hausdorff measure restricted to the unit sphere $S^{n-1}$. Under this substitution, Lebesgue measure transforms as \begin{align*} d\mathcal{L}^n(\xi) = \rho^{n-1}\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(\rho). \end{align*} Define the angular amplitude $b_\rho: S^{n-1} \to \mathbb{C}$ by \begin{align*} b_\rho(\omega) = \rho^{n-1}a(\rho\omega). \end{align*} Since $a$ is supported where $r_0 \leq |\xi| \leq r_1$, the radial variable $\rho$ only ranges over the compact interval $[r_0,r_1]$. Thus \begin{align*} K_t(z) = \frac{1}{(2\pi)^n}\int_{r_0}^{r_1} e^{it\rho}\int_{S^{n-1}} e^{i\rho z \cdot \omega}b_\rho(\omega)\, d\mathcal{H}^{n-1}(\omega)\, d\mathcal{L}^1(\rho). \end{align*} Now fix $\rho \in [r_0,r_1]$. The angular phase is \begin{align*} \psi_{\rho,z}: S^{n-1} \to \mathbb{R}, \qquad \psi_{\rho,z}(\omega) = \rho z \cdot \omega. \end{align*} Its stationary points occur exactly when $z$ is normal to the sphere at $\omega$, which means \begin{align*} \omega = \frac{z}{|z|} \qquad \text{or} \qquad \omega = -\frac{z}{|z|}. \end{align*} At each of these two points, the Hessian on the tangent space $T_\omega S^{n-1}$ is a nondegenerate quadratic form with eigenvalues of size comparable to $\rho |z|$. Because $\rho \in [r_0,r_1]$, this size is uniformly comparable to $|z|$. We use the following concrete stationary phase estimate on the sphere. Let $M_n = 2n+4$. There exists $C_n > 0$ such that, whenever $\lambda \geq 1$, $\nu \in S^{n-1}$, and $b \in C^{M_n}(S^{n-1})$, \begin{align*} \left|\int_{S^{n-1}} e^{i\lambda \nu \cdot \omega} b(\omega)\, d\mathcal{H}^{n-1}(\omega)\right| \leq C_n \lambda^{-(n-1)/2}\max_{0 \leq m \leq M_n}\|\nabla_{S^{n-1}}^m b\|_{L^\infty(S^{n-1})}. \end{align*} Here $\nabla_{S^{n-1}}^m b$ is the $m$th covariant derivative on the unit sphere. This estimate follows by rotating $\nu$ to a fixed coordinate direction, covering the sphere by finitely many charts, integrating by parts on charts where the tangential gradient of $\nu\cdot\omega$ is bounded below, and using the finite-dimensional normal form for nondegenerate critical points near the two critical points. In those critical charts, the phase becomes a constant plus a nondegenerate quadratic form with uniformly bounded coordinate changes; Taylor expansion of the amplitude to order $2n+4$ and integration by parts in the quadratic remainder give the local Euclidean oscillatory integral bound. The finite chart cover and rotation invariance make the constant independent of $\nu$. Apply this estimate with $\lambda=\rho |z|$, $\nu=z/|z|$, and $b=b_\rho$. The required derivative norms of $b_\rho$ are uniformly bounded for $\rho \in [r_0,r_1]$ because each spherical derivative of $b_\rho(\omega)=\rho^{n-1}a(\rho\omega)$ is a finite linear combination of derivatives of $a$ on the fixed annulus $r_0 \leq |\xi| \leq r_1$. If $\rho |z|<1$, the same estimate follows from the absolute-value bound after enlarging the constant. Therefore \begin{align*} \left|\int_{S^{n-1}} e^{i\rho z \cdot \omega}b_\rho(\omega)\, d\mathcal{H}^{n-1}(\omega)\right| \leq C_a(1+\rho |z|)^{-(n-1)/2}. \end{align*} This is the exact source of the dispersive exponent: the sphere has dimension $n-1$, and stationary phase contributes one factor of $|\rho z|^{-1/2}$ for each angular direction. In the conic region $\frac{|t|}{2} < |z| < 2|t|$ and with $\rho \in [r_0,r_1]$, we have \begin{align*} (1+\rho |z|)^{-(n-1)/2} \leq C_{r_0,r_1}|t|^{-(n-1)/2}. \end{align*} Thus \begin{align*} |K_t(z)| \leq \frac{1}{(2\pi)^n}\int_{r_0}^{r_1} C_a |t|^{-(n-1)/2}\, d\mathcal{L}^1(\rho) \leq C_a |t|^{-(n-1)/2}. \end{align*} The constant changes by the harmless factor $r_1-r_0$ and still depends only on the support annulus and finitely many derivatives of $a$. [/guided] [/step] [step:Handle the one-dimensional case directly] Assume $n=1$. Since $a \in C_c^\infty(\mathbb{R}\setminus\{0\})$, choose $0<r_0<r_1<\infty$ such that \begin{align*} \operatorname{supp} a \subset [-r_1,-r_0] \cup [r_0,r_1]. \end{align*} For $z \in \mathbb{R}$, \begin{align*} K_t(z)=\frac{1}{2\pi}\int_{0}^{\infty} e^{i(z+t)\xi}a(\xi)\, d\mathcal{L}^1(\xi) +\frac{1}{2\pi}\int_{-\infty}^{0} e^{i(z-t)\xi}a(\xi)\, d\mathcal{L}^1(\xi). \end{align*} Both integrals are over compact intervals. Therefore \begin{align*} |K_t(z)| \leq \frac{1}{2\pi}\int_{\mathbb{R}} |a(\xi)|\, d\mathcal{L}^1(\xi). \end{align*} Since $(n-1)/2=0$ when $n=1$, this is exactly \begin{align*} |K_t(z)| \leq C_a |t|^{0}. \end{align*} Thus the desired kernel estimate holds in dimension $1$. [/step] [step:Extend the estimate from Schwartz functions to all of $L^1(\mathbb{R}^n)$] Combining the previous steps gives, for every $t \in \mathbb{R}$ with $|t| \geq 1$, \begin{align*} \|K_t\|_{L^\infty(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}. \end{align*} Hence for every $\phi \in \mathcal{S}(\mathbb{R}^n)$, \begin{align*} \|P_t\phi\|_{L^\infty(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}\|\phi\|_{L^1(\mathbb{R}^n)}. \end{align*} We prove the approximation fact needed here: $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n)$. Let $f \in L^1(\mathbb{R}^n)$ and let $\varepsilon>0$. Choose $R>0$ so that \begin{align*} \int_{\mathbb{R}^n\setminus B(0,R)} |f|\, d\mathcal{L}^n < \frac{\varepsilon}{3}. \end{align*} Define $f_R=f\mathbb{1}_{B(0,R)}$. Choose a standard smooth compactly supported mollifier $\eta_\delta: \mathbb{R}^n \to [0,\infty)$ with support in $B(0,\delta)$ and total integral $1$. The standard $L^1$ continuity of translations gives $\|\eta_\delta*f_R-f_R\|_{L^1(\mathbb{R}^n)}<\varepsilon/3$ for sufficiently small $\delta>0$. Choose $\zeta\in C_c^\infty(\mathbb{R}^n)$ with $0\leq \zeta\leq 1$ and $\zeta=1$ on $B(0,R+1)$; for $\delta<1$, the function $\zeta(\eta_\delta*f_R)$ lies in $C_c^\infty(\mathbb{R}^n)\subset\mathcal{S}(\mathbb{R}^n)$ and differs from $f$ in $L^1$ by less than $\varepsilon$. Hence $\mathcal{S}(\mathbb{R}^n)$ is dense in $L^1(\mathbb{R}^n)$. This density implies that $P_t$ has a unique bounded extension \begin{align*} P_t: L^1(\mathbb{R}^n) \to L^\infty(\mathbb{R}^n). \end{align*} For $\phi \in L^1(\mathbb{R}^n)$, choose a sequence $(\phi_k)_{k=1}^\infty$ in $\mathcal{S}(\mathbb{R}^n)$ such that \begin{align*} \|\phi_k-\phi\|_{L^1(\mathbb{R}^n)} \to 0. \end{align*} Then $(P_t\phi_k)_{k=1}^\infty$ is Cauchy in $L^\infty(\mathbb{R}^n)$ because \begin{align*} \|P_t\phi_k-P_t\phi_m\|_{L^\infty(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}\|\phi_k-\phi_m\|_{L^1(\mathbb{R}^n)}. \end{align*} Define $P_t\phi$ to be its limit in $L^\infty(\mathbb{R}^n)$. Passing to the limit in the preceding inequality gives \begin{align*} \|P_t\phi\|_{L^\infty(\mathbb{R}^n)} \leq C_a |t|^{-(n-1)/2}\|\phi\|_{L^1(\mathbb{R}^n)}. \end{align*} The extension is unique because any two bounded linear extensions agree on the dense subspace $\mathcal{S}(\mathbb{R}^n) \subset L^1(\mathbb{R}^n)$ and are continuous in the $L^1$ norm. This proves the theorem. [/step]