Verify that a function is a solution to the 3-dimensional wave equation.

For $n=3$. How to verify that a smooth function $$u(.,t)=f*\partial_t\Phi_t+g*\Phi_t \tag{*}$$ is a solution to the 3-dimensional wave equation $$u_{tt}=\Delta{u} \quad u(x,0)=f(x) \quad u_t(x,0)=g(x) \tag{**}$$ where $\Phi_t$ is the tempered distribution given by $$\displaystyle \Phi_t (h) := \frac{t}{4\pi} \int_{S^2} h(t\omega)\ d\omega$$ for all Schwartz function $h$ and $\omega$ is just the Lebesgue measure on the sphere $S^2$?

For a tempered distribution $\,f\in\mathcal{S}'$, its Fourier transform $\,\hat{f}=F[f]\,$ is a linear functional $\,\hat{f}\in\mathcal{S}'\,$ defined by the identity $$\langle\hat{f},\varphi\rangle=\langle f,\hat{\varphi}\rangle\quad\forall\, \varphi\in\mathcal{S}$$ where $\,\hat{\varphi}\,$ designates a classical integral Fourier transform $$\hat{\varphi}(\xi)=\!\int\limits_{\mathbb{R}^3}\!\varphi(x)e^{-ix\cdot\xi}dx$$ of a rapidly decreasing test function $\,\varphi\in\mathcal{S}(\mathbb{R}^3)$.
Denote $\,\mathbb{R}_{+}=\{t\in\mathbb{R}\colon\;t>0\}$,  and let $\,u\in C^1(\mathbb{R}_{+};\mathcal{S}')\,$ be a solution of Cauchy problem $(\ast\ast)$. Notice that Fourier transform $\,\hat{u}\in C^1(\mathbb{R}_{+};\mathcal{S}')$ is to solve the Cauchy problem $$\begin{cases} \hat{u}_{tt}+|\xi|^2\hat{u}=0, \quad t>0,\\ \hat{u}|_{t=0}=\hat{f},\quad \hat{u}_t|_{t=0}=\hat{g}, \end{cases}$$ solution of which is of the form $$\hat{u}(\xi,t)=\hat{f}(\xi)\cos{(|\xi|t)}+ \hat{g}(\xi)\frac{\sin{(|\xi|t)}}{|\xi|}.\tag{1}$$ Consider a tempered distribution $$\Phi_t=F^{-1}\Bigl[\frac{\sin{(|\xi|t)}}{|\xi|}\Bigr]\tag{2}$$ and notice that Fourier transform $$\hat{\Phi}_t(\xi)=\frac{\sin{(|\xi|t)}}{|\xi|}= \sum_{n=0}^{\infty}(-1)^n t^{2n+1}\frac{(\xi_1^2+\xi_2^2+\xi_3^2)^{n}}{(2n+1)!}$$ is an entire function in $\,\xi=(\xi_1,\xi_2,\xi_3)\,$ on $\,\mathbb{C}^3$. Hence, a distribution $\,\Phi_t\in\mathcal{S}'\,$ has a compact support $\,{\rm supp\,}\Phi_t\Subset\mathbb{R}^3$. By the convolution theorem, representation $(1)$ immediately implies the required representation of solution $(\ast)$. Defintion $(2)$ of the tempered distribution $\Phi_t$ implies that $$\langle \Phi_t,h\rangle=\frac{1}{4\pi t}\langle\delta_{S_t},h\rangle= \frac{t}{4\pi} \int\limits_{S^2} h(t\omega)\, d\omega\quad \forall\,h\in \mathcal{S}$$ — for details see The issue of treating an inverse Fourier transform in terms of a tempered distribution.
• @ mkl314: Thanks! I think I understand (and wrote correctly) the definition of Fourier transform of a distribution. Could you please see this Note Equation 5 which is in Section 3: Tempered Distributions and then the paragraph right after Exercise 30: terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions where $\langle\hat{f},\varphi\rangle=\langle f,\check{\varphi}\rangle\quad\forall\, \varphi\in\mathcal{S}$. – user54992 May 9 '14 at 19:47
• @ mkl314: Would you mind guiding me to solve Exercise 40 based on the definition given in that Notes which is anti-$\mathcal{S}'$ as you called? – user54992 May 9 '14 at 21:45
• For tempered distributions, a definition of Fourier transform is subject to the definition of $\mathcal{S}'$ as being primary to everything else. Thus the classical definition of Fourier transform employed in the answer is subject to the classical definition of $\mathcal{S}'$ as a linear vector space of linear functionals on $\mathcal{S}$. Meanwhile, despite the fact that $\mathcal{S}'$ was defined in the "Notes"as an antilinear vector space of linear functionals on $\mathcal{S}$, the representation of $\Phi_t$ in Exercise 40 will be the same whatever definition of Fourier transform you take. – mkl314 May 14 '14 at 12:14