Surface measure and the wave equation

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I am having an annoying conceptual problem trying to solve problem 46 in Chapter 8 of Folland's "Real Analysis". I'll try to explain my problem as briefly as possible.

Consider the wave equation $\partial_{t}^{2}u-\Delta u=0$ in $\mathbb{R}^{3}$ with initial data $u(x,0)=f(x)$ and $\partial_{t}u(x,0)=g(x).$

Taking Fourier transform, we swiftly get: $\widehat{u}(\xi,t)=\widehat{f}(\xi)\cos (2 \pi \vert \xi \vert)+\widehat{g}(\xi) \dfrac{\sin (2 \pi \vert \xi \vert)}{2 \pi \vert \xi \vert}.$

Following Folland's notation, we write $u(x,t)=f \ast \partial_{t}W_{t}(x)+g \ast W_{t}(x),$ where $W_{t}$ is the inverse Fourier transform of $ \dfrac{\sin (2 \pi \vert \xi \vert)}{2 \pi \vert \xi \vert}.$

In the three-dimensional space, we have that the inverse transform of $\displaystyle \left(\dfrac{\sin (2 \pi \vert \xi \vert)}{2 \pi \vert \xi \vert} \right) $ is $\dfrac{\sigma_{t}(x)}{4 \pi t},$ with $\sigma_{t}(x)$ the surface measure in the sphere $\vert x \vert =t.$

My problem now is that I do not understand at all how to compute $f \ast \dfrac{\sigma_{t}(x)}{4 \pi t},$ for I do not know even where to integrate! What does it mean to take the convolution of $f$ with a surface measure?? Folland defines previously in his book the convolution of $f \in L^{1}\left( \mathbb{R}^{n} \right)$ with Borel measures in $\mathbb{R}^{n}.$

$f \ast W_{t}$ should be a integral over the whole space, but then I do not know if $\displaystyle \int_{\mathbb{R}^{3}} f(x-y)\dfrac{\sigma_{t}(y)}{4 \pi t}dy $ makes any sense....

I have also tried supposing that $f \ast W_{t}$ should be understood as $\displaystyle \dfrac{1}{4 \pi} \int_{\partial B(0,t} f(x-y) \dfrac{\sigma_{t}(y)}{t} $... but what do I do then? use spherical coordinates and write the measure as $t^{2}\sin \theta$....

I do not know how to interpret this convolution with a surface measure nor how to compute it, but even proceeding as I just indicated, I do not see how are we going to recover Kirchhoff's solution, which one tipically gets with the method of spherical means.

Thank you so much for any help!

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Folland defines previously in his book the convolution ... with Borel measures

And surface measure is an example of a Borel measure, so the aforementioned definition applies. Without having a book with me, I'll venture a guess: it is
$$(f*\mu) (x) = \int_{\mathbb R^n} f(x-y)\,d\mu(y)$$

Let's look at the one-dimensional case first. Here the surface of sphere of radius $t$ is the two-point set $\{-t,t\}$. The normalized measure on this "sphere" is $\mu=\frac12 (\delta_{-t}+\delta_t)$. The convolution of $f$ with this measure is $$(f*\mu)(x) = \frac12 (f(x+t)+f(x-t))$$ which you can recognize as a part of d'Alembert's formula.

Same in three dimensions. The convolution of $f$ with the normalized surface measure $\mu=\frac{1}{4\pi t}d\sigma_t(y)$ is the spherical mean of $f$: $$(f*\mu) (x) = \frac{1}{4\pi t}\int_{|y|=t} f(x-y)\,d\sigma_t(y) $$