Is the spatial-only convolution of two waves again a wave?

Call a function $f:\mathbb{R}^{1+n}\to \mathbb{R}$ a wave, if it satisfies the wave equation

$$\frac{1}{c^2} \frac{\partial^2 f}{dt^2} = \sum_{i=1}^n \frac{\partial^2 f}{dx_i^2} = div(grad\,f)$$

If we have two waves, $f$ and $g$, for every $t\in\mathbb{t}$ we can define their spatial convolution (i.e. not involving the time parameter)

$$(f\ast_s g)(t,\vec{x}) = \int_{\mathbb{R}^n} f(t,\vec{y})g(t,\vec{x}-\vec{y}) d\vec{y}\quad .$$

Will $f\ast_s g$ always be a wave?

What I did: according to the rules for derivatives of convolution, the right hand side is $$div(grad (f\ast_s g)) = (div(grad\, f))\ast_s g$$

For the 2nd time derivative on the left, using $f'$ and $f''$ as shorthand for the partial derivative by the time parameter and also leaving out the arguments from $f(t,\vec{y})$ and $g(t,\vec{x}-\vec{y})$ parameters for better readability, we get: \begin{aligned} \frac{\partial^2}{dt^2} (f \ast_s g) &= \int \frac{\partial^2}{dt^2}\; fg\; d\vec{y} \\ &= \int \frac{\partial}{dt}\left( f'g + fg'\right ) d\vec{y}\\ &= \int f''g + f'g' + f'g' + fg'' d\vec{y} \\ &= f''\ast_s g + 2 f'\ast_s g' + f\ast_s g'' \end{aligned}

If the last line would only contain the first term, I would have $\frac{1}{c^2}f''\ast_s g =(div (grad \,f))\ast_s g$ which would establish the proof. But I have these other two terms in there, which are in the way.

Does anyone see a chance to get rid of them generally, or under some mild extra requirements for $f$ and/or $g$?

ADDITION: the comments say, that yes, it is easy to show in the frequency domain that the convolution of 2 waves is a wave again. But here I only convolve the spatial parameters, not the time parameter and the derivation seems to show that in this case the result is not necessarily a wave, but I am looking how to fix the functions to get a wave again.

• It seems simpler to switch to Fourier transforms wrt the space variables since the Fourier transform of a convolution is the product of the Fourier transforms of the constituants. Then you find that the convolution does satisfy the wave equation. – Urgje Jan 24 '16 at 14:17
• I talked with a friend about it and his reaction was the same as yours until I specifically highlighted, that only the space coordinates shall be convolved, not the full 4-dimensional functions. And I just notice that my post might hide this with slightly wrong notation. – Harald Jan 30 '16 at 13:58
• Tried to improve the notation for "spatial convolution" to $\ast_s$, hoping this makes it clearer what I mean. – Harald Jan 30 '16 at 14:05